# On periodic functions like $\sin$ and $\cos$

A problem in Apostol's Calculus [Tom M. Apostol. Calculus, Volume 1, Second Edition (Wiley, 1967). Section 2.19 Exercises, Problem 21, page 125.]

Suppose the existence of a function $$f$$ with the properties that $$f$$ is odd, $$f'$$ is even, $$f(5)=7$$, and $$f'(x)=f(x+5)$$. This led me to modify this a bit and study it.

Suppose $$a$$ is real, $$f$$ is infinitely differentiable everywhere on $$\mathbb{R}, f$$ is an odd function, $$f(a)=1$$, and $$f'(x)=f(x+a)$$. It is easy to prove from the definition of the derivative that $$f'$$ is even. If $$a=0$$, there is no solution, since if there were, $$f(0)=-f(0)$$, so $$f(0)=0$$, but $$f(0)=1$$ by assumption.

In general, note that: $$f'(-x)=f(a-x)=f'(x)=f(x+a)$$ $$=-f(x-a)=-f(-x-a)$$ We also have: $$f(x+2a)=-f(x) \space \text{and }f(2a)=0$$ $$\text{and }f(x+4a)=-f(x+2a)=f(x)$$ so $$f$$ is periodic of period $$4a$$. $$f'$$ is the same as $$f$$ shifted $$|a|$$ units to the left or right, so $$f'(x+2a)=-f'(x)$$ and $$f'$$ is periodic of period $$4a$$. $$f'(0)=f(a)=1, f'(a)=f(2a)=0,$$ $$\text{ and }f'(2a)=-f'(0)=-1$$ Since $$f'(x) = f(x+a) = f(x-3a)$$, we can assume WLOG that $$a > 0$$ (or that $$a < 0$$). If $$a=\pi/2+2\pi n, f(x)=\sin x$$ is an obvious solution.

Now for my questions, none of which I can answer:

1. If $$a=\pi/2+2\pi n$$, is there any solution besides $$\sin x$$?

2. If $$a \ne \pi/2+2\pi n$$, is there a solution? Is it unique? Is there a simple formula for it?

The questions $$f'(x) = f(x-1)$$ and $$f'(x) = f(x+1)$$ may be pertinent, but so far they haven't helped me much. I have tried to apply Rolle's Theorem to prove uniqueness without success.

If $$f'(x) = f(x+a)$$ and $$f(x+4a) = f(x)$$, then $$f''''(x) = f(x+4a) = f(x)$$. The general solution of the differential equation $$f''''(x) = f(x)$$ is $$f(x) = c_1 e^x + c_2 e^{-x} + c_3 \cos(x) + c_4 \sin(x)$$ For $$f$$ to be periodic, we must have $$c_1 = c_2 = 0$$. For $$f$$ to be odd, $$c_3 = 0$$. So the only solutions can be $$c_4 \sin(x)$$, which is periodic with period $$2\pi$$. The only possibilities for $$a$$ are then $$\pi/2 + 2 \pi n$$, and if you want $$f(a) = 1$$ then $$c_4 = 1$$ and $$f(x) = \sin(x)$$.
• Thanks. Stupid of me not to realize this. It would be a little easier to consider $f''(x)=f'(x+a)=f(x+2a)=-f(x)$ with solutions $c_1 \cos x + c_2 \sin x$. This also means that Apostol's problem concerns a function that does not exist! Commented Mar 3, 2023 at 14:45