Use taylor series to evaluate #f^11 (0)# for #f(x) = x^3 cos(x^2)# ?
1 Answer
Aug 7, 2018
The answer is
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
#f^11(0) = 11!/(8!) x^0 = (11!)/(8!) = 990#
Hopefully this helps!
