# 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.

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

I have also found $a_{2n+1}$
$$f_n(x)=x-\frac{ x^{n+1}}{n(n+1)}+\frac{(1+2n-n^2) x^{2n+1}}{(2n+1)2n (n+1)n}+\frac{a_{3n+1} x^{3n+1}}{(3n+1)!}+...$$
 Note that typing "\left(...\right)" between dollars symbols withh automatically write correctly sized parentheses to the expression between them. – DonAntonio Nov 14 '12 at 13:41 @DonAntonio Thanks for advice. – Mathlover Nov 14 '12 at 13:48 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