How much makes $sumlimits_i=-infty^infty frac1i2pi+x$?How to prove a trigonometric identity $tan(A)=fracsin2A1+cos 2A$Testing the series $sumlimits_n=1^infty frac1n^k + cosn$Convergence of $sum limits_n=1^infty (1-fracsin a_na_n)$ when $sum limits_n=1^infty a_n$ convergesusing only power series to show $sumfrac1n^2=fracpi^26$How to prove $sumlimits_n=1^inftyfracsin nn=fracpi-12$Compute $sumlimits_n = 1^infty frac1n^4$.Does $sum_n=1^infty fraccos(ln(n))n$ converge?Does $a_0+sum limits_n=1^infty(a_n+a_-n)=sum limits_n=-infty^+inftya_n$?Find the sum of $sumlimits_k=1^inftyq^ksin kalpha$Given a convergent series $sumlimits_n=1^infty a_k$, show that $limlimits_ntoinftyfrac1nsumlimits_k=1^n a_k = 0$$sumlimits_n=1^inftyfracsin(nx)(n^4+x^4)^frac13$ How to major $sumlimits_n=1^inftysin(nx)$?

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How much makes $sumlimits_i=-infty^infty frac1i2pi+x$?


How to prove a trigonometric identity $tan(A)=fracsin2A1+cos 2A$Testing the series $sumlimits_n=1^infty frac1n^k + cosn$Convergence of $sum limits_n=1^infty (1-fracsin a_na_n)$ when $sum limits_n=1^infty a_n$ convergesusing only power series to show $sumfrac1n^2=fracpi^26$How to prove $sumlimits_n=1^inftyfracsin nn=fracpi-12$Compute $sumlimits_n = 1^infty frac1n^4$.Does $sum_n=1^infty fraccos(ln(n))n$ converge?Does $a_0+sum limits_n=1^infty(a_n+a_-n)=sum limits_n=-infty^+inftya_n$?Find the sum of $sumlimits_k=1^inftyq^ksin kalpha$Given a convergent series $sumlimits_n=1^infty a_k$, show that $limlimits_ntoinftyfrac1nsumlimits_k=1^n a_k = 0$$sumlimits_n=1^inftyfracsin(nx)(n^4+x^4)^frac13$ How to major $sumlimits_n=1^inftysin(nx)$?













2












$begingroup$


I believe $sumlimits_i=-infty^infty frac1i2pi+x = frac1+cos x2 sin x$ and that it is possible to prove it in a very indirect way (using filtering, Fourier Series and transforms. But is there a simpler way to get to this result ?










share|cite|improve this question











$endgroup$











  • $begingroup$
    You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^-y1_y > 0$'s Fourier transform is $frac1x+2i pi$ thus $e^-y 1_y>0$ is the FT of $frac1x-2i pi$ and $e^-n1_n > 0$ are the Fourier series coefficients of the $1$-periodization $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
    $endgroup$
    – reuns
    yesterday











  • $begingroup$
    Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
    $endgroup$
    – reuns
    yesterday










  • $begingroup$
    @reuns, I had to search to understand your notation «$1_x>0$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0)
    $endgroup$
    – Camion
    yesterday
















2












$begingroup$


I believe $sumlimits_i=-infty^infty frac1i2pi+x = frac1+cos x2 sin x$ and that it is possible to prove it in a very indirect way (using filtering, Fourier Series and transforms. But is there a simpler way to get to this result ?










share|cite|improve this question











$endgroup$











  • $begingroup$
    You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^-y1_y > 0$'s Fourier transform is $frac1x+2i pi$ thus $e^-y 1_y>0$ is the FT of $frac1x-2i pi$ and $e^-n1_n > 0$ are the Fourier series coefficients of the $1$-periodization $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
    $endgroup$
    – reuns
    yesterday











  • $begingroup$
    Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
    $endgroup$
    – reuns
    yesterday










  • $begingroup$
    @reuns, I had to search to understand your notation «$1_x>0$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0)
    $endgroup$
    – Camion
    yesterday














2












2








2





$begingroup$


I believe $sumlimits_i=-infty^infty frac1i2pi+x = frac1+cos x2 sin x$ and that it is possible to prove it in a very indirect way (using filtering, Fourier Series and transforms. But is there a simpler way to get to this result ?










share|cite|improve this question











$endgroup$




I believe $sumlimits_i=-infty^infty frac1i2pi+x = frac1+cos x2 sin x$ and that it is possible to prove it in a very indirect way (using filtering, Fourier Series and transforms. But is there a simpler way to get to this result ?







sequences-and-series proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









rtybase

11.4k31533




11.4k31533










asked yesterday









CamionCamion

265




265











  • $begingroup$
    You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^-y1_y > 0$'s Fourier transform is $frac1x+2i pi$ thus $e^-y 1_y>0$ is the FT of $frac1x-2i pi$ and $e^-n1_n > 0$ are the Fourier series coefficients of the $1$-periodization $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
    $endgroup$
    – reuns
    yesterday











  • $begingroup$
    Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
    $endgroup$
    – reuns
    yesterday










  • $begingroup$
    @reuns, I had to search to understand your notation «$1_x>0$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0)
    $endgroup$
    – Camion
    yesterday

















  • $begingroup$
    You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^-y1_y > 0$'s Fourier transform is $frac1x+2i pi$ thus $e^-y 1_y>0$ is the FT of $frac1x-2i pi$ and $e^-n1_n > 0$ are the Fourier series coefficients of the $1$-periodization $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
    $endgroup$
    – reuns
    yesterday











  • $begingroup$
    Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
    $endgroup$
    – reuns
    yesterday










  • $begingroup$
    @reuns, I had to search to understand your notation «$1_x>0$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0)
    $endgroup$
    – Camion
    yesterday
















$begingroup$
You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^-y1_y > 0$'s Fourier transform is $frac1x+2i pi$ thus $e^-y 1_y>0$ is the FT of $frac1x-2i pi$ and $e^-n1_n > 0$ are the Fourier series coefficients of the $1$-periodization $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
$endgroup$
– reuns
yesterday





$begingroup$
You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^-y1_y > 0$'s Fourier transform is $frac1x+2i pi$ thus $e^-y 1_y>0$ is the FT of $frac1x-2i pi$ and $e^-n1_n > 0$ are the Fourier series coefficients of the $1$-periodization $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
$endgroup$
– reuns
yesterday













$begingroup$
Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
$endgroup$
– reuns
yesterday




$begingroup$
Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
$endgroup$
– reuns
yesterday












$begingroup$
@reuns, I had to search to understand your notation «$1_x>0$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0)
$endgroup$
– Camion
yesterday





$begingroup$
@reuns, I had to search to understand your notation «$1_x>0$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0)
$endgroup$
– Camion
yesterday











2 Answers
2






active

oldest

votes


















1












$begingroup$

The product representation of the sine function is



$$sin( x)= xprod_n=1^infty left(1-fracx^2n^2pi^2right)tag1$$



Taking the logarithmic derivative of $(1)$ reveals



$$beginalign
cot(x)&=frac1x +sum_n=1^infty frac2xleft(n^2pi^2-x^2right)\\
&=frac1x+sum_n=1^infty left(frac1x+npi+frac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=1^Nfrac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=-N^-1frac1x+npiright)\\
&=lim_Ntoinftysum_n=-N^N frac1x+npi
endalign$$



Now note that



$$fraccos(x)+12sin(x)=frac12cot(x/2)$$



Can you wrap this up?






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
    $endgroup$
    – Camion
    yesterday











  • $begingroup$
    (for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
    $endgroup$
    – Camion
    yesterday










  • $begingroup$
    @Camion: Here is a proof without words.
    $endgroup$
    – robjohn
    11 hours ago











  • $begingroup$
    Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
    $endgroup$
    – Mark Viola
    10 hours ago


















0












$begingroup$

From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$

Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
    $endgroup$
    – robjohn
    yesterday










Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The product representation of the sine function is



$$sin( x)= xprod_n=1^infty left(1-fracx^2n^2pi^2right)tag1$$



Taking the logarithmic derivative of $(1)$ reveals



$$beginalign
cot(x)&=frac1x +sum_n=1^infty frac2xleft(n^2pi^2-x^2right)\\
&=frac1x+sum_n=1^infty left(frac1x+npi+frac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=1^Nfrac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=-N^-1frac1x+npiright)\\
&=lim_Ntoinftysum_n=-N^N frac1x+npi
endalign$$



Now note that



$$fraccos(x)+12sin(x)=frac12cot(x/2)$$



Can you wrap this up?






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
    $endgroup$
    – Camion
    yesterday











  • $begingroup$
    (for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
    $endgroup$
    – Camion
    yesterday










  • $begingroup$
    @Camion: Here is a proof without words.
    $endgroup$
    – robjohn
    11 hours ago











  • $begingroup$
    Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
    $endgroup$
    – Mark Viola
    10 hours ago















1












$begingroup$

The product representation of the sine function is



$$sin( x)= xprod_n=1^infty left(1-fracx^2n^2pi^2right)tag1$$



Taking the logarithmic derivative of $(1)$ reveals



$$beginalign
cot(x)&=frac1x +sum_n=1^infty frac2xleft(n^2pi^2-x^2right)\\
&=frac1x+sum_n=1^infty left(frac1x+npi+frac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=1^Nfrac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=-N^-1frac1x+npiright)\\
&=lim_Ntoinftysum_n=-N^N frac1x+npi
endalign$$



Now note that



$$fraccos(x)+12sin(x)=frac12cot(x/2)$$



Can you wrap this up?






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
    $endgroup$
    – Camion
    yesterday











  • $begingroup$
    (for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
    $endgroup$
    – Camion
    yesterday










  • $begingroup$
    @Camion: Here is a proof without words.
    $endgroup$
    – robjohn
    11 hours ago











  • $begingroup$
    Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
    $endgroup$
    – Mark Viola
    10 hours ago













1












1








1





$begingroup$

The product representation of the sine function is



$$sin( x)= xprod_n=1^infty left(1-fracx^2n^2pi^2right)tag1$$



Taking the logarithmic derivative of $(1)$ reveals



$$beginalign
cot(x)&=frac1x +sum_n=1^infty frac2xleft(n^2pi^2-x^2right)\\
&=frac1x+sum_n=1^infty left(frac1x+npi+frac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=1^Nfrac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=-N^-1frac1x+npiright)\\
&=lim_Ntoinftysum_n=-N^N frac1x+npi
endalign$$



Now note that



$$fraccos(x)+12sin(x)=frac12cot(x/2)$$



Can you wrap this up?






share|cite|improve this answer









$endgroup$



The product representation of the sine function is



$$sin( x)= xprod_n=1^infty left(1-fracx^2n^2pi^2right)tag1$$



Taking the logarithmic derivative of $(1)$ reveals



$$beginalign
cot(x)&=frac1x +sum_n=1^infty frac2xleft(n^2pi^2-x^2right)\\
&=frac1x+sum_n=1^infty left(frac1x+npi+frac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=1^Nfrac1x-npiright)\\
&=frac1x+lim_nto Nleft(sum_n=1^N frac1x+npi+sum_n=-N^-1frac1x+npiright)\\
&=lim_Ntoinftysum_n=-N^N frac1x+npi
endalign$$



Now note that



$$fraccos(x)+12sin(x)=frac12cot(x/2)$$



Can you wrap this up?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Mark ViolaMark Viola

133k1277176




133k1277176











  • $begingroup$
    Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
    $endgroup$
    – Camion
    yesterday











  • $begingroup$
    (for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
    $endgroup$
    – Camion
    yesterday










  • $begingroup$
    @Camion: Here is a proof without words.
    $endgroup$
    – robjohn
    11 hours ago











  • $begingroup$
    Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
    $endgroup$
    – Mark Viola
    10 hours ago
















  • $begingroup$
    Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
    $endgroup$
    – Camion
    yesterday











  • $begingroup$
    (for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
    $endgroup$
    – Camion
    yesterday










  • $begingroup$
    @Camion: Here is a proof without words.
    $endgroup$
    – robjohn
    11 hours ago











  • $begingroup$
    Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
    $endgroup$
    – Mark Viola
    10 hours ago















$begingroup$
Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
$endgroup$
– Camion
yesterday





$begingroup$
Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
$endgroup$
– Camion
yesterday













$begingroup$
(for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
$endgroup$
– Camion
yesterday




$begingroup$
(for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
$endgroup$
– Camion
yesterday












$begingroup$
@Camion: Here is a proof without words.
$endgroup$
– robjohn
11 hours ago





$begingroup$
@Camion: Here is a proof without words.
$endgroup$
– robjohn
11 hours ago













$begingroup$
Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
$endgroup$
– Mark Viola
10 hours ago




$begingroup$
Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
$endgroup$
– Mark Viola
10 hours ago











0












$begingroup$

From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$

Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
    $endgroup$
    – robjohn
    yesterday















0












$begingroup$

From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$

Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
    $endgroup$
    – robjohn
    yesterday













0












0








0





$begingroup$

From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$

Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$






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$endgroup$



From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$

Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









robjohnrobjohn

269k27310636




269k27310636











  • $begingroup$
    If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
    $endgroup$
    – robjohn
    yesterday
















  • $begingroup$
    If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
    $endgroup$
    – robjohn
    yesterday















$begingroup$
If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
$endgroup$
– robjohn
yesterday




$begingroup$
If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
$endgroup$
– robjohn
yesterday

















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