nyuuzyou/wikis huggingface dataset:
Generating Function by Power of Parameter

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $m \in \Z_{\ge 0}$ be a non-negative integer.
Then $z^m \map G z$ is the generating function for the sequence $\sequence {a_{n - m} }$.
== Proof ==
By letting $a_n = 0$ for all $n < 0$:
:$z^m \map G z = \ds \sum_{n \mathop \ge 0} a_{n - m} z^n$
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(3)$
Category:Generating Functions
Generating Function for Associated Legendre Function of the First Kind

== Theorem ==
Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.
Then the generating function for ${P_n}^m$ is:
:$\ds \frac {\paren {2 m}! \paren {1 - x^2}^{m / 2} t^m} {2^m m! \paren {1 - 2 t x + t^2}^{m + 1/2} } = \sum_{n \mathop = m}^\infty \map { {P_n}^m} x t^n$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Generating Function for $\map { {P_n}^m} x$: $26.11$
* : $\S 28$: Legendre and Associated Legendre Functions: Generating Function for $\map { {P_n}^m} x$: $28.59.$
Category:Associated Legendre Functions
Category:Examples of Generating Functions
Generating Function for Bernoulli Polynomials

== Theorem ==
Let $\map {B_n} x$ denote the $n$th Bernoulli polynomial.
Then the generating function for $B_n$ is:
:$\ds \frac {t e^{t x} } {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} x} {k!} t^k$
== Proof ==
By definition of the generating function for Bernoulli numbers:
:$\ds \frac t {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k$
By Power Series Expansion for Exponential Function:
:$\ds e^{t x} = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k$
Thus:
:$\ds \frac {t e^{t x} } {e^t - 1} = \paren {\sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k} \paren {\sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k}$
Combining like powers of $t$ we obtain: 
Category:Bernoulli Polynomials
Category:Examples of Generating Functions
Generating Function for Bessel Function of the First Kind of Order n of x

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
:$\map \exp {\dfrac {x \paren {t - \frac 1 t} } 2} = \ds \sum_{n \mathop = -\infty}^\infty \map {J_n} x t^n$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions: $4$
Category:Bessel Functions
Generating Function for Binomial Coefficients

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N: a_n = \begin{cases}
\dbinom m n & : n = 0, 1, 2, \ldots, m \\
0 & : \text{otherwise}\end{cases}$
where $\dbinom m n$ denotes a binomial coefficient.
Then the generating function for $\sequence {a_n}$ is given as:
:$\ds \map G z = \sum_{n \mathop = 0}^m \dbinom m n z^n = \paren {1 + z}^m$
== Proof ==
The result follows from the definition of a generating function.
== Sources ==
* : $\S 4.1$: Generating functions: Example $3$
Category:Examples of Generating Functions
Generating Function for Boubaker Polynomials

== Theorem ==
The Boubaker polynomials, defined as: 
:$\map {B_n} x = \begin{cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end{cases}$
have as an ordinary generating function:
:$\ds \map {f_{B_n, \operatorname {ORD} } } {x, t} = \sum_{n \mathop = 0}^{\infty} \map {B_n} x t^n = \frac {1 + 3 t^2} {1 + t \paren {t - x} }$
== Proof ==
Category:Boubaker Polynomials
Category:Examples of Generating Functions
Generating Function for Constant Sequence

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N: a_n = r$
for some $r \in \R$.
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac r {1 - z}$ for $\size z < 1$
== Proof ==
for $\size z < 1$.
== Examples ==
== Sources ==
* : $\text {3-4}$ Generating Functions: Example $\text {3-6}$
* : $\S 1.2.9$: Generating Functions: $(5)$
Category:Generating Function for Constant Sequence
Category:Examples of Generating Functions
Generating Function for Constant Sequence/Examples/a0=1, an=2

== Example of Generating Function for Constant Sequence ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \Z_{\ge 0}: a_n = \begin{cases} 1 & : n = 0 \\ 2 & : n > 0 \end{cases}$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac {1 + z} {1 - z}$
== Proof ==
Let $\map {H_1} z$ be the generating function for $\sequence {r_n}$ where:
:$r_n = 2$
Then from Generating Function for Constant Sequence:
:$\map H z = \dfrac 2 {1 - z}$
Then:
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $8$
Category:Generating Function for Sequence of Powers of Constant
Generating Function for Elementary Symmetric Function

== Theorem ==
Let $U$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the elementary symmetric function of degree $m$ on $U$:
Let $a_m := \map {e_m} U$ for $m = 0, 1, 2, \ldots$
Let $\map G z$ be a generating function for the sequence $\sequence {a_m}$:
:$\ds \map G z = \sum_{m \mathop = 0}^\infty a_m z^m$
Then:
:$\ds \map G z = \prod_{k \mathop = 1}^n \paren {1 + x_k z}$
=== Outline ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $10$
Category:Generating Function for Elementary Symmetric Function
Category:Elementary Symmetric Functions
Category:Examples of Generating Functions
Category:Summations
Generating Function for Elementary Symmetric Function/Outline

== Theorem ==
== Outline ==
Generating function discovery methods can find a formula for $\map G z$.
Let $n = 1$.
Then $U$ is a singleton:
:$U = \set {x_1}$.
Expand the formal series:
Product of Generating Functions and experience with elementary symmetric functions suggests:
:$\map G z = \paren {1 + x_1 z} \paren {1 + x_2 z} \cdots \paren {1 + x_n z}$
Category:Generating Function for Elementary Symmetric Function
Generating Function for Elementary Symmetric Function/Proof 1

== Theorem ==
== Proof ==
The summation for $\map G z$ is a finite sum $m = 0, 1, \ldots, n$, which settles convergence issues.
Begin with Viète's Formulas:
:$\ds \prod_{k \mathop = a}^b \paren {x - x_k} = x^n + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, x^m$
Change variables $x = -1 / z$:
Category:Generating Function for Elementary Symmetric Function
Generating Function for Elementary Symmetric Function/Proof 2

== Theorem ==
== Proof ==
Apply mathematical induction on $n$.
Let $\map P n$ be the statement:
Basis for the induction:
Set $U = \set {x_1}$ for $n = 1$.
Expand the formal series:
Then $\map P 1$ holds.
Induction step:
Assume $\map P n$ holds.
Let's prove $\map P {n + 1}$ holds.
The induction step uses a recursion relation:
Let $\map G z$ be defined by statement $\map P n$.
Let $\map {G^*} z$ be defined by statement $\map P {n + 1}$.
Then:
Then $\map P {n + 1}$ holds, completing the induction.
Category:Generating Function for Elementary Symmetric Function
Generating Function for Elementary Symmetric Function/Proof 3

== Theorem ==
== Proof ==
We have by definition of generating function that:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
We have that:
:$a_0 = 1$
Suppose $n = 1$.
Let $\map G z$ be the generating function for $\sequence {a_m}$ under this condition.
Then:
:$1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le 1$
can be fulfilled by only one set $\set {j_1, j_2, \ldots, j_m}$, that is:
:$j_1 = 1$
Thus in this case:
Then by Product of Generating Functions, it follows that:
:$\map G z = \paren {1 + x_1 z} \paren {1 + x_2 z} \cdots \paren {1 + x_n z}$
Category:Generating Function for Elementary Symmetric Function
Generating Function for Even Terms of Sequence

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Consider the subsequence $\sequence {b_n} := \tuple {a_0, a_2, a_4, \ldots}$
Then the generating function for $\sequence {b_n}$ is:
:$\dfrac 1 2 \paren {\map G z + \map G {-z} }$
== Proof ==
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(12)$
Category:Generating Functions
Generating Function for Fibonacci Numbers

== Theorem ==
Let $\map G z$ be the function defined as:
:$\map G z = \dfrac z {1 - z - z^2}$
Then $\map G z$ is a generating function for the Fibonacci numbers.
== Proof ==
Let the form of $\map G z$ be assumed as:
where $F_n$ denotes the $n$th Fibonacci number.
Then:
and so:
Hence the result:
:$\map G z = \dfrac z {1 - z - z^2}$
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $5$
* : $\S 1.2.8$: Fibonacci Numbers: $(11)$
Category:Fibonacci Numbers
Category:Examples of Generating Functions
Generating Function for Legendre Polynomials

== Theorem ==
Let $\map {P_n} x$ denote the $n$th Legendre polynomial.
Then the generating function for $P_n$ is:
:$\ds \frac 1 {\sqrt {1 - 2 x t + t^2} } = \sum_{n \mathop = 0}^\infty \map {P_n} x t^n$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Generating Function for Legendre Polynomials: $25.19$
* 
* 
* 
* 
* : $\S 28$: Legendre and Associated Legendre Functions: Generating Function for Legendre Polynomials: $28.19.$
Category:Legendre Polynomials
Category:Examples of Generating Functions
Generating Function for Linearly Recurrent Sequence

== Theorem ==
Let $\sequence {a_n}$ be a linearly recurrent sequence defined as:
:$a_n = \begin{cases} b_n & : 1 \le n \le m \\ c_1 a_{n - 1} + c_2 a_{n - 2} + \cdots + c_m a_{n - m} & : n > m \end{cases}$
where:
:$m \in \Z_{>0}$ is a (strictly) positive integer
:$b_1, \ldots, b_m$ are constants.
Then the generating function for $\sequence {a_n}$ is of the form:
:$\map G z = \dfrac {\map P z} {1 - c_1 z - c_2 z^2 - \cdots - c_m z^m}$
where $\map P z$ is a polynomial in $z$ given by $b_1 z + b_2 z^2 + \cdots + b_m z^m$.
== Proof ==
== Example ==
By this theorem, the generating function for $\sequence {F_n}$ is given by:
:$\map G z = \dfrac 1 {1 - z - z^2}$
== Sources ==
* : $\S 1.2.9$: Generating Functions
Category:Generating Functions
Generating Function for Lucas Numbers

== Theorem ==
Let $\map G z$ be the function defined as:
:$\map G z = \dfrac {2 - z} {1 - z - z^2}$
Then $\map G z$ is a generating function for the Lucas numbers.
== Proof ==
Let the form of $\map G z$ be assumed as:
where $L_n$ denotes the $n$th Lucas number.
Then:
and so:
Hence the result:
:$\map G z = \dfrac {2 - z} {1 - z - z^2}$
Category:Lucas Numbers
Category:Examples of Generating Functions
Generating Function for Natural Numbers

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = n - 1$
That is:
:$\sequence {a_n} = 0, 1, 2, 3, 4, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$G \paren z = \dfrac 1 {\paren {1 - z}^2}$
=== Corollary ===
== Proof ==
Take the sequence:
:$S_n = 1, 1, 1, \ldots$
From Generating Function for Constant Sequence, this has the generating function:
:$\ds G \paren z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$
By Derivative of Generating Function:
:$\ds \dfrac \d {\d z} G \paren z = 0 + 1 + 2 z + 3 z^2 \cdots = \sum_{n \mathop \ge 0} \paren {n + 1} z^n$
which is the generating function for the sequence $\sequence {a_n}$.
But:
:$G \paren z = \dfrac 1 {1 - z}$
and so by Power Rule for Derivatives and the Chain Rule for Derivatives:
:$\dfrac \d {\d z} G \paren z = \dfrac 1 {\paren {1 - z}^2}$
The result follows from the definition of a generating function.
== Sources ==
* : $\S 4.1$: Generating functions: $(2)$
* : $\S 1.2.9$: Generating Functions: $(16)$
Category:Examples of Generating Functions
Category:Natural Numbers
Generating Function for Natural Numbers/Corollary

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = n - 1$
That is:
:$\sequence {a_n} = 1, 2, 3, 4, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$H \paren z = \dfrac z {\paren {1 - z}^2}$
== Proof ==
From Generating Function for Natural Numbers:
:$\sequence {a_n} = 0, 1, 2, 3, 4, \ldots$
has the generating function:
:$G \paren z = \dfrac 1 {\paren {1 - z}^2}$
Then by Generating Function by Power of Parameter:
:$z G \paren z = \dfrac z {\paren {1 - z}^2}$
is the generating function for the sequence defined as:
:$\forall n \in \N_{> 0}: a_{n - 1} = n - 1$
that is:
:$0, a_1, a_2, \ldots$
where:
:$a_1 = 1, a_2 = 2, \ldots$
That is:
:$\sequence {a_n} = 1, 2, 3, 4, \ldots$
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $1$
Category:Examples of Generating Functions
Category:Natural Numbers
Generating Function for Odd Terms of Sequence

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Consider the subsequence $\sequence {b_n} := \tuple {a_1, a_3, a_5, \ldots}$
Then the generating function for $\sequence {b_n}$ is:
:$\dfrac 1 2 \paren {\map G z - \map G {-z} }$
== Proof ==
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(12)$
Category:Generating Functions
Generating Function for Powers of Two

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N: a_n = 2^n$
That is:
:$\sequence {a_n} = 1, 2, 4, 8, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - 2 z}$ for $\size z < \dfrac 1 2$
== Proof ==
This is valid for:
:$\size {2 z} < 1$
from which:
:$\size z < \dfrac 1 2$
follows directly by division by $2$.
The result follows from the definition of a generating function.
== Sources ==
* : $\S 4.1$: Generating functions
Category:Examples of Generating Functions
Generating Function for Sequence of Harmonic Numbers

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = H_n$
where $H_n$ denotes the $n$th harmonic number.
That is:
:$\sequence {a_n} = 1, 1 + \dfrac 1 2, 1 + \dfrac 1 2 + \dfrac 1 3, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} }$
== Proof ==
Take the sequence:
:$S_n = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$
From Generating Function for Sequence of Reciprocals of Natural Numbers, this has the generating function:
:$\map H z = \map \ln {\dfrac 1 {1 - z} }$
By definition, $\sequence {a_n}$ is the sequence of partial sums of $\sequence {a_n}$.
The result follows from Generating Function for Sequence of Partial Sums of Series.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(18)$
Category:Harmonic Numbers
Category:Examples of Generating Functions
Generating Function for Sequence of Partial Sums of Series

== Theorem ==
Let $s$ be the the series:
:$\ds s = \sum_{n \mathop = 1}^\infty a_n = a_0 + a_1 + a_2 + a_3 + \cdots$
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $\sequence {c_n}$ denote the sequence of partial sums of $s$.
Then the generating function for $\sequence {c_n}$ is given by:
:$\ds \dfrac 1 {1 - z} \map G z = \sum_{n \mathop \ge 0} c_n z^n$
== Proof ==
By definition of sequence of partial sums of $s$:
Consider the sequence $\sequence {b_n}$ defined as:
:$\forall n \in \Z_{\ge 0}: b_n = 1$
Let $\map H z$ be the generating function for $\sequence {b_n}$.
By Generating Function for Constant Sequence:
:$\map H z = \dfrac 1 {1 - z}$
Then:
Hence the result by definition of generating function.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(7)$
Category:Generating Functions
Category:Series
Generating Function for Sequence of Powers of Constant

== Theorem ==
Let $c \in \R$ be a constant.
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \Z_{\ge 0}: a_n = c^n$
That is:
:$\sequence {a_n} = 1, c, c^2, c^3, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - c z}$
== Proof ==
Consider the sequence $\sequence {b_n}$ defined as:
: $\forall n \in \Z_{\ge 0}: b_n = 1$
Let $\map H z$ be the generating function for $\sequence {b_n}$.
Then:
The result follows from the definition of a generating function.
== Examples ==
== Sources ==
* : $\S 1.2.9$: Generating Functions
Category:Generating Function for Sequence of Powers of Constant
Category:Examples of Generating Functions
Generating Function for Sequence of Powers of Constant/Examples

== Examples of Generating Function for Sequence of Powers of Constant ==
=== Example: $\sequence {2^n}$ ===
=== Example: $\sequence {2^n + 3^n}$ ===
=== Example: $\sequence {\paren {b + 1}^n - b^n}$ ===
Category:Generating Function for Sequence of Powers of Constant
Generating Function for Sequence of Powers of Constant/Examples/(b+1)^n - b^n

== Example of Generating Function for Sequence of Powers of Constant ==
Let $b \in \R_{>0}$ be a positive real number.
Let $\sequence {a_n}$ be the sequence defined as:
: $\forall n \in \Z_{\ge 0}: a_n = \paren {b + 1}^n - b^n$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac z {\paren {1 - b z} \paren {1 - b z - z} }$
== Proof ==
Let $\map {H_1} z$ be the generating function for $\sequence {r_n}$ where:
:$r_n = b^n$
Then from Generating Function for Sequence of Powers of Constant:
:$\map {H_1} z = \dfrac 1 {1 - b z}$
Let $\map {H_2} z$ be the generating function for $\sequence {s_n}$ where:
:$s_n = \paren {b + 1}^n$
Then again from Generating Function for Sequence of Powers of Constant:
From Linear Combination of Generating Functions:
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $7$
Category:Generating Function for Sequence of Powers of Constant
Generating Function for Sequence of Powers of Constant/Examples/2^n

== Example of Generating Function for Sequence of Powers of Constant ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \Z_{\ge 0}: a_n = 2^n$
That is:
:$\sequence {a_n} = 1, 2, 4, 8, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - 2 z}$
== Proof ==
A specific instance of Generating Function for Sequence of Powers of Constant:
:$\map G z = \dfrac 1 {1 - 2 z}$
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $6$
Category:Generating Function for Sequence of Powers of Constant
Generating Function for Sequence of Powers of Constant/Examples/2^n + 3^n

== Example of Generating Function for Sequence of Powers of Constant ==
Let $\sequence {a_n}$ be the sequence defined as:
: $\forall n \in \Z_{\ge 0}: a_n = 2^n + 3^n$
That is:
:$\sequence {a_n} = 2, 5, 13, 35, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - 2 z} + \dfrac 1 {1 - 3 z}$
== Proof ==
Let $\map {G_2} z$ be the generating function for $\sequence {2^n}$.
Let $\map {G_3} z$ be the generating function for $\sequence {3^n}$.
From Generating Function for Sequence of Powers of Constant:
:$\map {G_2} z = \dfrac 1 {1 - 2 z}$
:$\map {G_3} z = \dfrac 1 {1 - 3 z}$
From Linear Combination of Generating Functions:
:$\map G z = \map {G_2} z + \map {G_3} z$
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $1$
Category:Generating Function for Sequence of Powers of Constant
Generating Function for Sequence of Reciprocals of Natural Numbers

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = \dfrac 1 n$
That is:
:$\sequence {a_n} = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \map \ln {\dfrac 1 {1 - z} }$
== Proof ==
Take the sequence:
:$S_n = 1, 1, 1, \ldots$
From Generating Function for Constant Sequence, this has the generating function:
:$\ds \map G z = \sum_{n \mathop = 0}^\infty z^n = \frac 1 {1 - z}$
By Integral of Generating Function:
which is the power series whose coefficients are $\sequence {a_n}$.
But:
:$\map G z = \dfrac 1 {1 - z}$
and so by Primitive of Reciprocal and the Integration by Substitution:
The result follows from the definition of generating function.
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $2$
* : $\S 1.2.9$: Generating Functions: $(17)$
Category:Reciprocals
Category:Examples of Generating Functions
Generating Function for Sequence of Sum over k to n of Reciprocal of k by n-k

== Theorem ==
Let $\sequence {a_n}$ be the sequence whose terms are defined as:
:$\forall n \in \Z_{\ge 0}: a_n = \ds \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k \paren {n - k} }$
Then $\sequence {a_n}$ has the generating function $\map G z$ such that:
:$\map G z = \paren {\ln \dfrac 1 {1 - z} }^2$
and whose terms are:
:$a_n = \dfrac {2 H_{n - 1} } n$
== Proof ==
From Product of Generating Functions:
:$\map G z = \paren {\map {G_1} z}^2$
where $\map {G_1} z$ is the generating function for $\ds \sum_{k \mathop \ge 1} \dfrac 1 k$.
From Generating Function for Sequence of Reciprocals of Natural Numbers:
:$\map {G_1} z = \map \ln {\dfrac 1 {1 - z} }$
Hence:
:$\map G z = \paren {\ln \dfrac 1 {1 - z} }^2$
Differentiating $\map G z$ $z$ gives:
Integrating again $z$ gives:
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $6$
Category:Examples of Generating Functions
Generating Function for Triangular Numbers

== Theorem ==
Let $T_n$ denote the $n$th triangular number.
Then the generating function for $\sequence {T_n}$ is given as:
:$\ds \map G z = \frac z {\paren {1 - z}^3}$
=== Corollary ===
== Proof ==
Category:Triangular Numbers
Category:Examples of Generating Functions
Generating Function for Triangular Numbers/Corollary

== Theorem ==
Let $\sequence {b_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: b_n = \dfrac {\paren {n + 1} \paren {n + 2} } 2$
That is:
:$\sequence {b_n}_{n \mathop \ge 0} = 1, 3, 6, 10, \ldots, \dbinom {n + 2} 2, \ldots$
Then the generating function for $\sequence {b_n}$ is given as:
:$H \paren z = \dfrac 1 {\paren {1 - z}^3}$
== Proof ==
From Generating Function for Triangular Numbers:
:$H \paren z = \dfrac z {\paren {1 - z}^3}$
is the generating function for $\sequence {a_n}$ as given by:
:$\sequence {a_n} = 1, 3, 6, 10, \ldots, \dbinom {n + 1} 2, \ldots$
that is, such that:
:$a_1 = 1, a_2 = 3, a_3 = 6, \ldots$
The result follows from Generating Function Divided by Power of Parameter.
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $4$
Category:Examples of Generating Functions
Generating Function for mth Terms of Sequence

== Theorem ==
Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\omega = e^{2 i \pi / m} = \cos \dfrac {2 \pi} m + i \sin \dfrac {2 \pi} m$.
Then for $r \in \Z$ such that $0 \le r < m$:
:$\displaystyle \sum_{n \bmod m \mathop = r} a_n z^n = \dfrac 1 m \sum_{0 \mathop \le k \mathop < m} \omega^{-k r} G \left({\omega^k z}\right)$
== Proof ==
== Examples ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(13)$
Category:Generating Functions
Category:Generating Function for mth Terms of Sequence
Generating Function for mth Terms of Sequence/Examples/m = 3, r = 1

== Example of Generating Function for mth Terms of Sequence ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $\omega = e^{2 i \pi / 3} = -\dfrac 1 2 + \dfrac {\sqrt 3} 2 i$.
Then:
:$a_1 z + a_4 z^4 + a_7 z^7 + \cdots = \dfrac 1 3 \paren {\map G z + \omega^{-1} \map G {\omega z} + \omega^{-2} \map G {\omega^2 z} }$
== Proof ==
From Generating Function for mth Terms of Sequence, for $r \in \Z$ such that $0 \le r < m$:
:$\ds \sum_{n \bmod m \mathop = r} a_n z^n = \dfrac 1 m \sum_{0 \mathop \ge k \mathop < m} \omega^{-k r} \map G {\omega^k z}$
Setting $m = 3$ and $r - 1$:
== Sources ==
* : $\S 1.2.9$: Generating Functions
Category:Generating Function for mth Terms of Sequence
Generating Function for mth Terms of Sequence/Proof

== Theorem ==
== Proof ==
By definition of $\omega$:
:$\omega = \exp \dfrac {2 \pi i} m \implies \omega^m = 1$
and so:
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $14$
Category:Generating Function for mth Terms of Sequence