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$\displaystyle e^x= \sum_{j=0}^{\infty} \frac{x^j}{j!}$

The textbook says that when we differentiate this, we obtain the same series, so that $(e^x)'=e^x$.

But why is this? Isn't the derivative $\displaystyle \sum_{j=0}^{\infty}\frac{jx^{j-1}}{j!}$?

kiwifruit
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2 Answers2

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\begin{align*} \mathrm{e}^x &= \sum_{j=0}^\infty \frac{x^j}{j!} \\ &= \frac{x^0}{0!} + \sum_{j=1}^\infty \frac{x^j}{j!} \\ &= 1 + \sum_{j=1}^\infty \frac{x^j}{j!} \end{align*}

Differentiate. \begin{align*} \left( \mathrm{e}^x \right)' &= 1' + \sum_{j=1}^\infty \left(\frac{x^j}{j!} \right)' \\ &= 0 + \sum_{j=1}^\infty \frac{j x^{j-1}}{j!} \\ &= \sum_{j=1}^\infty \frac{x^{j-1}}{(j-1)!} \\ &= \sum_{j=0}^\infty \frac{x^{j}}{j!} \\ &= \mathrm{e}^x \end{align*}

Eric Towers
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  1. for $j=0$ this is a constant, hence your expression is incorrect (the sum's first term is for $j=1$).
  2. remember that $j! = j(j-1)!$
mookid
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  • Hmm, but my textbook starts with 0. So then we will get $\displaystyle \sum_{j=0}^{\infty} \frac{x^{j-1}}{j-1!}$? But that is also not the same result. – kiwifruit Apr 27 '14 at 23:51
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    the expression is correct. However, the derivative of $1 = x^ 0$ is $0$. Hence the expression of the derivative is a sum starting from 1/. – mookid Apr 27 '14 at 23:55
  • Ok, and can we say that $\displaystyle \sum_{j=1}^{\infty} \frac{x^{j-1}}{j-1!} = \sum_{j=0}^{\infty} \frac{x^j}{j!}$ That doesn't seem to work, since the exponent in the second equation will still be j – kiwifruit Apr 27 '14 at 23:57
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    @kiwifruit The index is a local variable; its scope exists only within the sum. So you can change it to whatever you like and the sum retains its identity. $\sum\limits_{j=1}^\infty f_j(x) = \sum\limits_{k=1}^\infty f_k(x) = \sum\limits_{\text{penguin}=1}^\infty f_{\text{penguin}}(x)$ That mean's you can replace the index $j$ with the index $j+1$ everywhere it occurs within the sum. $\sum\limits_{j=1}^{\infty} \frac{x^{j-1}}{(j-1)!} = \sum\limits_{(j+1)=1}^{\infty} \frac{x^{(j+1)-1}}{((j+1)-1)!} = \sum\limits_{j=0}^{\infty} \frac{x^{j}}{j!}$ – Graham Kemp Apr 28 '14 at 00:11