Suppose that $$f_1 = -\frac{x}{1+x}$$ and $$f_{i} = \frac{-x}{2+2x+xf_{i-1}}, \ \ i>1$$
What is the general form for $f_i ?$ (An approximate solution would be interesting too).
Context: this corresponds to determing the coefficients for the tridiagonal matrix algorithm for the following matrix which appears in the Crank-Nicolson method
$$A= \begin{bmatrix} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \mathstrut^{.^{.^{.}}} \\ \dotsc & -\alpha & 2(1+\alpha) & -\alpha & 0 & 0 & 0 & 0 & \dotsc \\ \dotsc & 0 & -\alpha & 2(1+\alpha) & -\alpha & 0 & 0 & 0 & \dotsc \\ \dotsc & 0 & 0 & -\alpha & 2(1+\alpha) & -\alpha & 0 & 0 & \dotsc \\ \dotsc & 0 & 0 & 0 & -\alpha & 2(1+\alpha) & -\alpha & 0 & \dotsc \\ \dotsc & 0 & 0 & 0 & 0 & -\alpha & 2(1+\alpha) & -\alpha & \dotsc \\ \mathstrut^{.^{.^{.}}} & \vdots &\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}$$ With top row of equal to $\begin{bmatrix} 2(1+\alpha) & -2\alpha & 0 & ... \end{bmatrix},$ and bottom row equal to $\begin{bmatrix} ... & 0 & -2\alpha & 2(1+\alpha)\end{bmatrix}.$
Here are the first few cases:
- $f_1 = -\frac{x}{x+1}$
- $f_2 = -\frac{x^2+x}{x^2 + 4x +2}$
- $f_3 = -\frac{x^3 +4x^2 + 2x}{x^3 + 9x^2 + 12x+4}$
- $f_4 = -\frac{x^4 + 9x^3 + 12x^2 + 4x}{x^4 + 16x^3 + 40x^2 + 32x + 8}$
So far, in general, it looks like we have $$f_i = - \frac{x P(x^{i-1})}{Q(x^i)}$$ Where both $P(x^i)$ and $Q(x^i)$ are polynomials of degree $i$ with strictly positive coefficients. It would be interesting to have an algorithm to determine the coefficients of the polynomials. It would also be interesting to have an approximation for $f_i$ that works for small $x.$
Letting $u=1+\frac1x,$ and $g_n=-\frac1{f_n},$ you get: $$g_0=u, g_{n+1}=2u-\frac{1}{g_n}.$$
We easily see that $g_n$ is the ratio of polynomials of $u,$ with the numerator of $g_n$ becoming the denominator of $g_{n+1}.$ So if we write: $g_{n}=\frac{p_{n+1}}{p_n},$ we have $$p_{0}=1, p_1=u, p_{n+2}=2up_{n+1}-p_{n}.$$
But this is a linear recurrence, and it can be written in terms of the roots od $v^2-2uv+1=0,$ $v_1,v_2=u\pm \sqrt{u^2-1}.$ Then there are constants (in terms of $u,$) $a,b,$ such that: $$p_n=av_1^n+bv_2^n.$$ Solving using $n=0,1$ we get $a=b=\frac12.$
So we actually get that $p_n$ is a fairly well-known polynomial sequence, the Chebyshev polynomials of the first kind, $T_n.$
Then you get $$f_n=-\frac{T_{n}(1+1/x)}{T_{n+1}(1+1/x)}$$
If you write it explicitly in terms of $v_1,v_2$ you can multiply numerator an denominator by $v_1^{n+1}$ and get:
$$f_n=-\left(u+\sqrt{u^2-1}\right)\frac{\left(u+\sqrt{u^2-1}\right)^{2n}+1}{\left(u+\sqrt{u^2-1}\right)^{2n+2}+1}$$
When $u>1,$ this will converge to $$-\frac1{u+\sqrt{u^2-1}}=-(u-\sqrt{u^2-1}).$$
When $u<-1,$ it will converge to $-\frac{1}{u-\sqrt{u^2-1}}=-(u+\sqrt{u^2-1}).$
In both cases, it converges to a value in $[-1,1].$
When $u\in[-1,1],$ we can solve $u=\cos \theta,$ and you get:
$$f_n=-\frac{\cos(n\theta)}{\cos((n+1)\theta)},$$ which, depending on $\theta,$ might not even always be defined for all $\theta.$