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superkillball
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Report Thread starter 15 years ago
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Would anyone help me to do the following question? Thanks
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Jedi Ezekiel
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Which module is this from?
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evariste
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(Original post by superkillball)
Would anyone help me to do the following question? Thanks
Its late check so please check for any silly errors.
f(x)=-1 if -L<x<0
f(x)=1 if 0<x<L
since f(-x)=-f(x) f is odd => a0 and an=0 for all n
piBn=int from {-L<x<0} of {-sin(nx)}dx+{0 to L} {sin(nx)}dx
piBn={1/ncos(nx)}at x=-L and x=0 + {-1/ncos(nx)} at x=0 to x=L
piBn=[1/n-{1/ncos(nl)}]+[-{1/ncos(nl)}-{-1/n}]
Bn=2{1-cosnl}/npi
when L=pi
Bn=2{1-cos(pi n)}/npi
so
Bn=0 n even
=4/pi n odd
so
f(x)=4/pi{(sinx)+(sin3x)/3+(sin5x)/5+...
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superkillball
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#4
Report Thread starter 15 years ago
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(Original post by evariste)
Its late check so please check for any silly errors.
f(x)=-1 if -L<x<0
f(x)=1 if 0<x<L
since f(-x)=-f(x) f is odd => a0 and an=0 for all n
piBn=int from {-L<x<0} of {-sin(nx)}dx+{0 to L} {sin(nx)}dx
piBn={1/ncos(nx)}at x=-L and x=0 + {-1/ncos(nx)} at x=0 to x=L
piBn=[1/n-{1/ncos(nl)}]+[-{1/ncos(nl)}-{-1/n}]
Bn=2{1-cosnl}/npi
when L=pi
Bn=2{1-cos(pi n)}/npi
so
Bn=0 n even
=4/pi n odd
so
f(x)=4/pi{(sinx)+(sin3x)/3+(sin5x)/5+...
thanks
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