1
$\begingroup$

I have troubles proving that the power series

$\sum_{n=1}^\infty n^3\frac{x^n}{n!}$

represents the function

$f(x)=e^x(x^3+3x^2+x)$.

My idea was using the identity theorem for power series and the definition of the Taylor series of a function to somehow show that the only function whose $n$-th derivative at $x=0$ equals $n^3$ is the aforementioned function, but right now I can't find the right arguments and I need some hints to point me in the right direction.

$\endgroup$
4
$\begingroup$

Hint: $$ \frac{n^3}{n!}=\frac{n^2}{(n-1)!}=\frac{1}{(n-1)!}+\frac{n+1}{(n-2)!}=\frac{1}{(n-1)!}+\frac{3}{(n-2)!}+\frac{1}{(n-3)!} $$

$\endgroup$
0
$\begingroup$

$\textbf{Using generating function method :}$

Define : $$G[x,\lambda]=\sum_{n=0}^{\infty}e^{i\lambda n}_{} \frac{x^{n}_{}}{n!}=e^{x e^{i\lambda}}_{}.$$ Now note $$\sum_{n=1}^{\infty}n^{3}_{}\frac{x^{n}_{}}{n!}=\sum_{n=0}^{\infty}n^{3}_{}\frac{x^{n}_{}}{n!}=\frac{\partial^{3}_{}}{\partial(i\lambda)^{3}_{}}G[x,\lambda]\Big|_{\lambda=0}^{}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.