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Question Number 28071 by abdo imad last updated on 20/Jan/18

let give A_p = ∫_0 ^π t^p cos(nx) with nand p from N 1) find a relation between A_p and A_(p−2) 2) find arelation between A_(2p) and A_(2p−2) 3) find a relation?betweer A_(2p+1) and A_(2p−1) 3) cslculat A_(0 ) , A_1 , A_2 , A_2 .

$${let}\:{give}\:\:{A}_{{p}} =\:\int_{\mathrm{0}} ^{\pi} \:{t}^{{p}} \:{cos}\left({nx}\right)\:\:{with}\:{nand}\:{p}\:{from}\:{N} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{relation}\:{between}\:\:{A}_{{p}} \:{and}\:{A}_{{p}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{arelation}\:{between}\:\:{A}_{\mathrm{2}{p}} \:\:{and}\:{A}_{\mathrm{2}{p}−\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{a}\:{relation}?{betweer}\:{A}_{\mathrm{2}{p}+\mathrm{1}} \:{and}\:\:{A}_{\mathrm{2}{p}−\mathrm{1}} \\ $$$$\left.\mathrm{3}\right)\:{cslculat}\:\:{A}_{\mathrm{0}\:} ,\:{A}_{\mathrm{1}} ,\:{A}_{\mathrm{2}} \:,\:{A}_{\mathrm{2}} . \\ $$

Commented by abdo imad last updated on 26/Jan/18

let integrate by parts 1)A_p = [(1/n)t^p sin(nt)]_0 ^π − ∫_0 ^π (p/n)t^(p−1) sin(nt)dx =−(p/n) ∫_0 ^π t^(p−1) sin(nt)dt =−(p/n)( [((−1)/n) t^(p−1) cos(nt)]_0 ^π −∫_0 ^π −((p−1)/n) t^(p−2) cos(nt)dt ) =−(p/n)( ((−π^(p−1) (−1)^n )/n) +((p−1)/n) ∫_0 ^π t^(p−2) cos(nt)dt) =(p/n^2 ) π^(p−1) (−1)^n −((p(p−1))/n^2 ) A_(p−2) so A_p = (1/n^2 )( p π^(p−1) (−1)^n −p(p−1) A_(p−2)) ) 2) A_(2p) = (1/n^2 )(2p π^(2p−1) (−1)^n −(2p)(2p−1) A_(2p−2) ) 3) A_(2p+1 ) ^( ) = (1/n^(2 ) )((2p+1)π^(2p) (−1)^n −2p(2p+1) A_(2p−1) ) 4) A_0 = ∫_0 ^π cos(nx)dx=[(1/n) sin(nx)]_0 ^π =0 A_1 = ∫_0 ^π t cos(nt)t = [ (t/n) sin(nt)]_0 ^π − ∫_0 ^π (1/n)sin(nt)dt = −(1/n) ∫_0 ^π sin(nt)dt= (1/n^2 )[ cos(nt)]_0 ^π =(1/n^2 )( (−1)^n −1) A_(2 ) =(1/n^2 )( 2π(−1)^n −2A_0 ) =((2π)/n^2 )(−1)^n ....be contunued...

$${let}\:{integrate}\:{by}\:{parts} \\ $$$$\left.\mathrm{1}\right){A}_{{p}} =\:\left[\frac{\mathrm{1}}{{n}}{t}^{{p}} {sin}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi} \:−\:\int_{\mathrm{0}} ^{\pi} \:\frac{{p}}{{n}}{t}^{{p}−\mathrm{1}} {sin}\left({nt}\right){dx} \\ $$$$=−\frac{{p}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{t}^{{p}−\mathrm{1}} \:{sin}\left({nt}\right){dt} \\ $$$$=−\frac{{p}}{{n}}\left(\:\left[\frac{−\mathrm{1}}{{n}}\:{t}^{{p}−\mathrm{1}} {cos}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi} \:\:−\int_{\mathrm{0}} ^{\pi} −\frac{{p}−\mathrm{1}}{{n}}\:{t}^{{p}−\mathrm{2}} \:{cos}\left({nt}\right){dt}\:\right) \\ $$$$=−\frac{{p}}{{n}}\left(\:\frac{−\pi^{{p}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\frac{{p}−\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{t}^{{p}−\mathrm{2}} \:{cos}\left({nt}\right){dt}\right) \\ $$$$=\frac{{p}}{{n}^{\mathrm{2}} }\:\pi^{{p}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \:\:−\frac{{p}\left({p}−\mathrm{1}\right)}{{n}^{\mathrm{2}} }\:{A}_{{p}−\mathrm{2}} \:\:{so} \\ $$$${A}_{{p}} =\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\:{p}\:\pi^{{p}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \:\:−{p}\left({p}−\mathrm{1}\right)\:{A}_{\left.{p}−\mathrm{2}\right)} \:\:\right) \\ $$$$\left.\mathrm{2}\right)\:{A}_{\mathrm{2}{p}} \:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\mathrm{2}{p}\:\pi^{\mathrm{2}{p}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \:−\left(\mathrm{2}{p}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)\:{A}_{\mathrm{2}{p}−\mathrm{2}} \:\right) \\ $$$$\left.\mathrm{3}\right)\:{A}_{\mathrm{2}{p}+\mathrm{1}\:} ^{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} =\:\frac{\mathrm{1}}{{n}^{\mathrm{2}\:} }\left(\left(\mathrm{2}{p}+\mathrm{1}\right)\pi^{\mathrm{2}{p}} \left(−\mathrm{1}\right)^{{n}} \:−\mathrm{2}{p}\left(\mathrm{2}{p}+\mathrm{1}\right)\:{A}_{\mathrm{2}{p}−\mathrm{1}} \right) \\ $$$$\left.\mathrm{4}\right)\:{A}_{\mathrm{0}} \:\:=\:\int_{\mathrm{0}} ^{\pi} {cos}\left({nx}\right){dx}=\left[\frac{\mathrm{1}}{{n}}\:{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} =\mathrm{0} \\ $$$${A}_{\mathrm{1}} =\:\int_{\mathrm{0}} ^{\pi} {t}\:{cos}\left({nt}\right){t}\:=\:\left[\:\frac{{t}}{{n}}\:{sin}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi} \:−\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{{n}}{sin}\left({nt}\right){dt} \\ $$$$=\:−\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\pi} {sin}\left({nt}\right){dt}=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left[\:{cos}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right) \\ $$$${A}_{\mathrm{2}\:} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} \:\:−\mathrm{2}{A}_{\mathrm{0}} \right)\:=\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} ....{be}\:{contunued}... \\ $$

Commented by abdo imad last updated on 26/Jan/18

A_3 = (1/n^2 )( 3 π^2 (−1)^n −6 A_1 ) = (1/n^2 )( 3 π^2 (−1)^n −(6/n^2 )( (−1)^n −1)) =((3 π^2 )/n^2 )(−1)^n −(6/n^4 )( (−1)^n −1).

$${A}_{\mathrm{3}} =\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{3}\:\pi^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} \:−\mathrm{6}\:{A}_{\mathrm{1}} \right) \\ $$$$=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{3}\:\pi^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} \:−\frac{\mathrm{6}}{{n}^{\mathrm{2}} }\left(\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{3}\:\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \:\:−\frac{\mathrm{6}}{{n}^{\mathrm{4}} }\left(\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right). \\ $$

Commented by abdo imad last updated on 26/Jan/18

A_p = ∫_0 ^π t^p cos(nt)dt.

$${A}_{{p}} =\:\int_{\mathrm{0}} ^{\pi} \:{t}^{{p}} {cos}\left({nt}\right){dt}. \\ $$

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