## 8.4. Taylor Series

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

Assume that

*f(x) = e*has a convergent power series expression centered at^{2x}*c = 0*, i.e.*e*. Find the coefficients^{2x}= a_{n}x^{n}*a*._{n}According to our theory we know that if our function can be represented as

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

then

a_{n}=

Thus, to find the coefficients *a _{n}* we bascially need to
find the

*n*-th derivative of the function

*f*at the center of the series. Taking derivatives on both sides and substituting 0 (the center in this example) we get:

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

and so on, so that by "poor man's induction" we get:

a_{n}=^{2n}/_{n!}

Therefore, if *e ^{2x}* had a series representation, it
would have to be:

e^{2x}=^{2n}/_{n!}x^{n}