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$n$ 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 :

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

share|improve this question
    
Going out on a limb here, but integrating the equation for $f$ will give you the inverse function of what you're looking for; calling $g := f^{-1}$, perhaps we could plug in $g(x)$ into $f$ instead of $x$, calculate the appropriate chain rules and powers, apply the inverse derivative rule, and solve the ODE for $g$? –  Eugene Shvarts Nov 14 '12 at 14:05
    
@EugeneShvarts Do you want me to write it as $x-\binom{-1/n}{1}\frac{x^{n+1}}{n+1}+\binom{-1/n}{2}\frac{x^{2n+1}}{2n+1}-....=‌​f^{-1}(x)=g(x)$? and then to find diff equation of $g(x)$. Did I understand correct? many thanks –  Mathlover Nov 14 '12 at 14:59
    
Ideally, you'd be able to find the ODE for $g$ directly from the ODE for $f$; the reason I make such a suggestion is that the function you end up integrating yields this result, which in particular is the inverse function in the first few easy cases. So, rephrasing the entire question to deal with $f$'s inverse may give you more direct results. –  Eugene Shvarts Nov 14 '12 at 23:54
1  
For larger $n$, I think the equation will have more than one solution. It isn't necessary that all of them will obey the same recurrence relation. –  dexter04 Nov 15 '12 at 13:19
1  
If you let $g = f'/f$, the equation $f^n+f'^n=1$ implies $g'+g^2+g^{2-n}=0$. –  Malper Sep 13 '13 at 3:39

1 Answer 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.

enter image description here

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 !

share|improve this answer
    
Thanks a lot for answer but I asked to find addition formula not to get the function itself. I know well that there is no simple closed form to express the $f_n(x)$ for $n>2$. I have been looking for a closed form of addition formula, maybe it has a closed form . –  Mathlover Apr 3 at 11:22

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