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What is the value of this product: $prod_n=1^infty ;frac31+2 cos(fracpi3^n) ;=;? $
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emphasized text$$left[, prod_n=1^infty ;frac31+2 cosleft(fracpi3^nright), right], =;? $$
Where $;[, .];$ denotes the integral part function.
$mathbf My Attempt$
I tried to confine the n-th term to get
$$frac12 le cosleft(fracpi3^nright) lt 1$$
$$ 1 lt ;frac31+2 cosleft(fracpi3^nright); le frac32 $$
But this didn't help as the product now is bounded below but unbounded above $rightarrow infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.
I tried also to bound product from above using
$$left(fracpi3^nright) lt left(fracpi2^nright)$$
$$cosleft(fracpi3^nright) gt cosleft(fracpi2^nright) quad(cos x text is decreasing on ]0, fracpi3]
)$$
And use the Telescoping product
$$cosleft(fracpi2^nright)=fracsinleft(fracpi2^n-1right)2sinleft(fracpi2^nright)$$
But this doesn't help much.
Any hint?
floor-function infinite-product
edited Jul 31 at 0:20
amWhy
189k25219431
asked Jul 15 at 17:34
Wolfdale
24919
add a comment |Â
up vote
5
down vote
favorite
emphasized text$$left[, prod_n=1^infty ;frac31+2 cosleft(fracpi3^nright), right], =;? $$
Where $;[, .];$ denotes the integral part function.
$mathbf My Attempt$
I tried to confine the n-th term to get
$$frac12 le cosleft(fracpi3^nright) lt 1$$
$$ 1 lt ;frac31+2 cosleft(fracpi3^nright); le frac32 $$
But this didn't help as the product now is bounded below but unbounded above $rightarrow infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.
I tried also to bound product from above using
$$left(fracpi3^nright) lt left(fracpi2^nright)$$
$$cosleft(fracpi3^nright) gt cosleft(fracpi2^nright) quad(cos x text is decreasing on ]0, fracpi3]
)$$
And use the Telescoping product
$$cosleft(fracpi2^nright)=fracsinleft(fracpi2^n-1right)2sinleft(fracpi2^nright)$$
But this doesn't help much.
Any hint?
floor-function infinite-product
edited Jul 31 at 0:20
amWhy
189k25219431
asked Jul 15 at 17:34
Wolfdale
24919
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
emphasized text$$left[, prod_n=1^infty ;frac31+2 cosleft(fracpi3^nright), right], =;? $$
Where $;[, .];$ denotes the integral part function.
$mathbf My Attempt$
I tried to confine the n-th term to get
$$frac12 le cosleft(fracpi3^nright) lt 1$$
$$ 1 lt ;frac31+2 cosleft(fracpi3^nright); le frac32 $$
But this didn't help as the product now is bounded below but unbounded above $rightarrow infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.
I tried also to bound product from above using
$$left(fracpi3^nright) lt left(fracpi2^nright)$$
$$cosleft(fracpi3^nright) gt cosleft(fracpi2^nright) quad(cos x text is decreasing on ]0, fracpi3]
)$$
And use the Telescoping product
$$cosleft(fracpi2^nright)=fracsinleft(fracpi2^n-1right)2sinleft(fracpi2^nright)$$
But this doesn't help much.
Any hint?
floor-function infinite-product
edited Jul 31 at 0:20
amWhy
189k25219431
asked Jul 15 at 17:34
Wolfdale
24919
emphasized text$$left[, prod_n=1^infty ;frac31+2 cosleft(fracpi3^nright), right], =;? $$
Where $;[, .];$ denotes the integral part function.
$mathbf My Attempt$
I tried to confine the n-th term to get
$$frac12 le cosleft(fracpi3^nright) lt 1$$
$$ 1 lt ;frac31+2 cosleft(fracpi3^nright); le frac32 $$
But this didn't help as the product now is bounded below but unbounded above $rightarrow infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.
I tried also to bound product from above using
$$left(fracpi3^nright) lt left(fracpi2^nright)$$
$$cosleft(fracpi3^nright) gt cosleft(fracpi2^nright) quad(cos x text is decreasing on ]0, fracpi3]
)$$
And use the Telescoping product
$$cosleft(fracpi2^nright)=fracsinleft(fracpi2^n-1right)2sinleft(fracpi2^nright)$$
But this doesn't help much.
Any hint?
floor-function infinite-product
edited Jul 31 at 0:20
amWhy
189k25219431
asked Jul 15 at 17:34
Wolfdale
24919
edited Jul 31 at 0:20
amWhy
189k25219431
edited Jul 31 at 0:20
amWhy
189k25219431
edited Jul 31 at 0:20
amWhy
189k25219431
189k25219431
asked Jul 15 at 17:34
Wolfdale
24919
asked Jul 15 at 17:34
Wolfdale
24919
asked Jul 15 at 17:34
Wolfdale
24919
24919
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
$$1+2cosalpha=e^ialpha+1+e^-ialpha=frace^3ialpha/2
-e^-3ialpha/2e^ialpha/2-e^ialpha/2=fracsin3alpha/2
sinalpha/2.$$
Therefore
$$prod_n=1^Nfrac31+2cos(pi/3^n)
=frac3^Nsin(3^-Npi/2)sin(pi/2)tofracpi2$$
as $Ntoinfty$.
The integer part of $pi/2$ is $1$.
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
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up vote
2
down vote
$$1+2cos2x=1+2(1-2sin^2x)=dfracsin3xsin x=dfracf(n+1)f(n)$$
where $f(m)=sin(3^mx)$
Here $3^mx=?$
Related:$cos x(2cos2x-1)=cos3x$
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
$$1+2cosalpha=e^ialpha+1+e^-ialpha=frace^3ialpha/2
-e^-3ialpha/2e^ialpha/2-e^ialpha/2=fracsin3alpha/2
sinalpha/2.$$
Therefore
$$prod_n=1^Nfrac31+2cos(pi/3^n)
=frac3^Nsin(3^-Npi/2)sin(pi/2)tofracpi2$$
as $Ntoinfty$.
The integer part of $pi/2$ is $1$.
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
add a comment |Â
up vote
6
down vote
accepted
$$1+2cosalpha=e^ialpha+1+e^-ialpha=frace^3ialpha/2
-e^-3ialpha/2e^ialpha/2-e^ialpha/2=fracsin3alpha/2
sinalpha/2.$$
Therefore
$$prod_n=1^Nfrac31+2cos(pi/3^n)
=frac3^Nsin(3^-Npi/2)sin(pi/2)tofracpi2$$
as $Ntoinfty$.
The integer part of $pi/2$ is $1$.
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$$1+2cosalpha=e^ialpha+1+e^-ialpha=frace^3ialpha/2
-e^-3ialpha/2e^ialpha/2-e^ialpha/2=fracsin3alpha/2
sinalpha/2.$$
Therefore
$$prod_n=1^Nfrac31+2cos(pi/3^n)
=frac3^Nsin(3^-Npi/2)sin(pi/2)tofracpi2$$
as $Ntoinfty$.
The integer part of $pi/2$ is $1$.
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
$$1+2cosalpha=e^ialpha+1+e^-ialpha=frace^3ialpha/2
-e^-3ialpha/2e^ialpha/2-e^ialpha/2=fracsin3alpha/2
sinalpha/2.$$
Therefore
$$prod_n=1^Nfrac31+2cos(pi/3^n)
=frac3^Nsin(3^-Npi/2)sin(pi/2)tofracpi2$$
as $Ntoinfty$.
The integer part of $pi/2$ is $1$.
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
answered Jul 15 at 17:43
Lord Shark the Unknown
85.7k951112
85.7k951112
add a comment |Â
add a comment |Â
up vote
2
down vote
$$1+2cos2x=1+2(1-2sin^2x)=dfracsin3xsin x=dfracf(n+1)f(n)$$
where $f(m)=sin(3^mx)$
Here $3^mx=?$
Related:$cos x(2cos2x-1)=cos3x$
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
add a comment |Â
up vote
2
down vote
$$1+2cos2x=1+2(1-2sin^2x)=dfracsin3xsin x=dfracf(n+1)f(n)$$
where $f(m)=sin(3^mx)$
Here $3^mx=?$
Related:$cos x(2cos2x-1)=cos3x$
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$1+2cos2x=1+2(1-2sin^2x)=dfracsin3xsin x=dfracf(n+1)f(n)$$
where $f(m)=sin(3^mx)$
Here $3^mx=?$
Related:$cos x(2cos2x-1)=cos3x$
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
$$1+2cos2x=1+2(1-2sin^2x)=dfracsin3xsin x=dfracf(n+1)f(n)$$
where $f(m)=sin(3^mx)$
Here $3^mx=?$
Related:$cos x(2cos2x-1)=cos3x$
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
answered Jul 15 at 18:09
lab bhattacharjee
215k14152264
215k14152264
add a comment |Â
add a comment |Â
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