RE: [math-fun] Re: A puzzle I read about
This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1. --Dan --------------------------------------------------------------------------------- << Dan Asimov asked:
For positive x<1, consider the alternating sum S(x) = x - x2 + x4 - x8 + x16 - x32 + - ... Does S(x) approach a limit as x approaches 1 from below, and if so what is this limit?
The limit does not exist. Since S(x) = x - S(x^2), the limit would have to be 1/2 if it existed. But I'll show that there are values of x arbitrarily close to 1 for which S(x) > 0.5008. Rewrite the definition as S(x) = SUM (x^4^k - x^(2*4^k)) k>=0 For 0 < x < 1, the summands are all positive, so any partial sum is a lower bound for S(x). In particular, calculation shows that, for x=199/200, the sum with k going from 0 to 5 is 0.50088158499... > 0.5008, so S(199/200) > 0.5008. Next note that S(x) = x - x^2 + S(x^4) > S(x^4). So S((199/200)^(1/4)) > S(199/200) > 0.5008, S((199/200)^(1/16)) > S((199/200)^(1/4)) > 0.5008, S((199/200)^(1/64)) > S((199/200)^(1/16)) > 0.5008, etc. Since the arguments of S approach 1, the limit can't be 1/2, so it doesn't exist.
Hello! Older references on this problem are: G. H. Hardy, On certain oscillating series, Quart. J. Pure Appl. Math. 38 (1907) 269-288 G. H. Hardy, Divergent Series, Oxford Univ. Press, 1949, p. 77 Steve Finch http://pauillac.inria.fr/algo/bsolve/
From: "" <dasimov@earthlink.net> Reply-To: dasimov@earthlink.net, math-fun <math-fun@mailman.xmission.com> To: "math-fun" <math-fun@mailman.xmission.com> Subject: RE: [math-fun] Re: A puzzle I read about Date: Wed, 18 May 2005 14:33:02 -0400
This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1.
--Dan ---------------------------------------------------------------------------------
<< Dan Asimov asked:
For positive x<1, consider the alternating sum S(x) = x - x2 + x4 - x8 + x16 - x32 + - ... Does S(x) approach a limit as x approaches 1 from below, and if so what is this limit?
The limit does not exist. Since S(x) = x - S(x^2), the limit would have to be 1/2 if it existed. But I'll show that there are values of x arbitrarily close to 1 for which S(x) > 0.5008.
Rewrite the definition as
S(x) = SUM (x^4^k - x^(2*4^k)) k>=0
For 0 < x < 1, the summands are all positive, so any partial sum is a lower bound for S(x). In particular, calculation shows that, for x=199/200, the sum with k going from 0 to 5 is 0.50088158499... > 0.5008, so
S(199/200) > 0.5008.
Next note that S(x) = x - x^2 + S(x^4) > S(x^4). So
S((199/200)^(1/4)) > S(199/200) > 0.5008,
S((199/200)^(1/16)) > S((199/200)^(1/4)) > 0.5008,
S((199/200)^(1/64)) > S((199/200)^(1/16)) > 0.5008,
etc. Since the arguments of S approach 1, the limit can't be 1/2, so it doesn't exist.
On Wednesday 18 May 2005 19:33, dasimov@earthlink.net wrote:
This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1.
So. Call an infinite set of non-negative integers "good" if the alternating power series F constructed from it in the same way has lim { x -> 1 from below } F(x) existing. We've established that the powers of 2 aren't good. It's well known that the full set of positive integers is good; the limit is 1/2. Which sets are good? Can the limit exist but not be 1/2? -- g
I've deleted most of the messages about this. Can someone repeat the Hardy references that were given in one message? Was there a page number given for Divergent Series? R. On Wed, 18 May 2005, Gareth McCaughan wrote:
On Wednesday 18 May 2005 19:33, dasimov@earthlink.net wrote:
This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1.
So. Call an infinite set of non-negative integers "good" if the alternating power series F constructed from it in the same way has lim { x -> 1 from below } F(x) existing. We've established that the powers of 2 aren't good. It's well known that the full set of positive integers is good; the limit is 1/2.
Which sets are good? Can the limit exist but not be 1/2?
-- g
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Well, how about 0 1 k k+1 2k 2k+1 ... Then (for x < 1) the alternating sum is just (1-x)/(1-x^k) which is 1/(1+x+x^2+ ... x^(k-1)) As x->1, this -> 1/k. --ms Gareth McCaughan wrote:
On Wednesday 18 May 2005 19:33, dasimov@earthlink.net wrote:
This is exactly the proof supplied by Noam Elkies, who includes the problem as #8 with some other cute problems and tidbits at http://www.math.harvard.edu/~elkies/Misc/. The given solution includes a nice graphic that can be zoomed in upon near x = 1.
So. Call an infinite set of non-negative integers "good" if the alternating power series F constructed from it in the same way has lim { x -> 1 from below } F(x) existing. We've established that the powers of 2 aren't good. It's well known that the full set of positive integers is good; the limit is 1/2.
Which sets are good? Can the limit exist but not be 1/2?
participants (5)
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dasimov＠earthlink.net -
Gareth McCaughan -
Mike Speciner -
Richard Guy -
Steven Finch