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Definition:

Let $$a_0 = a_1 = 1, \; a_{n+2} = a_{n+1} + (n+1) \cdot a_n, \; n \geq 0$$

Exercise:

Prove that $$\sum_{n\geq 0} \frac{a_n}{n!} x^n = \exp \left( x + \frac{1}{2} x^2 \right)$$


I don't know how to start with this. I do know that $$e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n, \; e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ but how do I get this wrapped up?

Thank you in advance!

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4 Answers

up vote 2 down vote accepted

Let $$S(x)= \sum_{n\ge 0} \frac{a^n}{n!}x^n$$

and note that

$$S'(x)= \sum_{n\ge 1} \frac{a^n}{(n-1)!}x^{n-1}$$

Replacing in the first equation the value of $a_n$ given by the recursion, $a_{n}=(a_{n+2}-a_{n+1})/(n+1)$ we get

$$S(x) = S_2(x) - S_1(x)$$

where

$$S_1(x) = \sum_{n\ge 0} \frac{a^{n+1}}{(n+1)!}x^n =\frac{1}{x} [ S(x)-a_0 ]$$

and

$$S_2(x) = \sum_{n\ge 0} \frac{a^{n+2}}{(n+1)!}x^n =\frac{1}{x} [ S'(x)-a_1 ]$$

What gives

$$S(x) \; \left( x+1 \right) = S'(x) + a_0 - a_1 = S'(x)$$

This first order differential equation is straightforward:

$$\frac{S'(x)}{S(x)} = \log'{S(x)} = x+1 \Rightarrow \log{S(x)} = \frac{1}{2}x^2+x +C$$

which, together with the initial condition $S(0)=1$, gives $S(x)=\exp(x+x^2/2)$.

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thank you! –  muffel Jun 13 '11 at 18:12
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HINT $\ $ The differential equation for the exponential yields a recurrence for the coefficients, for example see here and here.

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A very good method. Now, a more difficult question: Suppose we were not given the answer, merely asked to evaluate that generating function. –  GEdgar Jun 4 '11 at 16:12
    
@GEdgar Then simply proceed in reverse - transmute the recurrence to a differential equation then solve it - see the example in my second link. –  Bill Dubuque Jun 4 '11 at 16:28
    
@Buill-Dubuque Could you please give me another hint. How do I start best? –  muffel Jun 5 '11 at 22:00
    
@muffel The RHS is the unique solution of $\rm\: y' = (x+1)\ y,\ y(0) = 1\:.\:$ Show that this is equivalent to the given recurrence for the coefficients of the LHS series expansion. –  Bill Dubuque Jun 5 '11 at 22:35
    
@Bill-Dubuque I tried to understand the two examples but that didn't help much. The first one looks promising, but what is that y and y' and log all about? Sorry, but I've been ill and couldn't follow the lecture completely... –  muffel Jun 6 '11 at 6:40
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Here's a probabilistic interpretation/derivation of the result. Henceforth, we assume that $x \neq 0$ (the case $x=0$ being trivial).

Recall that the moment-generating function of $X \sim {\rm N}(\mu,\sigma^2)$ is given by $$ {\rm E}[e^{tX} ] = e^{\mu t + \sigma ^2 t^2 /2} . $$ Thus setting $t=1$, $\mu=x$, and $\sigma^2 = x^2$, we get $$ {\rm E}[e^{X} ] = e^{x + x^2 /2}. $$ On the other hand, $$ {\rm E}[e^{X} ] = \sum\limits_{n = 0}^\infty {\frac{{\mu _n }}{{n!}}} , $$ where $\mu_n = {\rm E}(X^n)$ is the $n$th moment of the ${\rm N}(x,x^2)$ distribution. Using integration by parts, we find $$ \mu_n = \frac{1}{{\sqrt {2\pi x^2 } }}\int_{ - \infty }^\infty {u^n \exp \bigg[ - \frac{{(u - x)^2 }}{{2x^2 }}\bigg]du} = \frac{1}{{\sqrt {2\pi x^2 } }}\int_{ - \infty }^\infty {\frac{{u^{n + 1} }}{{n + 1}}\exp \bigg[ - \frac{{(u - x)^2 }}{{2x^2 }}\bigg]\frac{{u - x}}{{x^2 }}\,du}. $$ From this we get $$ \mu _n = \frac{1}{{n + 1}}\bigg(\frac{{\mu _{n + 2} }}{{x^2 }} - \frac{{\mu _{n + 1} }}{x}\bigg). $$ Dividing both sides by $x^n$ yields $$ \frac{{\mu _n }}{{x^n }} = \frac{1}{{n + 1}}\bigg(\frac{{\mu _{n + 2} }}{{x^{n + 2} }} - \frac{{\mu _{n + 1} }}{{x^{n + 1} }}\bigg), $$ and hence $$ \frac{{\mu _{n + 2} }}{{x^{n + 2} }} = \frac{{\mu _{n + 1} }}{{x^{n + 1} }} + (n + 1)\frac{{\mu _n }}{{x^n }}, $$ leading to $$ e^{x + x^2 /2} = \sum\limits_{n = 0}^\infty {\frac{{\mu _n }}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{(\mu _n /x^n )x^n }}{{n!}}} . $$ Finally, letting $a_n = \mu_n / x^n$, we have that $$ e^{x + x^2 /2} = \sum\limits_{n = 0}^\infty {\frac{{a_n x^n }}{{n!}}}, $$ with $a_0 = a_1 = 1$, and $a_{n + 2} = a_{n + 1} + (n + 1)a_n $, as desired.

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To clarify, let $f(x)=\exp \left( x + \frac{1}{2} x^2 \right)$. Now differentiate $f$ twice, you should end up with a differential equation, something like $f''(x)=(2+2x+x^2)f(x)$ with the initial conditions $f(0)=f'(0)=1$ (see given values of $a_0$ and $a_1=1$). Now you need to solve this ordinary differential equation using the method of power series (most elementry books on ODEs have a chapter on this, you dont need to go too deep into the theory at all, no need for the method of frobenius etc. if you can follow any of the examples in the book you should be able to tackle the question).

Basically assume $f(x)=\sum_{n\geq 0} a_n x^n$ (noticed i've absorbed the n! term into $a_n$ for simplicity) then substitute into the ODE. By equating the coefficients you should get a recurrence relation for $a_n$. If it matches what you've written in your question, then your done.

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