## 8.4. Taylor Series

### Example 8.4.2 (b): Derivatives of Power Series

Assume again that

*f(x) = e*has a convergent power series expression, but this time centered at^{2x}*c = 1*. Find the coefficients of*this*power series.This time we assume our function can be represented as

e^{2x}= a_{n}(x-1)^{n}= = a_{0}+ a_{1}(x-1) + a_{2}(x-1)^{2}+ a_{3}(x-1)^{3}+ ...

Then, as before

a_{n}=

Again we need to find the *n*-th derivative of the function
*f* at the center of the series. Taking derivatives on both sides
and substituting 1 (the center in this example) we get:

n=0f(x)=e^{2x}f(1)=en=1f '(x)=2 e^{2x}f '(1)=2 e^{2}n=2f ''(x)=2^{2}e^{2x}f ''(1)=2^{2}e^{2}n=3f^{(3)}(x)=2^{3}e^{2x}f^{(3)}(1)=2^{3}e^{2}

Therefore, if *e ^{2x}* had a series representation
centered at

*c = 1*it would be:

e^{2x}=^{2n}/_{n!}e^{2}(x-1)^{n}