Optimal Degree for Approximating Cosine with Taylor Series

hadroneater
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Homework Statement


What degree Taylor Polynomial around a = 0(MacLaurin) is needed to approximate cos(0.25) to 5 decimals of accuracy?


Homework Equations


taylor series...to complicated to type out here

remainder of nth degree taylor polynomial = |R(x)| <= M/(n+1)! * |x - a|^(n+1)
where a = 0 in this case
and
M >= |f^(n+1)(t)|



The Attempt at a Solution


I don't really get this question at all. I know that |R(0.25)| = 0.00001 <= M/(n+1)! * |x - a|^(n+1)
But how do I get M when |f^(n+1)(t)| is unknown? I don't even know what |f^(n+1)(t)| means!
 
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hadroneater said:

The Attempt at a Solution


I don't really get this question at all. I know that |R(0.25)| = 0.00001 <= M/(n+1)! * |x - a|^(n+1)
But how do I get M when |f^(n+1)(t)| is unknown? I don't even know what |f^(n+1)(t)| means!

f(n+1)(t) is the n+1:th derivative of f(t). So is your plan to find the lowest upper bound for Mn? It might be easier (and more likely to be correct too) if you just calculated enough terms from the series until you have the desired accuracy.
 
But we aren't marked on that...it has to be through the remainder method.

Anyways, I know what f^(n+1)(t). I just don't know what to plug in for t. And after that, doesn't it just become a plug-and-check game for n until I get less than 0.00001?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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