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I'd like to find a closed-form expression for $$ x = \frac{1}{\displaystyle2 + \frac{3}{\displaystyle4 + \frac5{6+\cdots}}}$$

My own attempt implies $ x = \frac{1}{2} $. But by numerical approximation the sequence seems to converge to $ \approx 0.379731955 $.

Lorents
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gudetama cutey
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  • Hi and welcome to the site! It is advisable that you take a tour to see what we are about. Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. Here's a quick guide (if nothing else, read up the part on "avoiding no-clue questions"). – 5xum Mar 16 '23 at 13:30
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. – 5xum Mar 16 '23 at 13:30
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    Before you can talk about convergence, it is really helpful to be clear about exactly what the sequence is whose convergence is at issue. I.e., what exactly are the finite terms? Maybe $$\dfrac 12, \dfrac 1{2+\dfrac34}, \dfrac 1{2+\dfrac3{4+\dfrac56}}, \dots$$My attempt to calculate those terms matches your calculation. But I suspect your "solving" didn't examine this correctly. But since you didn't explain anything about what that solution was, there is no way to tell what might be wrong with it. – Paul Sinclair Mar 16 '23 at 13:46
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    To improve your post, let me suggest editing it to give more details behind this rather vague quote: When I tried solving this, it said it was equal to $\frac{1}{2}$. I'm particularly puzzled what the first "it" refers to. – Lee Mosher Mar 16 '23 at 13:52
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    Same question here found using formula searching tool approach0. – Jean Marie Mar 16 '23 at 13:52
  • Yes, show us your reasons for getting $\frac12.$ – Thomas Andrews Mar 16 '23 at 15:05
  • It is not clear why one would expect a closed formula for this. – Thomas Andrews Mar 16 '23 at 15:07
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    @JeanMarie That link suggests $\dfrac{1}{\sqrt{\frac{\pi}{2e}} \text{erfi}(\frac{1}{\sqrt{2}})} - 1 \approx 0.37973289588$ but my attempt seems closer to $0.37973195474$, as in the OP – Henry Mar 16 '23 at 15:52
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    @Henry Mathematica agrees and gives that result. In fact, it can give the partial continued fraction up to $k$ fwiw in terms of horrible $\Gamma$ function mess. It just converges very slowly. If anyone has Mathematica, run ContinuedFractionK[2n-1,2n,{n,1,k}] then take $k\to\infty$. – KStarGamer Mar 16 '23 at 15:59
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    @KStarGamer Thank you for the hints. I did it with the result $f=\frac{\sqrt{\frac{e}{\pi }} \left(\sqrt{2}-2 F\left(\frac{1}{\sqrt{2}}\right)\right)}{\text{erfi}\left(\frac{1}{\sqrt{2}}\right)}$, where $F(x)=e^{-x^2} \int_{0}^{x} e^{y^2};dy$ is the Dawson integral. The result is that obtained earlier by Henry. – Dr. Wolfgang Hintze Mar 16 '23 at 20:10

1 Answers1

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$$ x = \frac{1}{\displaystyle2 + \frac{3}{\displaystyle4 + \frac5{6+\cdots}}}= \frac {\int _0^1 x^\frac{1}{2}\cdot e^{\frac{1}{2}x} dx} {\int _0^1 x^{-\frac{1}{2}}\cdot e^{\frac{1}{2}x} dx} =0.37973195474099563280…$$

This is not difficult to prove by using Euler's Differential Method on the recursive equation $$(2n+1)I_n=(2n+2)I_{n+1}+I_{n+2}$$

My proof

Paul vdVeen
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  • What is Euler's Differential Method? – gist076923 Mar 16 '23 at 20:34
  • A quick google search brings up the method for numerically solving a differential equation – gist076923 Mar 16 '23 at 20:34
  • One of the best explanations of Euler's Differential Method is Khrushchev, S. (2008). Orthogonal Polynomials and Continued Fractions: From Euler's Point of View (Encyclopedia of Mathematics and its Applications). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511721403

    See also my book "Some Continued Fractions and proofs", https://www.vandeveen.nl/Wiskunde/Diversen/Continued_Fractions_and_Proofs.pdf

     – Paul vdVeen Mar 16 '23 at 20:39
  • +1. The proposer's $x=1/2$ is easily seen to be wrong. If $x=1/2$ then $3/(4+5/6+...)=0$ but then $\infty=4+5/(6+...)<4+5/6. $ BTW in older books the continued fraction is written $1+\frac {2}{3+}\frac {4}{5+}...$ – DanielWainfleet Mar 17 '23 at 01:08