# Equating coefficients of like terms involving a summation

I am working through this problem, and looking at the answers provided below, I can't seem to understand how the answers are arrived at.

$$u(x,1)=A_0+\sum_{n=1}^\infty A_n\sinh(n)\cos(nx)=1+\cos(2x).$$ Equating coefficients of like terms, $$A_n=\begin{cases} 1, & n=0,\\ 1/\sinh(2), & n=2,\\ 0 & \text{otherwise}. \end{cases}$$

For instance, for when $$n = 0$$ and equating coefficients of $$A_0$$:

I think that the summation on the left hand side is "gone" as the summation starts at $$n = 1$$, so to equate coefficients of $$A_0$$ it would be: $$1 = 0$$ as the RHS does not have any terms involving $$A_0$$. I can see that the way I am thinking is wrong, because how can $$1 = 0$$? But I can not seem to find any other way of equating these coefficients. I am also having trouble finding how they arrived at the given answer for $$n =2$$.

I appreciate your time and help,

thank you.

You are equating coefficients of $$\cos(nx)$$ (independent of $$x$$) in $$A_0+\sum_{n=1}^\infty A_n\sinh(n)\cos(nx) = 1+\cos(2x)$$
• When $$n=0$$, $$\cos(0x)\equiv1$$, so this is just the constant term on both sides, i.e., $$A_0=1$$.
• When $$n=1$$, the coefficients of $$\cos(x)$$ in the LHS is $$A_1\sinh(1)$$, and RHS has no $$\cos(x)$$ term, so $$A_1\sinh(1)=0$$.
• When $$n=2$$, the coefficients of $$\cos(2x)$$ in the LHS is $$A_2\sinh(2)$$, and RHS gives $$1$$, so $$A_2\sinh(2)=1$$.
• Similarly $$A_n\sinh(n)=0$$ for $$n\geq 3$$.