Mixing
A few conjectured limits of products involving the Thue–Morse sequence
$begingroup$
(related to my previous questions$^{[1]}$$!^{[2]}$)
Let's define the signed Thue–Morse sequence $t_n$ by the recurrence
$$t_0 = 1, quad t_n = (-1)^n , t_{lfloor n/2rfloor},tag1$$
or by the generating function
$$sum_{n=0}^infty t_n , x^n=prod_{n=0}^inftyleft(1-x^{2^n}right).tag{$1^prime$}$$
It seems that the following conjectures hold:
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}stackrel{color{gray}?}=frac12tag2$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{t_k}stackrel{color{gray}?}=frac1{sqrt2}tag3$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{(-1)^k,t_k}stackrel{color{gray}?}=frac1{2sqrt2}tag4$$
How can we prove these? Are there any other limits of products similar to these?
sequences-and-series number-theory limits products conjectures
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
$endgroup$
add a comment |
$begingroup$
(related to my previous questions$^{[1]}$$!^{[2]}$)
Let's define the signed Thue–Morse sequence $t_n$ by the recurrence
$$t_0 = 1, quad t_n = (-1)^n , t_{lfloor n/2rfloor},tag1$$
or by the generating function
$$sum_{n=0}^infty t_n , x^n=prod_{n=0}^inftyleft(1-x^{2^n}right).tag{$1^prime$}$$
It seems that the following conjectures hold:
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}stackrel{color{gray}?}=frac12tag2$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{t_k}stackrel{color{gray}?}=frac1{sqrt2}tag3$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{(-1)^k,t_k}stackrel{color{gray}?}=frac1{2sqrt2}tag4$$
How can we prove these? Are there any other limits of products similar to these?
sequences-and-series number-theory limits products conjectures
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
$endgroup$
$begingroup$
Related papers: algo.inria.fr/seminars/sem92-93/allouche.pdf, pdfs.semanticscholar.org/a4dc/…, arxiv.org/abs/1709.03398, arxiv.org/abs/1406.7407, arxiv.org/abs/1709.04104
$endgroup$
– Vladimir Reshetnikov
Apr 9 '18 at 1:58
$begingroup$
Related questions: math.stackexchange.com/q/29234/19661, math.stackexchange.com/q/924601/19661
$endgroup$
– Vladimir Reshetnikov
Apr 11 '18 at 18:52
add a comment |
$begingroup$
(related to my previous questions$^{[1]}$$!^{[2]}$)
Let's define the signed Thue–Morse sequence $t_n$ by the recurrence
$$t_0 = 1, quad t_n = (-1)^n , t_{lfloor n/2rfloor},tag1$$
or by the generating function
$$sum_{n=0}^infty t_n , x^n=prod_{n=0}^inftyleft(1-x^{2^n}right).tag{$1^prime$}$$
It seems that the following conjectures hold:
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}stackrel{color{gray}?}=frac12tag2$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{t_k}stackrel{color{gray}?}=frac1{sqrt2}tag3$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{(-1)^k,t_k}stackrel{color{gray}?}=frac1{2sqrt2}tag4$$
How can we prove these? Are there any other limits of products similar to these?
sequences-and-series number-theory limits products conjectures
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
$endgroup$
(related to my previous questions$^{[1]}$$!^{[2]}$)
Let's define the signed Thue–Morse sequence $t_n$ by the recurrence
$$t_0 = 1, quad t_n = (-1)^n , t_{lfloor n/2rfloor},tag1$$
or by the generating function
$$sum_{n=0}^infty t_n , x^n=prod_{n=0}^inftyleft(1-x^{2^n}right).tag{$1^prime$}$$
It seems that the following conjectures hold:
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}stackrel{color{gray}?}=frac12tag2$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{t_k}stackrel{color{gray}?}=frac1{sqrt2}tag3$$
$$lim_{ntoinfty}prod_{k=0}^{2^n-1}left(k+1right)^{(-1)^k,t_k}stackrel{color{gray}?}=frac1{2sqrt2}tag4$$
How can we prove these? Are there any other limits of products similar to these?
sequences-and-series number-theory limits products conjectures
sequences-and-series number-theory limits products conjectures
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
edited Apr 8 '18 at 23:47
Vladimir Reshetnikov
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
asked Apr 8 '18 at 1:33
Vladimir ReshetnikovVladimir Reshetnikov
24.3k5120233
24.3k5120233
$begingroup$
Related papers: algo.inria.fr/seminars/sem92-93/allouche.pdf, pdfs.semanticscholar.org/a4dc/…, arxiv.org/abs/1709.03398, arxiv.org/abs/1406.7407, arxiv.org/abs/1709.04104
$endgroup$
– Vladimir Reshetnikov
Apr 9 '18 at 1:58
$begingroup$
Related questions: math.stackexchange.com/q/29234/19661, math.stackexchange.com/q/924601/19661
$endgroup$
– Vladimir Reshetnikov
Apr 11 '18 at 18:52
add a comment |
$begingroup$
Related papers: algo.inria.fr/seminars/sem92-93/allouche.pdf, pdfs.semanticscholar.org/a4dc/…, arxiv.org/abs/1709.03398, arxiv.org/abs/1406.7407, arxiv.org/abs/1709.04104
$endgroup$
– Vladimir Reshetnikov
Apr 9 '18 at 1:58
$begingroup$
Related questions: math.stackexchange.com/q/29234/19661, math.stackexchange.com/q/924601/19661
$endgroup$
– Vladimir Reshetnikov
Apr 11 '18 at 18:52
$begingroup$
Related papers: algo.inria.fr/seminars/sem92-93/allouche.pdf, pdfs.semanticscholar.org/a4dc/…, arxiv.org/abs/1709.03398, arxiv.org/abs/1406.7407, arxiv.org/abs/1709.04104
$endgroup$
– Vladimir Reshetnikov
Apr 9 '18 at 1:58
$begingroup$
Related papers: algo.inria.fr/seminars/sem92-93/allouche.pdf, pdfs.semanticscholar.org/a4dc/…, arxiv.org/abs/1709.03398, arxiv.org/abs/1406.7407, arxiv.org/abs/1709.04104
$endgroup$
– Vladimir Reshetnikov
Apr 9 '18 at 1:58
$begingroup$
Related questions: math.stackexchange.com/q/29234/19661, math.stackexchange.com/q/924601/19661
$endgroup$
– Vladimir Reshetnikov
Apr 11 '18 at 18:52
$begingroup$
Related questions: math.stackexchange.com/q/29234/19661, math.stackexchange.com/q/924601/19661
$endgroup$
– Vladimir Reshetnikov
Apr 11 '18 at 18:52
add a comment |
1 Answer
1
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$begingroup$
2) For $nge 1$ we have
$$prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_{2k}}left(2k+1+tfrac12right)^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_k}left(2k+1+tfrac12right)^{-t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{2k+tfrac12}{2k+1+tfrac12}right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}.$$
So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals
$$prod_{k=0}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13prod_{k=1}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13fleft(frac14,frac34right)=frac13cdotfrac32=frac12.$$
3) Similarly to the previous case we can show that the left hand side of (3) equals $tfrac12 fleft(tfrac 12,1right)=tfrac 1{sqrt2}$.
1) Here preliminary calculations are a bit longer. For $nge 2$ we have
$$prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$
$$prod_{k=0}^{2^{n-2}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}left(frac{k+tfrac12}{k+1 }right)^{t_k}.$$
Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $tfrac 1{2sqrt2}.$
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
add a comment |
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1 Answer
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$begingroup$
2) For $nge 1$ we have
$$prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_{2k}}left(2k+1+tfrac12right)^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_k}left(2k+1+tfrac12right)^{-t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{2k+tfrac12}{2k+1+tfrac12}right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}.$$
So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals
$$prod_{k=0}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13prod_{k=1}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13fleft(frac14,frac34right)=frac13cdotfrac32=frac12.$$
3) Similarly to the previous case we can show that the left hand side of (3) equals $tfrac12 fleft(tfrac 12,1right)=tfrac 1{sqrt2}$.
1) Here preliminary calculations are a bit longer. For $nge 2$ we have
$$prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$
$$prod_{k=0}^{2^{n-2}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}left(frac{k+tfrac12}{k+1 }right)^{t_k}.$$
Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $tfrac 1{2sqrt2}.$
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
add a comment |
$begingroup$
2) For $nge 1$ we have
$$prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_{2k}}left(2k+1+tfrac12right)^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_k}left(2k+1+tfrac12right)^{-t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{2k+tfrac12}{2k+1+tfrac12}right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}.$$
So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals
$$prod_{k=0}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13prod_{k=1}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13fleft(frac14,frac34right)=frac13cdotfrac32=frac12.$$
3) Similarly to the previous case we can show that the left hand side of (3) equals $tfrac12 fleft(tfrac 12,1right)=tfrac 1{sqrt2}$.
1) Here preliminary calculations are a bit longer. For $nge 2$ we have
$$prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$
$$prod_{k=0}^{2^{n-2}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}left(frac{k+tfrac12}{k+1 }right)^{t_k}.$$
Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $tfrac 1{2sqrt2}.$
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
add a comment |
$begingroup$
2) For $nge 1$ we have
$$prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_{2k}}left(2k+1+tfrac12right)^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_k}left(2k+1+tfrac12right)^{-t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{2k+tfrac12}{2k+1+tfrac12}right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}.$$
So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals
$$prod_{k=0}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13prod_{k=1}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13fleft(frac14,frac34right)=frac13cdotfrac32=frac12.$$
3) Similarly to the previous case we can show that the left hand side of (3) equals $tfrac12 fleft(tfrac 12,1right)=tfrac 1{sqrt2}$.
1) Here preliminary calculations are a bit longer. For $nge 2$ we have
$$prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$
$$prod_{k=0}^{2^{n-2}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}left(frac{k+tfrac12}{k+1 }right)^{t_k}.$$
Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $tfrac 1{2sqrt2}.$
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
2) For $nge 1$ we have
$$prod_{k=0}^{2^n-1}left(k+tfrac12right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_{2k}}left(2k+1+tfrac12right)^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}left(2k+tfrac12right)^{t_k}left(2k+1+tfrac12right)^{-t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{2k+tfrac12}{2k+1+tfrac12}right)^{t_k}=$$
$$prod_{k=0}^{2^{n-1}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}.$$
So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals
$$prod_{k=0}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13prod_{k=1}^{infty}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}=$$
$$frac13fleft(frac14,frac34right)=frac13cdotfrac32=frac12.$$
3) Similarly to the previous case we can show that the left hand side of (3) equals $tfrac12 fleft(tfrac 12,1right)=tfrac 1{sqrt2}$.
1) Here preliminary calculations are a bit longer. For $nge 2$ we have
$$prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$
$$prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$
$$prod_{k=0}^{2^{n-2}-1}left(frac{k+tfrac14}{k+tfrac34}right)^{t_k}left(frac{k+tfrac12}{k+1 }right)^{t_k}.$$
Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $tfrac 1{2sqrt2}.$
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
answered Jan 20 at 8:10
Alex RavskyAlex Ravsky
41.4k32282
41.4k32282
add a comment |
add a comment |
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$begingroup$
Related papers: algo.inria.fr/seminars/sem92-93/allouche.pdf, pdfs.semanticscholar.org/a4dc/…, arxiv.org/abs/1709.03398, arxiv.org/abs/1406.7407, arxiv.org/abs/1709.04104
$endgroup$
– Vladimir Reshetnikov
Apr 9 '18 at 1:58
$begingroup$
Related questions: math.stackexchange.com/q/29234/19661, math.stackexchange.com/q/924601/19661
$endgroup$
– Vladimir Reshetnikov
Apr 11 '18 at 18:52