## Related questions with answers

Question

Differentiate the series

$E(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$

term-by-term to show that E(x) is equal to its derivative.

Solution

VerifiedStep 1

1 of 2$\begin{align*} &E(x)=\sum\limits_{n=0}\frac{x^n}{n!}\\ \Rightarrow&\dfrac{d\ }{dx}E(x)=\sum\limits_{n=0}\dfrac{d\ }{dx}\left(\frac{x^n}{n!}\right)\tag{differentiating both sides}\\ \Rightarrow&\dfrac{d\ }{dx}E(x)=\sum\limits_{n=1}\frac{nx^{n-1}}{n!}\\ \Rightarrow&\dfrac{d\ }{dx}E(x)=\sum\limits_{n=1}\frac{\cancel{n}x^{n-1}}{\cancel{n}\cdot(n-1)!}\\ \Rightarrow&\dfrac{d\ }{dx}E(x)=\sum\limits_{n=1}\frac{x^{n-1}}{(n-1)!}\\ \Rightarrow&\dfrac{d\ }{dx}E(x)=\sum\limits_{n=0}\frac{x^n}{n!}=E(x)\tag{re-indexing}\\ \end{align*}$

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