I've got an example and I wanted to know how it is expanded. Thanks for help.



Is converted to

$$h(x)={\sum _{n=0}^{\infty } \frac{5n{(-1)}^{n+1}x^n}{n!}}+{\sum _{n=0}^{\infty } \frac{2(-x)^n}{n!}}$$

I wanted to do this with


  • 4
    $\begingroup$ The main point is knowing the expansion for $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. Then manipulate. The first term in the series for h(x) looks wrong. $\endgroup$ – herb steinberg Jan 19 '18 at 0:49

This is an answer: $$5 e^{-x} + 7xe^x = \sum_{n=0}^{\infty} \frac{5 (-x)^n + 7 x^{n+1}}{n!}$$

  • $\begingroup$ I made a mistake. $\endgroup$ – Marianna Kalwat Jan 19 '18 at 14:02

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