Does this function exist/ is it a valid function?

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$\frac{d}{dx}\left[ f(x)^n \right] = f(nx)$

Basically I noticed that differentiating $\sin^2(x)$ gives $2\sin x\cos x \equiv \sin(2x)$ and wondered whether there is a function such that this is true for all values of

First off:
$\frac{d}{dx}\left[ f(x)^n \right] = nf'(x)f(x)^{n-1} = f(nx)$

$\frac{d}{dx}\left[ f(x)^{n+1} \right] = (n+1)f'(x)f(x)^n = f( nx+x )$

$\frac{f(nx+x)}{f(nx)} = \frac{(n+1)f'(x)f(x)^n}{nf'(x)f( x)^{n-1}} = \frac{n+1}{n}f(x)$

$f(x) = \frac{n}{n+1}\frac{f(nx+x)}{f(nx )}$

I honestly have no idea what I'm doing, but was wondering whether there was actually a function that satisfies the above equation; or how you'd even *begin* to formulate one that does (if it is, in fact, possible)

If someone could teach me a thing or two on how mathematicians go about inventing functions then I'd appreciate it a lot!

For example: How can you initially come up with a function

such that

, and then later determine that

is a solution?!

Random Equations:

Spoiler:

Show

$\frac{d}{dx}\left[ f(x)^0 \right] = f(0) = 0$

$\frac{d}{dx}\left[ f(-x)^n \right] = -nf'(-x)f(-x)^{n-1}$
$\frac{d}{dx}\left[ f(x)^{-n} \right] = -nf'(x)f(x)^{-n-1} =-f(x)^{-2n}\left( nf'(x)f(x)^{n-1}\right)$

$\frac{d}{dx}\left[ f(x)^{-n}\right] = -f(x)^{-2n} \frac{d}{dx}\left[ f(x)^n \right]$

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Euclidean

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(Original post by Callum Scott)
$\frac{d}{dx}\left[ f(x)^n \right] = f(nx)$

Basically I noticed that differentiating $\sin^2(x)$ gives $2\sin x\cos x \equiv \sin(2x)$ and wondered whether there is a function such that this is true for all values of

First off:
$\frac{d}{dx}\left[ f(x)^n \right] = nf'(x)f(x)^{n-1} = f(nx)$

$\frac{d}{dx}\left[ f(x)^{n+1} \right] = (n+1)f'(x)f(x)^n = f( nx+x )$

$\frac{f(nx+x)}{f(nx)} = \frac{(n+1)f'(x)f(x)^n}{nf'(x)f( x)^{n-1}} = \frac{n+1}{n}f(x)$

$f(x) = \frac{n}{n+1}\frac{f(nx+x)}{f(nx )}$

I honestly have no idea what I'm doing, but was wondering whether there was actually a function that satisfies the above equation; or how you'd even *begin* to formulate one that does (if it is, in fact, possible)

If someone could teach me a thing or two on how mathematicians go about inventing functions then I'd appreciate it a lot!

For example: How can you initially come up with a function

such that

, and then later determine that

is a solution?!

Random Equations:

Spoiler:

Show

$\frac{d}{dx}\left[ f(x)^0 \right] = f(0) = 0$

$\frac{d}{dx}\left[ f(-x)^n \right] = -nf'(-x)f(-x)^{n-1}$
$\frac{d}{dx}\left[ f(x)^{-n} \right] = -nf'(x)f(x)^{-n-1} =-f(x)^{-2n}\left( nf'(x)f(x)^{n-1}\right)$

$\frac{d}{dx}\left[ f(x)^{-n}\right] = -f(x)^{-2n} \frac{d}{dx}\left[ f(x)^n \right]$

This is actually a branch of mathematics called functional equations which constitutes a lot of Olympiad questions.

https://en.wikipedia.org/wiki/Functional_equation

In short, you solve functional equations by making a relevant substitution (commonly let x,y = 0) and working from there. However, there are probably more knowledgable users than me as I've only had a dabble in functional equations.

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Interrobang

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Moved to Maths

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DFranklin

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(Original post by Callum Scott)
$\frac{d}{dx}\left[ f(x)^n \right] = f(nx)$

Basically I noticed that differentiating $\sin^2(x)$ gives $2\sin x\cos x \equiv \sin(2x)$ and wondered whether there is a function such that this is true for all values of

First off:
$\frac{d}{dx}\left[ f(x)^n \right] = nf'(x)f(x)^{n-1} = f(nx)$

$\frac{d}{dx}\left[ f(x)^{n+1} \right] = (n+1)f'(x)f(x)^n = f( nx+x )$

$\frac{f(nx+x)}{f(nx)} = \frac{(n+1)f'(x)f(x)^n}{nf'(x)f( x)^{n-1}} = \frac{n+1}{n}f(x)$

$f(x) = \frac{n}{n+1}\frac{f(nx+x)}{f(nx )}$

f(x) = 0 obviously works (but isn't very interesting). I am fairly sure this is the only solution:

Spoiler:

Show

A slight rearrangement gives us:

$f(nx+x) = \dfrac{n+1}{n}f(x)f(nx)$ , Set x = 1: n = 1 gives us f(2) = 2f(1)^2, n=2 gives us f(3) = 3f(1)^3, n = 3 gives us f(4) = 3 f(1)^4.
Set x = 2: n = 1 gives us f(4) = 2 f(2)^2, and from the x=1 results we know this = 4 f(1)^4. So we've found f(4) = 3 f(1)^4 and f(4) = 4 f(1)^4 and so f(1) = 0.

Repeating the same argument but with x = y and x = 2y (instead of 1 and 2) will show f(y) = 0 for all y.

If someone could teach me a thing or two on how mathematicians go about inventing functions then I'd appreciate it a lot!

For example: How can you initially come up with a function

such that

, and then later determine that

is a solution?!

Very similar to the above: Setting x = 0 and assuming there's a y for which f(y) isn't 0 lets you deduce f(0) = 1. Setting x = 1 and y = 1, 2, 3, .. n. successively lets you deduce f(n) = f(1)^n. Then setting x = m/n and y = m/n, 2m/n, ..., m successively lets you deduce f(m/n)^n = f(m) = f(1)^n and so f(m/n) = f(1)^(m/n). Writing E = f(1), we've now shown f(q) = E^q for all rational q.

Couple of points to note:

Fairly obvious: There's nothing that requires E to equal e. It could be any (+ve real) number (requiring +ve real because we took nth roots in the arguments above and that's not a good thing to do with complex numbers).

Less obvious (actually quite deep): This is not enough to deduce f(x) = E^x for all real x. In fact, it doesn't have to be true that f(x) = E^x for all real x. We could have $f(\sqrt{2}) = F^{\sqrt{2}}$ where F is different from E, and the f(x+y)=f(x)f(y) relation will not ever let us find a contraiction.

[If we also require f to be continuous, then knowing f(q) = E^q for rational q is sufficient to deduce f(x) = E^x for all real x].

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