Finding coefficient of Fourier cosine series … [closed]
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This example from "Walter A Strauss-Partial differential equations _an introduction-Wiley(2009)" book page 108.
My question is : from where the nonzero coefficient come if $sin(mpi)=0$
fourier-series
asked Dec 28 '18 at 23:16
NawalNawal
223
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closed as unclear what you're asking by Kavi Rama Murthy, Andrew, Leucippus, Holo, KReiser Dec 29 '18 at 2:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
This example from "Walter A Strauss-Partial differential equations _an introduction-Wiley(2009)" book page 108.
My question is : from where the nonzero coefficient come if $sin(mpi)=0$
fourier-series
asked Dec 28 '18 at 23:16
NawalNawal
223
$endgroup$
closed as unclear what you're asking by Kavi Rama Murthy, Andrew, Leucippus, Holo, KReiser Dec 29 '18 at 2:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
This example from "Walter A Strauss-Partial differential equations _an introduction-Wiley(2009)" book page 108.
My question is : from where the nonzero coefficient come if $sin(mpi)=0$
fourier-series
asked Dec 28 '18 at 23:16
NawalNawal
223
$endgroup$
This example from "Walter A Strauss-Partial differential equations _an introduction-Wiley(2009)" book page 108.
My question is : from where the nonzero coefficient come if $sin(mpi)=0$
fourier-series
fourier-series
asked Dec 28 '18 at 23:16
NawalNawal
223
asked Dec 28 '18 at 23:16
NawalNawal
223
asked Dec 28 '18 at 23:16
NawalNawal
223
asked Dec 28 '18 at 23:16
NawalNawal
223
asked Dec 28 '18 at 23:16
NawalNawal
223
223
closed as unclear what you're asking by Kavi Rama Murthy, Andrew, Leucippus, Holo, KReiser Dec 29 '18 at 2:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Kavi Rama Murthy, Andrew, Leucippus, Holo, KReiser Dec 29 '18 at 2:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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For $m=0$, you can't take the integral normally since otherwise, you would get a division-by-zero error for $frac{2}{mx}$. Thus, instead, we have to plus in $m$ before we take the integral:
$$int_0^lcosleft(frac{0cdot pi x}{l}right)dx=int_0^l1dx=l$$
(Note that I used $cos 0=1$ in the first step to simplify.)
This leads to a non-zero coefficient for $m=0$.
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $m=0$, you can't take the integral normally since otherwise, you would get a division-by-zero error for $frac{2}{mx}$. Thus, instead, we have to plus in $m$ before we take the integral:
$$int_0^lcosleft(frac{0cdot pi x}{l}right)dx=int_0^l1dx=l$$
(Note that I used $cos 0=1$ in the first step to simplify.)
This leads to a non-zero coefficient for $m=0$.
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
For $m=0$, you can't take the integral normally since otherwise, you would get a division-by-zero error for $frac{2}{mx}$. Thus, instead, we have to plus in $m$ before we take the integral:
$$int_0^lcosleft(frac{0cdot pi x}{l}right)dx=int_0^l1dx=l$$
(Note that I used $cos 0=1$ in the first step to simplify.)
This leads to a non-zero coefficient for $m=0$.
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
For $m=0$, you can't take the integral normally since otherwise, you would get a division-by-zero error for $frac{2}{mx}$. Thus, instead, we have to plus in $m$ before we take the integral:
$$int_0^lcosleft(frac{0cdot pi x}{l}right)dx=int_0^l1dx=l$$
(Note that I used $cos 0=1$ in the first step to simplify.)
This leads to a non-zero coefficient for $m=0$.
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
For $m=0$, you can't take the integral normally since otherwise, you would get a division-by-zero error for $frac{2}{mx}$. Thus, instead, we have to plus in $m$ before we take the integral:
$$int_0^lcosleft(frac{0cdot pi x}{l}right)dx=int_0^l1dx=l$$
(Note that I used $cos 0=1$ in the first step to simplify.)
This leads to a non-zero coefficient for $m=0$.
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 28 '18 at 23:51
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
