Wikipedia:Reference desk/Archives/Mathematics/2015 June 22

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June 22

Generalization of a Certain Formula for Pi

Let $a_{0}=0$ and $a_{n+1}={\sqrt {A+Ba_{n}}}.$ Then, assuming convergence, we have $a_{\infty }=\ell ={\frac {B+{\sqrt {B^{2}+4A}}}{2}}.$ Thus, for $A=B=1$ we have $\ell =\phi ,$ for instance. Now, for $A=B={\frac {1}{2}}$ we have $\ell =1,$ and $\lim _{n\to \infty }{\color {red}2}^{n}{\sqrt {\frac {\ell -a_{n}}{2}}}={\frac {\pi }{4}}.$ My question would be with what constant to replace ${\color {red}2}$ in general, for different values of A and B, so that the limit in question should converge to a finite non-zero quantity. In other words, if $f\left({\frac {1}{2}},{\frac {1}{2}}\right)=2,$ what is the general formula for $f(A,B)$ ? Thank you. — 79.118.171.25 (talk) 22:57, 22 June 2015 (UTC)

Apparently, $f(1,1)={\sqrt {1+{\sqrt {5}}}}~,$ and the limit in question is the square root of the Paris constant. — 79.118.171.25 (talk) 03:07, 23 June 2015 (UTC)

I get $f(A,B)={\sqrt {2\ell /B}}$. My idea is to let $x_{n}=\ell -a_{n}$ and write its recurrence formula. The behavior for small $x$ is dictated by the linear term of its Maclaurin series. $\ell -{\sqrt {A+B(\ell -x)}}=0+{\frac {B}{2\ell }}x+...$ From there it's not hard to understand the behavior of ${\sqrt {\frac {x_{n}}{2}}}$ and find $f(A,B)$. Egnau (talk) 03:40, 23 June 2015 (UTC)

I arrived just these past few minutes at the same conclusion, and wanted to post it, but was unable to connect. :-) Thanks ! — 79.113.226.120 (talk) 03:53, 23 June 2015 (UTC)

And in general, for $a_{n+1}={\sqrt[{m}]{A+Ba_{n}}}$ we have $f_{m}(A,B)={\sqrt {\frac {m~\lambda ^{m-1}}{B}}}~,$ where $\lambda$ is a root of $t^{m}=A+Bt.$ — 79.113.226.120 (talk) 10:08, 23 June 2015 (UTC)

Resolved