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I'm required to solve the recurrence $y_{n+1} = 2y_n + n$ using an exponential generating series, and am getting stuck in a particular step. I solved it already using the OGS so I think I'm on the right track. Here's what I have so far:

$$ EGS: Y(x) = \sum_{n=0}^{\infty}\frac {y_nx^n} {n!} $$ $$ \sum_{n=0}^{\infty} \frac {y_{n+1}x^n} {n!} = \sum_{n=0}^{\infty} \frac{(2y_n+n)x^n} {n!} = 2\sum_{n=0}^{\infty} \frac {y_nx^n} { n!} + \sum_{n=0}^{\infty} \frac{n x^n} {n!} $$

Now the problem that I'm having is reducing this to a differential equation due to the last term. I know that $ e^x = \sum_{n=0}^{\infty} \frac{ x^n} {n!} $ , but I'm not sure what to do with the extra "n" term in this problem. I can reduce it to $$ \sum_{n=0}^{\infty} \frac{n x^n} {n!} = \sum_{n=0}^{\infty} \frac{ x^n} {(n-1)!} = \sum_{n=0}^{\infty} \frac{x* x^n} {n!} $$ ( I think ).

I'm not 100% sure what method to use here, possibly a Taylor Series or something? Any help is greatly appreciated. Thanks!

Dave McGuire
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  • $\sum_{n=0}^\infty \frac{nx^n}{n!} = \sum_{n=1}^\infty \frac{nx^n}{n!} = \sum_{n=1}^\infty \frac{x^n}{(n-1)!} = \sum_{n=0}^\infty \frac{x^{n+1}}{n!} = xe^x$. – Tunococ Sep 16 '13 at 23:19

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For your reduction from $\displaystyle\sum_{n=0}^\infty\frac{nx^n}{n!}$ to $\displaystyle\sum_{n=0}^\infty\frac{x^n}{(n-1)!}$, it looks like it runs into problems with the first term (when $n = 0$), since you're basically factoring out a 0 from the numerator and denominator.

So, first, note that $\displaystyle\sum_{n=0}^\infty\frac{nx^n}{n!} = \displaystyle\sum_{n=1}^\infty\frac{nx^n}{n!}$, and then you get

$\displaystyle\sum_{n=1}^\infty\frac{nx^n}{n!} = \displaystyle\sum_{n=1}^\infty\frac{x^n}{(n-1)!} = \displaystyle\sum_{n=0}^\infty\frac{x*x^n}{n!}$.

Now, to simplify, note that $x$ doesn't depend on $n$, so you can just bring one factor of $x$ out of the summation:

$\displaystyle\sum_{n=0}^\infty\frac{x*x^n}{n!} = x\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!} = xe^x$.

Dennis Meng
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