nyuuzyou/wikis huggingface dataset:
Summation Formula for Reciprocal of Binomial Coefficient

== Theorem ==
as long as the denominators are not zero.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $48$
Category:Binomial Coefficients
Category:Summation Formula for Reciprocal of Binomial Coefficient
Summation Formula for Reciprocal of Binomial Coefficient/Proof 1

== Theorem ==
== Proof ==
First note that:
The proof continues by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P m$ is true, for all $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop \ge 0} \binom m k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {m + x} m}$
from which it is to be shown that:
:$\ds \sum_{k \mathop \ge 0} \binom {m + 1} k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {m + 1 + x} {m + 1} }$
=== Induction Step ===
This is the induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $48$
Category:Summation Formula for Reciprocal of Binomial Coefficient
Summation Formula for Reciprocal of Binomial Coefficient/Proof 2

== Theorem ==
== Proof ==
Consider the value of $\map \Beta {x, n + 1}$, where $\Beta$ is the beta function.
We have:
which equates to the of the identity.
We also have:
which equates to the of the identity.
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $48$
Category:Summation Formula for Reciprocal of Binomial Coefficient
Summation Formula over Half-Integers

== Theorem == 
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane $\C$ with vertices $N \paren {\pm 1 \pm i}$.
Let $f$ be a meromorphic function on $\C$ with finitely many poles. 
Suppose that:
:$\ds \int_{C_N} \paren {\pi \tan \pi z} \map f z \rd z \to 0$
as $N \to \infty$. 
Let $X$ be the set of poles of $f$.
Let $Y$ be the set of poles of $\map f {\dfrac {2 z + 1} 2}$.
Then: 
:$\ds \sum_{n \mathop \in \Z \setminus Y} \map f {\frac {2 n + 1} 2} = \sum_{z_0 \mathop \in X} \Res {\pi \tan \paren {\pi z} \map f z} {z_0}$
If $Y \cap \Z = \O$, this becomes: 
:$\ds \sum_{n \mathop = -\infty}^\infty \map f {\frac {2 n + 1} 2} = \sum_{z_0 \mathop \in X} \Res {\pi \tan \paren {\pi z} \map f z} {z_0}$
 
== Proof == 
== Sources == 
* : $7.8$: Summation of Series
Category:Complex Analysis
Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r

== Theorem ==
:$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } } = 1$
where $1 \le m \le n$ and $x$ is arbitrary.
=== Example ===
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $34$
Category:Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r
Category:Summations
Category:Continued Products
Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r/Example

== Theorem ==
:$\dfrac {x \paren {x - 2} \paren {x - 3} } {\paren {-1} \paren {-2} \paren {-3} } + \dfrac {\paren {x + 1} \paren {x - 1} \paren {x - 2} } {\paren 1 \paren {-1} \paren {-2} } + \dfrac {\paren {x + 2} x \paren {x - 1} } {\paren 2 \paren 1 \paren {-1} } + \dfrac {\paren {x + 3} \paren {x + 1} x} {\paren 3 \paren 2 \paren 1 } = 1$
== Proof ==
This is an example of Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r:
:$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } } = 1$
where $n = 4$ and $m = 2$.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $34$
Category:Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r
Summation from k to m of 2k-1 Choose k by 2n-2k Choose n-k by -1 over 2k-1

== Theorem ==
:$\ds \sum_{k \mathop = 0}^m \binom {2 k - 1} k \binom {2 n - 2 k} {n - k} \dfrac {-1} {2 k - 1} = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m} + \dfrac 1 2 \dbinom {2 n} n$
== Proof ==
From Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Set $r - \dfrac 1 2$ and $s = -\dfrac 1 2$.
:$\ds \sum_{k \mathop = 0}^m \dbinom {1/2} k \dbinom {-1/2} {n - k} \paren {\frac n 2 - \paren {\frac 1 2 - \frac 1 2} k} = \paren {m + 1} \paren {n - m} \dbinom {1/2} {m + 1} \dbinom {-1/2} {n - m}$
Take the :
Now the :
Thus we have:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $53 \ \text{(b)}$
Category:Binomial Coefficients
Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k

== Theorem ==
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
== Proof ==
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop = 0}^j \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {j + 1} \paren {n - j} \dbinom r {j + 1} \dbinom s {n - j}$
from which it is to be shown that:
:$\ds \sum_{k \mathop = 0}^{j + 1} \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {j + 2} \paren {n - j - 1} \dbinom r {j + 2} \dbinom s {n - j - 1}$
=== Induction Step ===
This is the induction step:
So $\map P j \implies \map P {j + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $53 \ \text{(a)}$
Category:Binomial Coefficients
Summation is Linear/Scaling of Summations

== Theorem ==
Let $\tuple {x_1, \ldots, x_n}$ and $\tuple {y_1, \ldots, y_n}$ be finite sequences of numbers of equal length.
Let $\lambda$ be a number.
Then:
:$\ds \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
== Proof ==
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
=== Basis for the Induction ===
$\map P 1$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \lambda \sum_{i \mathop = 1}^k x_i = \sum_{i \mathop = 1}^k \lambda x_i$
from which it is to be shown that:
:$\ds \lambda \sum_{i \mathop = 1}^{k + 1} x_i = \sum_{i \mathop = 1}^{k + 1} \lambda x_i$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{> 0}: \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$
Category:Numbers
Category:Proofs by Induction
Summation is Linear/Sum of Summations

== Theorem ==
Let $\tuple {x_1, \ldots, x_n}$ and $\tuple {y_1, \ldots, y_n}$ be finite sequences of numbers of equal length.
Let $\lambda$ be a number.
Then:
:$\ds \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$
== Proof ==
The proof proceeds by mathematical induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$
=== Basis for the Induction ===
$\map P 1$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{i \mathop = 1}^k x_i + \sum_{i \mathop = 1}^k y_i = \sum_{i \mathop = 1}^k \paren {x_i + y_i}$
from which it is to be shown that:
:$\ds \sum_{i \mathop = 1}^{k + 1} x_i + \sum_{i \mathop = 1}^{k + 1} y_i = \sum_{i \mathop = 1}^{k + 1} \paren {x_i + y_i}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{> 0}: \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$
Category:Numbers
Category:Proofs by Induction
Summation of General Logarithms

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \prod_{\map R i} a_i$ denote a product over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$
== Proof ==
The proof proceeds by induction.
First let:
:$S := \set {a_i: \map R i}$
We have that $S$ is finite.
Hence the contents of $S$ can be well-ordered, by Finite Totally Ordered Set is Well-Ordered.
Let $S$ have $m$ elements, identified as:
:$S = \set {s_1, s_2, \ldots, s_m}$
For all $n \in \Z_{\ge 0}$ such that $n \le m$, let $\map P n$ be the proposition:
:$\ds \map {\log_b} {\prod_{i \mathop = 1}^n s_i} = \sum_{i \mathop = 1}^n \log_b s_i$
$\map P 0$ is the case:
=== Basis for the Induction ===
$\map P 1$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \map {\log_b} {\prod_{i \mathop = 1}^k s_i} = \sum_{i \mathop = 1}^k \log_b s_i$
from which it is to be shown that:
:$\ds \map {\log_b} {\prod_{i \mathop = 1}^{k + 1} s_i} = \sum_{i \mathop = 1}^{k + 1} \log_b s_i$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \map {\log_b} {\prod_{i \mathop = 1}^n s_i} = \sum_{i \mathop = 1}^n \log_b s_i$
Hence, by definition of $S$:
:$\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $24$
Category:Summations
Category:Continued Products
Category:Logarithms
Summation of Multiple of Mapping on Finite Set

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $\lambda \in \mathbb A$.
Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.
Then we have the equality of summations on finite sets:
:$\ds \sum_{s \mathop \in S} \map g s = \lambda \cdot \sum_{s \mathop \in S} \map f s$
== Outline of Proof ==
Using the definition of summation on a finite set, we reduce this to Indexed Summation of Multiple of Mapping.
== Proof ==
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following equality of indexed summations:
:$\ds \sum_{i \mathop = 0}^{n - 1} \map g {\map \sigma i} = \lambda \cdot \sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i}$
By Multiple of Mapping Composed with Mapping, $g \circ \sigma = \lambda \cdot \paren {f \circ \sigma}$.
The above equality now follows from Indexed Summation of Multiple of Mapping.
== Also see ==
* Indexed Summation of Multiple of Mapping
* General Distributivity Theorem
Category:Summations
Summation of Odd Reciprocals in terms of Harmonic Numbers

== Theorem ==
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n} - \dfrac {H_n} 2$
where $H_n$ denotes the $n$th harmonic number.
== Proof ==
== Historical Note ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $16$
Category:Harmonic Numbers
Summation of Odd Reciprocals in terms of Harmonic Numbers/Historical Note

== Historical Note on Summation of Odd Reciprocals in terms of Harmonic Numbers ==
 originally published the following result in ($1997$) as a solution to the exercise to evaluate $\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n - 1} - \dfrac {H_{n - 1} } 2$
The online errata sheet, found via https://www-cs-faculty.stanford.edu/~knuth/taocp.html, offers the simpler result as a replacement.
== Proof ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $16$ (Solution)
Category:Harmonic Numbers
Summation of Power Series by Harmonic Sequence

== Theorem ==
Consider the power series:
:$\map f x = \ds \sum_{k \mathop \ge 0} a_k x^k$
Let $\map f x$ converge for $x = x_0$.
Then:
:$\ds \sum_{k \mathop \ge 0} a_k {x_0}^k H_k = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$
where $H_n$ denotes the $n$th harmonic number.
== Proof ==
== Sources ==
* 
* 
* : $\S 1.2.7$: Harmonic Numbers: Exercise $20$
Category:Harmonic Numbers
Summation of Powers over Product of Differences

== Theorem ==
:$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
=== Example ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Historical Note ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $33$
Category:Summation of Powers over Product of Differences
Category:Summations
Category:Continued Products
Summation of Powers over Product of Differences/Example

== Theorem ==
== Proof ==
Instance of Summation of Powers over Product of Differences:
:$\ds \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin {cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end {cases}$
where:
:$n = 3$
:$x_1 = a$
:$x_2 = b$
:$x_3 = c$
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $33$
Category:Summation of Powers over Product of Differences
Summation of Powers over Product of Differences/Historical Note

== Historical Note on Summation of Powers over Product of Differences ==
The result Summation of Powers over Product of Differences was discussed by in a letter to in $1762$.
He subsequently published it in his of $1769$.
The proof involving complex analysis was devised in $1857$ by .
== Sources ==
* 
* : $\S 1.2.3$: Sums and Products: Exercise $33$
Category:Historical Notes
Summation of Powers over Product of Differences/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds S_n := \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
=== Edge Cases ===
$\map P 0$ is a vacuous summation:
:$\ds S_0 := \sum_{j \mathop = 1}^0 \paren {\dfrac { {x_j}^0} {\ds \prod_{\substack {1 \mathop \le k \mathop \le 0 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = 0 = \sum_{j \mathop = 1}^0 x_0$
which is seen to hold.
$\map P 1$ is the case:
This is also seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case where $n = 2$:
When $0 \le r < n - 1$, it must be that $r = 0$:
When $r = n - 1 = 1$:
When $r = n = 2$:
Thus in all cases, $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{j \mathop = 1}^m \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le m \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < m - 1 \\ 1 & : r = m - 1 \\ \ds \sum_{j \mathop = 1}^m x_j & : r = m \end{cases}$
from which it is to be shown that:
:$\ds \sum_{j \mathop = 1}^{m + 1} \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le m + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < m \\ 1 & : r = m \\ \ds \sum_{j \mathop = 1}^{m + 1} x_j & : r = {m + 1} \end{cases}$
=== Induction Step ===
This is the induction step:
For $n > 2$, let the formula be rewritten:
So:
When $r < n$, both parts are equal to $0$ or $1$ by the induction hypothesis.
Thus either:
:$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {1 - 1} = 0$
or:
:$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {0 - 0} = 0$
and so $\map P {m + 1}$ holds for $r < n$.
When $r = n$:
Thus:
:$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {x_{n + 1} - x_n} = 1$
and so $\map P {m + 1}$ holds for $r = n$.
Now:
Thus $\map P {m + 1}$ holds for $r = n + 1$.
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
== Historical Note ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $33$
Category:Summation of Powers over Product of Differences
Summation of Powers over Product of Differences/Proof 2

== Theorem ==
== Proof ==
By Cauchy's Residue Theorem:
:$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \dfrac 1 {2 \pi i} \int \limits_{\size z \mathop = R} \dfrac {z^r \rd z} {\paren {z - z_1} \cdots \paren {z - z_n} }$
where $R > \size {z_1}, \ldots, \size {z_n}$.
The Laurent series of the integrand converges uniformly on $\size z = R$:
On integrating term my term, everything vanishes except the coefficient of $z^{-1}$.
Thus:
:$\ds \sum_{\substack {j_1 \mathop + \mathop \cdots \mathop + j_n \mathop = r \mathop - n \mathop + 1 \\ j_1, \mathop \ldots j_n \mathop \ge 0} } {x_1}^{j_1} {x_2}^{j_2} \cdots {x_n}^{j_n}$
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.3$: Sums and Products: Exercise $33$
Category:Summation of Powers over Product of Differences
Summation of Powers over Product of Differences/Proof 3

== Theorem ==
== Proof ==
=== Definition ===
Symbol $\map {\mathbf {Cof } } {M, i, j}$ denotes cofactor $M_{i j}$ of matrix $M$
=== Lemma 1 ===
Proof of Lemma 1
By Effect of Elementary Row Operations on Determinant, the determinant of $A$ is unchanged by adding a linear combination of the first $n - 1$ columns to the last column.
Let $\map f x$ be the monic polynomial $\map {p_m} x = x^{n - 1} + $ lower power terms.
Then:
The last column on the right has all components zero except $m$th entry $\map {p_m} {x_m}$.
Apply cofactor expansion along column $n$.
=== Lemma 2 ===
 
 
Proof of Lemma 2:
Value of Vandermonde Determinant establishes $(1)$.
To prove (2), let:
:$\set {y_1, \ldots, y_{n - 1} } = \set {x_1, \ldots, x_n} \setminus \set {x_m}$
then apply $(1)$ with $\set {x_1, \ldots, x_n}$ replaced by $\set {y_1, \ldots, y_{n - 1} }$.
Theorem Details:
The Lemmas change the Theorem's equation to:
Cofactor expansion changes identity $(3)$ to:
The proof is divided into three cases.
Case 1: $0 \mathop \le r \mathop \le n - 2$:
The determinant on the left in $(4)$ is zero by Square Matrix with Duplicate Rows has Zero Determinant.
Case 2: $r = n - 1$:
The determinant on the left in $(4)$ is $\map \det A$.
Case 3: $r = n$:
Expand $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$ by Taylor's Theorem with remainder $R$ of degree $n - 2$:
:$\map p x = x^n - \paren {x_1 + \cdots + x_n} x^{n - 1} + \map R x$
Let $x = x_k$ for $1 \mathop \le k \mathop \le n$.
Then $x$ is a root of $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$:
Let:
Then:
The left side of $(4)$:
== Also see ==
* Sum of Elements in Inverse of Cauchy Matrix
* Vandermonde Matrix Identity for Cauchy Matrix
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $33$
* : Matrix Algebra Ch2 and Determinants Ch3
Category:Summation of Powers over Product of Differences
Summation of Product of Differences

== Theorem ==
:$\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {u_j - u_k} \paren {v_j - v_k} = n \sum_{j \mathop = 1}^n u_j v_j - \sum_{j \mathop = 1}^n u_j \sum_{j \mathop = 1}^n v_j$
== Proof ==
Take the Binet-Cauchy Identity:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
Make the following assignments:
Then we have:
:$\ds \paren {\sum_{i \mathop = 1}^n u_i v_i} \paren {\sum_{j \mathop = 1}^n 1 \times 1} = \paren {\sum_{i \mathop = 1}^n u_i \times 1} \paren {\sum_{j \mathop = 1}^n 1 \times v_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {u_i \times 1 - u_j \times 1} \paren {v_i \times 1 - v_j \times 1}$
and the result follows.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $31$
Category:Summations