f(x+y) g(f(x)f(y)) = f(x) + f(y) addition formula ?

[tommy1729]

I talked to my friend Mick about addition formula.

He wrote about it on math stackexchange.

Just like tanh(2^n arctanh(x)) is of interest to most of you for iteration reasons and the addition formula for tanh makes this a uniqueness criterion ( think about it if you want but I assume you already know that ) , I assume this addition formula and possible connection to dynamics interests you.

In a recent thread ( https://math.eretrandre.org/tetrationfor...p?tid=1652) James mentioned double periodic functions what motivated me to think about addition formulas again ( since double periodic ones can have nice ones ) 

Here is mick's question and some of our talk.

https://math.stackexchange.com/questions...fxfy-fx-fy

Im sure you have opinions or interest in it.

In case it gets deleted or such :

Consider real analytic functions \(f(x),g(x)\) such that the addition formula

\[f(x+y) g(f(x)f(y)) = f(x) + f(y)\]

is satisfied for any real \(x,y\).

Also \(f(-x) = - f(x)\) is satisfied.

The function \(g(xy)\) is given and we want to find a function f such that we have a formal group law on the interval \([-1,1]\).

For what functions \(g(x)\) can we find a function \(f(x)\) that satisfies the above ? And how do we find \(f(x)\) then ?

It is known that \(g(x)\) can not be a high degree polynomial , see the answer here

---

@MISC {3120824, TITLE = {Is \( \frac{x+y}{x^3 y^3 - x^3 y - x y^3 + 2 xy + 1} \) a formal group law?}, AUTHOR = {Mark S. (https://math.stackexchange.com/users/26369/mark-s)}, HOWPUBLISHED = {Mathematics Stack Exchange}, NOTE = {URL:https://math.stackexchange.com/q/3120824 (version: 2019-02-21)}, EPRINT = {https://math.stackexchange.com/q/3120824}, URL = {https://math.stackexchange.com/q/3120824} }

Is \( \frac{x+y}{x^3 y^3 - x^3 y - x y^3 + 2 xy + 1} \) a formal group law?

quote :

I remembered that sometimes mathematicians do work that goes beyond what naive manipulations can do (even with a computer), and so I decided to search online for 

Code:

rational formal group law

, and found a lot of good information.

Your question was answered very well in the mid 70's by Robert Bismuth in Corollary 2.5.4 of his Masters thesis: The [i]only[/i] rational (one-dimensional) formal group laws over a field are those of the form \(\boxed{F(x,y)=\dfrac{x+y+axy}{1+bxy}}\).

In particular, [i][b]no[/b][/i] rational functions with denominator of degree 3 are formal group laws.

That said, the proof in that thesis is long and somewhat hard to read. A much tidier proof for fields of characteristic zero is given in the 1991 paper "Rational formal group laws" by Robert Coleman and Francis Oisin McGuinness. More broadly, a (less explicit) characterization of [i]algebraic[/i] formal group laws was done in Coleman's 1986 paper One-dimensional algebraic formal groups.

All of this is summarized and presented with more context in MSE's very own Alex Walker's excellent blog post, Formal Groups and Where to Find Them.

end quote

---

So what classifies functions \(g(x)\) ?

A well known case is \(g(xy) = 1 + xy\) ( or if you want \(g(f(x)f(y)) = 1 + f(x)f(y)\) ) which corresponds to the tanh addition formula. (or equivalently \(g(x) = 1 + x\) )

I read about addition formulas for double periodic functions but most appear of the form

\[ t(x + y) = t(x) g(f(y),f(x)) + t(y) h(f(x),f(y)) \]

where \(g(f(x),f(y)),h(f(x),f(y))\) are not symmetric.

hence that is not the equivalent it appears. (correct me if wrong )

So I guess easy double periodic functions will not work ?

Some examples would be nice too.

A similar question occured here :

About the addition formula \(f(x+y) = f(x)g(y)+f(y)g(x)\)

with a nice answer.

 

---

[i][b]edit[/b][/i]

I talked about it and thought about it.

The ideas that came up are more or less like this.

Set \(x=y\)

then

\[f(2x) g(f(x)^2) = 2 f(x)\]

For a given \(g\) with a taylor series we can then set up a system of equations for the taylor coeff of \(f\).

However maybe taylor is not the best way , we have a radius.

Maybe fourier is better.

Or use integral transforms.

Not sure.

Since we need to property \(f((x+y)+z) = f(x+(y+z))\) this requirement reflects in the equation \(f(2x) g(f(x)^2) = 2 f(x)\) and the radii and expansion points of their taylor.

If this associativity cannot be satisfied then the radii are always " too small ".

If we can generalize a way to solve this or at least get the radii and conclude if this associativity is possible we have a nice tool to help solve the problem : We could test any \(g\).

A further generalization would then classify all \(g\).

The equation

\[f(2x) g(f(x)^2) = 2 f(x)\]

Looks a lot like complex dynamics so maybe tool from there can be used.

Just some ideas ...

My friend suggested solving the system of equations :

\[f(2x) g(f(x)^2) = 2 f(x)\]

\[f(3x) g(f(x)^3) = 3 f(x)\]

The existance of a solution \(f\) would then be sufficient assuming a nonzero radius when expanded around \(0\).

A bold statement ! Or not ??



mick addition

Consider real analytic functions \(f(x),g(x)\) such that the addition formula

\[f(x+y) g(f(x)f(y)) = f(x) + f(y)\]

is satisfied for any real \(x,y\).

Also \(f(-x) = - f(x)\) is satisfied.

The function \(g(xy)\) is given and we want to find a function f such that we have a formal group law on the interval \([-1,1]\).

For what functions \(g(x)\) can we find a function \(f(x)\) that satisfies the above ? And how do we find \(f(x)\) then ?

It is known that \(g(x)\) can not be a high degree polynomial , see the answer here

@MISC {3120824, TITLE = {Is \( \frac{x+y}{x^3 y^3 - x^3 y - x y^3 + 2 xy + 1} \) a formal group law?}, AUTHOR = {Mark S. (https://math.stackexchange.com/users/26369/mark-s)}, HOWPUBLISHED = {Mathematics Stack Exchange}, NOTE = {URL:https://math.stackexchange.com/q/3120824 (version: 2019-02-21)}, EPRINT = {https://math.stackexchange.com/q/3120824}, URL = {https://math.stackexchange.com/q/3120824} }

Is \( \frac{x+y}{x^3 y^3 - x^3 y - x y^3 + 2 xy + 1} \) a formal group law?

quote :

I remembered that sometimes mathematicians do work that goes beyond what naive manipulations can do (even with a computer), and so I decided to search online for rational formal group law, and found a lot of good information.

Your question was answered very well in the mid 70's by Robert Bismuth in Corollary 2.5.4 of his Masters thesis: The only rational (one-dimensional) formal group laws over a field are those of the form \(\boxed{F(x,y)=\dfrac{x+y+axy}{1+bxy}}\).

In particular, no rational functions with denominator of degree 3 are formal group laws.

That said, the proof in that thesis is long and somewhat hard to read. A much tidier proof for fields of characteristic zero is given in the 1991 paper "Rational formal group laws" by Robert Coleman and Francis Oisin McGuinness. More broadly, a (less explicit) characterization of algebraic formal group laws was done in Coleman's 1986 paper One-dimensional algebraic formal groups.

All of this is summarized and presented with more context in MSE's very own Alex Walker's excellent blog post, Formal Groups and Where to Find Them.

end quote

So what classifies functions \(g(x)\) ?

A well known case is \(g(xy) = 1 + xy\) ( or if you want \(g(f(x)f(y)) = 1 + f(x)f(y)\) ) which corresponds to the tanh addition formula. (or equivalently \(g(x) = 1 + x\) )

I read about addition formulas for double periodic functions but most appear of the form

\[ t(x + y) = t(x) g(f(y),f(x)) + t(y) h(f(x),f(y)) \]

where \(g(f(x),f(y)),h(f(x),f(y))\) are not symmetric.

hence that is not the equivalent it appears. (correct me if wrong )

So I guess easy double periodic functions will not work ?

Some examples would be nice too.

A similar question occured here :

About the addition formula \(f(x+y) = f(x)g(y)+f(y)g(x)\)

with a nice answer.


edit

I talked about it and thought about it.

The ideas that came up are more or less like this.

Set \(x=y\)

then

\[f(2x) g(f(x)^2) = 2 f(x)\]

For a given \(g\) with a taylor series we can then set up a system of equations for the taylor coeff of \(f\).

However maybe taylor is not the best way , we have a radius.

Maybe fourier is better.

Or use integral transforms.

Not sure.

Since we need to property \(f((x+y)+z) = f(x+(y+z))\) this requirement reflects in the equation \(f(2x) g(f(x)^2) = 2 f(x)\) and the radii and expansion points of their taylor.

If this associativity cannot be satisfied then the radii are always " too small ".

If we can generalize a way to solve this or at least get the radii and conclude if this associativity is possible we have a nice tool to help solve the problem : We could test any \(g\).

A further generalization would then classify all \(g\).

The equation

\[f(2x) g(f(x)^2) = 2 f(x)\]

Looks a lot like complex dynamics so maybe tool from there can be used.

Just some ideas ...

My friend suggested solving the system of equations :

\[f(2x) g(f(x)^2) = 2 f(x)\]

\[f(3x) g(f(x)^3) = 3 f(x)\]

The existance of a solution \(f\) would then be sufficient assuming a nonzero radius when expanded around \(0\).

A bold statement ! Or not ??


***

Where the bold statement came from me.

Then again maybe not so bold and an analogue to the uniqueness critertion mentioned before ;

" Just like tanh(2^n arctanh(x)) ... "

It is known that g(x) = 1 + 1/2 x + 1/4 x^2 has no solution as a formal group law. 
This might work as a working example to show radii restrictions though.

regards

tommy1729

[tommy1729]

Ok I had some ideas.

And it feels or looks like it is alot like the ideas from Leo W.
Which I realized only much later.

Lets redefine most things with new symbols to avoid confusion.

we want

f(x+y) g(f(x)f(y)) = f(x) + f(y)

and some additional properties.

The idea was that those look like tanh.

Why ? well we want uniqueness :

f(x + y) =/= f(x + y + 1) 

( f(x + y + z) gives complications when z being 0 or 1 gives the same.  )

or so, 

hence we want a function that is strictly increasing.
Strictly increasing from -1 to 1.

so 

f := R -> [-1,-1]

and having a formal group law.


So 

f(x) = Q(tanh(x))

Q(S(x)) = S(Q(x)) = id(x) = x

Q(-x) = - Q(x)

Then we "define the addition formula" for Q(tanh(x)) ;

Q(tanh(arctanh(S(x)) + arctanh(S(y)))

Now we let

Q(tanh(arctanh(S(x)) + arctanh(S(y))) = H(S(x),S(y))

Q(tanh(arctanh(S(x)) + arctanh(S(y))) = (x + y)/(1 + J(x,y))

substitute x,y with Q(x),Q(y) :

Q(tanh(arctanh(x) + arctanh(y))) = (Q(x) + Q(y))/( 1 + K(Q(x),Q(y)) )

Q(tanh(arctanh(x) + arctanh(y))) = (Q(x) + Q(y))/( 1 + L(Q(x)*Q(y)) )


We can simplify further :

Q( (x+y)/(1 + xy) ) = (Q(x) + Q(y))/( 1 + L(Q(x)*Q(y)) )

This is a very nice equation and it puts strong restrictions on Q.

This Q must be a solution to the ( more general ) equation for (a usually non-unique ) W ; what is just the above but setting x = y ;

W( 2x/(1 + x^2) ) = 2 W(x)/(1 + L(W(x)^2) )

Now define U(x) := 2 x / ( 1 + L(x^2) )

Then we arrive at 

W( 2x/(1 + x^2) ) = U(W(x))

which looks like a typical equation from continu iterations !
This relates to complex dynamics !

And the branches for this type of equation is what Leo talked about when generalizing the abel equation.

2x/(1 + x^2)  has the fixpoints -1,0,1.

So we can use many methods.

The methods with 2 fixpoints also work by symmetry of f(-x) = - f(x).
Hence the 3rd fixpoint is not a problem.

So our main focus to solving 

W( 2x/(1 + x^2) ) = U(W(x))

are the iterations of U ;

U^[z](x).

So we want

P(U(x)) = P(x)+1.

I conjecture that if this P satisfies the semi-group iso , and there is indeed a formal group law for the given Q , then it is exactly computable by THAT solution P as shown above ( up to branches and speaking locally ).



regards

tommy1729