maclaurin polynomial help

[treefairy2]

I'm asked to prove that the nth maclaurin polynomial for a function f(x^2) is equal to the 2*nth maclaurin polynomial for another function g(x). I have tried to write out the equation with for instance n=2 which gives me:

(g''''0/4!)*x^4 + (g'''0/3!)*x^3 + (g''0/2!)*x^2 + g'0*x + g(0) = (f''0/2!)*x^4 + f'0*x^2 + f(0)

Which is obviously not right. I fail to see how else I might go about proving this?

Thank you in advance for any help you might be able to provide,

 

[HallsofIvy]

Suppose the MacLaurin polynomial for f(x) is \(\displaystyle \sum_{i=0}^n f^{(i)}x^i\). The MacLaurin polynomial for \(\displaystyle f(x^2)\) is \(\displaystyle \sum_{i= 0}^n f^{(i)}(0)x^{2i}\)

You want to show that there is a function, g(x), such that the "2i"th derivative, at x= 0, is the ith derivative of f and all odd degree derivatives are 0 at x= 0

 

[treefairy2]

Suppose the MacLaurin polynomial for f(x) is \(\displaystyle \sum_{i=0}^n f^{(i)}x^i\). The MacLaurin polynomial for \(\displaystyle f(x^2)\) is \(\displaystyle \sum_{i= 0}^n f^{(i)}(0)x^{2i}\)

You want to show that there is a function, g(x), such that the "2i"th derivative, at x= 0, is the ith derivative of f and all odd degree derivatives are 0 at x= 0

Thank you.

I understand that I can do this with specific functions, for instance sin(x^2) but will it work generally for all maclaurin polynomials so that I can conclude that Tn g(x^m)=Tmn f(x) holds?