Prove that the Taylor series for f(x) = sin(x) centered at a = π/2 represents sin(x) for all x. In other words, show that limn→[infinity] Rn(x) = 0 for each x, where Rn(x) is the remainder between sin(x) and the nth degree Taylor polynomial for sin(x) centered at a = π/2.

Note that when you derive sin(x), you loop betweeen

cos(x)

-sin(x)

-cos(x)

and again sin(x)

Also, sin(π/2) = 1, -sin(π/2) = 0 and cos(π/2) = -cos(π/2) = 0

Therefore, Rn(x) will have an expression of this type

Note that this is the tail of an alternating series with the generic term being

(x-π/2)^{2k}/(2k)! (note that the series takes even entries thats why the 2k); which limit is equal to 0 for any fixed value of x because we have a factorial dividing. Then, for the Leibnitz criterion, the tail of the series tends to 0, and thus, we can conclude that the Taylor series represents sin(x) for all x.

the answer is 62