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22173
|
\section{Sum of Cosines of Arithmetic Sequence of Angles/Formulation 2}
Tags: Cosine 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 \cos {\theta + k \alpha}
| r = \map \cos {\theta + \alpha} + \map \cos {\theta + 2 \alpha} + \map \cos {\theta + 3 \alpha} + \dotsb
}}
{{eqn | r = \map \cos {\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 real parts:
:$\ds \sum_{k \mathop = 1}^n \map \cos {\theta + k \alpha} = \map \cos {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
{{qed}}
\end{proof}
|
22174
|
\section{Sum of Cosines of Fractions of Pi}
Tags: Polynomial Equations, Cosine Function
\begin{theorem}
Let $n \in \Z$ such that $n > 1$.
Then:
:$\ds \sum_{k \mathop = 1}^{n - 1} \cos \frac {2 k \pi} n = -1$
\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 real parts:
:$1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n = 0$
whence the result.
{{qed}}
\end{proof}
|
22175
|
\section{Sum of Cosines of Multiples of Angle}
Tags: Telescoping Series, Cosine Function
\begin{theorem}
{{begin-eqn}}
{{eqn | l = \frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k x}
| r = \frac 1 2 + \cos x + \cos 2 x + \cos 3 x + \cdots + \cos n x
| c =
}}
{{eqn| r = \frac {\map \sin {\paren {2 n + 1} x / 2} } {2 \map \sin {x / 2} }
| c =
}}
{{end-eqn}}
where $x$ is not an integer multiple of $2 \pi$.
\end{theorem}
\begin{proof}
By the Simpson's Formula for Cosine by Sine:
:$2 \cos \alpha \sin \beta = \map \sin {\alpha + \beta} - \map \sin {\alpha - \beta}$
Thus we establish the following sequence of identities:
{{begin-eqn}}
{{eqn | l = 2 \cdot \frac 1 2 \sin \frac x 2
| r = \sin \frac x 2
| c =
}}
{{eqn | l = 2 \cos x \sin \frac x 2
| r = \sin \frac {3 x} 2 - \sin \frac x 2
| c =
}}
{{eqn | l = 2 \cos 2 x \sin \frac x 2
| r = \sin \frac {5 x} 2 - \sin \frac {3 x} 2
| c =
}}
{{eqn | o = \cdots
| c =
}}
{{eqn | l = 2 \cos n x \sin \frac x 2
| r = \sin \frac {\paren {2 n + 1} x} 2 - \sin \frac {\paren {2 n - 1} x} 2
| c =
}}
{{end-eqn}}
Summing the above:
:$\ds 2 \sin \frac x 2 \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k x} } = \sin \frac {\paren {2 n + 1} x} 2$
as the sums on the {{RHS}} form a telescoping series.
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}
|
22176
|
\section{Sum of Cosines of k pi over 5}
Tags: Complex 5th Roots of Unity
\begin{theorem}
:$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 144 \degrees
| r = 180 \degrees - 36 \degrees
| c =
}}
{{eqn | ll= \leadsto
| l = \cos 36 \degrees
| r = -\cos 144 \degrees
| c = Cosine of Supplementary Angle
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = 108 \degrees
| r = 180 \degrees - 72 \degrees
| c =
}}
{{eqn | ll= \leadsto
| l = \cos 72 \degrees
| r = -\cos 108 \degrees
| c = Cosine of Supplementary Angle
}}
{{end-eqn}}
Thus:
:$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$
:500px
{{qed}}
\end{proof}
|
22177
|
\section{Sum of Cubes of 3 Consecutive Integers which is Square}
Tags: Sums of Cubes
\begin{theorem}
The following sequences of $3$ consecutive (strictly) positive integers have cubes that sum to a square:
:$1, 2, 3$
:$23, 24, 25$
No other such sequence of $3$ consecutive positive integers has the same property.
However, if we allow sequences containing zero and negative integers, we also have:
:$-1, 0, 1$
:$0, 1, 2$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 1^3 + 2^3 + 3^3
| r = 1 + 8 + 27
| c =
}}
{{eqn | r = 36
| c =
}}
{{eqn | r = 6^2
| c =
}}
{{eqn | l = 23^3 + 24^3 + 25^3
| r = 12 \, 167 + 13 \, 824 + 15 \, 625
| c =
}}
{{eqn | r = 41 \, 616
| c =
}}
{{eqn | r = 204^2
| c =
}}
{{eqn | l = \paren {-1}^3 + 0^3 + 1^3
| r = -1 + 0 + 1
| c =
}}
{{eqn | r = 0
| c =
}}
{{eqn | r = 0^2
| c =
}}
{{eqn | l = 0^3 + 1^3 + 2^3
| r = 0 + 1 + 8
| c =
}}
{{eqn | r = 9
| c =
}}
{{eqn | r = 3^2
| c =
}}
{{end-eqn}}
Any sequence of $3$ consecutive integers that have cubes that sum to a square would satisfy:
:$m^2 = \paren {n - 1}^3 + n^3 + \paren {n + 1}^3$
where $n$ is the middle number of the sequence, with $m, n \in \Z$.
Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | l = m^2
| r = \paren {n - 1}^3 + n^3 + \paren {n + 1}^3
}}
{{eqn | r = n^3 - 3 n^2 + 3 n - 27 + n^3 + n^3 + 3 n^2 + 3 n + 27
| c = Cube of Sum, Cube of Difference
}}
{{eqn | r = 3 n^3 + 6 n
}}
{{end-eqn}}
Substituting $y = 3 m$ and $x = 3 n$:
{{begin-eqn}}
{{eqn | l = \paren {\frac y 3}^2
| r = 3 \paren {\frac x 3}^3 + 6 \paren {\frac x 3}
}}
{{eqn | ll= \leadsto
| l = \frac {y^2} 9
| r = \frac {x^3} 9 + 2 x
}}
{{eqn | ll= \leadsto
| l = y^2
| r = x^3 + 18 x
}}
{{end-eqn}}
which is an elliptic curve.
According to [https://www.lmfdb.org/EllipticCurve/Q/2304/a/2 LMFDB], this elliptic curve has exactly $7$ lattice points:
:$\tuple {0, 0}, \tuple {3, \pm 9}, \tuple {6, \pm 18}, \tuple {72, \pm 612}$
which correspond to these values of $n$:
:$0, 1, 2, 24$
Hence there are no more solutions.
{{qed}}
\end{proof}
|
22178
|
\section{Sum of Cubes of 5 Consecutive Integers which is Square}
Tags: Sums of Cubes
\begin{theorem}
The following sequences of $5$ consecutive (strictly) positive integers have cubes that sum to squares:
:$1, 2, 3, 4, 5$
:$25, 26, 27, 28, 29$
:$96, 97, 98, 99, 100$
:$118, 119, 120, 121, 122$
No other such sequence of $5$ consecutive positive integers has the same property.
However, if we allow sequences containing zero and negative integers, we also have:
:$0, 1, 2, 3, 4$
:$-2, -1, 0, 1, 2$
{{OEIS|A126203}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 1^3 + 2^3 + 3^3 + 4^3 + 5^3
| r = 1 + 8 + 27 + 64 + 125
| c =
}}
{{eqn | r = 225
| c =
}}
{{eqn | r = 15^2
| c = also see Sum of Sequence of Cubes
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 25^3 + 26^3 + 27^3 + 28^3 + 29^3
| r = 15 \, 625 + 17 \, 576 + 19 \, 683 + 21 \, 952 + 24 \, 389
| c =
}}
{{eqn | r = 99 \, 225
| c =
}}
{{eqn | r = 315^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 96^3 + 97^3 + 98^3 + 99^3 + 100^3
| r = 884 \, 736 + 912 \, 673 + 941 \, 192 + 970 \, 299 + 1 \, 000 \, 000
| c =
}}
{{eqn | r = 4 \, 708 \, 900
| c =
}}
{{eqn | r = 2170^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 118^3 + 119^3 + 120^3 + 121^3 + 122^3
| r = 1 \, 643 \, 032 + 1 \, 685 \, 159 + 1 \, 728 \, 000 + 1 \, 771 \, 561 + 1 \, 815 \, 848
| c =
}}
{{eqn | r = 8 \, 643 \, 600
| c =
}}
{{eqn | r = 2940^2
| c =
}}
{{end-eqn}}
Then we also have:
{{begin-eqn}}
{{eqn | l = 0^3 + 1^3 + 2^3 + 3^3 + 4^3
| r = 0 + 1 + 8 + 27 + 64
| c =
}}
{{eqn | r = 100
| c =
}}
{{eqn | r = 10^2
| c = also see Sum of Sequence of Cubes
}}
{{end-eqn}}
and finally the degenerate case:
{{begin-eqn}}
{{eqn | l = \paren {-2}^3 + \paren {-1}^3 + 0^3 + 1^3 + 2^3
| r = \paren {-8} + \paren {-1} + 0 + 1 + 8
| c =
}}
{{eqn | r = 0
| c =
}}
{{eqn | r = 0^2
| c =
}}
{{end-eqn}}
Any sequence of $5$ consecutive integers that have cubes that sum to a square would satisfy:
:$m^2 = \paren {n - 2}^3 + \paren {n - 1}^3 + n^3 + \paren {n + 1}^3 + \paren {n + 2}^3$
where $n$ is the middle number of the sequence, with $m, n \in \Z$.
Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | l = m^2
| r = \paren {n - 2}^3 + \paren {n - 1}^3 + n^3 + \paren {n + 1}^3 + \paren {n + 2}^3
}}
{{eqn | r = n^3 - 6 n^2 + 12 n - 8 + n^3 - 3 n^2 + 3 n - 27 + n^3 + n^3 + 3 n^2 + 3 n + 27 + n^3 + 6 n^2 + 12 n + 8
| c = Cube of Sum, Cube of Difference
}}
{{eqn | r = 5 n^3 + 30 n
}}
{{end-eqn}}
Substituting $y = 5 m$ and $x = 5 n$:
{{begin-eqn}}
{{eqn | l = \paren {\frac y 5}^2
| r = 5 \paren {\frac x 5}^3 + 30 \paren {\frac x 5}
}}
{{eqn | ll= \leadsto
| l = \frac {y^2} {25}
| r = \frac {x^3} {25} + 6 x
}}
{{eqn | ll= \leadsto
| l = y^2
| r = x^3 + 150 x
}}
{{end-eqn}}
which is an elliptic curve.
According to [https://www.lmfdb.org/EllipticCurve/Q/57600/bt/2 LMFDB], this elliptic curve has exactly $13$ lattice points:
:$\tuple {0, 0}, \tuple {10, \pm 50}, \tuple {15, \pm 75}, \tuple {24, \pm 132}, \tuple {135, \pm 1575}, \tuple {490, \pm 10 \, 850}, \tuple {600, \pm 14 \, 700}$
which correspond to these values of $n$:
:$0, 2, 3, \dfrac {24} 5, 27, 98, 120$
Note that $\dfrac {24} 5$ is not an integer.
Hence there are no more solutions.
{{qed}}
\end{proof}
|
22179
|
\section{Sum of Cubes on Diagonals of Moessner's Order 4 Magic Square}
Tags: Magic Squares
\begin{theorem}
The sums of the cubes of the entries on the diagonals of Moessner's order $4$ magic square are equal.
\end{theorem}
\begin{proof}
Recall Moessner's order $4$ magic square:
{{:Definition:Moessner's Order 4 Magic Square}}
{{begin-eqn}}
{{eqn | l = 12^3 + 3^3 + 14^3 + 5^3
| r = 1728 + 27 + 2744 + 125
| c =
}}
{{eqn | r = 4624
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 9^3 + 2^3 + 15^3 + 8^3
| r = 729 + 8 + 3375 + 512
| c =
}}
{{eqn | r = 4624
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22180
|
\section{Sum of Cuts is Cut}
Tags: Cuts, Addition
\begin{theorem}
Let $\alpha$ and $\beta$ be cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p + q$
Then $\gamma$ is also a cut.
Thus the operation of addition on the set of cuts is closed.
\end{theorem}
\begin{proof}
By definition of cut, neither $\alpha$ nor $\beta$ are empty.
Hence there exist $p \in \alpha$ and $q \in \beta$.
Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty.
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.
Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$.
We have:
{{begin-eqn}}
{{eqn | l = p
| o = <
| r = s
| c = {{Defof|Cut (Analysis)|Cut}}: $s \notin \alpha$
}}
{{eqn | l = q
| o = <
| r = t
| c = {{Defof|Cut (Analysis)|Cut}}: $t \notin \beta$
}}
{{eqn | ll= \leadsto
| q = \forall p \in \alpha, q \in \beta
| l = p + q
| o = <
| r = s + t
| c = Rational Numbers form Ordered Field
}}
{{eqn | ll= \leadsto
| l = s + t
| o = \notin
| r = \gamma
| c = {{Defof|Cut (Analysis)|Cut}}
}}
{{end-eqn}}
Thus it is demonstrated that $\gamma$ does not contain every rational number.
Thus condition $(1)$ of the definition of a cut is fulfilled.
{{qed|lemma}}
Let $r \in \gamma$.
Let $s \in \Q$ such that $s < r$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
Let $t \in \Q$ such that $s = t + q$.
Then $t < p$.
Hence $t \in \alpha$.
Hence by definition of $\gamma$, $t + q = s \in \gamma$.
Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$.
Thus condition $(2)$ of the definition of a cut is fulfilled.
{{qed|lemma}}
{{AimForCont}} $r \in \gamma$ is the greatest element of $\gamma$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
By definition of a cut, $\alpha$ has no greatest element:
Hence:
:$\exists s \in \Q: s > p: s \in \alpha$
But then $s + q \in \gamma$ while $s + q > r$.
This contradicts the supposition that $r$ is the greatest element of $\gamma$.
Hence $\gamma$ itself can have no greatest element.
Thus condition $(3)$ of the definition of a cut is fulfilled.
{{qed|lemma}}
Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut.
{{qed}}
\end{proof}
|
22181
|
\section{Sum of Deviations from Mean}
Tags: 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:
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x} = 0$
\end{theorem}
\begin{proof}
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
{{begin-eqn}}
{{eqn | l = \sum \paren {x_i - \overline x}
| r = x_1 - \overline x + x_2 - \overline x + \cdots + x_n - \overline x
| c = {{Defof|Summation}}
}}
{{eqn | r = x_1 - \sum \frac {x_i} n + x_2 - \sum \frac {x_i} n + \cdots + x_n - \sum \frac {x_i} n
| c = {{Defof|Arithmetic Mean}}
}}
{{eqn | r = \paren {x_1 + x_2 + \cdots + x_n} - n \paren {\sum \frac {x_i} n}
}}
{{eqn | r = \sum x_i - \sum x_i
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22182
|
\section{Sum of Discrete Random Variables}
Tags: Discrete Random Variable, Probability Theory, Discrete Random Variables
\begin{theorem}
Let $X$ and $Y$ be discrete random variables on the probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $U: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map U \omega = \map X \omega + \map Y \omega$
Then $U$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
\end{theorem}
\begin{proof}
To show that $U$ is discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad$ The image of $U$ is a countable subset of $\R$;
:$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map U \omega = x} \in \Sigma$.
First we consider any $U_u = \set {\omega \in \Omega: \map U \omega = u}$ such that $U_u \ne \O$.
We have that $U_u = \set {\omega \in \Omega: \map X \omega + \map Y \omega = u}$.
Consider any $\omega \in U_u$.
Then:
:$\omega \in X_x \cap Y_x$
where:
:$X_x = \set{\omega \in \Omega: \map X \omega = x}, Y_x = \set {\omega \in \Omega: \map Y \omega = u - x}$
Because $X$ and $Y$ are discrete random variables, both $X_x \in \Sigma$ and $Y_x \in \Sigma$.
As $\struct {\Omega, \Sigma, \Pr}$ is a probability space, then $X_x \cap Y_x \in \Sigma$.
Now note that:
:$\ds U_u = \bigcup_{x \mathop \in \R} \paren {X_x \cap Y_x}$
That is, it is the union of all such intersections of sets whose discrete random variables add up to $u$.
As $X_x$ is a countable set it follows that $U_u$ is a countable union of countable sets.
From Countable Union of Countable Sets is Countable it follows that $X_x$ is a countable set.
And, by dint of $\struct {\Omega, \Sigma, \Pr}$ being a probability space, $U_u \in \Sigma$.
Thus $U$ is a discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
{{qed}}
{{explain|Might need to be more rigorous on the boundary instances.}}
{{proofread}}
\end{proof}
|
22183
|
\section{Sum of Elements in Inverse of Cauchy Matrix}
Tags: Products, Summations, Summation of Powers over Product of Differences, Cauchy Matrix
\begin{theorem}
Let $C_n$ be the Cauchy matrix of order $n$ given by:
:$C_n = \begin{bmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_m + y_1} & \dfrac 1 {x_m + y_2 } & \cdots & \dfrac 1 {x_m + y_n} \\
\end{bmatrix}$
Let $C_n^{-1}$ be its inverse, from Inverse of Cauchy Matrix:
:$b_{i j} = \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } }$
The sum of all the elements of $C_n^{-1}$ is:
:$\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$
\end{theorem}
\begin{proof}
It suffices to prove the Theorem for Cauchy matrix:
{{begin-eqn}}
{{eqn | l = C
| r = \begin{pmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2 } & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2 } & \cdots & \dfrac 1 {x_2 - y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2 } & \cdots & \dfrac 1 {x_n - y_n} \\
\end {pmatrix}
| c = Distinct values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ required.
}}
{{end-eqn}}
The sum $S$ of elements in $C^{-1}$ will be shown to be a scalar product of two vectors $\vec A$ and $\vec B$.
The statement of the theorem is obtained by replacing $y_k \to -y_k$ in $C$ and in the sum $S$.
Let:
{{begin-eqn}}
{{eqn | l = \vec \bsone
| r = \begin {pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix}
| c = vector of all ones
}}
{{end-eqn}}
The sum $S$ of elements in $C^{-1}$ is the matrix product:
{{begin-eqn}}
{{eqn | n = 1
| l = S
| r = \vec \bsone^T C^{-1} \vec \bsone
}}
{{end-eqn}}
To identify vectors $\vec A$ and $\vec B$, the tool is:
{{begin-eqn}}
{{eqn | l = C
| r = -P V_x^{-1} V_y Q^{-1}
| c = Vandermonde Matrix Identity for Cauchy Matrix
}}
{{end-eqn}}
Definitions of symbols:
{{begin-eqn}}
{{eqn | l = V_x
| r = \begin {pmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
\vdots & \vdots & \ddots & \vdots \\
x_1^{n - 1} & x_2^{n - 1} & \cdots & x_n^{n - 1} \\
\end {pmatrix},
\quad
V_y =
\begin {pmatrix}
1 & 1 & \cdots & 1 \\
y_1 & y_2 & \cdots & y_n \\
\vdots & \vdots & \ddots & \vdots \\
y_1^{n - 1} & y_2^{n - 1} & \cdots & y_n^{n - 1} \\
\end {pmatrix}
| c = {{Defof|Vandermonde Matrix|subdef = Formulation 1}}
}}
{{eqn | l = \map p x
| r = \prod_{i \mathop = 1}^n \paren {x - x_i},
\quad
\map {p_k} x =
\prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i}
,\quad
1 \mathop \le k \mathop \le n
| c = {{Defof|Polynomial over Complex Numbers}}
}}
{{eqn | l = P
| r = \begin {pmatrix}
\map {p_1} {x_1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map {p_n} {x_n} \\
\end {pmatrix},
\quad
Q = \begin {pmatrix}
\map p {y_1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \map p {y_n} \\
\end {pmatrix}
| c = {{Defof|Diagonal Matrix}}
}}
{{end-eqn}}
'''Compute the sum''' $S$:
{{begin-eqn}}
{{eqn | l = C^{-1}
| r = - Q V_y^{-1} V_x P^{-1}
| c = Inverse of Matrix Product
}}
{{eqn | l = S
| r = \vec \bsone^T C^{-1} \vec \bsone
| c = from $(1)$
}}
{{eqn | r = \paren {-\vec {\bsone}^T Q V_y^{-1} } \paren {V_x P^{-1} \vec \bsone}
}}
{{end-eqn}}
Define vectors $\vec A$ and $\vec B$:
{{begin-eqn}}
{{eqn | n = 2
| l = \vec A^T
| r = -\vec \bsone^T Q V_y^{-1}
}}
{{eqn | n = 3
| l = \vec B
| r = V_x P^{-1} \vec \bsone
| c = Sum $S$ equals scalar product $\vec A^T \vec B$
}}
{{end-eqn}}
'''Compute''' $\vec A$:
{{begin-eqn}}
{{eqn | l = \vec A
| r = -\paren {\vec \bsone^T Q V_y^{-1} }^T
| c = from $(2)$
}}
{{eqn | r = -\paren {V_y^T}^{-1} Q^T \vec \bsone
| c = Definition:Inverse Matrix and Transpose of Matrix Product
}}
{{end-eqn}}
Then $\vec A$ '''uniquely''' solves the problem
{{begin-eqn}}
{{eqn | l = \begin{pmatrix} 1 & \cdots & y_1^{n - 1} \\ \vdots & \cdots & \vdots \\ 1 & \cdots & y_n^{n - 1} \end {pmatrix} \vec A
| r = -\begin{pmatrix} \map p {y_1} \\ \vdots \\ \map p {y_n} \\ \end {pmatrix}
| c = coefficient matrix $V_y^T$
}}
{{end-eqn}}
The two polynomials
{{begin-eqn}}
{{eqn | l = \map p x
| r = \prod_{j \mathop = 1}^n \paren {x - x_j}
}}
{{eqn | r = x^n + \sum_{i \mathop = 1}^n a_i x^{i - 1}
}}
{{eqn | l = \map q y
| r = \prod_{j \mathop = 1}^n \paren {y - y_j}
}}
{{eqn | r = y^n + \sum_{i \mathop = 1}^n b_i y^{i - 1}
}}
{{end-eqn}}
have coefficients expressed by symmetric functions:
{{begin-eqn}}
{{eqn | l = a_i
| r = \paren {-1}^{n + 1 - i} e_{n + 1 - i} \paren {\set {x_1, \ldots, x_n} }
| c = Viète's Formulas
}}
{{eqn | l = b_i
| r = \paren {-1}^{n + 1 - i} e_{n + 1 - i} \paren {\set {y_1, \ldots, y_n} }
| c = Viète's Formulas
}}
{{end-eqn}}
Solve in equation $\map q {y_j} = 0$ for highest power $y_j^n$ and replace in the equation for $\map p {y_j}$:
{{begin-eqn}}
{{eqn | l = \map p {y_j}
| r = \sum_{i \mathop = 1}^n \paren {a_i - b_i } y_j^{i - 1}
| c = for $j = 1, \ldots, n$
}}
{{end-eqn}}
Vector $\begin {pmatrix} \map p {y_1} \\ \vdots \\ -\map p {y_n} \\ \end {pmatrix}$ is a linear combination of the column vectors of matrix $V_y^T$ with weights (coefficients) $b_i - a_i$.
Then:
{{begin-eqn}}
{{eqn | l = \vec A
| r = \begin {pmatrix} b_1 - a_1 \\ \vdots \\ b_n - a_n \\ \end{pmatrix}
}}
{{eqn | r = \begin{pmatrix} \paren {-1}^n \paren {e_n \paren {\set {y_1,\dots,y_n} } - e_n \paren {\set {x_1,\dots,x_n} } } \\
\vdots \\
\paren {-1}^1 \paren {e_1 \paren {\set {y_1, \dots, y_n} } - e_1 \paren {\set {x_1, \dots, x_n} } } \\
\end {pmatrix}
| c = The last entry contains the sum of the elements in $C^{-1}$: $S = x_1 + \cdots + x_n - \paren {y_1 + \cdots + y_n}$
}}
{{end-eqn}}
'''Compute''' $\vec B$:
{{begin-eqn}}
{{eqn | l = \vec B
| r = V_x P^{-1} \vec { \bsone }
| c = from $(3)$
}}
{{eqn | r = V_x \begin {pmatrix} \dfrac 1 {\map {p_1} {x_1} } \\ \vdots \\ \dfrac 1 {\map {p_n} {x_n} } \end{pmatrix}
}}
{{eqn | r = \begin {pmatrix} \ds \sum_{k \mathop = 1}^n \frac 1 {\map {p_k} {x_k} } \\ \vdots \\ \ds \sum_{k \mathop = 1}^n \frac {x_n^{n - 1} } { \map {p_n} {x_n} } \\ \end{pmatrix}
}}
{{eqn | r = \begin {pmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ \end{pmatrix}
| c = Summation of Powers over Product of Differences
}}
{{end-eqn}}
'''Evaluate''' $S$:
{{begin-eqn}}
{{eqn | l = S
| r = \vec A^T \vec B
}}
{{eqn | r = \begin {pmatrix} * \\ * \\ \vdots \\ x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n } \\ \end {pmatrix} ^T \begin {pmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ \end {pmatrix}
| c = Entries $*$ contribute nothing because of matching zero in $\vec B$.
}}
{{eqn | r = x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22184
|
\section{Sum of Elements in Inverse of Combinatorial Matrix}
Tags: Combinatorial Matrix
\begin{theorem}
Let $C_n$ be the combinatorial matrix of order $n$ given by:
:$C_n = \begin{bmatrix}
x + y & y & \cdots & y \\
y & x + y & \cdots & y \\
\vdots & \vdots & \ddots & \vdots \\
y & y & \cdots & x + y
\end{bmatrix}$
Let $C_n^{-1}$ be its inverse, from Inverse of Combinatorial Matrix:
:$b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$
where $\delta_{i j}$ is the Kronecker delta.
The sum of all the elements of $C_n^{-1}$ is:
:$\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \dfrac n {x + n y}$
\end{theorem}
\begin{proof}
All $n^2$ elements of $C_n^{-1}$ have a term $\dfrac {-y} {x \paren {x + n y} }$.
Further to this, the $n$ elements on the main diagonal contribute an extra $\dfrac {x + n y} {x \paren {x + n y} }$ to the total.
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}
| r = n^2 \dfrac {-y} {x \paren {x + n y} } + n \dfrac {x + n y} {x \paren {x + n y} }
| c =
}}
{{eqn | r = \dfrac {-n^2 y + n \paren {x + n y} } {x \paren {x + n y} }
| c =
}}
{{eqn | r = \dfrac {-n^2 y + n x + n^2 y} {x \paren {x + n y} }
| c =
}}
{{eqn | r = \dfrac {n x} {x \paren {x + n y} }
| c =
}}
{{eqn | r = \dfrac n {x + n y}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22185
|
\section{Sum of Elements in Inverse of Hilbert Matrix}
Tags: Cauchy Matrix
\begin{theorem}
Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Consider its inverse $H_n^{-1}$.
All the elements of $H_n^{-1}$ are integers.
The sum of all the elements $b_{i j}$ of $H_n^{-1}$ is:
:$\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = n^2$
\end{theorem}
\begin{proof}
From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
Then:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}
| r = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k
| c = Sum of Elements in Inverse of Cauchy Matrix
}}
{{eqn | r = \sum_{k \mathop = 1}^n k + \sum_{k \mathop = 1}^n \paren {k - 1}
| c =
}}
{{eqn | r = 2 \sum_{k \mathop = 1}^n k - n
| c =
}}
{{eqn | r = n \paren {n + 1} - n
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = n^2
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22186
|
\section{Sum of Elements in Inverse of Vandermonde Matrix}
Tags: Vandermonde Matrices, Vandermonde Matrix
\begin{theorem}
Let $V_n$ be the Vandermonde matrix of order $n$ given by:
:$V_n = \begin{pmatrix}
x_1 & x_2 & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^n & x_2^n & \cdots & x_n^n
\end{pmatrix}$
Let $V_n^{-1}$ be its inverse, from Inverse of Vandermonde Matrix:
:$b_{i j} = \begin {cases}
\paren {-1}^{j - 1} \paren {\dfrac {\ds \sum_{\substack {1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n - j} \mathop \le n \\ m_1, \ldots, m_{n - j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n - j} } } {x_i \ds \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_m - x_i} } } & : 1 \le j < n \\
\qquad \qquad \qquad \dfrac 1 {x_i \ds \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_i - x_m} } & : j = n
\end{cases}$
The sum of all the elements of $V_n^{-1}$ is:
:$\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = 1 - \prod_{k \mathop = 1}^n \paren {1 - \dfrac 1 {x_k} }$
\end{theorem}
\begin{proof}
From Sum of Elements of Invertible Matrix, the sum of elements in $V_n^{-1}$ is:
:$1 - \map \det {V_n^{-1} } \map \det {V_n - J_n}$
The plan is to expand $\map \det {V_n - J_n}$ and simplify.
The method is efficiently communicated in the $3 \times 3$ case:
{{begin-eqn}}
{{eqn | l = \map \det {V_3 - J_3}
| r = \map \det {\begin {matrix}
x_1 - 1 & x_2 - 1 & x_3 - 1 \\
x_1^2 - 1 & x_2^2 - 1 & x_3^2 - 1 \\
x_1^3 - 1 & x_2^3 - 1 & x_3^3 - 1 \\
\end {matrix} }
| c = where $J_3$ is the $3 \times 3$ ones matrix
}}
{{eqn | r = \paren {x_1 - 1} \paren {x_2 - 1} \paren {x_3 - 1} \map \det {\begin {matrix}
1 & 1 & 1 \\
x_1+1 & x_2+1 & x_3+1 \\
x_1^2+x+1+1 & x_2^2+x_2+1 & x_3^2+x_3+1 \\
\end {matrix} }
| c =
Use $a^n - b^n$ factorization.
Effect of Elementary Row Operations on Determinant collects common column factors outside the determinant.
}}
{{eqn | r =\paren {x_1 - 1} \paren {x_2 - 1} \paren {x_3 - 1}
\map \det {\begin {matrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
x_1^2 & x_2^2 & x_3^2 \\
\end {matrix} }
| c = Effect of Elementary Row Operations on Determinant simplifies the determinant.
}}
{{end-eqn}}
The proof is completed by '''induction''' using the lemmas below.
\end{proof}
|
22187
|
\section{Sum of Elements of Inverse of Matrix with Column of Ones}
Tags: Inverse Matrices, Matrix Inverses
\begin{theorem}
Let $\mathbf B = \sqbrk b_n$ denote the inverse of a square matrix $\mathbf A$ of order $n$.
Let $\mathbf A$ be such that it has a row or column of all ones.
Then the sum of elements in $\mathbf B$ is one:
:$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{ij} = 1$
\end{theorem}
\begin{proof}
If ones appear in a row of $\mathbf A$, then replace $\mathbf A$ by $\mathbf A^T$ and $\mathbf B$ by $\mathbf B^T$.
Assume $\mathbf A$ has a column of ones.
Apply Sum of Elements of Invertible Matrix to the inverse $\mathbf B = \mathbf A^{-1}$:
:$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$
where $\mathbf J_n$ denotes the square ones matrix of order $n$.
If $\mathbf A = \mathbf B^{-1}$ has a column of ones, then $\mathbf B^{-1} - \mathbf J_n$ has a column of zeros, implying determinant zero.
Substitute $\map \det {\mathbf B^{-1} - \mathbf J_n} = 0$ in Sum of Elements of Invertible Matrix:
:$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - 0$
which implies the statement.
{{qed}}
\end{proof}
|
22188
|
\section{Sum of Entries in Lesser Diagonal of Pascal's Triangle equal Fibonacci Number}
Tags: Fibonacci Numbers, Pascal's Triangle
\begin{theorem}
The sum of the entries in the $n$th lesser diagonal of Pascal's triangle equals the $n + 1$th Fibonacci number.
\end{theorem}
\begin{proof}
By definition, the entries in the $n$th lesser diagonal of Pascal's triangle are:
:$\dbinom n 0, \dbinom {n - 1} 1, \dbinom {n - 2} 2, \dbinom {n - 3} 3, \ldots$
and so the statement can be written:
:$F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$
The proof proceeds by strong induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \dbinom {0 - k} k
| r = \dbinom 0 0
| c =
}}
{{eqn | r = 1
| c =
}}
{{eqn | r = F_1
| c =
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22189
|
\section{Sum of Euler Numbers by Binomial Coefficients Vanishes}
Tags: Euler Numbers, Sum of Euler Numbers by Binomial Coefficients Vanishes, Binomial Coefficients
\begin{theorem}
$\forall n \in \Z_{>0}: \ds \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
where $E_k$ denotes the $k$th Euler number.
\end{theorem}
\begin{proof}
Take the definition of Euler numbers:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}
| r = \frac {2 e^x} {e^{2 x} + 1}
| c =
}}
{{eqn | r = \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }
| c = Multiply by $1$
}}
{{eqn | r = \paren {\frac 2 {e^x + e^{-x} } }
| c =
}}
{{end-eqn}}
From the definition of the exponential function:
{{begin-eqn}}
{{eqn | l = e^x
| r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
| c =
}}
{{eqn | r = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots
| c =
}}
{{eqn | l = e^{-x}
| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-x}^n} {n!}
| c =
}}
{{eqn | r = 1 - x + \frac {x^2} {2!} - \frac {x^3} {3!} + \frac {x^4} {4!} - \cdots
| c =
}}
{{eqn | l = \paren {\frac {e^x + e^{-x} } 2}
| r = \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }
| c =
}}
{{eqn | r = 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \cdots
| c = odd terms cancel in the sum.
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = 1
| r = \paren {\frac 2 {e^x + e^{-x} } } \paren {\frac {e^x + e^{-x} } 2}
| c =
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }
| c =
}}
{{end-eqn}}
By Product of Absolutely Convergent Series, we will let:
{{begin-eqn}}
{{eqn | l = a_n
| r = \frac {E_n x^n} {n!}
| c =
}}
{{eqn | l = b_n
| r = \frac {x^{2 n} } {\paren {2 n}!}
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty c_n
| r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}
| rr= = 1
| c =
}}
{{eqn | l = c_n
| r = \sum_{k \mathop = 0}^n a_k b_{n - k}
| c =
}}
{{eqn | l = c_0
| r = \frac {E_0 x^0} {0!} \frac {x^0 } {0!}
| rr= = 1
| c = $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$
}}
{{eqn | ll= \leadsto
| l = \sum_{n \mathop = 1}^\infty c_n
| r = \paren { \ds \sum_{n \mathop = 0}^\infty a_n } \paren {\ds \sum_{n \mathop = 0}^\infty b_n} - a_0 b_0
| rr= = 0
| c = Subtract $1$ from both sides
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn | l = c_1
| r = \frac {E_0 x^0} {0!} \frac {x^2} {2!} + \frac {E_1 x^1} {1!} \frac {x^0} {0!}
| c = $= a_0 b_1 + a_1 b_0$
}}
{{eqn | l = c_2
| r = \frac {E_0 x^0} {0!} \frac {x^4} {4!} + \frac {E_1 x^1} {1!} \frac {x^2} {2!} + \frac {E_2 x^2} {2!} \frac {x^0} {0!}
| c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$
}}
{{eqn | l = c_3
| r = \frac {E_0 x^0} {0!} \frac {x^6} {6!} + \frac {E_1 x^1} {1!} \frac {x^4} {4!} + \frac {E_2 x^2} {2!} \frac {x^2} {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!}
| c = $= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$
}}
{{eqn | l = c_4
| r = \frac {E_0 x^0} {0!} \frac {x^8} {8!} + \frac {E_1 x^1} {1!} \frac {x^6} {6!} + \frac {E_2 x^2} {2!} \frac {x^4} {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0} {0!}
| c = $= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$
}}
{{eqn | o = \cdots
}}
{{eqn | l = c_n
| r = \frac {E_0 x^0} {0!} \frac {x^{2 n} } {\paren {2 n}!} + \frac {E_1 x^1} {1!} \frac {x^{2 n - 2} } {\paren {2 n - 2}!} + \frac {E_2 x^2} {2!} \frac {x^{2 n - 4} } {\paren {2 n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0} {0!}
| c =
}}
{{end-eqn}}
Grouping terms with even exponents produces:
{{begin-eqn}}
{{eqn | l = \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2
| r = 0
| c = $x^2$ term from $c_1$ and $c_2$
}}
{{eqn | l = \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4
| r = 0
| c = $x^4$ term from $c_2$, $c_3$ and $c_4$
}}
{{eqn | o = \cdots
}}
{{eqn | l = \paren {\frac 1 {0! \paren {2 n}!} } E_0 + \paren {\frac 1 {2! \paren {2 n - 2 }!} } E_2 + \paren {\frac 1 {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\frac 1 {\paren {2 n}! 0!} } E_{2 n}
| r = 0
| c = $x^{2 n}$ term from $c_n$, $c_{n + 1} \cdots c_{2 n}$
}}
{{end-eqn}}
$\forall n \in \Z_{>0}$, multiplying the coefficients of $x^{2 n}$ through by $\paren {2 n}!$ gives:
:$\paren {\dfrac {\paren {2 n}! } {0! \paren {2 n}!} } E_0 + \paren {\dfrac {\paren {2 n}! } {2! \paren {2 n - 2 }!} } E_2 + \paren {\dfrac {\paren {2 n}! } {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\dfrac {\paren {2 n}! } {\paren {2 n}! 0!} } E_{2 n} = 0$
But those coefficients are the binomial coefficients:
:$\dbinom {2 n} 0 E_0 + \dbinom {2 n} 2 E_2 + \dbinom {2 n} 4 E_4 + \dbinom {2 n} 6 E_6 + \cdots + \dbinom {2 n} {2 n} E_{2 n} = 0$
Hence the result.
{{qed}}
\end{proof}
|
22190
|
\section{Sum of Euler Phi Function over Divisors}
Tags: Number Theory, Euler Phi Function
\begin{theorem}
Let $n \in \Z_{>0}$ be a strictly positive integer.
Then $\ds \sum_{d \mathop \divides n} \map \phi d = n$
where:
:$\ds \sum_{d \mathop \divides n}$ denotes the sum over all of the divisors of $n$
:$\map \phi d$ is the Euler $\phi$ function, the number of integers less than $d$ that are prime to $d$.
That is, the total of all the totients of all divisors of a number equals that number.
\end{theorem}
\begin{proof}
Let us define:
:$S_d = \set {m \in \Z: 1 \le m \le n, \gcd \set {m, n} = d}$.
That is, $S_d$ is all the numbers less than or equal to $n$ whose GCD with $n$ is $d$.
Now from Integers Divided by GCD are Coprime we have:
:$\gcd \set {m, n} = d \iff \dfrac m d, \dfrac n d \in \Z: \dfrac m d \perp \dfrac n d$
So the number of integers in $S_d$ equals the number of positive integers no bigger than $\dfrac n d$ which are prime to $\dfrac n d$.
That is, by definition of the Euler phi function:
:$\card {S_d} = \map \phi {\dfrac n d}$
From the definition of the $S_d$, it follows that for all $1 \le m \le n$:
:$\exists d \divides n: m \in S_d$
Therefore:
:$\ds \set {1, \ldots, n} = \bigcup_{d \mathop \divides n} S_d$
Moreover, it follows from the definition of the $S_d$ that they are pairwise disjoint.
Now from Corollary to Cardinality of Set Union, it follows that:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{d \mathop \divides n} \card {S_d}
}}
{{eqn | r = \sum_{d \mathop \divides n} \map \phi {\dfrac n d}
}}
{{end-eqn}}
But from Sum Over Divisors Equals Sum Over Quotients:
:$\ds \sum_{d \mathop \divides n} \map \phi {\dfrac n d} = \sum_{d \mathop \divides n} \map \phi d$
and hence the result.
{{qed}}
\end{proof}
|
22191
|
\section{Sum of Even Index Binomial Coefficients}
Tags: Proofs by Induction, Sum of Even Index Binomial Coefficients, Binomial Coefficients
\begin{theorem}
Let $n > 0$ be a (strictly) positive integer.
Then:
:$\ds \sum_{i \mathop \ge 0} \binom n {2 i} = 2^{n - 1}$
where $\dbinom n i$ is a binomial coefficient.
That is:
:$\dbinom n 0 + \dbinom n 2 + \dbinom n 4 + \dotsb = 2^{n - 1}$
\end{theorem}
\begin{proof}
From Sum of Binomial Coefficients over Lower Index we have:
:$\displaystyle \sum_{i \mathop \in \Z} \binom n i = 2^n$
That is:
:$\displaystyle \binom n 0 + \binom n 1 + \binom n 2 + \binom n 3 + \cdots + \binom n n = 2^n$
as $\displaystyle \binom n i = 0$ for $i < 0$ and $i > n$.
This can be written more conveniently as:
:$\displaystyle \binom n 0 + \binom n 1 + \binom n 2 + \binom n 3 + \binom n 4 + \cdots = 2^n$
Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have:
:$\displaystyle \sum_{i \mathop \in \Z} \left({-1}\right)^i \binom n i = 0$
That is:
:$\displaystyle \binom n 0 - \binom n 1 + \binom n 2 - \binom n 3 + \binom n 4 - \cdots = 0$
Adding them together, we get:
:$\displaystyle 2 \binom n 0 + 2 \binom n 2 + 2 \binom n 4 + \cdots = 2^n$
as the odd index coefficients cancel out.
Dividing by $2$ throughout gives us the result.
{{qed}}
\end{proof}
|
22192
|
\section{Sum of Even Integers is Even}
Tags: Euclidean Number Theory, Sum of Even Integers is Even, Even Integers
\begin{theorem}
The sum of any finite number of even integers is itself even.
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
:The sum of $n$ even integers is an even integer.
$P(1)$ is trivially true, as this just says:
:The sum of $1$ even integers is an even integer.
The sum of $0$ even integers is understood, from the definition of a vacuous summation, to be $0$, which is even.
So $P(0)$ is also true.
\end{proof}
|
22193
|
\section{Sum of Even Sequence of Products of Consecutive Fibonacci Numbers}
Tags: Sums of Sequences, Fibonacci Numbers
\begin{theorem}
Let $F_k$ be the $k$'th Fibonacci number.
Then:
:$\ds \sum_{j \mathop = 1}^{2 n} F_j F_{j + 1} = {F_{2 n + 1} }^2 - 1$
\end{theorem}
\begin{proof}
From Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers:
:$(1): \quad \ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^{2 n} F_j F_{j + 1}
| r = \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} + F_{2 n} F_{2 n + 1}
| c =
}}
{{eqn | r = {F_{2 n} }^2 + F_{2 n} F_{2 n + 1}
| c = from $(1)$
}}
{{eqn | r = F_{2 n} \paren {F_{2 n} + F_{2 n + 1} }
| c =
}}
{{eqn | r = F_{2 n} F_{2 n + 2}
| c =
}}
{{eqn | r = {F_{2 n + 1} }^2 - 1
| c = Cassini's Identity
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22194
|
\section{Sum of Expectations of Independent Trials}
Tags: Sum of Expectations of Independent Trials, 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 $\expect {X_j}$ denote the expectation of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then we have, whenever both sides are defined:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
That is, the sum of the expectations equals the expectation of the sum.
\end{theorem}
\begin{proof}
We will use induction on $n$, that is, on the number of terms in the sum.
\end{proof}
|
22195
|
\section{Sum of Exponential of i k x}
Tags: Exponential Function
\begin{theorem}
:$\ds \sum_{k \mathop = 0}^n \map \exp {i k x} = \paren {i \sin \frac {n x} 2 + \cos \frac {n x} 2} \frac {\map \sin {\frac {\paren {n + 1} x} 2} } {\sin \frac x 2}$
where $x$ is a complex number that is not an integer multiple of $2 \pi$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \map \exp {i k x}
| r = \frac {\map \exp {i x \paren {n + 1} } - 1} {\map \exp {i x} - 1}
| c = Sum of Geometric Sequence
}}
{{eqn | r = \frac {\map \exp {\frac {i x \paren {n + 1} } 2} \paren {\map \exp {\frac {i x \paren {n + 1} } 2} - \map \exp {\frac {-i x \paren {n + 1} } 2} } } {\map \exp {\frac {i x} 2} \paren {\map \exp {\frac {i x} 2} - \map \exp {-\frac {i x} 2} } }
| c = factoring $\dfrac {\map \exp {\frac {i x \paren {n + 1} } 2} } {\map \exp {\frac {i x} 2} }$
}}
{{eqn | r = \map \exp {\frac {i x n} 2} \frac {2 i \map \sin {\frac {\paren {n + 1} x} 2} } {2 i \sin \frac x 2}
| c = Exponential of Sum: Real Numbers: Corollary, Sine Exponential Formulation
}}
{{eqn | r = \paren {i \sin \frac {n x} 2 + \cos \frac {n x} 2} \frac {\map \sin {\frac {\paren {n + 1} x} 2} } {\sin \frac x 2}
| c = Euler's Formula
}}
{{end-eqn}}
{{qed}}
Category:Exponential Function
\end{proof}
|
22196
|
\section{Sum of Floor and Floor of Negative}
Tags: Floor and Ceiling, Analysis, Floor Function
\begin{theorem}
Let $x \in \R$. Then:
:$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$
where $\floor x$ denotes the floor of $x$.
\end{theorem}
\begin{proof}
Let $x \in \Z$.
Then from Real Number is Integer iff equals Floor:
:$x = \floor x$
Now $x \in \Z \implies -x \in \Z$, so:
:$\floor {-x} = -x$
Thus:
:$\floor x + \floor {-x} = x + \paren {-x} = x - x = 0$
Now let $x \notin \Z$.
From Real Number is Floor plus Difference:
:$x = n + t$
where $n = \floor x$ and $t \in \hointr 0 1$.
Thus:
:$-x = -\paren {n + t} = -n - t = -n - 1 + \paren {1 - t}$
As $t \in \hointr 0 1$, we have:
:$1 - t \in \hointr 0 1$
Thus:
:$\floor {-x} = -n - 1$
So:
:$\floor x + \floor {-x} = n + \paren {-n - 1} = n - n - 1 = -1$
{{qed}}
\end{proof}
|
22197
|
\section{Sum of Floors not greater than Floor of Sum}
Tags: Floor and Ceiling, Floor Function
\begin{theorem}
Let $\floor x$ denote the floor function.
Then:
:$\floor x + \floor y \le \floor {x + y}$
The equality holds:
:$\floor x + \floor y = \floor {x + y}$
{{iff}}:
:$x \bmod 1 + y \bmod 1 < 1$
where $x \bmod 1$ denotes the modulo operation.
\end{theorem}
\begin{proof}
From the definition of the modulo operation, we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which:
{{begin-eqn}}
{{eqn | l = \floor {x + y}
| r = \floor {\floor x + \paren {x \bmod 1} + \floor y + \paren {y \bmod 1} }
| c =
}}
{{eqn | r = \floor x + \floor y + \floor {\paren {x \bmod 1} + \paren {y \bmod 1} }
| c = Floor of Number plus Integer
}}
{{end-eqn}}
Hence the inequality.
The equality holds {{iff}}:
:$\floor {\paren {x \bmod 1} + \paren {y \bmod 1} } = 0$
that is, {{iff}}:
:$x \bmod 1 + y \bmod 1 < 1$
{{qed}}
\end{proof}
|
22198
|
\section{Sum of Fourth Powers with Product of Squares}
Tags: Square Function, Fourth Powers, Square Functions
\begin{theorem}
:$x^4 + x^2 y^2 + y^4 = \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x^6 - y^6
| r = \paren {x - y} \paren {x + y} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
| c = Difference of Two Sixth Powers
}}
{{eqn | r = \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
| c = Difference of Two Squares
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = x^6 - y^6
| r = \paren {x^2}^3 - \paren {y^2}^3
| c =
}}
{{eqn | r = \paren {x^2 - y^2} \paren {\paren {x^2}^2 + \paren {x^2} \paren {y^2} + \paren {y^2}^2}
| c = Difference of Two Cubes
}}
{{eqn | r = \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}
| c = simplifying
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}
| r = \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
| c = as both equal $x^6 - y^6$
}}
{{eqn | ll= \leadsto
| l = \paren {x^4 + x^2 y^2 + y^4}
| r = \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
| c = cancelling $\paren {x^2 - y^2}$ from both sides
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22199
|
\section{Sum of Functions of Bounded Variation is of Bounded Variation}
Tags: Bounded Variation, Total Variation
\begin{theorem}
Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $V_f$ and $V_g$ be the total variations of $f$ and $g$ respectively.
Then $f + g$ is of bounded variation with:
:$V_{f + g} \le V_f + V_g$
where $V_{f + g}$ denotes the total variation of $f + g$.
\end{theorem}
\begin{proof}
For each finite subdivision $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn | l = \map {V_{f + g} } P
| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {x_i} - \map {\paren {f + g} } {x_{i - 1} } }
| c = using the notation from the definition of bounded variation
}}
{{eqn | r = \sum_{i \mathop = 1}^n \size {\paren {\map f {x_i} - \map f {x_{i - 1} } } + \paren {\map g {x_i} - \map g {x_{i - 1} } } }
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } + \sum_{i \mathop = 1}^n \size {\map g {x_i} - \map g {x_{i - 1} } }
| c = Triangle Inequality
}}
{{eqn | r = \map {V_f} P + \map {V_g} P
}}
{{end-eqn}}
Note that since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that:
:$\map {V_f} P \le M$
:$\map {V_g} P \le K$
for all finite subdivisions $P$ of $\closedint a b$.
We therefore have:
:$\map {V_{f + g} } P \le M + K$
for all finite subdivisions $P$.
So $f + g$ is of bounded variation.
Note then that:
{{begin-eqn}}
{{eqn | l = V_{f + g}
| r = \sup_P \paren {\map {V_{f + g} } P}
| c = {{Defof|Total Variation}}
}}
{{eqn | o = \le
| r = \sup_P \paren {\map {V_f} P} + \sup_P \paren {\map {V_g} P}
}}
{{eqn | r = V_f + V_g
| c = {{Defof|Total Variation}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22200
|
\section{Sum of Functions of Exponential Order}
Tags: Exponential Order
\begin{theorem}
Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Suppose $f$ is of exponential order $a$ and $g$ is of exponential order $b$.
Then $f + g: t \mapsto \map f t + \map g t$ is of exponential order $\max \set {a, b}$.
\end{theorem}
\begin{proof}
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval.
By the definition of exponential order:
{{begin-eqn}}
{{eqn | l = \size {\map f t}
| o = <
| r = K_1 e^{a t}
}}
{{eqn | l = \size {\map g t}
| o = <
| r = K_2 e^{b t}
}}
{{eqn | l = \size {\map f t} + \size {\map g t}
| o = <
| r = K_1 e^{a t} + K_2 e^{b t}
| c = Real Number Inequalities can be Added
}}
{{eqn | ll= \leadsto
| l = \size {\map f t + \map g t}
| o = <
| r = K_1 e^{\max \size {a, b} t} + K_2 e^{\max \size {a, b} t}
| c = Triangle Inequality for Real Numbers, Exponential is Strictly Increasing
}}
{{eqn | r = K' e^{\max \size {a, b} t}
| c = $K' = K_1 + K_2$
}}
{{end-eqn}}
{{qed}}
Category:Exponential Order
\end{proof}
|
22201
|
\section{Sum of Geometric Sequence/Corollary 2}
Tags: Geometric Sequences, Sums of Sequences, Geometric Progressions, Sum of Geometric Progression, Sum of Geometric Sequence
\begin{theorem}
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}
| r = \sum_{j \mathop = 0}^{n - 1} D_x x^j
| c = Power Rule for Derivatives
}}
{{eqn | r = D_x \sum_{j \mathop = 0}^{n - 1} x^j
| c = Sum Rule for Derivatives
}}
{{eqn | r = D_x \frac {x^n - 1} {x - 1}
| c = Sum of Geometric Sequence
}}
{{eqn | r = \frac {\paren {n x^{n - 1} } \paren {x - 1} - \paren 1 \paren {x^n - 1} } {\paren {x - 1}^2}
| c = Quotient Rule for Derivatives
}}
{{eqn | r = \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}
}}
{{eqn | ll= \leadsto
| l = \sum_{j \mathop = 0}^{n - 1} j x^j
| r = x \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}
| c = Multiplication of Numbers Distributes over Addition
}}
{{eqn | r = x \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}
| c = by the above result
}}
{{eqn | r = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}
}}
{{end-eqn}}
{{qed}}
Category:Geometric Sequences
Category:Sum of Geometric Sequence
\end{proof}
|
22202
|
\section{Sum of Geometric Sequence/Examples/Common Ratio 1}
Tags: Sum of Geometric Sequence, Sum of Geometric Progression
\begin{theorem}
Consider the Sum of Geometric Sequence defined on the standard number fields for all $x \ne 1$.
:$\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
When $x = 1$, the formula reduces to:
:$\ds \sum_{j \mathop = 0}^n a 1^j = a \paren {n + 1}$
\end{theorem}
\begin{proof}
When $x = 1$, the {{RHS}} is undefined:
:$a \paren {\dfrac {1 - 1^{n + 1} } {1 - 1} } = a \dfrac 0 0$
However, the {{LHS}} degenerates to:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n a 1^j
| r = \sum_{j \mathop = 0}^n a
| c =
}}
{{eqn | r = a \paren {n + 1}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22203
|
\section{Sum of Geometric Sequence/Examples/Index to Minus 1}
Tags: Sum of Geometric Sequence, Sum of Geometric Progression
\begin{theorem}
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
still holds when $n = -1$:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = \frac {x^0 - 1} {x - 1}$
\end{theorem}
\begin{proof}
The summation on the {{LHS}} is vacuous:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = 0$
while on the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = \frac {x^{\paren {-1} + 1} - 1} {x - 1}
| r = \frac {x^0 - 1} {x - 1}
| c =
}}
{{eqn | r = \frac 0 {x - 1}
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
as long as $x \ne 1$.
However, the theorem itself is based on the assumption that $n \ge 0$, so while the result is correct, the derivation to achieve it is not.
{{qed}}
\end{proof}
|
22204
|
\section{Sum of Geometric Sequence/Examples/Index to Minus 2}
Tags: Sum of Geometric Sequence, Sum of Geometric Progression
\begin{theorem}
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
breaks down when $n = -2$:
:$\ds \sum_{j \mathop = 0}^{-2} x^j \ne \frac {x^{-1} - 1} {x - 1}$
\end{theorem}
\begin{proof}
The summation on the {{LHS}} is vacuous:
:$\ds \sum_{j \mathop = 0}^{-2} x^j = 0$
while on the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = \frac {x^{\paren {-2} + 1} - 1} {x - 1}
| r = \frac {x^{-1} - 1} {x - 1}
| c =
}}
{{eqn | r = \frac {1 / x - 1} {x - 1}
| c =
}}
{{eqn | r = \frac {\paren {1 - x} / x} {x - 1}
| c =
}}
{{eqn | r = \frac {1 - x} {x \paren {x - 1} }
| c =
}}
{{eqn | r = -\frac 1 x
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22205
|
\section{Sum of Geometric Sequence/Examples/One Seventh from 1 to n}
Tags: Sum of Geometric Sequence, Sum of Geometric Progression
\begin{theorem}
:$\ds \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \paren {1 - \frac 1 {7^{n + 1} } }$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n \dfrac 1 {7^j}
| r = \frac {1 - \frac 1 7^{n + 1} } {1 - \frac 1 7}
| c =
}}
{{eqn | r = \frac {7 - 7 \frac 1 7^{n + 1} } {7 - 1}
| c = multiplying top and bottom by $7$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22206
|
\section{Sum of Harmonic Numbers approaches Harmonic Number of Product of Indices}
Tags: Harmonic Numbers
\begin{theorem}
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let:
:$\map T {m, n} := H_m + H_n - H_{m n}$
where $H_n$ denotes the $n$th harmonic number.
Then as $m$ or $n$ increases, $\map T {m, n}$ never increases, and reaches its minimum when $m$ and $n$ approach infinity.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map T {m + 1, n} - \map T {m, n}
| r = \left({H_{m + 1} + H_n - H_{\paren {m + 1} n} }\right) - \paren {H_m + H_n - H_{m n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} + \paren {H_{\paren {m + 1} n} - H_{m n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} - \paren {\frac 1 {m n + 1} + \frac 1 {m n + 2} + \cdots + \frac 1 {m n + n} }
| c =
}}
{{eqn | o = \le
| r = \dfrac 1 {m + 1} - \paren {\frac 1 {m n + n} + \frac 1 {m n + n} + \cdots + \frac 1 {m n + n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} - \frac n {n \paren {m + 1} }
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
Similarly:
:$\map T {m, n + 1} - \map T {m, n} \le 0$
From Approximate Size of Sum of Harmonic Series, the limiting value of $\map T {m, n}$ is the Euler-Mascheroni constant:
:$\ds \lim_{m, n \mathop \to \infty} H_m + H_n - H_{m n} = \gamma$
{{qed}}
\end{proof}
|
22207
|
\section{Sum of Hyperbolic Tangent and Cotangent}
Tags: Hyperbolic Cotangent Function, Hyperbolic Tangent Function
\begin{theorem}
:$\tanh x + \coth x = 2 \coth 2 x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tanh x + \coth x
| r = \frac {\sinh x} {\cosh x} + \frac {\cosh x} {\sinh x}
| c = {{Defof|Hyperbolic Tangent}} and Hyperbolic Cotangent
}}
{{eqn | r = \frac {\cosh^2 x + \sinh^2 x} {\sinh x \cosh x}
| c =
}}
{{eqn | r = 2 \frac {\cosh^2 x + \sinh^2 x} {2 \sinh x \cosh x}
| c =
}}
{{eqn | r = 2 \frac {\cosh 2 x} {\sinh 2 x}
| c = Double Angle Formula for Hyperbolic Sine and Double Angle Formula for Hyperbolic Cosine
}}
{{eqn | r = 2 \coth 2 x
| c = {{Defof|Hyperbolic Cotangent}}
}}
{{end-eqn}}
{{qed}}
Category:Hyperbolic Tangent Function
Category:Hyperbolic Cotangent Function
\end{proof}
|
22208
|
\section{Sum of Ideals is Ideal}
Tags: Ideal Theory
\begin{theorem}
Let $J_1$ and $J_2$ be ideals of a ring $\struct{R, +, \circ}$.
Then:
: $J = J_1 + J_2$ is an ideal of $R$
where $J_1 + J_2$ is as defined in subset product.
\end{theorem}
\begin{proof}
By definition, $\struct {R, +}$ is an abelian group.
So from Subset Product of Abelian Subgroups, we have that:
:$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$
is itself a subgroup of $R$.
Now consider $a \circ b$ where $a \in J, b \in R$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| r = \paren {a_1 + a_2} \circ b
| c = by definition of $a$
}}
{{eqn | r = \paren {a_1 \circ b} + \paren {a_2 \circ b}
| c = as $\circ$ is distributive over $+$
}}
{{eqn | o = \in
| r = J_1 + J_2
| c = as $\paren {a_i \circ b} \in J_i$ because $J_i$ is an ideal of $R$
}}
{{end-eqn}}
Similarly, $b \circ a \in J_1 + J_2$
So by definition $J_1 + J_2$ is an ideal of $R$.
{{qed}}
\end{proof}
|
22209
|
\section{Sum of Ideals is Ideal/General Result}
Tags: Ideal Theory
\begin{theorem}
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Then:
:$J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.
\end{theorem}
\begin{proof}
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
$\map P 1$ is true, as this just says $J_1$ is an ideal of $R$.
\end{proof}
|
22210
|
\section{Sum of Ideals is Ideal/General Result/Corollary}
Tags: Ideal Theory
\begin{theorem}
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\left({R, +, \circ}\right)$.
Let $J = J_1 + J_2 + \cdots + J_n$ be an ideal of $R$ where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.
$J$ is contained in every subring of $R$ containing $\displaystyle \bigcup_{k \mathop = 1}^n {J_k}$.
\end{theorem}
\begin{proof}
From Sum of Ideals is Ideal/General Result we have that $J$ is an ideal of $R$.
The result follows directly from the definition of join of subgroups.
{{qed}}
\end{proof}
|
22211
|
\section{Sum of Identical Terms}
Tags: Numbers
\begin{theorem}
Let $x$ be a number.
Let $n \in \N$ be a natural number such that $n \ge 1$.
Then:
:$\ds \sum_{i \mathop = 1}^n x = n x$
{{explain|Why limit this to $n \ge 1$? It also works for zero.}}
\end{theorem}
\begin{proof}
{{finish|this could be actually nontrivial; induction on $n$ seems easiest}}
{{expand|generalize to $x$ an element of a vector space, or for that matter, any abelian group}}
Category:Numbers
\end{proof}
|
22212
|
\section{Sum of Independent Binomial Random Variables}
Tags: Probability Generating Functions, Binomial Distribution
\begin{theorem}
Let $X$ and $Y$ be discrete random variables with a binomial distribution:
:$X \sim \Binomial m p$
and
:$Y \sim \Binomial n p$
Let $X$ and $Y$ be independent.
Then their sum $Z = X + Y$ is distributed as:
:$Z \sim \Binomial {m + n} p$
\end{theorem}
\begin{proof}
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
:$\map {\Pi_X} s = \paren {q + p s}^m$
:$\map {\Pi_Y} s = \paren {q + p s}^n$
respectively.
Now because of their independence, we have:
{{begin-eqn}}
{{eqn | l = \map {\Pi_{X + Y} } s
| r = \map {\Pi_X} s \map {\Pi_Y} s
| c = PGF of Sum of Independent Discrete Random Variables
}}
{{eqn | r = \paren {q + p s}^m \paren {q + p s}^n
| c = Definition of $X$ and $Y$
}}
{{eqn | r = \paren {q + p s}^{m + n}
| c = Product of Powers
}}
{{end-eqn}}
This is the probability generating function for a discrete random variable with a binomial distribution:
:$\Binomial {m + n} p$
Therefore:
:$Z = X + Y \sim \Binomial {m + n} p$
{{qed}}
\end{proof}
|
22213
|
\section{Sum of Independent Poisson Random Variables is Poisson}
Tags: Probability Generating Functions, Sum of Independent Poisson Random Variables is Poisson, Poisson Distribution
\begin{theorem}
Let $X$ and $Y$ be independent discrete random variables with:
:$X \sim \Poisson {\lambda_1}$
and
:$Y \sim \Poisson {\lambda_2}$
for some $\lambda_1, \lambda_2 \in \R_{> 0}$.
Then their sum $X + Y$ is distributed:
:$X + Y \sim \Poisson {\lambda_1 + \lambda_2}$
\end{theorem}
\begin{proof}
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
:$\map {\Pi_X} s = e^{-\lambda_1 \paren {1 - s} }$
:$\map {\Pi_Y} s = e^{-\lambda_2 \paren {1 - s} }$
respectively.
Now because of their independence, we have:
{{begin-eqn}}
{{eqn | l = \map {\Pi_{X + Y} } s
| r = \map {\Pi_X} s \, \map {\Pi_Y} s
| c = PGF of Sum of Independent Discrete Random Variables
}}
{{eqn | r = e^{-\lambda_1 \paren {1 - s} } e^{-\lambda_2 \paren {1 - s} }
| c = Definition of $X$ and $Y$
}}
{{eqn | r = e^{-\paren {\lambda_1 + \lambda_2} \paren {1 - s} }
| c = Exponential of Sum
}}
{{end-eqn}}
This is the probability generating function for a discrete random variable with a Poisson distribution:
:$\Poisson {\lambda_1 + \lambda_2}$
Therefore:
:$Z = X + Y \sim \Poisson {\lambda_1 + \lambda_2}$
{{qed}}
\end{proof}
|
22214
|
\section{Sum of Indices of Real Number/Integers}
Tags: Powers, Sum of Indices of Real Number
\begin{theorem}
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z$ be integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$r^{n + m} = r^n \times r^m$
\end{theorem}
\begin{proof}
From Sum of Indices of Real Number: Positive Integers, we have that:
:$m \in \Z_{\ge 0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$
It remains to be shown that:
:$\forall m \in \Z_{<0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$
The proof will proceed by induction on $m$.
As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.
For all $p \in \Z_{>0}$, let $\map P p$ be the proposition:
:$\forall n \in \Z: r^{n + \paren {-p} } = r^n \times r^{-p}$
that is:
:$\forall n \in \Z: r^{n - p} = r^n \times r^{-p}$
\end{proof}
|
22215
|
\section{Sum of Indices of Real Number/Positive Integers}
Tags: Powers, Algebra, Sum of Indices of Real Number, Analysis
\begin{theorem}
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z_{\ge 0}$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
: $r^{n + m} = r^n \times r^m$
\end{theorem}
\begin{proof}
Proof by induction on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\forall n \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$
$\map P 0$ is true, as this just says:
:$r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$
\end{proof}
|
22216
|
\section{Sum of Indices of Real Number/Rational Numbers}
Tags: Powers, Sum of Indices of Real Number
\begin{theorem}
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
Let $x, y \in \Q$ be rational numbers.
Let $r^x$ be defined as $r$ to the power of $n$.
Then:
: $r^{x + y} = r^x \times r^y$
\end{theorem}
\begin{proof}
Let $x = \dfrac p q, y = \dfrac u v$.
Then:
{{begin-eqn}}
{{eqn | l = r^\paren {x + y}
| r = r^\paren {\paren {p / q} + \paren {u / v} }
| c =
}}
{{eqn | r = r^\paren {\paren {p v + u q} / q v}
| c =
}}
{{eqn | r = \paren {r^\paren {1 / q v} }^\paren {p v + u q}
| c = {{Defof|Rational Power}}
}}
{{eqn | r = \paren {r^\paren {1 / q v} }^\paren {p v} \times \paren {r^\paren {1 / q v} }^\paren {u q}
| c = Sum of Indices of Real Number: Integers
}}
{{eqn | r = r^\paren {p v / q v} \times r^\paren {u q / q v}
| c = {{Defof|Rational Power}}
}}
{{eqn | r = r^\paren {p / q} \times r^\paren {u / v}
| c =
}}
{{eqn | r = r^x \times r^y
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22217
|
\section{Sum of Infinite Arithmetic-Geometric Sequence}
Tags: Arithmetic-Geometric Sequences, Sums of Sequences
\begin{theorem}
Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $n = 0, 1, 2, \ldots$
Let:
:$\size r < 1$
where $\size r$ denotes the absolute value of $r$.
Then:
:$\ds \sum_{n \mathop = 0}^\infty \paren {a + k d} r^k = \frac a {1 - r} + \frac {r d} {\paren {1 - r}^2}$
\end{theorem}
\begin{proof}
From Sum of Arithmetic-Geometric Sequence, we have:
:$\ds s_n = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
We have that $\size r < 1$.
So by Sequence of Powers of Number less than One:
:$r^n \to 0$ as $n \to \infty$
and
:$r^{n - 1} \to 0$ as $n - 1 \to \infty$
Hence:
:$s_n \to \dfrac a {1 - r} + \dfrac {r d} {\paren {1 - r}^2}$
as $n \to \infty$.
The result follows.
{{qed}}
\end{proof}
|
22218
|
\section{Sum of Infinite Geometric Sequence}
Tags: Examples of Power Series, Geometric Sequences, Convergence Tests, Sum of Infinite Geometric Progression, Geometric Progressions, Sum of Geometric Progression, Sum of Infinite Geometric Sequence, Series, Sums of Sequences, Sum of Geometric Sequence
\begin{theorem}
Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$.
Let $z \in S$.
Let $\size z < 1$, where $\size z$ denotes:
:the absolute value of $z$, for real and rational $z$
:the complex modulus of $z$ for complex $z$.
Then $\ds \sum_{n \mathop = 0}^\infty z^n$ converges absolutely to $\dfrac 1 {1 - z}$.
\end{theorem}
\begin{proof}
From Sum of Geometric Progression, we have:
: $\displaystyle s_N = \sum_{n \mathop = 0}^N z^n = \frac {1 - z^{N+1}} {1 - z}$
We have that $\left \vert {z}\right \vert < 1$.
So by Power of Number less than One:
: $z^{N+1} \to 0$ as $N \to \infty$
Hence $s_N \to \dfrac 1 {1 - z}$ as $N \to \infty$.
The result follows.
{{qed}}
To demonstrate absolute convergence we note that the absolute value of $\left \vert {z}\right \vert$ is just $\left \vert {z}\right \vert$, and by assumption we have $\left \vert {z}\right \vert < 1$, so $\left \vert {z}\right \vert$ fulfils the same condition for convergence as $z$, and we get:
{{tidy|rewrite in house style}}
:$\displaystyle \sum_{n \mathop = 0}^\infty \left \vert {z}\right \vert^n = \frac 1 {1 - \left \vert {z}\right \vert}$.
{{qed}}
\end{proof}
|
22219
|
\section{Sum of Infinite 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}^\infty r^k \cos k x
| r = 1 + r \cos x + r^2 \cos 2 x + r^3 \cos 3 x + \cdots
| c =
}}
{{eqn| r = \dfrac {1 - r \cos 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 = 0}^\infty r^k \cos k x
| r = \map \Re {\sum_{k \mathop = 0}^\infty r^k e^{i k x} }
| c =
}}
{{eqn | r = \map \Re {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}
| c =
}}
{{eqn | r = \map \Re {\frac 1 {1 - r e^{i x} } }
| c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Re {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }
| c =
}}
{{eqn | r = \map \Re {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
| c =
}}
{{eqn | r = \map \Re {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + a^2} }
| c = Corollary to Euler's Formula
}}
{{eqn | r = \dfrac {1 - r \cos x} {1 - 2 r \cos x + r^2}
| c = after simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22220
|
\section{Sum of Infinite 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}^\infty r^k \sin k x
| r = r \sin x + r^2 \sin 2 x + r^3 \sin 3 x + \cdots
| c =
}}
{{eqn | r = \dfrac {r \sin 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}^\infty r^k \sin k x
| r = \map \Im {\sum_{k \mathop = 1}^\infty r^k e^{i k x} }
| c =
}}
{{eqn | r = \map \Im {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}
| c = as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$
}}
{{eqn | r = \map \Im {\frac 1 {1 - r e^{i x} } }
| c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Im {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }
| c =
}}
{{eqn | r = \map \Im {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
| c =
}}
{{eqn | r = \map \Im {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + r^2} }
| c = Corollary to Euler's Formula
}}
{{eqn | r = \dfrac {r \sin x} {1 - 2 r \cos x + r^2}
| c = after simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22221
|
\section{Sum of Infinite Series of Product of nth Power of Cosine by nth Multiple of Cosine}
Tags: Cosine Function
\begin{theorem}
Let $0 < \theta < \dfrac \pi 2$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \cos^n \theta \, \map \cos {n + 1} \theta
| r = \cos \theta + \cos \theta \cos 2 \theta + \cos^2 \theta \cos 3 \theta + \cos^3 \theta \cos 4 \theta + \cdots
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Let $0 < \theta < \dfrac \pi 2$.
Then $0 < \cos \theta < 1$.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty r^k \cos k \theta
| r = \dfrac {1 - r \cos \theta} {1 - 2 r \cos \theta + r^2}
| c = Sum of Infinite Series of Product of Power and Cosine: $\size r < 1$
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^\infty \cos^k \theta \cos k \theta
| r = \dfrac {1 - \cos^2 \theta } {1 - 2 \cos^2 \theta + \cos^2 \theta}
| c = setting $r = \cos \theta$
}}
{{eqn | ll= \leadsto
| l = \cos^0 \theta \cos 0 \theta + \sum_{k \mathop = 1}^\infty \cos^k \theta \cos k \theta
| r = \dfrac {1 - \cos^2 \theta } {1 - \cos^2 \theta}
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = 1 + \sum_{k \mathop = 1}^\infty \cos^k \theta \cos k \theta
| r = \dfrac {1 - \cos^2 \theta } {1 - \cos^2 \theta}
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = 1 + \sum_{k \mathop = 1}^\infty \cos^k \theta \cos k \theta
| r = 1
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \cos \theta \sum_{k \mathop = 1}^\infty \cos^{k - 1} \theta \, \map \cos {n + 1} \theta
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 1}^\infty \cos^{k - 1} \theta \cos k \theta
| r = 0
| c = dividing by $\cos \theta$
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^\infty \cos^k \theta \cos k \theta
| r = 0
| c = Translation of Index Variable of Summation
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22222
|
\section{Sum of Infinite Series of Product of nth Power of cos 2 theta by 2n+1th Multiple of Sine}
Tags: Cosine Function
\begin{theorem}
Let $\theta \in \R$ such that $\theta \ne m \pi$ for any $m \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \paren {\cos 2 \theta}^n \sin \paren {2 n + 1} \theta
| r = \sin \theta + \cos 2 \theta \sin 3 \theta + \paren {\cos 2 \theta}^2 \sin 5 \theta + \paren {\cos 2 \theta}^3 \sin 7 \theta + \cdots
| c =
}}
{{eqn | r = \dfrac {\csc \theta} 2
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Let $\theta \ne \dfrac {m \pi} 2$ for any $m \in \Z$.
Then $\size {\cos 2 \theta} < 1$.
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \sin \paren {2 k + 1} \theta r^k
| r = \dfrac {\paren {1 + r} \sin \theta} {1 - 2 r \cos 2 \theta + r^2}
| c = Power Series of Sine of Odd Theta: $\size r < 1$
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^\infty \paren {\cos 2 \theta}^k \sin \paren {2 k + 1} \theta
| r = \dfrac {\paren {1 + \cos 2 \theta} \sin \theta} {1 - 2 \cos 2 \theta \cos 2 \theta + \paren {\cos 2 \theta}^2}
| c = setting $r = \cos 2 \theta$
}}
{{eqn | r = \dfrac {\paren {1 + \cos 2 \theta} \sin \theta} {1 - \paren {\cos 2 \theta}^2}
| c = simplifying
}}
{{eqn | r = \dfrac {\paren {1 + 2 \cos^2 \theta - 1} \sin \theta} {1 - \paren {2 \cos^2 \theta - 1}^2}
| c = Double Angle Formula for Cosine
}}
{{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {1 - \paren {4 \cos^4 \theta - 4 \cos^2 \theta + 1} }
| c = simplifying and expanding
}}
{{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {4 \cos^2 \theta \paren {1 - \cos^2 \theta} }
| c = simplifying
}}
{{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {4 \cos^2 \theta \sin^2 \theta}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \dfrac 1 {2 \sin \theta}
| c = simplifying
}}
{{eqn | r = \dfrac {\csc \theta} 2
| c = {{Defof|Real Cosecant Function}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22223
|
\section{Sum of Integer Ideals is Greatest Common Divisor}
Tags: Lowest Common Multiple, Integers, Greatest Common Divisor, Ideal Theory
\begin{theorem}
Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal d = \ideal m + \ideal n$.
Then $d = \gcd \set {m, n}$.
\end{theorem}
\begin{proof}
By Sum of Ideals is Ideal we have that $\ideal d = \ideal m + \ideal n$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal d$ are all necessarily principal ideals.
By Subrings of Integers are Sets of Integer Multiples we have that:
:$\ideal m = m \Z, \ideal n = n \Z$
Thus:
:$\ideal d = \ideal m + \ideal n = \set {x \in \Z: \exists a, b \in \Z: x = a m + b n}$
That is, $\ideal d$ is the set of all integer combinations of $m$ and $n$.
The result follows by Bézout's Lemma.
{{qed}}
\end{proof}
|
22224
|
\section{Sum of Integrals on Adjacent Intervals for Continuous Functions}
Tags: Integral Calculus
\begin{theorem}
Let $f$ be a real function which is continuous on any closed interval $I$.
Let $a, b, c \in I$.
Then:
:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$
\end{theorem}
\begin{proof}
By Continuous Real Function is Darboux Integrable, $f$ is integrable on $I$.
The result follows by application of Sum of Integrals on Adjacent Intervals for Integrable Functions.
{{qed}}
\end{proof}
|
22225
|
\section{Sum of Integrals on Complementary Sets}
Tags: Measure Theory
\begin{theorem}
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A, E \in \Sigma$ with $A \subseteq E$.
Let $f$ be a $\mu$-integrable function on $X$.
Then
:$\ds \int_E f \rd \mu = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu$
\end{theorem}
\begin{proof}
Let $\chi_E$ be the characteristic function of $E$.
Because $A$ and $E \setminus A$ are disjoint:
:$A \cap \paren {E \setminus A} = \O$
By Characteristic Function of Union:
:$\chi_E = \chi_A + \chi_{E \mathop \setminus A}$
Integrating $f$ over $E$ gives:
{{begin-eqn}}
{{eqn | l = \int_E f \rd \mu
| r = \int \chi_E \cdot f \rd \mu
| c = {{Defof|Integral of Integrable Function over Measurable Set}}
}}
{{eqn | r = \int \paren {\chi_A + \chi_{E \mathop \setminus A} } \cdot f \rd \mu
| c = by the above argument
}}
{{eqn | r = \int \paren {\chi_A \cdot f + \chi_{E \mathop \setminus A} \cdot f} \rd \mu
| c = Pointwise Operation on Distributive Structure is Distributive
}}
{{eqn | r = \int \chi_A \cdot f \rd \mu + \int \chi_{E \mathop \setminus A} \cdot f \rd \mu
| c = Integral of Integrable Function is Additive
}}
{{eqn | r = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu
| c = {{Defof|Integral of Integrable Function over Measurable Set}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22226
|
\section{Sum of Internal Angles of Polygon}
Tags: Polygons
\begin{theorem}
The sum $S$ of all internal angles of a polygon with $n$ sides is given by the formula $S = \paren {n - 2} 180 \degrees$.
\end{theorem}
\begin{proof}
For convex polygons, name a vertex as $A_1$, go clockwise and name the vertices as $A_2, A_3, \ldots, A_n$.
By joining $A_1$ to every vertex except $A_2$ and $A_n$, one can form $\paren {n - 2}$ triangles.
From Sum of Angles of Triangle equals Two Right Angles, the sum of the internal angles of a triangle is $180 \degrees$.
Therefore, the sum of internal angles of a polygon with $n$ sides is $\paren {n - 2} 180 \degrees$.
{{qed}}
{{proof wanted|concave polygons?}}
\end{proof}
|
22227
|
\section{Sum of Little-O Estimates/Sequences}
Tags: Asymptotic Notation
\begin{theorem}
Let $\sequence {a_n}, \sequence {b_n}, \sequence {c_n}, \sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map o {b_n}$
:$c_n = \map o {d_n}$
where $o$ denotes little-o notation.
Then:
:$a_n + c_n = \map o {\size {b_n} + \size {d_n} }$
\end{theorem}
\begin{proof}
Let $\epsilon > 0$.
Then by definition of little-o notation:
:$\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {a_n} \le \epsilon \cdot \size {b_n}}$
:$\exists n_2 \in \N: \paren {n \ge n_2 \implies \size {c_n} \le \epsilon \cdot \size {d_n}}$
For $n \ge \max \set {n_1, n_2}$:
{{begin-eqn}}
{{eqn | l = \size {a_n + c_n}
| o = \le
| r = \size {a_n} + \size {c_n}
| c = Triangle Inequality
}}
{{eqn | o = \le
| r = \epsilon \cdot \size {b_n} + \epsilon \cdot \size {d_n}
}}
{{eqn | r = \epsilon \cdot \size {\size {b_n} + \size {d_n} }
| c = Absolute value is positive
}}
{{end-eqn}}
Hence by definition of little-o notation:
:$a_n + c_n = \map o {\size {b_n} + \size {d_n} }$
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
22228
|
\section{Sum of Logarithms/Natural Logarithm}
Tags: Logarithms, Sum of Logarithms, Analysis
\begin{theorem}
Let $x, y \in \R$ be strictly positive real numbers.
Then:
:$\ln x + \ln y = \map \ln {x y}$
where $\ln$ denotes the natural logarithm.
\end{theorem}
\begin{proof}
Let $y \in \R_{>0}$ be fixed.
Consider the function:
:$f \left({x}\right) = \ln xy - \ln x$.
From the definition of the natural logarithm, the Fundamental Theorem of Calculus and the Chain Rule:
:$\displaystyle \forall x > 0: f^{\prime} \left({x}\right) = \frac 1 {xy} y - \frac 1 x = \frac 1 x - \frac 1 x = 0$.
Thus from Zero Derivative implies Constant Function, $f$ is constant:
: $\forall x > 0: \ln xy - \ln x = c$
To determine the value of $c$, put $x = 1$.
From Logarithm of 1 is 0:
: $\ln 1 = 0$
Thus:
: $c = \ln y - \ln 1 = \ln y$
and hence the result.
{{qed}}
\end{proof}
|
22229
|
\section{Sum of Maximum and Minimum}
Tags: Algebra
\begin{theorem}
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b = \max \set {a, b} + \min \set {a, b}$
\end{theorem}
\begin{proof}
From the definitions of max and min:
:$\max \set {a, b} = \begin{cases}
b: & a \le b \\
a: & b \le a
\end{cases}$
and
:$\min \set {a, b} = \begin{cases}
a: & a \le b \\
b: & b \le a
\end{cases}$
* Let $a < b$.
Then:
: $\max \set {a, b} + \min \set {a, b} = b + a$
* Let $a > b$.
Then:
: $\max \set {a, b} + \min \set {a, b} = a + b$
* Finally, let $a = b$.
Then:
: $\max \set {a, b} = \min \set {a, b} = a = b$
Hence:
: $\max \set {a, b} + \min \set {a, b} = 2a = 2b = a + b$
{{qed}}
Note that this result does not apply to $a, b \in \C$ as there is no concept of ordering on the complex numbers $\C$.
Category:Algebra
\end{proof}
|
22230
|
\section{Sum of Non-Consecutive Fibonacci Numbers}
Tags: Fibonacci Numbers
\begin{theorem}
Let $S$ be a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
Let the largest element of $S$ be $F_j$.
Then:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
That is, the sum of all the elements of $S$ is strictly less than the next largest Fibonacci number.
That is, given some increasing sequence $\sequence {c_i}$ satisfying $c_i \ge 2$ and $c_{i + 1} \ge c_i + 1$:
:$\ds F_{c_k + 1} > \sum_{i \mathop = 0}^k F_{c_i}$
\end{theorem}
\begin{proof}
The proof proceeds by induction on $j$ for $j \ge 2$.
For all $j \in \N_{>0}$, let $\map P j$ be the proposition:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
Let the term '''allowable set''' be used to mean a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
\end{proof}
|
22231
|
\section{Sum of Odd Index Binomial Coefficients}
Tags: Binomial Coefficients
\begin{theorem}
:$\ds \sum_{i \mathop \ge 0} \binom n {2 i + 1} = 2^{n - 1}$
where $\dbinom n i$ is a binomial coefficient.
That is:
:$\dbinom n 1 + \dbinom n 3 + \dbinom n 5 + \dotsb = 2^{n - 1}$
\end{theorem}
\begin{proof}
From Sum of Binomial Coefficients over Lower Index we have:
:$\ds \sum_{i \mathop \in \Z} \binom n i = 2^n$
That is:
:$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \cdots + \dbinom n n = 2^n$
as $\dbinom n i = 0$ for $i < 0$ and $i > n$.
This can be written more conveniently as:
:$(1): \quad \dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \dbinom n 4 + \cdots = 2^n$
Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have:
:$\ds \sum_{i \mathop \in \Z} \paren {-1}^i \binom n i = 0$
That is:
:$(2): \quad \dbinom n 0 - \dbinom n 1 + \dbinom n 2 - \dbinom n 3 + \dbinom n 4 - \cdots = 0$
Subtracting $(2)$ from $(1)$, we get:
:$2 \dbinom n 1 + 2 \dbinom n 3 + 2 \dbinom n 5 + \cdots = 2^n$
as the even index coefficients cancel out.
Dividing by $2$ throughout gives us the result.
{{qed}}
\end{proof}
|
22232
|
\section{Sum of Odd Positive Powers}
Tags: Number Theory
\begin{theorem}
Let $n \in \N$ be an odd positive integer.
Let $x, y \in \Z_{>0}$ be (strictly) positive integers.
Then $x + y$ is a divisor of $x^n + y^n$.
\end{theorem}
\begin{proof}
Given that $n \in \N$ be odd, it can be expressed in the form:
:$n = 2 m + 1$
where $m \in \N$.
The proof proceeds by strong induction.
For all $m \in \N$, let $\map P m$ be the proposition:
: $x^{2 m + 1} + y^{2 m + 1} = \paren {x + y} \paren {x^{2 m} + \cdots + y^{2 m} }$
$\map P 0$ is the case:
: $x + y = x + y$
which is trivially an identity.
\end{proof}
|
22233
|
\section{Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers}
Tags: Sums of Sequences, Proofs by Induction, Fibonacci Numbers
\begin{theorem}
Let $F_k$ be the $k$'th Fibonacci number.
Then:
:$\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$
\end{proof}
|
22234
|
\section{Sum of Pair of Elements of Geometric Sequence with Three Elements in Lowest Terms is Coprime to other Element}
Tags: Geometric Progressions, Geometric Sequences
\begin{theorem}
Let $P = \tuple {a, b, c}$ be a geometric sequence of integers in its lowest terms.
Then $\paren {a + b}$, $\paren {b + c}$ and $\paren {a + c}$ are all coprime to each of $a$, $b$ and $c$.
{{:Euclid:Proposition/IX/15}}
\end{theorem}
\begin{proof}
Let the common ratio of $P$ in canonical form be $\dfrac q p$.
By Form of Geometric Sequence of Integers in Lowest Terms:
:$P = \tuple {p^2, p q, q^2}$
Then:
{{begin-eqn}}
{{eqn | l = p
| o = \perp
| r = q
| c = {{Defof|Canonical Form of Rational Number}}
}}
{{eqn | ll= \leadsto
| l = q
| o = \perp
| r = \paren {p + q}
| c = Numbers are Coprime iff Sum is Coprime to Both
}}
{{eqn | ll= \leadsto
| l = q
| o = \perp
| r = p \paren {p + q}
| c = Integer Coprime to all Factors is Coprime to Whole
}}
{{eqn | ll= \leadsto
| l = q^2
| o = \perp
| r = p \paren {p + q}
| c = Square of Coprime Number is Coprime
}}
{{eqn | ll= \leadsto
| l = q^2
| o = \perp
| r = p^2 + p q
| c = factorising
}}
{{eqn | ll= \leadsto
| l = c
| o = \perp
| r = a + b
| c =
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | ll= \leadsto
| l = p^2
| o = \perp
| r = p q + q^2
| c =
}}
{{eqn | ll= \leadsto
| l = a
| o = \perp
| r = b + c
| c = factorising
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = p + q
| o = \perp
| r = p
| c = Numbers are Coprime iff Sum is Coprime to Both
}}
{{eqn | lo= \land
| l = p + q
| o = \perp
| r = q
| c = Numbers are Coprime iff Sum is Coprime to Both
}}
{{eqn | ll= \leadsto
| l = \paren {p + q}^2
| o = \perp
| r = p q
| c = as above
}}
{{eqn | ll= \leadsto
| l = p^2 + q^2 + 2 p q
| o = \perp
| r = p q
| c = factorising
}}
{{eqn | ll= \leadsto
| l = p^2 + q^2
| o = \perp
| r = p q
| c =
}}
{{eqn | ll= \leadsto
| l = b
| o = \perp
| r = a + c
| c =
}}
{{end-eqn}}
{{qed}}
{{Euclid Note|15|IX}}
\end{proof}
|
22235
|
\section{Sum of Pandigital Triplet of 3-Digit Primes}
Tags: Prime Numbers, 999
\begin{theorem}
The smallest integer which is the sum of a set of $3$ three-digit primes using all $9$ digits from $1$ to $9$ once each is $999$:
:$149 + 263 + 587 = 999$
\end{theorem}
\begin{proof}
All three-digit primes end in $1, 3, 7, 9$.
Suppose $1$ is used as the units digit of a prime.
Since the digit $1$ cannot be used again, the sum of the primes is at least:
:$221 + 333 + 447 = 1001$
so $1$ cannot be used as a units digit .
The units digits of the primes are $3, 7, 9$.
To minimise the sum, the hundreds digits must be $1, 2, 4$.
This leaves $5, 6, 8$ be the tens digits.
The primes satisfying these conditions are:
:$157, 163, 167$
:$257, 263, 269, 283$
:$457, 463, 467, 487$
Only $269$ contain $9$, so we must choose it.
Only $157$ does not contain $6$, so must choose it next.
But then all primes beginning with $4$ have some digit coinciding with the above primes.
Hence the next minimal sum of the primes (if they exist) is:
:$10^2 \paren {1 + 2 + 5} + 10 \paren {4 + 6 + 8} + \paren {3 + 7 + 9} = 999$
and we have shown that these primes do exist.
{{qed}}
\end{proof}
|
22236
|
\section{Sum of Positive and Negative Parts}
Tags: Mapping Theory
\begin{theorem}
Let $X$ be a set, and let $f: X \to \overline \R$ be an extended real-valued function.
Let $f^+, f^-: X \to \overline \R$ be the positive and negative parts of $f$, respectively.
Then $\size {f} = f^+ + f^-$, where $\size {f}$ is the absolute value of $f$.
\end{theorem}
\begin{proof}
Let $x \in X$.
Suppose that $\map f x \ge 0$, where $\ge$ signifies the extended real ordering.
Then $\size {\map f x} = \map f x$, and:
:$\map {f^+} x = \map \max {\map f x, 0} = \map f x$
:$\map {f^-} x = - \map \min {\map f x, 0} = 0$
Hence $\map {f^+} x + \map {f^-} x = \map f x = \size {\map f x}$.
Next, suppose that $\map f x < 0$, again in the extended real ordering.
Then $\size {\map f x} = - \map f x$, and:
:$\map {f^+} x = \map \max {\map f x, 0} = 0$
:$\map {f^-} x = - \map \min {\map f x, 0} = - \map f x$
Hence $\map {f^+} x + \map {f^-} x = - \map f x = \size {\map f x}$.
Thus, for all $x \in X$:
:$\map {f^+} x + \map {f^-} x = \size {\map f x}$
That is, $f^+ + f^- = \size {f}$.
{{qed}}
\end{proof}
|
22237
|
\section{Sum of Powers of Primitive Complex Roots of Unity}
Tags: Roots of Unity, Complex Roots of Unity
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha = \exp \paren {\dfrac {2 k \pi i} n}$ denote a primitive complex $n$th root of unity.
Let $s \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^{n - 1} \alpha^{j s}
| r = 1 + \alpha^s + \alpha^{2 s} + \cdots + \alpha^{\paren {n - 1} s}
| c =
}}
{{eqn | r = \begin {cases} n & : n \divides s \\ 0 & : n \nmid s \end {cases}
| c =
}}
{{end-eqn}}
where:
:$n \divides s$ denotes that $n$ is a divisor of $s$
:$n \nmid s$ denotes that $n$ is not a divisor of $s$.
\end{theorem}
\begin{proof}
First we address the case where $n \divides s$.
Then:
{{begin-eqn}}
{{eqn | l = s
| r = q n
| c = for some $q \in \Z_{>0}$
}}
{{eqn | ll= \leadsto
| l = \alpha^{j s}
| r = \alpha^{j q n}
| c =
}}
{{eqn | r = \paren {\alpha^n}^{j q}
| c =
}}
{{eqn | r = 1^{j q}
| c =
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^{n - 1} \alpha^{j s}
| r = \sum_{j \mathop = 0}^{n - 1} 1
| c =
}}
{{eqn | r = n
| c =
}}
{{end-eqn}}
Now let $n \nmid S$.
Then:
{{begin-eqn}}
{{eqn | l = s
| r = q n + r
| c = for $0 < r < 1$
| cc= Division Theorem
}}
{{eqn | ll= \leadsto
| l = \alpha^s
| r = \alpha^{q n + r}
| c =
}}
{{eqn | r = \paren {\alpha^n}^q \alpha^r
| c =
}}
{{eqn | r = 1^q \alpha^r
| c =
}}
{{eqn | r = \alpha^r
| c =
}}
{{eqn | o = \ne
| r = 1
| c =
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^{n - 1} \alpha^{j s}
| r = \sum_{j \mathop = 0}^{n - 1} \alpha^{j r}
| c =
}}
{{eqn | r = \dfrac {\alpha^{n r} - 1} {\alpha^r - 1}
| c = valid because $\alpha^r \ne 1$
| cc= Sum of Geometric Sequence
}}
{{eqn | r = \dfrac {\paren {\alpha^n}^r - 1} {\alpha^r - 1}
| c =
}}
{{eqn | r = \dfrac {1^r - 1} {\alpha^r - 1}
| c =
}}
{{eqn | r = \dfrac {1 - 1} {\alpha^r - 1}
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22238
|
\section{Sum of Quaternion Conjugates}
Tags: Quaternions
\begin{theorem}
Let $\mathbf x, \mathbf y \in \mathbb H$ be quaternions.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
Then:
:$\overline {\mathbf x + \mathbf y} = \overline {\mathbf x} + \overline {\mathbf y}$
\end{theorem}
\begin{proof}
Let:
:$\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k$
Then:
{{begin-eqn}}
{{eqn | l = \overline {\mathbf x + \mathbf y}
| r = \overline {\paren {a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k} + \paren {e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k} }
| c =
}}
{{eqn | r = \overline {\paren {a + e} \mathbf 1 + \paren {b + f} \mathbf i + \paren {c + g} \mathbf j + \paren {d + h} \mathbf k}
| c = {{Defof|Quaternion Addition}}
}}
{{eqn | r = \paren {a + e} \mathbf 1 - \paren {b + f} \mathbf i - \paren {c + g} \mathbf j - \paren {d + h} \mathbf k
| c = {{Defof|Quaternion Conjugate}}
}}
{{eqn | r = \paren {a \mathbf 1 - b \mathbf i - c \mathbf j - d \mathbf k} + \paren {e \mathbf 1 - f \mathbf i - g \mathbf j - h \mathbf k}
| c = {{Defof|Quaternion Addition}}
}}
{{eqn | r = \overline {\mathbf x} + \overline {\mathbf y}
| c = {{Defof|Quaternion Conjugate}}
}}
{{end-eqn}}
{{qed}}
Category:Quaternions
\end{proof}
|
22239
|
\section{Sum of Rational Cuts is Rational Cut}
Tags: Cuts, Addition
\begin{theorem}
Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* + q^* = \paren {p + q}^*$
Thus the operation of addition on the set of rational cuts is closed.
\end{theorem}
\begin{proof}
From Sum of Cuts is Cut, $p^* + q^*$ is a cut.
Let $r \in p^* + q^*$.
Then:
:$r = s + t$
where $s < p$ and $t < q$
Thus:
:$r < p + q$
and so:
:$r \in \paren {p + q}^*$
Hence:
:$p^* + q^* \subseteq \paren {p + q}^*$
{{qed|lemma}}
Let $r \in \paren {p + q}^*$.
Then:
:$r < p + q$
Let:
:$h = p + q - r$
:$s = p - \dfrac h 2$
:$t = q - \dfrac h 2$
Then:
:$s \in p^*$
:$t \in q^*$
Hence:
:$r = s + t$
and so:
:$r \in p^* + q^*$
So:
:$\paren {p + q}^* \subseteq p^* + q^*$
{{qed|lemma}}
Hence the result by definition of set equality.
{{Qed}}
\end{proof}
|
22240
|
\section{Sum of Reciprocals in Base 10 with Zeroes Removed}
Tags: Reciprocals, Series
\begin{theorem}
The infinite series
:$\ds \sum_{\map P n} \frac 1 n$
where $\map P n$ is the propositional function:
:$\forall n \in \Z_{>0}: \map P n \iff$ the decimal representation of $n$ contains no instances of the digit $0$
converges to the approximate limit $23 \cdotp 10345 \ldots$
\end{theorem}
\begin{proof}
For each $k \in \N$, there are $9^k$ $k$-digit numbers containing no instances of the digit $0$.
Each of these numbers is at least $10^{k - 1}$.
Hence the reciprocals of each of these numbers is at most $\dfrac 1 {10^{k - 1}}$.
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{\map P n} \frac 1 n
| o = <
| r = \sum_{k \mathop = 1}^\infty \frac {9^k} {10^{k - 1} }
}}
{{eqn | r = \frac 9 {1 - \frac 9 {10} }
| c = Sum of Geometric Sequence
}}
{{eqn | r = 90
}}
{{end-eqn}}
showing that the sum converges.
{{finish|Finer approximations can be obtained (e.g. in virtue of Closed Form for Triangular Numbers/Direct Proof), but due to the slow growth of the harmonic series, many numbers must to summed to obtain the approximation in the theorem. Case in point: $H_{5 \times 10^9} < 23$}}
\end{proof}
|
22241
|
\section{Sum of Reciprocals of Divisors equals Abundancy Index}
Tags: Number Theory, Abundancy
\begin{theorem}
Let $n$ be a positive integer.
Let $\map {\sigma_1} n$ denote the divisor sum function of $n$.
Then:
:$\ds \sum_{d \mathop \divides n} \frac 1 d = \frac {\map {\sigma_1} n} n$
where $\dfrac {\map {\sigma_1} n} n$ is the abundancy index of $n$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{d \mathop \divides n} \frac 1 d
| r = \sum_{d \mathop \divides n} \frac 1 {\paren {\frac n d} }
| c = Sum Over Divisors Equals Sum Over Quotients
}}
{{eqn | r = \frac 1 n \sum_{d \mathop \divides n} d
| c =
}}
{{eqn | r = \frac {\map {\sigma_1} n} n
| c = {{Defof|Divisor Sum Function}}
}}
{{end-eqn}}
{{qed}}
Category:Number Theory
Category:Abundancy
\end{proof}
|
22242
|
\section{Sum of Reciprocals of Divisors of Perfect Number is 2}
Tags: Perfect Numbers
\begin{theorem}
Let $n$ be a perfect number.
Then:
:$\ds \sum_{d \mathop \divides n} \dfrac 1 d = 2$
That is, the sum of the reciprocals of the divisors of $n$ is equal to $2$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{d \mathop \divides n} d
| r = \map {\sigma_1} n
| c = {{Defof|Divisor Sum Function}}
}}
{{eqn | ll= \leadsto
| l = \dfrac 1 n \sum_{d \mathop \divides n} d
| r = \dfrac {\map {\sigma_1} n} n
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{d \mathop \divides n} \frac d n
| r = \dfrac {\map {\sigma_1} n} n
| c =
}}
{{eqn | ll= \leadsto
| l = \sum_{d \mathop \divides n} \frac 1 d
| r = \dfrac {\map {\sigma_1} n} n
| c =
}}
{{end-eqn}}
The result follows by definition of perfect number:
{{:Definition:Perfect Number/Definition 4}}{{qed}}
\end{proof}
|
22243
|
\section{Sum of Reciprocals of One Plus and Minus Sine}
Tags: Trigonometric Identities, Sine Function
\begin{theorem}
:$\dfrac 1 {1 - \sin x} + \dfrac 1 {1 + \sin x} = 2 \sec^2 x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \frac 1 {1 - \sin x} + \frac 1 {1 + \sin x}
| r = \frac {1 + \sin x + 1 - \sin x} {1 - \sin^2 x}
| c = Difference of Two Squares
}}
{{eqn | r = \frac 2 {\cos^2 x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 2 \sec^2 x
| c = {{Defof|Secant Function}}
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
Category:Sine Function
\end{proof}
|
22244
|
\section{Sum of Reciprocals of Powers of Odd Integers Alternating in Sign}
Tags: Sums of Sequences, Hyperbolic Secant Function, Sum of Reciprocals of Powers of Odd Integers Alternating in Sign, Definite Integrals
\begin{theorem}
:$\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} = \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x$
where:
:$\map \Re s > 0$
:$\Gamma$ is the gamma function
:$\sech$ is the hyperbolic secant function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \int_0^\infty x^{s - 1} \map \sech x \rd x
| r = 2 \int_0^\infty \frac { x^{s - 1} } {e^x + e^{-x} } \rd x
| c = {{Defof|Hyperbolic Secant}}
}}
{{eqn | r = 2 \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {- e^{-2 x} } } \rd x
}}
{{eqn | r = 2 \int_0^\infty x^{s - 1} e^{-x} \sum_{n \mathop = 0}^\infty \paren {-1}^n e^{-2 n x} \rd x
| c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x
| c = Fubini's Theorem
}}
{{eqn | r = 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty \paren {\frac t {2 n + 1} }^{s - 1} e^{-t} \frac {\rd t} {2 n + 1}
| c = substituting $t = \paren {2 n + 1} x$
}}
{{eqn | r = 2 \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} \int_0^\infty t^{s - 1} e^{-t} \rd t
}}
{{eqn | r = 2 \map \Gamma s \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}
| c = {{Defof|Gamma Function}}
}}
{{end-eqn}}
So:
:$\ds \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}$
{{qed}}
Category:Hyperbolic Secant Function
Category:Definite Integrals
Category:Sums of Sequences
\end{proof}
|
22245
|
\section{Sum of Reciprocals of Primes is Divergent/Lemma}
Tags: Number Theory
\begin{theorem}
Let $C \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = + \infty$
\end{theorem}
\begin{proof}
Fix $c \in \R$.
It is sufficient to show there exists $N \in \N$, such that:
:$(1): \quad n \ge N \implies \map \ln {\ln n} - C > c$
Proceed as follows:
{{begin-eqn}}
{{eqn | l = \map \ln {\ln n} - C
| o = >
| r = c
}}
{{eqn | ll= \leadstoandfrom
| l = \ln n
| o = >
| r = \map \exp {c + C}
| c = {{Defof|Exponential Function/Real|subdef = Power Series Expansion|Exponential}}
}}
{{eqn | ll= \leadstoandfrom
| l = n
| o = >
| r = \map \exp {\map \exp {c + C} }
| c = {{Defof|Exponential Function/Real|subdef = Power Series Expansion|Exponential}}
}}
{{end-eqn}}
Let $N \in \N$ such that $N > \map \exp {\map \exp {c + C} }$.
By Logarithm is Strictly Increasing it follows that $N$ satisfies condition $(1)$.
Hence the result.
{{qed}}
Category:Number Theory
\end{proof}
|
22246
|
\section{Sum of Reciprocals of Primes is Divergent/Proof 2}
Tags: Number Theory, Sum of Reciprocals of Primes is Divergent, Analytic Number Theory
\begin{theorem}
The series:
:$\ds \sum_{p \mathop \in \Bbb P} \frac 1 p$
where:
:$\Bbb P$ is the set of all prime numbers
is divergent.
\end{theorem}
\begin{proof}
Let $n \in \N$ be a natural number.
Let $p_n$ denote the $n$th prime number.
Consider the product:
:$\ds \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
By Sum of Infinite Geometric Sequence:
{{begin-eqn}}
{{eqn | l = \frac 1 {1 - \frac 1 2}
| r = 1 + \frac 1 2 + \frac 1 {2^2} + \cdots
| c =
}}
{{eqn | l = \frac 1 {1 - \frac 1 3}
| r = 1 + \frac 1 3 + \frac 1 {3^2} + \cdots
| c =
}}
{{eqn | l = \frac 1 {1 - \frac 1 5}
| r = 1 + \frac 1 5 + \frac 1 {5^2} + \cdots
| c =
}}
{{eqn | o = \cdots
| c =
}}
{{eqn | l = \frac 1 {1 - \frac 1 {p_n} }
| r = 1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots
| c =
}}
{{end-eqn}}
Consider what happens when all these series are multiplied together.
A new series will be generated whose terms consist of all possible products of one term selected from each of the series on the {{RHS}}.
This new series will converge in any order to the product of the terms on the {{LHS}}.
By the Fundamental Theorem of Arithmetic, every integer greater than $1$ is uniquely expressible as a product of powers of different primes.
Hence the product of these series is the series of reciprocals of all (strictly) positive integers whose prime factors are no greater than $p_n$.
In particular, all (strictly) positive integers up to $p_n$ have this property.
So:
:$\ds \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}
| o = \ge
| r = \sum_{k \mathop = 1}^{p_n} \frac 1 k
| c =
}}
{{eqn | o = >
| r = \int_1^{p_n + 1} \dfrac {\d x} x
| c =
}}
{{eqn | r = \map \ln {p_n + 1}
| c =
}}
{{eqn | r = \ln p_n
| c =
}}
{{end-eqn}}
It follows by taking reciprocals that:
:$\ds \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$
Taking logarithms of each side:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$
Next, note that the line $y = 2 x$ in the cartesian plane lies below the curve $y = \map \ln {1 + x}$ on the interval $\closedint {-\frac 1 2} 0$.
Also note that all primes are greater than or equal to $2$.
Thus it follows that:
:$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$
Applying this to $(1)$ yields:
:$\ds -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$
and so:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$
But:
:$\ds \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$
and so the series:
:$\ds \sum_{p \mathop \in \Bbb P} \frac 1 p$
is divergent.
{{qed}}
\end{proof}
|
22247
|
\section{Sum of Reciprocals of Squares plus 1}
Tags: Hyperbolic Cotangent Function, Infinite Series, Formulas for Pi
\begin{theorem}
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 + 1} = \frac 1 2 \paren {\pi \coth \pi - 1}$
\end{theorem}
\begin{proof}
Let $\map f x$ be the real function defined on $\hointl {-\pi} \pi$ as:
:$\map f x = e^x$
From Fourier Series: $e^x$ over $\openint {-\pi} \pi$, we have:
:$(1): \quad \ds \map f x \sim \map S x = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} } }$
Let $x = \pi$.
By Fourier's Theorem, we have:
{{begin-eqn}}
{{eqn | l = \map S \pi
| r = \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}
| c =
}}
{{eqn | r = \dfrac {e^\pi + e^{-\pi} } 2
| c =
}}
{{eqn | r = \cosh \pi
| c = {{Defof|Hyperbolic Cosine}}
}}
{{end-eqn}}
Thus setting $x = \pi$ in $(1)$, we have:
{{begin-eqn}}
{{eqn | l = \cosh \pi
| r = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos \pi} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} } }
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {\pi \cosh \pi} {\sinh \pi}
| r = 1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\paren {-1}^n} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} }
| c = Cosine of Multiple of Pi and multiplying both sides by $\dfrac \pi {\sinh \pi}$
}}
{{eqn | ll= \leadsto
| l = \dfrac {\pi \coth \pi - 1} 2
| r = \sum_{n \mathop = 1}^\infty \frac 1 {1 + n^2}
| c = {{Defof|Hyperbolic Cotangent|index = 2}}, Sine of Multiple of Pi and simplifying
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22248
|
\section{Sum of Replicative Functions is Replicative}
Tags: Replicative Functions
\begin{theorem}
Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.
Let $f$ and $g$ both be replicative functions.
Then the pointwise sum of $f$ and $g$ is also a replicative function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map {\paren {f + g} } {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \paren {\map f {x + \frac k n} + \map g {x + \frac k n} }
| c = {{Defof|Pointwise Sum}}
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} + \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n}
| c =
}}
{{eqn | r = \map f {n x} + \map g {n x}
| c = {{Defof|Replicative Function}}
}}
{{eqn | r = \map {\paren {f + g} } {n x}
| c = {{Defof|Pointwise Sum}}
}}
{{end-eqn}}
Hence the result by definition of replicative function.
{{qed}}
\end{proof}
|
22249
|
\section{Sum of Ring Products is Subring of Commutative Ring}
Tags: Rings, Subrings, Commutative Rings
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {S, +, \circ}$ and $\struct {T, +, \circ}$ be subrings of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then $S T$ is a subring of $\struct {R, +, \circ}$.
\end{theorem}
\begin{proof}
From Sum of All Ring Products is Additive Subgroup we have that $\struct {S T, +}$ is an additive subgroup of $R$.
Let $x_1, x_2 \in S T$.
Then:
:$\ds x_1 = \sum_{i \mathop = 1}^m s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^n s_j \circ t_j$
for some $s_i, t_i, s_j, t_j, m, n$, etc.
Then:
{{begin-eqn}}
{{eqn | l = x_1 \circ x_2
| r = \paren {\sum_{i \mathop = 1}^m s_i \circ t_i} \circ \paren {\sum_{j \mathop = 1}^n s'_j \circ t'_j}
| c =
}}
{{eqn | r = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {s_i \circ s'_j} \circ \paren {t_i \circ t'_j}
| c = as $\circ$ is commutative
}}
{{end-eqn}}
So $x_1 \circ x_2 \in S T$ and the result follows from the Subring Test.
{{qed}}
\end{proof}
|
22250
|
\section{Sum of Roots of Polynomial}
Tags: Polynomial Equations, Sum of Roots of Polynomial
\begin{theorem}
Let $P$ be the polynomial equation:
: $a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = 0$
such that $a_n \ne 0$.
The sum of the roots of $P$ is $-\dfrac {a_{n - 1} } {a_n}$.
\end{theorem}
\begin{proof}
Let the roots of $P$ be $z_1, z_2, \ldots, z_n$.
Then $P$ can be written in factored form as:
:$\displaystyle a_n \prod_{k \mathop = 1}^n \left({z - z_k}\right) = a_0 \left({z - z_1}\right) \left({z - z_2}\right) \cdots \left({z - z_n}\right)$
Multiplying this out, $P$ can be expressed as:
:$a_n \left({z^n - \left({z_1 + z_2 + \cdots + z_n}\right) z^{n-1} + \cdots + \left({-1}\right)^n z_1 z_2 \cdots z_n}\right) = 0$
where the coefficients of $z^{n-2}, z^{n-3}, \ldots$ are more complicated and irrelevant.
Equating powers of $z$, it follows that:
:$-a_n \left({z_1 + z_2 + \cdots + z_n}\right) = a_{n-1}$
from which:
:$z_1 + z_2 + \cdots + z_n = - \dfrac {a_{n-1}} {a_n}$
{{qed}}
\end{proof}
|
22251
|
\section{Sum of Sequence of Alternating Positive and Negative Factorials being Prime}
Tags: Prime Numbers, Factorials
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let:
{{begin-eqn}}
{{eqn | l = m
| r = \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \paren {n - k}!
| c =
}}
{{eqn | r = n! - \paren {n - 1}! + \paren {n - 2}! - \paren {n - 3}! + \cdots \pm 1
| c =
}}
{{end-eqn}}
The sequence of $n$ such that $m$ is prime begins:
:$3, 4, 5, 6, 7, 8, 10, 15, 19, 41, 59, 61, 105, 160, \ldots$
{{OEIS|A001272}}
The sequence of those values of $m$ begins:
:$5, 19, 101, 619, 4421, 35 \, 899, 3 \, 301 \, 819, 1 \, 226 \, 280 \, 710 \, 981, \ldots$
{{OEIS|A071828}}
\end{theorem}
\begin{proof}
Let $\map f n$ be defined as:
:$\map f n := \ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \paren {n - k}!$
First we observe that for $n > 1$:
:$\map f n := n! - \map f {n - 1}$
We have:
{{begin-eqn}}
{{eqn | l = \map f 1
| r = 1!
| c =
}}
{{eqn | r = 1
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 2
| r = 2! - \map f 1
| c =
}}
{{eqn | r = 2 - 1
| c = Examples of Factorials
}}
{{eqn | r = 1
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 3
| r = 3! - \map f 2
| c =
}}
{{eqn | r = 6 - 1
| c = Examples of Factorials
}}
{{eqn | r = 5
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 4
| r = 4! - \map f 3
| c =
}}
{{eqn | r = 24 - 5
| c = Examples of Factorials
}}
{{eqn | r = 19
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 5
| r = 5! - \map f 4
| c =
}}
{{eqn | r = 120 - 19
| c = Examples of Factorials
}}
{{eqn | r = 101
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 6
| r = 6! - \map f 5
| c =
}}
{{eqn | r = 720 - 101
| c = Examples of Factorials
}}
{{eqn | r = 619
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 7
| r = 7! - \map f 6
| c =
}}
{{eqn | r = 5040 - 619
| c = Examples of Factorials
}}
{{eqn | r = 4421
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 8
| r = 8! - \map f 7
| c =
}}
{{eqn | r = 40 \, 320 - 421
| c = Examples of Factorials
}}
{{eqn | r = 35 \, 899
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f 9
| r = 9! - \map f 8
| c =
}}
{{eqn | r = 362 \, 880 - 35 \, 899
| c = Examples of Factorials
}}
{{eqn | r = 326 \, 981 = 79 \times 4139
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f {10}
| r = 10! - \map f 9
| c =
}}
{{eqn | r = 3 \, 628 \, 800 - 326 \, 981
| c = Examples of Factorials
}}
{{eqn | r = 3 \, 301 \, 819
| c = which is prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f {11}
| r = 11! - \map f {10}
| c =
}}
{{eqn | r = 39 \, 916 \, 800 - 3 \, 301 \, 819
| c = Factorial of $11$
}}
{{eqn | r = 36 \, 614 \, 981 = 13 \times 2 \, 816 \, 537
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f {12}
| r = 12! - \map f {11}
| c =
}}
{{eqn | r = 479 \, 001 \, 600 - 36 \, 614 \, 981
| c = Factorial of $12$
}}
{{eqn | r = 442 \, 386 \, 619 = 29 \times 15 \, 254 \, 711
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f {13}
| r = 13! - \map f {12}
| c =
}}
{{eqn | r = 6 \, 227 \, 020 \, 800 - 36 \, 614 \, 981
| c = Factorial of $13$
}}
{{eqn | r = 5 \, 784 \, 634 \, 181 = 47 \times 1427 \times 86 \, 249
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f {14}
| r = 14! - \map f {13}
| c =
}}
{{eqn | r = 87 \, 178 \, 291 \, 200 - 5 \, 784 \, 634 \, 181
| c = Factorial of $14$
}}
{{eqn | r = 81 \, 393 \, 657 \, 019 = 23 \times 73 \times 211 \times 229 \, 751
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map f {15}
| r = 15! - \map f {14}
| c =
}}
{{eqn | r = 1 \, 307 \, 674 \, 368 \, 000 - 81 \, 393 \, 657 \, 019
| c = Factorial of $15$
}}
{{eqn | r = 1 \, 226 \, 280 \, 710 \, 981
| c = which is prime
}}
{{end-eqn}}
From here on in the numbers become unwieldy.
\end{proof}
|
22252
|
\section{Sum of Sequence of Binomial Coefficients by Powers of 2}
Tags: Sum of Sequence of Binomial Coefficients by Powers of 2, Binomial Coefficients
\begin{theorem}
Let $n \in \Z$ be an integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n 2^j \binom n j
| r = \dbinom n 0 + 2 \dbinom n 1 + 2^2 \dbinom n 2 + \dotsb + 2^n \dbinom n n
| c =
}}
{{eqn | r = 3^n
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\displaystyle \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^0 2^j \binom n j
| r = \dbinom 0 0
| c =
}}
{{eqn | r = 1
| c =
}}
{{eqn | r = 3^0
| c =
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22253
|
\section{Sum of Sequence of Binomial Coefficients by Sum of Powers of Integers}
Tags: Sums of Sequences
\begin{theorem}
Let $n, k \in \Z_{\ge 0}$ be positive integers.
Let $S_k = \ds \sum_{i \mathop = 1}^n i^k$.
Then:
:$\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$
\end{theorem}
\begin{proof}
Let $N$ be a positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \paren {N + 1}^{k + 1} - N^{k + 1}
| r = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i N^i - N^{k + 1}
| c = Binomial Theorem: Integral Index
}}
{{eqn | r = \binom {k + 1} 0 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i + \binom {k + 1} {k + 1} N^{k + 1} - N^{k + 1}
}}
{{eqn | r = 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i
| c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
Summing from $N = 1$ to $N = n$, we have on the {{LHS}}:
{{begin-eqn}}
{{eqn | l = \sum_{N \mathop = 1}^n \paren {\paren {N + 1}^{k + 1} - N^{k + 1} }
| r = \paren {n + 1}^{k + 1} - 1^{k + 1}
| c = Telescoping Series: Example 1
}}
{{eqn | r = \paren {n + 1}^{k + 1} - 1
}}
{{end-eqn}}
So, we have:
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^{k + 1} - 1
| r = \sum_{N \mathop = 1}^n \paren {1 + \sum_{i \mathop = 1}^k \binom {k + 1} i N^i}
}}
{{eqn | r = \sum_{N \mathop = 1}^n 1 + \sum_{i \mathop = 1}^k \binom {k + 1} i \sum_{N \mathop = 1}^n N^i
}}
{{eqn | r = n + \sum_{i \mathop = 1}^k \binom {k + 1} i S_i
}}
{{end-eqn}}
giving:
:$\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$
{{qed}}
\end{proof}
|
22254
|
\section{Sum of Sequence of Cubes}
Tags: Number Theory, Sum of Sequence of Cubes, Proofs by Induction, Sums of Sequences, Cube Numbers
\begin{theorem}
:$\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$
\end{theorem}
\begin{proof}
* First, from Closed Form for Triangular Numbers, we have that <math>\sum_{i=1}^n i = \frac {n \left({n+1}\right)} 2</math>.
So <math>\left({\sum_{i=1}^n i}\right)^2 = \frac{n^2(n+1)^2}{4}</math>.
* Next we use induction on <math>n\,</math> to show that <math>\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}</math>.
The base case holds since <math>1^3=\frac{1 (1+1)^2}{4}</math>.
Now we need to show that if it holds for <math>n\,</math>, it holds for <math>n+1\,</math>.
{{begin-equation}}
{{equation | l=<math>\sum_{i=1}^{n+1} i^3</math>
| r=<math>\sum_{i=1}^n i^3+(n+1)^3</math>
| c=
}}
{{equation | r=<math>\frac{n^2(n+1)^2}{4}+(n+1)^3</math>
| c=(by the induction hypothesis)
}}
{{equation | r=<math>\frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4}</math>
| c=
}}
{{equation | r=<math>\frac{n^4+6n^3+13n^2+12n+4}{4}</math>
| c=
}}
{{equation | r=<math>\frac{(n+1)^2(n+2)^2}{4}</math>
| c=
}}
{{end-equation}}
By the Principle of Mathematical Induction, the proof is complete.
{{qed}}
Category:Sums of Sequences
Category:Proofs by Induction
23552
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2010-01-01T10:42:16Z
Prime.mover
59
23552
wikitext
text/x-wiki
\end{proof}
|
22255
|
\section{Sum of Sequence of Even Index Fibonacci Numbers}
Tags: Sums of Sequences, Proofs by Induction, Fibonacci Numbers
\begin{theorem}
Let $F_k$ be the $k$th Fibonacci number.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \ge 1
| l = \sum_{j \mathop = 1}^n F_{2 j}
| r = F_2 + F_4 + F_6 + \cdots + F_{2 n}
| c =
}}
{{eqn | r = F_{2 n + 1} - 1
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n F_{2 j} = F_{2 n + 1} - 1$
\end{proof}
|
22256
|
\section{Sum of Sequence of Even Squares}
Tags: Sums of Sequences, Square Numbers
\begin{theorem}
:$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \paren {2 i}^2 = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren {2 i}^2
| r = 4 \sum_{i \mathop = 1}^n i^2
| c = adjustment of indices: $4 i^2 = 0$ when $i = 0$
}}
{{eqn | r = 4 \frac {n \paren {n + 1} \paren {2 n + 1} } 6
| c = Sum of Sequence of Squares
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3
| c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Square Numbers
Category:Sums of Sequences
\end{proof}
|
22257
|
\section{Sum of Sequence of Factorials}
Tags: Factorials, Sums of Sequences
\begin{theorem}
The sequence $S = \sequence {s_n}$ defined as:
:$\ds s_n = \sum_{k \mathop = 1}^n k!$
begins:
:$1, 3, 9, 33, 153, 873, 5913, 46 \, 233, 409 \, 113, 4 \, 037 \, 913, \ldots$
{{OEIS|A007489}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = s_1
| r = 1!
| c =
}}
{{eqn | r = 1
| c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_2
| r = s_1 + 2!
| c =
}}
{{eqn | r = 1 + 2
| c = {{Defof|Factorial}}
}}
{{eqn | r = 3
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_3
| r = s_2 + 3!
| c =
}}
{{eqn | r = 3 + 6
| c = {{Defof|Factorial}}
}}
{{eqn | r = 9
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_4
| r = s_3 + 4!
| c =
}}
{{eqn | r = 9 + 24
| c = {{Defof|Factorial}}
}}
{{eqn | r = 33
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_5
| r = s_4 + 5!
| c =
}}
{{eqn | r = 33 + 120
| c = {{Defof|Factorial}}
}}
{{eqn | r = 153
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_6
| r = s_5 + 6!
| c =
}}
{{eqn | r = 153 + 720
| c = {{Defof|Factorial}}
}}
{{eqn | r = 873
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_7
| r = s_6 + 7!
| c =
}}
{{eqn | r = 873 + 5040
| c = {{Defof|Factorial}}
}}
{{eqn | r = 5913
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_8
| r = s_7 + 8!
| c =
}}
{{eqn | r = 5913 + 40 \, 320
| c = {{Defof|Factorial}}
}}
{{eqn | r = 46 \, 223
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_9
| r = s_8 + 9!
| c =
}}
{{eqn | r = 46 \, 223 + 362 \, 880
| c = {{Defof|Factorial}}
}}
{{eqn | r = 409 \, 113
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = s_{10}
| r = s_9 + 10!
| c =
}}
{{eqn | r = 409 \, 113 + 3 \, 628 \, 800
| c = {{Defof|Factorial}}
}}
{{eqn | r = 4 \, 037 \, 913
| c =
}}
{{end-eqn}}
{{qed}}
Category:Factorials
Category:Sums of Sequences
\end{proof}
|
22258
|
\section{Sum of Sequence of Fibonacci Numbers}
Tags: Sums of Sequences, Proofs by Induction, Fibonacci Numbers
\begin{theorem}
Let $F_n$ denote the $n$th Fibonacci number.
Then:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
| c =
}}
{{eqn | r = 1 - 1
| c =
}}
{{eqn | r = F_2 - 1
| c =
}}
{{end-eqn}}
which is seen to hold.
\end{proof}
|
22259
|
\section{Sum of Sequence of Fifth Powers}
Tags: Fifth Powers, Triangular Numbers, Sums of Sequences
\begin{theorem}
:$\ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$
where $T_n$ denotes the $n$th triangular number.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$
\end{proof}
|
22260
|
\section{Sum of Sequence of Fourth Powers}
Tags: Fourth Powers, Sum of Sequence of Fourth Powers, Sums of Sequences
\begin{theorem}
:$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$
\end{theorem}
\begin{proof}
By definition of binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom k 4
| r = \frac {k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right)} {4 !}
| c =
}}
{{eqn | l = \binom k 3
| r = \frac {k \left({k - 1}\right) \left({k - 2}\right)} {3 !}
| c =
}}
{{eqn | l = \binom k 2
| r = \frac {k \left({k - 1}\right)} {2 !}
| c =
}}
{{eqn | l = \binom k 1
| r = k
| c =
}}
{{eqn | ll= \leadsto
| l = 24 \binom k 4
| r = k^4 - 6 k^3 + 11 k^2 - 6 k
| c =
}}
{{eqn | l = 36 \binom k 3
| r = 6 k^3 - 18 k^2 + 12 k
| c =
}}
{{eqn | l = 14 \binom k 2
| r = 7 k^2 - 7 k
| c =
}}
{{eqn | ll= \leadsto
| l = k^4
| r = 24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1
| c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n k^4
| r = \sum_{k \mathop = 0}^n \left({24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1}\right)
| c =
}}
{{eqn | r = 24 \binom {n + 1} 5 + 36 \binom {n + 1} 4 + 14 \binom {n + 1} 3 + \binom {n + 1} 2
| c = Sum of Binomial Coefficients over Upper Index
}}
{{eqn | r = 24 \frac {\left({n + 1}\right) n \left({n - 1}\right) \left({n - 2}\right) \left({n - 3}\right)} {5 !}
| c = Definition of Binomial Coefficient
}}
{{eqn | o =
| ro= +
| r = 36 \frac {\left({n + 1}\right) n \left({n - 1}\right) \left({n - 2}\right)} {4 !}
| c =
}}
{{eqn | o =
| ro= +
| r = 14 \frac {\left({n + 1}\right) n \left({n - 1}\right)} {3 !}
| c =
}}
{{eqn | o =
| ro= +
| r = \frac {\left({n + 1}\right) n } {2 !}
| c =
}}
{{eqn | r = \left({n + 1}\right) n \left({\frac {\left({n - 1}\right) \left({n - 2}\right) \left({n - 3}\right)} 5 + \frac {3 \left({n - 1}\right) \left({n - 2}\right)} 2 + \frac {7 \left({n - 1}\right)} 3 + \frac 1 2}\right)
| c =
}}
{{eqn | r = \frac {\left({n + 1}\right) n} {30} \left({6 \left({n^3 - 6 n^2 + 11 n - 6 }\right) + 45 \left({n^2 - 3 n + 2}\right) + 70 \left({n - 1}\right) + 15}\right)
| c =
}}
{{eqn | r = \frac {\left({n + 1}\right) n} {30} \left({6 n^3 - 36 n^2 + 66 n - 36 + 45 n^2 - 135 n + 90 + 70 n - 70 + 15}\right)
| c =
}}
{{eqn | r = \frac {\left({n + 1}\right) n} {30} \left({6 n^3 + 9 n^2 + n - 1}\right)
| c =
}}
{{eqn | r = \frac {\left({n + 1}\right) n} {30} \left({2 n + 1}\right) \left({3 n^2 + 3 n - 1}\right)
| c =
}}
{{eqn | r = \frac {\left({n + 1}\right) n \left({n + \frac 1 2}\right) \left({3 n^2 + 3 n - 1}\right)} {15}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22261
|
\section{Sum of Sequence of Odd Cubes}
Tags: Cube Numbers, Proofs by Induction, Sums of Sequences
\begin{theorem}
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$
\end{proof}
|
22262
|
\section{Sum of Sequence of Odd Index Fibonacci Numbers}
Tags: Sums of Sequences, Proofs by Induction, Fibonacci Numbers
\begin{theorem}
Let $F_k$ be the $k$th Fibonacci number.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \ge 1
| l = \sum_{j \mathop = 1}^n F_{2 j - 1}
| r = F_1 + F_3 + F_5 + \cdots + F_{2 n - 1}
| c =
}}
{{eqn | r = F_{2 n}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n F_{2 j - 1} = F_{2 n}$
\end{proof}
|
22263
|
\section{Sum of Sequence of Odd Squares/Formulation 1}
Tags: Sums of Sequences, Sum of Sequence of Odd Squares, Square Numbers
\begin{theorem}
:$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \paren {2 i + 1}^2 = \frac {\paren {n + 1} \paren {2 n + 1} \paren {2 n + 3} } 3$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren {2 i + 1}^2
| r = \sum_{i \mathop = 0}^n \paren {2 i}^2 + \sum_{i \mathop = 0}^n 4 i + \sum_{i \mathop = 0}^n 1
| c =
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \sum_{i \mathop = 0}^n i + \sum_{i \mathop = 0}^n 1
| c = Sum of Sequence of Even Squares
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 0}^n 1
| c = adjustment of indices: $4 i = 0$ when $i = 0$
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \frac {n \paren {n + 1} } 2 + \sum_{i \mathop = 0}^n 1
| c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 2 n \paren {n + 1} + \paren {n + 1}
| c = further simplification
}}
{{eqn | r = \frac {\paren {n + 1} \paren {2 n \paren {2 n + 1} + 6 n + 3} } 3
| c = factorising
}}
{{eqn | r = \frac {\paren {n + 1} \paren {4 n^2 + 8 n + 3} } 3
| c = multiplying out
}}
{{eqn | r = \frac {\paren {n + 1} \paren {2 n + 1} \paren {2 n + 3} } 3
| c = factorising
}}
{{end-eqn}}
{{qed}}
Category:Sum of Sequence of Odd Squares
\end{proof}
|
22264
|
\section{Sum of Sequence of Odd Squares/Formulation 2}
Tags: Sum of Sequence of Odd Squares
\begin{theorem}
:$\ds \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$
\end{proof}
|
22265
|
\section{Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient}
Tags: Proofs by Induction, Binomial Coefficients, Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient, Fibonacci Numbers
\begin{theorem}
Let $F_n$ denote the $n$th Fibonacci number.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
| l = F_{2 n}
| r = \sum_{k \mathop = 1}^n \dbinom n k F_k
| c =
}}
{{eqn | r = \dbinom n 1 F_1 + \dbinom n 2 F_2 + \dbinom n 3 F_3 + \dotsb + \dbinom n {n - 1} F_{n - 1} + \dbinom n n F_n
| c =
}}
{{end-eqn}}
where $\dbinom n k$ 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:
:$F_{2 n} = \ds \sum_{k \mathop = 1}^n \dbinom n k F_k$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^0 \dbinom 0 k F_k
| c = vacuously
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22266
|
\section{Sum of Sequence of Product of Lucas Numbers with Powers of 2}
Tags: Lucas Numbers, Fibonacci Numbers
\begin{theorem}
Let $L_k$ be the $k$th Lucas number.
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\ds \forall n \in \N_{>0}: \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$
That is:
:$2^0 L_1 + 2^1 L_2 + 2^2 L_3 + \cdots + 2^{n - 1} L^n = 2^n F_{n + 1} - 1$
\end{theorem}
\begin{proof}
Proof by induction:
For all $\forall n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$
\end{proof}
|
22267
|
\section{Sum of Sequence of Products of 3 Consecutive Reciprocals}
Tags: Sums of Sequences, Sum of Sequence of Products of 3 Consecutive Reciprocals, Reciprocals
\begin{theorem}
:$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }$
\end{theorem}
\begin{proof}
We observe that:
{{begin-eqn}}
{{eqn | l = \frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2}
| r = \frac {\paren {j + 1} \paren {j + 2} - 2 j \paren {j + 2} + j \paren {j + 1} } {j \paren {j + 1} \paren {j + 2} }
}}
{{eqn | r = \frac {j^2 + 3 j + 2 - 2 j^2 - 4 j + j^2 + j} {j \paren {j + 1} \paren {j + 2} }
}}
{{eqn | r = \frac 2 {j \paren {j + 1} \paren {j + 2} }
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} }
| r = \frac 1 2 \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2} }
}}
{{eqn | r = \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 1}^n \frac 2 {j + 1} + \sum_{j \mathop = 1}^n \frac 1 {j + 2} }
}}
{{eqn | r = \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 2}^{n + 1} \frac 2 j + \sum_{j \mathop = 3}^{n + 2} \frac 1 j}
| c = Translation of Index Variable of Summation
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 1 + \frac 1 2 + \sum_{j \mathop = 3}^n \frac 1 j - \frac 2 2 - \frac 2 {n + 1} - \sum_{j \mathop = 3}^n \frac 2 j + \frac 1 {n + 1} + \frac 1 {n + 2} + \sum_{j \mathop = 3}^n \frac 1 j}
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 1 + \frac 1 2 - \frac 2 2 - \frac 2 {n + 1} + \frac 1 {n + 1} + \frac 1 {n + 2} }
}}
{{eqn | r = \frac 1 2 \paren {\frac 1 2 - \frac 1 {n + 1} + \frac 1 {n + 2} }
}}
{{eqn | r = \frac {\paren {n + 1} \paren {n + 2} - 2 \paren {n + 2} + 2 \paren {n + 1} } {4 \paren {n + 1} \paren {n + 2} }
}}
{{eqn | r = \frac {n^2 + 3 n + 2 - 2 n - 4 + 2 n + 2} {4 \paren {n + 1} \paren {n + 2} }
}}
{{eqn | r = \frac {n^2 + 3 n} {4 \paren {n + 1} \paren {n + 2} }
}}
{{eqn | r = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }
}}
{{end-eqn}}
{{qed}}
Category:Sums of Sequences
Category:Reciprocals
Category:Sum of Sequence of Products of 3 Consecutive Reciprocals
\end{proof}
|
22268
|
\section{Sum of Sequence of Products of 3 Consecutive Reciprocals/Corollary}
Tags: Sum of Sequence of Products of 3 Consecutive Reciprocals, Limit of Series, Limits of Series, Reciprocals
\begin{theorem}
:$\ds \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac 1 4$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} }
| r = \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} }
}}
{{eqn | r = \lim_{n \mathop \to \infty} \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }
| c = Sum of Sequence of Products of 3 Consecutive Reciprocals
}}
{{eqn | r = \lim_{n \mathop \to \infty} \frac {1 + \frac 3 n} {4 \paren {1 + \frac 1 n} \paren {1 + \frac 2 n} }
| c = dividing top and bottom by $n^2$
}}
{{eqn | r = \frac 1 4
| c = Basic Null Sequences
}}
{{end-eqn}}
{{qed}}
Category:Limits of Series
Category:Reciprocals
Category:Sum of Sequence of Products of 3 Consecutive Reciprocals
\end{proof}
|
22269
|
\section{Sum of Sequence of Products of Consecutive Reciprocals}
Tags: Sums of Sequences, Sum of Sequence of Products of Consecutive Reciprocals, Proofs by Induction, Reciprocals
\begin{theorem}
:$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
: $\displaystyle \forall n \ge 1: \sum_{j = 1}^n \frac 1 {j \left({j+1}\right)} = \frac n {n+1}$
\end{proof}
|
22270
|
\section{Sum of Sequence of Seventh Powers}
Tags: Sums of Sequences, Seventh Powers
\begin{theorem}
:$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^0 j^7
| r = 0
| c =
}}
{{eqn | r = \frac {0^2 \paren {0 + 1}^2 \paren {3 \times 0^4 + 6 \times 0^3 - 0^2 - 4 \times 0 + 2} } {24}
| c =
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22271
|
\section{Sum of Sequence of Squares}
Tags: Sum of Sequence of Squares, Number Theory, Proofs by Induction, Induction, Sequences, Sums of Sequences, Algebra, Square Numbers
\begin{theorem}
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
: $\displaystyle \sum_{i=1}^n i^2 = \frac{n (n+1)(2n+1)} 6$
\end{proof}
|
22272
|
\section{Sum of Sequence of Squares of Fibonacci Numbers}
Tags: Sums of Sequences, Proofs by Induction, Fibonacci Numbers
\begin{theorem}
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
That is:
:${F_1}^2 + {F_2}^2 + {F_3}^2 + \cdots + {F_n}^2 = F_n F_{n + 1}$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
\end{proof}
|
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