Functional Equation (1 Viewer)

[KingOfActing]

So in my boredom these holidays I have thought up the following functional equation:


With the only restriction being that it is continuous on .

I first looked at constant solutions, which gave me

Then linear solutions were all

And in general, any polynomial solution is for some natural n. Just through inspection it becomes clear that this works for any real n.

I then went on to try rational functions, and after solving for general linear/linear and quadratic/quadratic, I noticed the pattern that the general rational function which is a solution to this functional equation is of the form . This solution also includes all previous solutions as well.

I've also found:


Is there any more general form for the solution than the one I've found? Any transcendental functions/non-elementary functions that satisfy the equation?

 

[Paradoxica]

So in my boredom these holidays I have thought up the following functional equation:


With the only restriction being that it is continuous on .

I first looked at constant solutions, which gave me

Then linear solutions were all

And in general, any polynomial solution is for some natural n. Just through inspection it becomes clear that this works for any real n.

I then went on to try rational functions, and after solving for general linear/linear and quadratic/quadratic, I noticed the pattern that the general rational function which is a solution to this functional equation is of the form . This solution also includes all previous solutions as well.

I've also found:


Is there any more general form for the solution than the one I've found? Any transcendental functions/non-elementary functions that satisfy the equation?

Transcendental functions probably don't work. e^x doesn't work, which means no trigonometric functions will work, hyperbolic or regular. None of the inverses work either. It seems transcendental functions won't work in general, but I have no proof of this.

Floor, Round and Ceiling don't work.

If g(x) is a solution, then |g(x)| is also a solution, but that trivially arises from the fact that the negative of a function works, if the function itself works, so that's a pointless observation.

Trying to see if any non-elementary functions work but Wolfram Mathematica is being a PITARN.

 

[glittergal96]

So in my boredom these holidays I have thought up the following functional equation:


With the only restriction being that it is continuous on .

I first looked at constant solutions, which gave me

Then linear solutions were all

And in general, any polynomial solution is for some natural n. Just through inspection it becomes clear that this works for any real n.

I then went on to try rational functions, and after solving for general linear/linear and quadratic/quadratic, I noticed the pattern that the general rational function which is a solution to this functional equation is of the form . This solution also includes all previous solutions as well.

I've also found:


Is there any more general form for the solution than the one I've found? Any transcendental functions/non-elementary functions that satisfy the equation?

There are LOTS of solutions. Let g(x) be an arbitrary nonvanishing continuous function on the set [-1,1]\{0}, such that g(1)^2=g(-1)^2=1.

Then g extends uniquely to a continuous solution of the functional equation on the nonzero reals. (Define it outside of [-1,1]\{0} using g(x):=1/g(1/x).)

 

[Paradoxica]

There are LOTS of solutions. Let g(x) be an arbitrary nonvanishing continuous function on the set [-1,1]\{0}, such that g(1)^2=g(-1)^2=1.

Then g extends uniquely to a continuous solution of the functional equation on the nonzero reals. (Define it outside of [-1,1]\{0} using g(x):=1/g(1/x).)

Note this does not necessarily yield smooth functions, although there was never any say on that.

It does satisfy the requirements of the functional equation, so that's completely fine to me.

 

[Shadowdude]

There are LOTS of solutions. Let g(x) be an arbitrary nonvanishing continuous function on the set [-1,1]\{0}, such that g(1)^2=g(-1)^2=1.

Then g extends uniquely to a continuous solution of the functional equation on the nonzero reals. (Define it outside of [-1,1]\{0} using g(x):=1/g(1/x).)

Note this does not necessarily yield smooth functions, although there was never any say on that.

It does satisfy the requirements of the functional equation, so that's completely fine to me.

omg why didn't you two talk to each other yesterday


you guys would've had tons of fun discussing such asian topics like these


(@glittergal: the above sentence is indeed racist, not 'racial')

 

[KingOfActing]

There are LOTS of solutions. Let g(x) be an arbitrary nonvanishing continuous function on the set [-1,1]\{0}, such that g(1)^2=g(-1)^2=1.

Then g extends uniquely to a continuous solution of the functional equation on the nonzero reals. (Define it outside of [-1,1]\{0} using g(x):=1/g(1/x).)

Note this does not necessarily yield smooth functions, although there was never any say on that.

It does satisfy the requirements of the functional equation, so that's completely fine to me.

I honestly don't have a reason for asking this, I was just bored and thought it up. Thank you for the replies!

 

[glittergal96]

Note this does not necessarily yield smooth functions, although there was never any say on that.

It does satisfy the requirements of the functional equation, so that's completely fine to me.

No it does not, but we can easily find all smooth solutions to the functional equation by putting in slightly more work.

Note that if we find all solutions f defined on the domain of positive reals, then it is trivial to find all solutions on the full domain of nonzero reals. (Let f,g be arbitrary solutions to the problem on the positive reals, and define h(x)=f(x) for x>0 and h(x)=g(-x) for all x<0.)

With this in mind we set out trying to solve the functional equation on the domain R+.

Note the any solution f cannot vanish anywhere, from the functional equation. As f is also assumed to be continuous, this means that f must be of fixed sign. We assume f is positive, keeping in mind that taking negatives everywhere would still give us a solution.

Then we can define g(t)=log(f(e^t)). This function g is well defined on the reals by our assumptions on f, and the functional equation transforms to g(t)+g(-t)=0, that is g must be a smooth odd function.

Conversely, if we take an arbitrary smooth odd function g defined on the reals, then the function f implicitly defined by e^(g(log(t)))=f(t) (and the negative of this) are smooth solutions to the original functional equation over the positive reals.

So this completely resolves the solution set of the original functional equation over the nonzero reals.

 

[glittergal96]

omg why didn't you two talk to each other yesterday


you guys would've had tons of fun discussing such asian topics like these


(@glittergal: the above sentence is indeed racist, not 'racial')

?

 

[KingOfActing]

No it does not, but we can easily find all smooth solutions to the functional equation by putting in slightly more work.

Note that if we find all solutions f defined on the domain of positive reals, then it is trivial to find all solutions on the full domain of nonzero reals. (Let f,g be arbitrary solutions to the problem on the positive reals, and define h(x)=f(x) for x>0 and h(x)=g(-x) for all x<0.)

With this in mind we set out trying to solve the functional equation on the domain R+.

Note the any solution f cannot vanish anywhere, from the functional equation. As f is also assumed to be continuous, this means that f must be of fixed sign. We assume f is positive, keeping in mind that taking negatives everywhere would still give us a solution.

Then we can define g(t)=log(f(e^t)). This function g is well defined on the reals by our assumptions on f, and the functional equation transforms to g(t)+g(-t)=0, that is g must be a smooth odd function.

Conversely, if we take an arbitrary smooth odd function g defined on the reals, then the function f implicitly defined by e^(g(log(t)))=f(t) (and the negative of this) are smooth solutions to the original functional equation over the positive reals.

So this completely resolves the solution set of the original functional equation over the nonzero reals.

Mirin' knowledge

 

[InteGrand]

?

There was a BOS Beach Meet yesterday. Shadowdude seems to have been informed that you were there.

 

[glittergal96]

Mirin' knowledge

Its not as fancy as it might look. If you learn a little about functional equations then the tricks I used were very standard.

 

[glittergal96]

There was a BOS Beach Meet yesterday. Shadowdude seems to have been informed that you were there.

Not sure what is leading him to believe that...

 

[Librah]

Not sure what is leading him to believe that...

Yeah so me, Realisenothing and 2 friends were taking the piss out of Shadowdude yesterday lol.