Find solution to multivariate equation in the infinite limit for one variable.

Question

I have the below equation (where $\{x|x\in\mathbb{R},0\lt x\lt \frac{1}{2}\}$ and $\{n|n\in\mathbb{N},n\geq5\}$):

$-(\frac{x}{1-x})^{n-1}=\frac{-x(n+2)+2}{-x(n+2)+n}$

A) Ideally, I want to determine a closed form solution for $x$ (in terms of $n$).

B) I want to determine the value of $x$ that satisfies the equation in the limit as $n\rightarrow\infty$.

I have the following results so far.

Bounds for the LHS:

$0\lt x\lt\frac{1}{2}\Rightarrow 0\lt\frac{x}{1-x}\lt 1\Rightarrow -1\lt-(\frac{x}{1-x})^{n-1}\lt 0$

Thus, the RHS satisfies:

$\frac{-x(n+2)+2}{-x(n+2)+n}\lt 0\Rightarrow -x(n+2)+2\lt 0\Rightarrow x\gt\frac{2}{n+2}$

My hypothesis is that $x\rightarrow\frac{2}{n+2}$ as $n\rightarrow\infty$. Empirically, $x\rightarrow\frac{2}{n+2}$ very fast (within $10^{-5}$ for $n$ as small as $10$).

But how do I prove this?

Answer 1

Instead of the equation, consider that you look for the zero of function $$f_n(x)=\left(\frac{x}{1-x}\right)^{n-1}+\frac{2-(n+2) x}{n-(n+2) x}$$ If you croos multiply, this reduces to a polynomial of degree $n$ in $x$; so no hope to obtain a closed form solution.

However, we can have rather good approximations since you already found that $x > \frac 2{n+2}$.

So, the idea is to make one single iteration of Newton method using $x_0=\frac 2{n+2}$. I shall let you playing with the derivatives. We have $$f_n(x_0)=2^{n-1} n^{1-n}\qquad f_n'(x_0)=\frac{1}{4} (n+2) \left(2^n (n-1) (n+2)n^{-n}-\frac{4}{n-2}\right)$$ $$f_n''(x_0)=(n+2)^2 \left(2^{n-3} (n-1) (n+2)n^{1-n}-\frac{2}{(n-2)^2}\right)$$

The second derivative is always negative (its limit, reached from below is $-2$). Since $f(x_0)\,f''(x0)<0$, by Darboux theorem, the first iterate of Newton method $$x_1=x_0-\frac{f_n(x_0)}{f'_n(x_0)}$$ will be an overestimate of the solution.

The table below shows that the approximation is quite decent $$\left( \begin{array}{ccc} n & \text{estimate}& \text{solution} \\ 5 & 0.299691340050 & 0.299597236208 \\ 6 & 0.252177068215 & 0.252170390583 \\ 7 & 0.222527643713 & 0.222527489709 \\ 8 & 0.200036679861 & 0.200036677659 \\ 9 & 0.181821967060 & 0.181821967060 \\ 10 & 0.166667008008 & 0.166667008008 \end{array} \right)$$ The result would be better using Halley method; $$x_1=x_0- \frac {2 f_n(x_0) f_n'(x_0)} {2 {[f_n'(x_0)]}^2 - f_n(x_0) f_n''(x_0)}$$ For example, for $n=5$ it would give as an estimate $x_1=0.299579811359$.

The formula given by Newton method is $$\color{red}{x_1=\frac 2{n+2}\Bigg[1+\frac{2^n (n-2) n}{4 n^n-2^n (n-2) (n-1) (n+2)}\Bigg]}$$

For large values of $n$, using equivalents $$\frac{2^n (n-2) n}{4 n^n-2^n (n-2) (n-1) (n+2)} \sim \frac{2^n n^2}{4n^n-2^n n^3}=\frac{1}{\left(\frac{n}{2}\right)^{n-2}-n}\sim \left(\frac{2}{n}\right)^{n-2}$$ So, a shortcut approximation could simply write $$\color{blue}{x \sim\frac 2{n+2}\Bigg[1+\left(\frac{2}{n}\right)^{n-2}\Bigg]}$$

Edit (for your curiosity)

If you accept to go from Halley to Householder method, you have "almost" the analytical solution. Repeating the same calculations as before (with more figures)

$$\left( \begin{array}{ccc} 5 & \color{red}{0.29959}621836908618302 & 0.29959723620765433801 \\ 6 & \color{red}{0.252170390}31936431722 & 0.25217039058345961901 \\ 7 & \color{red}{0.2225274897091}0826671 & 0.22252748970915718360 \\ 8 & \color{red}{0.2000366776586031}5836 & 0.20003667765860316213 \\ 9 & \color{red}{0.18182196703765741544} & 0.18182196703765741544 \\ 10 & \color{red}{0.16666700800737521733} & 0.16666700800737521733 \end{array} \right)$$

Update

If you search for a very high accuracy, I suggest to let

$$x \sim\frac 2{n+2}\Bigg[1+\left(\frac{2}{n}\right)^{n-2}\,\exp \bigg[-\sum _{i=0}^p \frac{a_i}{n^i} \bigg]\Bigg]$$ and to identify the $a_i$.

It seems that $(0,2,2,3)$ is not bad.

For $n=100$, it gives an absolute error equal to $\color{red}{1.22\times 10^{-168}}$.

Answer 2

A)

For the exact solution of $x$ in terms of $n$ for this $n^{th}$ degree polynomial equation, it seems it is always possible. At least WolframAlpha gives such closed forms for high $n$ as well (e.g. $n=20$: https://www.wolframalpha.com/input?i=-%28x%2F%281%E2%88%92x%29%29%5E%2820%E2%88%921%29%3D%28-x*%2820%2B2%29%2B2%29%2F%28-x*%2820%2B2%29%2B20%29).

How does WolframAlpha do this? How can I get the closed form myself?

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B)

The following approach to try to show the limit seems to have a flaw, but maybe it's on the right track.

Since we know $x\gt\frac{2}{n+2}$, we can substitute $x=\frac{2}{n+2}+\epsilon$ into the equation (where $\epsilon>0$ can be arbitrarily small).

So the equation becomes

$-(\frac{\frac{2}{n+2}+\epsilon}{1-(\frac{2}{n+2}+\epsilon)})^{n-1}=\frac{-(\frac{2}{n+2}+\epsilon)(n+2)+2}{-(\frac{2}{n+2}+\epsilon)(n+2)+n}$

$-(\frac{2+\epsilon n+2\epsilon}{n-\epsilon n-2\epsilon})^{n-1}=\frac{-\epsilon n-2\epsilon}{n-2-\epsilon n-2\epsilon}$

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We first evaluate the limit for the LHS: $\lim\limits_{n\rightarrow\infty}{LHS}=\lim\limits_{n\rightarrow\infty}{-(\frac{2+\epsilon n+2\epsilon}{n-\epsilon n-2\epsilon})^{n-1}}$

Defining $f(n)=\frac{2+\epsilon n+2\epsilon}{n-\epsilon n-2\epsilon}$ and $g(n)=n-1$, we're trying to find $\lim\limits_{n\rightarrow\infty}{-f(n)^{g(n)}}=-\lim\limits_{n\rightarrow\infty}{\exp{(\ln{f(n)^{g(n)}})}}=-\lim\limits_{n\rightarrow\infty}{\exp{(g(n)\ln{f(n)})}}$.

Since they're continuous, this is $-\exp{(\lim\limits_{n\rightarrow\infty}{g(n)\ln{f(n)}})}=-\exp{(\lim\limits_{n\rightarrow\infty}{g(n)}\lim\limits_{n\rightarrow\infty}{\ln{f(n)}})}=-\exp{(\infty \ln{\frac{\epsilon}{1-\epsilon}})}$. Since $0\lt\epsilon\lt 1$, $\ln{\frac{\epsilon}{1-\epsilon}}<0$. So we end up with $-\exp{(-\infty)}=0$.

We now evaluate the limit for the RHS and set it to equal to $0$ (since it must match the LHS): $\lim\limits_{n\rightarrow\infty}{RHS}=\lim\limits_{n\rightarrow\infty}{\frac{-\epsilon n-2\epsilon}{n-2-\epsilon n -2\epsilon}}=\frac{-\epsilon}{1-\epsilon}=0$.

Thus, as $n\rightarrow\infty$, $\epsilon=0\Rightarrow x=\frac{2}{n+2}$.

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The problem with this approach is that it gives $\epsilon=0$ upon substituting other values of $x$ as well (e.g. $x=\frac{3}{n+2}+\epsilon$ or $x=\frac{2}{n+2}+\frac{1}{n^2}+\epsilon$). Can I make the following argument somehow: We already know that $x\gt\frac{2}{n+2}$, so since the approach works upon substituting any value $x\gt\frac{2}{n+2}$, the limiting value must be $\frac{2}{n+2}$?

Answer 3

Answering your own answer

• Wolfram Alpha does not give you the exact solution : it just solve numerically for all roots of a polynomial of degree $20$. Just try this one in which I changed the $20$ to $20.1$. In what I wrote in my previous answer, you can treat $n$ as a real.

• If we follow your idea $$x=\frac 2 {n+2}+\epsilon \quad \implies \quad \left(\frac{2+(n+2) \epsilon }{n-(n+2) \epsilon }\right)^{n-1}+\frac{(n+2) \epsilon }{(2-n)+(n+2) \epsilon }=0$$ Expanding as Taylor series, we end with $$0=2^{n-1} n^{1-n}+(n+2) \left(2^{n-2} (n-1) (n+2) n^{-n}+\frac{1}{2-n}\right) \epsilon +O\left(\epsilon ^2\right)$$ Ignoring the higher order terms $$\epsilon=\frac{ n\,(n-2)} {n+2 }\,\,\frac{2^{(n+1)}}{4 n^n-2^n (n-2) (n-1) (n+2) }$$ where you see a kind of exponential trend