Is this formula for $\frac{e^2-3}{e^2+1}$ known? How to prove it?

sequences-and-series exponential-function irrational-numbers continued-fractions

Solution 1:

For the first one, you could write \begin{equation} f(n) = \frac{n-f(n+1)}{n+f(n+1)} \end{equation} Then you suggest \begin{equation} f(2) = \frac{e^2-3}{e^2+1} \end{equation} But this then gives \begin{align} f(1) = \frac{2}{e^2-1}\\ f(3) = \frac{4}{e^2-1} \\ f(4) = \frac{3e^2-15}{e^2+3}\\ f(5) = \frac{-2e^2+18}{e^2-3} \end{align} I don't know if there is a recurrence relation that solves this, but you have a few more closed forms...

The second one we have \begin{equation} g(2) = \frac{2 - g(2)}{2+g(2)} \end{equation} so we can solve the quadratic $x^2+3x-2$ to get $(\sqrt{17}-3)/2$.

For the third one, we have \begin{equation} h(n) = \frac{n-h(2n)}{n+h(2n)} \end{equation} using the trial version of $h(2)=1/2$, we get \begin{align} h(4)=\frac{2}{3}\\ h(8)=\frac{4}{5}\\ h(16)=\frac{8}{9} \end{align} then it is likely that \begin{equation} h(n)=\frac{n}{n+2} \end{equation} as this satisfies the recurrence and that $h(2)=1/2$.

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