0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $n\geq 3$.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.