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Question Number 131275 by Engr_Jidda last updated on 03/Feb/21
Orthogonal and Orthonormal sets given that y_n (t)=ϱ^t B_n sin((nπ)/4)t and y_m (t)=ϱ^t B_m sin((mπ)/4)t show that ∫_0 ^4 y_n (t)y_m (t)dt is orthogonal and orthonormal
OrthogonalandOrthonormalsetsgiventhatyn(t)=ϱtBnsinnπ4tandym(t)=ϱtBmsinmπ4tshowthat∫04yn(t)ym(t)dtisorthogonalandorthonormal
Answered by Ar Brandon last updated on 03/Feb/21
℧=∫_0 ^4 ϱ^(2t) B_m B_n sin(((mπ)/4)t)sin(((nπ)/4)t)dt =−((B_m B_n )/2)∫_0 ^4 ϱ^(2t) {cos((((m+n)π)/4)t)−cos((((m−n)π)/4)t)}dt =−((2B_m B_n )/π)∫_0 ^π ϱ^((8u)/π) [cos((m+n)u)−cos((m−n)u)]du =−((2B_m B_n )/π){[cos((m+n)u)−cos((m−n)u)]ϱ^((8u)/π) ∙(π/8) −∫[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]((πϱ^((8u)/π) )/8)}_0 ^π =((2B_m B_n )/π)∫[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]((πϱ^((8u)/π) )/8)du =((2B_m B_n )/π){[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]((π^2 ϱ^((8u)/π) )/(64)) −∫{(m−n)^2 cos((m−n)u)−(m+n)^2 cos((m+n)u)}((π^2 ϱ^((8u)/π) )/(64))du}_0 ^π =0
℧=∫04ϱ2tBmBnsin(mπ4t)sin(nπ4t)dt=−BmBn2∫04ϱ2t{cos((m+n)π4t)−cos((m−n)π4t)}dt=−2BmBnπ∫0πϱ8uπ[cos((m+n)u)−cos((m−n)u)]du=−2BmBnπ{[cos((m+n)u)−cos((m−n)u)]ϱ8uπ⋅π8−∫[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]πϱ8uπ8}0π=2BmBnπ∫[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]πϱ8uπ8du=2BmBnπ{[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]π2ϱ8uπ64−∫{(m−n)2cos((m−n)u)−(m+n)2cos((m+n)u)}π2ϱ8uπ64du}0π=0
Commented by Engr_Jidda last updated on 03/Feb/21
Thank you so much sir. is this for orthogonal or orthonormal?
Thankyousomuchsir.isthisfororthogonalororthonormal?
Commented by Ar Brandon last updated on 03/Feb/21
��I don't really know Sir. I'm a student. I just solved the calculus and left the conclusion to be made by you. Assuming you have studied the topic��
Commented by Engr_Jidda last updated on 03/Feb/21
yes sir, thank you once again
yessir,thankyouonceagain

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