Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 145634 by mathmax by abdo last updated on 06/Jul/21

let s(x)=Σ_(n=1) ^∞  (((−1)^n )/((2x^2 +2x(√(1+x^2 ))+1)^n ))  1) explicite s(x)  2) calculate ∫_0 ^1 s(x)dx

lets(x)=n=1(1)n(2x2+2x1+x2+1)n1)explicites(x)2)calculate01s(x)dx

Answered by Olaf_Thorendsen last updated on 06/Jul/21

S(x) = Σ_(n=1) ^∞ (((−1)^n )/((2x^2 +2x(√(1+x^2 ))+1)^n ))  S(x) = Σ_(n=1) ^∞ (((−1)^n )/((x^2 +2x(√(1+x^2 ))+(1+x^2 ))^n ))  S(x) = Σ_(n=1) ^∞ (((−1)^n )/((x+(√(1+x^2 )))^(2n) ))  S(x) = Σ_(n=0) ^∞ (((−1)^n )/((x+(√(1+x^2 )))^(2n) ))−1  S(x) = (1/(1+(1/((x+(√(1+x^2 )))^2 ))))−1  S(x) = −(1/((x+(√(1+x^2 )))^2 +1))  ∫_0 ^1 S(x)dx = −∫_0 ^1 (dx/((x+(√(1+x^2 )))^2 +1))  = −∫_0 ^(sh^(−1) (1)) ((chu)/((shu+(√(1+sh^2 u)))^2 +1))du  = −∫_0 ^(sh^(−1) (1)) ((chu)/((shu+chu)^2 +1))du  = −∫_0 ^(sh^(−1) (1)) ((chu)/(2ch^2 u+2shuchu))du  = −(1/2)∫_0 ^(sh^(−1) (1)) (du/(chu+shu))  = −(1/2)∫_0 ^(sh^(−1) (1)) (du/e^u )  = (1/2)[e^(−u) ]_0 ^(sh^(−1) (1))   = (1/2)(e^(−sh^(−1) (1)) −1)  = (1/2)((1/(1+(√(1+1^2 ))))−1)  = (1/2)((1/(1+(√2)))−1)  = (1/2)((√2)−2) = (1/( (√2)))−1

S(x)=n=1(1)n(2x2+2x1+x2+1)nS(x)=n=1(1)n(x2+2x1+x2+(1+x2))nS(x)=n=1(1)n(x+1+x2)2nS(x)=n=0(1)n(x+1+x2)2n1S(x)=11+1(x+1+x2)21S(x)=1(x+1+x2)2+101S(x)dx=01dx(x+1+x2)2+1=0sh1(1)chu(shu+1+sh2u)2+1du=0sh1(1)chu(shu+chu)2+1du=0sh1(1)chu2ch2u+2shuchudu=120sh1(1)duchu+shu=120sh1(1)dueu=12[eu]0sh1(1)=12(esh1(1)1)=12(11+1+121)=12(11+21)=12(22)=121

Answered by mathmax by abdo last updated on 07/Jul/21

1)we have (x/( (√(1+x^2 ))))=_(x=sht)    ((sht)/(cht))=((e^t  −e^(−t) )/(e^t +e^(−t) ))=((1−e^(−2t) )/(1+e^(−2t) ))  =(1−e^(−2t) )Σ_(n=0) ^∞ (−1)^n  e^(−2nt)    =Σ_(n=0) ^(∞ )  (−1)^n  e^(−2nt) −Σ_(n=0) ^∞ (−1)^n  e^(−(2n+2)t)   =Σ_(n=0) ^∞  (−1)^n  e^(−2nt) +Σ_(n=1) ^∞  (−1)^n e^(−2nt)   =1+2Σ_(n=0) ^∞  (((−1)^n )/e^(2nt) )=1+2Σ_(n=1) ^∞  (((−1)^n )/e^(2nlog(x+(√(1+x^2 )))) )  =1+2Σ_(n=1) ^∞  (((−1)^n )/((x+(√(1+x^2 )))^(2n) ))=1+2Σ_(n=1) ^∞  (((−1)^n )/((x^2 +2x(√(1+x^2 ))+1+x^2 )^n ))  =1+2s(x) ⇒2s(x)=(x/( (√(1+x^2 ))))−1 ⇒s(x)=(1/2)((x/( (√(1+x^2 ))))−1)

1)wehavex1+x2=x=shtshtcht=etetet+et=1e2t1+e2t=(1e2t)n=0(1)ne2nt=n=0(1)ne2ntn=0(1)ne(2n+2)t=n=0(1)ne2nt+n=1(1)ne2nt=1+2n=0(1)ne2nt=1+2n=1(1)ne2nlog(x+1+x2)=1+2n=1(1)n(x+1+x2)2n=1+2n=1(1)n(x2+2x1+x2+1+x2)n=1+2s(x)2s(x)=x1+x21s(x)=12(x1+x21)

Commented by mathmax by abdo last updated on 07/Jul/21

2)∫_0 ^1 s(x)dx=(1/2)∫_0 ^1  ((xdx)/( (√(1+x^2 ))))−(1/2)  =(1/2)[(√(1+x^2 ))]_0 ^1 −(1/2) =(1/2)((√2)−1)−(1/2)=((√2)/2)−1

2)01s(x)dx=1201xdx1+x212=12[1+x2]0112=12(21)12=221

Terms of Service

Privacy Policy

Contact: info@tinkutara.com