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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)$?
$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 ?
sequences-and-series proof-writing
edited yesterday
rtybase
11.4k31533
asked yesterday
CamionCamion
265
$endgroup$
add a comment |
$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 ?
sequences-and-series proof-writing
edited yesterday
rtybase
11.4k31533
asked yesterday
CamionCamion
265
$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
add a comment |
$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 ?
sequences-and-series proof-writing
edited yesterday
rtybase
11.4k31533
asked yesterday
CamionCamion
265
$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
sequences-and-series proof-writing
edited yesterday
rtybase
11.4k31533
asked yesterday
CamionCamion
265
edited yesterday
rtybase
11.4k31533
asked yesterday
CamionCamion
265
edited yesterday
rtybase
11.4k31533
edited yesterday
rtybase
11.4k31533
edited yesterday
rtybase
11.4k31533
11.4k31533
asked yesterday
CamionCamion
265
asked yesterday
CamionCamion
265
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
add a comment |
$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
add a comment |
2 Answers
2
active
oldest
votes
$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)$$
Can you wrap this up?
answered yesterday
Mark ViolaMark Viola
133k1277176
$endgroup$
$begingroup$
Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
$endgroup$
– Camion
yesterday
$begingroup$
(for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
$endgroup$
– Camion
yesterday
$begingroup$
@Camion: Here is a proof without words.
$endgroup$
– robjohn♦
11 hours ago
$begingroup$
Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
$endgroup$
– Mark Viola
10 hours ago
add a comment |
$begingroup$
From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$
Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$
answered yesterday
robjohn♦robjohn
269k27310636
$endgroup$
$begingroup$
If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
$endgroup$
– robjohn♦
yesterday
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The product representation of the sine function is
$$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)$$
Can you wrap this up?
answered yesterday
Mark ViolaMark Viola
133k1277176
$endgroup$
$begingroup$
Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
$endgroup$
– Camion
yesterday
$begingroup$
(for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
$endgroup$
– Camion
yesterday
$begingroup$
@Camion: Here is a proof without words.
$endgroup$
– robjohn♦
11 hours ago
$begingroup$
Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
$endgroup$
– Mark Viola
10 hours ago
add a comment |
$begingroup$
The product representation of the sine function is
$$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)$$
Can you wrap this up?
answered yesterday
Mark ViolaMark Viola
133k1277176
$endgroup$
$begingroup$
Shame on me for not having been able to see that $fraccos(x)+12sin(x)=frac12cot(x/2)$
$endgroup$
– Camion
yesterday
$begingroup$
(for future reference : $fraccos(x)+12sin(x)=fraccos²(x/2)-sin²(x/2)+14sin(x/2)cos(x/2)=frac2cos²(x/2)4sin(x/2)cos(x/2)=frac12cot(x/2)$)
$endgroup$
– Camion
yesterday
$begingroup$
@Camion: Here is a proof without words.
$endgroup$
– robjohn♦
11 hours ago
$begingroup$
Hi Rob. In the wordless proof, how does one know that the angles involved differ by a multiplicative factor of 2 without words?
$endgroup$
– Mark Viola
10 hours ago
add a comment |
$begingroup$
The product representation of the sine function is
$$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)$$
Can you wrap this up?
answered yesterday
Mark ViolaMark Viola
133k1277176
$endgroup$
The product representation of the sine function is
$$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?
answered yesterday
Mark ViolaMark Viola
133k1277176
answered yesterday
Mark ViolaMark Viola
133k1277176
answered yesterday
Mark ViolaMark Viola
133k1277176
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
add a comment |
$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
add a comment |
$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
$$
answered yesterday
robjohn♦robjohn
269k27310636
$endgroup$
$begingroup$
If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
$endgroup$
– robjohn♦
yesterday
add a comment |
$begingroup$
From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$
Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$
answered yesterday
robjohn♦robjohn
269k27310636
$endgroup$
$begingroup$
If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series).
$endgroup$
– robjohn♦
yesterday
add a comment |
$begingroup$
From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$
Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$
answered yesterday
robjohn♦robjohn
269k27310636
$endgroup$
From this answer we get
$$
sum_kinmathbbZfrac1k+x=picot(pi x)
$$
Thus,
$$
beginalign
sum_kinmathbbZfrac12pi k+x
&=frac12pisum_kinmathbbZfrac1k+fracx2pi\
&=frac12cotleft(frac x2right)
endalign
$$
answered yesterday
robjohn♦robjohn
269k27310636
answered yesterday
robjohn♦robjohn
269k27310636
answered yesterday
robjohn♦robjohn
269k27310636
answered yesterday
robjohn♦robjohn
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
add a comment |
$begingroup$
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
add a comment |
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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 $lim_N to inftysum_n=-N^Nfrac1x+n-2i pi$
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– reuns
yesterday
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Otherwise look at the Fourier series of $sum_n h(x+n)$ where $h(x)=e^-ax1_x> 0$
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– reuns
yesterday
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@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)
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– Camion
yesterday