Question Number 131275 by Engr_Jidda last updated on 03/Feb/21
$${Orthogonal}\:{and}\:{Orthonormal}\:{sets} \\ $$$${given}\:{that}\:\:{y}_{{n}} \left({t}\right)=\varrho^{{t}} {B}_{{n}} {sin}\frac{{n}\pi}{\mathrm{4}}{t}\: \\ $$$${and}\:{y}_{{m}} \left({t}\right)=\varrho^{{t}} {B}_{{m}} {sin}\frac{{m}\pi}{\mathrm{4}}{t} \\ $$$${show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{4}} {y}_{{n}} \left({t}\right){y}_{{m}} \left({t}\right){dt} \\ $$$${is}\:{orthogonal}\:{and}\:{orthonormal} \\ $$
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](Q131278.png)
$$\mho=\int_{\mathrm{0}} ^{\mathrm{4}} \varrho^{\mathrm{2t}} {B}_{\mathrm{m}} {B}_{\mathrm{n}} \mathrm{sin}\left(\frac{\mathrm{m}\pi}{\mathrm{4}}\mathrm{t}\right)\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{4}}\mathrm{t}\right)\mathrm{dt} \\ $$$$\:\:\:\:=−\frac{{B}_{\mathrm{m}} {B}_{\mathrm{n}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{4}} \varrho^{\mathrm{2t}} \left\{\mathrm{cos}\left(\frac{\left(\mathrm{m}+\mathrm{n}\right)\pi}{\mathrm{4}}\mathrm{t}\right)−\mathrm{cos}\left(\frac{\left(\mathrm{m}−\mathrm{n}\right)\pi}{\mathrm{4}}\mathrm{t}\right)\right\}\mathrm{dt} \\ $$$$\:\:\:\:=−\frac{\mathrm{2}{B}_{\mathrm{m}} {B}_{\mathrm{n}} }{\pi}\int_{\mathrm{0}} ^{\pi} \varrho^{\frac{\mathrm{8u}}{\pi}} \left[\mathrm{cos}\left(\left(\mathrm{m}+\mathrm{n}\right)\mathrm{u}\right)−\mathrm{cos}\left(\left(\mathrm{m}−\mathrm{n}\right)\mathrm{u}\right)\right]\mathrm{du} \\ $$$$\:\:\:\:=−\frac{\mathrm{2}{B}_{\mathrm{m}} {B}_{\mathrm{n}} }{\pi}\left\{\left[\mathrm{cos}\left(\left(\mathrm{m}+\mathrm{n}\right)\mathrm{u}\right)−\mathrm{cos}\left(\left(\mathrm{m}−\mathrm{n}\right)\mathrm{u}\right)\right]\varrho^{\frac{\mathrm{8u}}{\pi}} \centerdot\frac{\pi}{\mathrm{8}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\int\left[\left(\mathrm{m}−\mathrm{n}\right)\mathrm{sin}\left(\left(\mathrm{m}−\mathrm{n}\right)\mathrm{u}\right)−\left(\mathrm{m}+\mathrm{n}\right)\mathrm{sin}\left(\left(\mathrm{m}+\mathrm{n}\right)\mathrm{u}\right)\right]\frac{\pi\varrho^{\frac{\mathrm{8u}}{\pi}} }{\mathrm{8}}\right\}_{\mathrm{0}} ^{\pi} \\ $$$$\:\:\:\:=\frac{\mathrm{2}{B}_{\mathrm{m}} {B}_{\mathrm{n}} }{\pi}\int\left[\left(\mathrm{m}−\mathrm{n}\right)\mathrm{sin}\left(\left(\mathrm{m}−\mathrm{n}\right)\mathrm{u}\right)−\left(\mathrm{m}+\mathrm{n}\right)\mathrm{sin}\left(\left(\mathrm{m}+\mathrm{n}\right)\mathrm{u}\right)\right]\frac{\pi\varrho^{\frac{\mathrm{8u}}{\pi}} }{\mathrm{8}}\mathrm{du} \\ $$$$\:\:\:\:=\frac{\mathrm{2}{B}_{\mathrm{m}} {B}_{\mathrm{n}} }{\pi}\left\{\left[\left(\mathrm{m}−\mathrm{n}\right)\mathrm{sin}\left(\left(\mathrm{m}−\mathrm{n}\right)\mathrm{u}\right)−\left(\mathrm{m}+\mathrm{n}\right)\mathrm{sin}\left(\left(\mathrm{m}+\mathrm{n}\right)\mathrm{u}\right)\right]\frac{\pi^{\mathrm{2}} \varrho^{\frac{\mathrm{8u}}{\pi}} }{\mathrm{64}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:−\int\left\{\left(\mathrm{m}−\mathrm{n}\right)^{\mathrm{2}} \mathrm{cos}\left(\left(\mathrm{m}−\mathrm{n}\right)\mathrm{u}\right)−\left(\mathrm{m}+\mathrm{n}\right)^{\mathrm{2}} \mathrm{cos}\left(\left(\mathrm{m}+\mathrm{n}\right)\mathrm{u}\right)\right\}\frac{\pi^{\mathrm{2}} \varrho^{\frac{\mathrm{8u}}{\pi}} }{\mathrm{64}}\mathrm{du}\right\}_{\mathrm{0}} ^{\pi} \\ $$$$\:\:\:\:\:=\mathrm{0} \\ $$
Commented by Engr_Jidda last updated on 03/Feb/21
$${Thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$$${is}\:{this}\:{for}\:{orthogonal}\:{or}\:{orthonormal}? \\ $$
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
$${yes}\:{sir},\:{thank}\:{you}\:{once}\:{again} \\ $$