Difference between revisions of "Derivative of the exponential function"
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==Proof== | ==Proof== | ||
| + | By definition, | ||
$$e^z = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}.$$ | $$e^z = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}.$$ | ||
[[Term-by-term differentiation]] of this sum shows | [[Term-by-term differentiation]] of this sum shows | ||
| Line 16: | Line 17: | ||
==References== | ==References== | ||
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=findme}}: 4.2.5 | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
| + | [[Category:Justify]] | ||
Latest revision as of 00:10, 23 December 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} e^z = e^z,$$ where $e^z$ denotes the exponential function.
Proof
By definition, $$e^z = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}.$$ Term-by-term differentiation of this sum shows $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} e^z &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\mathrm{d}}{\mathrm{d}z} \left[ \dfrac{z^k}{k!} \right] \\ &=\displaystyle\sum_{k=1}^{\infty} \dfrac{z^{k-1}}{(k-1)!} \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!} \\ &=e^z, \end{array}$$ as was to be shown. █
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 4.2.5
