Generalizing a Trigonometric Infinite Product of Vieta
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The second exercise in "Statistical Independence in Probability, Analysis and Number Theory," by Mark Kac is to prove that $$
sin xover x=prod_k=1^inftyfrac13left(1+2cos2xover3^kright)tag1
$$
and generalize it. This is a generalization of Vieta's formula $$
sin xover x=prod_k=1^inftycosxover 2^ktag2
$$
which is proved in the text.
It's not hard to prove $(1)$. You just write $sin x = sin(xover3+2xover3)$ and plug away, but I'm having trouble seeing what the generalization is supposed to be. The only thing I've been able to come up with for the next case is $$
sin xover x=prod_k=1^infty
frac12left(cosxover4^k+cos3xover4^kright)tag3
$$
Again, this isn't hard to prove, but I'm having trouble seeing a pattern in $(1), (2), textand (3).$ To derive $(3)$ I used the triple angle formula for cosine, and it seems to me that as you go forward you'll need multiple angle formulas for increasing large multiples, so I foresee a lot of complication.
Can you see a formula for the $n=4$ case that is more clearly a generalization of $(1)$ than $(3)$ is? Or do you know what generalization Kac had in mind? I'm assuming that he intends for you to come up with a formula for general $n$.
trigonometry infinite-product
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The second exercise in "Statistical Independence in Probability, Analysis and Number Theory," by Mark Kac is to prove that $$
sin xover x=prod_k=1^inftyfrac13left(1+2cos2xover3^kright)tag1
$$
and generalize it. This is a generalization of Vieta's formula $$
sin xover x=prod_k=1^inftycosxover 2^ktag2
$$
which is proved in the text.
It's not hard to prove $(1)$. You just write $sin x = sin(xover3+2xover3)$ and plug away, but I'm having trouble seeing what the generalization is supposed to be. The only thing I've been able to come up with for the next case is $$
sin xover x=prod_k=1^infty
frac12left(cosxover4^k+cos3xover4^kright)tag3
$$
Again, this isn't hard to prove, but I'm having trouble seeing a pattern in $(1), (2), textand (3).$ To derive $(3)$ I used the triple angle formula for cosine, and it seems to me that as you go forward you'll need multiple angle formulas for increasing large multiples, so I foresee a lot of complication.
Can you see a formula for the $n=4$ case that is more clearly a generalization of $(1)$ than $(3)$ is? Or do you know what generalization Kac had in mind? I'm assuming that he intends for you to come up with a formula for general $n$.
trigonometry infinite-product
2
I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help?
– mjqxxxx
Jul 28 at 5:07
@mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/pi,$ I proved it by converting to a Riemann sum. Thanks.
– saulspatz
Jul 28 at 5:13
I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way.
– copper.hat
Jul 28 at 5:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The second exercise in "Statistical Independence in Probability, Analysis and Number Theory," by Mark Kac is to prove that $$
sin xover x=prod_k=1^inftyfrac13left(1+2cos2xover3^kright)tag1
$$
and generalize it. This is a generalization of Vieta's formula $$
sin xover x=prod_k=1^inftycosxover 2^ktag2
$$
which is proved in the text.
It's not hard to prove $(1)$. You just write $sin x = sin(xover3+2xover3)$ and plug away, but I'm having trouble seeing what the generalization is supposed to be. The only thing I've been able to come up with for the next case is $$
sin xover x=prod_k=1^infty
frac12left(cosxover4^k+cos3xover4^kright)tag3
$$
Again, this isn't hard to prove, but I'm having trouble seeing a pattern in $(1), (2), textand (3).$ To derive $(3)$ I used the triple angle formula for cosine, and it seems to me that as you go forward you'll need multiple angle formulas for increasing large multiples, so I foresee a lot of complication.
Can you see a formula for the $n=4$ case that is more clearly a generalization of $(1)$ than $(3)$ is? Or do you know what generalization Kac had in mind? I'm assuming that he intends for you to come up with a formula for general $n$.
trigonometry infinite-product
The second exercise in "Statistical Independence in Probability, Analysis and Number Theory," by Mark Kac is to prove that $$
sin xover x=prod_k=1^inftyfrac13left(1+2cos2xover3^kright)tag1
$$
and generalize it. This is a generalization of Vieta's formula $$
sin xover x=prod_k=1^inftycosxover 2^ktag2
$$
which is proved in the text.
It's not hard to prove $(1)$. You just write $sin x = sin(xover3+2xover3)$ and plug away, but I'm having trouble seeing what the generalization is supposed to be. The only thing I've been able to come up with for the next case is $$
sin xover x=prod_k=1^infty
frac12left(cosxover4^k+cos3xover4^kright)tag3
$$
Again, this isn't hard to prove, but I'm having trouble seeing a pattern in $(1), (2), textand (3).$ To derive $(3)$ I used the triple angle formula for cosine, and it seems to me that as you go forward you'll need multiple angle formulas for increasing large multiples, so I foresee a lot of complication.
Can you see a formula for the $n=4$ case that is more clearly a generalization of $(1)$ than $(3)$ is? Or do you know what generalization Kac had in mind? I'm assuming that he intends for you to come up with a formula for general $n$.
trigonometry infinite-product
asked Jul 28 at 5:04


saulspatz
10.4k21323
10.4k21323
2
I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help?
– mjqxxxx
Jul 28 at 5:07
@mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/pi,$ I proved it by converting to a Riemann sum. Thanks.
– saulspatz
Jul 28 at 5:13
I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way.
– copper.hat
Jul 28 at 5:36
add a comment |Â
2
I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help?
– mjqxxxx
Jul 28 at 5:07
@mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/pi,$ I proved it by converting to a Riemann sum. Thanks.
– saulspatz
Jul 28 at 5:13
I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way.
– copper.hat
Jul 28 at 5:36
2
2
I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help?
– mjqxxxx
Jul 28 at 5:07
I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help?
– mjqxxxx
Jul 28 at 5:07
@mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/pi,$ I proved it by converting to a Riemann sum. Thanks.
– saulspatz
Jul 28 at 5:13
@mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/pi,$ I proved it by converting to a Riemann sum. Thanks.
– saulspatz
Jul 28 at 5:13
I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way.
– copper.hat
Jul 28 at 5:36
I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way.
– copper.hat
Jul 28 at 5:36
add a comment |Â
1 Answer
1
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votes
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You want to find an expression for
$ a_n(x) := sin(nx)/sin(x) $
written in terms of cosines of multiple angles. The first examples are:
$$ a_2(x) = 2cos(x), $$ $$ a_3(x) = 1 + 2cos(2x), $$
$$ a_4(x) = 2(cos(x) + cos(3x)), $$ $$ a_5(x) = 1 + 2cos(2x) + 2cos(4x), $$
$$ a_6(x) = 2(cos(x) + cos(3x) + cos(5x)). $$ The pattern is now obvious.
The general infinite product is:
$$ fracsin xx = prod_k=1^ infty frac1n a_nBig(fracxn^kBig). $$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You want to find an expression for
$ a_n(x) := sin(nx)/sin(x) $
written in terms of cosines of multiple angles. The first examples are:
$$ a_2(x) = 2cos(x), $$ $$ a_3(x) = 1 + 2cos(2x), $$
$$ a_4(x) = 2(cos(x) + cos(3x)), $$ $$ a_5(x) = 1 + 2cos(2x) + 2cos(4x), $$
$$ a_6(x) = 2(cos(x) + cos(3x) + cos(5x)). $$ The pattern is now obvious.
The general infinite product is:
$$ fracsin xx = prod_k=1^ infty frac1n a_nBig(fracxn^kBig). $$
add a comment |Â
up vote
1
down vote
accepted
You want to find an expression for
$ a_n(x) := sin(nx)/sin(x) $
written in terms of cosines of multiple angles. The first examples are:
$$ a_2(x) = 2cos(x), $$ $$ a_3(x) = 1 + 2cos(2x), $$
$$ a_4(x) = 2(cos(x) + cos(3x)), $$ $$ a_5(x) = 1 + 2cos(2x) + 2cos(4x), $$
$$ a_6(x) = 2(cos(x) + cos(3x) + cos(5x)). $$ The pattern is now obvious.
The general infinite product is:
$$ fracsin xx = prod_k=1^ infty frac1n a_nBig(fracxn^kBig). $$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You want to find an expression for
$ a_n(x) := sin(nx)/sin(x) $
written in terms of cosines of multiple angles. The first examples are:
$$ a_2(x) = 2cos(x), $$ $$ a_3(x) = 1 + 2cos(2x), $$
$$ a_4(x) = 2(cos(x) + cos(3x)), $$ $$ a_5(x) = 1 + 2cos(2x) + 2cos(4x), $$
$$ a_6(x) = 2(cos(x) + cos(3x) + cos(5x)). $$ The pattern is now obvious.
The general infinite product is:
$$ fracsin xx = prod_k=1^ infty frac1n a_nBig(fracxn^kBig). $$
You want to find an expression for
$ a_n(x) := sin(nx)/sin(x) $
written in terms of cosines of multiple angles. The first examples are:
$$ a_2(x) = 2cos(x), $$ $$ a_3(x) = 1 + 2cos(2x), $$
$$ a_4(x) = 2(cos(x) + cos(3x)), $$ $$ a_5(x) = 1 + 2cos(2x) + 2cos(4x), $$
$$ a_6(x) = 2(cos(x) + cos(3x) + cos(5x)). $$ The pattern is now obvious.
The general infinite product is:
$$ fracsin xx = prod_k=1^ infty frac1n a_nBig(fracxn^kBig). $$
edited Jul 28 at 11:57
answered Jul 28 at 11:48


Somos
11k1831
11k1831
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2
I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help?
– mjqxxxx
Jul 28 at 5:07
@mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/pi,$ I proved it by converting to a Riemann sum. Thanks.
– saulspatz
Jul 28 at 5:13
I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way.
– copper.hat
Jul 28 at 5:36