Taylor Series: Simple Homework Statement, Find 1st & 2nd Terms

[clandarkfire]

Homework Statement

"Determine the first two non-vanishing terms in the Taylor series of $$\frac{1-\cos(x)}{x^2}$$ about x = 0 using the definition of the Taylor series (i.e. compute the derivatives of the function)."

So I know how compute the Taylor series about x=0; it involves finding f(0), f'(0), f''(0), etc. But In this particular case, it seems that f(0) and all the derivatives are undefined at x=0. This presents a problem.

I know that I can just replace cos(x) with it's Taylor series, which would make this easy, but the question specifically tells me not to..

What am I missing?

 

[STEMucator]

Homework Statement

"Determine the first two non-vanishing terms in the Taylor series of $$\frac{1-\cos(x)}{x^2}$$ about x = 0 using the definition of the Taylor series (i.e. compute the derivatives of the function)."

So I know how compute the Taylor series about x=0; it involves finding f(0), f'(0), f''(0), etc. But In this particular case, it seems that f(0) and all the derivatives are undefined at x=0. This presents a problem.

I know that I can just replace cos(x) with it's Taylor series, which would make this easy, but the question specifically tells me not to..

What am I missing?

I believe the question would want you to use the Taylor expansion of cos(x). The question doesn't say not to anywhere.

 

[clandarkfire]

I don't think so; it says to compute the derivatives of the function. Also, part b of the question asks me to use the Taylor expansion of cos(x) and compare it with the result from this part.

 

[STEMucator]

I don't think so; it says to compute the derivatives of the function. Also, part b of the question asks me to use the Taylor expansion of cos(x) and compare it with the result from this part.

Ah that's an interesting approach then. Start taking a few derivatives and rather than considering what's happening precisely at $x=0$, use limits to your advantage ( You'll notice a pattern by the 3rd and 5th derivatives ).