Equating coefficients of like terms involving a summation

Question

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.

Answer 1

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 $n\geq 3$.