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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!

4 Answers 4

1

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.

3

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.2011-06-04
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    @GEdgar Then simply proceed in reverse - transmute the recurrence to a differential equation then solve it - see the example in my second link.2011-06-04
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    @Buill-Dubuque Could you please give me another hint. How do I start best?2011-06-05
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    @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.2011-06-05
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    @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...2011-06-06
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    The log arises via [logarithmic differentiation.](http://en.wikipedia.org/wiki/Logarithmic_differentiation) You don't really need to do that here since the DE for the exponential is obvious. Substitute the purported power series into the DE and do some rearranging to derive the stated recurrence.2011-06-06
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    OK. We need a solution for someone who doesn't know what a differential equation is. Maybe a good recommendation is to see the instructor for help. He/she knows what the background is supposed to be, we don't.2011-06-12
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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.

2

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 $S(x)=\exp(x+x^2/2)$.