Question

I found this ridiculous formula, anyone want to play around with it with me?

Answer 1

\[\scriptsize \frac{d}{dx}(f(x)) = \frac{ g(h(x))h(g(x))\frac{d}{dx}[\int\limits_{g(x)}^{h(x)}f(x)dx] + f(g(x))g(h(x)) \frac{d}{dx}[\int\limits_{f(x)}^{g(x)}h(x)dx] + f(h(x))h(g(x))\frac{d}{dx}[\int\limits_{h(x)}^{f(x)}g(x)dx]}{f(h(x))h(g(x))g(f(x))-f(g(x))g(h(x))h(f(x))}\]

Answer 2

If you can't read that, then I'll show you my shorthand notation where I represent composition of functions as subscripts: \[\Large f(g(x))=f_g\]

\[ \frac{d}{dx}(f(x)) = \frac{ g_h h_g \frac{d}{dx}[\int\limits_{g(x)}^{h(x)}f(x)dx] + f_g g_h \frac{d}{dx}[\int\limits_{f(x)}^{g(x)}h(x)dx] + g_h h_g \frac{d}{dx}[\int\limits_{h(x)}^{f(x)}g(x)dx]}{f_h h_g g_f - f_g g_h h_f}\]

Answer 3

Compactifying further, let's call thse separate derivatives of integrals A, B, and C: \[\Large \frac{d}{dx}(f(x)) = \frac{ g_h h_g A+ f_g g_h B + g_h h_g C}{f_h h_g g_f - f_g g_h h_f}\] and now it is easier to play with... uhhh... ok this is probably too scary for us mortals, so I'll just let it go I guess. lol But it looks cool.

Answer 4

\[
A = f_h\frac{dh}{dx}-f_g\frac{dg}{dx}
\]

Answer 5

True, that's how I derived this equation actually.

Answer 6

Does it simplify or something?

Answer 7

Well, it's just 3 equations that look just like that. Honestly it just looks like there is a hint of a determinant here. I don't know, just thought I'd type it out. Right now I'm trying to see what it implies if you have f(g(x))=h(x), g(h(x))=f(x), and h(f(x))=g(x).

Answer 8

\[
f(g(x))=h(x),\ g(h(x))=f(x)\implies g(f(g(x))) = f(x)
\]

Answer 9

\[
g\circ f \circ g = f
\]

Answer 10

Is it possible to have functions where compositions are commutative?

Answer 11

The identity function is commutative, but they often are not.

Answer 12

The trivial solution is that \(g\) is the identity function.

Answer 13

f(g(h(x)))=h(h(x))=f(f(x))

Answer 14

But that leads to them all having to be the identity function.

Answer 15

Can we somehow suspend belief that these are "regular" functions somehow and keep going?

Answer 16

I don't know what you mean by regular.

Answer 17

If you are going to integrate or differentiate them, they need to be continuous on particular intervals.

Answer 18

Another solution is if \(f(x) = 0\).

Answer 19

So \(f(x)=x\) and \(f(x)=0\) work.

Answer 20

What if we invent some kind of algebraic element that isn't a number but changes depending on what it's multiplied by or something.

Answer 21

Yeah that won't really work, it's kind of nonsense. Oh well, just wandering around.

Answer 22

You want only the multiplication operation?

Answer 23

No, just playing around. Not really looking for anything in particular.

Answer 24

If you let \(g(x) = x+a\) then: \[
f(x+a) +a = f(x)
\]

Answer 25

So it's almost a periodic function.

Answer 26

\(g(x) = ax\) leads to: \[
af(ax) =f(x)
\]

Answer 27

I like the periodic thing, we could just define it to be a function on mod 10 or something maybe

Answer 28

You can, but you have to drop the integration differentiation or extend the function to be continuous.

Answer 29

Or not exactly, but like just defined to have a periodic range that's continuous.

Answer 30

If you let \(g(x) = x+a\) then: \[
f(x+a) +a = f(x)
\]Differentiate it and: \[
f'(x+a) = f'(x)
\]So it's derivative is clearly periodic.

Answer 31

It might not be an elementary function though.

Answer 32

For the initial value: \[
f(a) +a = f(0)\\
f(0)+a = f(-a)
\]

Answer 33

Actually, now that I think about it, the way to solve this is to use recursion.

Answer 34

\[
f(x) = f(x+a)+a = f(x+2a)+2a = f(x+na)+na
\]Let \(x=-na\) and: \[
f(-na) = f_0+na
\]