Mixing
On calculating the limit of the infinite product $prod_{k=3}^n (1-tan^4frac{pi}{2^k})$
$begingroup$
Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?
What I attempted:-
$log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.
Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$
Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $
Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.
Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
sequences-and-series
edited Jan 3 at 16:52
Robert Z
94.5k1063134
asked Jan 3 at 15:39
BhargobBhargob
663415
$endgroup$
add a comment |
$begingroup$
Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?
What I attempted:-
$log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.
Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$
Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $
Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.
Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
sequences-and-series
edited Jan 3 at 16:52
Robert Z
94.5k1063134
asked Jan 3 at 15:39
BhargobBhargob
663415
$endgroup$
add a comment |
$begingroup$
Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?
What I attempted:-
$log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.
Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$
Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $
Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.
Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
sequences-and-series
edited Jan 3 at 16:52
Robert Z
94.5k1063134
asked Jan 3 at 15:39
BhargobBhargob
663415
$endgroup$
Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?
What I attempted:-
$log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.
Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$
Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $
Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.
Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
sequences-and-series
sequences-and-series
edited Jan 3 at 16:52
Robert Z
94.5k1063134
asked Jan 3 at 15:39
BhargobBhargob
663415
edited Jan 3 at 16:52
Robert Z
94.5k1063134
asked Jan 3 at 15:39
BhargobBhargob
663415
edited Jan 3 at 16:52
Robert Z
94.5k1063134
edited Jan 3 at 16:52
Robert Z
94.5k1063134
edited Jan 3 at 16:52
Robert Z
94.5k1063134
94.5k1063134
asked Jan 3 at 15:39
BhargobBhargob
663415
asked Jan 3 at 15:39
BhargobBhargob
663415
asked Jan 3 at 15:39
BhargobBhargob
663415
663415
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
Hence, as $n$ goes to infinity,
$$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
where we used the known fact that
$$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
frac{1}{2^{n-1}sin(pi/2^n)}$$
(see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
$endgroup$
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
add a comment |
$begingroup$
What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$begin{split}
S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
&=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
&=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
&=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
&= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
&=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
end{split}$$
Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
you can verify that, for any $0<theta<pi$,
$$
prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
$$
Which yields, for $theta=frac pi 4$,
$$
prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
$$
So finally,
$$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
In other words,
$$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
Hence, as $n$ goes to infinity,
$$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
where we used the known fact that
$$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
frac{1}{2^{n-1}sin(pi/2^n)}$$
(see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
$endgroup$
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
add a comment |
$begingroup$
You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
Hence, as $n$ goes to infinity,
$$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
where we used the known fact that
$$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
frac{1}{2^{n-1}sin(pi/2^n)}$$
(see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
$endgroup$
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
add a comment |
$begingroup$
You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
Hence, as $n$ goes to infinity,
$$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
where we used the known fact that
$$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
frac{1}{2^{n-1}sin(pi/2^n)}$$
(see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
$endgroup$
You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
Hence, as $n$ goes to infinity,
$$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
where we used the known fact that
$$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
frac{1}{2^{n-1}sin(pi/2^n)}$$
(see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
edited Jan 3 at 16:51
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
answered Jan 3 at 15:56
Robert ZRobert Z
94.5k1063134
94.5k1063134
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
add a comment |
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
$begingroup$
Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
$endgroup$
– Robert Z
Jan 3 at 16:50
add a comment |
$begingroup$
What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$begin{split}
S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
&=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
&=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
&=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
&= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
&=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
end{split}$$
Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
you can verify that, for any $0<theta<pi$,
$$
prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
$$
Which yields, for $theta=frac pi 4$,
$$
prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
$$
So finally,
$$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
In other words,
$$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
$endgroup$
add a comment |
$begingroup$
What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$begin{split}
S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
&=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
&=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
&=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
&= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
&=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
end{split}$$
Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
you can verify that, for any $0<theta<pi$,
$$
prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
$$
Which yields, for $theta=frac pi 4$,
$$
prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
$$
So finally,
$$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
In other words,
$$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
$endgroup$
add a comment |
$begingroup$
What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$begin{split}
S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
&=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
&=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
&=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
&= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
&=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
end{split}$$
Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
you can verify that, for any $0<theta<pi$,
$$
prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
$$
Which yields, for $theta=frac pi 4$,
$$
prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
$$
So finally,
$$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
In other words,
$$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
$endgroup$
What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$begin{split}
S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
&=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
&=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
&=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
&= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
&=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
end{split}$$
Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
you can verify that, for any $0<theta<pi$,
$$
prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
$$
Which yields, for $theta=frac pi 4$,
$$
prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
$$
So finally,
$$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
In other words,
$$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
edited Jan 3 at 16:18
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
answered Jan 3 at 16:12
Stefan LafonStefan Lafon
1,07616
1,07616
add a comment |
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
