help with this calc prob?

Question

3) The series is the value of the Maclaurin series of a function f(x) at a particular point. What function and what point? What is the sum of the series?

π - (π ^3) / 3! + (π ^5) / 5! - ◦◦◦ + ((-1)^n)* (π^(2n+1)) / (2n+1)! + ◦◦◦

Answer 1

You should recognise the series
x - x^3/3! + x^5/5! - ... + (-1)^nx^(2n+1)/(2n+1)! + ...
as the expansion for sin x around x = 0.
Hence the given series is the expansion for sin π, and therefore has sum 0.

You should also recognise the following series:

1 - x^2/2! + x^4/4! - ... + (-1)^nx^(2n)/(2n)! + ...
= cos x (valid for all x)

1 + x + x^2/2! + ... + x^n/n!
= e^x (valid for all x)

x - x^2/2 + x^3/3 - ... + (-1)^(n+1)x^n/n
= ln (1 + x) (valid for |x| < 1).

Answer 2

thats the maclaurin's theorem which says sinx=(x - (x ^3) / 3! + (x ^5) / 5! - ◦◦◦ + ((-1)^n)* (x^(2n+1)) / (2n+1)! + ◦◦◦)
so your answer is sin(pi)=0