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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)) ......

$${i}\:{got} \\ $$$${a}_{{n}} =\mathrm{cot}\:\left\{\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\frac{\pi}{\mathrm{8}}\right]\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{8}}\right]\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right\} \\ $$$${which}\:{delivers}: \\ $$$${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${a}_{\mathrm{4}} =−\frac{\mathrm{11}}{\mathrm{2}} \\ $$$${a}_{\mathrm{5}} =\frac{\mathrm{3}}{\mathrm{79}} \\ $$$$...... \\ $$

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

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

$${has}\:{somebody}\:{got}\:{a}\:{more}\:{simple} \\ $$$${formula}\:{for}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}? \\ $$

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)

$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} {a}_{{n}−\mathrm{1}} −\mathrm{1}}{{a}_{{n}} +{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{{n}} ={cot}\left({b}_{{n}} \right)\Rightarrow{cot}\left({b}_{{n}+\mathrm{1}} \right)=\frac{{cot}\left({b}_{{n}} \right){cot}\left({b}_{{n}−\mathrm{1}} \right)−\mathrm{1}}{{cot}\left({b}_{{n}} \right)+\mathrm{cot}\left(\mathrm{b}_{\mathrm{n}−\mathrm{1}} \right)}= \\ $$$$\mathrm{cot}\left(\mathrm{b}_{\mathrm{n}} +\mathrm{b}_{\mathrm{n}−\mathrm{1}} \right)\Rightarrow{b}_{{n}+\mathrm{1}} ={b}_{{n}} +{b}_{{n}−\mathrm{1}} \\ $$$${b}_{{n}+\mathrm{1}} −{b}_{{n}} −{b}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${b}_{\mathrm{0}} =\frac{\pi}{\mathrm{4}};{b}_{\mathrm{1}} =\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$

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 }

$${a}_{{n}−\mathrm{1}} {a}_{{n}} −{a}_{{n}−\mathrm{1}} {a}_{{n}+\mathrm{1}} =\mathrm{1}+{a}_{{n}} {a}_{{n}+\mathrm{1}} \\ $$$${a}_{{n}−\mathrm{1}} \left({a}_{{n}} −{a}_{{n}+\mathrm{1}} \right)=\mathrm{1}+{a}_{{n}} {a}_{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}+{a}_{{n}} {a}_{{n}+\mathrm{1}} }{{a}_{{n}} −{a}_{{n}+\mathrm{1}} }={a}_{{n}−\mathrm{1}} \\ $$$${let}\:{a}_{{n}} =\mathrm{cot}\:\theta_{{n}} \\ $$$$\frac{\mathrm{1}+\mathrm{cot}\:\theta_{{n}} \mathrm{cot}\:\theta_{{n}+\mathrm{1}} }{\mathrm{cot}\:\theta_{{n}} −\mathrm{cot}\:\theta_{{n}+\mathrm{1}} }=\mathrm{cot}\:\theta_{{n}−\mathrm{1}} \\ $$$$\mathrm{cot}\:\left(\theta_{{n}+\mathrm{1}} −\theta_{{n}} \right)=\mathrm{cot}\:\theta_{{n}−\mathrm{1}} \\ $$$$\Rightarrow\theta_{{n}+\mathrm{1}} −\theta_{{n}} =\theta_{{n}−\mathrm{1}} \: \\ $$$$\Rightarrow\theta_{{n}+\mathrm{1}} −\theta_{{n}} −\theta_{{n}−\mathrm{1}} =\mathrm{0}\:\:\:\leftarrow\:{Fibonacci} \\ $$$${r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\theta_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$$\theta_{\mathrm{0}} ={A}+{B}=\mathrm{cot}^{−\mathrm{1}} {a}_{\mathrm{0}} =\mathrm{cot}^{−\mathrm{1}} \mathrm{1}=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\frac{{A}+{B}}{\mathrm{2}}=\frac{\pi}{\mathrm{8}} \\ $$$$\theta_{\mathrm{1}} =\frac{{A}+{B}}{\mathrm{2}}+\frac{\:\sqrt{\mathrm{5}}\left({A}−{B}\right)}{\mathrm{2}}=\mathrm{cot}^{−\mathrm{1}} {a}_{\mathrm{1}} =\mathrm{cot}^{−\mathrm{1}} \mathrm{2} \\ $$$$\Rightarrow\frac{\:{A}−{B}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow{A}=\frac{\pi}{\mathrm{8}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow{B}=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\theta_{{n}} =\left[\frac{\pi}{\mathrm{8}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right)\right]\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left[\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right)\right]\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$$\Rightarrow{a}_{{n}} =\mathrm{cot}\:\left\{\left[\frac{\pi}{\mathrm{8}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right)\right]\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +\left[\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{cot}^{−\mathrm{1}} \mathrm{2}−\frac{\pi}{\mathrm{8}}\right)\right]\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right\} \\ $$

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.

$$\frac{{a}_{{n}−\mathrm{1}} }{{a}_{{n}+\mathrm{1}} }−\frac{{a}_{{n}−\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{1}}{{a}_{{n}} {a}_{{n}+\mathrm{1}} +\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }−\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{r}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\left({r}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{1}−{r}\right)={r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{4}} −{r}^{\mathrm{3}} +{r}^{\mathrm{2}} +{r}−\mathrm{1}=\mathrm{0} \\ $$$$\left({r}^{\mathrm{2}} −\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\right)−\left({r}−\frac{\mathrm{1}}{{r}}\right)=−\mathrm{1} \\ $$$${say}\:{tbe}\:{two}\:{real}\:{roots}\:{are}\:{r}_{\mathrm{1}} \:\&\:{r}_{\mathrm{2}} . \\ $$$${a}_{{n}} ={Ar}_{\mathrm{1}} ^{{n}} +{Br}_{\mathrm{2}} ^{{n}} \\ $$$${can}\:{we}\:{do}\:{this}?\:\:{i}'{ve}\:{forgotten}\:{this} \\ $$$${sequence}\:{and}\:{series}\:{topic}.\:{been}\:\mathrm{20}{y}. \\ $$

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.

$${no}\:{sir}. \\ $$$${if}\:{we}\:{assume}\:{a}_{{n}} ={Ar}^{{n}} \\ $$$$\frac{{a}_{{n}−\mathrm{1}} }{{a}_{{n}+\mathrm{1}} }−\frac{{a}_{{n}−\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{1}}{{a}_{{n}} {a}_{{n}+\mathrm{1}} +\mathrm{1}} \\ $$$$\Rightarrow\frac{{Ar}^{{n}−\mathrm{1}} }{{Ar}^{{n}+\mathrm{1}} }−\frac{{Ar}^{{n}−\mathrm{1}} }{{Ar}^{{n}} }=\frac{\mathrm{1}}{{A}^{\mathrm{2}} {r}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{r}^{\mathrm{2}} }−\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{A}^{\mathrm{2}} {r}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}}\:\:\:!!! \\ $$$${we}\:{can}'{t}\:{find}\:{r}\:{which}\:{is}\:{independent} \\ $$$${from}\:{n}\:{and}\:{A}. \\ $$$${that}\:{means}\:{a}_{{n}} \:{can}\:{not}\:{be}\:{expressed}\: \\ $$$${as}\:{type}\:{Ar}^{{n}} . \\ $$$${generally}\:{characteristic}\:{equation} \\ $$$${exists}\:{only}\:{for}\:{linear}\:{recurrence} \\ $$$${relations} \\ $$$${a}_{{n}} ={c}_{\mathrm{1}} {a}_{{n}−\mathrm{1}} +{c}_{\mathrm{2}} {a}_{{n}−\mathrm{2}} +...+{c}_{{k}} {a}_{{n}−{k}} \\ $$$${with}\:{c}_{{i}} ={constants}. \\ $$

Commented by ajfour last updated on 24/Dec/24

Thanks, i ll revise.

$${Thanks},\:{i}\:{ll}\:{revise}. \\ $$

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