General solution or approximate solution

Question

Is there a known general or approximate explicit solution for $\xi$ in $$(1+\xi)^m (1-\xi)^n = C$$ where $m$ and $n$ positive fractions and $C$ being constant?

Answer 1

THEOREM.

The equation $$ aqx^p+x^q=1\tag 1 $$ admits root $x$ such that $$ x^n=\frac{n}{q}\sum^{\infty}_{k=0}\frac{\Gamma(\{n+pk\}/q)(-q a)^k}{\Gamma(\{n+pk\}/q-k+1)k!}\textrm{, }n>0\tag 2 $$ where $\Gamma(x)$ is Euler's the Gamma function. (Note that the root of $(1)$, for no confusion is given from $(2)$ with $n=1$).

Here the sum $(2)$ is valid for all real numbers $n,p$ and $q$ and for complex $a$ with $$ |a|\leq |p|^{p/q}|p-q|^{(p-q)/q}.\tag 3 $$

$\ldots$ You want to solve the equation $$ (1+\xi)^m(1-\xi)^n=C.\tag 4 $$ Set $1+\xi=x$, then $(4)$ becomes $x^m(1-(x-1))^n=C$ or $x^m(2-x)^n=C$. Hence $x^{m/n}(2-x)=C^{1/m}$ or $2-x=C^{1/m}x^{-m/n}$. Hence $2x^{-1}-1=C^{1/m}x^{-m/n-1}$. Setting $2x^{-1}=y$ we have $y-1=\frac{C^{1/m}}{2^{m/n+1}}y^{m/n+1}$ or $$ -C_1y^A+y=1,\tag 5 $$ where $C_1=-C^{1/m}2^{-m/n-1}$, $A=m/n+1$ and $$ \xi=2/y-1.\tag 6 $$ Hence using $(1),(2)$ we can extract a solution of $(5)$ and hence of $(4)$.

Answer 2

First I thought, by deriving the general solution was $\xi=\frac{m-n}{m+n}$.

But for a solution with given $C$, I'd first reformulate the problem to $(1+\xi)(1-\xi)^a=c'$ with $a=n/m>0$ and $c'=c^{\frac{1}{m}}$...

Answer 3

Since $m$ and $n$ are fractions, you can take the LCM of their denominators and write $$ \left\{ \matrix{ m = {p \over L}\quad n = {q \over L}\quad \left| {\;p,q,L \in N} \right. \hfill \cr \left( {1 + x} \right)^{\,p} \left( {1 - x} \right)^{\,q} = C^{\,L} \hfill \cr} \right. $$ so that we can always reduce the problem to integral powers.

The approach I deem might work better is the following

Let's change the sign of the second term, so as to have a monic polynomial $$ \bbox[lightyellow] { \left( {x + 1} \right)^{\,p} \left( {x - 1} \right)^{\,q} = \left( { - 1} \right)^{\,q} C^{\,L} = c\quad \left| {\;p,q,L \in N} \right. }\tag{1}$$

Let's then consider that $$ \left( {x + 1} \right)^{\,p} \left( {x - 1} \right)^{\,q} = \left\{ {\matrix{ {\left( {x^{\,2} - 1} \right)^{\,p} \left( {x - 1} \right)^{\,q - p} } & {0 \le q - p} \cr {\left( {x^{\,2} - 1} \right)^{\,q} \left( {x + 1} \right)^{\,p - q} } & {0 < p - q} \cr } } \right. $$

We can therefore concentrate to examine $$ \left\{ \matrix{ f(x,y) = \left( {x^{\,2} - 1} \right)^{\,s} \left( {y - 1} \right)^{\,t} = c \hfill \cr y = x \hfill \cr} \right.\quad \left| {\;s,t \in N} \right. $$ that is $$ \left\{ \matrix{ y = 1 + \left( {{c \over {\left( {x^{\,2} - 1} \right)^{\,s} }}} \right)^{\,1\,/\,t} \hfill \cr y = x \hfill \cr} \right. $$ where it is easy to define the domain of existence according to the sign of $x^2-1,\; s, \; c, \; t$, and a sketch surely helps to focus the situation.
For example

The sketch clearly indicates that in this case (and for other $s$ and $t$ with same parity) we will have only one solution for positive $c$, and one or three for negative $c$.

It also tells us how we can compute the solution(s) by any of the classical methods of iterative approximations, and where it is appropriate to place the starting point.

Answer 4

Let WLOG $$(m,n)\in\mathbb N^2\tag1$$ and $$(1+\xi)^m(1-\xi)^n=C.\tag2$$

So $$m\log(1+\xi)+n\log(1-\xi)=\log C,\quad x\in(-1,1).\tag3$$ Note that $$\lim_{x\to\pm1}LHS(3) = -\infty.$$ Maximum of $LHS(3)$ achieves when $$\dfrac m{1+\xi} -\dfrac n{1-\xi}=0,$$ and that gives $$\xi_m=\dfrac{m-n}{m+n}, \quad \max LHS(2) = \left(\dfrac {2m}{m+n}\right)^m\left(\dfrac {2n}{m+n}\right)^n.$$ Thus, the solution of $(2)$ exists only if $$C \in\left[0,\dfrac{2^{m+n}m^mn^n}{(m+n)^{m+n}}\right].$$ If $\xi\not = \xi_m,$ then there are two solutions of $(2),$ which can be calculated using the iteration formulas $$\xi_0 = -1,\quad \xi_{i+1}=1-C(1+\xi_i)^{-\frac mn}$$ for the less root and $$\xi_0=1,\quad \xi_{i+1}=C(1-\xi_i)^{-\frac nm}-1$$ for the greater root.

Answer 5

Hint.

As $-1 < \xi < 1$ consider $\xi = \cos x$ then following

$$ 1-\cos x = 2\sin^2(\frac x2)\\ 1+\cos x = 2\cos^2(\frac x2) $$

we have

$$ \sin^{2n}(\frac x2)\cos^{2m}(\frac x2) = \frac{C}{2^{n+m}} $$

now making considerations over the $\sin^{2n}(\frac x2)\cos^{2m}(\frac x2)$ extrema ...