If the series representation of $e^{-x}$ is:
$$\sum_{k=0}^{\infty} \frac{(-x)^k}{k!} $$
Then what is for $e^{-kx}$?
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Do you know the basic rules for constructing new power series from old ones? Also, slight nitpick but you don't want to use k as a summation index and a constant. – Alex Zorn Apr 10 '14 at 23:19
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You should use a different index of summation than variables used outside the scope of the summation. $$ e^{-kx}=\sum_{j=0}^\infty\frac{(-kx)^j}{j!} $$
robjohn
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Consider the summation $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \text{.}$$ You demonstrated that you already know that you're allowed to replace $x$ with $-x$ to obtain the summation $$e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \text{.}$$
Intuitively you should be able to throw a k in there as well: $$e^{-kx} = \sum_{n=0}^{\infty} \frac{(-kx)^n}{n!} \text{.}$$
syusim
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Are these approximations good only in a small region around x= 0 ? or would be true for all x ? – whisperer May 11 '20 at 20:22