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Since $$f(x+y)=f(y+x)$$, So an addition formula must be symmetric.

If we define $$f(x+y)=U(f(x),f(y))$$

If we define $f(x)=p$ and $f(y)=q$

$$f(x+y)=U(p,q)$$

and because of $f(x+y)=f(y+x)$,

$$f(x+y)=U(f(x),f(y))=U(f(y),f(x))$$

Thus $U(p,q)$ is a two variable symmetric function

$$U(p,q)=U(q,p)$$

An example:

$$f(x+y)=f(x)f(y)(f(x)+f(y))$$

$U(x,y)=xy(x+y)$ kernel can be a candidate of an addition formula of $f(x)$ because it is a symmetric function.

But If we extend it for 3 components $f(x+y+z)$

$$f(f^{-1}(x)+f^{-1}(y)+f^{-1}(z))=zxy(x+y)(z+xy(x+y))$$ The result is asymmetrical, so $$f(x+y)=f(x)f(y)(f(x)+f(y))$$ cannot be an addition formula

Other example is $tan(x)$ addition formula , It has symmetric kernel too $U(x,y)=\frac{x+y}{1-xy}$ and after 3 component adding, the result is also symmetric.

$$tan(x+y+z)=\frac{tan x+tan y+tan z -\tan x\tan y\tan z}{1-(\tan x \tan y+\tan x \tan z +\tan y \tan z)}$$

Is there a formula to determine if a symmetric function $U(x,y)$ is kernel of an addition formula of a function without testing as I made above to add 3 components to determine it manually ?

Thanks

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    $\begingroup$ Sorry, didn't you forget to define what a kernel of an addition formula is? $\endgroup$ Feb 4, 2016 at 14:36
  • $\begingroup$ In both your examples $U(x,y)$ is a rational function. Are you mainly interested in rational $U$? $\endgroup$ Feb 4, 2016 at 16:25
  • $\begingroup$ @JulianRosen I am interested in any $U(x,y)$, not only rationals. I just wanted to show 2 simple examples to demonstrate the situation. I would like to find the formula of the classify which condition is required to to be an addition formula of a symmetric $U(x,y)$ $\endgroup$
    – Mathlover
    Feb 4, 2016 at 16:39
  • $\begingroup$ Is the question equivalent to the following? Given $U(x, y)$ determine if there exists a function $f$ such that $U(f(x), f(c-x)) = const$ for $\forall c$. $\endgroup$
    – dxiv
    Feb 4, 2016 at 18:25
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    $\begingroup$ @AlexandreEremenko, perhaps you are referring to this paper: ams.org/journals/tran/1927-029-02/S0002-9947-1927-1501393-4/… $\endgroup$
    – Nick Gill
    Mar 11, 2016 at 13:47

2 Answers 2

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The relevant result is a theorem of Weierstrass which says that if $f$ is meromorphic (of one variable), and satisfies an addition theorem $$f(x+y)=F(f(x),f(y)),$$ then $f$ is elliptic (possibly degenerate). The converse is also true. The a priori assumption that $f$ is meromorphic can be substantially relaxed with the same conclusion. Thus we are reduced to describing all possible addition formulas for elliptic curves, and I do not think that there is an explicit answer.

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  • $\begingroup$ Hi , Have you seen my answer and the condition? Is it vaild proof? $f(x+y)=F(f(x),f(y))$ We can write a condition for 2 variable function $F(x,y)$ $\frac{\partial^2}{\partial x \partial y} (\ln [\frac{\partial {F(x,y)}}{\partial x}]) = \frac{\partial^2}{\partial x \partial y} (\ln[ \frac{\partial {F(x,y)}}{\partial y}])$ $\cfrac{\partial}{\partial y} (\cfrac{\frac{\partial^2 {F(x,y)}}{\partial x^2}}{\frac{\partial {F(x,y)}}{\partial x}}) = \cfrac{\partial}{\partial x} (\cfrac{\frac{\partial^2 {F(x,y)}}{\partial y^2}}{\frac{\partial {F(x,y)}}{\partial y}})$ $\endgroup$
    – Mathlover
    Jul 14, 2016 at 7:24
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I found out a way to determine if a two variable symmetric function is an addition function of one variable function or not.

Let's assume that we know $$\frac{df(x)}{dx}=G(f(x))$$

We can write that

$$\int \frac{d(f(x))}{G(f(x))}=\int dx$$

$$\int \frac{d(f(x))}{G(f(x))}=x+c$$

$$\int_{f(0)}^{f(x)} \frac{dz}{G(z)}=x$$

$$\int_{f(0)}^{f(y)} \frac{dz}{G(z)}=y$$

$$\int_{f(0)}^{f(x+y)} \frac{dz}{G(z)}=x+y$$

$$\int_{f(0)}^{f(x+y)} \frac{dz}{G(z)}=\int_{f(0)}^{f(x)} \frac{dz}{G(z)}+\int_{f(0)}^{f(y)} \frac{dz}{G(z)} $$

If we define $$f(x+y)=U(f(x),f(y))$$

$$\int_{f(0)}^{U(f(x),f(y))} \frac{dz}{G(z)}=\int_{f(0)}^{f(x)} \frac{dz}{G(z)}+\int_{f(0)}^{f(y)} \frac{dz}{G(z)} $$

If we define $f(x)=p$ and $f(y)=q$

$$\int_{f(0)}^{U(p,q)} \frac{dz}{G(z)}=\int_{f(0)}^{p} \frac{dz}{G(z)}+\int_{f(0)}^{q} \frac{dz}{G(z)} $$

If we derivate both sides over $p$ $$\frac{\partial {U(p,q)}}{\partial p} \frac{1}{G(U(p,q))}=\frac{1}{G(p)} $$ $$\frac{\partial {U(p,q)}}{\partial p} =\frac{G(U(p,q))}{G(p)} $$ If we derivate both sides over $q$ $$\frac{\partial {U(p,q)}}{\partial q} \frac{1}{G(U(p,q))}=\frac{1}{G(q)} $$ $$\frac{\partial {U(p,q)}}{\partial q} =\frac{G(U(p,q))}{G(q)} $$

And Finally divide two results that we got

$$\frac{\frac{\partial {U(p,q)}}{\partial p} }{\frac{\partial {U(p,q)}}{\partial q} }=\frac{G(q)}{G(p)} $$

We can continue to cancel $G(x)$ $$\ln [\frac{\partial {U(p,q)}}{\partial p}] -\ln[ \frac{\partial {U(p,q)}}{\partial q}] =\ln [G(q)]-\ln [G(p)] $$

If we derivative both sides over $p$ $$\frac{\frac{\partial^2 {U(p,q)}}{\partial p^2}}{\frac{\partial {U(p,q)}}{\partial p}} -\frac{\frac{\partial^2 {U(p,q)}}{\partial p\partial q}}{\frac{\partial {U(p,q)}}{\partial q}} =-\frac{G'(p)}{G(p)} $$

If we derivative both sides over $q$, we get a condition without $G(x)$

$$\frac{\partial}{\partial q}[\frac{\frac{\partial^2 {U(p,q)}}{\partial p^2}}{\frac{\partial {U(p,q)}}{\partial p}} -\frac{\frac{\partial^2 {U(p,q)}}{\partial p\partial q}}{\frac{\partial {U(p,q)}}{\partial q}}] =0 $$


Let's test the example :

$$f(x+y)=f(x)f(y)(f(x)+f(y))$$

$$f(x+y)=U(f(x),f(y))=f(x)f(y)(f(x)+f(y))$$ If we define $f(x)=p$ and $f(y)=q$

$$f(x+y)=U(p,q)=p q(p+q))$$

$$\frac{\partial {U(p,q)}}{\partial p}=q(p+q)+p q=q^2+2 p q$$

$$\frac{\partial {U(p,q)}}{\partial q}=p(p+q)+p q=p^2+2 p q$$

To check condition

$$\frac{\frac{\partial {U(p,q)}}{\partial p} }{\frac{\partial {U(p,q)}}{\partial q} }=\frac{G(q)}{G(p)} $$

$$\frac{G(q)}{G(p)}=\frac{q^2+2 p q}{p^2+2 p q} $$

It is impossible to find a $G(x)$ that satisfies the relation.


If we test other example :

$$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$$

$$f(x+y)=U(f(x),f(y))=\frac{f(x)+f(y)}{1-f(x)f(y)}$$ If we define $f(x)=p$ and $f(y)=q$

$$f(x+y)=U(p,q)=\frac{p+q}{1-pq}$$

$$\frac{\partial {U(p,q)}}{\partial p}=\frac{1+q^2}{(1-pq)^2}$$

$$\frac{\partial {U(p,q)}}{\partial q}=\frac{1+p^2}{(1-pq)^2}$$

To check condition

$$\frac{\frac{1+q^2}{(1-pq)^2}}{\frac{1+p^2}{(1-pq)^2}}=\frac{G(q)}{G(p)}$$

$$G(x)=c(1+x^2) $$ where $c$ is constant

Thus we can write that

$$f'(x)=c(1+f^2(x))$$

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