Eikonal Blog

2012.11.26

Generating functions

Recently I have been rereading the Herbert S. Wilf’s free online book Generatingfunctionologyhttp://www.math.upenn.edu/~wilf/DownldGF.html. It is choke-full of interesting results.

Definitions

For a number series ${\bf a} :\equiv \{a_n|n\in\{0,\infty\}\}$ one defines following generating functions (GFs):

• Ordinary Power Series GF (OPSGF): $OPSGF[a](x) :\equiv \sum_{n=0}^\infty a_n x^n \leftrightarrow_{opsgf} {\bf a}$;
• Exponential Series GF (EGF): $EGF[a](x) :\equiv \sum_{n=0}^\infty a_n \frac{x^n}{n!} \leftrightarrow_{egf} {\bf a}$;
• Dirichlet GF (DGF): $DGF(x;s) :\equiv \sum_n \frac{a_n}{n^s} x^n \leftrightarrow_{dgf} {\bf a}$;
• Lambert GF (LGF): $LGF(x) :\equiv \sum_n \frac{a_n}{1-x^n} x^n \leftrightarrow_{lgf} {\bf a}$;
• Bell GF (BGF): $BGF(x;s) :\equiv \sum_n a_{p^n} x^n \leftrightarrow_{bgf} {\bf a}$;
• Poisson GF (PGF): $PGF(x;s) :\equiv \sum_n \frac{a_n}{n!} x^n e^{-x} = e^{-x} EGF(x) \leftrightarrow_{pgf} {\bf a}$;

Simple results

• $\{a_n=n\} \leftrightarrow_{opsgf} f(x) = \frac{x}{(1-x)^2}$.
• $\{a_n=n^2\} \leftrightarrow_{opsgf} f(x) = \frac{x(x+1)}{(1-x)^3}$.
• $\{a_n=b^n\} \leftrightarrow_{opsgf} f(x) = \frac{1}{1-b x}$.
• $\{a_n=n\} \leftrightarrow_{egf} f(x) = x e^{x}$.
• $\{a_n=n^2\} \leftrightarrow_{egf} f(x) = x(x+1)e^{x}$.
• $\{a_n=b^n\} \leftrightarrow_{egf} f(x) = e^{b x}$.
• For Fibonacci numbers $\{F_n|F_{n+1} = F_n + F_{n-1} | f_0 = F_1 = 1\}$ one has: $OPSGF(x) = \frac{x}{1-x-x^2}$, leading to $F_n = \frac{r_+^n-(r_-^n}{\sqrt{5}}$ ($r_\pm = \frac{1\pm\sqrt{5}}{2}$).
• $\{a_n|a_{n+1}=2 a_n + 1| a_0=0\} \leftrightarrow_{opsgf} OPSGF = \frac{x}{(1-x)(1-2x)} = \frac{1}{1-2x} - \frac{1}{1-x}$; which leads to $a_n=2^n - 1$.
• For $a_{n+1} = 2 a_n + n$, $alatex _0=1$ one has: $OPSGF = \frac{2}{1-2x} - \frac1{(1-x)^2}$ leading to
$a_n = 2^{n-1} - n - 1$.

Define $[x^n]f(x)$ as the coefficient next to $x^n$ in power series $f(x)$. Examples and properties:

• $[x^n] e^x = \frac1{n!}$,
• $[x^n] \frac1{1-ax} = a^n$,
• $[x^n] (1+x)^s = \binom{s}{n}$,
• $[x^n] \{x^m f(x)\} = [x^{n-m}] f(x)$,
• $[\lambda x^n] f(x) = \frac1{\lambda} [x^n]f(x)$,
• $\left[\frac{x^n}{n!}\right] e^x = 1$.

For binomial coeficients:

• $\sum_{k=0}^{\infty} \binom{n}{k} x^k = (1+x)^n$,
• $\sum_{n=0}^{\infty} \binom{n}{k} y^n = \frac{y^k}{(1-y)^{k+1}}$,
• $\binom{n}{k} = [x^k y^n] \frac1{1 - y (1+x)}$.

Some orthogonal polynomials:

• Tchebitshev polynomials generating function: $\sum_{n=0}^\infty \frac{z^n}{n!} T_n(x) = e^{zx}\cos(z\sqrt{1-x^2})$.
• Legendre polynomials generating function: $\frac1{(1-2tz+z^2)^{\frac12}} = \sum_{n=0}^\infty P_n(t)z^n$.
• Generating function for associated Legendre polynomials: $(1-2tc+t^2)^{\alpha} = |1-t e^{i\theta}|^{2\alpha} = \sum_{n=0}^\infty t^n P_n^{\alpha}(c)$.

Dirichlet series generating functions

• For $a_n=1$: $DGF = \zeta(s)$.
• For $a_n=\mu(n)$ (the Moebius function): $DGF = \frac1{\zeta(s)}$.
• For $a_n=d(n)=\sigma_0(n)$ (the zeroth-order divisor function): $DGF = \zeta(s)^2$.
• For $a_n=\sigma_k(n)$ (the $k$th-order divisor function): $DGF = \zeta(s)\zeta(s-k)$.
• For $a_n=\phi(n)$ (the totient function): $DGF = \frac{\zeta(s-1)}{\zeta(s)}$.
• For $a_n=H(n)$ (the number of ordered factorizations): $DGF = \frac{1}{2 - \zeta(s)}$.
• For $a_n=\frac12 [1-(-1)^n]$: $DGF = \lambda(s)$ (the Dirichlet lambda function).

Moebius inversion formula:

• If two DGF series $A(s)$ and $B(s)$ have coefficient relation $a_n = \sum_{d|n} b_d$, then $A(s) = B(s) \zeta(s)$, and $b_n = \sum_{d|n} \mu\left(\frac{n}{d}\right) a_d$.
• If $a_n = \sum_{d|n} b_d$, then $b_n = \sum_{d|n} \mu\left(\frac{n}{d}\right) a_d$.

2010.03.12

Fibonacci numbers – simple generalizations

Filed under: mathematics — Tags: , , , — sandokan65 @ 17:32

Source: 2009.02.04 Wed

The Fibonacci series $\{f_n | n\in {\Bbb N}_0\}$ are defined by the recursive equation

$f_{n} = f_{n-1} + f_{n-2}$

and the “initial conditions” (i.e. known) $f_0$ and $f_1$.

The case $f_0=f_1=1$ is the original Fibonacci’s sequence, and traditionally the members of the sequence are symbolized by capital case F: $\{F_0=1, F_1=1, F_2=2, \cdots\}$.

The exponential-type generating function ${\cal F}(z) :\equiv \sum_{n=0}^\infty \frac{z^n}{n!} f_n$ satisfies the simple IVP (Initial value Problem):

• ${\cal F}''(z) = {\cal F}'(z) + {\cal F}(z)$,
• ${\cal F}(0) = f_0$,
• ${\cal F}'(0) = f_1$,

with the unique solution

${\cal F}(z) = \frac{f_0}{\sqrt{5}} [(\varphi-1) e^{+\varphi z} + \varphi e^{(1-\varphi) z} ] + \frac{f_1}{\sqrt{5}} [e^{+\varphi z} - e^{(1-\varphi) z}]$.

Here $\varphi :\equiv \frac{\sqrt{5}+1}2$ is the Golden Section, a root of the characteristic polynomial $P(\lambda) = \lambda^2 – \lambda-1$ (the other solution being $1-\varphi = \frac{\sqrt{5}-1}2$).

Having the explicit expression for the generating function, one gets the explicit expression for individual sequence members:

$f_n = \varphi^n \frac{(1-\varphi) f_0 + f_1}{\sqrt{5}} + (1-\varphi)^n \frac{\varphi f_0 - f_1}{\sqrt{5}}$.

First slight generalization

Use the recursive equation $f_{n+2} = m f_{n+1} + k f_n$. It leads to the following ODE for the generating function ${\cal F}''(z) = m {\cal F}'(z) + k {\cal F}(z)$, the characteristic root equation $\lambda^2 = m \lambda + k$ with the solutions $\lambda_\pm = \frac{m\pm \kappa}2$ (where $\kappa :\equiv \sqrt{m^2+4 k}$). The solution for the generating function is

${\cal F}(z) = f_0 \left[\frac{1-m}2 e^{\lambda_+ z} - \frac{1+m}2 e^{\lambda_- z}\right] + f_1 \left[e^{\lambda_+ z} - e^{\lambda_- z}\right]$

and the explicit expression for the sequence members is

$f_n = f_0 \left[\frac{1-m}2 (\lambda_+)^n - \frac{1+m}2 (\lambda_-)^n\right] + f_1 \left[(\lambda_+)^n - (\lambda_-)^n\right]$.

Second slight generalization

Use the recursive equation $f_{n+3} = f_{n+3} + f_{n+1} + f_n$. Here:

• ${\cal F}'''(z) = {\cal F}''(z) + {\cal F}'(z) + {\cal F}(z)$
• $\lambda^3 = \lambda^2 + \lambda + 1 = \frac{\lambda^4-1}{\lambda-1}$, i.e. $\lambda^4 = 2\lambda^3 - 1$ (with $\lambda\ne 1$), with solutions (thanks to the Wolfram Alpha online calculator http://www.wolframalpha.com/input/?i=x3%3Dx^2%2Bx%2B1):
• $\lambda_0 = \frac13 (1+ \sqrt[3]{19-3 \sqrt{33}}+ \sqrt[3]{19+3 \sqrt{33}}) \simeq 1.83929$
• $\lambda_\pm = \frac16 (2- (1 \mp i \sqrt{3}) \sqrt[3]{19-3 \sqrt{33}} - (1 \pm i \sqrt{3}) \sqrt[3]{19+3 \sqrt{33}})\simeq -0.419643 \mp 0.606291 i$, (note: $|\lambda_\pm|=0.737353...$)
• $\lambda_0 + + \lambda_+ + \lambda_- = - 1$, $\lambda_0 \lambda_+ + \lambda_0 \lambda_- + \lambda_+ \lambda_- = -1$, $\lambda_0 + \lambda_+ \lambda_- = +1$.
• $\lim_{n\rightarrow \infty} \frac{f_{n+2}+f_{n+1}+f_{n}}{f_{n+2}} = \lambda_0$.