Prove series is bounded on $mathbb{C}$ (Complex Analysis)












1












$begingroup$



Define a mapping $$f(z)=cot(z)-frac {1}{z}-sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$$



Show that $f$ is bounded on $mathbb{C}$




I am trying to derive the Euler factorization of $sin(z)=zprod_{n=1}^{infty}(1-frac {z^2}{n^2pi^2})$



I was able to show that $sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$ converges locally uniformly (i.e. compactly converges on ${zin mathbb{C}:znotin npi, nin mathbb{Z}}$), and that $f(z)$ has removable singularities at $npi$ for $nin mathbb{Z}$, which I will use later to derive the Euler factorization.



I am not really sure how to bound $f$ though, and I have a suspicion that $f(z)=0$.



Any help would be much appreciated. Thanks in advance!






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I suppose that you meant $cot(z)$ instead of $cot(t)$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 13:46










  • $begingroup$
    My apologies, yes I do. Edited.
    $endgroup$
    – Math is Life
    Nov 28 '18 at 13:47
















1












$begingroup$



Define a mapping $$f(z)=cot(z)-frac {1}{z}-sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$$



Show that $f$ is bounded on $mathbb{C}$




I am trying to derive the Euler factorization of $sin(z)=zprod_{n=1}^{infty}(1-frac {z^2}{n^2pi^2})$



I was able to show that $sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$ converges locally uniformly (i.e. compactly converges on ${zin mathbb{C}:znotin npi, nin mathbb{Z}}$), and that $f(z)$ has removable singularities at $npi$ for $nin mathbb{Z}$, which I will use later to derive the Euler factorization.



I am not really sure how to bound $f$ though, and I have a suspicion that $f(z)=0$.



Any help would be much appreciated. Thanks in advance!






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I suppose that you meant $cot(z)$ instead of $cot(t)$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 13:46










  • $begingroup$
    My apologies, yes I do. Edited.
    $endgroup$
    – Math is Life
    Nov 28 '18 at 13:47














1












1








1





$begingroup$



Define a mapping $$f(z)=cot(z)-frac {1}{z}-sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$$



Show that $f$ is bounded on $mathbb{C}$




I am trying to derive the Euler factorization of $sin(z)=zprod_{n=1}^{infty}(1-frac {z^2}{n^2pi^2})$



I was able to show that $sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$ converges locally uniformly (i.e. compactly converges on ${zin mathbb{C}:znotin npi, nin mathbb{Z}}$), and that $f(z)$ has removable singularities at $npi$ for $nin mathbb{Z}$, which I will use later to derive the Euler factorization.



I am not really sure how to bound $f$ though, and I have a suspicion that $f(z)=0$.



Any help would be much appreciated. Thanks in advance!






complex-analysis







share|cite|improve this question











$endgroup$





Define a mapping $$f(z)=cot(z)-frac {1}{z}-sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$$



Show that $f$ is bounded on $mathbb{C}$




I am trying to derive the Euler factorization of $sin(z)=zprod_{n=1}^{infty}(1-frac {z^2}{n^2pi^2})$



I was able to show that $sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$ converges locally uniformly (i.e. compactly converges on ${zin mathbb{C}:znotin npi, nin mathbb{Z}}$), and that $f(z)$ has removable singularities at $npi$ for $nin mathbb{Z}$, which I will use later to derive the Euler factorization.



I am not really sure how to bound $f$ though, and I have a suspicion that $f(z)=0$.



Any help would be much appreciated. Thanks in advance!






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 14:00







Math is Life

















asked Nov 28 '18 at 13:39









Math is LifeMath is Life

574




574












  • $begingroup$
    I suppose that you meant $cot(z)$ instead of $cot(t)$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 13:46










  • $begingroup$
    My apologies, yes I do. Edited.
    $endgroup$
    – Math is Life
    Nov 28 '18 at 13:47


















  • $begingroup$
    I suppose that you meant $cot(z)$ instead of $cot(t)$.
    $endgroup$
    – José Carlos Santos
    Nov 28 '18 at 13:46










  • $begingroup$
    My apologies, yes I do. Edited.
    $endgroup$
    – Math is Life
    Nov 28 '18 at 13:47
















$begingroup$
I suppose that you meant $cot(z)$ instead of $cot(t)$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 13:46




$begingroup$
I suppose that you meant $cot(z)$ instead of $cot(t)$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 13:46












$begingroup$
My apologies, yes I do. Edited.
$endgroup$
– Math is Life
Nov 28 '18 at 13:47




$begingroup$
My apologies, yes I do. Edited.
$endgroup$
– Math is Life
Nov 28 '18 at 13:47










1 Answer
1






active

oldest

votes


















2












$begingroup$

The formula summation (together with 1/z) is in fact the Mittag Leffler expansion of $cot z$. So your guess is right, $f$ is indeed zero. To show this, let $g(z):=frac{pi}{z^2-a^2}cot pi z$ where $anotin mathbb{Z}$ and $gamma_n$ be the square contour of length $n$ centered at the origin, i.e. $gamma_n$ has vertices $[(n+frac{1}{2})+i(n+frac{1}{2}), (n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})+i(n+frac{1}{2})]$. First show the following:



a) $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$



b) Now compute the integral $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$ directly using the residue theorem, i.e. find the residues of $g(z)$. The poles of $g(z)$ are exactly at $z=pm a$ and the integers.



With this, you can get the desired summation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
    $endgroup$
    – Math is Life
    Nov 29 '18 at 0:41










  • $begingroup$
    I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
    $endgroup$
    – thedilated
    Nov 29 '18 at 6:07











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1 Answer
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1 Answer
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2












$begingroup$

The formula summation (together with 1/z) is in fact the Mittag Leffler expansion of $cot z$. So your guess is right, $f$ is indeed zero. To show this, let $g(z):=frac{pi}{z^2-a^2}cot pi z$ where $anotin mathbb{Z}$ and $gamma_n$ be the square contour of length $n$ centered at the origin, i.e. $gamma_n$ has vertices $[(n+frac{1}{2})+i(n+frac{1}{2}), (n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})+i(n+frac{1}{2})]$. First show the following:



a) $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$



b) Now compute the integral $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$ directly using the residue theorem, i.e. find the residues of $g(z)$. The poles of $g(z)$ are exactly at $z=pm a$ and the integers.



With this, you can get the desired summation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
    $endgroup$
    – Math is Life
    Nov 29 '18 at 0:41










  • $begingroup$
    I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
    $endgroup$
    – thedilated
    Nov 29 '18 at 6:07
















2












$begingroup$

The formula summation (together with 1/z) is in fact the Mittag Leffler expansion of $cot z$. So your guess is right, $f$ is indeed zero. To show this, let $g(z):=frac{pi}{z^2-a^2}cot pi z$ where $anotin mathbb{Z}$ and $gamma_n$ be the square contour of length $n$ centered at the origin, i.e. $gamma_n$ has vertices $[(n+frac{1}{2})+i(n+frac{1}{2}), (n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})+i(n+frac{1}{2})]$. First show the following:



a) $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$



b) Now compute the integral $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$ directly using the residue theorem, i.e. find the residues of $g(z)$. The poles of $g(z)$ are exactly at $z=pm a$ and the integers.



With this, you can get the desired summation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
    $endgroup$
    – Math is Life
    Nov 29 '18 at 0:41










  • $begingroup$
    I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
    $endgroup$
    – thedilated
    Nov 29 '18 at 6:07














2












2








2





$begingroup$

The formula summation (together with 1/z) is in fact the Mittag Leffler expansion of $cot z$. So your guess is right, $f$ is indeed zero. To show this, let $g(z):=frac{pi}{z^2-a^2}cot pi z$ where $anotin mathbb{Z}$ and $gamma_n$ be the square contour of length $n$ centered at the origin, i.e. $gamma_n$ has vertices $[(n+frac{1}{2})+i(n+frac{1}{2}), (n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})+i(n+frac{1}{2})]$. First show the following:



a) $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$



b) Now compute the integral $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$ directly using the residue theorem, i.e. find the residues of $g(z)$. The poles of $g(z)$ are exactly at $z=pm a$ and the integers.



With this, you can get the desired summation.






share|cite|improve this answer











$endgroup$



The formula summation (together with 1/z) is in fact the Mittag Leffler expansion of $cot z$. So your guess is right, $f$ is indeed zero. To show this, let $g(z):=frac{pi}{z^2-a^2}cot pi z$ where $anotin mathbb{Z}$ and $gamma_n$ be the square contour of length $n$ centered at the origin, i.e. $gamma_n$ has vertices $[(n+frac{1}{2})+i(n+frac{1}{2}), (n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})-i(n+frac{1}{2}), -(n+frac{1}{2})+i(n+frac{1}{2})]$. First show the following:



a) $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$



b) Now compute the integral $lim_{nrightarrow infty} int_{gamma_n}g(z)dz=0$ directly using the residue theorem, i.e. find the residues of $g(z)$. The poles of $g(z)$ are exactly at $z=pm a$ and the integers.



With this, you can get the desired summation.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 6:02

























answered Nov 28 '18 at 14:15









thedilatedthedilated

1,0641615




1,0641615












  • $begingroup$
    I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
    $endgroup$
    – Math is Life
    Nov 29 '18 at 0:41










  • $begingroup$
    I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
    $endgroup$
    – thedilated
    Nov 29 '18 at 6:07


















  • $begingroup$
    I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
    $endgroup$
    – Math is Life
    Nov 29 '18 at 0:41










  • $begingroup$
    I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
    $endgroup$
    – thedilated
    Nov 29 '18 at 6:07
















$begingroup$
I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
$endgroup$
– Math is Life
Nov 29 '18 at 0:41




$begingroup$
I'm not sure I understand, I haven't done any Mittlag Leffler expansions. Could you please explain?
$endgroup$
– Math is Life
Nov 29 '18 at 0:41












$begingroup$
I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
$endgroup$
– thedilated
Nov 29 '18 at 6:07




$begingroup$
I have edited my post adding more details on the exercise. Of course this is just one of the many methods on showing the expansion of $cot pi z$. Alternatively, you can check out proofwiki.org/wiki/…
$endgroup$
– thedilated
Nov 29 '18 at 6:07


















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