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Question Number 2358 by 123456 last updated on 18/Nov/15
a_(n+1) =−((a_n (∣a_n ∣+1))/(∣a_n ∣)),a_1 =1 b_(n+1) =b_n +a_(n+1) ,b_1 =a_1 b_(10) =?
an+1=−an(∣an∣+1)∣an∣,a1=1bn+1=bn+an+1,b1=a1b10=?
Answered by Filup last updated on 18/Nov/15
continued (without working) (working is simple) a_1 =1 b_1 =1 a_2 =−2 b_2 =−1 a_3 =3 b_3 =2 a_4 =−4 b_4 =−2 a_5 =5 b_5 =3 a_6 =−6 b_6 =−3 a_7 =7 b_7 =4 a_8 =−8 b_8 =−4 a_9 =9 b_9 =5 a_(10) =−10 b_(10) =−5 ∴b_(10) =−5
continued(withoutworking)(workingissimple)a1=1b1=1a2=−2b2=−1a3=3b3=2a4=−4b4=−2a5=5b5=3a6=−6b6=−3a7=7b7=4a8=−8b8=−4a9=9b9=5a10=−10b10=−5∴b10=−5
Commented by Filup last updated on 18/Nov/15
Interestingly: a_n =(−1)^(n+1) n={1, −2, 3, −4, ...} or: a_(n+1) =(−1)^n n ∴b_(n+1) =b_n +(−1)^n n, b_1 =1
Interestingly:an=(−1)n+1n={1,−2,3,−4,…}or:an+1=(−1)nn∴bn+1=bn+(−1)nn,b1=1
Commented by Filup last updated on 18/Nov/15
a_(n+1) =−((a_n (a_n +1))/(∣a_n ∣)), (a_1 =1) a_(n+1) =−sgn(a_n )(a_n +1) a_(n+1) =sgn(−a_n )(a_n +1) ∴ if a_k =+ve, a_(k+1) =−ve Now, knowing the sequence gives: 1, −2, 3, −4, ... We can evaluate that: a_(n+1) =(−1)^n n ∴a_(n+1) =(−1)^n n=sgn(−a_n )(a_n +1) Thought that was interesting as it is able to show any a_(n+1) when given a_n . Or the other way around e.g. a_(10) =−10 ∴−10=sgn(−a_9 )(a_9 +1) ∵a_(n+1) =sgn(−a_9 )a_9 ±a_(n+1) =∓a_n ↓ RHS shows that if a_k =+ve, a_(k±1) =−ve ∴−10=sgn(−a_9 )(a_9 +1) ∴−10=−sgn(a_9 )(a_9 +1) LHS=−ve, ∴RHS=+ve sgn(x)=±1→sgn(a_9 )=+ve ∴−10=−(+(a_9 +1)) ∴−10=−a_9 −1 ∴10=a_9 +1 ∴a_9 =9 Which is true from above
an+1=−an(an+1)∣an∣,(a1=1)an+1=−sgn(an)(an+1)an+1=sgn(−an)(an+1)∴ifak=+ve,ak+1=−veNow,knowingthesequencegives:1,−2,3,−4,…Wecanevaluatethat:an+1=(−1)nn∴an+1=(−1)nn=sgn(−an)(an+1)Thoughtthatwasinterestingasitisabletoshowanyan+1whengivenan.Ortheotherwayarounde.g.a10=−10∴−10=sgn(−a9)(a9+1)∵an+1=sgn(−a9)a9±an+1=∓an↓RHSshowsthatifak=+ve,ak±1=−ve∴−10=sgn(−a9)(a9+1)∴−10=−sgn(a9)(a9+1)LHS=−ve,∴RHS=+vesgn(x)=±1→sgn(a9)=+ve∴−10=−(+(a9+1))∴−10=−a9−1∴10=a9+1∴a9=9Whichistruefromabove
Commented by RasheedAhmad last updated on 18/Nov/15
G_(OO) D Discovery!
GOODDiscovery!
Answered by Filup last updated on 18/Nov/15
a_(n+1) =−a_n (1+(1/(∣a_n ∣))) a_(n+1) =−a_n −(a_n /(∣a_n ∣)) (a_n /(∣a_n ∣))= { ((1 if a_n >0)),((−1 if a_n <0)) :} =sgn(a_n ) if a_n =0, (a_n /(∣a_n ∣))=undefined −sgn(x)=sgn(−x) a_(n+1) =−(a_n +sgn(a_n )) a_(n+1) =sgn(−a_n )−a_n ∴b_(n+1) =b_n +sgn(−a_n )−a_n continue
an+1=−an(1+1∣an∣)an+1=−an−an∣an∣an∣an∣={1ifan>0−1ifan<0=sgn(an)ifan=0,an∣an∣=undefined−sgn(x)=sgn(−x)an+1=−(an+sgn(an))an+1=sgn(−an)−an∴bn+1=bn+sgn(−an)−ancontinue
Commented by Filup last updated on 18/Nov/15
only way i can tell to solve for b_(10) is to solve a_2 , a_3 , ..., a_(10) and plugging them into b. I′m lazy so i havent done it. If this goes unanswered, i will do it.
onlywayicantelltosolveforb10istosolvea2,a3,…,a10andpluggingthemintob.I′mlazysoihaventdoneit.Ifthisgoesunanswered,iwilldoit.
Commented by RasheedAhmad last updated on 18/Nov/15
Like your approach! −−−−−−−−−−−− There is a mistake(which however doesn′t affect your logic allover). (a_n /(∣a_n ∣))= { ((1 if a_n >0)),((0 if a_n =0_(−) )),((−1 if a_n <0)) :} If a_n =0, (a_n /(∣a_n ∣)) is undefined.
Likeyourapproach!−−−−−−−−−−−−Thereisamistake(whichhoweverdoesn′taffectyourlogicallover).an∣an∣={1ifan>00ifan=0−−1ifan<0Ifan=0,an∣an∣isundefined.
Commented by Filup last updated on 18/Nov/15
Thank You <3
ThankYou<3
Commented by Filup last updated on 18/Nov/15
And, yes. I excluded the undefined possibility because a_n ≠0,∀n>1 where a_1 =1
And,yes.Iexcludedtheundefinedpossibilitybecausean≠0,∀n>1wherea1=1
Answered by RasheedAhmad last updated on 18/Nov/15
Trying a different approach a_(n+1) =−((a_n (∣a_n ∣+1))/(∣a_n ∣)),a_1 =1 b_(n+1) =b_n +a_(n+1) ,b_1 =a_1 b_(10) =? −−−−−−−−−− Case−I : When a_n >0⇒∣a_n ∣=a_n a_(n+1) =−((a_n (∣a_n ∣+1))/(∣a_n ∣)) ⇒ a_(n+1) =−((a_n (a_n +1))/a_n )=−(a_n +1) { (( a_(n+1) =−(a_n +1))),((b_(n+1) =b_n −(a_n +1))) :} Case−II :When a_n =0⇒∣a_n ∣=a_n =0 a_(n+1) is undefined in this case which means if any of a_n is 0 all the next a_n ′s will be undefined. Case−III:When a_n <0⇒∣a_n ∣=−a_n a_(n+1) =−((a_n (−a_n +1))/(−a_n ))=−a_n +1 { ((a_(n+1) =−a_n +1)),((b_(n+1) =b_n −a_n +1)) :} −−−−−−−−−−−−−−− { (( { (( a_(n+1) =−(a_n +1))),((b_(n+1) =b_n −(a_n +1))) :} when a_n >0 )),((a_(n+1) is undefined when a_n =0)),(( { ((a_(n+1) =−a_n +1)),((b_(n+1) =b_n −a_n +1)) :} when a_n <0)) :} Since a_1 =1>0, for a_2 Case−I is applied. The remaining process of mine is similar to that of Mr. Filup.
Tryingadifferentapproachan+1=−an(∣an∣+1)∣an∣,a1=1bn+1=bn+an+1,b1=a1b10=?−−−−−−−−−−Case−I:Whenan>0⇒∣an∣=anan+1=−an(∣an∣+1)∣an∣⇒an+1=−an(an+1)an=−(an+1){an+1=−(an+1)bn+1=bn−(an+1)Case−II:Whenan=0⇒∣an∣=an=0an+1isundefinedinthiscasewhichmeansifanyofanis0allthenextan′swillbeundefined.Case−III:Whenan<0⇒∣an∣=−anan+1=−an(−an+1)−an=−an+1{an+1=−an+1bn+1=bn−an+1−−−−−−−−−−−−−−−{{an+1=−(an+1)bn+1=bn−(an+1)whenan>0an+1isundefinedwhenan=0{an+1=−an+1bn+1=bn−an+1whenan<0Sincea1=1>0,fora2Case−Iisapplied.TheremainingprocessofmineissimilartothatofMr.Filup.
Commented by Filup last updated on 18/Nov/15
I didn′t even think about evaluating a_n the way you did. Nice work!
Ididn′teventhinkaboutevaluatinganthewayyoudid.Nicework!
Commented by RasheedAhmad last updated on 19/Nov/15
THANK^S
THANKS

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