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I am trying to solve the following recurrence :

$T_n=\frac{1}{4-T_{n-1}}$

I tried various methods using range transformation but still can't figure it out.

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    The *homographic transformation* $$T\mapsto\frac1{4-T}$$ has two fixed points $u$ and $v$, such that $u=1/(4-u)$ and $v=1/(4-v)$, whose values are $u,v=2\pm\sqrt3$ and such that $u=4-1/u$ and $v=4-1/v$. In such cases, it is a theorem that the reduced variable $$x_n=\frac{T_n-u}{T_n-v}$$ solves a much simpler recursion, here $$x_n=wx_{n-1}\qquad w=\frac{u}v$$ It follows that $x_n=w^nx_0$ for every $n$, that is, $$\frac{T_n-u}{T_n-v}=w^n\frac{T_0-u}{T_0-v}$$ that is, $$T_n=\frac{u^nv(T_0-u)-v^nu(T_0-v)}{u^n(T_0-u)-v^n(T_0-v)}$$ for every $n$ and $T_0$ such that the denominator is not zero. ... – 2017-02-03
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    ... If $T_0=0$, one gets $$T_n=\frac{uv^{n+1}-vu^{n+1}}{v^{n+1}-u^{n+1}}$$ hence $T_n$ is well defined for every $n\geqslant0$. Additionally the formula for $T_n$ shows that the values of $T_0$ such that the sequence is not fully defined are exactly $$T_0=\frac{u^{n+1}-v^{n+1}}{u^n-v^n}\qquad (n\geqslant1)$$ – 2017-02-03

2 Answers 2

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If it has a limit $L$, $T_n=\frac{1}{4-T_{n-1}} $ becomes $L=\frac{1}{4-L} $ or $L^2-4L+1 = 0$ so that $L =\dfrac{4\pm\sqrt{16-4}}{2} =2\pm\sqrt{3} $. Let $L_1 =2+\sqrt{3} $ and $L_2 =2-\sqrt{3} $.

If $0 < T_{n-1} < 1$, then $\frac14 < T_n < \frac13 $. After this, $\frac1{4-1/3} < T_{n+1} < \frac1{4-1/4} $ or $\frac{3}{13} < T_{n+1} < \frac{4}{15} $.

If $0 < x < \frac14$, then $1+x <\frac1{1-x} < 1+2x $, so that, if $0 < T_{n-1} < 1$, then $T_n =\frac{1}{4-T_{n-1}} =\frac14\frac{1}{1-T_{n-1}/4} \gt\frac14(1+T_{n-1}/4) $ and $T_n =\frac14\frac{1}{1-T_{n-1}/4} \lt\frac14(1+2T_{n-1}/4) =\frac14+\frac{T_{n-1}}{8} \lt \frac38 $.

Let $u_n = T_n-L_2 $. Then $u_n+L_2 =\dfrac1{4-(u_{n-1}+L_2)} $ or

$\begin{array}\\ 1 &=(u_n+L_2)(4-L_2-u_{n-1})\\ &=u_n(4-L_2)-L_2u_{n-1}+L_2(4-L_2)\\ &=u_n(4-L_2)-L_2u_{n-1}+1 \qquad\text{since } L_2(4-L_2) = 1\\ \text{so}\\ u_n(4-L_2) &=L_2u_{n-1}\\ \text{or}\\ u_n &=\frac{L_2}{4-L_2}u_{n-1}\\ &=L_2^2u_{n-1}\\ \end{array} $

Therefore, if $0 < u_n < 1$, $u_{n+k} =L_2^{2k}u_n $.

Since $L_2^2 =\frac{7-4\sqrt{3}}{4} \approx 0.07179676972449 $, this converges quickly.

We then have $T_{n+k}-L_2 =L_2^{2k}(T_n-L_2) $ or $T_{n+k} =L_2^{2k}(T_n-L_2)+L_2 $.

Therefore, if $0 \le T_0 < 1$, $T_{k} =L_2^{2k}(T_0-L_2)+L_2 $.

In particular, if $T_0 = 0$, $T_{k} =L_2-L_2^{2k+1} $.

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We have,

$$4T_n-T_{n-1}T_n=1$$

Let $\frac{f(n-1)}{f(n)}=T_{n}$. So that $T_{n-1}=\frac{f(n-2)}{f(n-1)}$.

This gives,

$$4\frac{f(n-1)}{f(n)}-\frac{f(n-2)}{f(n)}=1$$

$$4f(n-1)-f(n-2)-f(n)=0$$

$$f(n-2)-4(n-1)+f(n)=0$$

This has characteristic equation,

$$r^2-4r+1=0$$

Whose solutions are $r=2 \pm \sqrt{3}$. Hence,

$$f(n)=c_1(2+\sqrt{3})^n+c_2(2-\sqrt{3})^n$$

$$T_{n}=\frac{c_1(2+\sqrt{3})^{n-1}+c_2(2-\sqrt{3})^{n-1}}{c_1(2+\sqrt{3})^n+c_2(2-\sqrt{3})^n}$$

The constants $c_1,c_2$ can be deduced by the initial condition.

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    How do you find the constants in that case ? – 2017-02-03
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    Given say $T_1$ you can get $T_2$. Then we plug the two values into the formula to get a linear system of equations for $c_1$ and $c_2$. @Zok – 2017-02-03
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    That's extremely complicated btw, given that T(0)= 0 * he forgot to mention *. You will end up having an extremely long equation – 2017-02-03
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    everytime I try to solve 2 linear equations, I don't get an answer, it ends up having something like: c1( ...) = 0 which leads to c2 = 0 hence c1 =0 lol – 2017-02-03
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    Okay sir I'm waiting your answet habibi. – 2017-02-03
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    I see the issue with $f(0)=0$ division by $0$ occurs, neither $T_n$ nor $T_{n-1}$ can $0$ at any point or $0=1$. And so our formula does not necessarily work for that case. – 2017-02-03
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    So in that case it's impossible to define c1 and c2. – 2017-02-03
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    Then your solution isn't correct..? – 2017-02-03
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    It correct for $T_0 or T_1,... \neq 0$ – 2017-02-03
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    But t(0) = 0 in the main question. That's the tricky point lol – 2017-02-03
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    How do you know @Zok – 2017-02-03
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    https://cerocks.files.wordpress.com/2011/03/fundamentals-of-algorithmics-brassard_ingles.pdf book page 145 ( 161 pdf ), problem 4.42 :) – 2017-02-03
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    http://math.stackexchange.com/questions/2127093/closed-form-t-n-frac14-t-n-1-for-t-0-0 @Zok – 2017-02-03
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    in today's class the professor who made the book solved it. The answer will be the same formula u provided with c1=1, c2=-1. – 2017-02-03