A Taylor series for f(x) = sin x, expanded about x = 2π, will have the form (below) .... ?

Question

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∑ [g(k)] (x - 2π)^2k+1 .
k=0

Given this, g(k) =

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possible solutions....


a. (-1)^k / (2k + 1)!

b. (-1)^k+1 / (2k + 1)!

c. (-1)^k / (2k)!

d. (-1)^k+1 / (2k)!

e. none of these

Answer 1

It should be the same as for sin x expanded about 0, because all the derivatives will be the same at 2π as at 0. The corresponding series for the expansion about 0 has g(k) = (-1)^k / (2k + 1)!, so this is what we want here too.

If you don't want to do it that way, we know the denominator will be (2k+1)! to go with the power of (x-2π), and the numerator will be the (2k+1)'th derivative of sin x at x = 2π; that is, when k is even (first, fifth, etc. derivative) it is cos 2π = 1, and when k is odd it is -cos 2π = -1. So it is (-1)^k. So answer (a) is correct.