Derivative of a periodic function with respect to a parameter

Question

Assume we have a real periodic function $f(x)$ with some fundamental period $P$. Let us introduce a coordinate transformation $x=ay$, where $a \in \mathbb{R}$ and $a \neq 0$. Then, on the one hand,

$$ \frac{\partial f}{\partial a} = \frac{\partial f(ay)}{\partial (ay)} \frac{\partial(ay)}{\partial a} = \frac{\partial f(x)}{\partial x} y = a^{-1} \frac{\partial f(x)}{\partial x} x $$

which is clearly not periodic (due to the $x$ term). On the other hand, I expect $\partial_a f$ to also be periodic in $x$ (see this question for example). Perhaps the latter statement is invalid in case of multivariable functions? What kind of requirement should one impose on coordinate transformations to make $\partial_a f$ periodic? Is there a general statement that is valid for any kind of coordinate transformation that depends on a parameter (i.e. $y=g_a (x)$)?

Answer 1

Problem

Suppose we have a real periodic function $f(x)$ with fundamental period $P$. Then when you start writing $\dfrac{\partial f}{\partial a}$, you're not really working with $f$ anymore, since you're writing a partial derivative.

Instead, we have a new function of two variables $g(a,y)=f(ay)$. Then certainly, if $x=bz$, then we have (just as you calculated): $$\left.\dfrac{\partial g}{\partial a}\right|_{(a,y)=(b,z)}=f'(bz)*\left.\dfrac{\partial (ay)}{\partial a}\right|_{(a,y)=(b,z)}=f'(bz)*z=\dfrac{xf'(x)}{b}\text{.}$$

A short answer would be "you wouldn't expect $g(a,y)$ to be periodic".

---

Solution

That said, if you want to say something like $g(a,y)=f(ay)$ is "periodic in $ay$", then you need to calculate with $a$, $y$, and $ay$ a bit more carefully.

Specifically, if $ay$ increases by $P$, and $y$ stays constant (as happens in a calculation of $\dfrac{\partial g}{\partial a}$), then we have something like $(a+\Delta a)y=ay+P$, so that $\Delta a=\dfrac{P}{y}$.

This means that what we would expect of $\dfrac{\partial g}{\partial a}$ is that if we increase $x=ay$ by $P$ and increase $a$ by $\dfrac{P}{y}=\dfrac{P}{x/a}=\dfrac{aP}{x}$ then we should get the same result. Indeed, this is the case: $$\dfrac{(x+P)f'(x+P)}{a+\dfrac{aP}{x}}=\dfrac{(x+P)f'(x)}{a+\dfrac{aP}{x}}=\dfrac{x(x+P)f'(x)}{ax+aP}=\dfrac{xf'(x)}{a}\checkmark$$

---

Generalizing

Is there a general statement that is valid for any kind of coordinate transformation

Well, if you have $g(a,y)=f(h(a,y))$, and $h(a,y)$ increases by $P$ and $y$ stays constant, then you'd want to try to solve $h(a+\Delta a,y)=h(a,y)+P$ for $\Delta a$ as a function of $y$ (and $P$) alone (necessary for $(\star)$ below), which may not be possible (it could necessarily depend on $a$).

But if you can do that, then increasing $x$ by $P$ and $a$ by $\Delta a$ will work just as before. We have $$\left.\dfrac{\partial g}{\partial a}\right|_{(a,y)=(b,z)}=f'(h(b,z))*\left.\dfrac{\partial h}{\partial a}\right|_{(a,y)=(b,z)}=f'(x)*\left.\dfrac{\partial h}{\partial a}\right|_{(a,y)=(b,z)}\text{.}$$

And we can then calculate: $$f'(x+P)*\left.\dfrac{\partial h}{\partial a}\right|_{(a,y)=(b+\Delta b,z)}$$ $$=f'(x)*\lim_{t\to0}\dfrac{h\left(b+\Delta b+t,z\right)-h\left(b+\Delta b,z\right)}{t}$$ $$=f'(x)*\lim_{t\to0}\dfrac{\left(P+h\left(b+t,z\right)\right)-\left(P+h\left(b,z\right)\right)}{t}\tag{$\star$}$$ $$=f'(x)*\lim_{t\to0}\dfrac{h\left(b+t,z\right)-h\left(b,z\right)}{t}=f'(x)*\left.\dfrac{\partial h}{\partial a}\right|_{(a,y)=(b,z)}\checkmark$$