Partial Fractions Expansion of $tanh(z)/z$












8












$begingroup$


I have seen the following formula in papers (without citations) and in Mathematica's documentation about Tanh:



$$
frac{tanh(z)}{8z}=sum_{k=1}^{infty} frac{1}{(2k-1)^2 pi^2+4z^2}
$$



I have no idea how to prove it and I have also encountered in my research similar sums involving, for instance, $mathrm{coth}$. It would be nice to have a general method for working with these problems; any suggestions?






sequences-and-series trigonometry partial-fractions







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$endgroup$












  • $begingroup$
    Maybe you can try the Taylor Series?
    $endgroup$
    – A. Chu
    Jul 27 '12 at 14:14






  • 1




    $begingroup$
    As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $tan z$, and then a simple change of variables $iz$ will give you a similar one for $tanh z$.
    $endgroup$
    – GEdgar
    Jul 28 '12 at 12:36
















8












$begingroup$


I have seen the following formula in papers (without citations) and in Mathematica's documentation about Tanh:



$$
frac{tanh(z)}{8z}=sum_{k=1}^{infty} frac{1}{(2k-1)^2 pi^2+4z^2}
$$



I have no idea how to prove it and I have also encountered in my research similar sums involving, for instance, $mathrm{coth}$. It would be nice to have a general method for working with these problems; any suggestions?






sequences-and-series trigonometry partial-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe you can try the Taylor Series?
    $endgroup$
    – A. Chu
    Jul 27 '12 at 14:14






  • 1




    $begingroup$
    As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $tan z$, and then a simple change of variables $iz$ will give you a similar one for $tanh z$.
    $endgroup$
    – GEdgar
    Jul 28 '12 at 12:36














8












8








8


3



$begingroup$


I have seen the following formula in papers (without citations) and in Mathematica's documentation about Tanh:



$$
frac{tanh(z)}{8z}=sum_{k=1}^{infty} frac{1}{(2k-1)^2 pi^2+4z^2}
$$



I have no idea how to prove it and I have also encountered in my research similar sums involving, for instance, $mathrm{coth}$. It would be nice to have a general method for working with these problems; any suggestions?






sequences-and-series trigonometry partial-fractions







share|cite|improve this question











$endgroup$




I have seen the following formula in papers (without citations) and in Mathematica's documentation about Tanh:



$$
frac{tanh(z)}{8z}=sum_{k=1}^{infty} frac{1}{(2k-1)^2 pi^2+4z^2}
$$



I have no idea how to prove it and I have also encountered in my research similar sums involving, for instance, $mathrm{coth}$. It would be nice to have a general method for working with these problems; any suggestions?






sequences-and-series trigonometry partial-fractions




sequences-and-series trigonometry partial-fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 27 '12 at 15:09









J. M. is not a mathematician

60.9k5151290




60.9k5151290










asked Jul 27 '12 at 13:56









Gabriel LandiGabriel Landi

477214




477214












  • $begingroup$
    Maybe you can try the Taylor Series?
    $endgroup$
    – A. Chu
    Jul 27 '12 at 14:14






  • 1




    $begingroup$
    As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $tan z$, and then a simple change of variables $iz$ will give you a similar one for $tanh z$.
    $endgroup$
    – GEdgar
    Jul 28 '12 at 12:36


















  • $begingroup$
    Maybe you can try the Taylor Series?
    $endgroup$
    – A. Chu
    Jul 27 '12 at 14:14






  • 1




    $begingroup$
    As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $tan z$, and then a simple change of variables $iz$ will give you a similar one for $tanh z$.
    $endgroup$
    – GEdgar
    Jul 28 '12 at 12:36
















$begingroup$
Maybe you can try the Taylor Series?
$endgroup$
– A. Chu
Jul 27 '12 at 14:14




$begingroup$
Maybe you can try the Taylor Series?
$endgroup$
– A. Chu
Jul 27 '12 at 14:14




1




1




$begingroup$
As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $tan z$, and then a simple change of variables $iz$ will give you a similar one for $tanh z$.
$endgroup$
– GEdgar
Jul 28 '12 at 12:36




$begingroup$
As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $tan z$, and then a simple change of variables $iz$ will give you a similar one for $tanh z$.
$endgroup$
– GEdgar
Jul 28 '12 at 12:36










3 Answers
3






active

oldest

votes


















11












$begingroup$

There is the infinite product representation



$$cosh,z=prod_{k=1}^infty left(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



Taking logarithms gives



$$logcosh,z=sum_{k=1}^infty logleft(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



If we differentiate both sides, we have



$$tanh,z=sum_{k=1}^infty frac{frac{8z}{pi^2(2k-1)^2}}{1+frac{4z^2}{pi^2(2k-1)^2}}$$



which simplifies to



$$tanh,z=sum_{k=1}^infty frac{8z}{4z^2+pi^2(2k-1)^2}$$



Note that the infinite product that we started with is the factorization of $cosh$ over its (imaginary) zeroes.



Here is a related question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:29










  • $begingroup$
    Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 1:35










  • $begingroup$
    Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:41










  • $begingroup$
    Our answers to this question seem to be paralleled here :-)
    $endgroup$
    – robjohn♦
    Jul 28 '12 at 13:23










  • $begingroup$
    @rob, hence the upvote. ;)
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 13:49





















3












$begingroup$

In this answer, it is shown that for all $zinmathbb{C}setminusmathbb{Z}$,
$$
picot(pi z)=frac1z+sum_{k=1}^inftyfrac{2z}{z^2-k^2}tag{1}
$$
Applying the identity $tan(x)=cot(x)-2cot(2x)$ to $(1)$ gives
$$
begin{align}
pitan(pi z)
&=sum_{k=1}^inftyfrac{2z}{z^2-k^2}-sum_{k=1}^inftyfrac{2z}{z^2-frac{k^2}{4}}\
&=sum_{k=1}^inftyfrac{8z}{4z^2-(2k)^2}-sum_{k=1}^inftyfrac{8z}{4z^2-k^2}\
&=-sum_{k=1}^inftyfrac{8z}{4z^2-(2k-1)^2}\
&=sum_{k=1}^inftyfrac{8z}{(2k-1)^2-4z^2}tag{2}
end{align}
$$
Applying the identity $tanh(x)=-itan(ix)$ to $(2)$ yields
$$
pitanh(pi z)=sum_{k=1}^inftyfrac{8z}{(2k-1)^2+4z^2}tag{3}
$$
Finally, applying the change variables $zmapsto z/pi$ to $(3)$ reveals
$$
frac{tanh(z)}{8z}=sum_{k=1}^inftyfrac{1}{(2k-1)^2pi^2+4z^2}tag{4}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you get equation $(2)$ from the series sum?
    $endgroup$
    – Danny
    May 4 '17 at 4:40










  • $begingroup$
    @Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
    $endgroup$
    – robjohn♦
    May 4 '17 at 12:57





















0












$begingroup$

A general approach to find such expansions is a technique based on expanding a meromorphic function in terms of rational fractions by exploiting its poles. A theorem related to this work is called the Mittag-Leffler theorem in complex analysis. See here (starting from page (96)).
You can find examples of such expansions for the functions $tan(z),,sec(z), cot(z),, csc(z) ,. $






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
    $endgroup$
    – Did
    Aug 24 '12 at 11:43










  • $begingroup$
    I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
    $endgroup$
    – Mhenni Benghorbal
    Sep 11 '12 at 22:35











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

There is the infinite product representation



$$cosh,z=prod_{k=1}^infty left(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



Taking logarithms gives



$$logcosh,z=sum_{k=1}^infty logleft(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



If we differentiate both sides, we have



$$tanh,z=sum_{k=1}^infty frac{frac{8z}{pi^2(2k-1)^2}}{1+frac{4z^2}{pi^2(2k-1)^2}}$$



which simplifies to



$$tanh,z=sum_{k=1}^infty frac{8z}{4z^2+pi^2(2k-1)^2}$$



Note that the infinite product that we started with is the factorization of $cosh$ over its (imaginary) zeroes.



Here is a related question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:29










  • $begingroup$
    Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 1:35










  • $begingroup$
    Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:41










  • $begingroup$
    Our answers to this question seem to be paralleled here :-)
    $endgroup$
    – robjohn♦
    Jul 28 '12 at 13:23










  • $begingroup$
    @rob, hence the upvote. ;)
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 13:49


















11












$begingroup$

There is the infinite product representation



$$cosh,z=prod_{k=1}^infty left(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



Taking logarithms gives



$$logcosh,z=sum_{k=1}^infty logleft(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



If we differentiate both sides, we have



$$tanh,z=sum_{k=1}^infty frac{frac{8z}{pi^2(2k-1)^2}}{1+frac{4z^2}{pi^2(2k-1)^2}}$$



which simplifies to



$$tanh,z=sum_{k=1}^infty frac{8z}{4z^2+pi^2(2k-1)^2}$$



Note that the infinite product that we started with is the factorization of $cosh$ over its (imaginary) zeroes.



Here is a related question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:29










  • $begingroup$
    Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 1:35










  • $begingroup$
    Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:41










  • $begingroup$
    Our answers to this question seem to be paralleled here :-)
    $endgroup$
    – robjohn♦
    Jul 28 '12 at 13:23










  • $begingroup$
    @rob, hence the upvote. ;)
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 13:49
















11












11








11





$begingroup$

There is the infinite product representation



$$cosh,z=prod_{k=1}^infty left(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



Taking logarithms gives



$$logcosh,z=sum_{k=1}^infty logleft(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



If we differentiate both sides, we have



$$tanh,z=sum_{k=1}^infty frac{frac{8z}{pi^2(2k-1)^2}}{1+frac{4z^2}{pi^2(2k-1)^2}}$$



which simplifies to



$$tanh,z=sum_{k=1}^infty frac{8z}{4z^2+pi^2(2k-1)^2}$$



Note that the infinite product that we started with is the factorization of $cosh$ over its (imaginary) zeroes.



Here is a related question.






share|cite|improve this answer











$endgroup$



There is the infinite product representation



$$cosh,z=prod_{k=1}^infty left(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



Taking logarithms gives



$$logcosh,z=sum_{k=1}^infty logleft(1+frac{4z^2}{pi^2(2k-1)^2}right)$$



If we differentiate both sides, we have



$$tanh,z=sum_{k=1}^infty frac{frac{8z}{pi^2(2k-1)^2}}{1+frac{4z^2}{pi^2(2k-1)^2}}$$



which simplifies to



$$tanh,z=sum_{k=1}^infty frac{8z}{4z^2+pi^2(2k-1)^2}$$



Note that the infinite product that we started with is the factorization of $cosh$ over its (imaginary) zeroes.



Here is a related question.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 13 '17 at 12:20









Community♦

1




1










answered Jul 27 '12 at 15:05









J. M. is not a mathematicianJ. M. is not a mathematician

60.9k5151290




60.9k5151290












  • $begingroup$
    @jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:29










  • $begingroup$
    Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 1:35










  • $begingroup$
    Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:41










  • $begingroup$
    Our answers to this question seem to be paralleled here :-)
    $endgroup$
    – robjohn♦
    Jul 28 '12 at 13:23










  • $begingroup$
    @rob, hence the upvote. ;)
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 13:49




















  • $begingroup$
    @jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:29










  • $begingroup$
    Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 1:35










  • $begingroup$
    Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
    $endgroup$
    – Gabriel Landi
    Jul 28 '12 at 1:41










  • $begingroup$
    Our answers to this question seem to be paralleled here :-)
    $endgroup$
    – robjohn♦
    Jul 28 '12 at 13:23










  • $begingroup$
    @rob, hence the upvote. ;)
    $endgroup$
    – J. M. is not a mathematician
    Jul 28 '12 at 13:49


















$begingroup$
@jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
$endgroup$
– Gabriel Landi
Jul 28 '12 at 1:29




$begingroup$
@jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support.
$endgroup$
– Gabriel Landi
Jul 28 '12 at 1:29












$begingroup$
Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
$endgroup$
– J. M. is not a mathematician
Jul 28 '12 at 1:35




$begingroup$
Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts.
$endgroup$
– J. M. is not a mathematician
Jul 28 '12 at 1:35












$begingroup$
Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
$endgroup$
– Gabriel Landi
Jul 28 '12 at 1:41




$begingroup$
Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M.
$endgroup$
– Gabriel Landi
Jul 28 '12 at 1:41












$begingroup$
Our answers to this question seem to be paralleled here :-)
$endgroup$
– robjohn♦
Jul 28 '12 at 13:23




$begingroup$
Our answers to this question seem to be paralleled here :-)
$endgroup$
– robjohn♦
Jul 28 '12 at 13:23












$begingroup$
@rob, hence the upvote. ;)
$endgroup$
– J. M. is not a mathematician
Jul 28 '12 at 13:49






$begingroup$
@rob, hence the upvote. ;)
$endgroup$
– J. M. is not a mathematician
Jul 28 '12 at 13:49













3












$begingroup$

In this answer, it is shown that for all $zinmathbb{C}setminusmathbb{Z}$,
$$
picot(pi z)=frac1z+sum_{k=1}^inftyfrac{2z}{z^2-k^2}tag{1}
$$
Applying the identity $tan(x)=cot(x)-2cot(2x)$ to $(1)$ gives
$$
begin{align}
pitan(pi z)
&=sum_{k=1}^inftyfrac{2z}{z^2-k^2}-sum_{k=1}^inftyfrac{2z}{z^2-frac{k^2}{4}}\
&=sum_{k=1}^inftyfrac{8z}{4z^2-(2k)^2}-sum_{k=1}^inftyfrac{8z}{4z^2-k^2}\
&=-sum_{k=1}^inftyfrac{8z}{4z^2-(2k-1)^2}\
&=sum_{k=1}^inftyfrac{8z}{(2k-1)^2-4z^2}tag{2}
end{align}
$$
Applying the identity $tanh(x)=-itan(ix)$ to $(2)$ yields
$$
pitanh(pi z)=sum_{k=1}^inftyfrac{8z}{(2k-1)^2+4z^2}tag{3}
$$
Finally, applying the change variables $zmapsto z/pi$ to $(3)$ reveals
$$
frac{tanh(z)}{8z}=sum_{k=1}^inftyfrac{1}{(2k-1)^2pi^2+4z^2}tag{4}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you get equation $(2)$ from the series sum?
    $endgroup$
    – Danny
    May 4 '17 at 4:40










  • $begingroup$
    @Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
    $endgroup$
    – robjohn♦
    May 4 '17 at 12:57


















3












$begingroup$

In this answer, it is shown that for all $zinmathbb{C}setminusmathbb{Z}$,
$$
picot(pi z)=frac1z+sum_{k=1}^inftyfrac{2z}{z^2-k^2}tag{1}
$$
Applying the identity $tan(x)=cot(x)-2cot(2x)$ to $(1)$ gives
$$
begin{align}
pitan(pi z)
&=sum_{k=1}^inftyfrac{2z}{z^2-k^2}-sum_{k=1}^inftyfrac{2z}{z^2-frac{k^2}{4}}\
&=sum_{k=1}^inftyfrac{8z}{4z^2-(2k)^2}-sum_{k=1}^inftyfrac{8z}{4z^2-k^2}\
&=-sum_{k=1}^inftyfrac{8z}{4z^2-(2k-1)^2}\
&=sum_{k=1}^inftyfrac{8z}{(2k-1)^2-4z^2}tag{2}
end{align}
$$
Applying the identity $tanh(x)=-itan(ix)$ to $(2)$ yields
$$
pitanh(pi z)=sum_{k=1}^inftyfrac{8z}{(2k-1)^2+4z^2}tag{3}
$$
Finally, applying the change variables $zmapsto z/pi$ to $(3)$ reveals
$$
frac{tanh(z)}{8z}=sum_{k=1}^inftyfrac{1}{(2k-1)^2pi^2+4z^2}tag{4}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how did you get equation $(2)$ from the series sum?
    $endgroup$
    – Danny
    May 4 '17 at 4:40










  • $begingroup$
    @Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
    $endgroup$
    – robjohn♦
    May 4 '17 at 12:57
















3












3








3





$begingroup$

In this answer, it is shown that for all $zinmathbb{C}setminusmathbb{Z}$,
$$
picot(pi z)=frac1z+sum_{k=1}^inftyfrac{2z}{z^2-k^2}tag{1}
$$
Applying the identity $tan(x)=cot(x)-2cot(2x)$ to $(1)$ gives
$$
begin{align}
pitan(pi z)
&=sum_{k=1}^inftyfrac{2z}{z^2-k^2}-sum_{k=1}^inftyfrac{2z}{z^2-frac{k^2}{4}}\
&=sum_{k=1}^inftyfrac{8z}{4z^2-(2k)^2}-sum_{k=1}^inftyfrac{8z}{4z^2-k^2}\
&=-sum_{k=1}^inftyfrac{8z}{4z^2-(2k-1)^2}\
&=sum_{k=1}^inftyfrac{8z}{(2k-1)^2-4z^2}tag{2}
end{align}
$$
Applying the identity $tanh(x)=-itan(ix)$ to $(2)$ yields
$$
pitanh(pi z)=sum_{k=1}^inftyfrac{8z}{(2k-1)^2+4z^2}tag{3}
$$
Finally, applying the change variables $zmapsto z/pi$ to $(3)$ reveals
$$
frac{tanh(z)}{8z}=sum_{k=1}^inftyfrac{1}{(2k-1)^2pi^2+4z^2}tag{4}
$$






share|cite|improve this answer











$endgroup$



In this answer, it is shown that for all $zinmathbb{C}setminusmathbb{Z}$,
$$
picot(pi z)=frac1z+sum_{k=1}^inftyfrac{2z}{z^2-k^2}tag{1}
$$
Applying the identity $tan(x)=cot(x)-2cot(2x)$ to $(1)$ gives
$$
begin{align}
pitan(pi z)
&=sum_{k=1}^inftyfrac{2z}{z^2-k^2}-sum_{k=1}^inftyfrac{2z}{z^2-frac{k^2}{4}}\
&=sum_{k=1}^inftyfrac{8z}{4z^2-(2k)^2}-sum_{k=1}^inftyfrac{8z}{4z^2-k^2}\
&=-sum_{k=1}^inftyfrac{8z}{4z^2-(2k-1)^2}\
&=sum_{k=1}^inftyfrac{8z}{(2k-1)^2-4z^2}tag{2}
end{align}
$$
Applying the identity $tanh(x)=-itan(ix)$ to $(2)$ yields
$$
pitanh(pi z)=sum_{k=1}^inftyfrac{8z}{(2k-1)^2+4z^2}tag{3}
$$
Finally, applying the change variables $zmapsto z/pi$ to $(3)$ reveals
$$
frac{tanh(z)}{8z}=sum_{k=1}^inftyfrac{1}{(2k-1)^2pi^2+4z^2}tag{4}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 4 '17 at 12:40

























answered Jul 28 '12 at 13:18









robjohn♦robjohn

266k27304626




266k27304626












  • $begingroup$
    how did you get equation $(2)$ from the series sum?
    $endgroup$
    – Danny
    May 4 '17 at 4:40










  • $begingroup$
    @Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
    $endgroup$
    – robjohn♦
    May 4 '17 at 12:57




















  • $begingroup$
    how did you get equation $(2)$ from the series sum?
    $endgroup$
    – Danny
    May 4 '17 at 4:40










  • $begingroup$
    @Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
    $endgroup$
    – robjohn♦
    May 4 '17 at 12:57


















$begingroup$
how did you get equation $(2)$ from the series sum?
$endgroup$
– Danny
May 4 '17 at 4:40




$begingroup$
how did you get equation $(2)$ from the series sum?
$endgroup$
– Danny
May 4 '17 at 4:40












$begingroup$
@Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
$endgroup$
– robjohn♦
May 4 '17 at 12:57






$begingroup$
@Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
$endgroup$
– robjohn♦
May 4 '17 at 12:57













0












$begingroup$

A general approach to find such expansions is a technique based on expanding a meromorphic function in terms of rational fractions by exploiting its poles. A theorem related to this work is called the Mittag-Leffler theorem in complex analysis. See here (starting from page (96)).
You can find examples of such expansions for the functions $tan(z),,sec(z), cot(z),, csc(z) ,. $






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
    $endgroup$
    – Did
    Aug 24 '12 at 11:43










  • $begingroup$
    I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
    $endgroup$
    – Mhenni Benghorbal
    Sep 11 '12 at 22:35
















0












$begingroup$

A general approach to find such expansions is a technique based on expanding a meromorphic function in terms of rational fractions by exploiting its poles. A theorem related to this work is called the Mittag-Leffler theorem in complex analysis. See here (starting from page (96)).
You can find examples of such expansions for the functions $tan(z),,sec(z), cot(z),, csc(z) ,. $






share|cite|improve this answer











$endgroup$









  • 5




    $begingroup$
    For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
    $endgroup$
    – Did
    Aug 24 '12 at 11:43










  • $begingroup$
    I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
    $endgroup$
    – Mhenni Benghorbal
    Sep 11 '12 at 22:35














0












0








0





$begingroup$

A general approach to find such expansions is a technique based on expanding a meromorphic function in terms of rational fractions by exploiting its poles. A theorem related to this work is called the Mittag-Leffler theorem in complex analysis. See here (starting from page (96)).
You can find examples of such expansions for the functions $tan(z),,sec(z), cot(z),, csc(z) ,. $






share|cite|improve this answer











$endgroup$



A general approach to find such expansions is a technique based on expanding a meromorphic function in terms of rational fractions by exploiting its poles. A theorem related to this work is called the Mittag-Leffler theorem in complex analysis. See here (starting from page (96)).
You can find examples of such expansions for the functions $tan(z),,sec(z), cot(z),, csc(z) ,. $







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 '13 at 4:38

























answered Jul 27 '12 at 15:44









Mhenni BenghorbalMhenni Benghorbal

43.2k63574




43.2k63574








  • 5




    $begingroup$
    For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
    $endgroup$
    – Did
    Aug 24 '12 at 11:43










  • $begingroup$
    I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
    $endgroup$
    – Mhenni Benghorbal
    Sep 11 '12 at 22:35














  • 5




    $begingroup$
    For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
    $endgroup$
    – Did
    Aug 24 '12 at 11:43










  • $begingroup$
    I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
    $endgroup$
    – Mhenni Benghorbal
    Sep 11 '12 at 22:35








5




5




$begingroup$
For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
$endgroup$
– Did
Aug 24 '12 at 11:43




$begingroup$
For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
$endgroup$
– Did
Aug 24 '12 at 11:43












$begingroup$
I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
$endgroup$
– Mhenni Benghorbal
Sep 11 '12 at 22:35




$begingroup$
I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research.
$endgroup$
– Mhenni Benghorbal
Sep 11 '12 at 22:35


















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