Mixing
Proving $pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$ using the Fourier series for $cosh ax$
$begingroup$
The exercise wants me to prove the identity
$$pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
using the Fourier series of $cosh ax, ; x in [-pi, pi], ; a neq 0$.
Evaluating the coefficients ($b_n$ is actually zero as $cosh ax$ is even) we have that:
$displaystyle {color{gray} blacksquare} ;; a_0 = frac{1}{pi}int_{-pi}^{pi}cosh ax , {rm d}x= frac{2sinh pi a}{pi a}$
$require{cancel}displaystyle {color{gray} blacksquare} ;; a_n = frac{1}{pi}int_{-pi}^{pi}cosh ax cos n x, {rm d}x= frac{2[a sinh pi a cos pi n+ cancelto{0}{n cosh pi a sin pi n}]}{pi (a^2+n^2)}$
So far so good except one little problem. I get that $cos pi n$ in the nominator which is $(-1)^n$ so I get the alternating series not the wanted one. That is , this way I evaluated the series:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n}{n^2 +a^2} implies \
implies picoth pi a = frac{1}{a}+ sum_{n=1}^{infty}frac{2(-1)^n}{n^2+a^2}$$
and not what I want. Calculations of the coefficients were done by Wolfram because they were too tedious to be done by hand.
However, I know that using contour integration using the kernel $pi cot pi z$ that this series indeed evaluates to the LHS. What is wrong here?
real-analysis fourier-series
edited Jan 1 at 20:19
Blue
47.7k870151
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
$endgroup$
|
show 2 more comments
$begingroup$
The exercise wants me to prove the identity
$$pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
using the Fourier series of $cosh ax, ; x in [-pi, pi], ; a neq 0$.
Evaluating the coefficients ($b_n$ is actually zero as $cosh ax$ is even) we have that:
$displaystyle {color{gray} blacksquare} ;; a_0 = frac{1}{pi}int_{-pi}^{pi}cosh ax , {rm d}x= frac{2sinh pi a}{pi a}$
$require{cancel}displaystyle {color{gray} blacksquare} ;; a_n = frac{1}{pi}int_{-pi}^{pi}cosh ax cos n x, {rm d}x= frac{2[a sinh pi a cos pi n+ cancelto{0}{n cosh pi a sin pi n}]}{pi (a^2+n^2)}$
So far so good except one little problem. I get that $cos pi n$ in the nominator which is $(-1)^n$ so I get the alternating series not the wanted one. That is , this way I evaluated the series:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n}{n^2 +a^2} implies \
implies picoth pi a = frac{1}{a}+ sum_{n=1}^{infty}frac{2(-1)^n}{n^2+a^2}$$
and not what I want. Calculations of the coefficients were done by Wolfram because they were too tedious to be done by hand.
However, I know that using contour integration using the kernel $pi cot pi z$ that this series indeed evaluates to the LHS. What is wrong here?
real-analysis fourier-series
edited Jan 1 at 20:19
Blue
47.7k870151
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
$endgroup$
$begingroup$
Hint: variable $x$ does not appear in the RHS of your result, as it should.
$endgroup$
– Aretino
Jul 27 '15 at 20:54
$begingroup$
Yes... but I cannot see where the mistake is. As for $x$ you are right. Then I plug in $x=pi$.
$endgroup$
– Tolaso
Jul 27 '15 at 20:56
$begingroup$
Oups... I forgot the $cos nx$ is the RHS.
$endgroup$
– Tolaso
Jul 27 '15 at 20:57
1
$begingroup$
Yes, and then you must put $x=pi$.
$endgroup$
– Aretino
Jul 27 '15 at 20:58
1
$begingroup$
Similar work with $a:=iz$.
$endgroup$
– Raymond Manzoni
Jul 27 '15 at 21:07
|
show 2 more comments
$begingroup$
The exercise wants me to prove the identity
$$pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
using the Fourier series of $cosh ax, ; x in [-pi, pi], ; a neq 0$.
Evaluating the coefficients ($b_n$ is actually zero as $cosh ax$ is even) we have that:
$displaystyle {color{gray} blacksquare} ;; a_0 = frac{1}{pi}int_{-pi}^{pi}cosh ax , {rm d}x= frac{2sinh pi a}{pi a}$
$require{cancel}displaystyle {color{gray} blacksquare} ;; a_n = frac{1}{pi}int_{-pi}^{pi}cosh ax cos n x, {rm d}x= frac{2[a sinh pi a cos pi n+ cancelto{0}{n cosh pi a sin pi n}]}{pi (a^2+n^2)}$
So far so good except one little problem. I get that $cos pi n$ in the nominator which is $(-1)^n$ so I get the alternating series not the wanted one. That is , this way I evaluated the series:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n}{n^2 +a^2} implies \
implies picoth pi a = frac{1}{a}+ sum_{n=1}^{infty}frac{2(-1)^n}{n^2+a^2}$$
and not what I want. Calculations of the coefficients were done by Wolfram because they were too tedious to be done by hand.
However, I know that using contour integration using the kernel $pi cot pi z$ that this series indeed evaluates to the LHS. What is wrong here?
real-analysis fourier-series
edited Jan 1 at 20:19
Blue
47.7k870151
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
$endgroup$
The exercise wants me to prove the identity
$$pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
using the Fourier series of $cosh ax, ; x in [-pi, pi], ; a neq 0$.
Evaluating the coefficients ($b_n$ is actually zero as $cosh ax$ is even) we have that:
$displaystyle {color{gray} blacksquare} ;; a_0 = frac{1}{pi}int_{-pi}^{pi}cosh ax , {rm d}x= frac{2sinh pi a}{pi a}$
$require{cancel}displaystyle {color{gray} blacksquare} ;; a_n = frac{1}{pi}int_{-pi}^{pi}cosh ax cos n x, {rm d}x= frac{2[a sinh pi a cos pi n+ cancelto{0}{n cosh pi a sin pi n}]}{pi (a^2+n^2)}$
So far so good except one little problem. I get that $cos pi n$ in the nominator which is $(-1)^n$ so I get the alternating series not the wanted one. That is , this way I evaluated the series:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n}{n^2 +a^2} implies \
implies picoth pi a = frac{1}{a}+ sum_{n=1}^{infty}frac{2(-1)^n}{n^2+a^2}$$
and not what I want. Calculations of the coefficients were done by Wolfram because they were too tedious to be done by hand.
However, I know that using contour integration using the kernel $pi cot pi z$ that this series indeed evaluates to the LHS. What is wrong here?
real-analysis fourier-series
real-analysis fourier-series
edited Jan 1 at 20:19
Blue
47.7k870151
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
edited Jan 1 at 20:19
Blue
47.7k870151
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
edited Jan 1 at 20:19
Blue
47.7k870151
edited Jan 1 at 20:19
Blue
47.7k870151
edited Jan 1 at 20:19
Blue
47.7k870151
47.7k870151
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
asked Jul 27 '15 at 20:47
TolasoTolaso
3,4101231
3,4101231
$begingroup$
Hint: variable $x$ does not appear in the RHS of your result, as it should.
$endgroup$
– Aretino
Jul 27 '15 at 20:54
$begingroup$
Yes... but I cannot see where the mistake is. As for $x$ you are right. Then I plug in $x=pi$.
$endgroup$
– Tolaso
Jul 27 '15 at 20:56
$begingroup$
Oups... I forgot the $cos nx$ is the RHS.
$endgroup$
– Tolaso
Jul 27 '15 at 20:57
1
$begingroup$
Yes, and then you must put $x=pi$.
$endgroup$
– Aretino
Jul 27 '15 at 20:58
1
$begingroup$
Similar work with $a:=iz$.
$endgroup$
– Raymond Manzoni
Jul 27 '15 at 21:07
|
show 2 more comments
$begingroup$
Hint: variable $x$ does not appear in the RHS of your result, as it should.
$endgroup$
– Aretino
Jul 27 '15 at 20:54
$begingroup$
Yes... but I cannot see where the mistake is. As for $x$ you are right. Then I plug in $x=pi$.
$endgroup$
– Tolaso
Jul 27 '15 at 20:56
$begingroup$
Oups... I forgot the $cos nx$ is the RHS.
$endgroup$
– Tolaso
Jul 27 '15 at 20:57
1
$begingroup$
Yes, and then you must put $x=pi$.
$endgroup$
– Aretino
Jul 27 '15 at 20:58
1
$begingroup$
Similar work with $a:=iz$.
$endgroup$
– Raymond Manzoni
Jul 27 '15 at 21:07
$begingroup$
Hint: variable $x$ does not appear in the RHS of your result, as it should.
$endgroup$
– Aretino
Jul 27 '15 at 20:54
$begingroup$
Hint: variable $x$ does not appear in the RHS of your result, as it should.
$endgroup$
– Aretino
Jul 27 '15 at 20:54
$begingroup$
Yes... but I cannot see where the mistake is. As for $x$ you are right. Then I plug in $x=pi$.
$endgroup$
– Tolaso
Jul 27 '15 at 20:56
$begingroup$
Yes... but I cannot see where the mistake is. As for $x$ you are right. Then I plug in $x=pi$.
$endgroup$
– Tolaso
Jul 27 '15 at 20:56
$begingroup$
Oups... I forgot the $cos nx$ is the RHS.
$endgroup$
– Tolaso
Jul 27 '15 at 20:57
$begingroup$
Oups... I forgot the $cos nx$ is the RHS.
$endgroup$
– Tolaso
Jul 27 '15 at 20:57
1
1
$begingroup$
Yes, and then you must put $x=pi$.
$endgroup$
– Aretino
Jul 27 '15 at 20:58
$begingroup$
Yes, and then you must put $x=pi$.
$endgroup$
– Aretino
Jul 27 '15 at 20:58
1
1
$begingroup$
Similar work with $a:=iz$.
$endgroup$
– Raymond Manzoni
Jul 27 '15 at 21:07
$begingroup$
Similar work with $a:=iz$.
$endgroup$
– Raymond Manzoni
Jul 27 '15 at 21:07
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Actually I forgot to add the $cos nx$ in the RHS. Hence:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n cos nx}{n^2 +a^2} overset{x=pi}{implies}$$
$$overset{x=pi}{implies} pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
and the identity is proven.
@Aretino, thanks for the heads up.
edited Jan 1 at 20:15
clathratus
3,521332
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
$endgroup$
add a comment |
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$begingroup$
Actually I forgot to add the $cos nx$ in the RHS. Hence:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n cos nx}{n^2 +a^2} overset{x=pi}{implies}$$
$$overset{x=pi}{implies} pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
and the identity is proven.
@Aretino, thanks for the heads up.
edited Jan 1 at 20:15
clathratus
3,521332
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
$endgroup$
add a comment |
$begingroup$
Actually I forgot to add the $cos nx$ in the RHS. Hence:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n cos nx}{n^2 +a^2} overset{x=pi}{implies}$$
$$overset{x=pi}{implies} pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
and the identity is proven.
@Aretino, thanks for the heads up.
edited Jan 1 at 20:15
clathratus
3,521332
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
$endgroup$
add a comment |
$begingroup$
Actually I forgot to add the $cos nx$ in the RHS. Hence:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n cos nx}{n^2 +a^2} overset{x=pi}{implies}$$
$$overset{x=pi}{implies} pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
and the identity is proven.
@Aretino, thanks for the heads up.
edited Jan 1 at 20:15
clathratus
3,521332
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
$endgroup$
Actually I forgot to add the $cos nx$ in the RHS. Hence:
$$cosh ax = frac{sinh pi a}{pi a}+ frac{2 asinh pi a}{pi}sum_{n=1}^{infty}frac{ (-1)^n cos nx}{n^2 +a^2} overset{x=pi}{implies}$$
$$overset{x=pi}{implies} pi coth pi a= frac{1}{a}+ sum_{n=1}^{infty}frac{2a}{n^2+a^2}$$
and the identity is proven.
@Aretino, thanks for the heads up.
edited Jan 1 at 20:15
clathratus
3,521332
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
edited Jan 1 at 20:15
clathratus
3,521332
edited Jan 1 at 20:15
clathratus
3,521332
edited Jan 1 at 20:15
clathratus
3,521332
3,521332
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
answered Jul 27 '15 at 21:05
TolasoTolaso
3,4101231
3,4101231
add a comment |
add a comment |
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$begingroup$
Hint: variable $x$ does not appear in the RHS of your result, as it should.
$endgroup$
– Aretino
Jul 27 '15 at 20:54
$begingroup$
Yes... but I cannot see where the mistake is. As for $x$ you are right. Then I plug in $x=pi$.
$endgroup$
– Tolaso
Jul 27 '15 at 20:56
$begingroup$
Oups... I forgot the $cos nx$ is the RHS.
$endgroup$
– Tolaso
Jul 27 '15 at 20:57
1
$begingroup$
Yes, and then you must put $x=pi$.
$endgroup$
– Aretino
Jul 27 '15 at 20:58
1
$begingroup$
Similar work with $a:=iz$.
$endgroup$
– Raymond Manzoni
Jul 27 '15 at 21:07