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Square root in suqare root in square root in square root in

  1. Aug 31, 2011 #1
    square root in suqare root in square root in square root in....

    1. The problem statement, all variables and given/known data
    Find the value of:
    [tex]\sqrt{1001\sqrt{1002\sqrt{1003\sqrt{1004...}}}}[/tex]


    2. Relevant equations
    exponential and surd


    3. The attempt at a solution
    I tried to square it but didn't find any patterns because the radicands are not the same. Help...

    Thanks
     
  2. jcsd
  3. Aug 31, 2011 #2

    lanedance

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    Re: square root in suqare root in square root in square root in....

    not sure if it will work, but how about writing as
    [tex] (1001(1002...)^{\frac{1}{2}})^{\frac{1}{2}}[/tex]

    then maybe see if you can simplify and take a logarithm?
     
  4. Aug 31, 2011 #3

    Ray Vickson

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    Re: square root in suqare root in square root in square root in....

    Here is a suggestion which I have actually tried---and it works. Look at successive approximations: let S[1](x) = sqrt(x), S[2](x) = sqrt(x*S[1](x)),... S[n+1](x) = sqrt(x*S[n](x)),.. and compute S[1](x),..., S[n](x) numerically for a few values of x and for n up to about 20-30; for example, you could use a spreadsheet. The results may surprise you, as they did me. The answer for lim_{n->infinity}S[n](1001) is then clear, which may then help in finding a way to prove it. It is always easier to prove something when you know what you are looking for.

    RGV
     
  5. Aug 31, 2011 #4

    Dick

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    Re: square root in suqare root in square root in square root in....

    That's always a good first strategy. The numerical approach at first APPEARS to converge to 1002 to a good approximation. Which would be surprising! But I don't think that's actually the limit. It's a bit less than 1002. I really don't think this is a function you can compute analytically. It would be interesting to be shown it could, though. You CAN show it does converge using lanedance's suggestion. And you CAN write a numerical approximation to the answer. But I think that's all you can do.
     
    Last edited: Aug 31, 2011
  6. Sep 1, 2011 #5

    ehild

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    Re: square root in suqare root in square root in square root in....

    As Lanedance suggested, the expression can be written in the form

    [tex]\prod_1^n{(1000+k)^{\frac{1}{2^k}}}[/tex].

    Taking the logarithm, it is a series and some of the convergence-criteria can be applied to show that it is convergent or not, and can find a lower limit.

    ehild
     
  7. Sep 1, 2011 #6
    Re: square root in suqare root in square root in square root in....

    Sir, shouldn't the exponent be ( 1 / 2k ) ?

    If not so, could you please elaborate.

    Thanks.
     
  8. Sep 1, 2011 #7

    NascentOxygen

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    Re: square root in suqare root in square root in square root in....

    We need powers of 1/2, 1/4, 1/8, 1/16, 1/32, etc.

    sqrt(sqrt(sqrt(x))) = x to the power 1/8
     
  9. Sep 1, 2011 #8
    Re: square root in suqare root in square root in square root in....

    I think the answer lies in the finding of f(x)

    where

    f**2(x-1) = ( x-1 ) * f(x)

    f squared of (x-1 ) equals ( x minus 1 ) times ( f of x )

    f(x-1) * f(x-1) = (x-1) * f(x)

    if we can solve this equation and find f(x)

    We can then plug in the value x=1001

    and find out what f(1001) is.
     
  10. Sep 1, 2011 #9

    Ray Vickson

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    Re: square root in suqare root in square root in square root in....

    If S(x) = sqrt(x*sqrt(x+1)*sqrt(... ))), taking logarithms gives L(x) = lnS(x) = sum_{k=0..infinity} ln(k+x)/2^(k+1). The ratio test show that the series is convergent for any x > 0. We can get good numerical values for L(x) just by taking many terms, in a simple spreadsheet or programmable hand-held calculator. For example, by performing an error analysis on the infinite series we are guaranteed that taking 41 terms (sum_{k=0..40}) gives more than full 6-decimal accuracy for S(1001) = exp(L(1001)).

    RGV
     
    Last edited: Sep 1, 2011
  11. Sep 2, 2011 #10

    uart

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    Re: square root in suqare root in square root in square root in....

    Just for fun I download the mpmath package for Python. The following is the evaluation for [strike]7000[/strike] 6999 terms at 1200 dp precision. :smile:

    Code (Text):
    from mpmath import mp

    mp.dps=1200
    s=mp.mpf(0)
    p2=mp.mpf(2)

    for k in range(1,7000) :
        x=mp.mpf(1000+k)
        s=s+mp.log(x)/p2
        p2=p2+p2

    print mp.exp(s)
    Code (Text):
    1001.99900397913691898234439715244344613729998738717041977274676200148199509558817554545417076124318480940610680361311614854981603236455318607100162785113625170638256422574541347581096615035978851664139085456338883594740422731293953208208144172243868905270361292059529263885996278587885993677025105711750619813487545990403024502836716136675785661195184541807980681958180327723191234228986168561284824245851495221985327190918407653825883156881071095089445088055616086340359845081535782146671200814749246453040284862549740844129680298559371650374132490575305619530188927496517152722612153193833757754310883347426813571188377508484422925408646686318012394719790714830903494186462930528915274650314653701307950669845027675076978273020036611482127244665921724832927562221042065147872776648580806362571880958572257635842221805174250500907725156069329281841466085876234699317471711743482677662082232151927374650792614369602453252418264601925148261436388726832874441122942437367419289549047172944608662444481832971555133410691363743174251382431045383895821360549236143488774247447599308457363870075777157976111773915678002918789458376189266258999122256218899692315961897948080517046667533746089790257098403498
     
  12. Sep 2, 2011 #11

    Dick

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    Re: square root in suqare root in square root in square root in....

    You can also use the power series expansion of log(1+x) to start writing the expression in terms of powers of n=1000. n*exp(2/n-6/(2n^2)+...). The n*exp(2/n) part shows you why the result is close to 1002.
     
  13. Sep 2, 2011 #12

    ehild

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    Re: square root in suqare root in square root in square root in....

    It is easy to find a lower and an upper bound for the original expression, S.

    The logarithm of the original expression is

    [itex]L=ln(S)=\sum_1^{\infty}ln(1000+k)(\frac{1}{2})^k[/itex]

    [itex]L>ln(1000+1)\sum_1^{\infty}(\frac{1}{2})^k=ln(1001)[/itex], so S>1001.

    To find an upper bound, multiply L by 2 and subtract L.

    [itex]2L=ln(1001)+\sum_1^{\infty}ln(1001+k)(\frac{1}{2})^k[/itex]

    [itex]2L-L=ln(1001)+\sum_1^{\infty}ln(\frac{1001+k}{1000+k})(\frac{1}{2})^k[/itex]

    [itex]ln(\frac{1001+k}{1000+k})=ln(1+\frac{1}{1000+k})<ln(1.001)[/itex],

    [itex]L<ln1001+ln1.001\sum_1^{\infty}(\frac{1}{2})^k=ln1001+ln1.001=ln(1002.001)[/itex]

    S<1002.001

    ehild
     
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