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I am having troubles understanding how to operate with generating functions to obtain a final formula. I have been looking to the answer to this question: Exponential generating function and number of balls .

$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$ This expands to $$\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + \sum_{n=2}^\infty \frac1{16}2^nn(n-1)\frac{x^n}{n!} + \sum_{n=4}^\infty\frac1{768}2^nn(n-1)(n-2)(n-3)\frac{x^n}{n!}. $$ Simplification yields $$\frac{x^2}{2!}+3\frac{x^3}{3!}+13\frac{x^4}{4!} + \sum_{n=5}^\infty \frac1{768} 2^nn(n-1)(n^2-5n+54)\frac{x^n}{n!}. $$

What are the steps to obtain this and the simplification? I can't see how to get from one to another I have tried a different approach, since $$ e^x + e^{-x} = \sum_{n=0}^\infty {2x^{2n}\over (2n)!}; \; e^x = \sum_{n=0}^\infty {x^{n}\over n!}, $$ Multiplying everything, I obtain: $$ \frac{1}{2}\frac{x^2}{2!}\left(\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}\right)\cdot \left(\sum_{n=0}^\infty \frac{x^n}{n!}\right) + \frac{1}{2}\frac{x^4}{4!}\left(\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}\right)\cdot \left(\sum_{n=0}^\infty \frac{x^n}{n!}\right) $$

Removing the 2 of the numerator of the fractions and simplifying $x^{2n}/2n!$ for $x^{n}/n!$; and if I understand multiplication correctly, I should have: $$ \sum_{k=0}^{n}\binom{n}{k}\frac{x^n}{n!} $$ as a result of both the multiplications, but doesn't look like any form of the right answer.

PS: Decided to make another question instead of commenting the answer because it is 2 years old already.

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    $\begingroup$ You do know that $(e^x+e^{-x})e^x=e^{2x}+1$, don't you? $\endgroup$ Commented Dec 8, 2017 at 21:25
  • $\begingroup$ Yeah, I know that, did not see that simple thing, that's embarrasing. But I still don't know how to get to the second step from:$$ \frac{1}{2}\frac{x^2}{2!}+ \frac{1}{2}\frac{x^4}{4!} + \frac{1}{2}\frac{x^2}{2!}\cdot\sum_{n=0}^{\infty}\frac{(2x)^n}{n!} + \frac{1}{2}\frac{x^4}{4!}\cdot \sum_{n=0}^{\infty}\frac{(2x)^n}{n!}; $$ To start the sum from n=2 and n=4 I understand it is needed to multiply by $n(n-1)$ and $n(n-1)(n-2)(n-3)$ but I don't know where the $1/16$ and $1/768$ come from. $\endgroup$
    – Geid
    Commented Dec 8, 2017 at 21:57

1 Answer 1

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The term with $\frac{1}{16}$ arises as follows: \begin{align*} \frac{x^2}{2}\cdot\frac{1}{2}e^{2x} &= \frac{1}{4}x^2\sum_{n=0}^\infty \frac{2^nx^n}{n!} \\ &= \frac{1}{4}\sum_{n=0}^\infty \frac{2^nx^{n+2}}{n!} \\ &= \frac{1}{16}\sum_{n=0}^\infty \frac{2^{n+2}x^{n+2}}{n!} \\ &= \frac{1}{16}\sum_{n=2}^\infty \frac{2^nx^n}{(n-2)!} \\ &= \frac{1}{16}\sum_{n=2}^\infty n(n-1)\frac{2^nx^n}{n!}. \end{align*} The other term arises similarly, starting from $\frac{x^4}{24}\cdot\frac{1}{2}e^{2x}$.

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  • $\begingroup$ OK, I see it now, thanks a lot. I didn't know how to operate with the terms of the sum, but now it's clear. $\endgroup$
    – Geid
    Commented Dec 8, 2017 at 22:39

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