Menu Close

let-u-n-0-1-x-n-sin-pix-dx-1-prove-that-u-n-converges-2-prove-that-u-n-0-pi-sint-t-dt-

Tinku Tara 0 Comments
Pin




Question Number 32042 by abdo imad last updated on 18/Mar/18
let u_n =∫_0 ^1 x^n sin(πx)dx 1) prove that Σ u_n converges 2) prove that Σ u_n = ∫_0 ^π ((sint)/t)dt .
letun=∫01xnsin(πx)dx1)provethatΣunconverges2)provethatΣun=∫0πsinttdt.
Commented by abdo imad last updated on 20/Mar/18
let put S_n =Σ_(k=0) ^n ∫_0 ^1 x^k sin(πx)dx S_n = ∫_0 ^1 (Σ_(k=0) ^n x^k )sin(πx)dx =∫_0 ^1 ((1−x^(n+1) )/(1−x)) sin(πx)dx ⇒ S_(n ) − ∫_0 ^1 ((sin(πx))/(1−x))dx = −∫_0 ^1 (x^(n+1) /(1−x)) sin(πx)dx and ∃m>0/ ∣ S_n − ∫_0 ^1 ((sin(πx))/(1−x))dx∣ ≤ m ∫_0 ^1 x^(n+1) dx =(m/(n+2)) →_(n→∞) 0 ⇒ S_n converged and lim S_n = ∫_0 ^1 ((sin(πx))/(1−x)) dx 2) ch .1−x=t give ∫_0 ^1 ((sin(πx))/(1−x)) dx =∫_0 ^1 ((sin(π(1−t)))/t)= dt = ∫_0 ^1 ((sin(πt))/t) dt after ch. πt =u give lim _(n→∞) S_n = ∫_0 ^π ((sin(u))/(u/π)) (du/π) =∫_0 ^π ((sinu)/u) du . so Σ_(n=0) ^∞ u_n = ∫_0 ^π ((sinu)/u) du .
letputSn=∑k=0n∫01xksin(πx)dxSn=∫01(∑k=0nxk)sin(πx)dx=∫011−xn+11−xsin(πx)dx⇒Sn−∫01sin(πx)1−xdx=−∫01xn+11−xsin(πx)dxand∃m>0/∣Sn−∫01sin(πx)1−xdx∣⩽m∫01xn+1dx=mn+2→n→∞0⇒SnconvergedandlimSn=∫01sin(πx)1−xdx2)ch.1−x=tgive∫01sin(πx)1−xdx=∫01sin(π(1−t))t=dt=∫01sin(πt)tdtafterch.πt=ugivelimn→∞Sn=∫0πsin(u)uπduπ=∫0πsinuudu.so∑n=0∞un=∫0πsinuudu.

Pin

Leave a Reply

Your email address will not be published.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