Does this function exist/ is it a valid function?

Watch
Callum Scott
Badges: 2
Rep:
?
#1
Report Thread starter 5 years ago
#1
\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 n

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 f(x) such that f(x+y) = f(x)f(y), and then later determine that e^x 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]
0
reply
Euclidean
Badges: 10
Rep:
?
#2
Report 5 years ago
#2
(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 n

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 f(x) such that f(x+y) = f(x)f(y), and then later determine that e^x 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.
0
reply
Interrobang
Badges: 20
Rep:
?
#3
Report 5 years ago
#3
Moved to Maths
1
reply
DFranklin
Badges: 18
Rep:
?
#4
Report 5 years ago
#4
(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 n

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 f(x) such that f(x+y) = f(x)f(y), and then later determine that e^x 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].
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (188)
57.85%
I don't have everything I need (137)
42.15%

Watched Threads

View All