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22273
|
\section{Sum of Sequence of k x k!}
Tags: Factorials, Sum of Sequence of k x k!
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n j \times j!
| r = 1 \times 1! + 2 \times 2! + 3 \times 3! + \dotsb + n \times n!
| c =
}}
{{eqn | r = \paren {n + 1}! - 1
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\displaystyle \sum_{j \mathop = 1}^n j \times j! = \paren {n + 1}! - 1$
\end{proof}
|
22274
|
\section{Sum of Sequences of Fifth and Seventh Powers}
Tags: Fifth Powers, Seventh Powers, Sums of Sequences
\begin{theorem}
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
\end{proof}
|
22275
|
\section{Sum of Series of Product of Power and Cosine}
Tags: Trigonometric Series, Cosine Function
\begin{theorem}
Let $r \in \R$ such that $\size r < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n r^k \map \cos {k x}
| r = 1 + r \cos x + r^2 \cos 2 x + r^3 \cos 3 x + \cdots + r^n \cos n x
| c =
}}
{{eqn | r = \dfrac {r^{n + 2} \cos n x - r^{n + 1} \map \cos {n + 1} x - r \cos x + 1} {1 - 2 r \cos x + r^2}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n r^k \map \cos {k x}
| r = \map \Re {\sum_{k \mathop = 0}^n r^k e^{i k x} }
| c =
}}
{{eqn | r = \map \Re {\sum_{k \mathop = 0}^n \paren {r e^{i x} }^k}
| c =
}}
{{eqn | r = \map \Re {\frac {1 - r^{n + 1} e^{i \paren {n + 1} x} } {1 - r e^{i x} } }
| c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Re {\frac {\paren {1 - r^{n + 1} e^{i \paren {n + 1} x} } \paren {1 - r e^{-i x} } } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }
| c =
}}
{{eqn | r = \map \Re {\frac {r^{n + 2} e^{i n x} - r^{n + 1} e^{i \paren {n + 1} x} - r e^{-i x} + 1} {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
| c =
}}
{{eqn | r = \map \Re {\frac {r^{n + 2} \paren {\cos n x + i \sin n x} - r^{n + 1} \paren {\map \cos {n + 1} x + i \map \sin {n + 1} x} - r \paren {\cos x - i \sin x} + 1} {1 - 2 r \cos x + a^2} }
| c = Euler's Formula and Corollary to Euler's Formula
}}
{{eqn | r = \dfrac {r^{n + 2} \cos n x - r^{n + 1} \map \cos {n + 1} x - r \cos x + 1} {1 - 2 r \cos x + r^2}
| c = after simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22276
|
\section{Sum of Series of Product of Power and Sine}
Tags: Trigonometric Series, Sine Function
\begin{theorem}
Let $r \in \R$ such that $\size r < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n r^k \map \sin {k x}
| r = r \sin x + r^2 \sin 2 x + r^3 \sin 3 x + \cdots + r^n \sin n x
| c =
}}
{{eqn | r = \dfrac {r \sin x - r^{n + 1} \map \sin {n + 1} x + r^{n + 2} \sin n x} {1 - 2 r \cos x + r^2}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n r^k \map \sin {k x}
| r = \map \Im {\sum_{k \mathop = 1}^n r^k e^{i k x} }
| c =
}}
{{eqn | r = \map \Im {\sum_{k \mathop = 0}^n \paren {r e^{i x} }^n}
| c = as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$
}}
{{eqn | r = \map \Im {\frac {1 - r^{n + 1} e^{i \paren {n + 1} x} } {1 - r e^{i x} } }
| c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Im {\frac {\paren {1 - r^{n + 1} e^{i \paren {n + 1} x} } \paren {1 - r e^{-i x} } } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }
| c =
}}
{{eqn | r = \map \Im {\frac {r^{n + 2} e^{i n x} - r^{n + 1} e^{i \paren {n + 1} x} - r e^{-i x} + 1} {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
| c =
}}
{{eqn | r = \map \Im {\frac {r^{n + 2} \paren {\cos n x + i \sin n x} - r^{n + 1} \paren {\map \cos {n + 1} x + i \map \sin {n + 1} x} - r \paren {\cos x - i \sin x} + 1} {1 - 2 r \cos x + a^2} }
| c = Euler's Formula and Corollary to Euler's Formula
}}
{{eqn | r = \dfrac {r \sin x - r^{n + 1} \map \sin {n + 1} x + r^{n + 2} \sin n x} {1 - 2 r \cos x + r^2}
| c = after simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22277
|
\section{Sum of Sines of Arithmetic Sequence of Angles}
Tags: Sine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
\end{theorem}
\begin{proof}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles:
:$\displaystyle \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
Equating imaginary parts:
:$\displaystyle \sum_{k \mathop = 0}^n \map \sin {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \sin {\theta + \frac {n \alpha} 2}$
{{qed}}
\end{proof}
|
22278
|
\section{Sum of Sines of Arithmetic Sequence of Angles/Formulation 1}
Tags: Sine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \map \sin {\theta + k \alpha}
| r = \sin \theta + \map \sin {\theta + \alpha} + \map \sin {\theta + 2 \alpha} + \map \sin {\theta + 3 \alpha} + \dotsb
}}
{{eqn | r = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \sin {\theta + \frac {n \alpha} 2}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1:
:$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$
Equating imaginary parts:
:$\ds \sum_{k \mathop = 0}^n \map \sin {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \sin {\theta + \frac {n \alpha} 2}$
{{qed}}
\end{proof}
|
22279
|
\section{Sum of Sines of Arithmetic Sequence of Angles/Formulation 2}
Tags: Sine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \map \sin {\theta + k \alpha}
| r = \map \sin {\theta + \alpha} + \map \sin {\theta + 2 \alpha} + \map \sin {\theta + 3 \alpha} + \dotsb
}}
{{eqn | r = \map \sin {\theta + \frac {n + 1} 2 \alpha}\frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2:
:$\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the {{RHS}} is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 1}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
Equating imaginary parts:
:$\ds \sum_{k \mathop = 1}^n \map \sin {\theta + k \alpha} = \map \sin {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
{{qed}}
\end{proof}
|
22280
|
\section{Sum of Sines of Fractions of Pi}
Tags: Polynomial Equations, Sine Function
\begin{theorem}
Let $n \in \Z$ such that $n > 1$.
Then:
:$\ds \sum_{k \mathop = 1}^{n - 1} \sin \frac {2 k \pi} n = 0$
\end{theorem}
\begin{proof}
Consider the equation:
:$z^n - 1 = 0$
whose solutions are the complex roots of unity:
:$1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \paren {n - 1} \pi i / n}$
By Sum of Roots of Polynomial:
:$1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \paren {n - 1} \pi i / n} = 0$
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
from which comes:
:$\paren {1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n} + i \paren {\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n} = 0$
Equating imaginary parts:
:$\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n = 0$
{{qed}}
\end{proof}
|
22281
|
\section{Sum of Sines of Multiples of Angle}
Tags: Sum of Sines of Multiples of Angle, Telescoping Series, Sine Function
\begin{theorem}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \sin k x
| r = \sin x + \sin 2 x + \sin 3 x + \cdots + \sin n x
| c =
}}
{{eqn | r = \frac {\sin \frac {\paren {n + 1} x} 2 \sin \frac {n x} 2} {\sin \frac x 2}
| c =
}}
{{end-eqn}}
where $x$ is not an integer multiple of $2 \pi$.
\end{theorem}
\begin{proof}
By Simpson's Formula for Sine by Sine:
:$2 \sin \alpha \sin \beta = \cos \left({\alpha - \beta}\right) - \cos \left({\alpha + \beta}\right)$
Thus we establish the following sequence of identities:
{{begin-eqn}}
{{eqn | l = 2 \sin x \sin \frac x 2
| r = \cos \frac x 2 - \cos \frac {3 x} 2
| c =
}}
{{eqn | l = 2 \sin 2 x \sin \frac x 2
| r = \cos \frac {3 x} 2 - \cos \frac {5 x} 2
| c =
}}
{{eqn | o = \cdots
| c =
}}
{{eqn | l = 2 \sin n x \sin \frac x 2
| r = \cos \frac {\left({2 n - 1}\right) x} 2 - \cos \frac {\left({2 n + 1}\right) x} 2
| c =
}}
{{end-eqn}}
Summing the above:
{{begin-eqn}}
{{eqn | l = 2 \sin \frac x 2 \left({\sum_{k \mathop = 1}^n \sin k x}\right)
| r = \cos \frac x 2 - \cos \frac {\left({2 n + 1}\right) x} 2
| c = Sums on {{RHS}} form Telescoping Series
}}
{{eqn | r = -2 \sin \left({\dfrac {\frac x 2 + \frac {\left({2 n + 1}\right) x} 2} 2}\right) \sin \left({\dfrac {\frac x 2 - \frac {\left({2 n + 1}\right) x} 2} 2}\right)
| c = Prosthaphaeresis Formula for Cosine minus Cosine
}}
{{eqn | r = -2 \sin \dfrac {\left({n + 1}\right) x} 2 \sin \dfrac {-n x} 2
| c =
}}
{{eqn | r = 2 \sin \dfrac {\left({n + 1}\right) x} 2 \sin \dfrac {n x} 2
| c = Sine Function is Odd
}}
{{end-eqn}}
The result follows by dividing both sides by $2 \sin \dfrac x 2$.
It is noted that when $x$ is a multiple of $2 \pi$ then:
:$\sin \dfrac x 2 = 0$
leaving the {{RHS}} undefined.
{{qed}}
\end{proof}
|
22282
|
\section{Sum of Square Roots as Square Root of Sum}
Tags: Algebra
\begin{theorem}
:$\sqrt a + \sqrt b = \sqrt {a + b + \sqrt {4 a b} }$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sqrt a + \sqrt b
| r = \sqrt {\paren {\sqrt a + \sqrt b}^2}
}}
{{eqn | r = \sqrt {\sqrt a^2 + \sqrt b^2 + 2 \sqrt a \sqrt b}
| c = Square of Sum
}}
{{eqn | r = \sqrt {a + b + \sqrt {4 a b} }
| c = Power of Product
}}
{{end-eqn}}
{{qed}}
Category:Algebra
\end{proof}
|
22283
|
\section{Sum of Squared Deviations from Mean}
Tags: Sum of Squared Deviations from Mean, Descriptive Statistics, Arithmetic Mean
\begin{theorem}
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $\overline x$ denote the arithmetic mean of $S$.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n \paren {x_i^2 - \overline x^2}$
\end{theorem}
\begin{proof}
For brevity, let us write $\displaystyle \sum$ for $\displaystyle \sum_{i \mathop = 1}^n$.
Then:
{{begin-eqn}}
{{eqn|l = \sum \left({x_i - \overline{x} }\right)^2
|r = \sum \left({x_i - \overline{x} }\right)\left({x_i - \overline{x} }\right)
}}
{{eqn|r = \sum x_i\left({x_i - \overline{x} }\right) - \overline{x}\sum \left({x_i - \overline{x} }\right)
|c = Summation is Linear
}}
{{eqn|r = \sum x_i\left({x_i - \overline{x} }\right) - 0
|c = Sum of Deviations from Mean
}}
{{eqn|r = \sum x_i\left({x_i - \overline{x} }\right) + 0
}}
{{eqn|r = \sum x_i\left({x_i - \overline{x} }\right) + \overline{x}\sum \left({x_i - \overline{x} }\right)
|c = Sum of Deviations from Mean
}}
{{eqn|r = \sum \left({x_i + \overline{x} }\right)\left({x_i - \overline{x} }\right)
|c = Summation is Linear
}}
{{eqn|r = \sum \left({x_i^2 - \overline{x}^2 }\right)
}}
{{end-eqn}}
{{qed}}
Category:Descriptive Statistics
110686
110653
2012-10-12T19:01:51Z
Prime.mover
59
110686
wikitext
text/x-wiki
\end{proof}
|
22284
|
\section{Sum of Squares as Product of Factors with Square Roots}
Tags: Sums of Squares, Algebra
\begin{theorem}
:$x^2 + y^2 = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}
| r = x^2 - x \sqrt {2 x y} + x y + x \sqrt {2 x y} - \sqrt {2 x y} \sqrt {2 x y} + y \sqrt {2 x y} + x y - y \sqrt {2 x y} + y^2
| c =
}}
{{eqn | r = x^2 + \paren {x - x} \sqrt {2 x y} + 2 x y - 2 x y + \paren {y - y} \sqrt {2 x y} + y^2
| c =
}}
{{eqn | r = x^2 + y^2
| c =
}}
{{end-eqn}}
{{qed}}
Category:Algebra
Category:Sums of Squares
\end{proof}
|
22285
|
\section{Sum of Squares of Binomial Coefficients}
Tags: Sum of Squares of Binomial Coefficients, Binomial Coefficients
\begin{theorem}
:$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
where $\dbinom n i$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
:$\displaystyle \sum_{i=0}^n \binom n i^2 = \binom {2 n} n$
$P(0)$ is true, as this just says $\displaystyle \binom 0 0^2 = 1 = \binom {2 \times 0} 0$. This holds by definition.
\end{proof}
|
22286
|
\section{Sum of Squares of Complex Moduli of Sum and Differences of Complex Numbers}
Tags: Complex Modulus
\begin{theorem}
Let $\alpha, \beta \in \C$ be complex numbers.
Then:
:$\cmod {\alpha + \beta}^2 + \cmod {\alpha - \beta}^2 = 2 \cmod \alpha^2 + 2 \cmod \beta^2$
\end{theorem}
\begin{proof}
Let:
:$\alpha = x_1 + i y_1$
:$\beta = x_2 + i y_2$
Then:
{{begin-eqn}}
{{eqn | o =
| r = \cmod {\alpha + \beta}^2 + \cmod {\alpha - \beta}^2
| c =
}}
{{eqn | r = \cmod {\paren {x_1 + i y_1} + \paren {x_2 + i y_2} }^2 + \cmod {\paren {x_1 + i y_1} - \paren {x_2 + i y_2} }^2
| c = Definition of $\alpha$ and $\beta$
}}
{{eqn | r = \cmod {\paren {x_1 + x_2} + i \paren {y_1 + y_2} }^2 + \cmod {\paren {x_1 - x_2} + i \paren {y_1 - y_2} }^2
| c = {{Defof|Complex Addition}}, {{Defof|Complex Subtraction}}
}}
{{eqn | r = \paren {x_1 + x_2}^2 + \paren {y_1 + y_2}^2 + \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2
| c = {{Defof|Complex Modulus}}
}}
{{eqn | r = {x_1}^2 + 2 x_1 x_2 + {x_2}^2 + {y_1}^2 + 2 y_1 y_2 + {y_2}^2 + {x_1}^2 - 2 x_1 x_2 + {x_2}^2 + {y_1}^2 - 2 y_1 y_2 + {y_2}^2
| c = Square of Sum, Square of Difference
}}
{{eqn | r = 2 {x_1}^2 + 2 {x_2}^2 + 2 {y_1}^2 + 2 {y_2}^2
| c = simplifying
}}
{{eqn | r = 2 \paren { {x_1}^2 + {y_1}^2} + 2 \paren { {x_2}^2 + {y_2}^2}
| c = simplifying
}}
{{eqn | r = 2 \cmod {x_1 + i y_1}^2 + 2 \cmod {x_2 + i y_2}^2
| c = {{Defof|Complex Modulus}}
}}
{{eqn | r = 2 \cmod \alpha^2 + 2 \cmod \beta^2
| c = Definition of $\alpha$ and $\beta$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22287
|
\section{Sum of Squares of Divisors of 24 and 26 are Equal}
Tags: 24, Divisors, 26, Square Numbers
\begin{theorem}
The sum of the squares of the divisors of $24$ equals the sum of the squares of the divisors of $26$:
:$\map {\sigma_2} {24} = \map {\sigma_2} {26}$
where $\sigma_\alpha$ denotes the divisor function.
\end{theorem}
\begin{proof}
The divisors of $24$ are:
:$1, 2, 3, 4, 6, 8, 12, 24$
The divisors of $26$ are:
:$1, 2, 13, 26$
Then we have:
{{begin-eqn}}
{{eqn | r = 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 8^2 + 12^2 + 24^2
| o =
| c =
}}
{{eqn | r = 1 + 4 + 9 + 16 + 36 + 64 + 144 + 576
| c =
}}
{{eqn | r = 850
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | r = 1^2 + 2^2 + 13^2 + 26^2
| o =
| c =
}}
{{eqn | r = 1 + 4 + 169 + 676
| c =
}}
{{eqn | r = 850
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22288
|
\section{Sum of Squares of Secant and Cosecant}
Tags: Trigonometric Identities
\begin{theorem}
:$\sec^2 x + \csc^2 x = \sec^2 x \csc^2 x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sec^2 x + \csc^2 x
| r = \frac 1 {\cos^2 x} + \frac 1 {\sin^2 x}
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{eqn | r = \frac {\sin^2 x + \cos^2 x} {\cos^2 x \sin^2 x}
| c =
}}
{{eqn | r = \frac 1 {\cos^2 x \sin^2 x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sec^2 x \csc^2 x
| c = {{Defof|Secant Function}} and {{Defof|Cosecant}}
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
22289
|
\section{Sum of Squares of Sine and Cosine/Corollary 1}
Tags: Cotangent Function, Trigonometric Identities, Tangent Function, Secant Function, Sum of Squares of Sine and Cosine
\begin{theorem}
:$\sec^2 x - \tan^2 x = 1 \quad \text{(when $\cos x \ne 0$)}$
where $\sec$, $\tan$ and $\cos$ are secant, tangent and cosine respectively.
\end{theorem}
\begin{proof}
When $\cos x \ne 0$:
{{begin-eqn}}
{{eqn | l = \cos^2 x + \sin^2 x
| r = 1
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll= \leadsto
| l = 1 + \frac {\sin^2 x} {\cos^2 x}
| r = \frac 1 {\cos^2 x}
| c = dividing both sides by $\cos^2 x$, as $\cos x \ne 0$
}}
{{eqn | ll= \leadsto
| l = 1 + \tan^2 x
| r = \sec^2 x
| c = {{Defof|Tangent Function}} and {{Defof|Secant Function}}
}}
{{eqn | ll= \leadsto
| l = \sec^2 x - \tan^2 x
| r = 1
| c = rearranging
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22290
|
\section{Sum of Squares of Standard Gaussian Random Variables has Chi-Squared Distribution}
Tags: Chi-Squared Distribution
\begin{theorem}
Let $X_1, X_2, \ldots, X_n$ be independent random variables.
Let $X_i \sim \Gaussian 0 1$ for $1 \le i \le n$ where $\Gaussian 0 1$ is the standard Gaussian Distribution.
Then:
:$\ds \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_n$
where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.
\end{theorem}
\begin{proof}
By Square of Standard Gaussian Random Variable has Chi-Squared Distribution, we have:
:$X^2_i \sim \chi^2_1$
for $1 \le i \le n$.
So, by Sum of Chi-Squared Random Variables, we have:
:$\ds \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_{1 + 1 + 1 \ldots} = \chi^2_n$
{{qed}}
Category:Chi-Squared Distribution
\end{proof}
|
22291
|
\section{Sum of Squares on Pairs of Rows and Columns of Moessner's Order 4 Magic Square}
Tags: Magic Squares
\begin{theorem}
The sums of the squares of the entries are equal on the following pairs of rows and columns of Moessner's order $4$ magic square:
:Rows $1$ and $4$
:Rows $2$ and $3$
:Columns $1$ and $4$
:Columns $2$ and $3$.
\end{theorem}
\begin{proof}
Recall Moessner's order $4$ magic square:
{{:Definition:Moessner's Order 4 Magic Square}}
\end{proof}
|
22292
|
\section{Sum of Successive Powers in 2 ways}
Tags: Powers, 8191, 31, Sums of Sequences
\begin{theorem}
$31$ and $8191$ can be expressed as the sum of successive powers starting from $1$ in in $2$ different ways.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 31
| r = 1 + 5 + 5^2
| c =
}}
{{eqn | r = 1 + 2 + 2^2 + 2^3 + 2^4
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8191
| r = 1 + 90 + 90^2
| c =
}}
{{eqn | r = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^{10}+ 2^{11}+ 2^{12}
| c =
}}
{{end-eqn}}
These are the only two examples known.
{{qed}}
\end{proof}
|
22293
|
\section{Sum of Summations equals Summation of Sum}
Tags: Summations
\begin{theorem}
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
\end{theorem}
\begin{proof}
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{\map R i} \paren {b_i + c_i}
| r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }
| c = Definition of $b_i$ and $c_i$
}}
{{eqn | r = \sum_{\map R i} \paren {\sum_{j \mathop = 1}^2 a_{i j} }
| c = {{Defof|Summation}}
}}
{{eqn | r = \sum_{j \mathop = 1}^2 \paren {\sum_{\map R i} a_{i j} }
| c = Exchange of Order of Summation
}}
{{eqn | r = \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}
| c = {{Defof|Summation}}
}}
{{eqn | r = \sum_{\map R i} b_i + \sum_{\map R i} c_i
| c = Definition of $b_i$ and $c_i$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22294
|
\section{Sum of Summations equals Summation of Sum/Infinite Sequence}
Tags: Summations
\begin{theorem}
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers $\Z$.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be infinite.
Let $\ds \sum_{\map R i} b_i$ and $\ds \sum_{\map R i} c_i$ be convergent.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
\end{theorem}
\begin{proof}
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{R \left({i}\right)} \left({b_i + c_i}\right)
| r = \sum_{R \left({i}\right)} \left({a_{i 1} + a_{i 2} }\right)
| c = by definition
}}
{{eqn | r = \sum_{R \left({i}\right)} \left({\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }\right)
| c = Definition of Summation
}}
{{eqn | r = \sum_{1 \mathop \le j \mathop \le 2} \left({\sum_{R \left({i}\right)} a_{i j} }\right)
| c = Exchange of Order of Summation: Finite and Infinite Series
}}
{{eqn | r = \sum_{R \left({i}\right)} a_{i 1} + \sum_{R \left({i}\right)} a_{i 2}
| c = Definition of Summation
}}
{{eqn | r = \sum_{R \left({i}\right)} b_i + \sum_{R \left({i}\right)} c_i
| c = by definition
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22295
|
\section{Sum of Summations over Overlapping Domains/Example}
Tags: Sum of Summations over Overlapping Domains, Summations
\begin{theorem}
:$\displaystyle \sum_{1 \mathop \le j \mathop \le m} a_j + \sum_{m \mathop \le j \mathop \le n} a_j = \left({\sum_{1 \mathop \le j \mathop \le n} a_j}\right) + a_m$
\end{theorem}
\begin{proof}
Let $\map R j$ be the propositional function $1 \mathop \le j \mathop \le m$.
Let $\map S j$ be the propositional function $m \mathop \le j \mathop \le n$.
Then we have:
{{begin-eqn}}
{{eqn | l = \map R j \lor \map S j
| r = \paren {1 \mathop \le j \mathop \le m} \lor \paren {m \mathop \le j \mathop \le n}
| c =
}}
{{eqn | r = \paren {1 \mathop \le j \mathop \le n}
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map R j \land \map S j
| r = \paren {1 \mathop \le j \mathop \le m} \land \paren {m \mathop \le j \mathop \le n}
| c =
}}
{{eqn | r = \paren {j = m}
| c =
}}
{{end-eqn}}
The result follows from Sum of Summations over Overlapping Domains.
{{qed}}
\end{proof}
|
22296
|
\section{Sum of Tangent and Cotangent}
Tags: Trigonometric Identities
\begin{theorem}
:$\tan x + \cot x = \sec x \csc x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan x + \cot x
| r = \frac {\sin x} {\cos x} + \cot x
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\sin x} {\cos x} + \frac {\cos x} {\sin x}
| c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\sin^2 x + \cos^2x} {\cos x \sin x}
| c =
}}
{{eqn | r = \frac 1 {\cos x \sin x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sec x \frac 1 {\sin x}
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \sec x \csc x
| c = Cosecant is Reciprocal of Sine
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
22297
|
\section{Sum of Terms of Magic Cube}
Tags: Magic Cubes
\begin{theorem}
The total of all the entries in a magic cube of order $n$ is given by:
:$T_n = \dfrac {n^3 \paren {n^3 + 1} } 2$
\end{theorem}
\begin{proof}
Let $M_n$ denote a magic cube of order $n$.
$M_n$ is by definition an arrangement of the first $n^3$ (strictly) positive integers into an $n \times n \times n$ cubic array containing the positive integers from $1$ upwards.
Thus there are $n^3$ entries in $M_n$, going from $1$ to $n^3$.
Thus:
{{begin-eqn}}
{{eqn | l = T_n
| r = \sum_{k \mathop = 1}^{n^3} k
| c =
}}
{{eqn | r = \frac {n^3 \paren {n^3 + 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22298
|
\section{Sum of Terms of Magic Square}
Tags: Magic Squares
\begin{theorem}
The total of all the entries in a magic square of order $n$ is given by:
:$T_n = \dfrac {n^2 \paren {n^2 + 1} } 2$
\end{theorem}
\begin{proof}
Let $M_n$ denote a magic square of order $n$.
$M_n$ is by definition a square matrix of order $n$ containing the positive integers from $1$ upwards.
Thus there are $n^2$ entries in $M_n$, going from $1$ to $n^2$.
Thus:
{{begin-eqn}}
{{eqn | l = T_n
| r = \sum_{k \mathop = 1}^{n^2} k
| c =
}}
{{eqn | r = \frac {n^2 \paren {n^2 + 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22299
|
\section{Sum of Triangular Matrices}
Tags: Matrix Entrywise Addition, Matrix Algebra, Triangular Matrices, Matrix Entrywise Sum
\begin{theorem}
Let $\mathbf A = \left[{a}\right]_{n}, \mathbf B = \left[{b}\right]_{n}$ be square matrices of order $n$.
Let $\mathbf C = \mathbf A + \mathbf B$ be the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
If $\mathbf A$ and $\mathbf B$ are upper triangular matrices, then so is $\mathbf C$.
If $\mathbf A$ and $\mathbf B$ are lower triangular matrices, then so is $\mathbf C$.
\end{theorem}
\begin{proof}
From the definition of matrix addition, we have:
:$\forall i, j \in \left[{1 .. n}\right]: c_{ij} = a_{ij} + b_{ij}$
If $\mathbf A$ and $\mathbf B$ are upper triangular matrices, we have:
:$\forall i > j: a_{ij} = b_{ij} = 0$
Hence:
:$\forall i > j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
and so $\mathbf C$ is itself upper triangular.
Similarly, if $\mathbf A$ and $\mathbf B$ are lower triangular matrices, we have:
:$\forall i < j: a_{ij} = b_{ij} = 0$
Hence:
:$\forall i < j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
and so $\mathbf C$ is itself lower triangular.
{{Qed}}
Category:Triangular Matrices
Category:Matrix Entrywise Addition
\end{proof}
|
22300
|
\section{Sum of Two Cubes in Complex Domain}
Tags: Cube Roots of Unity, Algebra
\begin{theorem}
:$a^3 + b^3 = \paren {a + b} \paren {a \omega + b \omega^2} \paren {a \omega^2 + b \omega}$
where:
: $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
\end{theorem}
\begin{proof}
From Sum of Cubes of Three Indeterminates Minus 3 Times their Product:
:$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$
Setting $x \gets 0, y \gets a, z \gets b$:
:$0^3 + a^3 + b^3 - 3 \times 0 \times a b = \paren {0 + a + b} \paren {0 + \omega a + \omega^2 b} \paren {0 + \omega^2 a + \omega b}$
The result follows.
{{qed}}
\end{proof}
|
22301
|
\section{Sum of Two Fifth Powers}
Tags: Fifth Powers, Examples of Use of Sum of Two Odd Powers
\begin{theorem}
:$x^5 + y^5 = \paren {x + y} \paren {x^4 - x^3 y + x^2 y^2 - x y^3 + y^4}$
\end{theorem}
\begin{proof}
From Sum of Two Odd Powers:
:$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$
We have that $5 = 2 \times 2 + 1$.
The result follows directly by setting $n = 2$.
{{qed}}
\end{proof}
|
22302
|
\section{Sum of Two Fourth Powers}
Tags: Fourth Powers, Algebra
\begin{theorem}
:$x^4 + y^4 = \paren {x^2 + \sqrt 2 x y + y^2} \paren {x^2 - \sqrt 2 x y + y^2}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | r = \paren {x^2 + \sqrt 2 x y + y^2} \paren {x^2 - \sqrt 2 x y + y^2}
| o =
| c =
}}
{{eqn | r = x^2 \paren {x^2 - \sqrt 2 x y + y^2} + \sqrt 2 x y \paren {x^2 - \sqrt 2 x y + y^2} + y^2 \paren {x^2 - \sqrt 2 x y + y^2}
| c =
}}
{{eqn | r = x^4 - \sqrt 2 x^3 y + x^2 y^2 + \sqrt 2 x^3 y - 2 x^2 y^2 + \sqrt 2 x y^3 + x^2 y^2 - \sqrt 2 x y^3 + y^4
| c =
}}
{{eqn | r = x^4 + y^4
| c = gathering terms
}}
{{end-eqn}}
{{qed}}
Category:Fourth Powers
\end{proof}
|
22303
|
\section{Sum of Two Odd Powers/Examples/Sum of Two Cubes}
Tags: Third Powers, Algebra, Examples of Use of Sum of Two Odd Powers, Cubes, Sum of Two Cubes
\begin{theorem}
:$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$
\end{theorem}
\begin{proof}
From Difference of Two Powers:
:$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n-1} a^{n - j - 1} b^j$
Let $x = a$ and $y = -b$.
Then:
{{begin-eqn}}
{{eqn | l = x^3 + y^3
| r = x^3 - \paren {-y^3}
| c =
}}
{{eqn | r = x^3 - \paren {-y}^3
| c =
}}
{{eqn | r = \paren {x - \paren {-y} } \sum_{j \mathop = 0}^2 x^{n - j - 1} \paren {-y}^j
| c =
}}
{{eqn | r = \paren {x + y} \paren {x^2 + x \paren {-y} + \paren {-y}^2}
| c =
}}
{{eqn | r = \paren {x + y} \paren {x^2 - x y + y^2}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22304
|
\section{Sum of Two Rational 4th Powers but not Two Integer 4th Powers}
Tags: Fourth Powers, 5906
\begin{theorem}
$5906$ is the smallest integer which can be expressed as the sum of two rational $4$th powers, but not two integer $4$th powers.
\end{theorem}
\begin{proof}
:$5906 = \paren {\dfrac {149} {17} }^4 + \paren {\dfrac {25} {17} }^4$
Suppose $5906$ is a sum of two integer $4$th powers.
We have:
:$9^4 = 6561 > 5906$
which shows that no $4$th power greater than $8^4$ is in the sum.
:$7^4 + 7^4 = 4802 < 5906$
which shows that some $4$th power greater than $7^4$ is in the sum.
So the sum must contain $8^4$.
We have:
:$5906 - 8^4 = 1810$
but $1810$ is not an integer $4$th power.
Therefore $5906$ is not a sum of two integer $4$th powers.
{{finish|It remains to show that it is the smallest such. It is, acccording to A NEW CHARACTERIZATION OF THE INTEGER 5906 by A. Bremner and P. Morton}}
\end{proof}
|
22305
|
\section{Sum of Two Sides of Triangle Greater than Third Side}
Tags: Triangles, Triangle Inequality, Euclid Book I
\begin{theorem}
Given a triangle $ABC$, the sum of the lengths of any two sides of the triangle is greater than the length of the third side.
{{:Euclid:Proposition/I/20}}
\end{theorem}
\begin{proof}
:350 px
Let $ABC$ be a triangle
We can extend $BA$ past $A$ into a straight line.
There exists a point $D$ such that $DA = CA$.
Therefore, from Isosceles Triangle has Two Equal Angles:
:$\angle ADC = \angle ACD$
Thus by Euclid's fifth common notion:
:$\angle BCD > \angle BDC$
Since $\triangle DCB$ is a triangle having $\angle BCD$ greater than $\angle BDC$, this means that $BD > BC$.
But:
:$BD = BA + AD$
and:
:$AD = AC$
Thus:
:$BA + AC > BC$
A similar argument shows that $AC + BC > BA$ and $BA + BC > AC$.
{{qed}}
{{Euclid Note|20|I|It is a geometric interpretation of the Triangle Inequality.}}
\end{proof}
|
22306
|
\section{Sum of Two Squares not Congruent to 3 modulo 4}
Tags: Sum of Squares, Sums of Squares
\begin{theorem}
Let $n \in \Z$ such that $n = a^2 + b^2$ where $a, b \in \Z$.
Then $n$ is not congruent modulo $4$ to $3$.
\end{theorem}
\begin{proof}
Let $n \equiv 3 \pmod 4$.
{{AimForCont}} $n$ can be expressed as the sum of two squares:
:$n = a^2 + b^2$.
From Square Modulo 4, either $a^2 \equiv 0$ or $a^2 \equiv 1 \pmod 4$.
Similarly for $b^2$.
So $a^2 + b^2 \not \equiv 3 \pmod 4$ whatever $a$ and $b$ are.
Thus $n$ cannot be the sum of two squares.
The result follows by Proof by Contradiction.
{{qed}}
Category:Sums of Squares
\end{proof}
|
22307
|
\section{Sum of Unitary Divisors is Multiplicative}
Tags: Multiplicative Functions, Sum of Unitary Divisors
\begin{theorem}
Let $\map {\sigma^*} n$ denote the sum of unitary divisors of $n$.
Then the function:
:$\ds \sigma^*: \Z_{>0} \to \Z_{>0}: \map {\sigma^*} n = \sum_{\substack d \mathop \divides n \\ d \mathop \perp \frac n d} d$
is multiplicative.
\end{theorem}
\begin{proof}
Let $a, b$ be coprime integers.
Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $b$.
That is, any divisor $d$ of $a b$ is in the form:
:$d = a_i b_j$
in a unique way, where $a_i \divides a$ and $b_j \divides b$.
First we show that:
:$d$ is an unitary divisor of $a b$ {{iff}} $a_i, b_j$ are unitary divisors of $a, b$ respectively
In the forward implication we are given $d \perp \dfrac {a b} d$.
By Divisor of One of Coprime Numbers is Coprime to Other:
:$a_i, b_j \perp \dfrac {a b} d$
By Divisor of One of Coprime Numbers is Coprime to Other again:
:$a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$
In the backward implication we are given $a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$.
By Divisor of One of Coprime Numbers is Coprime to Other:
:$a \perp b \implies \paren {a_i \perp b_j \land \dfrac a {a_i} \perp \dfrac b {b_j} }$
By Product of Coprime Pairs is Coprime:
:$d = a_i b_j \perp \dfrac a {a_i} \dfrac b {b_j} = \dfrac {a b} d$
{{qed|lemma}}
We can list the unitary divisors of $a$ and $b$ as
:$1, a_1, a_2, \ldots, a$
and:
:$1, b_1, b_2, \ldots, b$
and thus the sum of their unitary divisors are:
:$\ds \map {\sigma^*} a = \sum_{i \mathop = 1}^r a_i$
:$\ds \map {\sigma^*} b = \sum_{j \mathop = 1}^s b_j$
Consider all unitary divisors of $a b$ with the same $a_i$.
Their sum is:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^s a_i b_j
| r = a_i \sum_{j \mathop = 1}^s b_j
| c =
}}
{{eqn | r = a_i \map {\sigma^*} b
| c =
}}
{{end-eqn}}
Summing over all $a_i$:
{{begin-eqn}}
{{eqn | l = \map {\sigma^*} {a b}
| r = \sum_{i \mathop = 1}^r \paren {a_i \map {\sigma^*} b}
| c =
}}
{{eqn | r = \paren {\sum_{i \mathop = 1}^r a_i} \map {\sigma^*} b
| c =
}}
{{eqn | r = \map {\sigma^*} a \map {\sigma^*} b
| c =
}}
{{end-eqn}}
{{qed}}
Category:Sum of Unitary Divisors
Category:Multiplicative Functions
\end{proof}
|
22308
|
\section{Sum of Unitary Divisors of Integer}
Tags: Sum of Unitary Divisors
\begin{theorem}
Let $n$ be an integer such that $n \ge 2$.
Let $\map {\sigma^*} n$ be the sum of all positive unitary divisors of $n$.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Then:
:$\ds \map {\sigma^*} n = \prod_{1 \mathop \le i \mathop \le r} \paren {1 + p_i^{k_i} }$
\end{theorem}
\begin{proof}
We have that the Sum of Unitary Divisors is Multiplicative.
From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have:
:$\map {\sigma^*} n = \map {\sigma^*} {p_1^{k_1} } \map {\sigma^*} {p_2^{k_2} } \ldots \map {\sigma^*} {p_r^{k_r} }$
From Sum of Unitary Divisors of Power of Prime, we have:
:$\ds \map {\sigma^*} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
Hence the result.
{{qed}}
Category:Sum of Unitary Divisors
\end{proof}
|
22309
|
\section{Sum of Unitary Divisors of Power of Prime}
Tags: Prime Numbers, Sum of Unitary Divisors
\begin{theorem}
Let $n = p^k$ be the power of a prime number $p$.
Then the sum of all positive unitary divisors of $n$ is $1 + n$.
\end{theorem}
\begin{proof}
Let $d \divides n$.
By Divisors of Power of Prime, $d = p^a$ for some positive integer $a \le k$.
We have $\dfrac n d = p^{k - a}$.
Suppose $d$ is a unitary divisor of $n$.
Then $d$ and $\dfrac n d$ are coprime.
If both $a, k - a \ne 0$, $p^a$ and $p^{k - a}$ have a common divisor: $p$.
Hence either $a = 0$ or $k - a = 0$.
This leads to $d = 1$ or $p^k$.
Hence the sum of all positive unitary divisors of $n$ is:
:$1 + p^k = 1 + n$
{{qed}}
Category:Prime Numbers
Category:Sum of Unitary Divisors
\end{proof}
|
22310
|
\section{Sum of Variances of Independent Trials}
Tags: Expectation
\begin{theorem}
Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of experiments whose outcomes are independent of each other.
Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\EE_1, \EE_2, \ldots, \EE_n$ respectively.
Let $\var {X_j}$ be the variance of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then:
:$\ds \var {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \var {X_j}$
That is, the sum of the variances equals the variance of the sum.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \var {\sum_{j \mathop = 1}^n X_j}
| r = \expect {\paren {\sum_{j \mathop = 1}^n X_j}^2} - \expect {\sum_{j \mathop = 1}^n X_j}^2
| c = Variance as Expectation of Square minus Square of Expectation/Discrete
}}
{{eqn | r = \expect {\sum_{0 \mathop < i \mathop < j \mathop \le n} 2 X_i X_j + \sum_{j \mathop = 1}^n X_j^2} - 2 \sum_{0 \mathop < i \mathop < j \mathop \le n} \expect {X_i} \expect {X_j} - \sum_{j \mathop = 1}^n \expect{X_j}^2
| c =
}}
{{eqn | r = \sum_{0 \mathop < i \mathop < j \mathop \le n} 2 \expect {X_i X_j} + \sum_{j \mathop = 1}^n \expect {X_j^2} - 2 \sum_{0 \mathop < i \mathop < j \mathop \le n} \expect {X_i} \expect {X_j} - \sum_{j \mathop = 1}^n \expect{X_j}^2
| c = Linearity of Expectation Function/Discrete
}}
{{eqn | r = 2 \sum_{0 \mathop < i \mathop < j \mathop \le n} \expect {X_i} \expect {X_j} + \sum_{j \mathop = 1}^n \expect {X_j^2} - 2 \sum_{0 \mathop < i \mathop < j \mathop \le n} \expect {X_i} \expect {X_j} - \sum_ {j \mathop = 1}^n \expect {X_j}^2
| c = Condition for Independence from Product of Expectations/Corollary
}}
{{eqn | r = \sum_{j \mathop = 1}^n \expect {X_j^2} - \sum_{j \mathop = 1}^n \expect {X_j}^2
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^n \expect {X_j^2} - \expect {X_j}^2
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^n \var {X_j}
| c = Variance as Expectation of Square minus Square of Expectation/Discrete
}}
{{end-eqn}}
{{qed}}
Category:Expectation
\end{proof}
|
22311
|
\section{Sum of Wholly Real Numbers is Wholly Real}
Tags: Complex Addition
\begin{theorem}
Let $x = \tuple {a, 0}$ and $y = \tuple {b, 0}$ be wholly real complex numbers.
Then $x + y$ is also wholly real.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = x + y
| r = \tuple {a, 0} + \tuple {b, 0}
| c =
}}
{{eqn | r = \tuple {a + b, 0 + 0}
| c = {{Defof|Complex Addition}}
}}
{{eqn | r = \tuple {a + b, 0}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22312
|
\section{Sum of r+k Choose k up to n}
Tags: Binomial Coefficients
\begin{theorem}
Let $r \in \R$ be a real number.
Then:
:$\ds \forall n \in \Z: n \ge 0: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
where $\dbinom r k$ is a binomial coefficient.
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition
:$\ds \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
\end{proof}
|
22313
|
\section{Sum of r Powers is between Power of Maximum and r times Power of Maximum}
Tags: Inequalities
\begin{theorem}
Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n \le r a^n$
\end{theorem}
\begin{proof}
This proof is divided into $2$ parts:
\end{proof}
|
22314
|
\section{Sum of r Powers is not Greater than r times Power of Maximum}
Tags: Inequalities
\begin{theorem}
Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a_1^n + a_2^n + \cdots + a_r^n \le r a^n$
\end{theorem}
\begin{proof}
By definition of the $\max$ operation:
:$\exists k \in \set {1, 2, \ldots, r}: a_k = a$
Then:
:$\forall i \in \set {1, 2, \ldots, r}: a_i \le a_k$
Hence:
{{begin-eqn}}
{{eqn | q = \forall i \in \set {1, 2, \ldots, r}
| l = a_i
| r = a_k
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{i \mathop = 1}^r a_i^n
| o = \le
| r = \sum_{i \mathop = 1}^r a_k^n
| c =
}}
{{eqn | r = r a_k^n
| c =
}}
{{eqn | r = r a^n
| c = Definition of $a_k$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22315
|
\section{Sum of two Fourth Powers cannot be Fourth Power}
Tags: Number Theory
\begin{theorem}
$\forall a, b, c \in \Z_{>0}$, the equation $a^4 + b^4 = c^4$ has no solutions.
\end{theorem}
\begin{proof}
This is a direct consequence of Fermat's Right Triangle Theorem.
{{qed}}
\end{proof}
|
22316
|
\section{Sum over Complement of Finite Set}
Tags: Summation, Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $T \subseteq S$ be a subset.
Let $S \setminus T$ be its relative complement.
Then we have the equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S \setminus T} \map f s = \sum_{s \mathop \in S} \map f s - \sum_{t \mathop \in T} \map f t$
\end{theorem}
\begin{proof}
Note that by Subset of Finite Set is Finite, $T$ is indeed finite.
By Set is Disjoint Union of Subset and Relative Complement, $S$ is the disjoint union of $S \setminus T$ and $T$.
The result now follows from Sum over Disjoint Union of Finite Sets.
{{qed}}
\end{proof}
|
22317
|
\section{Sum over Disjoint Union of Finite Sets}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ and $T$ be finite disjoint sets.
Let $S \cup T$ be their union.
Let $f: S \cup T \to \mathbb A$ be a mapping.
Then we have the equality of summations over finite sets:
:$\ds \sum_{u \mathop \in S \mathop \cup T} \map f u = \sum_{s \mathop \in S} \map f s + \sum_{t \mathop \in T} \map f t$
\end{theorem}
\begin{proof}
Note that by Union of Finite Sets is Finite, the union $S \cup T$ is finite.
Let $m$ be the cardinality of $S$ and $n$ be the cardinality of $T$.
Let $\N_{< m}$ denote an initial segment of the natural numbers.
Let $\sigma: \N_{<m} \to S$ and $\tau: \N_{<n} \to T$ be bijections.
Let $\alpha: \N_{< n} \to \closedint m {m + n - 1}$ be the mapping defined as:
:$\map \alpha k = k + m$
By Translation of Integer Interval is Bijection, $\alpha$ is a bijection.
By Composite of Bijections is Bijection and Disjoint Union of Bijections is Bijection, the union:
:$\sigma \cup \paren {\tau \circ \alpha} : \N_{<m} \cup \closedint m {m + n - 1} \to S \cup T$ is a bijection.
By Union of Integer Intervals, $\N_{<m} \cup \closedint m {m + n - 1} = \N_{< m + n}$.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{u \mathop \in S \mathop \cup T} \map f u
| r = \sum_{i \mathop = 0}^{m + n - 1} \map {f \circ \paren {\sigma \cup \paren {\tau \circ \alpha} } } i
| c = {{Defof|Summation}}
}}
{{eqn | r = \sum_{i \mathop = 0}^{m - 1} \map {f \circ \paren {\sigma \cup \paren {\tau \circ \alpha} } } i + \sum_{i \mathop = m}^{m + n - 1} \map {f \circ \paren {\sigma \cup \paren {\tau \circ \alpha} } } i
| c = Indexed Summation over Adjacent Intervals
}}
{{eqn | r = \sum_{i \mathop = 0}^{m - 1} \map {\paren {f \circ \sigma} } i + \sum_{i \mathop = m}^{m + n - 1} \map {\paren {g \circ \tau \circ \alpha} } i
| c = Restriction of Union of Mappings to Component of Domain
}}
{{eqn | r = \sum_{i \mathop = 0}^{m - 1} \map {\paren {f \circ \sigma} } i + \sum_{i \mathop = 0}^{n - 1} \map {\paren {g \circ \tau} } i
| c = Definition of $\alpha$, Indexed Summation over Translated Interval
}}
{{eqn | r = \sum_{s \mathop \in S} \map f s + \sum_{t \mathop \in T} \map g t
| c = {{Defof|Summation}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22318
|
\section{Sum over Integers of Cosine of n + alpha of theta over n + alpha}
Tags: Cosine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number which is specifically not an integer.
For $0 \le \theta < 2 \pi$:
:$\ds \dfrac 1 \alpha + \sum_{n \mathop \ge 1} \dfrac {2 \alpha} {\alpha^2 - n^2} = \sum_{n \mathop \in \Z} \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}$
\end{theorem}
\begin{proof}
First we establish the following, as they will be needed later.
{{begin-eqn}}
{{eqn | o =
| r = \cos \paren {\alpha + n} \theta + \cos \paren {\alpha - n} \theta
| c =
}}
{{eqn | r = 2 \map \cos {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \cos {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}
| c = Prosthaphaeresis Formula for Cosine plus Cosine
}}
{{eqn | n = 1
| r = 2 \cos \alpha \theta \cos n \theta
| c = simplification
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | o =
| r = \cos \paren {\alpha + n} \theta - \cos \paren {\alpha - n} \theta
| c =
}}
{{eqn | r = -2 \map \sin {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \sin {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}
| c = Prosthaphaeresis Formula for Cosine minus Cosine
}}
{{eqn | n = 2
| r = -2 \sin \alpha \theta \sin n \theta
| c = simplification
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{n \mathop = -\infty}^\infty \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}
| c =
}}
{{eqn | r = \dfrac {\cos \paren {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^\infty \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -\infty}^{-1} \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}
| c =
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^\infty \dfrac {\cos \paren {-n + \alpha} \theta} {-n + \alpha}
| c =
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\cos \paren {\alpha + n} \theta} {\alpha + n} + \dfrac {\cos \paren {\alpha - n} \theta} {\alpha - n} }
| c =
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\paren {\alpha - n} \paren {\cos \paren {\alpha + n} \theta} + \paren {\alpha + n} \paren {\cos \paren {\alpha - n} \theta} }{\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\alpha \cos \paren {\alpha + n} \theta - n \cos \paren {\alpha + n} \theta + \alpha \cos \paren {\alpha - n} \theta + n \cos \paren {\alpha - n} \theta} {\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\alpha \paren {\cos \paren {\alpha + n} \theta + \cos \paren {\alpha - n} \theta} - n \paren {\cos \paren {\alpha + n} \theta - \cos \paren {\alpha - n} \theta} } {\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {2 \alpha \cos \alpha \theta \cos n \theta + 2 n \sin \alpha \theta \sin n \theta} {\alpha^2 - n^2} }
| c = from $(1)$ and $(2)$
}}
{{eqn | r = \dfrac {\cos \alpha \theta} \alpha + 2 \alpha \cos \alpha \theta \sum_{n \mathop = 1}^\infty \paren {\dfrac {\cos n \theta} {\alpha^2 - n^2} } + 2 \sin \alpha \theta \sum_{n \mathop = 1}^\infty \paren {\dfrac {n \sin n \theta} {\alpha^2 - n^2} }
| c = Linear Combination of Indexed Summations
}}
{{end-eqn}}
Setting $\theta = 0$:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = -\infty}^\infty \dfrac {\map \cos {\paren {n + \alpha} 0} } {n + \alpha}
| r = \dfrac {\cos \alpha 0} \alpha + 2 \alpha \cos \alpha 0 \sum_{n \mathop = 1}^\infty \paren {\dfrac {\cos n 0 } {\alpha^2 - n^2} } + 2 \sin \alpha 0 \sum_{n \mathop = 1}^\infty \paren {\dfrac {n \sin n 0} {\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac 1 \alpha + \sum_{n \mathop = 1}^\infty \dfrac {2 \alpha} {\alpha^2 - n^2} + 0
| c = Sine of Zero is Zero and Cosine of Zero is One
}}
{{eqn | r = \dfrac 1 \alpha + \sum_{n \mathop = 1}^\infty \dfrac {2 \alpha} {\alpha^2 - n^2}
| c =
}}
{{end-eqn}}
To establish this identity for all other values of $\theta$ on the interval $0 \le \theta < 2\pi$, we will demonstrate that the sum is a constant function.
We will do this by showing that the derivative of the function is zero everywhere which by Zero Derivative implies Constant Function will complete the proof.
We have:
{{begin-eqn}}
{{eqn | l = \map f \theta
| r = \sum_{n \mathop \in \Z} \dfrac {\map \cos {n + \alpha} \theta} {n + \alpha}
| c =
}}
{{eqn | l = \map {f'} \theta
| r = \sum_{n \mathop \in \Z} -\map \sin {n + \alpha} \theta
| c = Derivative of Cosine Function/Corollary
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map {f'} \theta
| r = -\sum_{n \mathop \in \Z} \paren {\map \sin {\alpha \theta} \map \cos {n \theta} + \map \cos {\alpha \theta } \map \sin {n \theta} }
| c = Sine of Sum
}}
{{eqn | r = -\map \sin {\alpha \theta } \sum_{n \mathop \in \Z} \map \cos {n \theta} - \map \cos {\alpha \theta} \sum_{n \mathop \in \Z} \map \sin {n \theta}
| c = Linear Combination of Indexed Summations
}}
{{end-eqn}}
From Cosine Function is Even and Sine Function is Odd, we have:
:$\map \cos {-n \theta} = \map \cos {n \theta}$
and:
:$\map \sin {-n \theta} = -\map \sin {n \theta}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {f'} \theta
| r = -\map \sin {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \map \cos {n \theta} } - \map \cos {\alpha \theta} \times 0
| c =
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \map \cos {n \theta} }
| c =
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {\dfrac {e^{i n \theta } + e^{-i n \theta} } 2} }
| c = Cosine Exponential Formulation/Real Domain
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {1 + \sum_{n \mathop = 1}^\infty e^{i n \theta} + \sum_{n \mathop = 1}^\infty e^{-i n \theta} }
| c = Linear Combination of Indexed Summations
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {\sum_{n \mathop = 0}^\infty e^{i n \theta} + \sum_{n \mathop = 0}^\infty e^{-i n \theta} - 1}
| c = re-indexing the sum
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {\dfrac 1 {1 - e^{i \theta} } + \dfrac 1 {1 - e^{-i \theta} } - 1}
| c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {\dfrac {\paren {1 - e^{-i \theta} } + \paren {1 - e^{i \theta} } - \paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } {\paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } }
| c =
}}
{{eqn | r = -\map \sin {\alpha \theta} \paren {\dfrac {\paren {1 - e^{-i \theta} } + \paren {1 - e^{i \theta} } - \paren {1 - e^{i \theta} - e^{-i \theta} + 1} } {\paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } }
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22319
|
\section{Sum over Integers of Sine of n + alpha of theta over n + alpha}
Tags: Sine Function
\begin{theorem}
Let $\alpha \in \R$ be a real number which is specifically not an integer.
For $0 < \theta < 2\pi$:
:$\ds \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} = \pi$
\end{theorem}
\begin{proof}
First we establish the following, as they will be needed later.
{{begin-eqn}}
{{eqn | o =
| r = \map \sin {\alpha + n} \theta + \map \sin {\alpha - n} \theta
| c =
}}
{{eqn | r = 2 \map \sin {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \cos {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}
| c = Prosthaphaeresis Formula for Sine plus Sine
}}
{{eqn | n = 1
| r = 2 \sin \alpha \theta \cos n \theta
| c = simplification
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | o =
| r = \map \sin {\alpha + n} \theta - \map \sin {\alpha - n} \theta
| c =
}}
{{eqn | r = 2 \map \cos {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \sin {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}
| c = Prosthaphaeresis Formula for Sine minus Sine
}}
{{eqn | n = 2
| r = 2 \cos \alpha \theta \sin n \theta
| c = simplification
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | o =
| r = \sum_{n \mathop = -\infty}^\infty \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}
| c =
}}
{{eqn | r = \dfrac {\map \sin {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^\infty \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -\infty}^{-1} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^\infty \dfrac {\map \sin {-n + \alpha} \theta} {-n + \alpha}
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\map \sin {\alpha + n} \theta} {\alpha + n} + \dfrac {\map \sin {\alpha - n} \theta} {\alpha - n} }
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\paren {\alpha - n} \map \sin {\alpha + n} \theta + \paren {\alpha + n} \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\alpha \map \sin {\alpha + n} \theta - n \map \sin {\alpha + n} \theta + \alpha \map \sin {\alpha - n} \theta + n \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty\paren {\dfrac {\alpha \paren {\map \sin {\alpha + n} \theta + \map \sin {\alpha - n} \theta} - n \paren {\map \sin {\alpha + n} \theta - \map \sin {\alpha - n} \theta} } {\alpha^2 - n^2} }
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {2 \alpha \sin \alpha \theta \cos n \theta - 2 n \cos \alpha \theta \sin n \theta} {\alpha^2 - n^2} }
| c = from $(1)$ and $(2)$
}}
{{eqn | r = \dfrac {\sin \alpha \theta} \alpha + 2 \sin \alpha \theta \sum_{n \mathop = 1}^\infty \dfrac {\alpha \cos n \theta } {\alpha^2 - n^2} - 2 \cos \alpha \theta \sum_{n \mathop = 1}^\infty \dfrac {n \sin n \theta} {\alpha^2 - n^2}
| c = Linear Combination of Indexed Summations
}}
{{end-eqn}}
Setting $\theta = \pi$:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = -\infty}^\infty \dfrac {\map \sin {n + \alpha} \pi} {n + \alpha}
| r = \dfrac {\sin \alpha \pi} \alpha + 2 \sin \alpha \pi \sum_{n \mathop = 1}^\infty \dfrac {\alpha \cos n \pi } {\alpha^2 - n^2} - 2 \cos \alpha \pi \sum_{n \mathop = 1}^\infty \dfrac {n \sin n \pi} {\alpha^2 - n^2}
| c =
}}
{{eqn | r = \dfrac {\sin \alpha \pi} \alpha + 2 \sin \alpha \pi \sum_{n \mathop = 1}^\infty \dfrac {\alpha \cos n \pi } {\alpha^2 - n^2} - 0
| c = $\sin n \pi = 0$
}}
{{eqn | n = 3
| r = \dfrac {\sin \pi \alpha} \alpha + 2 \sin \pi \alpha \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac \alpha {\alpha^2 - n^2}
| c = $\cos n \pi = \paren {-1}^n$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \pi \cosec \pi \alpha
| r = \dfrac 1 \alpha + 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}
| c = Mittag-Leffler Expansion for the Cosecant Function
}}
{{eqn | ll= \leadsto
| l = \pi
| r = \sin \pi \alpha \paren{\dfrac 1 \alpha + 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2} }
| c = multiplying both sides by $\sin \pi \alpha$
}}
{{eqn | n = 4
| r = \dfrac {\sin \pi \alpha} \alpha + 2 \sin \pi \alpha \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}
| c =
}}
{{eqn | ll= \leadsto
| l = \pi
| r = \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \pi} {n + \alpha}
| c = equating $(3)$ and $(4)$
}}
{{end-eqn}}
To establish this identity for all other values of $\theta$ on the interval $0 < \theta < 2 \pi$, we will demonstrate that the sum is a constant function.
We will do this by showing that the derivative of the function is zero everywhere which by Zero Derivative implies Constant Function will complete the proof.
We have:
{{begin-eqn}}
{{eqn | l = \map f \theta
| r = \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}
| c =
}}
{{eqn | l = \map {f'} \theta
| r = \sum_{n \mathop \in \Z} \map \cos {n + \alpha} \theta
| c = Derivative of Sine Function/Corollary
}}
{{eqn | ll= \leadsto
| l = \map {f'} \theta
| r = \sum_{n \mathop \in \Z} \paren {\map \cos {\alpha \theta } \map \cos {n \theta} - \map \sin {\alpha \theta } \map \sin {n \theta} }
| c = Cosine of Sum
}}
{{eqn | r = \map \cos {\alpha \theta} \sum_{n \mathop \in \Z} \map \cos {n \theta} - \map \sin {\alpha \theta} \sum_{n \mathop \in \Z} \map \sin {n \theta}
| c = Linear Combination of Indexed Summations
}}
{{end-eqn}}
From Cosine Function is Even and Sine Function is Odd, we have:
:$\map \cos {-n \theta} = \map \cos {n \theta}$
and:
:$\map \sin {-n \theta} = -\map \sin {n \theta}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {f'} \theta
| r = \map \cos {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \map \cos {n \theta} } - \map \sin {\alpha \theta} \times 0
| c =
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \map \cos {n \theta} }
| c =
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {\dfrac {e^{i n \theta} + e^{-i n \theta} } 2} }
| c = Cosine Exponential Formulation/Real Domain
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {1 + \sum_{n \mathop = 1}^\infty e^{i n \theta} + \sum_{n \mathop = 1}^\infty e^{-i n \theta} }
| c = Linear Combination of Indexed Summations
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {\sum_{n \mathop = 0}^\infty e^{i n \theta} + \sum_{n \mathop = 0}^\infty e^{-i n \theta} - 1 }
| c = Re-index the sum
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {\dfrac 1 {1 - e^{i \theta} } + \dfrac 1 {1 - e^{-i \theta} } - 1}
| c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {\dfrac {\paren {1 - e^{-i \theta} } + \paren {1 - e^{i \theta} } - \paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } {\paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } }
| c =
}}
{{eqn | r = \map \cos {\alpha \theta} \paren {\dfrac {\paren {1 - e^{-i \theta} } + \paren {1 - e^{i \theta} } - \paren {1 - e^{i \theta} - e^{-i \theta} + 1} } {\paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } }
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22320
|
\section{Sum over Union of Finite Sets}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ and $T$ be finite sets.
Let $f: S \cup T \to \mathbb A$ be a mapping.
Then we have the equality of summations over finite sets:
:$\ds \sum_{u \mathop \in S \mathop \cup T} \map f u = \sum_{s \mathop \in S} \map f s + \sum_{t \mathop \in T} \map f t - \sum_{v \mathop \in S \mathop \cap T} \map f v$
\end{theorem}
\begin{proof}
Follows from:
:Mapping Defines Additive Function of Subalgebra of Power Set
:Power Set is Algebra of Sets
:Inclusion-Exclusion Principle
{{qed}}
\end{proof}
|
22321
|
\section{Sum over j of Function of Floor of mj over n}
Tags: Summations, Floor Function
\begin{theorem}
Let $f$ be a real function.
Then:
:$\ds \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} } = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} - \map f r} + n \map f {m - 1}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = r
| m = \floor {\dfrac {m j} n}
| c =
}}
{{eqn | ll= \leadsto
| l = r
| o = \le
| m = \dfrac {m j} n
| mo= <
| r = r + 1
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac {r n} m
| o = \le
| m = j
| mo= <
| r = \dfrac {\paren {r + 1} n} m
| c =
}}
{{eqn | ll= \leadsto
| l = \ceiling {\dfrac {r n} m}
| o = \le
| m = j
| mo= <
| r = \ceiling {\dfrac {\paren {r + 1} n} m}
| c = as $j$ is an integer
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} }
| c =
}}
{{eqn | r = \sum_{0 \mathop \le \ceiling {\frac {r n} m} \mathop < n} \map f r
| c =
}}
{{eqn | r = \sum_{0 \mathop \le r \mathop < m} \map f r \paren {\ceiling {\dfrac {\paren {r + 1} n} m} - \ceiling {\dfrac {r n} m} }
| c =
}}
{{eqn | r = \map f 0 \ceiling {\dfrac n m} + \map f 1 \paren {\ceiling {\dfrac {2 n} m} - \ceiling {\dfrac n m} } + \cdots + \map f {m - 1} \paren {\ceiling {\dfrac {m n} m} - \ceiling {\dfrac {\paren {m - 1} n} m} }
| c =
}}
{{eqn | r = \ceiling {\dfrac n m} \paren {\map f 0 + \map f 1} + \ceiling {\dfrac {2 n} m} \paren {\map f 0 + \map f 1} + \cdots + \ceiling {\dfrac {\paren {m - 1} n} m} \paren {\map f {m - 2} + \map f {m - 1} } + n \map f {m - 1}
| c =
}}
{{eqn | r = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} + \map f r} + n \map f {m - 1}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22322
|
\section{Sum over k from 1 to n of n Choose k by Sine of n Theta}
Tags: Sine Function, Binomial Coefficients
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \dbinom n k \sin k \theta = \paren {2 \cos \dfrac \theta 2}^n \sin \dfrac {n \theta} 2$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {1 + e^{i \theta} }^n
| r = \sum_{k \mathop = 0}^n \dbinom n k e^{i k \theta}
| c = Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^n \dbinom n k \paren {\cos k \theta + i \sin k \theta}
| c = Euler's Formula
}}
{{eqn | ll= \leadsto
| l = \map \Im {\paren {1 + e^{i \theta} }^n}
| r = \map \Im {\sum_{k \mathop = 0}^n \dbinom n k \paren {\cos k \theta + i \sin k \theta} }
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^n \dbinom n k \sin k \theta
| c = taking imaginary parts
}}
{{eqn | r = \sum_{k \mathop = 1}^n \dbinom n k \sin k \theta
| c = as the zeroth term vanishes: $\sin 0 = 0$
}}
{{end-eqn}}
At the same time:
{{begin-eqn}}
{{eqn | l = \map \Im {\paren {1 + e^{i \theta} }^n}
| r = \map \Im {e^{i n \theta / 2} \paren {e^{i \theta / 2} + e^{-i \theta / 2} }^n}
| c =
}}
{{eqn | r = \map \Im {e^{i n \theta / 2} \paren {2 \cos \dfrac \theta 2}^n}
| c = Cosine Exponential Formulation
}}
{{eqn | r = \map \Im {\paren {\cos \dfrac {n \theta} 2 + i \sin \dfrac {n \theta} 2} \paren {2 \cos \dfrac \theta 2}^n}
| c = Euler's Formula
}}
{{eqn | r = \sin \dfrac {n \theta} 2 \paren {2 \cos \dfrac \theta 2}^n
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22323
|
\section{Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt}
Tags: Binomial Coefficients
\begin{theorem}
:$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
where $\delta_{n 0}$ is the Kronecker delta.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $zmsp P n$ be the proposition:
:$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
\end{proof}
|
22324
|
\section{Sum over k of -2 Choose k}
Tags: Binomial Coefficients
\begin{theorem}
:$\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
where:
:$\dbinom {-2} k$ is a binomial coefficient
:$\ceiling x$ denotes the ceiling of $x$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom {-2} k
| r = \sum_{k \mathop = 0}^n \paren {-1} \binom {k - \paren {-2} - 1} k
| c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \sum_{k \mathop = 0}^n \paren {-1} \binom {k + 1} k
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^n \paren {-1} \paren {k + 1}
| c = Binomial Coefficient with Self minus One
}}
{{eqn | r = 1 - 2 + 3 - 4 + \cdots \pm \paren {n + 1}
| c =
}}
{{eqn | r = \paren {1 + 2 + 3 + 4 + \cdots + \paren {n + 1} } - 2 \times \paren {2 + 4 + 6 + 8 + \cdots + m}
| c = where $m = n$ or $m = n + 1$ according to whether $n$ is odd or even
}}
{{end-eqn}}
When $n$ is even, we have:
{{begin-eqn}}
{{eqn | r = \paren {1 + 2 + 3 + 4 + \cdots + \paren {n + 1} } - 2 \times \paren {2 + 4 + 6 + 8 + \cdots + n}
| o =
| c =
}}
{{eqn | r = \paren {1 + 2 + 3 + 4 + \cdots + \paren {n + 1} } - 4 \paren {1 + 2 + 3 + 4 + \cdots + \frac n 2}
| c =
}}
{{eqn | r = \sum_{k \mathop = 2}^{n + 1} k - 4 \sum_{k \mathop = 1}^{\frac n 2} k
| c =
}}
{{eqn | r = \frac {\paren {n + 1} \paren {n + 2} } 2 - 4 \frac {\frac n 2 \paren {\frac n 2 + 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {\paren {n + 1} \paren {n + 2} } 2 - \frac {n \paren {n + 2} } 2
| c = simplifying
}}
{{eqn | r = \frac {\paren {n + 2} \paren {n + 1 - n} } 2
| c = simplifying
}}
{{eqn | r = \frac {n + 2} 2
| c = simplifying
}}
{{end-eqn}}
As $n$ is even, $n + 1$ is odd, and so:
:$\dfrac {n + 2} 2 = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
{{qed|lemma}}
When $n$ is odd, we have:
{{begin-eqn}}
{{eqn | r = \paren {1 + 2 + 3 + 4 + \cdots + \paren {n + 1} } - 2 \times \paren {2 + 4 + 6 + 8 + \cdots + n + 1}
| o =
| c =
}}
{{eqn | r = \paren {1 + 2 + 3 + 4 + \cdots + \paren {n + 1} } - 4 \paren {1 + 2 + 3 + 4 + \cdots + \frac {n + 1} 2}
| c =
}}
{{eqn | r = \sum_{k \mathop = 2}^{n + 1} k - 4 \sum_{k \mathop = 1}^{\frac {n + 1} 2} k
| c =
}}
{{eqn | r = \frac {\paren {n + 1} \paren {n + 2} } 2 - 4 \frac {\frac {n + 1} 2 \paren {\frac {n + 1} 2 + 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {\paren {n + 1} \paren {n + 2} } 2 - \frac {\paren {n + 1} \paren {n + 3} } 2
| c = simplifying
}}
{{eqn | r = \frac {\paren {n + 1} \paren {\paren {n + 2} - \paren {n + 3} } } 2
| c = simplifying
}}
{{eqn | r = \paren {-1} \frac {n + 1} 2
| c = simplifying
}}
{{end-eqn}}
As $n$ is odd, $n + 1$ is even, and so $\dfrac {n + 1} 2$ is an integer.
Thus from Real Number is Integer iff equals Ceiling:
:$\paren {-1} \dfrac {n + 1} 2 = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
{{qed|lemma}}
Thus:
:$\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
whether $n$ is odd or even.
Hence the result.
{{qed}}
\end{proof}
|
22325
|
\section{Sum over k of Floor of Log base b of k}
Tags: Logarithms, Summations, Floor Function
\begin{theorem}
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\log_b k} = \paren {n + 1} \floor {\log_b n} - \dfrac {b^{\floor {\log_b n} + 1} - b} {b - 1}$
\end{theorem}
\begin{proof}
From Sum of Sequence as Summation of Difference of Adjacent Terms:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \floor {\log_b k} = n \floor {\log_b n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\map {\log_b} {k + 1} } - \floor {\log_b k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\map {\log_b} {k + 1} } - \floor {\log_b k} }$
As $b \ge 2$, we have that:
:$\map {\log_b} {k + 1} - \log_b k < 1$
As $b$ is an integer:
:$\lfloor {\map {\log_b} {k + 1} } - \floor {\log_b k} = 1$
{{iff}} $k + 1$ is a power of $b$.
So:
{{begin-eqn}}
{{eqn | l = S
| r = \sum_{k \mathop = 1}^{n - 1} k \sqbrk {k + 1 \text{ is a power of } b}
| c = where $\sqbrk {\, \cdot \,}$ is Iverson's convention
}}
{{eqn | r = \sum_{1 \mathop \le t \mathop \le \log_b n} \paren {b^t - 1}
| c =
}}
{{eqn | r = \sum_{t \mathop = 1}^{\floor {\log_b n} } \paren {b^t - 1}
| c =
}}
{{eqn | r = \sum_{t \mathop = 1}^{\floor {\log_b n} } b^t - \sum_{t \mathop = 1}^{\floor {\log_b n} } 1
| c =
}}
{{eqn | r = \sum_{t \mathop = 1}^{\floor {\log_b n} } b^t - \floor {\log_b n}
| c =
}}
{{eqn | r = b \sum_{t \mathop = 0}^{\floor {\log_b n} - 1} b^t - \floor {\log_b n}
| c = extracting $b$ as a factor
}}
{{eqn | r = b \frac {b^{\floor {\log_b n} } - 1} {b - 1} - \floor {\log_b n}
| c = Sum of Geometric Progression
}}
{{eqn | r = \frac {b^{\floor {\log_b n} + 1} - b} {b - 1} - \floor {\log_b n}
| c =
}}
{{end-eqn}}
The result follows by substituting $S$ back into $(1)$ and factoring out $\floor {\log_b n}$.
{{qed}}
\end{proof}
|
22326
|
\section{Sum over k of Floor of Root k}
Tags: Summations, Floor Function
\begin{theorem}
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = \floor {\sqrt n} \paren {n - \dfrac {\paren {2 \floor {\sqrt n} + 5} \paren {\floor {\sqrt n} - 1} } 6}$
\end{theorem}
\begin{proof}
From Sum of Sequence as Summation of Difference of Adjacent Terms:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = n \floor {\sqrt n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - \floor {\sqrt k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - \floor {\sqrt k} }$
We have that:
:$\sqrt {k + 1} - \sqrt k < 1$
and so:
:$\floor {\sqrt {k + 1} } - \floor {\sqrt k} = 1$
{{iff}} $k + 1$ is a square number.
So:
{{begin-eqn}}
{{eqn | l = S
| r = \sum_{k \mathop = 1}^{n - 1} k \sqbrk {\text {$k + 1$ is a square number} }
| c =
}}
{{eqn | l = S
| r = \sum_{k \mathop = 2}^n (k - 1) \sqbrk {\text {$k$ is a square number} }
| c =
}}
{{eqn | r = \sum_{t \mathop = 2}^{\floor {\sqrt n} } \paren {t^2 - 1}
| c = $1$ is not included in this summation
}}
{{eqn | r = \sum_{t \mathop = 1}^{\floor {\sqrt n} } t^2 - 1 - \sum_{t \mathop = 2}^{\floor {\sqrt n} } 1
| c =
}}
{{eqn | r = \sum_{t \mathop = 1}^{\floor {\sqrt n} } t^2 - \sum_{t \mathop = 1}^{\floor {\sqrt n} } 1
| c =
}}
{{eqn | r = \frac {\floor {\sqrt n} \paren {\floor {\sqrt n} + 1} \paren {2 \floor {\sqrt n} + 1} } 6 - \floor {\sqrt n}
| c = Sum of Sequence of Squares
}}
{{eqn | r = \frac {\floor {\sqrt n} \paren {\floor {\sqrt n} + 1} \paren {2 \floor {\sqrt n} + 1} - 6 \floor {\sqrt n} } 6
| c =
}}
{{eqn | r = \floor {\sqrt n} \frac {\paren {\floor {\sqrt n} + 1} \paren {2 \floor {\sqrt n} + 1} - 6} 6
| c =
}}
{{eqn | r = \floor {\sqrt n} \frac {2 \floor {\sqrt n}^2 + 3 \floor {\sqrt n} - 5} 6
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 1}^n \floor {\sqrt k}
| r = n \floor {\sqrt n} - \floor {\sqrt n} \frac {\paren {2 \floor {\sqrt n} + 5} \paren {\floor {\sqrt n} - 1} } 6
| c =
}}
{{eqn | r = \floor {\sqrt n} \paren {n - \dfrac {\paren {2 \floor {\sqrt n} + 5} \paren {\floor {\sqrt n} - 1} } 6}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22327
|
\section{Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
where:
:$\dbinom n m$ denotes a binomial coefficient
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$\ds {n + 1 \brace k + 1}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
\end{proof}
|
22328
|
\section{Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
where:
:$\ds {k + 1 \brace m + 1}$ and so on denotes a Stirling number of the second kind
:$\dbinom n k$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
\end{proof}
|
22329
|
\section{Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
where:
:$\ds {k \brace m}$ denotes a Stirling number of the second kind
:$\dbinom n k$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
\end{proof}
|
22330
|
\section{Sum over k of Sum over j of Floor of n + jb^k over b^k+1}
Tags: Summations, Floor Function
\begin{theorem}
Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.
Then:
:$\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = n$
where $\floor {\, \cdot \,}$ denotes the floor function.
\end{theorem}
\begin{proof}
We have that $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$ is in the form $\floor {\dfrac {m k + x} n}$ so that:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }
| r = \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b}
| c =
}}
{{eqn | r = \sum_{0 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b} - \floor {\dfrac n {b^{k + 1} } }
| c =
}}
{{eqn | r = \dfrac {\paren {1 - 1} \paren {b - 1} } 2 + \dfrac {\paren {1 - 1} } 2 + 1 \floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } }
| c = Summation over k of Floor of mk+x over n
}}
{{eqn | r = \floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } }
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }
| r = \sum_{k \mathop \ge 0} \paren {\floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } } }
| c =
}}
{{eqn | r = \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } }
| c = {{Defof|Telescoping Series}}
}}
{{eqn | r = n
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22331
|
\section{Sum over k of Unsigned Stirling Number of the First Kind of n+1 with k+1 by Stirling Number of the Second Kind of k with m by -1^k-m}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
where:
:$\sqbrk {n \ge m}$ is Iverson's convention
:$\ds {n + 1 \brack k + 1}$ denotes an unsigned Stirling number of the first kind
:$\ds {k \brace m}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
\end{proof}
|
22332
|
\section{Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
where:
:$\ds {n + 1 \brack k + 1}$ etc. denotes an unsigned Stirling number of the first kind
:$\dbinom k m$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
\end{proof}
|
22333
|
\section{Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\dbinom k m$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
\end{proof}
|
22334
|
\section{Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
where:
:$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
:$\ds {m + k \brace k}$ denotes a Stirling number of the second kind
:$\ds {n \brack n - m}$ denotes an unsigned Stirling number of the first kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | r = \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brace k}
| o =
| c =
}}
{{eqn | r = \sum_k \binom {- n} k \binom n {n + k}
| c = Stirling Number of the Second Kind of Number with Self
}}
{{eqn | r = \sum_k \binom {- n} k \delta_{0 k}
| c = as $\dbinom n {n + k} = 0$ for $k = 0$, and Binomial Coefficient with Self
}}
{{eqn | r = \binom {- n} 0
| c = All terms but where $k = 0$ vanish
}}
{{eqn | r = 1
| c = Binomial Coefficient with Zero
}}
{{eqn | r = {n \brack n - 0}
| c = Unsigned Stirling Number of the First Kind of Number with Self
}}
{{end-eqn}}
So $\map P 0$ is seen to hold.
\end{proof}
|
22335
|
\section{Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
where:
:$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
:$\ds {m + k \brack k}$ denotes an unsigned Stirling number of the first kind
:$\ds {n \brace n - m}$ denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall n \in \Z_{\ge 0}: \ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
\end{proof}
|
22336
|
\section{Sum over k of m choose k by -1^m-k by k to the n}
Tags: Stirling Numbers, Binomial Coefficients
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
where:
:$\dbinom m k$ denotes a binomial coefficient
:$\ds {n \brace m}$ etc. denotes a Stirling number of the second kind
:$m!$ denotes a factorial.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
\end{proof}
|
22337
|
\section{Sum over k of n Choose k by Fibonacci Number with index m+k}
Tags: Binomial Coefficients, Fibonacci Numbers
\begin{theorem}
:$\ds \sum_{k \mathop \ge 0} \binom n k F_{m + k} = F_{m + 2 n}$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$F_n$ denotes the $n$th Fibonacci number.
\end{theorem}
\begin{proof}
From Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k:
:$(1): \quad \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
Letting $t = 2$ in $(1)$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k {F_2}^k {F_{2 - 1} }^{n - k} F_{m + k}
| r = \sum_{k \mathop \ge 0} \binom n k 1^k 1^{n - k} F_{m + k}
| c = {{Defof|Fibonacci Number}}: $F_1 = 1, F_2 = 1$
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom n k F_{m + k}
| c =
}}
{{eqn | r = F_{m + 2 n}
| c = from $(1)$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22338
|
\section{Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k}
Tags: Binomial Coefficients, Fibonacci Numbers
\begin{theorem}
:$\ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$F_n$ denotes the $n$th Fibonacci number.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall m, t \in \N: \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
$\map P 0$ is the case:
:$\ds \binom 0 0 {F_t}^0 {F_{t - 1} }^0 F_{m + 0} = F_m$
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22339
|
\section{Sum over k of n Choose k by p^k by (1-p)^n-k by Absolute Value of k-np}
Tags: Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_{k \mathop \in \Z} \dbinom n k p^k \paren {1 - p}^{n - k} \size {k - n p} = 2 \ceiling {n p} \dbinom n {\ceiling {n p} } p^{\ceiling {n p} } \paren {1 - p}^{n - 1 - \ceiling {n p} }$
\end{theorem}
\begin{proof}
Let $t_k = k \dbinom n k p^k \paren {1 - p}^{n + 1 - k}$.
Then:
:$t_k - t_{k + 1} = \dbinom n k p^k \paren {1 - p}^{n - k} \paren {k - n p}$
Thus the stated summation is:
:$\ds \sum_{k \mathop < \ceiling {n p} } \paren {t_{k + 1} - t_k} + \sum_{k \mathop \ge \ceiling {n p} } \paren {t_k - t_{k + 1} } = 2 t_{\ceiling {n p} }$
{{qed}}
\end{proof}
|
22340
|
\section{Sum over k of n Choose k by x to the k by kth Harmonic Number}
Tags: Harmonic Numbers, Binomial Coefficients
\begin{theorem}
Let $x \in \R_{> 0}$ be a real number.
Then:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \paren {x + 1}^n \paren {H_n - \map \ln {1 + \frac 1 x} } + \epsilon$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$H_k$ denotes the $k$th harmonic number
:$0 < \epsilon < \dfrac 1 {x \paren {n + 1} }$
\end{theorem}
\begin{proof}
Let $S_n := \ds \sum_{k \mathop \in \Z} \binom n k x^k H_k$.
Then:
{{begin-eqn}}
{{eqn | l = S_{n + 1}
| r = \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \paren {\binom n k + \binom n {k - 1} } x^k H_k
| c = Pascal's Rule
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom n k x^k H_k + \sum_{k \mathop \in \Z} \binom n {k - 1} x^k H_k
| c =
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} \paren {H_{k - 1} + \frac 1 k}
| c = Definition of $S_n$
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} H_{k - 1} + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} \frac 1 k
| c =
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 0} \binom n k x^k H_k + \sum_{k \mathop \ge 1} \binom n {k - 1} x^k \frac 1 k
| c = Translation of Index Variable of Summation
}}
{{eqn | r = S_n + x S_n + \sum_{k \mathop \ge 1} \frac k {n + 1} \binom {n + 1} k x^k \frac 1 k
| c = Definition of $S_n$, Factors of Binomial Coefficient
}}
{{eqn | r = \paren {x + 1} S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k x^k
| c = simplifying
}}
{{eqn | r = \paren {x + 1} S_n + \frac 1 {n + 1} \paren {\sum_{k \mathop \ge 0} \binom {n + 1} k x^k - \binom {n + 1} 0 x^0}
| c = forcing the summation to include the $k = 0$ term
}}
{{eqn | r = \paren {x + 1} S_n + \frac {\paren {x + 1}^{n + 1} - 1} {n + 1}
| c = Binomial Theorem and Binomial Coefficient with $0$
}}
{{eqn | ll= \leadsto
| l = \frac {S_{n + 1} } {\paren {x + 1}^{n + 1} }
| r = \frac {S_n} {\paren {x + 1}^n} + \frac 1 {n + 1} - \frac 1 {\paren {n + 1} \paren {x + 1}^{n + 1} }
| c = dividing through by $\paren {x + 1}^{n + 1}$
}}
{{end-eqn}}
As $S_1 = x$, we have that:
{{begin-eqn}}
{{eqn | l = \dfrac {S_n} {\paren {x + 1}^n}
| r = \frac {S_{n - 1} } {\paren {x + 1}^{n - 1} } + \frac 1 n - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = \frac {S_{n - 2} } {\paren {x + 1}^{n - 2} } + \frac 1 {n - 1} - \frac 1 {\paren {n - 1} \paren {x + 1}^{n - 1} } + \frac 1 n - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = \dots}}
{{eqn | r = \frac {S_1} {\paren {x + 1}^1} + \frac 1 2 + \frac 1 3 + \dots + \frac 1 n - \frac 1 {2 \paren {x + 1}^2} - \frac 1 {3 \paren {x + 1}^3} - \dots - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = \frac x {x + 1} - 1 + \frac 1 {x + 1} + 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 n - \frac 1 {x + 1} - \frac 1 {2 \paren {x + 1}^2} - \frac 1 {3 \paren {x + 1}^3} - \dots - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 n - \frac 1 {x + 1} - \frac 1 {2 \paren {x + 1}^2} - \frac 1 {3 \paren {x + 1}^3} - \dots - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = H_n - \ds \sum_{k \mathop = 1}^n \frac 1 {k \paren {x + 1}^k}
| c = {{Defof|Harmonic Number}}
}}
{{end-eqn}}
From Corollary to Power Series Expansion for $\map \ln {1 + x}$ we have:
:$\ds \map \ln {1 - z} = -\sum_{k \mathop \ge 1} \frac {z^k} k$
Hence by Logarithm of Reciprocal we have:
:$\ds \ln \frac 1 {1 - z} = -\map \ln {1 - z} = \sum_{k \mathop \ge 1} \frac {z^k} k$
Thus:
{{begin-eqn}}
{{eqn | l = \map \ln {1 + \frac 1 x}
| r = \ln \dfrac 1 {1 - \dfrac 1 {x + 1} }
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 1} \frac 1 {k \paren {x + 1}^k}
| c =
}}
{{end-eqn}}
So as $n \to \infty$, $\ds \sum_{k \mathop \ge 1} \frac 1 {k \paren {x + 1}^k}$ converges, and so:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \frac 1 {k \paren {x + 1}^k}
| o = <
| r = \frac 1 {\paren {n + 1} \paren {x + 1}^{n + 1} } \sum_{k \mathop \ge 0} \frac 1 {\paren {x + 1}^k}
| c =
}}
{{eqn | r = \frac 1 {\paren {n + 1} \paren {x + 1}^n x}
| c =
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
22341
|
\section{Sum over k of n Choose k by x to the k by kth Harmonic Number/x = -1}
Tags: Harmonic Numbers, Binomial Coefficients
\begin{theorem}
While for $x \in \R_{>0}$ be a real number:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \paren {x + 1}^n \paren {H_n - \map \ln {1 + \frac 1 x} } + \epsilon$
when $x = -1$ we have:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \dfrac {-1} n$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$H_k$ denotes the $k$th harmonic number.
\end{theorem}
\begin{proof}
When $x = -1$ we have that $1 + \dfrac 1 x = 0$, so $\map \ln {1 + \dfrac 1 x}$ is undefined.
Let $S_n = \ds \sum_{k \mathop \in \Z} \binom n k x^k H_k$
Then:
{{begin-eqn}}
{{eqn | l = S_{n + 1}
| r = \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \paren {\binom n k + \binom n {k - 1} } x^k H_k
| c = Pascal's Rule
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 1} \paren {\binom n {k - 1} } x^{k - 1} \paren {H_{k - 1} + \frac 1 k}
| c =
}}
{{eqn | r = S_n + x S_n + \sum_{k \mathop \ge 1} \binom n {k - 1} x^k \frac 1 k
| c =
}}
{{eqn | r = \paren {1 + x} S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k x^k
| c = (look up result for this)
}}
{{end-eqn}}
{{LinkWanted|for the above}}
Setting $x = -1$:
{{begin-eqn}}
{{eqn | l = S_{n + 1}
| r = \paren {1 + -1} S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k \paren {-1}^k
| c =
}}
{{eqn | r = \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k \paren {-1}^k
| c =
}}
{{eqn | ll= \leadsto
| l = S_n
| r = \frac 1 n \sum_{k \mathop \ge 1} \binom n k \paren {-1}^k
| c =
}}
{{eqn | r = \frac 1 n \paren {\sum_{k \mathop \ge 0} \binom n k \paren {-1}^k - \binom n 0}
| c =
}}
{{eqn | r = \frac 1 n \paren {0 - \binom n 0}
| c = Alternating Sum and Difference of Binomial Coefficients for Given n
}}
{{eqn | r = \frac {-1} n
| c = Binomial Coefficient with Zero
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22342
|
\section{Sum over k of r+tk choose k by s-tk choose n-k}
Tags: Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} = \sum_{k \mathop \ge 0} \dbinom {r + s - k} {n - k} t^k$
where $\dbinom {r + t k} k$ and so on denotes a binomial coefficient.
\end{theorem}
\begin{proof}
Let $\map f {r, s, t, n}$ be the function defined as:
:$\ds \map f {r, s, t, n} := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$
We have:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}
| c =
}}
{{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {r + t k} {r + t k}
| c =
}}
{{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac r {r + t k} + \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {t k} {r + t k}
| c =
}}
{{end-eqn}}
{{proof wanted|in progress}}
\end{proof}
|
22343
|
\section{Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t}
Tags: Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t, Binomial Coefficients
\begin{theorem}
For $n \in \Z_{\ge 0}$:
:$\ds \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t} = \map {A_n} {r + s, t}$
where $\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
:$\map {A_n} {x, t} = \dbinom {x - n t} n \dfrac x {x - n t}$
where $x \ne n t$.
\end{theorem}
\begin{proof}
Let:
:$\displaystyle S = \sum_k A_k \left({r, t}\right) A_{n - k} \left({s, t}\right)$
Both sides of the statement of the theorem are polynomials in $r$, $s$ and $t$.
Therefore it can be assumed that $r \ne k t \ne s$ for $0 \le k \le n$ or something will become undefined.
By replacing the polynomials $A_n$ with their binomial coefficient definitions, the theorem can be expressed as:
:$\displaystyle S = \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} \dfrac s {s - \left({n - k}\right) t}$
Using the technique of partial fractions:
:$\dfrac 1 {r - k t} \dfrac 1 {s - \left({n - k}\right) t} = \dfrac 1 {r + s - n t} \left({\dfrac 1 {r - k t} + \dfrac 1 {s - \left({n - k}\right) t} }\right)$
Thus:
:$\displaystyle S = \frac s {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} + \frac r {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac s {s - \left({n - k}\right) t}$
From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$:
:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
for $r, s, t \in \R, n \in \Z$.
Thus we have:
:$S = \dfrac s {r + s - n t} \dbinom {r + s - t n} n + \dfrac r {r + s - n t} \dbinom {s + r - t n} n$
after changing $k$ to $n - k$ in the second term.
That is:
:$S = \dbinom {r + s - n t} n \dfrac {r + s} {r + s - n t}$
which is $A_n \left({r + s, t}\right)$.
{{qed}}
\end{proof}
|
22344
|
\section{Sum over k of r-kt choose k by r over r-kt by z^k}
Tags: Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\map {A_n} {x, t}$ be the polynomial of degree $n$ defined as:
:$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
for $x \ne n t$.
Let $z = x^{t + 1} - x^t$.
Then:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
for sufficiently small $z$.
\end{theorem}
\begin{proof}
From Sum over $k$ of $\paren {-1}^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$ and renaming variables:
:$\ds \sum_j \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$
where $\delta_{k 0}$ is the Kronecker delta.
Thus:
:$\ds \sum_{j, k} \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k = 1$
We have:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{j, k} \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k
| c =
}}
{{eqn | r = \sum_j \paren {-1}^j \dfrac r {r - j t} \sum_k \dbinom k j \dbinom {r - j t} k w^k
| c =
}}
{{eqn | r = \sum_j \paren {-1}^j \dfrac r {r - j t} \sum_k \dbinom {r - j t} j \dbinom {r - j t - j} {k - j} w^k
| c = Product of $\dbinom r m$ with $\dbinom m k$
}}
{{eqn | r = \sum_j \paren {-1}^j \dfrac r {r - j t} \dbinom {r - j t} j \sum_k \dbinom {r - j t - j} {k - j} w^k
| c =
}}
{{eqn | r = \sum_j \paren {-1}^j \map {A_j} {r, t} \sum_k \dbinom {r - j t - j} {k - j} w^k
| c = Definition of $\map {A_j} {r, t}$
}}
{{eqn | r = \sum_j \paren {-1}^j \map {A_j} {r, t} \paren {1 + w}^{r - j t - j} w^j
| c = Binomial Theorem
}}
{{end-eqn}}
Now let:
:$x = \dfrac 1 {1 + w}$
{{begin-eqn}}
{{eqn | l = x
| o = :=
| r = \dfrac 1 {1 + w}
| c =
}}
{{eqn | ll= \leadsto
| l = z
| r = -\frac w {\paren {1 + w}^{1 + t} }
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {1 + w}^{r - j k - j} w^j \paren {-1}^j
| r = z^r z^j
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \sum_j \map {A_j} {r, t} z^j \paren {1 + w}^r
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_j \map {A_j} {r, t} z^j
| r = \paren {1 + w}^{- r}
| c =
}}
{{eqn | r = x^r
| c =
}}
{{end-eqn}}
{{finish|Establish the fact that we have convergence by ratio test and estimates for large $k$. Or use complex variable theory to establish that the function is analytic around $x {{=}} 1$.}}
{{qed}}
\end{proof}
|
22345
|
\section{Sum over k of r-kt choose k by z^k}
Tags: Sum over k of r-kt choose k by z^k, Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$
where $\dbinom {r - t k} k$ denotes a binomial coefficient.
\end{theorem}
\begin{proof}
From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$:
:$\displaystyle (1): \quad \sum_k A_k \left({r, t}\right) z^k = x^r$
where:
:$A_n \left({x, t}\right)$ is the polynomial of degree $n$ defined as:
::$A_n \left({x, t}\right) := \dbinom {x - n t} n \dfrac x {x - n t}$
:for $x \ne n t$
:$z = x^{t + 1} - x^t$.
Differentiating $(1)$ {{WRT|Differentiation}} $z$:
{{begin-eqn}}
{{eqn | l = \sum_k A_k \left({r, t}\right) k z^{k - 1}
| r = \frac {\mathrm d} {\mathrm d z} x^r
| c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
| l = \sum_k A_k \left({r, t}\right) k z^k
| r = z \frac {\mathrm d} {\mathrm d z} x^r
| c =
}}
{{eqn | r = \left({x^{t + 1} - x^t}\right) r x^{r - 1} \frac {\mathrm d x} {\mathrm d z}
| c = Chain Rule
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = z
| r = x^{t + 1} - x^t
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {\mathrm d z} {\mathrm d x}
| r = \left({t + 1}\right) x^t - t x^{t - 1}
| c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
| l = \frac {\mathrm d x} {\mathrm d z}
| r = \frac 1 {\left({t + 1}\right) x^t - t x^{t - 1} }
| c = Derivative of Inverse Function
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sum_k k A_k \left({r, t}\right) z^k
| r = \frac {\left({x^{t + 1} - x^t}\right) r x^{r - 1} } {\left({t + 1}\right) x^t - t x^{t - 1} }
| c =
}}
{{eqn | r = \frac {\left({x - 1}\right) r x^r} {\left({t + 1}\right) x - t}
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \sum_k A_k \left({r, t}\right) z^k - \frac t r \sum_k k A_k \left({r, t}\right) z^k
| r = x^r - \frac t r \frac {\left({x - 1}\right) r x^r} {\left({t + 1}\right) x - t}
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_k \left({1 - \frac t r k}\right) A_k \left({r, t}\right) z^k
| r = \frac {\left({\left({t + 1}\right) x - t}\right) x^r - t \left({x - 1}\right) x^r} {\left({t + 1}\right) x - t}
| c =
}}
{{eqn | r = \frac {x^{r + 1} } {\left({t + 1}\right) x - t}
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \sum_k \frac {r - t k} r \dbinom {r - k t} k \dfrac r {r - k t} z^k
| r = \frac {x^{r + 1} } {\left({t + 1}\right) x - t}
| c = substituting for $A_k \left({r, t}\right)$
}}
{{eqn | ll= \leadsto
| l = \sum_k \dbinom {r - k t} k z^k
| r = \frac {x^{r + 1} } {\left({t + 1}\right) x - t}
| c = simplifying
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22346
|
\section{Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction}
Tags: Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk, Binomial Coefficients
\begin{theorem}
Let $r, s, t \in \R, n \in \Z$.
Consider the equation:
:$\ds (1): \quad \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
where $\dbinom {r - t k} k$ etc. are binomial coefficients.
Then equation $(1)$ holds for the special case where $s = n - 1 - r + n t$.
\end{theorem}
\begin{proof}
Substituting $n - 1 - r + n t$ for $s$ in the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \binom {r + s - t n} n
| r = \binom {r + \paren {n - 1 - r + n t} - t n} n
| c =
}}
{{eqn | r = \binom {r + n - 1 - r + n t - t n} n
| c =
}}
{{eqn | r = \binom {n - 1} n
| c =
}}
{{end-eqn}}
Substituting $n - 1 - r + n t$ for $s$ in the {{LHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - 1 - r + t k} {n - k} \frac r {r - t k}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \dfrac {\paren {r - t k}! \paren {n - 1 - r + t k}! \, r} {k! \paren {r - t k - k}! \paren {n - k}! \paren {k - 1 - r + t k}! \paren {r - t k} }
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \dfrac {\paren {r - t k - 1}! \paren {n - 1 - r + t k}!} {\paren {r - t k - k}! \paren {k - 1 - r + t k}!}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \prod_{0 \mathop < j \mathop < k} \paren {r - t k - j} \prod_{0 \mathop < j \mathop < n \mathop - k} \paren {n - 1 - r + t k - j}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \paren {-1}^{k - 1} \prod_{0 \mathop < j \mathop < k} \paren {-r + t k + j} \prod_{k \mathop \le j \mathop < n} \paren {- r + t k + j}
| c =
}}
{{end-eqn}}
The two products give a polynomial of degree $n - 1$ in $k$.
Hence the sum for all $k$ is $0$.
Thus we have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - 1 - r + t k} {n - k} \frac r {r - t k}
| r = 0
| c =
}}
{{eqn | r = \binom {n - 1} n
| c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
Thus the equation indeed holds for the special case where $s = n - 1 - r + n t$.
{{qed}}
\end{proof}
|
22347
|
\section{Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma}
Tags: Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk, Binomial Coefficients
\begin{theorem}
Let this hold for $\tuple {r, s, t, n}$:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
and also for $\tuple {r, s - t, t, n - 1}$.
Then it also holds for $\tuple {r, s + 1, t, n}$.
\end{theorem}
\begin{proof}
Evaluating the equation for $\tuple {r, s - t, t, n - 1}$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {s - t} - t \paren {\paren {n - 1} - k} } {\paren {n - 1} - k} \frac r {r - t k}
| r = \binom {r + \paren {s - t} - t \paren {n - 1} } {n - 1}
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t - t n + t + t k} {n - k - 1} \frac r {r - t k}
| r = \binom {r + s - t - t n + 1} {n - 1}
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k - 1} \frac r {r - t k}
| r = \binom {r + s - t n} {n - 1}
| c =
}}
{{end-eqn}}
Adding the equation in $\tuple {r, s, t, n}$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \paren {\binom {s - t \paren {n - k} } {n - k - 1} + \binom {s - t \paren {n - k} } {n - k} } \frac r {r - t k}
| r = \binom {r + s - t n} {n - 1} + \binom {r + s - t n} n
| c =
}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + 1 - t \paren {n - k} } {n - k } \frac r {r - t k}
| r = \binom {r + s + 1 - t n} n
| c = Pascal's Rule
}}
{{end-eqn}}
Hence the equation holds for $\tuple {r, s + 1, t, n}$
{{qed}}
\end{proof}
|
22348
|
\section{Sum over k of r Choose k by -1^r-k by Polynomial}
Tags: Factorials, Sum over k of r Choose k by -1^r-k by Polynomial, Binomial Coefficients
\begin{theorem}
Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
\end{theorem}
\begin{proof}
From the corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \left({-1}\right)^{r - k}$:
:$\displaystyle \sum_k \binom r k \binom k n \left({-1}\right)^{r - k} = \delta_{n r}$
where $\delta_{n r}$ denotes the Kronecker delta.
Thus when $n \ne r$:
:$\displaystyle \sum_k \binom r k \binom k n \left({-1}\right)^{r - k} = 0$
and so:
:$\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} \left({c_0 \binom k 0 + c_1 \binom k 1 + \cdots + c_m \binom k m}\right) = c_r$
as the only term left standing is the $r$th one.
Choosing the coefficients $c_i$ as appropriate, a polynomial in $k$ can be expressed as a summation of binomial coefficients in the form:
:$c_0 \dbinom k 0 + c_1 \dbinom k 1 + \cdots + c_m \dbinom k m$
Thus we can rewrite such a polynomial in $k$ as:
:$b_0 + b_1 k + \cdots + b_r k^r$
{{explain|Why is the parameter of $b_r$ multiplied by $r!$?}
Hence the result.
{{qed}}
\end{proof}
|
22349
|
\section{Sum over k of r Choose k by Minus r Choose m Minus 2k}
Tags: Binomial Coefficients
\begin{theorem}
Let $r \in \R$, $m \in \Z$.
:$\ds \sum_{k \mathop \in \Z} \binom r k \binom {-r} {m - 2 k} \paren {-1}^{m + k} = \binom r m$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \paren {1 - x^2}
| r = \paren {1 - x} \paren {1 + x}
| c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
| l = \paren {1 - x}^r
| r = \dfrac {\paren {1 - x^2}^r} {\paren {1 + x}^r}
| c =
}}
{{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}
| c =
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = \sum_{m \mathop \in \mathop \Z} \binom r m x^m
| r = \paren {1 - x}^r
| c = Binomial Theorem
}}
{{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}
| c =
}}
{{eqn | r = \sum_{k \mathop \in \mathop \Z} \binom r k \paren {-x^2}^k \sum_{j \mathop \in \mathop \Z} \binom {-r} j x^j
| c =
}}
{{end-eqn}}
Comparing the coefficients of $x^m$ on both sides yields the result.
{{qed}}
\end{proof}
|
22350
|
\section{Sum over k of r Choose k by s-kt Choose r by -1^k}
Tags: Binomial Coefficients
\begin{theorem}
Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \binom {s - k t} r \paren {-1}^k = t^r$
where $\dbinom r k$ etc. are binomial coefficients.
\end{theorem}
\begin{proof}
From Sum over $k$ of $\dbinom r k \paren {-1}^k$ by Polynomial:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
{{proof wanted}}
\end{proof}
|
22351
|
\section{Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k}
Tags: Stirling Numbers, Factorials
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
where $\ds {k \brace m}$ etc. denotes a Stirling number of the second kind.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
\end{proof}
|
22352
|
\section{Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial}
Tags: Stirling Numbers, Factorials
\begin{theorem}
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
where:
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$ n!$ denotes a factorial.
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
\end{proof}
|
22353
|
\section{Sum over k to n of k Choose m by kth Harmonic Number}
Tags: Harmonic Numbers, Binomial Coefficients
\begin{theorem}
:$\ds \sum_{k \mathop = 1}^n \binom k m H_k = \binom {n + 1} {m + 1} \paren {H_{n + 1} - \frac 1 {m + 1} }$
where:
:$\dbinom k m$ denotes a binomial coefficient
:$H_k$ denotes the $k$th harmonic number.
\end{theorem}
\begin{proof}
First we note that by Pascal's Rule:
:$\dbinom k m = \dbinom {k + 1} {m + 1} - \dbinom k {m + 1}$
Thus:
{{begin-eqn}}
{{eqn | l = \dbinom k m H_k
| r = \dbinom k {m + 1} \paren {H_{k + 1} - \dfrac 1 {k + 1} } - \dfrac k {m + 1} H_k
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 1}^n \binom k m H_k
| r = \paren {\binom 2 {m + 1} H_2 - \binom 1 {m + 1} H_1}
| c =
}}
{{eqn | o =
| ro= +
| r = \cdots
| c =
}}
{{eqn | o =
| ro= +
| r = \paren {\binom {n + 1} {m + 1} H_{n + 1} - \binom n {m + 1} H_1}
| c =
}}
{{eqn | o =
| ro= -
| r = \sum_{k \mathop = 1}^n \binom {k + 1} {m + 1} \frac 1 {k + 1}
| c =
}}
{{eqn | r = \binom {n + 1} {m + 1} H_{n + 1} - \binom n {m + 1} H_1 - \frac 1 {m + 1} \sum_{k \mathop = 0}^n \binom k n + \frac 1 {k + 1} \binom 0 m
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 1}^n \binom k m H_k
| r = \binom {n + 1} {m + 1} \paren {H_{n + 1} - \frac 1 {m + 1} }
| c = Sum of Binomial Coefficients over Upper Index
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22354
|
\section{Sum over k to p over 2 of Floor of 2kq over p}
Tags: Prime Numbers
\begin{theorem}
Let $p \in \Z$ be an odd prime.
Let $q \in \Z$ be an odd integer and $p \nmid q$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {2 k q} p} \equiv \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {k q} p} \pmod 2$
\end{theorem}
\begin{proof}
When $k < \dfrac p 4$ we have:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac {\paren {p - 1 - 2 k} q} p}
| r = \floor {q - \dfrac {\paren {2 k + 1} q} p}
}}
{{eqn | r = q + \floor {-\dfrac {\paren {2 k + 1} q} p}
| c = Floor of Number plus Integer
}}
{{eqn | r = q - \ceiling {\dfrac {\paren {2 k + 1} q} p}
| c = Floor of Negative equals Negative of Ceiling
}}
{{eqn | r = q - 1 - \floor {\dfrac {\paren {2 k + 1} q} p}
| c = Floor equals Ceiling iff Integer
}}
{{eqn | o = \equiv
| r = \floor {\dfrac {\paren {2 k + 1} q} p}
| rr= \pmod 2
| c =
}}
{{end-eqn}}
Here it is noted that $\dfrac {\paren {2 k + 1} q} p$ is not an integer, since we have:
:$p \nmid q$
:$p > \dfrac p 2 + 1 > 2 k + 1$
Thus it is possible to replace the last terms:
:$\floor {\dfrac {\paren {p - 1} q} p}, \floor {\dfrac {\paren {p - 3} q} p}, \ldots$
by:
:$\floor {\dfrac q p}, \floor {\dfrac {3 q} p}, \ldots$
The result follows.
{{qed}}
\end{proof}
|
22355
|
\section{Sum to Infinity of 2x^2n over n by 2n Choose n}
Tags: Central Binomial Coefficients
\begin{theorem}
For $\cmod x < 1$:
:$\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} } = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}$
\end{theorem}
\begin{proof}
By Gregory Series:
:$\ds \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$
Let $t = \dfrac x {\sqrt {1 - x^2} }$.
Let $y = \arcsin x$.
Then:
{{begin-eqn}}
{{eqn | l = t
| r = \frac {\sin y} {\sqrt {1 - \sin^2 y} }
}}
{{eqn | r = \frac {\sin y} {\cos y}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \tan y
}}
{{end-eqn}}
Hence $\arctan t = \arcsin x$.
We have:
{{begin-eqn}}
{{eqn | l = \frac {2 x \arcsin x} {\sqrt {1 - x^2} }
| r = 2 t \arctan t
}}
{{eqn | r = 2 t \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}
| c = Gregory Series
}}
{{eqn | r = 2 t \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m - 1} } {2 m - 1}
| c = Translation of Index Variable of Summation
}}
{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m} } {2 m - 1}
}}
{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {\paren {2 m - 1} \paren {1 - x^2}^m}
}}
{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {2 m - 1} \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}
| c = Binomial Theorem for Negative Index and Negative Parameter
}}
{{end-eqn}}
It remains to show the the coefficient of $x^{2 n}$ on the {{RHS}} is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is:
:$\ds 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$
The {{LHS}} above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\ds \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$.
We have:
{{begin-eqn}}
{{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {n - 1} {r - 1}
| o =
}}
{{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 0}^{n - 1} \frac {\paren {-1}^r} {2 r + 1} \dbinom {n - 1} r
| c = Translation of Index Variable of Summation
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \sum_{r \mathop = 0}^{n - 1} \paren {-1}^r \dbinom {n - 1} r y^{2 r} \d y
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \paren {1 - y^2}^{n - 1} \d y
| c = Binomial Theorem
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_{\frac \pi 2}^0 \sin^{2 n - 2} \theta \, \frac {\d y} {\d \theta} \d \theta
| c = by substitution of $y = \cos \theta$
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_0^{\frac \pi 2} \sin^{2 n - 1} \theta \, \d \theta
}}
{{eqn | r = 2 n \dbinom {2 n} n \frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!}
| c = Definite Integral from 0 to Half Pi of Odd Power of Sine x
}}
{{eqn | r = 2 n \paren {\frac {\paren {2 n}!} {n! \, n!} } \paren {\frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!} }
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = 2 n \paren {\frac {2 n} {n^2} } \paren {2^{2 n - 2} }
}}
{{eqn | r = 2^{2 n}
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22356
|
\section{Sum with Maximum is Maximum of Sum}
Tags: Max and Min Operations, Max Operation
\begin{theorem}
Let $a, b, c \in \R$ be real numbers.
Then:
:$a + \max \set {b, c} = \max \set {a + b, a + c}$
\end{theorem}
\begin{proof}
{{WLOG}}, there are two cases to consider:
:$(1): \quad b \ge c$
:$(2): \quad b < c$
First let $b \ge c$.
We have:
{{begin-eqn}}
{{eqn | l = b
| o = \ge
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = a + b
| o = \ge
| r = a + c
| c = Addition of Real Numbers is Compatible with Usual Ordering
}}
{{eqn | ll= \leadsto
| l = \max \set {a + b, a + c}
| r = a + b
| c = {{Defof|Maximum Element}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = a + \max \set {b, c}
| r = a + b
| c = {{Defof|Maximum Element}}
}}
{{eqn | r = \max \set {a + b, a + c}
| c =
}}
{{end-eqn}}
{{qed|lemma}}
Now let $b < c$.
We have:
{{begin-eqn}}
{{eqn | l = b
| o = <
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = a + b
| o = <
| r = a + c
| c = Addition of Real Numbers is Compatible with Usual Ordering
}}
{{eqn | ll= \leadsto
| l = \max \set {a + b, a + c}
| r = a + c
| c = {{Defof|Maximum Element}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = a + \max \set {b, c}
| r = a + c
| c = {{Defof|Maximum Element}}
}}
{{eqn | r = \max \set {a + b, a + c}
| c =
}}
{{end-eqn}}
{{qed|lemma}}
Thus the result holds in both cases.
Hence the result, by Proof by Cases.
{{qed}}
\end{proof}
|
22357
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\section{Sum with One is Immediate Successor in Naturally Ordered Semigroup}
Tags: Naturally Ordered Semigroup
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $1$ be the one of $S$.
Let $n \in S$.
Then $n \circ 1$ is the immediate successor of $n$.
That is, for all $m \in S$:
:$n \prec m \iff n \circ 1 \preceq m$
\end{theorem}
\begin{proof}
By Zero Strictly Precedes One, $0 \prec 1$, where $0$ is the zero of $S$.
Hence from Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible:
:$n \circ 0 \prec n \circ 1$
and by Zero is Identity in Naturally Ordered Semigroup, $n \circ 0 = n$.
Now suppose that $n \prec m$.
Then by {{NOSAxiom|3}}, there exists $p \in S$ such that:
:$n \circ p = m$
Moreover, since $n \ne m$, it follows that $p \ne 0$.
Hence $0 \prec p$ by definition of zero.
Therefore, by definition of one:
:$1 \preceq p$
Now by compatibility of $\preceq$ with $\circ$:
:$n \circ 1 \preceq n \circ p = m$
as desired.
Conversely, if $n \circ 1 \preceq m$, it is immediate from:
:$n \prec n \circ 1$
that $n \prec m$.
{{qed}}
\end{proof}
|
22358
|
\section{Summary of Topology on P-adic Numbers}
Tags: P-adic Number Theory, Topology of P-adic Numbers
\begin{theorem}
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\tau_p$ be the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$.
Then $\struct{\Q_p, \tau_p}$ is:
:$(1): \quad$ Hausdorff
:$(2): \quad$ second-countable
:$(3): \quad$ totally disconnected
:$(4): \quad$ locally compact
\end{theorem}
\begin{proof}
Follows from:
:P-adic Numbers is Hausdorff Topological Space
:P-adic Numbers is Second Countable Topological Space
:P-adic Numbers is Totally Disconnected Topological Space
:P-adic Numbers is Locally Compact Topological Space
{{qed}}
\end{proof}
|
22359
|
\section{Summation Formula (Complex Analysis)}
Tags: Complex Analysis
\begin{theorem}
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane $\C$ with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Let $f$ be a meromorphic function on $\C$ with finitely many poles.
Suppose that:
:$\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
as $N \to \infty$.
Let $X$ be the set of poles of $f$.
Then:
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
If $X \cap \Z = \O$, this becomes:
:$\ds \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
\end{theorem}
\begin{proof}
By Summation Formula: Lemma, there exists a constant $A$ such that:
:$\cmod {\map \cot {\pi z} } < A$
for all $z$ on $C_N$.
Since $f$ has only finitely many poles, we can take $N$ large enough so that no poles of $f$ lie on $C_N$.
Let $X_N$ be the set of poles of $f$ contained in the region bounded by $C_N$.
From Poles of Cotangent Function, $\map \cot {\pi z}$ has poles at $z \in \Z$.
Let $A_N = \set {n \in \Z : -N \le n \le N}$
We then have:
{{begin-eqn}}
{{eqn | l = \oint_{C_N} \pi \map \cot {\pi z} \map f z \rd z
| r = 2 \pi i \sum_{z_0 \mathop \in X_N \mathop \cap A_N} \Res {\pi \map \cot {\pi z} \map f z} {z_0}
| c = Residue Theorem
}}
{{eqn | r = 2 \pi i \paren {\sum_{n \mathop \in A_N \mathop \setminus X_N} \Res {\pi \map \cot {\pi z} \map f z} n + \sum_{z_0 \mathop \in X_N} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }
}}
{{end-eqn}}
We then have, for each integer $n$:
{{begin-eqn}}
{{eqn | l = \Res {\pi \map \cot {\pi z} \map f z} n
| r = \lim_{z \mathop \to n} \paren {\paren {z - n} \pi \map \cot {\pi z} \map f z}
| c = Residue at Simple Pole
}}
{{eqn | r = \map f n \lim_{z \mathop \to n} \paren {\frac {z - n} z + 2 \sum_{k \mathop = 1}^\infty \frac {z \paren {z - n} } {z^2 - k^2} }
| c = Mittag-Leffler Expansion for Cotangent Function
}}
{{eqn | r = \map f n \cdot 2 \lim_{z \mathop \to n} \paren {\frac z {z + n} }
}}
{{eqn | r = \map f n \frac {2 n} {2 n}
}}
{{eqn | r = \map f n
}}
{{end-eqn}}
Note that by hypothesis:
:$\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
So, taking $N \to \infty$:
:$\ds 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \mathop \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }$
which gives:
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
{{qed}}
\end{proof}
|
22360
|
\section{Summation Formula (Complex Analysis)/Lemma}
Tags: Complex Analysis
\begin{theorem}
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Then there exists a constant real number $A$ independent of $N$ such that:
:$\cmod {\map \cot {\pi z} } < A$
for all $z \in C_N$.
\end{theorem}
\begin{proof}
Let $z = x + iy$ for real $x, y$.
\end{proof}
|
22361
|
\section{Summation Formula for Polygonal Numbers}
Tags: Proofs by Induction, Polygonal Numbers
\begin{theorem}
Let $\map P {k, n}$ be the $n$th $k$-gonal number.
Then:
:$\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$
\end{theorem}
\begin{proof}
We have that:
$\map P {k, n} = \begin{cases}
0 & : n = 0 \\
\map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0
\end{cases}$
Proof by induction:
For all $n \in \N_{>0}$, let $\map \Pi n$ be the proposition:
:$\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$
\end{proof}
|
22362
|
\section{Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r/Example}
Tags: Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r, Summations, Products
\begin{theorem}
:$\dfrac {x \paren {x - 2} \paren {x - 3} } {\paren {-1} \paren {-2} \paren {-3} } + \dfrac {\paren {x + 1} \paren {x - 1} \paren {x - 2} } {\paren 1 \paren {-1} \paren {-2} } + \dfrac {\paren {x + 2} x \paren {x - 1} } {\paren 2 \paren 1 \paren {-1} } + \dfrac {\paren {x + 3} \paren {x + 1} x} {\paren 3 \paren 2 \paren 1 } = 1$
\end{theorem}
\begin{proof}
This is an example of Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r:
:$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } } = 1$
where $n = 4$ and $m = 2$.
{{qed}}
\end{proof}
|
22363
|
\section{Summation from k to m of 2k-1 Choose k by 2n-2k Choose n-k by -1 over 2k-1}
Tags: Binomial Coefficients
\begin{theorem}
:$\ds \sum_{k \mathop = 0}^m \binom {2 k - 1} k \binom {2 n - 2 k} {n - k} \dfrac {-1} {2 k - 1} = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m} + \dfrac 1 2 \dbinom {2 n} n$
\end{theorem}
\begin{proof}
From Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Set $r - \dfrac 1 2$ and $s = -\dfrac 1 2$.
:$\ds \sum_{k \mathop = 0}^m \dbinom {1/2} k \dbinom {-1/2} {n - k} \paren {\frac n 2 - \paren {\frac 1 2 - \frac 1 2} k} = \paren {m + 1} \paren {n - m} \dbinom {1/2} {m + 1} \dbinom {-1/2} {n - m}$
Take the {{LHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop = 0}^m \dbinom {1/2} k \dbinom {-1/2} {n - k} \paren {\frac n 2 - \paren {\frac 1 2 - \frac 1 2} k}
| c =
}}
{{eqn | r = \dfrac n 2 \sum_{k \mathop = 0}^m \dbinom {1/2} k \dbinom {-1/2} {n - k}
| c = immediate simplification
}}
{{eqn | r = \dfrac n 2 \sum_{k \mathop = 0}^m \dbinom {1/2} k \dfrac {\paren {-1}^{n - k} } {4^{n - k} } \dbinom {2 n - 2 k} {n - k}
| c = Binomial Coefficient of Minus Half
}}
{{eqn | r = \dfrac n 2 \sum_{k \mathop = 0}^m \paren {\dfrac {\paren {-1}^{k - 1} } {2^{2 k - 1} \paren {2 k - 1} } \dbinom {2 k - 1} k - \delta_{k 0} } \dfrac {\paren {-1}^{n - k} } {2^{2 n - 2 k} } \dbinom {2 n - 2 k} {n - k}
| c = Binomial Coefficient of Half: Corollary
}}
{{eqn | r = \dfrac n 2 \sum_{k \mathop = 0}^m \dfrac {\paren {-1}^{n - 1} } {2^{2 n - 1} \paren {2 k - 1} } \dbinom {2 k - 1} k \dbinom {2 n - 2 k} {n - k} - \dfrac n 2 \dfrac {\paren {-1}^n} {2^{2 n} } \dbinom {2 n} n
| c = separating out the $\delta_{k 0}$
}}
{{eqn | r = \dfrac {\paren {-1}^n n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^m \dbinom {2 k - 1} k \dbinom {2 n - 2 k} {n - k} \dfrac {-1} {\paren {2 k - 1} } - \dfrac 1 2 \dbinom {2 n} n}
| c = simplifying
}}
{{end-eqn}}
Now the {{RHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \paren {m + 1} \paren {n - m} \dbinom {1/2} {m + 1} \dbinom {-1/2} {n - m}
| c =
}}
{{eqn | r = \paren {m + 1} \paren {n - m} \dbinom {1/2} {m + 1} \dfrac {\paren {-1}^{n - m} } {2^{2 n - 2 m} } \dbinom {2 n - 2 m} {n - m}
| c = Binomial Coefficient of Minus Half
}}
{{eqn | r = \paren {m + 1} \paren {n - m} \dfrac {\paren {-1}^m} {\paren {2 m + 1} 2^{2 m + 2} } \dbinom {2 m + 2} {m + 1} \dfrac {\paren {-1}^{n - m} } {2^{2 n - 2 m} } \dbinom {2 n - 2 m} {n - m}
| c = Binomial Coefficient of Half
}}
{{eqn | r = \dfrac 1 {2^{2 n + 2} } \paren {m + 1} \paren {n - m} \dfrac {\paren {-1}^n} {2 m + 1} \dbinom {2 m + 2} {m + 1} \dbinom {2 n - 2 m} {n - m}
| c = simplification
}}
{{eqn | r = \dfrac 1 {2^{2 n + 2} } \paren {m + 1} \paren {n - m} \dfrac {\paren {-1}^n} {2 m + 1} \dfrac {2 m + 2} {m + 1} \dbinom {2 m + 1} m \dbinom {2 n - 2 m} {n - m}
| c = Factors of Binomial Coefficient
}}
{{eqn | r = \dfrac {\paren {-1}^n} {2^{2 n + 1} } \paren {n - m} \dfrac {m + 1} {2 m + 1} \dbinom {2 m + 1} m \dbinom {2 n - 2 m} {n - m}
| c = simplification
}}
{{eqn | r = \dfrac {\paren {-1}^n} {2^{2 n + 1} } \paren {n - m} \dfrac {m + 1} {2 m + 1} \dfrac {2 m + 1} {2 m + 1 - m} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m}
| c = Factors of Binomial Coefficient: Corollary 1
}}
{{eqn | r = \dfrac {\paren {-1}^n} {2^{2 n + 1} } \paren {n - m} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m}
| c = simplification
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {-1}^n n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^m \dbinom {2 k - 1} k \dbinom {2 n - 2 k} {n - k} \dfrac {-1} {\paren {2 k - 1} } - \dfrac 1 2 \dbinom {2 n} n}
| r = \dfrac {\paren {-1}^n} {2^{2 n + 1} } \paren {n - m} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m}
| c = as the {{RHS}} equals the {{LHS}}
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^m \dbinom {2 k - 1} k \dbinom {2 n - 2 k} {n - k} \dfrac {-1} {\paren {2 k - 1} } - \dfrac 1 2 \dbinom {2 n} n
| r = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m}
| c = dividing both sides by $\dfrac {\paren {-1}^n n} {2^{2 n} }$
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^m \dbinom {2 k - 1} k \dbinom {2 n - 2 k} {n - k} \dfrac {-1} {\paren {2 k - 1} }
| r = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m} + \dfrac 1 2 \dbinom {2 n} n
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22364
|
\section{Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k}
Tags: Binomial Coefficients
\begin{theorem}
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
\end{theorem}
\begin{proof}
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
\end{proof}
|
22365
|
\section{Summation is Linear}
Tags: Numbers, Sums
\begin{theorem}
Let $\tuple {x_1, \ldots, x_n}$ and $\tuple {y_1, \ldots, y_n}$ be finite sequences of numbers of equal length.
Let $\lambda$ be a number.
Then:
\end{theorem}
\begin{proof}
{{handwaving|Dots are not to be used in the formal part of a proof on this site. Induction is the way to go.}}
{{begin-eqn}}
{{eqn|o=|r=\sum_{i=1}^n x_i+\sum_{i=1}^n y_i}}
{{eqn|r=(x_1+x_2+...+x_n)+(y_1+y_2+...+y_n)|c=by definition of summation}}
{{eqn|r=(x_1+y_1)+(x_2+y_2)+...+(x_n+y_n)|c=because addition is commutative}}
{{eqn|r=\sum_{i=1}^n (x_i+y_i)|c=by definition of summation}}
{{eqn|o=}}
{{eqn|o=|r=\lambda\cdot\sum_{i=1}^n x_i}}
{{eqn|r=\lambda(x_1+x_2+...+x_n)|c=by definition of summation}}
{{eqn|r=\lambda x_1+\lambda x_2+...+\lambda x_n|c=by the distributive property of multiplication}}
{{eqn|r=\sum_{i=1}^n \lambda\cdot x_i|c=by definition of summation}}
{{end-eqn}}
{{MissingLinks|Scalar Product with Sum ought to save time}}
Category:Numbers
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\end{proof}
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22366
|
\section{Summation is Linear/Scaling of Summations}
Tags: Numbers, Proofs by Induction
\begin{theorem}
Let $\tuple {x_1, \ldots, x_n}$ and $\tuple {y_1, \ldots, y_n}$ be finite sequences of numbers of equal length.
Let $\lambda$ be a number.
Then:
:$\ds \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
\end{theorem}
\begin{proof}
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
\end{proof}
|
22367
|
\section{Summation is Linear/Sum of Summations}
Tags: Numbers, Proofs by Induction
\begin{theorem}
Let $\tuple {x_1, \ldots, x_n}$ and $\tuple {y_1, \ldots, y_n}$ be finite sequences of numbers of equal length.
Let $\lambda$ be a number.
Then:
:$\ds \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$
\end{theorem}
\begin{proof}
The proof proceeds by mathematical induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$
\end{proof}
|
22368
|
\section{Summation of General Logarithms}
Tags: Logarithms, Summations, Products
\begin{theorem}
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \prod_{\map R i} a_i$ denote a product over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$
\end{theorem}
\begin{proof}
The proof proceeds by induction.
First let:
:$S := \set {a_i: \map R i}$
We have that $S$ is finite.
Hence the contents of $S$ can be well-ordered, by Finite Totally Ordered Set is Well-Ordered.
Let $S$ have $m$ elements, identified as:
:$S = \set {s_1, s_2, \ldots, s_m}$
For all $n \in \Z_{\ge 0}$ such that $n \le m$, let $\map P n$ be the proposition:
:$\ds \map {\log_b} {\prod_{i \mathop = 1}^n s_i} = \sum_{i \mathop = 1}^n \log_b s_i$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \map {\log_b} {\prod_{i \mathop = 1}^0 s_i}
| r = \log_b 1
| c = {{Defof|Vacuous Product}}
}}
{{eqn | r = 0
| c = Logarithm of 1 is 0
}}
{{eqn | r = \sum_{i \mathop = 1}^n \log_b s_i
| c = {{Defof|Vacuous Summation}}
}}
{{end-eqn}}
\end{proof}
|
22369
|
\section{Summation of Multiple of Mapping on Finite Set}
Tags: Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $\lambda \in \mathbb A$.
Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.
Then we have the equality of summations on finite sets:
:$\ds \sum_{s \mathop \in S} \map g s = \lambda \cdot \sum_{s \mathop \in S} \map f s$
\end{theorem}
\begin{proof}
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following equality of indexed summations:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map g {\map \sigma i} = \lambda \cdot \sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i}$
By Multiple of Mapping Composed with Mapping, $g \circ \sigma = \lambda \cdot \paren {f \circ \sigma}$.
The above equality now follows from Indexed Summation of Multiple of Mapping.
{{qed}}
\end{proof}
|
22370
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\section{Summation of Powers over Product of Differences}
Tags: Summations, Summation of Powers over Product of Differences, Products
\begin{theorem}
:$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\displaystyle S_n := \sum_{j \mathop = 1}^n \left({\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} }\right) = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
\end{proof}
|
22371
|
\section{Summation of Product of Differences}
Tags: Summations
\begin{theorem}
:$\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {u_j - u_k} \paren {v_j - v_k} = n \sum_{j \mathop = 1}^n u_j v_j - \sum_{j \mathop = 1}^n u_j \sum_{j \mathop = 1}^n v_j$
\end{theorem}
\begin{proof}
Take the Binet-Cauchy Identity:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
Make the following assignments:
{{begin-eqn}}
{{eqn | q = 1 \le i \le n
| l = a_i
| o = :=
| r = u_i
}}
{{eqn | q = 1 \le i \le n
| l = c_i
| o = :=
| r = v_i
}}
{{eqn | q = 1 \le j \le n
| l = b_j
| o = :=
| r = 1
}}
{{eqn | q = 1 \le j \le n
| l = d_j
| o = :=
| r = 1
}}
{{end-eqn}}
Then we have:
:$\ds \paren {\sum_{i \mathop = 1}^n u_i v_i} \paren {\sum_{j \mathop = 1}^n 1 \times 1} = \paren {\sum_{i \mathop = 1}^n u_i \times 1} \paren {\sum_{j \mathop = 1}^n 1 \times v_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {u_i \times 1 - u_j \times 1} \paren {v_i \times 1 - v_j \times 1}$
and the result follows.
{{qed}}
\end{proof}
|
22372
|
\section{Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2}
Tags: Summation to n of Summation to Index, Summation of Products of n Numbers taken m at a time with Repetitions, Summations
\begin{theorem}
Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \dotsc, x_b}$.
Then:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$
\end{theorem}
\begin{proof}
Let:
{{begin-eqn}}
{{eqn | l = S_1
| o = :=
| r = \sum_{i \mathop = m}^n \sum_{j \mathop = m}^i a_i a_j
| c =
}}
{{eqn | r = \sum_{j \mathop = m}^n \sum_{i \mathop = j}^n a_i a_j
| c = Summation of i from 1 to n of Summation of j from 1 to i
}}
{{eqn | r = \sum_{i \mathop = m}^n \sum_{j \mathop = i}^n a_i a_j
| c = Change of Index Variable of Summation
}}
{{eqn | o = =:
| r = S_2
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 2 S_1
| r = S_1 + S_2
| c =
}}
{{eqn | r = \sum_{i \mathop = m}^n \left({\sum_{j \mathop = m}^i a_i a_j + \sum_{j \mathop = i}^n a_i a_j}\right)
| c =
}}
{{eqn | r = \sum_{i \mathop = m}^n \left({\left({\sum_{j \mathop = m}^n a_i a_j}\right) + a_i a_i}\right)
| c = Sum of Summations over Overlapping Domains: Example
}}
{{eqn | r = \sum_{i \mathop = m}^n \sum_{j \mathop = m}^n a_i a_j + \sum_{i \mathop = m}^n a_i a_i
| c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = \left({\sum_{i \mathop = m}^n a_i}\right) \left({\sum_{j \mathop = m}^n a_j}\right) + \left({\sum_{i \mathop = m}^n {a_i}^2}\right)
| c = Change of Index Variable of Summation
}}
{{end-eqn}}
Hence the result on multiplying by $\dfrac 1 2$.
{{qed}}
\end{proof}
|
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