Menu Close

let-A-p-0-pi-x-p-cos-nx-dx-1-calculate-A-0-A-1-A-2-2-determine-a-relation-of-recurrence-between-A-p-




Question Number 67795 by mathmax by abdo last updated on 31/Aug/19
let  A_p =∫_0 ^π  x^p  cos(nx)dx  1) calculate A_0 ,A_1 ,A_2   2)determine a relation of recurrence between  A_p
$${let}\:\:{A}_{{p}} =\int_{\mathrm{0}} ^{\pi} \:{x}^{{p}} \:{cos}\left({nx}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{\mathrm{0}} ,{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){determine}\:{a}\:{relation}\:{of}\:{recurrence}\:{between}\:\:{A}_{{p}} \\ $$
Commented by mathmax by abdo last updated on 03/Sep/19
1) A_0 =∫_0 ^π  cos(nx)dx =[(1/n)sin(nx)]_0 ^π  =0  A_1 =∫_0 ^π  x cos(nx)dx   by parts   A_1 =[(x/n)sin(nx)]_0 ^(π ) −∫_0 ^π (1/n)sin(nx)dx  =−(1/n) ∫_0 ^π  sin(nx)dx =−(1/n)[−(1/n)cos(nx)]_0 ^π =(1/n^2 ){ (−1)^n −1}  A_2 =∫_0 ^π  x^2 cos(nx)dx =_(by parts)    [(x^2 /n)sin(nx)]_0 ^π  −∫_0 ^π  ((2x)/n)sin(nx)dx  =−(1/n)∫_0 ^π  xsin(nx)dx =−(1/n){ [−(x/n)cos(nx)]_0 ^π −∫_0 ^π −(1/n)cos(nx)dx}  =−(1/n){−(1/n)(π(−1)^n ) +(1/n^2 )[sin(nx)]_0 ^π }  =(π/n^2 )(−1)^n
$$\left.\mathrm{1}\right)\:{A}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\pi} \:{cos}\left({nx}\right){dx}\:=\left[\frac{\mathrm{1}}{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:=\mathrm{0} \\ $$$${A}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\pi} \:{x}\:{cos}\left({nx}\right){dx}\:\:\:{by}\:{parts}\: \\ $$$${A}_{\mathrm{1}} =\left[\frac{{x}}{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi\:} −\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{sin}\left({nx}\right){dx}\:=−\frac{\mathrm{1}}{{n}}\left[−\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\} \\ $$$${A}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {cos}\left({nx}\right){dx}\:=_{{by}\:{parts}} \:\:\:\left[\frac{{x}^{\mathrm{2}} }{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}{x}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\pi} \:{xsin}\left({nx}\right){dx}\:=−\frac{\mathrm{1}}{{n}}\left\{\:\left[−\frac{{x}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} −\int_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right){dx}\right\} \\ $$$$=−\frac{\mathrm{1}}{{n}}\left\{−\frac{\mathrm{1}}{{n}}\left(\pi\left(−\mathrm{1}\right)^{{n}} \right)\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left[{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \right\} \\ $$$$=\frac{\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \\ $$
Commented by mathmax by abdo last updated on 04/Sep/19
2) by parts  u^′ =x^p  and v=cos(nx) ⇒  A_p =[(x^(p+1) /(p+1)) cos(nx)]_0 ^π  −∫_0 ^π  (x^(p+1) /(p+1))(−nsin(nx))dx  =(π^(p+1) /(p+1))(−1)^n   +(n/(p+1)) ∫_0 ^π   x^(p+1)  sin(nx)dx again by parts  u^′  =x^(p+1)  and v =sin(nx) ⇒  ∫_0 ^π   x^(p+1)  sin(nx)dx =[(x^(p+2) /(p+2)) sin(nx)]_0 ^π  −∫_0 ^π  (x^(p+2) /(p+2))(1/n)cos(nx)dx  =−(1/(n(p+2))) ∫_0 ^π  x^(p+2)  cos(nx)dx =−(1/(n(p+2))) A_(p+2)  ⇒  A_p =(π^(p+1) /(p+1))(−1)^n  +(n/(p+1))(−(1/(n(p+2))))A_(p+2)  ⇒  A_p =(π^(p+1) /(p+1))(−1)^n −(1/((p+1)(p+2))) A_(p+2)    ⇒  (1/((p+1)(p+2))) A_(p+2) =(π^(p+1) /(p+1))(−1)^n  −A_p  ⇒  A_(p+2) =(p+1)(p+2)(π^(p+1) /(p+1))(−1)^n − (p+1)(p+2)A_p  ⇒  A_(p+2)  =(p+2)(−1)^n π^(p+1)  −(p+1)(p+2)A_p
$$\left.\mathrm{2}\right)\:{by}\:{parts}\:\:{u}^{'} ={x}^{{p}} \:{and}\:{v}={cos}\left({nx}\right)\:\Rightarrow \\ $$$${A}_{{p}} =\left[\frac{{x}^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\:{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{{x}^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\left(−{nsin}\left({nx}\right)\right){dx} \\ $$$$=\frac{\pi^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} \:\:+\frac{{n}}{{p}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\pi} \:\:{x}^{{p}+\mathrm{1}} \:{sin}\left({nx}\right){dx}\:{again}\:{by}\:{parts} \\ $$$${u}^{'} \:={x}^{{p}+\mathrm{1}} \:{and}\:{v}\:={sin}\left({nx}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:{x}^{{p}+\mathrm{1}} \:{sin}\left({nx}\right){dx}\:=\left[\frac{{x}^{{p}+\mathrm{2}} }{{p}+\mathrm{2}}\:{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{{x}^{{p}+\mathrm{2}} }{{p}+\mathrm{2}}\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}\left({p}+\mathrm{2}\right)}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{{p}+\mathrm{2}} \:{cos}\left({nx}\right){dx}\:=−\frac{\mathrm{1}}{{n}\left({p}+\mathrm{2}\right)}\:{A}_{{p}+\mathrm{2}} \:\Rightarrow \\ $$$${A}_{{p}} =\frac{\pi^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} \:+\frac{{n}}{{p}+\mathrm{1}}\left(−\frac{\mathrm{1}}{{n}\left({p}+\mathrm{2}\right)}\right){A}_{{p}+\mathrm{2}} \:\Rightarrow \\ $$$${A}_{{p}} =\frac{\pi^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} −\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}\:{A}_{{p}+\mathrm{2}} \:\:\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}\:{A}_{{p}+\mathrm{2}} =\frac{\pi^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} \:−{A}_{{p}} \:\Rightarrow \\ $$$${A}_{{p}+\mathrm{2}} =\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\frac{\pi^{{p}+\mathrm{1}} }{{p}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} −\:\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right){A}_{{p}} \:\Rightarrow \\ $$$${A}_{{p}+\mathrm{2}} \:=\left({p}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}} \pi^{{p}+\mathrm{1}} \:−\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right){A}_{{p}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *