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[songoku]

square root in suqare root in square root in square root in...

Homework Statement

Find the value of:
\sqrt{1001\sqrt{1002\sqrt{1003\sqrt{1004...}}}}

Homework Equations

exponential and surd

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

 

[lanedance]

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

then maybe see if you can simplify and take a logarithm?

 

[Ray Vickson]

Homework Statement

Find the value of:
\sqrt{1001\sqrt{1002\sqrt{1003\sqrt{1004...}}}}

Homework Equations

exponential and surd

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

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

 

[Dick]

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

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.

 

[ehild]

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

\prod_1^n{(1000+k)^{\frac{1}{2^k}}}.

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

 

[stallionx]

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

\prod_1^n{(1000+k)^{\frac{1}{2^k}}}.

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

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

If not so, could you please elaborate.

Thanks.

 

[NascentOxygen]

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

 

[stallionx]

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.

 

[Ray Vickson]

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.

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

 

[uart]

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.

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)

1001.99900397913691898234439715244344613729998738717041977274676200148199509558817554545417076124318480940610680361311614854981603236455318607100162785113625170638256422574541347581096615035978851664139085456338883594740422731293953208208144172243868905270361292059529263885996278587885993677025105711750619813487545990403024502836716136675785661195184541807980681958180327723191234228986168561284824245851495221985327190918407653825883156881071095089445088055616086340359845081535782146671200814749246453040284862549740844129680298559371650374132490575305619530188927496517152722612153193833757754310883347426813571188377508484422925408646686318012394719790714830903494186462930528915274650314653701307950669845027675076978273020036611482127244665921724832927562221042065147872776648580806362571880958572257635842221805174250500907725156069329281841466085876234699317471711743482677662082232151927374650792614369602453252418264601925148261436388726832874441122942437367419289549047172944608662444481832971555133410691363743174251382431045383895821360549236143488774247447599308457363870075777157976111773915678002918789458376189266258999122256218899692315961897948080517046667533746089790257098403498

 

[Dick]

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.

 

[ehild]

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

The logarithm of the original expression is

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

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

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

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

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

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

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

S<1002.001

ehild