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Simple Taylor series

  • #1

Homework Statement


"Determine the first two non-vanishing terms in the Taylor series of [tex]\frac{1-\cos(x)}{x^2}[/tex] 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?
 

Answers and Replies

  • #2
Zondrina
Homework Helper
2,065
136

Homework Statement


"Determine the first two non-vanishing terms in the Taylor series of [tex]\frac{1-\cos(x)}{x^2}[/tex] 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.
 
  • #3
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.
 
  • #4
Zondrina
Homework Helper
2,065
136
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 ).
 

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