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I was looking for a closed form solution to $$1 - \left(\frac{n-x}{n+1}\right)^n = \left(\frac{n(x+1)}{n+1}\right)^n,$$ Where we fix a $n \in \mathbb{N}$ and want to find some $x \in (-1, n).$ I know there exists exactly one solution for all $n$, but I was wondering if anyone had any ideas/tricks to find the closed form solution for some particular $n$. If it is helpful, the limit as $n\to\infty$ has $x\to0$ as a solution and the value is $1 - \frac{1}{e}$. Furthermore, if we extend $n$ to be negative we get that as $n\to-1: x\to -1$ is a solution and the value is $1 - \phi$ where $\phi$ is the golden ratio.

So far, I have shown also that the solution for any $x$ lies between $\left(0, \frac1n\right]$. If anyone has experienced a similar problem and could also direct me to any papers those would be very much appreciated.

Please let me know if you have any ideas. Everything helps. Thank you so much for your taking your time to read this.

TShiong
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Brayden
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  • Could you clarify what is the specific question you are asking? Not clear to me – Zima Jun 19 '23 at 08:45
  • Is the numerator on the right-hand side $x+1$, $(x+1)^n$ or $n(x+1)$? Asking because the expression on the LHS is quite different from the expression on the RHS, I wonder if it is intentional or a typo. –  Jun 19 '23 at 08:48

2 Answers2

2

Considering that, for $n>0$, you look for the zero of function $$f_n(x)= \left(\frac{n(x+1)}{n+1}\right)^n+\left(\frac{n-x}{n+1}\right)^n -1$$ you can obtain good and explicit approximations of the solution using $\color{red}{\text{the first iterate } x_1}$ of any Newton-like method starting with $x_0=\frac 1{2n}$ which is the mid point of the range you properly found.

$$\left( \begin{array}{ccccc} n & \text{Newton} & \text{Halley} & \text{Householder} & \text{solution}\\ 2 & 0.20192307692 & 0.20007680492 & 0.20000307201 & 0.20000000001 \\ 3 & 0.14786159467 & 0.14742915877 & 0.14742113081 & 0.14742098978 \\ 4 & 0.11640467089 & 0.11628024514 & 0.11627903269 & 0.11627902258 \\ 5 & 0.09591899676 & 0.09588305657 & 0.09588286839 & 0.09588286762 \\ 6 & 0.08154173322 & 0.08153326724 & 0.08153324548 & 0.08153324563 \\ 7 & 0.07090308560 & 0.07090222401 & 0.07090222329 & 0.07090222329 \\ 8 & 0.06271554490 & 0.06271537743 & 0.06271537749 & 0.06271537749 \\ 9 & 0.05622081012 & 0.05621899986 & 0.05621900215 & 0.05621900216 \\ 10 & 0.05094369255 & 0.05093961650 & 0.05093962441 & 0.05093962448 \\ 15 & 0.03466687967 & 0.03465440239 & 0.03465444977 & 0.03465445080 \\ 20 & 0.02627067957 & 0.02625540494 & 0.02625547564 & 0.02625547775 \\ 25 & 0.02114808974 & 0.02113239644 & 0.02113247642 & 0.02113247926 \\ 30 & 0.01769709234 & 0.01768185314 & 0.01768193537 & 0.01768193528 \\ 35 & 0.01521430619 & 0.01519982012 & 0.01519990129 & 0.01519990122 \\ 40 & 0.01334241629 & 0.01332875729 & 0.01332883591 & 0.01332883584 \\ 45 & 0.01188066213 & 0.01186781431 & 0.01186788976 & 0.01186788970 \\ 50 & 0.01070756392 & 0.01069547703 & 0.01069554913 & 0.01069554908 \\ \end{array} \right)$$

Edit

We could do better onsidering instead that we look for the zero of function $$g_n(x)=\log\left( \left(\frac{n(x+1)}{n+1}\right)^n+\left(\frac{n-x}{n+1}\right)^n\right) $$ which is more linear than $f(x)$.

Expanding as series around $x=0$ gives $$g(x)=\log \left(2 \left(\frac{n}{n+1}\right)^n\right)+2\sum_{k=1}^\infty \frac{(n-1)\,P_k(n)}{a_k\,\,n^{k-1}}x^k $$ where the first coefficients $a_k$ form the (unknown) sequence $$\{1,4,12,96,240,2880,6720,322560,80640,14515200\}$$ The very first polynomials $P_k(n)$ being $$\left( \begin{array}{cc} k & P_k(n) \\ 1 & 1 \\ 2 & n^2+n+2 \\ 3 & -3 n^3-2 n^2+n+4 \\ 4 & -n^6-5 n^5+11 n^4-n^3-14 n^2+2 n+24 \\ 5 & 5 n^7+20 n^6-20 n^5-12 n^4-7 n^3-32 n^2-2 n+48 \\ \end{array} \right)$$ We can generate as many terms as required and, using power series reversion with the explicit formula for the $n^{\text{th}}$ term given by Morse and Feshbach, we could find the exact solution.

Claude Leibovici
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0

Assuming $-1<x<n\in\Bbb N$ is wanted, define $x=\sqrt[n]w-1$ and use Lagrange reversion:

$$w=\left(\frac1n+1\right)^n-\left(\frac1n+1-\frac{\sqrt[n]w} n\right)^n\implies x=\frac 1n+\frac1n\sum_{k=1}^\infty\frac{(-1)^k}{k!}\left.\frac{d^{k-1}}{dw^{k-1}}w^{\frac1n-1}\left(\frac1n+1-\frac{\sqrt[n]w}n\right)^{nk}\right|_{\left(\frac1n+1\right)^n}$$

Binomial series gives

$$\frac{d^{k-1}}{dw^{k-1}}w^{\frac1n-1}\left(\frac1n+1-\frac{\sqrt[n]w}n\right)^{nk}=\sum_{m=0}^{nk}\binom{nk}m \left(\frac1n+1\right)^m \left(-\frac1n\right)^{kn-m}\frac{d^{k-1}}{dw^{k-1}}w^{k+\frac{1-m}n-1}$$

where factorial power $u^{(v)}$ appears. Therefore:

$$\bbox[4px,border: 5px ridge #AECCE4 ]{1 - \left(\frac{n-x}{n+1}\right)^n = \left(\frac{n(x+1)}{n+1}\right)^n \implies x=\frac1n+\frac1n\left(\frac1n+1\right)\sum_{k=1}^\infty\sum_{m=0}^{nk}\binom{nk}m\left(\frac{1-m}n+k-1\right)^{(k-1)}\frac{(-1)^{m+(n-1)k}n^{m-nk}}{k!}}$$

shown here. If $k$ goes up to $\infty$, then both sums are interchangeable. Just maybe one can use the Gauss multiplication formula and double hypergeometric series function for a “closed form”. The Fox wright function, a special case of Fox H gives

$$\sum_{m=0}^\infty\frac{\Gamma\left(k+\frac1n-\frac mn\right)\Gamma(1+kn) \left(-(-n)^{-n}\right)^k(-n)^m}{\Gamma\left(1+\frac1n-\frac mn\right)\Gamma(1-m+kn)m!k!}= \frac{\Gamma(1+kn) \left(-(-n)^{-n}\right)^k}{k!}\,_1\Psi_2\left(^{\qquad\left(k+\frac1n,-\frac1n\right)}_{\left(1+\frac1n,-\frac1n\right),(1+kn,-1)};-n\right)$$

so

$$\bbox[4px,border: 5px ridge #AECCE4 ]{1 - \left(\frac{n-x}{n+1}\right)^n = \left(\frac{n(x+1)}{n+1}\right)^n \implies x=\frac1n+\frac1n\left(\frac1n+1\right)\sum_{k=1}^\infty \frac{(-1)^k (-n)^{-kn}(kn)!}{k!}\,_1\Psi_2\left(^{\qquad\left(k+\frac1n,-\frac1n\right)}_{\left(1+\frac1n,-\frac1n\right),(1+kn,-1)};-n\right)}$$

Тyma Gaidash
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