Rate of Convergence for sin(1/x^2) with Maclaurin is undefined?by krittis Tags: maclaurin, rate of convergence, sin(1/x^2) 

#1
Sep1411, 12:41 AM

P: 2

I decided to put my attempt at a solution before the question, because the "solution" is what my question is about.
1. The problem statement, all variables and given/known data Find the rate of convergence for the following as n>infinity: lim [sin(1/n^2)] n>inf Let f(n) = sin(1/n^2) for simplicity. 2. The attempt at a solution I was searching through other forums and resources, and finally found a solution. It said to use the Maclaurin Series (thus x0 = 0), but this would make every term to look like: f(0) + f'(0)*(n^1) + (1/2)*f''(0)*(n^2) + (1/4)*f'''(0)*(n^3) + ... + remainder 3. Relevant equations How can we solve when 1/0 is undefined? For example, in *every* term we have a sin(1/0^x) somewhere. This doesn't work, obviously. This is solvable using the first few terms of the Maclaurin polynomial according to other sources, but I do not understand how. Did I overlook something? Or am I fundamentally misunderstanding the question? Thanks! 



#2
Sep1411, 12:51 AM

P: 2

Oh dear. I just figured it out.
We shouldn't be finding the Maclaurin series for sin(1/x^2) at all. It should be for sin(x) first (with x0=0), then we let x=1/x^2 afterwards. Silly me... 


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