deduce the derivative of Fourier series when the series is not continuous.
$begingroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
edited Dec 30 '18 at 23:06
Bernard
119k639112
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
$endgroup$
add a comment |
$begingroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
edited Dec 30 '18 at 23:06
Bernard
119k639112
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
$endgroup$
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
add a comment |
$begingroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
edited Dec 30 '18 at 23:06
Bernard
119k639112
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
$endgroup$
If $f'(x)$ is soft piecewise then the Fourier series of a continuous function $f$ in $[-L,L]$
$$f(x)=a_0+sum_{n=1}^infty[a_ncos(frac{npi x}{L})+b_nsin(frac{npi x}{L})]$$
can't be derivate term to term, but we can deduce this:
$$f'(x)=frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L})),tag1$$
I want to prove $(1)$ but I don't have a clear idea of how prove this, I think of proving
$$f(x)=int_{-L}^{L}[frac{1}{2L}(f(L)-f(-L))+sum_{n=1}^{infty}(frac{npi}{L}b_n+frac{(-1)^n}{L}(f(L)-f(-L))cos(frac{npi x}{L})+sum_{n=1}^infty(-frac{npi}{L}a_nsin(frac{npi x}{L}))]$$
because it is equivalent, but can someone give me a hint of how proceed with this exercise?
fourier-series
fourier-series
edited Dec 30 '18 at 23:06
Bernard
119k639112
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
edited Dec 30 '18 at 23:06
Bernard
119k639112
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
edited Dec 30 '18 at 23:06
Bernard
119k639112
edited Dec 30 '18 at 23:06
Bernard
119k639112
edited Dec 30 '18 at 23:06
Bernard
119k639112
119k639112
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
asked Dec 30 '18 at 22:54
Bvss12Bvss12
1,785618
1,785618
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
add a comment |
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36
add a comment |
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$begingroup$
What does it mean to say $f'$ is "soft piecewise"?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:36