Menu Close

let-S-n-0-n-x-1-x-x-1-x-3-dx-1-calculate-S-n-2-find-lim-n-S-n-

Tinku Tara 0 Comments
Pin




Question Number 39660 by abdo mathsup 649 cc last updated on 09/Jul/18
let S_n = ∫_0 ^n ((x(−1)^([x]) )/((x+1 −[x])^3 ))dx 1) calculate S_n 2) find lim_(n→+∞) S_n
letSn=∫0nx(−1)[x](x+1−[x])3dx1)calculateSn2)findlimn→+∞Sn
Commented by maxmathsup by imad last updated on 11/Jul/18
1) S_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) ((x(−1)^k )/((x+1−k)^3 ))dx=Σ_(k=0) ^(n−1) (−1)^k ∫_k ^(k+1) (x/((x+1−k)^3 ))dx but ∫_k ^(k+1) (x/((x+1−k)^2 ))dx = ∫_k ^(k+1) ((x+1 −k +k−1)/((x+1−k)^2 ))dx =∫_k ^(k+1) (dx/(x+1−k)) +(k−1)[−(1/(x+1−k))]_k ^(k+1) =[ln∣x+1−k∣]_k ^(k+1) +(k−1) {1 −(1/2)}=ln(2) +(1/2)(k−1) S_n = Σ_(k=0) ^(n−1) (−1)^k {ln(2) +(1/2)(k−1)} =ln(2) Σ_(k=0) ^(n−1) (−1)^k +(1/2) Σ_(k=0) ^(n−1) (−1)^k (k−1) but we have Σ_(k=0) ^(n−1) (−1)^k =((1−(−1)^n )/2) Σ_(k=0) ^(n−1) (−1)^k (k−1)^ =−1 +Σ_(k=1) ^(n−1) (−1)^k (k−1) =−1 +Σ_(k=0) ^(n−2) (−1)^(k+1) k =−1 −Σ_(k=0) ^(n−2) k(−1)^k =−1−Σ_(k=1) ^(n−2) k(−1)^k Σ_(k=0) ^N x^k =((x^(N+1) −1)/(x−1)) ⇒Σ_(k=1) ^N k x^(k−1) =((Nx^(N+1) −(N+1)x^n +1)/((x−1)^2 )) ⇒ Σ_(k=1) ^N k x^k =(x/((x−1)^2 )){ Nx^(N+1) −(N+1)x^N +1} ⇒ Σ_(k=1) ^(n−2) k(−1)^k = ((−1)/4){(n−2)(−1)^(n−1) −(n−1)(−1)^(n−2) +1} ⇒ S_n =((ln(2))/2){1−(−1)^n } +(1/2){ −1 +(1/4){(n−2)(−1)^(n−1) −(n−1)(−1)^(n−2) +1}}
1)Sn=∑k=0n−1∫kk+1x(−1)k(x+1−k)3dx=∑k=0n−1(−1)k∫kk+1x(x+1−k)3dxbut∫kk+1x(x+1−k)2dx=∫kk+1x+1−k+k−1(x+1−k)2dx=∫kk+1dxx+1−k+(k−1)[−1x+1−k]kk+1=[ln∣x+1−k∣]kk+1+(k−1){1−12}=ln(2)+12(k−1)Sn=∑k=0n−1(−1)k{ln(2)+12(k−1)}=ln(2)∑k=0n−1(−1)k+12∑k=0n−1(−1)k(k−1)butwehave∑k=0n−1(−1)k=1−(−1)n2∑k=0n−1(−1)k(k−1)=−1+∑k=1n−1(−1)k(k−1)=−1+∑k=0n−2(−1)k+1k=−1−∑k=0n−2k(−1)k=−1−∑k=1n−2k(−1)k∑k=0Nxk=xN+1−1x−1⇒∑k=1Nkxk−1=NxN+1−(N+1)xn+1(x−1)2⇒∑k=1Nkxk=x(x−1)2{NxN+1−(N+1)xN+1}⇒∑k=1n−2k(−1)k=−14{(n−2)(−1)n−1−(n−1)(−1)n−2+1}⇒Sn=ln(2)2{1−(−1)n}+12{−1+14{(n−2)(−1)n−1−(n−1)(−1)n−2+1}}
Commented by maxmathsup by imad last updated on 11/Jul/18
2) we have S_(2n) =−(1/2) +(1/8){−(2n−2)−(2n−1) +1)} =−(1/2) +(1/8){−4n +4} =(1/2) +(1/2){−n+1}=1−(n/2) S_(2n+1) =ln(2) +(1/2){ −1 +(1/4)((2n−1)−(2n)(−1) +1)} =ln(2) +(1/2){−1 +(1/4)(4n +2)} and its clear that (S_n ) is not convergent!
2)wehaveS2n=−12+18{−(2n−2)−(2n−1)+1)}=−12+18{−4n+4}=12+12{−n+1}=1−n2S2n+1=ln(2)+12{−1+14((2n−1)−(2n)(−1)+1)}=ln(2)+12{−1+14(4n+2)}anditsclearthat(Sn)isnotconvergent!

Pin

Leave a Reply

Your email address will not be published.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