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If $f_0(x)=e^x$ and define $f_{n+1}(x)=xf_n'(x)$, what is the exact value of $$\sum_{n=0}^\infty\frac{f_n(1)}{n!}$$ I tried finding general formula of $f_n(x)$ for $n$.

But it is not of the form that i already know.

For another way, i set $g(x)=$taht sigma.l, get its derivative and relation between $g(x)$ and $g'(x)$

But it was not useful.

Jack D'Aurizio
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LeeHanWoong
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  • Please use mathjax to format https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Shuri2060 Jul 19 '17 at 09:21
  • Try and use mathjax syntax, this is not readable. – Furrane Jul 19 '17 at 09:21
  • Not sure, but maybe $f_{n+1}(x)=(xf_n(x))'-f_n(x)$ might help. You then get $f_{n+1}(x)=(x(f_n(x)-f_{n-1}(x)+f_{n-2}(x)-...\pm f_1(x)))'\pm e^x$. Looking at the series expansion may also help. – Shuri2060 Jul 19 '17 at 09:30

1 Answers1

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You may notice that $e^{x}=\sum_{n\geq 0}\frac{x^n}{n!}$ is an entire function and study how the operator $xD:f\mapsto xf'$ acts on the power series. We have $$ (xD)^m e^x = \sum_{n\geq 0}\frac{x^n}{n!}n^m \tag{1}$$ hence $$ \sum_{m\geq 0}\frac{f_m(1)}{m!} = e+\sum_{m\geq 1}\sum_{n\geq 0}\frac{n^m}{m! n!}=e+\sum_{n\geq 0}\frac{e^n-1}{n!}=\color{red}{\large e^e}.\tag{2} $$

Jack D'Aurizio
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