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been trying to solve the non-linear recurrence $a_{n+1}=\frac{10}{7-a_{n}}$ with $a_1=4$ I know I'm supposed to be getting $a_{n}=\frac{15(2)^n}{5(2)^n+5^n}+2$, but I just don't see how, I've tried an iterative approach expanding out a few terms in both the forward or backwards direction but don't seem to get anywhere, I've also considered the possibility of generating functions but given the non-linear nature of the problem I figured there wouldn't be much success to find in that area. any help would be appreciated

Reuben Miller
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    You have a Mobius transformation with matrix $M = \left( \begin{array}{rr} 0 & 10 \ -1 & 7 \ \end{array} \right) $ You then need $M^k.$ You find a matrix $P$ such that $ PDP^{-1} = M$ where $D$ is diagonal (the eigenvalues). Then $M^k = P D^k P^{-1} $ and $M^k$ is the coefficient matrix for the Mobius transformation applied $k$ times...... – Will Jagy Apr 30 '24 at 03:38
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    Have a look at https://math.stackexchange.com/questions/3885418/finding-the-closed-form-of-a-recursive-sequence/3885693#3885693 – Claude Leibovici Apr 30 '24 at 04:55
  • ok @WillJagy I don't have as much experience with Mobius Transformations, but I know through the process of diagonalization we find $PD^nP^{-1}=\begin{pmatrix}2&5\ ::1&1\end{pmatrix}\begin{pmatrix}5^n&0\ ::0&2^n\end{pmatrix}\begin{pmatrix}-\frac{1}{3}&\frac{5}{3}\ ::\frac{1}{3}&-\frac{2}{3}\end{pmatrix}=\begin{pmatrix}0&10\-1&7\end{pmatrix}^n$ which implies $M^n=\begin{pmatrix}\frac{-2\cdot ::5^n+5\cdot ::2^n}{3}&\frac{2\cdot ::5^{1+n}-5\cdot ::2^{1+n}}{3}\ :\frac{-5^n+2^n}{3}&\frac{5^{1+n}-2^{1+n}}{3}\end{pmatrix}$ so where do I go from here to find $a_{n}$ – Reuben Miller Apr 30 '24 at 16:09
  • A coefficient matrix $ \left( \begin{array}{rr} s & t \ u & v \ \end{array} \right) $ refers to the Mobius Transformation $f(z) = \frac{sz + t}{uz + v}.$ Since you start with $z=a_1=4,$ to get $a_n$ you actually need $M^{n-1}.$ If you prefer, calculate $a_0$ to use $M^n.$ One mercy, the transformation is a fraction, you are free to multiply all four entries of the coefficient matrix by any useful constant, in your case $3.$ That's all, more calculation. – Will Jagy Apr 30 '24 at 17:34
  • Given a Mobius transformation $f(z)$ with coefficient matrix $M,$ the coefficient matrix for $f^{-1}$ is just $M^{-1}.$ Homework, lots of letters: given transformations $f(w) = \frac{mw+n}{ow +p}$ and $g(z) = \frac{qz+r}{sz+t}$ Find the transformation $f \circ g$ and especially its coefficient matrix. Let's see, with my $f(w)$ and $g(z)$ set $w = g(z)$ so as to calculate $f(g(z))$ and confirm the matrix of coefficients. That's what makes these useful for problems, the matrix of the composition is the product of the matrices, and in the same order. – Will Jagy Apr 30 '24 at 18:30

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