Equations Like f(2x)=4f(x)

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exfret
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Equations Like f(2x)=4f(x)

Postby exfret » Fri May 01, 2015 10:33 pm UTC

I was wondering: Is there a studied way to solve equations for an unknown function when the argument of the function isn't always 'x'. Some examples of 'solvable' equations of the form I'm talking about:

1. f(x) + f(1/x) = 1
2. f(2x) = 4f(x)
3. f(x^2) = x + 1
4. f(f(sqrt(x-1))) = f(sqrt(x))

Possible answers, if you want them:

Spoiler:
1. 1/(x^2+1)
2. f(x) = x^2
3. f(x) = ln(ln(x))/ln(2)
4. f(x) = sqrt(x^2 + 1)


Also, here are some equations that I was curious about (I don't know of their solutions or even if they exist):
1. f'(x) = f(f(x))
2. f(e^x) = f(x) + 1
3. f(x^2) + f(2x) + f(1) = 0
4. f(x) + f(x) = f(3x)

I was just wondering because I thought it would be interesting to find functions which followed certain properties, such as when their argument is raised to the power of two, the function returns a value equivalent to twice its current value plus one or something of the like. Also, I apologize for I have little experience in abstract algebra or any other 'higher' subject areas, so there could be some things I might overlook (also for my lack of latex). Hopefully y'all can help. :)

>-)
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Re: Equations Like f(2x)=4f(x)

Postby >-) » Fri May 01, 2015 11:52 pm UTC

f(x) = 0 seems to work for a surprising number of these.

I don't know if there's any general technique for solving these, but I suspect a lot of these have no solution.

For example, take f(x^2)+f(2x)+f(1) = 0
We know f(1) is constant, so f(x^2)+f(2x) is also constant.
this seems very unlikely for any nontrivial functions, though i can't pinpoint any reason

edit: since f(x^2)+f(2x) is constant, you know that
d/dx f(x^2) = - d/dx f(2x)
2x f'(x^2) = -2f'(2x)
-x = f(2x)/f(x^2)

if you let f(x) be 1/x, you get on the rhs
(1/(2x))/(1/x^2)
= x^2/(2x)
= x/2

which is, interestingly, a linear factor away from -x
this might actually lead somewhere.

edit 2: related math overflow question https://mathoverflow.net/questions/17614/solving-ffx-gx
edit 3: this page might help https://en.wikipedia.org/wiki/Functiona ... _equations
http://www.imomath.com/index.php?options=339
Last edited by >-) on Sat May 02, 2015 12:08 am UTC, edited 3 times in total.

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notzeb
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Re: Equations Like f(2x)=4f(x)

Postby notzeb » Sat May 02, 2015 12:06 am UTC

These sorts of equations are called functional equations. They come up often on math competitions (they are the most fun type of competition problem, in my opinion). One of my friends might have coauthered a book about them.
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Tirian
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Re: Equations Like f(2x)=4f(x)

Postby Tirian » Sat May 02, 2015 7:42 pm UTC

exfret wrote:1. f'(x) = f(f(x))


These are called idempotent functions, and they're pretty important in linear algebra.

Let y be a point in the range of f. Then there is an x in the domain such that f(x)=y. Then the idempotent condition means that y = f(y). Since y was an arbitrary member of the range, it follows that f restricted to its range is the identity mapping. The interested reader can show that all such mappings satisfy the idempotent condition, so that is a complete categorization.

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jaap
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Re: Equations Like f(2x)=4f(x)

Postby jaap » Sat May 02, 2015 7:56 pm UTC

Tirian wrote:
exfret wrote:1. f'(x) = f(f(x))


These are called idempotent functions, and they're pretty important in linear algebra.

Let y be a point in the range of f. Then there is an x in the domain such that f(x)=y. Then the idempotent condition means that y = f(y). Since y was an arbitrary member of the range, it follows that f restricted to its range is the identity mapping. The interested reader can show that all such mappings satisfy the idempotent condition, so that is a complete categorization.

I think you overlooked the apostrophe in the statement: f '(x) = f(f(x))

Tirian
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Re: Equations Like f(2x)=4f(x)

Postby Tirian » Sat May 02, 2015 8:05 pm UTC

I didn't ignore it, I spent quite a lot of time convincing myself that it was either a speck on my monitor or a weirdness of the font.

Carry on then.

pichutarius
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Joined: Sun May 03, 2015 3:06 am UTC

Re: Equations Like f(2x)=4f(x)

Postby pichutarius » Sun May 03, 2015 8:21 am UTC

Non exhaustive solutions:

1. f'[x] = ff[x]
let n = 1/2 + i Sqrt[3]/2 , a = n^(1/n)
f[x] = a x^n
*However this doesnt hold for whole complex plane, only a subset of it.
Spoiler:
To see the domain of f[x] on a complex plane, use wolframalpha dot com and enter this:
RegionPlot[pi+3ArcTan[x,y]+3Sqrt[3]Log[Sqrt[x^2+y^2]]<6pi,{x,-130,30},{y,-100,60}]


2. f[e^x] = f[x] + 1
Define e^e^e^...^e = Tower[x] where x is the number of layers of e^e^e...^e.
Define InverseTower[x] as the inverse function of Tower[x].
f[x] = InverseTower[x]

3. f[x^2] + f[2x] + f[1] = 0
Define log_2[x] = log[x]/log[2]
f[x] = Cos[pi log_2[-2 + log_2[x]]] when x>4
f[x] = 0 when x==1
f[x] is left undefined elsewhere

4. f[x] + f[x] = f[3x]
f[x] = x^(log[2]/log[3])

exfret
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Re: Equations Like f(2x)=4f(x)

Postby exfret » Sun May 03, 2015 2:11 pm UTC

Well, I guess I should have made the apostrophe more clear, but indempotent functions do sound pretty important. I was wondering, though, if there could exist a function such that when composed with itself it returns it's derivative. But I was just curious; such a function's existence could mean absolutely nothing. That makes me wonder if there's a function such that it's derivative is it's inverse, too. That would be pretty cool. I also feel pretty stupid for not knowing that the types of equations I was looking at were called "functional equations". It seems like I should have found the answer pretty easily...

Cauchy
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Re: Equations Like f(2x)=4f(x)

Postby Cauchy » Mon May 04, 2015 2:10 am UTC

Putnam 2010 B-5 asks the question: Is there a strictly increasing function f : R -> R such that f ' (x) = f(f(x)) for all x?
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.


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