27
$\begingroup$

I was just playing around with a calculator, and came to the conclusion that:

$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$

Now I'm curious. Is it possible to evaluate the exact value of the following?

$\endgroup$
  • 13
    $\begingroup$ In general, expressions like this that are easy to evaluate usually have some sort of symmetry such that we can relate the value to its self via e.g. a square like for example $x^2 = ax + b$. One such example is $x = \sqrt{\frac{1}{2^2} + \sqrt{\frac{1}{2^4} + \ldots + \frac{1}{2^{2^n}}}}$ since then $x^2 = \frac{1}{2^2} + \frac{1}{2}x \to x = \frac{1+\sqrt{5}}{4}$. This one does not have such a symmetry - which does not rule out an analytic solution, but it makes less likely / much harder to find it if it exist. $\endgroup$ – Winther Dec 28 '14 at 13:23
  • 2
    $\begingroup$ Well, proving convergence is at least not too difficult. $\endgroup$ – Gabriel Romon Dec 28 '14 at 13:27
  • 4
    $\begingroup$ $1.2857367633569968672336\ldots$ does not possess a known closed form. $\endgroup$ – Lucian Dec 28 '14 at 15:14
  • 3
    $\begingroup$ The two-letter answer "No" is not allowed. Had it been allowed, it would have got the bounty! $\endgroup$ – GEdgar Dec 31 '14 at 16:14
  • $\begingroup$ How are people numerically evaluating this? It seems like it also suffers from some bad cancellation when using the obvious way (for loop from n = big to n = 0). $\endgroup$ – Y. S. Jan 4 '15 at 23:35
5
+100
$\begingroup$

This question is related to at least five others:

This makes a short answer possible (and desirable).
For a numerical calculation backward recursion is proposed (again): $$ a_{n-1} = \sqrt{1/2^n+a_n} \qquad \mbox{with} \quad \lim_{n\to\infty} a_n = 0 $$ Here comes the Pascal program snippet that is supposed to do the job:

program apart;
procedure again(n : integer); var a,two : double; k : integer; begin two := 1; for k := n downto 2 do two := two/2; a := 0; for k := n downto 2 do begin a := sqrt(two+a); two := two*2; end; Writeln(a); end;
begin again(52); end.
Note that an error analysis is not implemented in the program. This has not much sense because the accuracy is determined by the smallest $1/2^n$ that can be represented with some significance; that is for $n\approx 52$ in double precision Pascal. The outcome is, of course, in concordance with the value already found by Lucian:
 1.28573676335699E+0000
Disclaimer. I certainly would have tried the closed form - whatever that means in modern times - if I only could believe that such a thing does indeed exist here.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

It is easy to show, assuming the limit exists:

$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$

It is also easy to see the following set of inequalities:

$$\sqrt{\frac{1}{2}+1} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}$$

$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+1}} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+...}}}$$

$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{8}+1}}} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+...}}}}$$

And so on, getting better and better approximations.

Calculating the nested radicals:

$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}=\frac{1+\sqrt{3}}{2}$$

$$\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+...}}}=\frac{1+\sqrt{2}}{2}$$

$$\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+...}}}=\frac{1+\sqrt{1.5}}{2}$$

We get following set of boundaries for the value we need:

$$R=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}=1.285737\dots$$

$$1.22474<R<1.36603$$

$$1.27202<R<1.30656$$

$$1.28251<R<1.29120$$

And so on. There is no closed form, but it's not hard to evaluate this nested radical with good precision.

| cite | improve this answer | |
$\endgroup$
-6
$\begingroup$

$$y=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}\equiv\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$

$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$ But the Let the term in bracket be x therefore:$$x=\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\equiv\sqrt{\frac12+x}$$ squaring both sides $$x^2=\frac12+x$$ $$x^2-x-\frac12=0$$solving the equation gives: $$x=\frac{1+\sqrt{3}}{2}$$ but we have $$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}=\sqrt{\frac12+\frac{1}{\sqrt{2}}x}$$ Putting the value of x into y gives$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\frac{1+\sqrt{3}}{2}}=1.211$$ Therefore $$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}=1.211$$ This is an exact value.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @Adriandmen I just posted a solution for $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ $\endgroup$ – shola Otitoju Jan 1 '15 at 19:53
  • $\begingroup$ Please note that $$ \sqrt{\frac 1 2 + \frac{1}{\sqrt{2}}*\sqrt{\frac 1 2+\sqrt{\frac 1 4+\sqrt{...}}}} = \sqrt{\frac 1 2 + \sqrt{\frac 1 4+\frac 1 2\sqrt{\frac 1 4+\sqrt{...}}}} \neq \sqrt{\frac 1 2 + \sqrt{\frac 1 4+\sqrt{\frac 1 8+\sqrt{...}}}}$$ $\endgroup$ – cirpis Jan 1 '15 at 20:02
  • $\begingroup$ I see the mistake. thanks $\endgroup$ – shola Otitoju Jan 1 '15 at 20:16
  • $\begingroup$ will correct it now $\endgroup$ – shola Otitoju Jan 1 '15 at 20:20
  • 1
    $\begingroup$ @sholaOtitoju Note that $\sqrt{1/2 + \sqrt{1/4+\sqrt{1/8+\sqrt{1/16+\ldots}}}} = \sqrt{1/2 + 1/\sqrt{2}\sqrt{1/2 + \sqrt{1/2+\sqrt{1+\sqrt{8+\ldots}}}}}$ so your formula is not correct. This is also easily checked numerically as the solution is $x = 1.2857\ldots$ while you get $x= 1.211\ldots$. $\endgroup$ – Winther Jan 6 '15 at 5:10

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.