## 8.4. Taylor Series

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

Find the 10th and 11th derivative at zero for the function

*f(x) =*. As a hint, try to get the geometric series function involved somehow.^{x3}/_{1-x2}The obvious first try is to take 10 derivatives, then substitute
*x=0*. Let's take a few derivatives:

This is getting out of hand - the derivatives are getting more and more complicated. Being notoriously lazy, the good mathematician looks for a short-cut. The hint mentions the geometric series ... recall

^{1}/_{1-x}= 1 + x + x^{2}+ x^{3}+ ...

Substituting *x ^{2}* and multiplying by

*x*gives us:

^{3}^{x3}/_{1-x2}= x^{3}(1 + x^{2}+ x^{4}+ x^{6}+ ... ) =

= x^{3}+ x^{5}+ x^{7}+ x^{9}...

Comparing this with the coefficients of a power series we get:

a_{0}= a_{1}= a_{2}= 0

a_{3}= a_{5}= a_{7}= ... = 1

a_{4}= a_{6}= a_{8}= ... = 0

Since *a _{n} = *, or
equivalently

*f*, we have our answer:

^{(n)}(c) = n! a_{n}

f^{(10)}(0) = 10! a_{10}= 0f^{(11)}(0) = 11! a_{11}= 11! = 39,916,800