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let-x-0-1-and-x-n-1-ln-e-x-n-x-n-1-prove-that-x-n-0-2-prove-that-x-n-converges-and-ddyermine-its-sum-




Question Number 99839 by mathmax by abdo last updated on 23/Jun/20
let x_0 =1 and x_(n+1) =ln(e^x_n  −x_n )  1) prove that x_n  →0  2)prove that Σ x_n  converges and ddyermine its sum
letx0=1andxn+1=ln(exnxn)1)provethatxn02)provethatΣxnconvergesandddyermineitssum
Commented by bachamohamed last updated on 23/Jun/20
do you have the solution ? spead the solution for us.
doyouhavethesolution?speadthesolutionforus.
Commented by abdomathmax last updated on 23/Jun/20
you must take a try sir first...
youmusttakeatrysirfirst
Commented by bachamohamed last updated on 23/Jun/20
yes i tried and did not succed   we await your anser thank you sir
yesitriedanddidnotsuccedweawaityouranserthankyousir
Commented by mathmax by abdo last updated on 23/Jun/20
nevermind we are here for help...
nevermindwearehereforhelp
Answered by maths mind last updated on 23/Jun/20
1)  shiw x_n  cv   let show x_n ≥0   we havef(x)=e^x −x−1 is   increasing for x≥0  1≥x_0 =1≥0  supose≥1 x_n ≥0⇒≥f(1)≥f(x_n )≥f(0)=0⇒e^x_n  −x_n ≥1⇒ln(e^x_n  −x_n )≥0  f(1)=e^1 −1−1≤1⇒x_(n+1) ≤1  ⇒1≥x_(n+1) ≥0⇒by induction ∀n∈N 1≥ x_n ≥0  x_(n+1) −x_n =ln(e^x_n  −x_n )−x_n =ln(e^x_n  −x_n )−ln(e^x_n  )  since e^x_n  −x_n ≤e^(xn) ⇒x_(n+1) ≤x_n   ⇒x_n   is bounded and deacrease ⇒cv   l=lim_(n→∞) X_n ⇒ln(e^l −l)=l⇒e^l −l=e^l ⇒l=0  x_(n+1) =ln(e^x_n  −x_n )  n→∞ x_n →0  e^x_n  =1+x_n +(((x_n )^2 )/2)+0(x_n ^2 )⇒e^(xn) −x_n =1+(x_n ^2 /2)  x_(n+1) =ln(1+(x_n ^2 /2)+o(x_n ^2 ))  =(x_n ^2 /2)+0(x_n ^2 )⇒∃c∈]0,1[  such  ⇒x_(n+1) ≤cx_n ^2 =(cx_n )x_n   since x_n →0⇒∃N ∀n≥N   x_n <(1/2)  ⇒x_(n+1) ≤(c/2)x_n ⇒x_n ≤((c/2))^(n−N) ,∀n≥N  ⇒Σx_n =Σ_(n≤N) x_n +Σ_(n>N) x_n   first finite 2nd by comparaison withe cv geoemtry serie  ⇒Σx_n   Cv  it Sum i worcking on it
1)shiwxncvletshowxn0wehavef(x)=exx1isincreasingforx01x0=10supose1xn0⇒⩾f(1)f(xn)f(0)=0exnxn1ln(exnxn)0f(1)=e1111xn+111xn+10byinductionnN1xn0xn+1xn=ln(exnxn)xn=ln(exnxn)ln(exn)sinceexnxnexnxn+1xnxnisboundedanddeacreasecvl=limnXnln(ell)=lell=ell=0xn+1=ln(exnxn)nxn0exn=1+xn+(xn)22+0(xn2)exnxn=1+xn22xn+1=ln(1+xn22+o(xn2))=xn22+0(xn2)c]0,1[suchxn+1cxn2=(cxn)xnsincexn0NnNxn<12xn+1c2xnxn(c2)nN,nNΣxn=nNxn+n>Nxnfirstfinite2ndbycomparaisonwithecvgeoemtryserieΣxnCvitSumiworckingonit
Commented by abdomathmax last updated on 23/Jun/20
thanks sir.
thankssir.
Answered by abdomathmax last updated on 23/Jun/20
2) let complete the solution we have  x_n →0 (n→∞) and e^x_(n+1)   =e^x_n  −x_n  ⇒  x_n =e^x_n  −e^x_(n+1)    let S_n =Σ_(k=1) ^n  x_k  ⇒  S_n =Σ_(k=0) ^n  (e^x_k  −e^x_(k+1)  ) =e^x_0  −e^x_1  +e^x_1  −e^x_2  +....  +e^x_n  −e^x_(n+1)   =e−e^x_(n+1)   →e−1 ⇒  Σ_(n=0) ^∞  x^n  =e−1
2)letcompletethesolutionwehavexn0(n)andexn+1=exnxnxn=exnexn+1letSn=k=1nxkSn=k=0n(exkexk+1)=ex0ex1+ex1ex2+.+exnexn+1=eexn+1e1n=0xn=e1
Commented by abdomathmax last updated on 23/Jun/20
Σ_(n=0) ^∞  x_n =e−1
n=0xn=e1
Commented by bachamohamed last updated on 23/Jun/20
thank′s sir
thankssir
Commented by mathmax by abdo last updated on 23/Jun/20
you are welcome.
youarewelcome.

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