Addition formula for $f_n(x+y)$ in closed form.

$n$ is a positive integer. $$f_n(x)^n+\left(\frac{df_n(x)}{dx}\right)^n=1$$

$f_n(0)=0$, $f_n'(0)=1$ then

I am looking for the addition formula for $f_n(x+y)$ in closed form.

if $n=1$ then

$$f_1(x)=1-e^{-x}=x-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+....$$

and

$f_1(x+y)=f_1(x)+f_1(y)-f_1(x)f_1(y)$

if $n=2$ then

$$f_2(x)=\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$$ and

we know that $$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)=\sin(x)\sin'(y)+\sin(y)\sin'(x) $$

thus $f_2(x+y)=f_2(x)f_2'(y)+f_2(y)f_2'(x)$

My attempts to solve the problem:

$$f_n(x)^n+\left(\frac{df_n(x)}{dx}\right)^n=1$$

$$\int \frac{df_n(x)}{\sqrt[n]{1-f_n(x)^n}}=x$$

$$f_n(x)-\binom{-1/n}{1}\frac{f_n(x)^{n+1}}{n+1}+\binom{-1/n}{2}\frac{f_n(x)^{2n+1}}{2n+1}-....=x$$

$$f_n(x)+\frac{f_n(x)^{n+1}}{n(n+1)}+\frac{(n+1)f_n(x)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)(2n+1)f_n(x)^{3n+1}}{3!n^3(3n+1)}+....=x$$

$$f_n(y)+\frac{f_n(y)^{n+1}}{n(n+1)}+\frac{(n+1)f_n(y)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)(2n+1)f_n(y)^{3n+1}}{3!n^3(3n+1)}+....=y$$

$$f_n(x+y)+\frac{f_n(x+y)^{n+1}}{n(n+1)}+\frac{(n+1)f_n(x+y)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)(2n+1)f_n(x+y)^{3n+1}}{3!n^3(3n+1)}+....=x+y$$

$$f_n(x+y)+\frac{f_n(x+y)^{n+1}}{n(n+1)}+\frac{(n+1)f_n(x+y)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)(2n+1)f_n(x+y)^{3n+1}}{3!n^3(3n+1)}+....=f_n(x)+f_n(y)+\frac{f_n(x)^{n+1}}{n(n+1)}+\frac{f_n(y)^{n+1}}{n(n+1)}+\frac{(n+1)f_n(x)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)f_n(y)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)(2n+1)f_n(x)^{3n+1}}{3!n^3(3n+1)}+\frac{(n+1)(2n+1)f_n(y)^{3n+1}}{3!n^3(3n+1)}+....$$

But I could not find $f_n(x+y)$ as alone in one side. I need your hand for ideas and references how can be found the addition formula.

Many thanks for answers

EDIT: (Added on Nov 15)

I want to add my results about power series of $f_n(x)$ I thought that power series can give me a way to find the addition formula.

$$f_n(x)+\frac{f_n(x)^{n+1}}{n(n+1)}+\frac{(n+1)f_n(x)^{2n+1}}{2!n^2(2n+1)}+\frac{(n+1)(2n+1)f_n(x)^{3n+1}}{3!n^3(3n+1)}+....=x$$ Because of that result above, only $x,x^{n+1},x^{2n+1},x^{3n+1},...$ terms will not be zero. Thus we can write, $$f_n(x)=x+\frac{a_{n+1} x^{n+1}}{(n+1)!}+\frac{a_{2n+1} x^{2n+1}}{(2n+1)!}+...$$

$$f_n(x)^n+\left(\frac{df_n(x)}{dx}\right)^n=1$$

$$\left(x+\frac{a_{n+1} x^{n+1}}{(n+1)!}+\frac{a_{2n+1} x^{2n+1}}{(2n+1)!}+... \right)^n+\left(1+\frac{a_{n+1} x^{n}}{n!}+\frac{a_{2n+1} x^{2n}}{(2n)!}+... \right)^n=1$$

$$x^n\left(1+\frac{a_{n+1} x^{n}}{(n+1)!}+\frac{a_{2n+1} x^{2n}}{(2n+1)!}+... \right)^n+\left(1+\frac{a_{n+1} x^{n}}{n!}+\frac{a_{2n+1} x^{2n}}{(2n)!}+... \right)^n=1$$

we can find easly the result below if we check only $x^n$ terms.

$$1+\frac{n a_{n+1} }{n!}=0$$ then

$$a_{n+1}=-(n-1)!$$

$$f_n(x)=x-\frac{ x^{n+1}}{n(n+1)}+\frac{a_{2n+1} x^{2n+1}}{(2n+1)!}+...$$

Yet I have not found an easy way to find $a_{2n+1},a_{3n+1},...$ terms

If someone sees an easy way how to find the pattern of $a_{2n+1},a_{3n+1},...$ , please write it to me

Thanks for advice

Update: Added on Nov 16

I have also found $a_{2n+1}$

Now my last result is :

$$f_n(x)=x-\frac{ x^{n+1}}{n(n+1)}+\frac{(1+2n-n^2) x^{2n+1}}{(2n+1)2n (n+1)n}+\frac{a_{3n+1} x^{3n+1}}{(3n+1)!}+...$$

Solution 1:

The general solution is the inverse function of the Incomplete Beta function (attachment)

Simple closed form exist in cases $n=1$ and $n=2$. In case of $n>2$, there is no simpler closed form made only with elementary functions.

If you want an approach of the solution in the general case, expressed on the form of a limited series, first expand the incomplete beta function ( formula in handbooks of special function) and invert the series. This would be an hard work !