# What is the function you get when you double the arguments of sin and cos in the Fourier Series of another function?

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 $$f(2x)$$.

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

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.