# Which functions $f$ satisfies the equation $f’(x)=\dfrac{1}{x}\left(f(\sqrt{1-x^2})- f(x)\right)$?

What functions $$f$$ satisfy the equation $$f'(x)=\frac{f(\sqrt{1-x^2}) - f(x)}x$$

Could you kindly show how you arrived at the function?

• Thanks for editing it. I didn’t really know how😅 Mar 21, 2020 at 19:05
• This might be help you in future. Mar 21, 2020 at 19:11
• Hint: Note that $$f'(\sqrt{1-x^2})=-\dfrac{x}{\sqrt{1-x^2}}f'(x).$$ Use this to find $f'$ and then integrate. Mar 21, 2020 at 19:15
• Thanks again 🙏🏽 Mar 21, 2020 at 19:17
• When u say f’((1-x^2)^1/2) are u referring to the derivative with respect to x or to (1-x^2)^1/2 Mar 21, 2020 at 21:20

$$f'(x)=\frac{f(\sqrt{1-x^2}) - f(x)}x$$ $$f(x)=g(x^2)$$

$$f'(x)=2xg'(x^2)$$ $$2xg'(x^2)=\frac{g(1-x^2) - g(x^2)}{x}$$ $$2x^2g'(x^2)=g(1-x^2) - g(x^2)$$ $$x^2=s$$ $$2sg'(s)=g(1-s) - g(s)$$ $$Dg(s)=g'(s)$$

$$Ag(s)=g(1-s)$$ $$(2sD-A+1)g=0$$

Operator Analysis

$$\left(2sD-\exp\left(\pi isD\right)\exp(D)+1\right)g=0$$ Baker-Campbell-Hausdorff Formula

This solves $$\ln(\exp(X)\exp(Y))$$ which is usually $$X+Y$$ except when $$XY-YX\neq 0$$

$$\pi isD*D-D*\pi isD=\pi isD^2-\pi i(1+sD)D=\pi isD^2-\pi iD-\pi isD^2=-\pi iD$$

$$\exp(\pi isD)\exp(D)=\exp(\pi i(s-\frac{1}{2})D)$$

$$\left(2sD-\exp\left(\pi i\left(s-\frac{1}{2}\right)D\right)+1\right)g=0$$

There are infinitely many solutions to this but the easiest is $$g=0$$.

• I’ll have to read up the links you gave as I’m a high school student and I got lost from “Operator Analysis”. What I do know though is that g=k satisfies the equation where k is any real number. Are u saying there are still infinite solutions other than g=k Mar 21, 2020 at 22:38
• Yep. Although why is a bit complicated. It partially has to do with how f(x+1)=f(x) has infinitely many solutions. Mar 21, 2020 at 23:20
• f(x) for f(x+1)=f(x) is a periodic function of a periodicity( don’t know if that’s the word) of 1 right? So something like f(x) = sin(2pix) or f(x)=k. How does this relate to my question? If it’s still too complex then it’s okay. I’ve found out what I wanted to. But I’d still like to get it though. Mar 21, 2020 at 23:31