Mixing
Calculating the vlaue of a sum using a fourier series
$begingroup$
For
$$f(t)=t,quad tin[0,2pi) tag{1}$$
we find the fourier series
$$hat{hat{f}}(x)=pi -2 sum_{k=1}^infty frac{sin(kx)}{k}tag{2}$$
I want to calculate the value of
$$sum_{n=1}^inftyfrac{(-1)^n}{2n-1}tag{3}$$
using (2).
Now for $x=pi/2$ we get
$$hat{hat{f}}(pi/2)=pi -2 sum_{k=1}^infty frac{sin(frac{pi}{2}k)}{k}$=pi -2 sum_{k=1}^infty frac{(-1)^k}{k} tag{3}$$
So that looks a bit like (3). I should now be able to somehow reform the sum or "choose" k s.t. I actualyl get (3). But I just don't see it.
fourier-series
asked Jan 31 at 13:43
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
For
$$f(t)=t,quad tin[0,2pi) tag{1}$$
we find the fourier series
$$hat{hat{f}}(x)=pi -2 sum_{k=1}^infty frac{sin(kx)}{k}tag{2}$$
I want to calculate the value of
$$sum_{n=1}^inftyfrac{(-1)^n}{2n-1}tag{3}$$
using (2).
Now for $x=pi/2$ we get
$$hat{hat{f}}(pi/2)=pi -2 sum_{k=1}^infty frac{sin(frac{pi}{2}k)}{k}$=pi -2 sum_{k=1}^infty frac{(-1)^k}{k} tag{3}$$
So that looks a bit like (3). I should now be able to somehow reform the sum or "choose" k s.t. I actualyl get (3). But I just don't see it.
fourier-series
asked Jan 31 at 13:43
xotixxotix
291411
$endgroup$
2
$begingroup$
in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $sin(pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$.
$endgroup$
– Hayk
Jan 31 at 14:40
$begingroup$
Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now.
$endgroup$
– xotix
Jan 31 at 15:13
add a comment |
$begingroup$
For
$$f(t)=t,quad tin[0,2pi) tag{1}$$
we find the fourier series
$$hat{hat{f}}(x)=pi -2 sum_{k=1}^infty frac{sin(kx)}{k}tag{2}$$
I want to calculate the value of
$$sum_{n=1}^inftyfrac{(-1)^n}{2n-1}tag{3}$$
using (2).
Now for $x=pi/2$ we get
$$hat{hat{f}}(pi/2)=pi -2 sum_{k=1}^infty frac{sin(frac{pi}{2}k)}{k}$=pi -2 sum_{k=1}^infty frac{(-1)^k}{k} tag{3}$$
So that looks a bit like (3). I should now be able to somehow reform the sum or "choose" k s.t. I actualyl get (3). But I just don't see it.
fourier-series
asked Jan 31 at 13:43
xotixxotix
291411
$endgroup$
For
$$f(t)=t,quad tin[0,2pi) tag{1}$$
we find the fourier series
$$hat{hat{f}}(x)=pi -2 sum_{k=1}^infty frac{sin(kx)}{k}tag{2}$$
I want to calculate the value of
$$sum_{n=1}^inftyfrac{(-1)^n}{2n-1}tag{3}$$
using (2).
Now for $x=pi/2$ we get
$$hat{hat{f}}(pi/2)=pi -2 sum_{k=1}^infty frac{sin(frac{pi}{2}k)}{k}$=pi -2 sum_{k=1}^infty frac{(-1)^k}{k} tag{3}$$
So that looks a bit like (3). I should now be able to somehow reform the sum or "choose" k s.t. I actualyl get (3). But I just don't see it.
fourier-series
fourier-series
asked Jan 31 at 13:43
xotixxotix
291411
asked Jan 31 at 13:43
xotixxotix
291411
asked Jan 31 at 13:43
xotixxotix
291411
asked Jan 31 at 13:43
xotixxotix
291411
asked Jan 31 at 13:43
xotixxotix
291411
291411
2
$begingroup$
in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $sin(pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$.
$endgroup$
– Hayk
Jan 31 at 14:40
$begingroup$
Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now.
$endgroup$
– xotix
Jan 31 at 15:13
add a comment |
2
$begingroup$
in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $sin(pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$.
$endgroup$
– Hayk
Jan 31 at 14:40
$begingroup$
Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now.
$endgroup$
– xotix
Jan 31 at 15:13
2
2
$begingroup$
in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $sin(pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$.
$endgroup$
– Hayk
Jan 31 at 14:40
$begingroup$
in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $sin(pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$.
$endgroup$
– Hayk
Jan 31 at 14:40
$begingroup$
Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now.
$endgroup$
– xotix
Jan 31 at 15:13
$begingroup$
Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now.
$endgroup$
– xotix
Jan 31 at 15:13
add a comment |
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$begingroup$
in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $sin(pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$.
$endgroup$
– Hayk
Jan 31 at 14:40
$begingroup$
Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now.
$endgroup$
– xotix
Jan 31 at 15:13