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Question Number 214888 by Emmanuel07 last updated on 22/Dec/24

Commented by mr W last updated on 23/Dec/24

i got  a_n =cot {[(1/( (√5)))(tan^(−1) 2−((3π)/8))+(π/8)](((1−(√5))/2))^n −[(1/( (√5)))(tan^(−1) 2−((3π)/8))−(π/8)](((1+(√5))/2))^n }  which delivers:  a_0 =1  a_1 =2  a_2 =(1/3)  a_3 =−(1/7)  a_4 =−((11)/2)  a_5 =(3/(79))  ......

igotan=cot{[15(tan123π8)+π8](152)n[15(tan123π8)π8](1+52)n}whichdelivers:a0=1a1=2a2=13a3=17a4=112a5=379......

Commented by mr W last updated on 23/Dec/24

has somebody got a more simple  formula for a_n  in terms of n?

hassomebodygotamoresimpleformulaforanintermsofn?

Answered by maths2 last updated on 23/Dec/24

a_(n+1) =((a_n a_(n−1) −1)/(a_n +a_(n−1) ))  a_n =cot(b_n )⇒cot(b_(n+1) )=((cot(b_n )cot(b_(n−1) )−1)/(cot(b_n )+cot(b_(n−1) )))=  cot(b_n +b_(n−1) )⇒b_(n+1) =b_n +b_(n−1)   b_(n+1) −b_n −b_(n−1) =0  b_0 =(π/4);b_1 =cot^(−1) (2)

an+1=anan11an+an1an=cot(bn)cot(bn+1)=cot(bn)cot(bn1)1cot(bn)+cot(bn1)=cot(bn+bn1)bn+1=bn+bn1bn+1bnbn1=0b0=π4;b1=cot1(2)

Answered by mr W last updated on 24/Dec/24

a_(n−1) a_n −a_(n−1) a_(n+1) =1+a_n a_(n+1)   a_(n−1) (a_n −a_(n+1) )=1+a_n a_(n+1)   ((1+a_n a_(n+1) )/(a_n −a_(n+1) ))=a_(n−1)   let a_n =cot θ_n   ((1+cot θ_n cot θ_(n+1) )/(cot θ_n −cot θ_(n+1) ))=cot θ_(n−1)   cot (θ_(n+1) −θ_n )=cot θ_(n−1)   ⇒θ_(n+1) −θ_n =θ_(n−1)    ⇒θ_(n+1) −θ_n −θ_(n−1) =0   ← Fibonacci  r^2 −r−1=0  ⇒r=((1±(√5))/2)  ⇒θ_n =A(((1+(√5))/2))^n +B(((1−(√5))/2))^n   θ_0 =A+B=cot^(−1) a_0 =cot^(−1) 1=(π/4)  ⇒((A+B)/2)=(π/8)  θ_1 =((A+B)/2)+(( (√5)(A−B))/2)=cot^(−1) a_1 =cot^(−1) 2  ⇒(( A−B)/2)=(1/( (√5)))(cot^(−1) 2−(π/8))  ⇒A=(π/8)+(1/( (√5)))(cot^(−1) 2−(π/8))  ⇒B=(π/8)−(1/( (√5)))(cot^(−1) 2−(π/8))  ⇒θ_n =[(π/8)+(1/( (√5)))(cot^(−1) 2−(π/8))](((1+(√5))/2))^n +[(π/8)−(1/( (√5)))(cot^(−1) 2−(π/8))](((1−(√5))/2))^n   ⇒a_n =cot {[(π/8)+(1/( (√5)))(cot^(−1) 2−(π/8))](((1+(√5))/2))^n +[(π/8)−(1/( (√5)))(cot^(−1) 2−(π/8))](((1−(√5))/2))^n }

an1anan1an+1=1+anan+1an1(anan+1)=1+anan+11+anan+1anan+1=an1letan=cotθn1+cotθncotθn+1cotθncotθn+1=cotθn1cot(θn+1θn)=cotθn1θn+1θn=θn1θn+1θnθn1=0Fibonaccir2r1=0r=1±52θn=A(1+52)n+B(152)nθ0=A+B=cot1a0=cot11=π4A+B2=π8θ1=A+B2+5(AB)2=cot1a1=cot12AB2=15(cot12π8)A=π8+15(cot12π8)B=π815(cot12π8)θn=[π8+15(cot12π8)](1+52)n+[π815(cot12π8)](152)nan=cot{[π8+15(cot12π8)](1+52)n+[π815(cot12π8)](152)n}

Commented by ajfour last updated on 24/Dec/24

(a_(n−1) /a_(n+1) )−(a_(n−1) /a_n )=(1/(a_n a_(n+1) +1))  (1/r^2 )−(1/r)=(1/(r^3 +1))  (r^3 +1)(1−r)=r^2   r^4 −r^3 +r^2 +r−1=0  (r^2 −(1/r^2 ))−(r−(1/r))=−1  say tbe two real roots are r_1  & r_2 .  a_n =Ar_1 ^n +Br_2 ^n   can we do this?  i′ve forgotten this  sequence and series topic. been 20y.

an1an+1an1an=1anan+1+11r21r=1r3+1(r3+1)(1r)=r2r4r3+r2+r1=0(r21r2)(r1r)=1saytbetworealrootsarer1&r2.an=Ar1n+Br2ncanwedothis?iveforgottenthissequenceandseriestopic.been20y.

Commented by mr W last updated on 24/Dec/24

no sir.  if we assume a_n =Ar^n   (a_(n−1) /a_(n+1) )−(a_(n−1) /a_n )=(1/(a_n a_(n+1) +1))  ⇒((Ar^(n−1) )/(Ar^(n+1) ))−((Ar^(n−1) )/(Ar^n ))=(1/(A^2 r^(2n+1) +1))  ⇒(1/r^2 )−(1/r)=(1/(A^2 r^(2n+1) +1))   !!!  we can′t find r which is independent  from n and A.  that means a_n  can not be expressed   as type Ar^n .  generally characteristic equation  exists only for linear recurrence  relations  a_n =c_1 a_(n−1) +c_2 a_(n−2) +...+c_k a_(n−k)   with c_i =constants.

nosir.ifweassumean=Arnan1an+1an1an=1anan+1+1Arn1Arn+1Arn1Arn=1A2r2n+1+11r21r=1A2r2n+1+1!!!wecantfindrwhichisindependentfromnandA.thatmeansancannotbeexpressedastypeArn.generallycharacteristicequationexistsonlyforlinearrecurrencerelationsan=c1an1+c2an2+...+ckankwithci=constants.

Commented by ajfour last updated on 24/Dec/24

Thanks, i ll revise.

Thanks,illrevise.

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