0
$\begingroup$

Suppose $$f(x) =\frac{a_0}{2}+\sum_{k=1}^\infty a_k\cos(kx)+b_k\sin(kx) $$

Then what is $$S(x)=\frac{a_0}{2}+\sum_{k=1}^\infty a_k\cos(2kx)+b_k\sin(2kx)$$ in terms of $f(x)$. I tried writing down things in terms of the inner products $$\begin{aligned}a_k&=\left\langle f(x),\cos(kx)\right\rangle=\left\langle S(x),\cos(2kx)\right\rangle, \\ b_k&=\left\langle f(x),\sin(kx)\right\rangle=\left\langle S(x),\sin(2kx)\right\rangle\end{aligned}$$ and this did not help. I also wrote down the integrals$$b_k=\frac{1}{\pi}\int_{-\pi}^\pi S(x)\sin(2kx)\, dx=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(kx)\, dx$$ then equating integrands and using the double angle formula for sin you get $$S(x)=\frac{f(x)}{2\cos(kx)}$$ but this doesn't seem to make sense. I also separated the even and odd coefficients of $f$ and that did not help me either. The answer is not $f(2x)$.

$\endgroup$
3
  • 1
    $\begingroup$ $S(x) = f(2\cdot x)$ and I am not convinced there exists a better expression. Also you can't equate integrands if the integrals are equal. $\endgroup$ Feb 17 '19 at 11:02
  • $\begingroup$ "The answer is not $f(2x)$." Really? Why not? $\endgroup$
    – user856
    Feb 17 '19 at 11:23
  • $\begingroup$ @Rahul that's just what the book says ¯_(ツ)_/¯ doesn't make sense to me either $\endgroup$
    – N Dizzle
    Feb 17 '19 at 17:11
1
$\begingroup$

If $f(x) = \sum_{k \in \mathbb Z} c_k e^{ikx}$, then $f(2x) = \sum_{k \in \mathbb Z} c_k e^{i2kx}$. If we define $d_k$ to be the coefficients of $f(2x)$, $$ f(2x) =: \sum_{k \in \mathbb Z} d_k e^{ikx},$$ we find that $c_k \neq d_k$, but instead $d_{2k} = c_k$, and $d_{2k-1}=0$. That is, $$ d_k = \begin{cases} c_{k/2} & k \ \text{even,} \\ 0 & k \ \text{odd.} \end{cases}$$ Hence (as mentioned in comments) you cannot equate the coefficients like that. The equality $$ \sum_{k \in \mathbb Z} c_k e^{i2kx} = \sum_{k \in \mathbb Z} d_k e^{ikx} $$ is valid (by its very definition) but one has to be more careful when equating the coefficients.

$\endgroup$

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.