Closed form for the series $sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)$series involving $log left(tanhfracpi k2 right)$Show that $int_0^1ln(-lnx)cdotmathrm dxover 1+x^2=-sumlimits_n=0^infty1over 2n+1cdot2piover e^pi(2n+1)+1$ and evaluate itAn arctan series with a parameter $sum_n=1^infty arctan left(frac2a^2n^2}right)$Closed form for $prod_n=1^inftysqrt[2^n]{fracGamma(2^n+frac12)Gamma(2^n)$Closed form for $prod_n=1^inftysqrt[2^n]tanh(2^n),$A closed form for the infinite series $sum_n=1^infty (-1)^n+1arctan left( frac 1 n right)$Conjectured closed form for $sum_n=-infty^inftyfrac1coshpi n+frac1sqrt2$Closed form of infinite product $prodlimits_k=0^infty 2 left(1-fracx^1/2^k+11+x^1/2^k right)$About $prod_ninmathbb Zleft[tanhBig(fracpi2sqrta^2n^2+abBig)overtanhBig(fracpi2sqrtn^2/a^2+b/aBig)right]^(-1)^n=1$A closed form of $sum_n=1^inftyleft[ H_n^2-left(ln n+gamma+frac12n right)^2right]$Closed form for $sum_n=1^infty left(e-left(1+frac1nright)^n right)^2$?closed-form of series
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Closed form for the series $sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)$
series involving $log left(tanhfracpi k2 right)$Show that $int_0^1ln(-lnx)cdotmathrm dxover 1+x^2=-sumlimits_n=0^infty1over 2n+1cdot2piover e^pi(2n+1)+1$ and evaluate itAn arctan series with a parameter $sum_n=1^infty arctan left(frac2a^2n^2right)$Closed form for $prod_n=1^inftysqrt[2^n]fracGamma(2^n+frac12)Gamma(2^n)$Closed form for $prod_n=1^inftysqrt[2^n]tanh(2^n),$A closed form for the infinite series $sum_n=1^infty (-1)^n+1arctan left( frac 1 n right)$Conjectured closed form for $sum_n=-infty^inftyfrac1coshpi n+frac1sqrt2$Closed form of infinite product $prodlimits_k=0^infty 2 left(1-fracx^1/2^k+11+x^1/2^k right)$About $prod_ninmathbb Zleft[tanhBig(fracpi2sqrta^2n^2+abBig)overtanhBig(fracpi2sqrtn^2/a^2+b/aBig)right]^(-1)^n=1$A closed form of $sum_n=1^inftyleft[ H_n^2-left(ln n+gamma+frac12n right)^2right]$Closed form for $sum_n=1^infty left(e-left(1+frac1nright)^n right)^2$?closed-form of series
$begingroup$
Is there a closed form for: $$f(x)=sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)=2sum_n=0^infty frac12n+1frac1e^pi (2n+1) x+1$$
This sum originated from a recent question, where we have:
$$f(1)= -frac1piint_0^1 ln left( ln frac1x right) fracdx1+x^2=ln fracGamma (3/4)pi^1/4$$
If we differentiate w.r.t. $x$, we obtain:
$$f'(x)=sum_k=1^infty (-1)^k fracpi ksinh pi k x$$
There is again a closed form for $x=1$ (obtained numerically):
$$f'(1)=-frac14$$
So, is there a closed form or at least an integral definition for arbitrary $x>0$?
The series converges absolutely (numerically at least):
$$sum_k=1^infty ln left( tanh fracpi k x2 right)< infty$$
Thus, this series can also be expressed as a logarithm of an infinite product:
$$f(x)=ln prod_k=1^infty tanh (pi k x) - ln prod_k=1^infty tanh left( pi (k-1/2) x right)$$
$$e^f(x)= prod_k=1^infty fractanh (pi k x)tanh left( pi (k-1/2) x right)$$
This by the way leads to:
$$prod_k=1^infty fractanh (pi k)tanh left( pi (k-1/2) right)=fracpi^1/4Gamma(3/4)$$
I feel like there is a way to use the infinite product form for $sinh$ and $cosh$:
$$sinh (pi x)=pi x prod_n=1^infty left(1+fracx^2n^2 right)$$
$$cosh (pi x)=prod_n=1^infty left(1+fracx^2(n-1/2)^2 right)$$
sequences-and-series definite-integrals closed-form infinite-product
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
$endgroup$
add a comment |
$begingroup$
Is there a closed form for: $$f(x)=sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)=2sum_n=0^infty frac12n+1frac1e^pi (2n+1) x+1$$
This sum originated from a recent question, where we have:
$$f(1)= -frac1piint_0^1 ln left( ln frac1x right) fracdx1+x^2=ln fracGamma (3/4)pi^1/4$$
If we differentiate w.r.t. $x$, we obtain:
$$f'(x)=sum_k=1^infty (-1)^k fracpi ksinh pi k x$$
There is again a closed form for $x=1$ (obtained numerically):
$$f'(1)=-frac14$$
So, is there a closed form or at least an integral definition for arbitrary $x>0$?
The series converges absolutely (numerically at least):
$$sum_k=1^infty ln left( tanh fracpi k x2 right)< infty$$
Thus, this series can also be expressed as a logarithm of an infinite product:
$$f(x)=ln prod_k=1^infty tanh (pi k x) - ln prod_k=1^infty tanh left( pi (k-1/2) x right)$$
$$e^f(x)= prod_k=1^infty fractanh (pi k x)tanh left( pi (k-1/2) x right)$$
This by the way leads to:
$$prod_k=1^infty fractanh (pi k)tanh left( pi (k-1/2) right)=fracpi^1/4Gamma(3/4)$$
I feel like there is a way to use the infinite product form for $sinh$ and $cosh$:
$$sinh (pi x)=pi x prod_n=1^infty left(1+fracx^2n^2 right)$$
$$cosh (pi x)=prod_n=1^infty left(1+fracx^2(n-1/2)^2 right)$$
sequences-and-series definite-integrals closed-form infinite-product
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
$endgroup$
1
$begingroup$
In my answer to the linked question I have shown that the series on right equals $$frac12logvartheta_3(q)$$ where $q = e^-pi x$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:14
2
$begingroup$
One can get a closed form if one considers the function $$g(x) = sum_n = 1^inftylogtanh fracnpi x2 = logvartheta_4(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = frac14logfracvartheta_4^4(q)vartheta_3^4(q) = frac14log(1 - k^2)$$ so that if $x = sqrtr, r inmathbbQ^+$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:42
$begingroup$
@ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form
$endgroup$
– Yuriy S
Feb 26 '17 at 16:45
2
$begingroup$
The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^-pi$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$).
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:51
add a comment |
$begingroup$
Is there a closed form for: $$f(x)=sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)=2sum_n=0^infty frac12n+1frac1e^pi (2n+1) x+1$$
This sum originated from a recent question, where we have:
$$f(1)= -frac1piint_0^1 ln left( ln frac1x right) fracdx1+x^2=ln fracGamma (3/4)pi^1/4$$
If we differentiate w.r.t. $x$, we obtain:
$$f'(x)=sum_k=1^infty (-1)^k fracpi ksinh pi k x$$
There is again a closed form for $x=1$ (obtained numerically):
$$f'(1)=-frac14$$
So, is there a closed form or at least an integral definition for arbitrary $x>0$?
The series converges absolutely (numerically at least):
$$sum_k=1^infty ln left( tanh fracpi k x2 right)< infty$$
Thus, this series can also be expressed as a logarithm of an infinite product:
$$f(x)=ln prod_k=1^infty tanh (pi k x) - ln prod_k=1^infty tanh left( pi (k-1/2) x right)$$
$$e^f(x)= prod_k=1^infty fractanh (pi k x)tanh left( pi (k-1/2) x right)$$
This by the way leads to:
$$prod_k=1^infty fractanh (pi k)tanh left( pi (k-1/2) right)=fracpi^1/4Gamma(3/4)$$
I feel like there is a way to use the infinite product form for $sinh$ and $cosh$:
$$sinh (pi x)=pi x prod_n=1^infty left(1+fracx^2n^2 right)$$
$$cosh (pi x)=prod_n=1^infty left(1+fracx^2(n-1/2)^2 right)$$
sequences-and-series definite-integrals closed-form infinite-product
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
$endgroup$
Is there a closed form for: $$f(x)=sum_k=1^infty (-1)^k ln left( tanh fracpi k x2 right)=2sum_n=0^infty frac12n+1frac1e^pi (2n+1) x+1$$
This sum originated from a recent question, where we have:
$$f(1)= -frac1piint_0^1 ln left( ln frac1x right) fracdx1+x^2=ln fracGamma (3/4)pi^1/4$$
If we differentiate w.r.t. $x$, we obtain:
$$f'(x)=sum_k=1^infty (-1)^k fracpi ksinh pi k x$$
There is again a closed form for $x=1$ (obtained numerically):
$$f'(1)=-frac14$$
So, is there a closed form or at least an integral definition for arbitrary $x>0$?
The series converges absolutely (numerically at least):
$$sum_k=1^infty ln left( tanh fracpi k x2 right)< infty$$
Thus, this series can also be expressed as a logarithm of an infinite product:
$$f(x)=ln prod_k=1^infty tanh (pi k x) - ln prod_k=1^infty tanh left( pi (k-1/2) x right)$$
$$e^f(x)= prod_k=1^infty fractanh (pi k x)tanh left( pi (k-1/2) x right)$$
This by the way leads to:
$$prod_k=1^infty fractanh (pi k)tanh left( pi (k-1/2) right)=fracpi^1/4Gamma(3/4)$$
I feel like there is a way to use the infinite product form for $sinh$ and $cosh$:
$$sinh (pi x)=pi x prod_n=1^infty left(1+fracx^2n^2 right)$$
$$cosh (pi x)=prod_n=1^infty left(1+fracx^2(n-1/2)^2 right)$$
sequences-and-series definite-integrals closed-form infinite-product
sequences-and-series definite-integrals closed-form infinite-product
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
edited Apr 13 '17 at 12:21
Community♦
1
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
asked Feb 26 '17 at 13:01
Yuriy SYuriy S
15.9k433118
15.9k433118
1
$begingroup$
In my answer to the linked question I have shown that the series on right equals $$frac12logvartheta_3(q)$$ where $q = e^-pi x$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:14
2
$begingroup$
One can get a closed form if one considers the function $$g(x) = sum_n = 1^inftylogtanh fracnpi x2 = logvartheta_4(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = frac14logfracvartheta_4^4(q)vartheta_3^4(q) = frac14log(1 - k^2)$$ so that if $x = sqrtr, r inmathbbQ^+$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:42
$begingroup$
@ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form
$endgroup$
– Yuriy S
Feb 26 '17 at 16:45
2
$begingroup$
The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^-pi$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$).
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:51
add a comment |
1
$begingroup$
In my answer to the linked question I have shown that the series on right equals $$frac12logvartheta_3(q)$$ where $q = e^-pi x$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:14
2
$begingroup$
One can get a closed form if one considers the function $$g(x) = sum_n = 1^inftylogtanh fracnpi x2 = logvartheta_4(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = frac14logfracvartheta_4^4(q)vartheta_3^4(q) = frac14log(1 - k^2)$$ so that if $x = sqrtr, r inmathbbQ^+$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:42
$begingroup$
@ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form
$endgroup$
– Yuriy S
Feb 26 '17 at 16:45
2
$begingroup$
The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^-pi$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$).
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:51
1
1
$begingroup$
In my answer to the linked question I have shown that the series on right equals $$frac12logvartheta_3(q)$$ where $q = e^-pi x$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:14
$begingroup$
In my answer to the linked question I have shown that the series on right equals $$frac12logvartheta_3(q)$$ where $q = e^-pi x$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:14
2
2
$begingroup$
One can get a closed form if one considers the function $$g(x) = sum_n = 1^inftylogtanh fracnpi x2 = logvartheta_4(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = frac14logfracvartheta_4^4(q)vartheta_3^4(q) = frac14log(1 - k^2)$$ so that if $x = sqrtr, r inmathbbQ^+$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:42
$begingroup$
One can get a closed form if one considers the function $$g(x) = sum_n = 1^inftylogtanh fracnpi x2 = logvartheta_4(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = frac14logfracvartheta_4^4(q)vartheta_3^4(q) = frac14log(1 - k^2)$$ so that if $x = sqrtr, r inmathbbQ^+$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$.
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:42
$begingroup$
@ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form
$endgroup$
– Yuriy S
Feb 26 '17 at 16:45
$begingroup$
@ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form
$endgroup$
– Yuriy S
Feb 26 '17 at 16:45
2
2
$begingroup$
The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^-pi$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$).
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:51
$begingroup$
The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^-pi$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$).
$endgroup$
– Paramanand Singh
Feb 26 '17 at 16:51
add a comment |
1 Answer
1
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votes
$begingroup$
Let’s use $~displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1~$ . $enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^-pi x~$ with $~x>0~$ we get
$displaystyle e^f(x) = prodlimits_k=1^inftyfractanh(kpi x)tanh((k-frac12)pi x) = prodlimits_k=1^inftyfrac fracq^-k-q^kq^-k+q^k fracq^frac12-k-q^k-frac12 q^frac12-k+q^k-frac12 = prodlimits_k=1^inftyfrac(1-q^2k)(1+q^2k-1)(1+q^2k)(1-q^2k-1) =$
$displaystyle = prodlimits_k=1^infty (1-q^2k)(1+q^2k-1)^2 = sumlimits_k=-infty^+infty q^k^2 = vartheta(0;ix)$
The “closed form” for $,f,$ is:
$$f(x) = lnvartheta(0;ix)$$
Please see e.g. Theta function .
Note:
$displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1$
$Leftrightarrowhspace2cm$ (logarithm)
$displaystyle sumlimits_k=1^infty sumlimits_v=1^infty frac(-1)^v-1z^kvv = sumlimits_k=1^infty ln(1+z^k) = -sumlimits_k=1^infty ln(1-z^2k-1) = sumlimits_k=1^infty sumlimits_v=1^infty fracz^(2k-1)vv$
$Leftrightarrowhspace2cm$ (exchanging the sum symbols which is valid for $~|z|<1~$
$hspace2.7cm$ and using $~displaystylefracx1-x=sumlimits_k=1^infty x^k~$)
$displaystylesumlimits_v=1^infty frac(-1)^v-1vfracz^v1-z^v = sumlimits_v=1^infty frac1vfracz^v1-z^v - 2sumlimits_v=1^infty frac12vfracz^2v1-z^2v = sumlimits_v=1^infty frac1vfracz^v1-z^2v$
answered Mar 22 at 10:54
user90369user90369
8,525926
$endgroup$
$begingroup$
Where does the first identity come from? "Let's use..."
$endgroup$
– Diger
Mar 22 at 11:27
1
$begingroup$
@Diger : I've added a note . ;)
$endgroup$
– user90369
Mar 22 at 12:54
$begingroup$
Very nice answer, thank you! I haven't checked numerically yet, but I trust you
$endgroup$
– Yuriy S
Mar 25 at 8:03
$begingroup$
@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let’s use $~displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1~$ . $enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^-pi x~$ with $~x>0~$ we get
$displaystyle e^f(x) = prodlimits_k=1^inftyfractanh(kpi x)tanh((k-frac12)pi x) = prodlimits_k=1^inftyfrac fracq^-k-q^kq^-k+q^k fracq^frac12-k-q^k-frac12 q^frac12-k+q^k-frac12 = prodlimits_k=1^inftyfrac(1-q^2k)(1+q^2k-1)(1+q^2k)(1-q^2k-1) =$
$displaystyle = prodlimits_k=1^infty (1-q^2k)(1+q^2k-1)^2 = sumlimits_k=-infty^+infty q^k^2 = vartheta(0;ix)$
The “closed form” for $,f,$ is:
$$f(x) = lnvartheta(0;ix)$$
Please see e.g. Theta function .
Note:
$displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1$
$Leftrightarrowhspace2cm$ (logarithm)
$displaystyle sumlimits_k=1^infty sumlimits_v=1^infty frac(-1)^v-1z^kvv = sumlimits_k=1^infty ln(1+z^k) = -sumlimits_k=1^infty ln(1-z^2k-1) = sumlimits_k=1^infty sumlimits_v=1^infty fracz^(2k-1)vv$
$Leftrightarrowhspace2cm$ (exchanging the sum symbols which is valid for $~|z|<1~$
$hspace2.7cm$ and using $~displaystylefracx1-x=sumlimits_k=1^infty x^k~$)
$displaystylesumlimits_v=1^infty frac(-1)^v-1vfracz^v1-z^v = sumlimits_v=1^infty frac1vfracz^v1-z^v - 2sumlimits_v=1^infty frac12vfracz^2v1-z^2v = sumlimits_v=1^infty frac1vfracz^v1-z^2v$
answered Mar 22 at 10:54
user90369user90369
8,525926
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$begingroup$
Where does the first identity come from? "Let's use..."
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– Diger
Mar 22 at 11:27
1
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@Diger : I've added a note . ;)
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– user90369
Mar 22 at 12:54
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Very nice answer, thank you! I haven't checked numerically yet, but I trust you
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– Yuriy S
Mar 25 at 8:03
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@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
add a comment |
$begingroup$
Let’s use $~displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1~$ . $enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^-pi x~$ with $~x>0~$ we get
$displaystyle e^f(x) = prodlimits_k=1^inftyfractanh(kpi x)tanh((k-frac12)pi x) = prodlimits_k=1^inftyfrac fracq^-k-q^kq^-k+q^k fracq^frac12-k-q^k-frac12 q^frac12-k+q^k-frac12 = prodlimits_k=1^inftyfrac(1-q^2k)(1+q^2k-1)(1+q^2k)(1-q^2k-1) =$
$displaystyle = prodlimits_k=1^infty (1-q^2k)(1+q^2k-1)^2 = sumlimits_k=-infty^+infty q^k^2 = vartheta(0;ix)$
The “closed form” for $,f,$ is:
$$f(x) = lnvartheta(0;ix)$$
Please see e.g. Theta function .
Note:
$displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1$
$Leftrightarrowhspace2cm$ (logarithm)
$displaystyle sumlimits_k=1^infty sumlimits_v=1^infty frac(-1)^v-1z^kvv = sumlimits_k=1^infty ln(1+z^k) = -sumlimits_k=1^infty ln(1-z^2k-1) = sumlimits_k=1^infty sumlimits_v=1^infty fracz^(2k-1)vv$
$Leftrightarrowhspace2cm$ (exchanging the sum symbols which is valid for $~|z|<1~$
$hspace2.7cm$ and using $~displaystylefracx1-x=sumlimits_k=1^infty x^k~$)
$displaystylesumlimits_v=1^infty frac(-1)^v-1vfracz^v1-z^v = sumlimits_v=1^infty frac1vfracz^v1-z^v - 2sumlimits_v=1^infty frac12vfracz^2v1-z^2v = sumlimits_v=1^infty frac1vfracz^v1-z^2v$
answered Mar 22 at 10:54
user90369user90369
8,525926
$endgroup$
$begingroup$
Where does the first identity come from? "Let's use..."
$endgroup$
– Diger
Mar 22 at 11:27
1
$begingroup$
@Diger : I've added a note . ;)
$endgroup$
– user90369
Mar 22 at 12:54
$begingroup$
Very nice answer, thank you! I haven't checked numerically yet, but I trust you
$endgroup$
– Yuriy S
Mar 25 at 8:03
$begingroup$
@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
add a comment |
$begingroup$
Let’s use $~displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1~$ . $enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^-pi x~$ with $~x>0~$ we get
$displaystyle e^f(x) = prodlimits_k=1^inftyfractanh(kpi x)tanh((k-frac12)pi x) = prodlimits_k=1^inftyfrac fracq^-k-q^kq^-k+q^k fracq^frac12-k-q^k-frac12 q^frac12-k+q^k-frac12 = prodlimits_k=1^inftyfrac(1-q^2k)(1+q^2k-1)(1+q^2k)(1-q^2k-1) =$
$displaystyle = prodlimits_k=1^infty (1-q^2k)(1+q^2k-1)^2 = sumlimits_k=-infty^+infty q^k^2 = vartheta(0;ix)$
The “closed form” for $,f,$ is:
$$f(x) = lnvartheta(0;ix)$$
Please see e.g. Theta function .
Note:
$displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1$
$Leftrightarrowhspace2cm$ (logarithm)
$displaystyle sumlimits_k=1^infty sumlimits_v=1^infty frac(-1)^v-1z^kvv = sumlimits_k=1^infty ln(1+z^k) = -sumlimits_k=1^infty ln(1-z^2k-1) = sumlimits_k=1^infty sumlimits_v=1^infty fracz^(2k-1)vv$
$Leftrightarrowhspace2cm$ (exchanging the sum symbols which is valid for $~|z|<1~$
$hspace2.7cm$ and using $~displaystylefracx1-x=sumlimits_k=1^infty x^k~$)
$displaystylesumlimits_v=1^infty frac(-1)^v-1vfracz^v1-z^v = sumlimits_v=1^infty frac1vfracz^v1-z^v - 2sumlimits_v=1^infty frac12vfracz^2v1-z^2v = sumlimits_v=1^infty frac1vfracz^v1-z^2v$
answered Mar 22 at 10:54
user90369user90369
8,525926
$endgroup$
Let’s use $~displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1~$ . $enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^-pi x~$ with $~x>0~$ we get
$displaystyle e^f(x) = prodlimits_k=1^inftyfractanh(kpi x)tanh((k-frac12)pi x) = prodlimits_k=1^inftyfrac fracq^-k-q^kq^-k+q^k fracq^frac12-k-q^k-frac12 q^frac12-k+q^k-frac12 = prodlimits_k=1^inftyfrac(1-q^2k)(1+q^2k-1)(1+q^2k)(1-q^2k-1) =$
$displaystyle = prodlimits_k=1^infty (1-q^2k)(1+q^2k-1)^2 = sumlimits_k=-infty^+infty q^k^2 = vartheta(0;ix)$
The “closed form” for $,f,$ is:
$$f(x) = lnvartheta(0;ix)$$
Please see e.g. Theta function .
Note:
$displaystyleprodlimits_k=1^infty (1+z^k)(1-z^2k-1) =1$
$Leftrightarrowhspace2cm$ (logarithm)
$displaystyle sumlimits_k=1^infty sumlimits_v=1^infty frac(-1)^v-1z^kvv = sumlimits_k=1^infty ln(1+z^k) = -sumlimits_k=1^infty ln(1-z^2k-1) = sumlimits_k=1^infty sumlimits_v=1^infty fracz^(2k-1)vv$
$Leftrightarrowhspace2cm$ (exchanging the sum symbols which is valid for $~|z|<1~$
$hspace2.7cm$ and using $~displaystylefracx1-x=sumlimits_k=1^infty x^k~$)
$displaystylesumlimits_v=1^infty frac(-1)^v-1vfracz^v1-z^v = sumlimits_v=1^infty frac1vfracz^v1-z^v - 2sumlimits_v=1^infty frac12vfracz^2v1-z^2v = sumlimits_v=1^infty frac1vfracz^v1-z^2v$
answered Mar 22 at 10:54
user90369user90369
8,525926
edited Mar 22 at 15:44
answered Mar 22 at 10:54
user90369user90369
8,525926
answered Mar 22 at 10:54
user90369user90369
8,525926
answered Mar 22 at 10:54
user90369user90369
8,525926
8,525926
$begingroup$
Where does the first identity come from? "Let's use..."
$endgroup$
– Diger
Mar 22 at 11:27
1
$begingroup$
@Diger : I've added a note . ;)
$endgroup$
– user90369
Mar 22 at 12:54
$begingroup$
Very nice answer, thank you! I haven't checked numerically yet, but I trust you
$endgroup$
– Yuriy S
Mar 25 at 8:03
$begingroup$
@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
add a comment |
$begingroup$
Where does the first identity come from? "Let's use..."
$endgroup$
– Diger
Mar 22 at 11:27
1
$begingroup$
@Diger : I've added a note . ;)
$endgroup$
– user90369
Mar 22 at 12:54
$begingroup$
Very nice answer, thank you! I haven't checked numerically yet, but I trust you
$endgroup$
– Yuriy S
Mar 25 at 8:03
$begingroup$
@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
$begingroup$
Where does the first identity come from? "Let's use..."
$endgroup$
– Diger
Mar 22 at 11:27
$begingroup$
Where does the first identity come from? "Let's use..."
$endgroup$
– Diger
Mar 22 at 11:27
1
1
$begingroup$
@Diger : I've added a note . ;)
$endgroup$
– user90369
Mar 22 at 12:54
$begingroup$
@Diger : I've added a note . ;)
$endgroup$
– user90369
Mar 22 at 12:54
$begingroup$
Very nice answer, thank you! I haven't checked numerically yet, but I trust you
$endgroup$
– Yuriy S
Mar 25 at 8:03
$begingroup$
Very nice answer, thank you! I haven't checked numerically yet, but I trust you
$endgroup$
– Yuriy S
Mar 25 at 8:03
$begingroup$
@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
$begingroup$
@YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above).
$endgroup$
– user90369
Mar 25 at 9:08
add a comment |
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1
$begingroup$
In my answer to the linked question I have shown that the series on right equals $$frac12logvartheta_3(q)$$ where $q = e^-pi x$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals.
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– Paramanand Singh
Feb 26 '17 at 16:14
2
$begingroup$
One can get a closed form if one considers the function $$g(x) = sum_n = 1^inftylogtanh fracnpi x2 = logvartheta_4(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = frac14logfracvartheta_4^4(q)vartheta_3^4(q) = frac14log(1 - k^2)$$ so that if $x = sqrtr, r inmathbbQ^+$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$.
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– Paramanand Singh
Feb 26 '17 at 16:42
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@ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form
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– Yuriy S
Feb 26 '17 at 16:45
2
$begingroup$
The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^-pi$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$).
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– Paramanand Singh
Feb 26 '17 at 16:51