Functional Equation

- Kainui

Functional Equation

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- Kainui

\[f(x+\tfrac{1}{2}) - f(x-\tfrac{1}{2}) = f(x)\]

- Kainui

This one is kinda difficult and I don't really know but I was also looking at these two other related problems that might or might not help,
\[f(x+1)-f(x)=f(x)\]
and
\[f(x)-f(x-1)=f(x)\]
All three of these things are like finite difference equations versions of the differential equation \(y'=y\), which has \(y=e^x\) as an answer, so this is sorta how I came up with this question.

- freckles

so are you trying to find explicit form for f?

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## More answers

- freckles

you know that last equation you wrote is the same as -f(x-1)=0
or f(x-1)=0

- freckles

or i guess since f is constant then you could really say f(x)=0

- anonymous

i have no idea, but i bet there is a fibonacci somewhere

- Kainui

Yeah, these three functional equations are all pretty similar but all seem to have different answers, so yeah that last one is pretty straightforward @freckles haha xD

- anonymous

\[f(x+1)-f(x)=f(x)\] is an example? maybe you can mimic it to do \[f(x+1)-f(x)=f(x+\frac{1}{2})\]

- SolomonZelman

if f is linear

- SolomonZelman

and with positive slope

- freckles

\[f(x+1)-f(x)=f(x) \\ \text{ the second one gives } \\ f(x+1)-2f(x)=0 \\ \text{ and the characteristic equation is } r-2=0 \implies r=2 \\ \text{ so } f(x)=2^x\]

- freckles

just wanted to play with the little guys before I look at the pretty clown in the room

- Kainui

Yeah one little extra thing is we can throw an arbitrary constant on that second one as well @freckles but I am not sure if this is the most general solution or not to this one either.

- freckles

oh yeah

- Kainui

@SolomonZelman Good thinking about the linearity, I think if g(x) and h(x) are solutions to any of these then f(x)=g(x)+h(x) is also a solution can be proven for most of these I think.

- SolomonZelman

Yeah if f is a polynomial of power greater than 1, it seems as though that f(x) is wlays ahead in power at least by 1 from f(x+1/2)-f(x-1/2).

- Kainui

Supose \( g(x+1)=2g(x) \), then \(f(x)=C2^x+g(x)\) is a solution to \(f(x+1)=2f(x)\),
\[f(x+1)=C2^{x+1}+g(x+1) = 2*C2^x + 2*g(x) = 2 f(x)\]
Is there some way to show that \(g(x)=0\) is the only thing or are there other solutions?

- freckles

\[f(x)=c \cdot r^x \\ f(x+\frac{1}{2})=c \cdot r^{x+\frac{1}{2}} \\ f(x-\frac{1}{2})=c \cdot r^{x-\frac{1}{2}} \\ c \cdot r^{x} r^{\frac{1}{2}}-c \cdot r^{x} r^{-\frac{1}{2}}-c \cdot r^x=0 \\ r^\frac{1}{2}-r^{\frac{-1}{2}}-1=0 \\ r-1-r^\frac{1}{2}=0 \\ u=r^\frac{1}{2} \\ u^2=r \\ u^2-u-1=0 \\ u=\frac{1 \pm \sqrt{1-4(-1}}{2} \\ u=\frac{1 \pm \sqrt{5}}{2} \\ r^\frac{1}{2}=\frac{1 \pm \sqrt{5}}{2}\]

- freckles

one sec checking my work

- Kainui

Looks nice! :D

- SolomonZelman

where did you get ...-1 in line 5?

- freckles

I divided both sides by c*r^x

- SolomonZelman

oh

- freckles

\[r^\frac{1}{2}>0 \text{ for real } r \\ r^\frac{1}{2}=\frac{1+\sqrt{5}}{2} \\ r=(\frac{1+\sqrt{5}}{2})^2=\frac{1+2 \sqrt{5}+5}{4}=\frac{6+2 \sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}\]

- freckles

and just so you know I recently learned how to solve recursive sequences using particular solution guess method or whatever it is called :p

- freckles

\[f(x)=c \cdot (\frac{3+\sqrt{5}}{2})^x\]

- Kainui

@freckles Actually @satellite73 has some sorta magical knack for this, cause this is the golden ratio lol xD
\[r^{1/2} = \frac{1 + \sqrt{5}}{2} = \varphi\]

- Kainui

Also I am trying to think about the other root, can't the negative one also work as well since squaring it will make r positive?

- Kainui

Also @freckles I have never heard of this method before where did you learn this, that was really impressive :O

- freckles

let me see if i can dig up my sources

- freckles

i think it was from a power point

- freckles

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CCkQFjACahUKEwj7nv33r5HJAhUC5iYKHR4sCjc&url=http%3A%2F%2Fwww.eecs.yorku.ca%2Fcourse_archive%2F2008-09%2FS%2F1019%2FWebsite_files%2F21-linear-recurrences.pdf&usg=AFQjCNHp1UWwxkhMP4JUWxakVmiy7hj6-A&sig2=ouSdohecc5XQbWhSW39Ceg

- Kainui

Ok so here's the most general solution I think we have for \(f(x+\tfrac{1}{2}) - f(x-\tfrac{1}{2}) = f(x)\)
\[f(x) = a \varphi^{2x} + b \psi^{2x}\]
With:
\[\varphi = \frac{1+\sqrt{5}}{2}\]
and
\[\psi = \frac{1-\sqrt{5}}{2}\]
and a,b are arbitrary constants.

- freckles

it was about recursive relations

- freckles

finding the homogeneous solution for recursive relation

- freckles

and this functional equation is a homogeneous recursive relation

- freckles

and so is the others

- Kainui

Awesome, thanks, I've been interested in finite difference equations lately and trying to rediscover the calculus analogs on my own for fun, like 'indefinite summation' is the inverse of a finite difference , basically like telescoping series and such. I think this will help me a lot.

- freckles

http://web.cs.wpi.edu/~cs504/s00m/notes/recurrence/solve/step2/step2.html
don't know if you care but I thought this was the most amazing source on the non-homogeneous solution part which we don't have here

- freckles

like the combination of those two links really help me understood how to solve recurrence relations

- Kainui

Ohhhh that looks cool, you might also be interested in this book called Generatinfunctionology, it's really pretty good, it uses generating functions to derive some pretty nice closed forms for recurrence relations, let me find the link. If you think of more stuff that you wanna share go for it.

- Kainui

Here's the book, you can download the pdf from this page, I think the book was published in like 1990 and the author died so that's why the webpage looks pretty dated but it's an awesome book, very cool and easy to read which is rare for math books covering cool stuff like this. https://www.math.upenn.edu/~wilf/DownldGF.html

- freckles

http://openstudy.com/users/myininaya#/updates/56312856e4b0b6bc456ca33a myininaya was my inspiration for learning this
:p

- freckles

she actually tried the series way

- Kainui

Hahaha I am doing this cause I want to find Green's theorem for difference equations! Captivating or sleep-inducing? I guess I don't know haha.

- Kainui

@freckles Do you know if there's some kind of "Wronskian" we can use to show that we have a complete set of solutions? It isn't even clear to me that this has two, I would have only expected 1 solution since really:
\[\Delta f(x)= f(x)\] is what I was solving in all three cases

- freckles

I honestly did not go that far in my research

- Kainui

Yeah I forget how the Wronskian even works in regular calculus, I just know how to use it. I guess this is good, gives me something to look into and some reading material you just gave me.

- freckles

hey did you check to see if that other solution I got worked or did you just assume it did?

- Astrophysics

It tells you whether your solutions are part of a fundamental set, and if it's linearly independent or dependant. If the wronskian = 0 it's dependant and if it does not equal 0 it's linearly independent and the solutions are part of a fundamental set.
Just some info if you forgot err yeah.

- Kainui

Yeah it works, since you can have negative square roots @freckles, like what I mean is:
\[\sqrt{4}= -2\]
That's true, cause
\[4=(-2)^2\]
So there wasn't really an issue there.

- freckles

oh ok I was using the other principal square root thingy
should have used the other square root thingy

- Kainui

Yeah I think you were thinking it was like \(\sqrt{-r}\) then it would be imaginary for positive r, I had to like think through this myself cause it didn't seem right to me either haha

- freckles

I consider the range of to the square root function to be [0,inf)
instead of (-inf,inf)
is what I did...
\[r^\frac{1}{2}>0\]
and that one number was positive while the other one was negative

- freckles

but yeah it makes sense why the other one works too

- Kainui

Basically right now I'm looking at \(\Delta f(x) = f(x+1)-f(x)\) and looking at this summation:
\[\sum_{k=a}^{n-1}\Delta f(k) =f(n)+C\]
And since we have just shown that \(\Delta f(x)=f(x)\) is satisfied by \(f(x)=a 2^x\) we have a kinda neat thing if we plug it in, for finite differences \(2^x\) is like \(e^x\) in calculus. :D

- Kainui

Also we could just plug it in and test it if we're still having doubts on that negative root, which is really what I did and it works out fine:
\[\psi^{2x+1}-\psi^{2x-1} = \psi^{2x}\]
divide out \(\psi^{2x}\) since it is never 0
\[\psi-\psi^{-1} =1\]
Multiply this by \(\psi\) and rearrange little:
\[\psi^2=\psi+1\]
That's true, cause that's a root to that quadratic equation, so we're done I think!

- freckles

cool then

- freckles

http://math.stackexchange.com/questions/1525015/functional-equation-fx3-fx3-2-fx2-12
I just found this functional equation
I wonder if we can use our knowledge of solving recurrence relations to solve this one
I honestly haven't looked at that answer that was posted as reply much

- freckles

not yet
i will when I give up

- Kainui

Ok sounds fun let's play around with it

- Kainui

My instinct is that the square on the \((f(x))^2\) is gonna mess us up haha

- freckles

\[f(x^3)-f(x^3-2)-(f(x))^2=12 \\ \text{ first we solve } f(x^3)-f(x^3-2)-(f(x))^2=0 \\ \text{ Let } f(x)=c r^x \\ f(x^3)=c r^{x^3} \\ f(x^3-2)=c r^{x^3-2} \\ c r^{x^3}-c r^{x^3} r^{-2}-c^2 r^{2x}=0\]
I think your right :(

- Kainui

I plugged in x=0 and solved for
\[f(0)= \frac{1\pm \sqrt{1-48 f(-2)}}{2}\]
I don't know just sorta playing around, this one looks like it will be kinda tough hmmm

- freckles

\[1-r^{-2}-c r^{2x-x^3}=0\]
yep it looks like this solution will not be accepted by this equation definitely

- Kainui

Maybe we can modify the strategy a bit to work though maybe? I don't know

- freckles

ok i'm going to look at what they did lol

- Kainui

Haha ok

- Kainui

I don't believe that their solution that they found is the most general one

- freckles

someone suggested the solution was of cubic form
definitely no exponential form involved it seems
i guess they went with a cubic solution because of the x^3 inside the function thingy

- freckles

I wonder how that one guy knew it was
\[f(x)=ax^3+b\]
and not
\[f(x)=ax^3+cx^2+bx+d\]

- freckles

like I know the first form is a subset of the other form

- freckles

but the first form is easier to work with if you know it is of that form

- Kainui

Well it looks like they went through with it and after they looked at all the terms, then it ended up working out to that smaller thing... But I'm still pretty doubtful, there could be some Taylor series that work maybe.
Actually I wonder, what if they replace x with \(x^2\) does it still work? One sec...

- Kainui

Like it seems that if f(x) is a solution then \(f(x^n)\) would also have to be a solution too huh. hmmm

- freckles

just wanted to try to see if I could get the f(x)=6(x^3-1) solution
\[f(x)=ax^3+b \\ f(x^3)=ax^9+b \\ f(x^3-2)=a(x^3-2)^3+b =a((x^3)^3+3(x^{3})^2(-2)+3(x^3)(-2)^2+1)+b \\ f(x^3-2)=ax^9-6ax^6+12ax^3+a+ab \\ f(x^3)-f(x^3-2)-(f(x))^2=0 \\ (ax^9+b)-(ax^9-6ax^6+12ax^3+a+ab)-(ax^3+b)^2=0 \\ x^6(6a)+x^3(-12a)+(b-a-ab)-(a^2x^6-2abx^3+b^2)=0 \\ x^6(6a-a^2)+x^3(-12a+2ab)+(b-a-ab-b^2)=0 \\ 6a-a^2=0 \implies a=0 \text{ or } a=6 \\ \text{ if } a=6 \text{ then } -12(6)+2(6)b=0 \implies b=\frac{12(6)}{2(6)}=6 \\ \text{ now checking to see if we have } b-a-ab-b^2=0 \\ 6-6-6\cdot6-6^2=0 \text{ and good! } \\ a=b=6\]
they got b=-6 though

- freckles

oops nevermind

- freckles

I see what I did wrong

- freckles

also i don't know how to subtract anymore

- freckles

I said -6^2-6^2 is zero what!!!

- Kainui

Hahaha xD

- freckles

so they found the homogeneous solution it looks like
I wonder if there is a particular one

- freckles

and if it is a constant form

- freckles

\[f(x)=c \\ f(x^3)=c \\f(x^3-2)=c \\ \\ c-c=c^2+12\]

- freckles

\[-12=c^2 \\ c=\pm i \sqrt{ 12 }\]

- freckles

\[f(x)=6x^3-6 \pm i \sqrt{12}\]

- Kainui

Ooooooh ok I really want to solve the general case of this now this seems quite interesting to see what complex numbers pop up.

- Kainui

Wait wait I am sort of confused now how you found this, I was thinking something different and now that I read what you wrote again I think differently about it, let me think this looks interesting but I don't quite understand it.

- Kainui

So we could do this again and find a particular solution where it might equal cx instead of just c, is that right?

- freckles

I chose f(x)=c
\[f(x^3)-f(x^3-2)-(f(x))^2=12\]
because the sum/difference of those functions is a constant

- Kainui

Ah, I see, this is an interesting insight.

- freckles

I wonder if there is particular solution with real coefficients

- freckles

dumb question
there can be more than one particular solution if we consider complex numbers?

- Kainui

No I don't think so, I think the first time I read it I thought you did something differently that I didn't understand.

- freckles

i guess wolfram doesn't include particular solution because it is complex

- freckles

i guess i need to do more research
maybe instead of researching just recursive relations I should search functional equations though they look the same to me well except functional equations have given me an impression that they are just a bit more fancier in their appearance

- Kainui

Yeah, I think it's all pretty related and I really want to bring all this stuff together. Really recurrence relations are basically like differential equations it seems so a lot of different strategies seem to have similar ones. I'll probably be playing with these all tonight and tomorrow too since these are pretty fun, tell me if you find anything cool if you end up doing more research tonight or tomorrow or whatever

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