8.4. Taylor Series
Example 8.4.2 (a): Derivatives of Power Series
According to our theory we know that if our function can be represented as
e2x = anxn = = a0 + a1x + a2x2 + a3x3 + ...
Thus, to find the coefficients an 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=0 f(x)=e2x f(0)=1 n=1 f '(x)=2 e2x f '(0)=2 n=2 f ''(x)=22e2x f ''(0)=22 n=3 f (3)(x)=23e2x f (3)(0)=23
and so on, so that by "poor man's induction" we get:
an = 2n/n!
Therefore, if e2x had a series representation, it would have to be:
e2x = 2n/n! xn