Finding the value of $prod_n=0^infty a_n$ with $a_0=1/2$ and $a_n=1+(a_n-1-1)^2$
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2
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Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
sequences-and-series algebra-precalculus infinite-product
edited Sep 7 at 11:15
M. Winter
18.2k62764
asked Sep 7 at 9:35
ami_ba
1,0071619
add a comment |Â
up vote
2
down vote
favorite
Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
sequences-and-series algebra-precalculus infinite-product
edited Sep 7 at 11:15
M. Winter
18.2k62764
asked Sep 7 at 9:35
ami_ba
1,0071619
Not a sum. A product. The "summation " symbol is $Pi$ instead of $Sigma$; the latter means "sum" but the former means "product".
– Oscar Lanzi
Sep 7 at 9:57
oh....thank you
– ami_ba
Sep 7 at 11:33
Looks like denominators (starting at $n=0$) are $2^2^n$ and numerators are 1 more than denominators. Just a guess... and I've no idea about the product of all these. Maybe this should be made into a comment. OK if someone wants to, o just inform me and I'll do that.
– coffeemath
Sep 8 at 1:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
sequences-and-series algebra-precalculus infinite-product
edited Sep 7 at 11:15
M. Winter
18.2k62764
asked Sep 7 at 9:35
ami_ba
1,0071619
Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
sequences-and-series algebra-precalculus infinite-product
sequences-and-series algebra-precalculus infinite-product
edited Sep 7 at 11:15
M. Winter
18.2k62764
asked Sep 7 at 9:35
ami_ba
1,0071619
edited Sep 7 at 11:15
M. Winter
18.2k62764
asked Sep 7 at 9:35
ami_ba
1,0071619
edited Sep 7 at 11:15
M. Winter
18.2k62764
edited Sep 7 at 11:15
M. Winter
18.2k62764
edited Sep 7 at 11:15
M. Winter
18.2k62764
18.2k62764
asked Sep 7 at 9:35
ami_ba
1,0071619
asked Sep 7 at 9:35
ami_ba
1,0071619
asked Sep 7 at 9:35
ami_ba
1,0071619
1,0071619
Not a sum. A product. The "summation " symbol is $Pi$ instead of $Sigma$; the latter means "sum" but the former means "product".
– Oscar Lanzi
Sep 7 at 9:57
oh....thank you
– ami_ba
Sep 7 at 11:33
Looks like denominators (starting at $n=0$) are $2^2^n$ and numerators are 1 more than denominators. Just a guess... and I've no idea about the product of all these. Maybe this should be made into a comment. OK if someone wants to, o just inform me and I'll do that.
– coffeemath
Sep 8 at 1:56
add a comment |Â
Not a sum. A product. The "summation " symbol is $Pi$ instead of $Sigma$; the latter means "sum" but the former means "product".
– Oscar Lanzi
Sep 7 at 9:57
oh....thank you
– ami_ba
Sep 7 at 11:33
Looks like denominators (starting at $n=0$) are $2^2^n$ and numerators are 1 more than denominators. Just a guess... and I've no idea about the product of all these. Maybe this should be made into a comment. OK if someone wants to, o just inform me and I'll do that.
– coffeemath
Sep 8 at 1:56
Not a sum. A product. The "summation " symbol is $Pi$ instead of $Sigma$; the latter means "sum" but the former means "product".
– Oscar Lanzi
Sep 7 at 9:57
Not a sum. A product. The "summation " symbol is $Pi$ instead of $Sigma$; the latter means "sum" but the former means "product".
– Oscar Lanzi
Sep 7 at 9:57
oh....thank you
– ami_ba
Sep 7 at 11:33
oh....thank you
– ami_ba
Sep 7 at 11:33
Looks like denominators (starting at $n=0$) are $2^2^n$ and numerators are 1 more than denominators. Just a guess... and I've no idea about the product of all these. Maybe this should be made into a comment. OK if someone wants to, o just inform me and I'll do that.
– coffeemath
Sep 8 at 1:56
Looks like denominators (starting at $n=0$) are $2^2^n$ and numerators are 1 more than denominators. Just a guess... and I've no idea about the product of all these. Maybe this should be made into a comment. OK if someone wants to, o just inform me and I'll do that.
– coffeemath
Sep 8 at 1:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
13
down vote
accepted
Let $b_n=a_n-1$ so that $b_0=-frac 12$ and $b_n=b_n-1^2$.
Consequently, $b_n=b_0^2^n$ hence
$$a_n=1+(-2)^-2^n$$
Let $x=frac 12$, then
beginalign
(1+x)prod_n=0^Na_n
&=(1+x)prod_n=0^N(1+(-x)^2^n)\
&=(1+x)(1-x)(1+x^2)cdots(1+x^2^N)\
&=(1-x^2)(1+x^2)cdots(1+x^2^N)\
&=(1-x^4)(1+x^4)cdots(1+x^2^N)\
&=(1-x^8)(1+x^8)cdots(1+x^2^N)\
&=cdots\
&=(1-x^2^N)(1+x^2^N)\
&=1-x^2^N+1\
&xrightarrowNtoinfty1
endalign
so that
$$prod_n=0^infty a_n=frac 23$$
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
2
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
Let $b_n=a_n-1$ so that $b_0=-frac 12$ and $b_n=b_n-1^2$.
Consequently, $b_n=b_0^2^n$ hence
$$a_n=1+(-2)^-2^n$$
Let $x=frac 12$, then
beginalign
(1+x)prod_n=0^Na_n
&=(1+x)prod_n=0^N(1+(-x)^2^n)\
&=(1+x)(1-x)(1+x^2)cdots(1+x^2^N)\
&=(1-x^2)(1+x^2)cdots(1+x^2^N)\
&=(1-x^4)(1+x^4)cdots(1+x^2^N)\
&=(1-x^8)(1+x^8)cdots(1+x^2^N)\
&=cdots\
&=(1-x^2^N)(1+x^2^N)\
&=1-x^2^N+1\
&xrightarrowNtoinfty1
endalign
so that
$$prod_n=0^infty a_n=frac 23$$
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
2
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
add a comment |Â
up vote
13
down vote
accepted
Let $b_n=a_n-1$ so that $b_0=-frac 12$ and $b_n=b_n-1^2$.
Consequently, $b_n=b_0^2^n$ hence
$$a_n=1+(-2)^-2^n$$
Let $x=frac 12$, then
beginalign
(1+x)prod_n=0^Na_n
&=(1+x)prod_n=0^N(1+(-x)^2^n)\
&=(1+x)(1-x)(1+x^2)cdots(1+x^2^N)\
&=(1-x^2)(1+x^2)cdots(1+x^2^N)\
&=(1-x^4)(1+x^4)cdots(1+x^2^N)\
&=(1-x^8)(1+x^8)cdots(1+x^2^N)\
&=cdots\
&=(1-x^2^N)(1+x^2^N)\
&=1-x^2^N+1\
&xrightarrowNtoinfty1
endalign
so that
$$prod_n=0^infty a_n=frac 23$$
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
2
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
add a comment |Â
up vote
13
down vote
accepted
up vote
13
down vote
accepted
Let $b_n=a_n-1$ so that $b_0=-frac 12$ and $b_n=b_n-1^2$.
Consequently, $b_n=b_0^2^n$ hence
$$a_n=1+(-2)^-2^n$$
Let $x=frac 12$, then
beginalign
(1+x)prod_n=0^Na_n
&=(1+x)prod_n=0^N(1+(-x)^2^n)\
&=(1+x)(1-x)(1+x^2)cdots(1+x^2^N)\
&=(1-x^2)(1+x^2)cdots(1+x^2^N)\
&=(1-x^4)(1+x^4)cdots(1+x^2^N)\
&=(1-x^8)(1+x^8)cdots(1+x^2^N)\
&=cdots\
&=(1-x^2^N)(1+x^2^N)\
&=1-x^2^N+1\
&xrightarrowNtoinfty1
endalign
so that
$$prod_n=0^infty a_n=frac 23$$
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
Let $b_n=a_n-1$ so that $b_0=-frac 12$ and $b_n=b_n-1^2$.
Consequently, $b_n=b_0^2^n$ hence
$$a_n=1+(-2)^-2^n$$
Let $x=frac 12$, then
beginalign
(1+x)prod_n=0^Na_n
&=(1+x)prod_n=0^N(1+(-x)^2^n)\
&=(1+x)(1-x)(1+x^2)cdots(1+x^2^N)\
&=(1-x^2)(1+x^2)cdots(1+x^2^N)\
&=(1-x^4)(1+x^4)cdots(1+x^2^N)\
&=(1-x^8)(1+x^8)cdots(1+x^2^N)\
&=cdots\
&=(1-x^2^N)(1+x^2^N)\
&=1-x^2^N+1\
&xrightarrowNtoinfty1
endalign
so that
$$prod_n=0^infty a_n=frac 23$$
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
answered Sep 7 at 9:49
Fabio Lucchini
6,38411126
6,38411126
2
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
add a comment |Â
2
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
2
2
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
Beautiful solution, for sure ! Thanks for providing it and $+1$.
– Claude Leibovici
Sep 7 at 9:53
add a comment |Â
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Not a sum. A product. The "summation " symbol is $Pi$ instead of $Sigma$; the latter means "sum" but the former means "product".
– Oscar Lanzi
Sep 7 at 9:57
oh....thank you
– ami_ba
Sep 7 at 11:33
Looks like denominators (starting at $n=0$) are $2^2^n$ and numerators are 1 more than denominators. Just a guess... and I've no idea about the product of all these. Maybe this should be made into a comment. OK if someone wants to, o just inform me and I'll do that.
– coffeemath
Sep 8 at 1:56