Mixing
Computing $sum_{n = 1}^{+ infty} frac{1}{n^2 + 1}$ using Fourier series.
$begingroup$
$f$ is a $2 pi$ periodic function on $ ] - pi,pi[$ defined as :
$$f(x) = e^{-x}$$
Using Fourier series, compute the sums :
$$ sum_{n = 1}^{+ infty} frac{1}{n^2 + 1} , sum_{- infty}^{+ infty} frac{1}{n^2 + 1} $$
How do I compute $sum_{n = 1}^{+ infty} frac{1}{n^2 + 1}$ ?
I have computed the fourier coefficient and found that:
$$a_0 = frac{2}{pi} (1 - e^{- pi}) $$
$$a_n = frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
$$b_n = frac{2n}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
Using Dirichlet theorem and taking $x = 0$ to make the $b_n$ disappear, I get:
$$S_f (1) = frac{2}{pi} (1 - e^{- pi}) + sum_{n = 1}^{+ infty} frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) . (-1)^n = 1 $$
I do not see how to proceed to get the value of the sum?
The same problem for the second sum.
real-analysis sequences-and-series fourier-analysis fourier-series
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
$endgroup$
|
show 6 more comments
$begingroup$
$f$ is a $2 pi$ periodic function on $ ] - pi,pi[$ defined as :
$$f(x) = e^{-x}$$
Using Fourier series, compute the sums :
$$ sum_{n = 1}^{+ infty} frac{1}{n^2 + 1} , sum_{- infty}^{+ infty} frac{1}{n^2 + 1} $$
How do I compute $sum_{n = 1}^{+ infty} frac{1}{n^2 + 1}$ ?
I have computed the fourier coefficient and found that:
$$a_0 = frac{2}{pi} (1 - e^{- pi}) $$
$$a_n = frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
$$b_n = frac{2n}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
Using Dirichlet theorem and taking $x = 0$ to make the $b_n$ disappear, I get:
$$S_f (1) = frac{2}{pi} (1 - e^{- pi}) + sum_{n = 1}^{+ infty} frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) . (-1)^n = 1 $$
I do not see how to proceed to get the value of the sum?
The same problem for the second sum.
real-analysis sequences-and-series fourier-analysis fourier-series
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
$endgroup$
1
$begingroup$
"same problem for the second sum" - not exactly; once you find the first sum, multiply it by $2$ and add $1$.
$endgroup$
– user170231
Jan 30 at 16:20
$begingroup$
@user170231 Thank you. I'm still stuck with the first sun though.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:21
1
$begingroup$
Your coefficients are not correct: $$a_0=frac1piint_{-pi}^pi e^{-x},mathrm dx=frac{e^pi-e^{-pi}}pi=frac{2sinhpi}pi$$ Similarly you should have ended up with $$a_n=frac{2(-1)^nsinhpi}{pi(n^2+1)}text{ and }b_n=frac{2n(-1)^nsinhpi}{pi(n^2+1)}$$ I think you are just missing a factor of $e^pi$ in certain places. (NB: $2sinhpi=e^pi-e^{-pi}$.)
$endgroup$
– user170231
Jan 30 at 17:04
$begingroup$
@user170231 To compute the coefficients, I used the formula: $$ a_n = frac{2}{pi} int_{0}^{pi} f(x) cos (nx)$$ and : $$ b_n = frac{2}{pi} int_{0}^{pi} f(x) sin (nx)$$ Isn't it correct?
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 17:18
1
$begingroup$
The formula I cited specifically works only for functions that are $2pi$-periodic over the interval $(-pi,pi)$. There is a formula for $P$-periodic functions over the interval $(x_0,x_0+P)$ listed on the page I linked under the section "A more general definition" (Eq. 4) for the more general case. It suggests$$a_n=frac2piint_0^pi f(x)cos nx,mathrm dx$$works for a function $f(x)$ that is $pi$-periodic over $(0,pi)$.
$endgroup$
– user170231
Jan 30 at 18:33
|
show 6 more comments
$begingroup$
$f$ is a $2 pi$ periodic function on $ ] - pi,pi[$ defined as :
$$f(x) = e^{-x}$$
Using Fourier series, compute the sums :
$$ sum_{n = 1}^{+ infty} frac{1}{n^2 + 1} , sum_{- infty}^{+ infty} frac{1}{n^2 + 1} $$
How do I compute $sum_{n = 1}^{+ infty} frac{1}{n^2 + 1}$ ?
I have computed the fourier coefficient and found that:
$$a_0 = frac{2}{pi} (1 - e^{- pi}) $$
$$a_n = frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
$$b_n = frac{2n}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
Using Dirichlet theorem and taking $x = 0$ to make the $b_n$ disappear, I get:
$$S_f (1) = frac{2}{pi} (1 - e^{- pi}) + sum_{n = 1}^{+ infty} frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) . (-1)^n = 1 $$
I do not see how to proceed to get the value of the sum?
The same problem for the second sum.
real-analysis sequences-and-series fourier-analysis fourier-series
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
$endgroup$
$f$ is a $2 pi$ periodic function on $ ] - pi,pi[$ defined as :
$$f(x) = e^{-x}$$
Using Fourier series, compute the sums :
$$ sum_{n = 1}^{+ infty} frac{1}{n^2 + 1} , sum_{- infty}^{+ infty} frac{1}{n^2 + 1} $$
How do I compute $sum_{n = 1}^{+ infty} frac{1}{n^2 + 1}$ ?
I have computed the fourier coefficient and found that:
$$a_0 = frac{2}{pi} (1 - e^{- pi}) $$
$$a_n = frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
$$b_n = frac{2n}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) $$
Using Dirichlet theorem and taking $x = 0$ to make the $b_n$ disappear, I get:
$$S_f (1) = frac{2}{pi} (1 - e^{- pi}) + sum_{n = 1}^{+ infty} frac{2}{pi(n^2 + 1)} (1 - (-1)^n . e^{- pi} ) . (-1)^n = 1 $$
I do not see how to proceed to get the value of the sum?
The same problem for the second sum.
real-analysis sequences-and-series fourier-analysis fourier-series
real-analysis sequences-and-series fourier-analysis fourier-series
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
asked Jan 30 at 15:45
Zouhair El YaagoubiZouhair El Yaagoubi
538411
538411
1
$begingroup$
"same problem for the second sum" - not exactly; once you find the first sum, multiply it by $2$ and add $1$.
$endgroup$
– user170231
Jan 30 at 16:20
$begingroup$
@user170231 Thank you. I'm still stuck with the first sun though.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:21
1
$begingroup$
Your coefficients are not correct: $$a_0=frac1piint_{-pi}^pi e^{-x},mathrm dx=frac{e^pi-e^{-pi}}pi=frac{2sinhpi}pi$$ Similarly you should have ended up with $$a_n=frac{2(-1)^nsinhpi}{pi(n^2+1)}text{ and }b_n=frac{2n(-1)^nsinhpi}{pi(n^2+1)}$$ I think you are just missing a factor of $e^pi$ in certain places. (NB: $2sinhpi=e^pi-e^{-pi}$.)
$endgroup$
– user170231
Jan 30 at 17:04
$begingroup$
@user170231 To compute the coefficients, I used the formula: $$ a_n = frac{2}{pi} int_{0}^{pi} f(x) cos (nx)$$ and : $$ b_n = frac{2}{pi} int_{0}^{pi} f(x) sin (nx)$$ Isn't it correct?
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 17:18
1
$begingroup$
The formula I cited specifically works only for functions that are $2pi$-periodic over the interval $(-pi,pi)$. There is a formula for $P$-periodic functions over the interval $(x_0,x_0+P)$ listed on the page I linked under the section "A more general definition" (Eq. 4) for the more general case. It suggests$$a_n=frac2piint_0^pi f(x)cos nx,mathrm dx$$works for a function $f(x)$ that is $pi$-periodic over $(0,pi)$.
$endgroup$
– user170231
Jan 30 at 18:33
|
show 6 more comments
1
$begingroup$
"same problem for the second sum" - not exactly; once you find the first sum, multiply it by $2$ and add $1$.
$endgroup$
– user170231
Jan 30 at 16:20
$begingroup$
@user170231 Thank you. I'm still stuck with the first sun though.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:21
1
$begingroup$
Your coefficients are not correct: $$a_0=frac1piint_{-pi}^pi e^{-x},mathrm dx=frac{e^pi-e^{-pi}}pi=frac{2sinhpi}pi$$ Similarly you should have ended up with $$a_n=frac{2(-1)^nsinhpi}{pi(n^2+1)}text{ and }b_n=frac{2n(-1)^nsinhpi}{pi(n^2+1)}$$ I think you are just missing a factor of $e^pi$ in certain places. (NB: $2sinhpi=e^pi-e^{-pi}$.)
$endgroup$
– user170231
Jan 30 at 17:04
$begingroup$
@user170231 To compute the coefficients, I used the formula: $$ a_n = frac{2}{pi} int_{0}^{pi} f(x) cos (nx)$$ and : $$ b_n = frac{2}{pi} int_{0}^{pi} f(x) sin (nx)$$ Isn't it correct?
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 17:18
1
$begingroup$
The formula I cited specifically works only for functions that are $2pi$-periodic over the interval $(-pi,pi)$. There is a formula for $P$-periodic functions over the interval $(x_0,x_0+P)$ listed on the page I linked under the section "A more general definition" (Eq. 4) for the more general case. It suggests$$a_n=frac2piint_0^pi f(x)cos nx,mathrm dx$$works for a function $f(x)$ that is $pi$-periodic over $(0,pi)$.
$endgroup$
– user170231
Jan 30 at 18:33
1
1
$begingroup$
"same problem for the second sum" - not exactly; once you find the first sum, multiply it by $2$ and add $1$.
$endgroup$
– user170231
Jan 30 at 16:20
$begingroup$
"same problem for the second sum" - not exactly; once you find the first sum, multiply it by $2$ and add $1$.
$endgroup$
– user170231
Jan 30 at 16:20
$begingroup$
@user170231 Thank you. I'm still stuck with the first sun though.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:21
$begingroup$
@user170231 Thank you. I'm still stuck with the first sun though.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:21
1
1
$begingroup$
Your coefficients are not correct: $$a_0=frac1piint_{-pi}^pi e^{-x},mathrm dx=frac{e^pi-e^{-pi}}pi=frac{2sinhpi}pi$$ Similarly you should have ended up with $$a_n=frac{2(-1)^nsinhpi}{pi(n^2+1)}text{ and }b_n=frac{2n(-1)^nsinhpi}{pi(n^2+1)}$$ I think you are just missing a factor of $e^pi$ in certain places. (NB: $2sinhpi=e^pi-e^{-pi}$.)
$endgroup$
– user170231
Jan 30 at 17:04
$begingroup$
Your coefficients are not correct: $$a_0=frac1piint_{-pi}^pi e^{-x},mathrm dx=frac{e^pi-e^{-pi}}pi=frac{2sinhpi}pi$$ Similarly you should have ended up with $$a_n=frac{2(-1)^nsinhpi}{pi(n^2+1)}text{ and }b_n=frac{2n(-1)^nsinhpi}{pi(n^2+1)}$$ I think you are just missing a factor of $e^pi$ in certain places. (NB: $2sinhpi=e^pi-e^{-pi}$.)
$endgroup$
– user170231
Jan 30 at 17:04
$begingroup$
@user170231 To compute the coefficients, I used the formula: $$ a_n = frac{2}{pi} int_{0}^{pi} f(x) cos (nx)$$ and : $$ b_n = frac{2}{pi} int_{0}^{pi} f(x) sin (nx)$$ Isn't it correct?
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 17:18
$begingroup$
@user170231 To compute the coefficients, I used the formula: $$ a_n = frac{2}{pi} int_{0}^{pi} f(x) cos (nx)$$ and : $$ b_n = frac{2}{pi} int_{0}^{pi} f(x) sin (nx)$$ Isn't it correct?
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 17:18
1
1
$begingroup$
The formula I cited specifically works only for functions that are $2pi$-periodic over the interval $(-pi,pi)$. There is a formula for $P$-periodic functions over the interval $(x_0,x_0+P)$ listed on the page I linked under the section "A more general definition" (Eq. 4) for the more general case. It suggests$$a_n=frac2piint_0^pi f(x)cos nx,mathrm dx$$works for a function $f(x)$ that is $pi$-periodic over $(0,pi)$.
$endgroup$
– user170231
Jan 30 at 18:33
$begingroup$
The formula I cited specifically works only for functions that are $2pi$-periodic over the interval $(-pi,pi)$. There is a formula for $P$-periodic functions over the interval $(x_0,x_0+P)$ listed on the page I linked under the section "A more general definition" (Eq. 4) for the more general case. It suggests$$a_n=frac2piint_0^pi f(x)cos nx,mathrm dx$$works for a function $f(x)$ that is $pi$-periodic over $(0,pi)$.
$endgroup$
– user170231
Jan 30 at 18:33
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
With the correct coefficients (see comments), you have for $x in]-pi,pi[$:
$$
e^{-x} = frac{2}{pi} sinh(pi) + 2 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n (nsin (nx) + cos(nx)) }{pi(n^2 + 1)}
$$
so
$$
e^{x} + e^{-x} = frac{4}{pi} sinh(pi) + 4 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n cos(nx) }{pi(n^2 + 1)}
$$
Now we can evaluate this for $x to pmpi$. To be more precise (see the answer by @kvantour), the evaluation $x to pmpi$ corresponds to the average of the two values at $pmpi$ which is exactly what the Fourier series converges to at this point of discontinuity.
$$
cosh (pm pi) = frac{2}{pi} sinh(pi)+ 2 sinh(pi) sum_{n=1}^{infty}frac{1 }{pi(n^2 + 1)}
$$
and in turn
$$
sum_{n=1}^{infty}frac{1 }{(n^2 + 1)} = frac{pi coth(pi) -1}{2 }
$$
Then the second question follows:
$$
sum_{- infty}^{+ infty} frac{1}{n^2 + 1} = 1 + 2 sum_{1}^{+ infty} frac{1}{n^2 + 1} = pi coth(pi) simeq 1.0037 ; pi
$$
Comment: for comparison,
$$
int_{- infty}^{+ infty} frac{1}{x^2 + 1} {rm{dx}} = pi
$$
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
$endgroup$
1
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
1
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
1
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
add a comment |
$begingroup$
When you state Using Fourier series, it does not necessarily mean that you have to use the Fourier Cosine and Sine series. You can also write it in exponential form:
Any periodic function $tilde{f}(t)$ with period $2pi$ can be written
as:
$$ f(t) = sum_{n=-infty}^infty c_n,e^{int} $$ with $$
c_n=frac{1}{2pi}int_{-pi}^pi tilde f(t),e^{-int},textrm{d}t
$$
note: we make a distinction between the fourier series $f(t)$ and the periodic function $tilde f(t)$ to indicate that they are different when $tilde f(t)$ is discontiniuous.
When you apply this for $tilde f(t)$, which is periodic over $2pi$, discontinuous and defined in the region $]-pi,pi[$ as $exp(-t)$, you obtain
$$
c_n = frac{sinh(pi)}picdotfrac{(-1)^n,(1-in)}{1+n^2}
$$
giving you:
$$
f(t)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(-1)^n,(1-in)}{1+n^2},e^{int}
$$
You already notice the familiar part in the sum which is of interest, the problem is the $(-1)^n$. This we can get rid of with the proper choice of $t$. If $t=pmpi$ then $exp(pm inpi)=(-1)^n$. But be advised $f(t)$ is periodic with a period of $2pi$ and is not continuous in the points $t=npi$. The value the Fourier Series will return is:
$$ f(pi)=f(-pi)= frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(1-in)}{1+n^2} = frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
The latter reduction of the sum is straightforward as the imaginary part is zero.
But we cannot use this, as we do not know what $f(pi)$ is since $tilde f(t)$ is discontinuous at these points. Luckily, some smart people solved this conundrum and showed that at a discontinuity, the Fourier series converges to the average of the two values, i.e.
$$ f(pi)=f(-pi)=frac{lim_{trightarrowpi^-}tilde f(t) + lim_{trightarrow pi^+}tilde f(t)}{2} = frac{exp(-pi)+exp(pi)}{2} = cosh(pi)$$
See: Fourier series at discontinuities
So in the end, we have the solution:
$$f(pi)=cosh(pi)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
or
$$bbox[5px,border:2px solid #00A000]{pi coth(pi)=sum_{n=-infty}^inftyfrac{1}{1+n^2}=1+2sum_{n=1}^inftyfrac{1}{1+n^2}}$$
remark: computing $f(0)$ gives directly, without any fuss:
$$bbox[5px,border:2px solid #000000]{frac{pi}{sinh(pi)}=sum_{n=-infty}^inftyfrac{(-1)^n}{1+n^2}}$$
answered Jan 31 at 11:58
kvantourkvantour
36829
$endgroup$
1
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
add a comment |
$begingroup$
Try using Parseval's theorem,
$$frac{1}{pi}int_{-pi}^pi |f(x)|^2 dx = frac{a_0}{2} + sum_{n=1}^{infty} a_n^2+b_n^2$$
If it comes out too nasty, then I think your coeffecients might be off.
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
$endgroup$
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
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$begingroup$
With the correct coefficients (see comments), you have for $x in]-pi,pi[$:
$$
e^{-x} = frac{2}{pi} sinh(pi) + 2 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n (nsin (nx) + cos(nx)) }{pi(n^2 + 1)}
$$
so
$$
e^{x} + e^{-x} = frac{4}{pi} sinh(pi) + 4 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n cos(nx) }{pi(n^2 + 1)}
$$
Now we can evaluate this for $x to pmpi$. To be more precise (see the answer by @kvantour), the evaluation $x to pmpi$ corresponds to the average of the two values at $pmpi$ which is exactly what the Fourier series converges to at this point of discontinuity.
$$
cosh (pm pi) = frac{2}{pi} sinh(pi)+ 2 sinh(pi) sum_{n=1}^{infty}frac{1 }{pi(n^2 + 1)}
$$
and in turn
$$
sum_{n=1}^{infty}frac{1 }{(n^2 + 1)} = frac{pi coth(pi) -1}{2 }
$$
Then the second question follows:
$$
sum_{- infty}^{+ infty} frac{1}{n^2 + 1} = 1 + 2 sum_{1}^{+ infty} frac{1}{n^2 + 1} = pi coth(pi) simeq 1.0037 ; pi
$$
Comment: for comparison,
$$
int_{- infty}^{+ infty} frac{1}{x^2 + 1} {rm{dx}} = pi
$$
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
$endgroup$
1
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
1
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
1
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
add a comment |
$begingroup$
With the correct coefficients (see comments), you have for $x in]-pi,pi[$:
$$
e^{-x} = frac{2}{pi} sinh(pi) + 2 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n (nsin (nx) + cos(nx)) }{pi(n^2 + 1)}
$$
so
$$
e^{x} + e^{-x} = frac{4}{pi} sinh(pi) + 4 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n cos(nx) }{pi(n^2 + 1)}
$$
Now we can evaluate this for $x to pmpi$. To be more precise (see the answer by @kvantour), the evaluation $x to pmpi$ corresponds to the average of the two values at $pmpi$ which is exactly what the Fourier series converges to at this point of discontinuity.
$$
cosh (pm pi) = frac{2}{pi} sinh(pi)+ 2 sinh(pi) sum_{n=1}^{infty}frac{1 }{pi(n^2 + 1)}
$$
and in turn
$$
sum_{n=1}^{infty}frac{1 }{(n^2 + 1)} = frac{pi coth(pi) -1}{2 }
$$
Then the second question follows:
$$
sum_{- infty}^{+ infty} frac{1}{n^2 + 1} = 1 + 2 sum_{1}^{+ infty} frac{1}{n^2 + 1} = pi coth(pi) simeq 1.0037 ; pi
$$
Comment: for comparison,
$$
int_{- infty}^{+ infty} frac{1}{x^2 + 1} {rm{dx}} = pi
$$
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
$endgroup$
1
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
1
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
1
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
add a comment |
$begingroup$
With the correct coefficients (see comments), you have for $x in]-pi,pi[$:
$$
e^{-x} = frac{2}{pi} sinh(pi) + 2 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n (nsin (nx) + cos(nx)) }{pi(n^2 + 1)}
$$
so
$$
e^{x} + e^{-x} = frac{4}{pi} sinh(pi) + 4 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n cos(nx) }{pi(n^2 + 1)}
$$
Now we can evaluate this for $x to pmpi$. To be more precise (see the answer by @kvantour), the evaluation $x to pmpi$ corresponds to the average of the two values at $pmpi$ which is exactly what the Fourier series converges to at this point of discontinuity.
$$
cosh (pm pi) = frac{2}{pi} sinh(pi)+ 2 sinh(pi) sum_{n=1}^{infty}frac{1 }{pi(n^2 + 1)}
$$
and in turn
$$
sum_{n=1}^{infty}frac{1 }{(n^2 + 1)} = frac{pi coth(pi) -1}{2 }
$$
Then the second question follows:
$$
sum_{- infty}^{+ infty} frac{1}{n^2 + 1} = 1 + 2 sum_{1}^{+ infty} frac{1}{n^2 + 1} = pi coth(pi) simeq 1.0037 ; pi
$$
Comment: for comparison,
$$
int_{- infty}^{+ infty} frac{1}{x^2 + 1} {rm{dx}} = pi
$$
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
$endgroup$
With the correct coefficients (see comments), you have for $x in]-pi,pi[$:
$$
e^{-x} = frac{2}{pi} sinh(pi) + 2 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n (nsin (nx) + cos(nx)) }{pi(n^2 + 1)}
$$
so
$$
e^{x} + e^{-x} = frac{4}{pi} sinh(pi) + 4 sinh(pi) sum_{n=1}^{infty}frac{ (-1)^n cos(nx) }{pi(n^2 + 1)}
$$
Now we can evaluate this for $x to pmpi$. To be more precise (see the answer by @kvantour), the evaluation $x to pmpi$ corresponds to the average of the two values at $pmpi$ which is exactly what the Fourier series converges to at this point of discontinuity.
$$
cosh (pm pi) = frac{2}{pi} sinh(pi)+ 2 sinh(pi) sum_{n=1}^{infty}frac{1 }{pi(n^2 + 1)}
$$
and in turn
$$
sum_{n=1}^{infty}frac{1 }{(n^2 + 1)} = frac{pi coth(pi) -1}{2 }
$$
Then the second question follows:
$$
sum_{- infty}^{+ infty} frac{1}{n^2 + 1} = 1 + 2 sum_{1}^{+ infty} frac{1}{n^2 + 1} = pi coth(pi) simeq 1.0037 ; pi
$$
Comment: for comparison,
$$
int_{- infty}^{+ infty} frac{1}{x^2 + 1} {rm{dx}} = pi
$$
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
edited Jan 31 at 15:25
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
answered Jan 30 at 17:48
AndreasAndreas
8,4161137
8,4161137
1
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
1
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
1
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
add a comment |
1
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
1
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
1
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
1
1
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
$begingroup$
You cannot make the statement $xrightarrowpi$ and just know its value. $f(x)$ is discontinuous in that point.
$endgroup$
– kvantour
Jan 31 at 11:52
1
1
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
$begingroup$
@kvantour Thank you. I have adressed that point and edited my answer accordingly.
$endgroup$
– Andreas
Jan 31 at 15:13
1
1
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
$begingroup$
Good point. done.
$endgroup$
– Andreas
Jan 31 at 15:25
add a comment |
$begingroup$
When you state Using Fourier series, it does not necessarily mean that you have to use the Fourier Cosine and Sine series. You can also write it in exponential form:
Any periodic function $tilde{f}(t)$ with period $2pi$ can be written
as:
$$ f(t) = sum_{n=-infty}^infty c_n,e^{int} $$ with $$
c_n=frac{1}{2pi}int_{-pi}^pi tilde f(t),e^{-int},textrm{d}t
$$
note: we make a distinction between the fourier series $f(t)$ and the periodic function $tilde f(t)$ to indicate that they are different when $tilde f(t)$ is discontiniuous.
When you apply this for $tilde f(t)$, which is periodic over $2pi$, discontinuous and defined in the region $]-pi,pi[$ as $exp(-t)$, you obtain
$$
c_n = frac{sinh(pi)}picdotfrac{(-1)^n,(1-in)}{1+n^2}
$$
giving you:
$$
f(t)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(-1)^n,(1-in)}{1+n^2},e^{int}
$$
You already notice the familiar part in the sum which is of interest, the problem is the $(-1)^n$. This we can get rid of with the proper choice of $t$. If $t=pmpi$ then $exp(pm inpi)=(-1)^n$. But be advised $f(t)$ is periodic with a period of $2pi$ and is not continuous in the points $t=npi$. The value the Fourier Series will return is:
$$ f(pi)=f(-pi)= frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(1-in)}{1+n^2} = frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
The latter reduction of the sum is straightforward as the imaginary part is zero.
But we cannot use this, as we do not know what $f(pi)$ is since $tilde f(t)$ is discontinuous at these points. Luckily, some smart people solved this conundrum and showed that at a discontinuity, the Fourier series converges to the average of the two values, i.e.
$$ f(pi)=f(-pi)=frac{lim_{trightarrowpi^-}tilde f(t) + lim_{trightarrow pi^+}tilde f(t)}{2} = frac{exp(-pi)+exp(pi)}{2} = cosh(pi)$$
See: Fourier series at discontinuities
So in the end, we have the solution:
$$f(pi)=cosh(pi)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
or
$$bbox[5px,border:2px solid #00A000]{pi coth(pi)=sum_{n=-infty}^inftyfrac{1}{1+n^2}=1+2sum_{n=1}^inftyfrac{1}{1+n^2}}$$
remark: computing $f(0)$ gives directly, without any fuss:
$$bbox[5px,border:2px solid #000000]{frac{pi}{sinh(pi)}=sum_{n=-infty}^inftyfrac{(-1)^n}{1+n^2}}$$
answered Jan 31 at 11:58
kvantourkvantour
36829
$endgroup$
1
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
add a comment |
$begingroup$
When you state Using Fourier series, it does not necessarily mean that you have to use the Fourier Cosine and Sine series. You can also write it in exponential form:
Any periodic function $tilde{f}(t)$ with period $2pi$ can be written
as:
$$ f(t) = sum_{n=-infty}^infty c_n,e^{int} $$ with $$
c_n=frac{1}{2pi}int_{-pi}^pi tilde f(t),e^{-int},textrm{d}t
$$
note: we make a distinction between the fourier series $f(t)$ and the periodic function $tilde f(t)$ to indicate that they are different when $tilde f(t)$ is discontiniuous.
When you apply this for $tilde f(t)$, which is periodic over $2pi$, discontinuous and defined in the region $]-pi,pi[$ as $exp(-t)$, you obtain
$$
c_n = frac{sinh(pi)}picdotfrac{(-1)^n,(1-in)}{1+n^2}
$$
giving you:
$$
f(t)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(-1)^n,(1-in)}{1+n^2},e^{int}
$$
You already notice the familiar part in the sum which is of interest, the problem is the $(-1)^n$. This we can get rid of with the proper choice of $t$. If $t=pmpi$ then $exp(pm inpi)=(-1)^n$. But be advised $f(t)$ is periodic with a period of $2pi$ and is not continuous in the points $t=npi$. The value the Fourier Series will return is:
$$ f(pi)=f(-pi)= frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(1-in)}{1+n^2} = frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
The latter reduction of the sum is straightforward as the imaginary part is zero.
But we cannot use this, as we do not know what $f(pi)$ is since $tilde f(t)$ is discontinuous at these points. Luckily, some smart people solved this conundrum and showed that at a discontinuity, the Fourier series converges to the average of the two values, i.e.
$$ f(pi)=f(-pi)=frac{lim_{trightarrowpi^-}tilde f(t) + lim_{trightarrow pi^+}tilde f(t)}{2} = frac{exp(-pi)+exp(pi)}{2} = cosh(pi)$$
See: Fourier series at discontinuities
So in the end, we have the solution:
$$f(pi)=cosh(pi)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
or
$$bbox[5px,border:2px solid #00A000]{pi coth(pi)=sum_{n=-infty}^inftyfrac{1}{1+n^2}=1+2sum_{n=1}^inftyfrac{1}{1+n^2}}$$
remark: computing $f(0)$ gives directly, without any fuss:
$$bbox[5px,border:2px solid #000000]{frac{pi}{sinh(pi)}=sum_{n=-infty}^inftyfrac{(-1)^n}{1+n^2}}$$
answered Jan 31 at 11:58
kvantourkvantour
36829
$endgroup$
1
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
add a comment |
$begingroup$
When you state Using Fourier series, it does not necessarily mean that you have to use the Fourier Cosine and Sine series. You can also write it in exponential form:
Any periodic function $tilde{f}(t)$ with period $2pi$ can be written
as:
$$ f(t) = sum_{n=-infty}^infty c_n,e^{int} $$ with $$
c_n=frac{1}{2pi}int_{-pi}^pi tilde f(t),e^{-int},textrm{d}t
$$
note: we make a distinction between the fourier series $f(t)$ and the periodic function $tilde f(t)$ to indicate that they are different when $tilde f(t)$ is discontiniuous.
When you apply this for $tilde f(t)$, which is periodic over $2pi$, discontinuous and defined in the region $]-pi,pi[$ as $exp(-t)$, you obtain
$$
c_n = frac{sinh(pi)}picdotfrac{(-1)^n,(1-in)}{1+n^2}
$$
giving you:
$$
f(t)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(-1)^n,(1-in)}{1+n^2},e^{int}
$$
You already notice the familiar part in the sum which is of interest, the problem is the $(-1)^n$. This we can get rid of with the proper choice of $t$. If $t=pmpi$ then $exp(pm inpi)=(-1)^n$. But be advised $f(t)$ is periodic with a period of $2pi$ and is not continuous in the points $t=npi$. The value the Fourier Series will return is:
$$ f(pi)=f(-pi)= frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(1-in)}{1+n^2} = frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
The latter reduction of the sum is straightforward as the imaginary part is zero.
But we cannot use this, as we do not know what $f(pi)$ is since $tilde f(t)$ is discontinuous at these points. Luckily, some smart people solved this conundrum and showed that at a discontinuity, the Fourier series converges to the average of the two values, i.e.
$$ f(pi)=f(-pi)=frac{lim_{trightarrowpi^-}tilde f(t) + lim_{trightarrow pi^+}tilde f(t)}{2} = frac{exp(-pi)+exp(pi)}{2} = cosh(pi)$$
See: Fourier series at discontinuities
So in the end, we have the solution:
$$f(pi)=cosh(pi)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
or
$$bbox[5px,border:2px solid #00A000]{pi coth(pi)=sum_{n=-infty}^inftyfrac{1}{1+n^2}=1+2sum_{n=1}^inftyfrac{1}{1+n^2}}$$
remark: computing $f(0)$ gives directly, without any fuss:
$$bbox[5px,border:2px solid #000000]{frac{pi}{sinh(pi)}=sum_{n=-infty}^inftyfrac{(-1)^n}{1+n^2}}$$
answered Jan 31 at 11:58
kvantourkvantour
36829
$endgroup$
When you state Using Fourier series, it does not necessarily mean that you have to use the Fourier Cosine and Sine series. You can also write it in exponential form:
Any periodic function $tilde{f}(t)$ with period $2pi$ can be written
as:
$$ f(t) = sum_{n=-infty}^infty c_n,e^{int} $$ with $$
c_n=frac{1}{2pi}int_{-pi}^pi tilde f(t),e^{-int},textrm{d}t
$$
note: we make a distinction between the fourier series $f(t)$ and the periodic function $tilde f(t)$ to indicate that they are different when $tilde f(t)$ is discontiniuous.
When you apply this for $tilde f(t)$, which is periodic over $2pi$, discontinuous and defined in the region $]-pi,pi[$ as $exp(-t)$, you obtain
$$
c_n = frac{sinh(pi)}picdotfrac{(-1)^n,(1-in)}{1+n^2}
$$
giving you:
$$
f(t)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(-1)^n,(1-in)}{1+n^2},e^{int}
$$
You already notice the familiar part in the sum which is of interest, the problem is the $(-1)^n$. This we can get rid of with the proper choice of $t$. If $t=pmpi$ then $exp(pm inpi)=(-1)^n$. But be advised $f(t)$ is periodic with a period of $2pi$ and is not continuous in the points $t=npi$. The value the Fourier Series will return is:
$$ f(pi)=f(-pi)= frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{(1-in)}{1+n^2} = frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
The latter reduction of the sum is straightforward as the imaginary part is zero.
But we cannot use this, as we do not know what $f(pi)$ is since $tilde f(t)$ is discontinuous at these points. Luckily, some smart people solved this conundrum and showed that at a discontinuity, the Fourier series converges to the average of the two values, i.e.
$$ f(pi)=f(-pi)=frac{lim_{trightarrowpi^-}tilde f(t) + lim_{trightarrow pi^+}tilde f(t)}{2} = frac{exp(-pi)+exp(pi)}{2} = cosh(pi)$$
See: Fourier series at discontinuities
So in the end, we have the solution:
$$f(pi)=cosh(pi)=frac{sinh(pi)}pi sum_{n=-infty}^inftyfrac{1}{1+n^2}$$
or
$$bbox[5px,border:2px solid #00A000]{pi coth(pi)=sum_{n=-infty}^inftyfrac{1}{1+n^2}=1+2sum_{n=1}^inftyfrac{1}{1+n^2}}$$
remark: computing $f(0)$ gives directly, without any fuss:
$$bbox[5px,border:2px solid #000000]{frac{pi}{sinh(pi)}=sum_{n=-infty}^inftyfrac{(-1)^n}{1+n^2}}$$
answered Jan 31 at 11:58
kvantourkvantour
36829
edited Jan 31 at 15:20
answered Jan 31 at 11:58
kvantourkvantour
36829
answered Jan 31 at 11:58
kvantourkvantour
36829
answered Jan 31 at 11:58
kvantourkvantour
36829
36829
1
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
add a comment |
1
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
1
1
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
$begingroup$
I computed the sum using $f(0)$, but I will pay more attention to the points of discontinuity, so thank you so much for your additional important information.
$endgroup$
– Zouhair El Yaagoubi
Jan 31 at 12:11
add a comment |
$begingroup$
Try using Parseval's theorem,
$$frac{1}{pi}int_{-pi}^pi |f(x)|^2 dx = frac{a_0}{2} + sum_{n=1}^{infty} a_n^2+b_n^2$$
If it comes out too nasty, then I think your coeffecients might be off.
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
$endgroup$
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
add a comment |
$begingroup$
Try using Parseval's theorem,
$$frac{1}{pi}int_{-pi}^pi |f(x)|^2 dx = frac{a_0}{2} + sum_{n=1}^{infty} a_n^2+b_n^2$$
If it comes out too nasty, then I think your coeffecients might be off.
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
$endgroup$
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
add a comment |
$begingroup$
Try using Parseval's theorem,
$$frac{1}{pi}int_{-pi}^pi |f(x)|^2 dx = frac{a_0}{2} + sum_{n=1}^{infty} a_n^2+b_n^2$$
If it comes out too nasty, then I think your coeffecients might be off.
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
$endgroup$
Try using Parseval's theorem,
$$frac{1}{pi}int_{-pi}^pi |f(x)|^2 dx = frac{a_0}{2} + sum_{n=1}^{infty} a_n^2+b_n^2$$
If it comes out too nasty, then I think your coeffecients might be off.
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
answered Jan 30 at 16:08
Nicholas ParrisNicholas Parris
1563
1563
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
add a comment |
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
$begingroup$
I do not see how Parseval's theorem will give me the sum, can you elaborate more? Although, I have checked the coefficient well. They seem to be correct.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:12
add a comment |
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$begingroup$
"same problem for the second sum" - not exactly; once you find the first sum, multiply it by $2$ and add $1$.
$endgroup$
– user170231
Jan 30 at 16:20
$begingroup$
@user170231 Thank you. I'm still stuck with the first sun though.
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 16:21
1
$begingroup$
Your coefficients are not correct: $$a_0=frac1piint_{-pi}^pi e^{-x},mathrm dx=frac{e^pi-e^{-pi}}pi=frac{2sinhpi}pi$$ Similarly you should have ended up with $$a_n=frac{2(-1)^nsinhpi}{pi(n^2+1)}text{ and }b_n=frac{2n(-1)^nsinhpi}{pi(n^2+1)}$$ I think you are just missing a factor of $e^pi$ in certain places. (NB: $2sinhpi=e^pi-e^{-pi}$.)
$endgroup$
– user170231
Jan 30 at 17:04
$begingroup$
@user170231 To compute the coefficients, I used the formula: $$ a_n = frac{2}{pi} int_{0}^{pi} f(x) cos (nx)$$ and : $$ b_n = frac{2}{pi} int_{0}^{pi} f(x) sin (nx)$$ Isn't it correct?
$endgroup$
– Zouhair El Yaagoubi
Jan 30 at 17:18
1
$begingroup$
The formula I cited specifically works only for functions that are $2pi$-periodic over the interval $(-pi,pi)$. There is a formula for $P$-periodic functions over the interval $(x_0,x_0+P)$ listed on the page I linked under the section "A more general definition" (Eq. 4) for the more general case. It suggests$$a_n=frac2piint_0^pi f(x)cos nx,mathrm dx$$works for a function $f(x)$ that is $pi$-periodic over $(0,pi)$.
$endgroup$
– user170231
Jan 30 at 18:33