Question

in fourier series cos(npi) is (-1)^n and sin(npi)=0,
what is cos(npi/2) and sin(npi/2)?

and when the answer is 0 when n is odd i change n into 2n, while if answer is 0 when n is even i change n into 2n-1.
do i apply the same thing for npi/2?

Answer 1

im always stucked here when im doing the cosine and sine series in the PDE, especially when its stepfunction with pi as interval.

Answer 2

\[\cos(\pi/2n) =
\begin{cases}
(-1)^{n/2} &\;\;\;\;\;\;\text{if }n\text{ is even}\\
0 &\;\;\;\;\;\;\text{if }n\text{ is odd}
\end{cases}\]\[\sin(\pi/2n) =
\begin{cases}
0 &\;\text{if }n\text{ is even}\\
(-1)^{(n-1)/2} &\;\text{if }n\text{ is odd}
\end{cases}\]

Answer 3

oh. so if i want n to represent for all n, change
n into 2n if n is odd gives 0, and change
n into 2n-1 if n is even gives 0?