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Let ex-a=∑an(x-a)n=a0+a1(x-a)+a2(x-a)2+a3(x-a)3+... for integer n≥0.

When we differentiate this as a series we get:

a1+a2(x-a)+a3(x-a)2+...=∑nan(x-a)n-1 (n≥1) and when we differentiate ex-a we get ex-a. In fact, every derivative of ex-a is going to be the same, but the derivative of the series augments the coefficient of x-a like this:

n(n-1)(n-2)...an etc., while the power of x-a decreases. The first term in the differentiated series will always be a constant (that is, won't contain x), so by setting x to a particular value on both sides of the equation so that all other terms become zero, it's easy (in this case) to find the coefficients an.

To find an we need an equation we can solve. 

Take the second derivative as an example: n(n-1)an(x-a)n-2. (When n=2, this reduces to n(n-1)a2=2!a2.)

Go back to ex-a=∑an(x-a)n=a0+a1(x-a)+a2(x-a)2+a3(x-a)3+... When x=a, ex-a becomes e0=1 and the series becomes a0 (because all the other terms become zero). Therefore a0=1.

Similarly, the first derivative becomes 1 when x=a, so we have a1=1. The second derivative is 2a2=1. The third derivative is 6a3=3!a3=1. And so on. From this we get an=1/n!.

So ex-a=∑(x-a)n/n!. (Note that 0!=1.) But ex-a can be written ex/ea, therefore ex/ea=∑(x-a)n/n! and ex=ea∑(x-a)n/n!=ea(1+(x-a)+(x-a)2/2!+(x-a)3/3!+..., so proving the given statement.

I've probably given you more than you asked for, because the question assumes you know the Taylor series for ex=1+x+x2/2!+x3/3!+... and by replacing x with x-a on both sides you come up with same proof. However, I wanted to show you how straightforward it is to derive the series for ex. It's not as frightening as you might have thought and it may help you to understand better the basic concepts behind deriving Taylor or Maclaurin series coefficients.

by Top Rated User (1.1m points)

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