Taylor Polynomial of Polynomials

[robbiedoes]

In an exercise the following function f(x) = x^4 has to be written in terms of a 3th order Taylor polynomial around the point a = 2.

From my understanding this couldn't be possible, because in order to form a Taylor series, f(x) has to be differentiable (n+1)-times or in other words has to be infinitely differentiable. Now x^4 is differentiable 4 times, this would fit within the approximation of the third order, but is this true?

[HallsofIvy]

Where did you get the idea that polynomials were NOT infinitely differentiable?
Any polynomial IS infinitely differentible! The "n+1" derivative of a n degree polynomial certainly exist- it is 0.

For this problem, one way to find the "Taylor
series around the point a = 2" is to use the definition:
f(x)= x^4 so f(a)= a^4.
f'(x)= 4x^3 so f'(a)= 4a^3
f"(x)= 12x^2 so f"(a)= 12a^2
f"'(x)= 24x so f"'(a)= 24a
f""(x)= 24 so f""(a)= 24
and all further derivatives are 0.

The complete (infinite) Taylor series for x^4 about x= a is a^4+ 4a^3(x-a)+ (12a^2/2)(x- a)^2+ 24a/(3!)(x- a)^3+ (24/4!)(x- a)^4+ 0(x-a)^5+ ...= a^4+ 4a^3(x-a)+ 6a^2(x- a)^2+ 4a(x- a)^3+ (x- a)^4+ 0(x-a)^5+ ...

You can get the "3rd degree Taylor polynomial" by just dropping the terms of degree 4 or higher: 4a^3(x-a)+ 6a^2(x- a)^2+ 4a(x- a)^3.

Since the Taylors series for a polynomial is just that polynomial "shifted", we can also get this by letting y= x-a so that a= y+a. Then x^4= (y+a)^4= y^4+ 4ay^3+ 6a^2y^2+ 4a^3y+ a^4 (by the binomial theorem) so, replacing y by x- a again, x^4= (x-a)^4+ 4a(x-a)^3+ 6a^2(x-a)^2+ 4a^3(x-a)+ a^4. That is exactly what we got before. And the "third degree Taylor polynomial about x= a" is just that with the fourth degree term dropped.

[HallsofIvy]

I accidently dropped the constant term- the third degree Taylor polynomia is

a^4+ 4a^3(x-a)+ 6a^2(x- a)^2+ 4a(x- a)^3

[robbiedoes]

Thanks for your help and your notion about polynomials that are actually infinitely differentiable. It helped me a lot!