1
$\begingroup$

I want to build this infinite continued fraction

$$F_{n}(x)= \frac{1}{1-x\frac{(n+1)^2}{4(n+1)^2-1}F_{n+1}(x)} $$

which gives for $n=0$

$$F_{0}(x)=\dfrac{1}{1-\dfrac{(1/3)x}{1-\dfrac{(4/15)x}{1-\dfrac{(9/35)x}{1-\ddots}}}}$$ I took inspiration from this Post (@Michael E2), the problem is that when I transform it as a list representation 

{b0,{a1, b1},{a2, b2},...}

Clear[F2,iF2];

iF2[0]=0;
iF2[1]={1,1};
iF2[2]={-x/3,1};
iF2[n_]:={-x(n+1)^2/(4(n+1)^2-1),1};
F2[n_]:=Table[iF2[k],{k,0,n}];

I can't find all the terms, so I find for 5 terms

Block[{n=5},F2[n]]
(*{0,{1,1},{-x/3,1},{-16x/63,1},{-25x/99,1},{-36x/143,1}}*)

it lacks after {$-x/3,1$} the terms {$-4x/15,1$} and {$-9x/35,1$}

What is wrong please?

|improve this question
$\endgroup$
  • 1
    $\begingroup$ In your definition you should change + to -, that is iF2[n_]:={-x(n-1)^2/(4(n-1)^2-1),1}; $\endgroup$ – yarchik Jul 6 '19 at 14:18
1
$\begingroup$

You could use Nest to compute the continued fraction

Nest [# /. F[n_] :> 1/(1 - x (n + 1)^2/(4 (n + 1)^2 - 1) F[n + 1]) &, F[0], 5] 
(*1/(1 - x/(3 (1 - (4 x)/(15 (1 - (9 x)/(35 (1 - (16x)/(63 (1 - 25/99 x F[5])))))))))*)
|improve this answer
$\endgroup$
  • $\begingroup$ @Ulrich_Neumann, how to get the value for n=200000? $\endgroup$ – Gallagher Aug 31 '19 at 18:55
  • $\begingroup$ @Gallagher Just replace "5" by "200000" ...But I'm afraid, wether Mathematica ever finishs the analytical calculation $\endgroup$ – Ulrich Neumann Sep 1 '19 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.