Solve Recurrence Relation $a_k=(a_k-1)^2-2$

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$$a_k=left(a_k-1right)^2-2$$



$a_0=frac52$



Then find $$P=prod_k=0^infty left(1-frac1a_kright)$$



My try:



I rewrote the Recurrence equation as



$$a_k+1=(a_k-1-1)(a_k-1+1)$$ $implies$



$$frac1a_k-1-1=fraca_k-1+1a_k+1$$ $implies$



$$fraca_k-1a_k-1-1=frac(a_k-1)^2+a_k-1a_k+1=fraca_k+a_k-1+2a_k+1$$



any hint here?




sequences-and-series recurrence-relations infinite-product




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  • Related: math.stackexchange.com/q/121586, math.stackexchange.com/q/781024.
    – Martin R
    Jul 19 at 9:56










  • Correct me if I'm wrong, but I think you can just substitute $a_k-1=sqrta_k+2$ into your last equation to get $fracsqrta_k+2sqrta_k+2-1=fraca_k+sqrta_k+2+1a_k+1$
    – Shrey Joshi
    Jul 19 at 13:09














up vote
4
down vote

favorite
1












$$a_k=left(a_k-1right)^2-2$$



$a_0=frac52$



Then find $$P=prod_k=0^infty left(1-frac1a_kright)$$



My try:



I rewrote the Recurrence equation as



$$a_k+1=(a_k-1-1)(a_k-1+1)$$ $implies$



$$frac1a_k-1-1=fraca_k-1+1a_k+1$$ $implies$



$$fraca_k-1a_k-1-1=frac(a_k-1)^2+a_k-1a_k+1=fraca_k+a_k-1+2a_k+1$$



any hint here?




sequences-and-series recurrence-relations infinite-product




share|cite|improve this question



















  • Related: math.stackexchange.com/q/121586, math.stackexchange.com/q/781024.
    – Martin R
    Jul 19 at 9:56










  • Correct me if I'm wrong, but I think you can just substitute $a_k-1=sqrta_k+2$ into your last equation to get $fracsqrta_k+2sqrta_k+2-1=fraca_k+sqrta_k+2+1a_k+1$
    – Shrey Joshi
    Jul 19 at 13:09












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





$$a_k=left(a_k-1right)^2-2$$



$a_0=frac52$



Then find $$P=prod_k=0^infty left(1-frac1a_kright)$$



My try:



I rewrote the Recurrence equation as



$$a_k+1=(a_k-1-1)(a_k-1+1)$$ $implies$



$$frac1a_k-1-1=fraca_k-1+1a_k+1$$ $implies$



$$fraca_k-1a_k-1-1=frac(a_k-1)^2+a_k-1a_k+1=fraca_k+a_k-1+2a_k+1$$



any hint here?




sequences-and-series recurrence-relations infinite-product




share|cite|improve this question











$$a_k=left(a_k-1right)^2-2$$



$a_0=frac52$



Then find $$P=prod_k=0^infty left(1-frac1a_kright)$$



My try:



I rewrote the Recurrence equation as



$$a_k+1=(a_k-1-1)(a_k-1+1)$$ $implies$



$$frac1a_k-1-1=fraca_k-1+1a_k+1$$ $implies$



$$fraca_k-1a_k-1-1=frac(a_k-1)^2+a_k-1a_k+1=fraca_k+a_k-1+2a_k+1$$



any hint here?





sequences-and-series recurrence-relations infinite-product





share|cite|improve this question










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asked Jul 19 at 9:41









Ekaveera Kumar Sharma

5,20111122




5,20111122











  • Related: math.stackexchange.com/q/121586, math.stackexchange.com/q/781024.
    – Martin R
    Jul 19 at 9:56










  • Correct me if I'm wrong, but I think you can just substitute $a_k-1=sqrta_k+2$ into your last equation to get $fracsqrta_k+2sqrta_k+2-1=fraca_k+sqrta_k+2+1a_k+1$
    – Shrey Joshi
    Jul 19 at 13:09
















  • Related: math.stackexchange.com/q/121586, math.stackexchange.com/q/781024.
    – Martin R
    Jul 19 at 9:56










  • Correct me if I'm wrong, but I think you can just substitute $a_k-1=sqrta_k+2$ into your last equation to get $fracsqrta_k+2sqrta_k+2-1=fraca_k+sqrta_k+2+1a_k+1$
    – Shrey Joshi
    Jul 19 at 13:09















Related: math.stackexchange.com/q/121586, math.stackexchange.com/q/781024.
– Martin R
Jul 19 at 9:56




Related: math.stackexchange.com/q/121586, math.stackexchange.com/q/781024.
– Martin R
Jul 19 at 9:56












Correct me if I'm wrong, but I think you can just substitute $a_k-1=sqrta_k+2$ into your last equation to get $fracsqrta_k+2sqrta_k+2-1=fraca_k+sqrta_k+2+1a_k+1$
– Shrey Joshi
Jul 19 at 13:09




Correct me if I'm wrong, but I think you can just substitute $a_k-1=sqrta_k+2$ into your last equation to get $fracsqrta_k+2sqrta_k+2-1=fraca_k+sqrta_k+2+1a_k+1$
– Shrey Joshi
Jul 19 at 13:09










1 Answer
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Hint: There are three steps. First, for each $k=0,1,2,ldots$, show that $$a_k=2^2^k+frac12^2^k,.$$
Second, write $$1-frac1a_k=fraca_k-1a_k=left(fraca_k+1+1a_k+1right)frac1a_k,,$$
for all $k=0,1,2,ldots$.
Finally, show that
$$prod_k=0^n,a_k=frac23left(2^2^n+1-frac12^2^n+1right),,$$ using the identity
$$(x-y),prod_k=0^n,left(x^2^k+y^2^kright)=x^2^n+1-y^2^n+1,,$$
for all $n=0,1,2,ldots$.




You should in the end obtain $$prod_k=0^infty,left(1-frac1a_kright)=lim_ntoinfty,frac37left(frac2^2^n+1+1+frac12^2^n+1 2^2^n+1-frac12^2^n+1 right)=frac37,.$$







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  • @OscarLanzi Not sure what you are talking about? Did you really read what I had written?
    – Batominovski
    Jul 19 at 14:13











  • Not what I first saw, I saw an error. Must be a bug on my end.
    – Oscar Lanzi
    Jul 19 at 20:28











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1 Answer
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up vote
4
down vote













Hint: There are three steps. First, for each $k=0,1,2,ldots$, show that $$a_k=2^2^k+frac12^2^k,.$$
Second, write $$1-frac1a_k=fraca_k-1a_k=left(fraca_k+1+1a_k+1right)frac1a_k,,$$
for all $k=0,1,2,ldots$.
Finally, show that
$$prod_k=0^n,a_k=frac23left(2^2^n+1-frac12^2^n+1right),,$$ using the identity
$$(x-y),prod_k=0^n,left(x^2^k+y^2^kright)=x^2^n+1-y^2^n+1,,$$
for all $n=0,1,2,ldots$.




You should in the end obtain $$prod_k=0^infty,left(1-frac1a_kright)=lim_ntoinfty,frac37left(frac2^2^n+1+1+frac12^2^n+1 2^2^n+1-frac12^2^n+1 right)=frac37,.$$







share|cite|improve this answer























  • @OscarLanzi Not sure what you are talking about? Did you really read what I had written?
    – Batominovski
    Jul 19 at 14:13











  • Not what I first saw, I saw an error. Must be a bug on my end.
    – Oscar Lanzi
    Jul 19 at 20:28















up vote
4
down vote













Hint: There are three steps. First, for each $k=0,1,2,ldots$, show that $$a_k=2^2^k+frac12^2^k,.$$
Second, write $$1-frac1a_k=fraca_k-1a_k=left(fraca_k+1+1a_k+1right)frac1a_k,,$$
for all $k=0,1,2,ldots$.
Finally, show that
$$prod_k=0^n,a_k=frac23left(2^2^n+1-frac12^2^n+1right),,$$ using the identity
$$(x-y),prod_k=0^n,left(x^2^k+y^2^kright)=x^2^n+1-y^2^n+1,,$$
for all $n=0,1,2,ldots$.




You should in the end obtain $$prod_k=0^infty,left(1-frac1a_kright)=lim_ntoinfty,frac37left(frac2^2^n+1+1+frac12^2^n+1 2^2^n+1-frac12^2^n+1 right)=frac37,.$$







share|cite|improve this answer























  • @OscarLanzi Not sure what you are talking about? Did you really read what I had written?
    – Batominovski
    Jul 19 at 14:13











  • Not what I first saw, I saw an error. Must be a bug on my end.
    – Oscar Lanzi
    Jul 19 at 20:28













up vote
4
down vote










up vote
4
down vote









Hint: There are three steps. First, for each $k=0,1,2,ldots$, show that $$a_k=2^2^k+frac12^2^k,.$$
Second, write $$1-frac1a_k=fraca_k-1a_k=left(fraca_k+1+1a_k+1right)frac1a_k,,$$
for all $k=0,1,2,ldots$.
Finally, show that
$$prod_k=0^n,a_k=frac23left(2^2^n+1-frac12^2^n+1right),,$$ using the identity
$$(x-y),prod_k=0^n,left(x^2^k+y^2^kright)=x^2^n+1-y^2^n+1,,$$
for all $n=0,1,2,ldots$.




You should in the end obtain $$prod_k=0^infty,left(1-frac1a_kright)=lim_ntoinfty,frac37left(frac2^2^n+1+1+frac12^2^n+1 2^2^n+1-frac12^2^n+1 right)=frac37,.$$







share|cite|improve this answer















Hint: There are three steps. First, for each $k=0,1,2,ldots$, show that $$a_k=2^2^k+frac12^2^k,.$$
Second, write $$1-frac1a_k=fraca_k-1a_k=left(fraca_k+1+1a_k+1right)frac1a_k,,$$
for all $k=0,1,2,ldots$.
Finally, show that
$$prod_k=0^n,a_k=frac23left(2^2^n+1-frac12^2^n+1right),,$$ using the identity
$$(x-y),prod_k=0^n,left(x^2^k+y^2^kright)=x^2^n+1-y^2^n+1,,$$
for all $n=0,1,2,ldots$.




You should in the end obtain $$prod_k=0^infty,left(1-frac1a_kright)=lim_ntoinfty,frac37left(frac2^2^n+1+1+frac12^2^n+1 2^2^n+1-frac12^2^n+1 right)=frac37,.$$








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edited Jul 19 at 10:01


























answered Jul 19 at 9:45









Batominovski

23.2k22777




23.2k22777











  • @OscarLanzi Not sure what you are talking about? Did you really read what I had written?
    – Batominovski
    Jul 19 at 14:13











  • Not what I first saw, I saw an error. Must be a bug on my end.
    – Oscar Lanzi
    Jul 19 at 20:28

















  • @OscarLanzi Not sure what you are talking about? Did you really read what I had written?
    – Batominovski
    Jul 19 at 14:13











  • Not what I first saw, I saw an error. Must be a bug on my end.
    – Oscar Lanzi
    Jul 19 at 20:28
















@OscarLanzi Not sure what you are talking about? Did you really read what I had written?
– Batominovski
Jul 19 at 14:13





@OscarLanzi Not sure what you are talking about? Did you really read what I had written?
– Batominovski
Jul 19 at 14:13













Not what I first saw, I saw an error. Must be a bug on my end.
– Oscar Lanzi
Jul 19 at 20:28





Not what I first saw, I saw an error. Must be a bug on my end.
– Oscar Lanzi
Jul 19 at 20:28













 

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