Question

Use taylor series to evaluate `f^11 (0)` for `f(x) = x^3 cos(x^2)` ?

Answer 1

The answer is `990`.

Explanation:

We can rewrite as follows:

`f(x) = x^3(1 - x^2/2 + x^4/(4!)- x^6/(6!) + x^8/(8!) - x^10/(10!) + ...)`

`f(x) = x^3 - x^5/2 + x^7/(4!) - x^9/(6!) + x^11/(8!) - x^13/(10!)`

Now note that we don't need any further terms because when we compute the 11th derivative all the terms except that of original degree `11` will cancel to `0`.

`f^11(0) = 11!/(8!) x^0 = (11!)/(8!) = 990`

Hopefully this helps!