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


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Question Number 131275 by Engr_Jidda last updated on 03/Feb/21

$O r t h o g o n a l a n d O r t h o n o r m a l s e t s$ $g i v e n t h a t y_{n} (t) = ϱ^{t} B_{n} s i n \frac{n π}{4} t$ $a n d y_{m} (t) = ϱ^{t} B_{m} s i n \frac{m π}{4} t$ $s h o w t h a t \int_{0}^{4} y_{n} (t) y_{m} (t) d t$ $i s o r t h o g o n a l a n d o r t h o n o r m a l$
	Answered by Ar Brandon last updated on 03/Feb/21
	$℧=∫_0 ^4 ϱ^(2t) B_m B_n sin(((mπ)/4)t)sin(((nπ)/4)t)dt =−((B_m B_n )/2)∫_0 ^4 ϱ^(2t) {cos((((m+n)π)/4)t)−cos((((m−n)π)/4)t)}dt =−((2B_m B_n )/π)∫_0 ^π ϱ^((8u)/π) [cos((m+n)u)−cos((m−n)u)]du =−((2B_m B_n )/π){[cos((m+n)u)−cos((m−n)u)]ϱ^((8u)/π) ∙(π/8) −∫[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]((πϱ^((8u)/π) )/8)}_0 ^π =((2B_m B_n )/π)∫[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]((πϱ^((8u)/π) )/8)du =((2B_m B_n )/π){[(m−n)sin((m−n)u)−(m+n)sin((m+n)u)]((π^2 ϱ^((8u)/π) )/(64)) −∫{(m−n)^2 cos((m−n)u)−(m+n)^2 cos((m+n)u)}((π^2 ϱ^((8u)/π) )/(64))du}_0 ^π =0$
	$℧ = \int_{0}^{4} ϱ^{2 t} B_{m} B_{n} \sin (\frac{m π}{4} t) \sin (\frac{n π}{4} t) dt$ $= - \frac{B_{m} B_{n}}{2} \int_{0}^{4} ϱ^{2 t} {\cos (\frac{(m + n) π}{4} t) - \cos (\frac{(m - n) π}{4} t)} dt$ $= - \frac{2 B_{m} B_{n}}{π} \int_{0}^{π} ϱ^{\frac{8 u}{π}} [\cos ((m + n) u) - \cos ((m - n) u)] du$ $= - \frac{2 B_{m} B_{n}}{π} {[\cos ((m + n) u) - \cos ((m - n) u)] ϱ^{\frac{8 u}{π}} \cdot \frac{π}{8}$ ${- \int [(m - n) \sin ((m - n) u) - (m + n) \sin ((m + n) u)] \frac{π ϱ^{\frac{8 u}{π}}}{8}}}_{0}^{π}$ $= \frac{2 B_{m} B_{n}}{π} \int [(m - n) \sin ((m - n) u) - (m + n) \sin ((m + n) u)] \frac{π ϱ^{\frac{8 u}{π}}}{8} du$ $= \frac{2 B_{m} B_{n}}{π} {[(m - n) \sin ((m - n) u) - (m + n) \sin ((m + n) u)] \frac{π^{2} ϱ^{\frac{8 u}{π}}}{64}$ ${- \int {{(m - n)}^{2} \cos ((m - n) u) - {(m + n)}^{2} \cos ((m + n) u)} \frac{π^{2} ϱ^{\frac{8 u}{π}}}{64} du}}_{0}^{π}$ $= 0$
	Commented by Engr_Jidda last updated on 03/Feb/21

	$T h a n k y o u s o m u c h s i r .$ $i s t h i s f o r o r t h o g o n a l o r o r t h o n o r m a l ?$
	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

	$y e s s i r, t h a n k y o u o n c e a g a i n$
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