An equality about sin function?A gamma function identityInterpolating a sum of binomial coefficients using a...
An equality about sin function?
A gamma function identityInterpolating a sum of binomial coefficients using a sin functionIntegral inequality for convex functionWhat is known about this series?A Bessel function integral identity involving $int_0^pi frac{K_{j-1/2}(w)}{w^{j-1/2}}sin^{2p-1}(theta), dtheta$Sum of square roots of binomial coefficientsA question on the sine functionquestion about equality series containing hypergeometric term and a simple term
$begingroup$
Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
begin{equation*}
prod_{s=1}^{2n}sum_{k=1}^{2n}(-i)^ksinfrac{skpi}{2n+1}=(-1)^nfrac{2n+1}{2^n},
end{equation*}
where $i=sqrt{-1}$.
Is it a known equality? If it is true, would you please give me some insights on how to derive this equality?
ca.classical-analysis-and-odes sequences-and-series
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
$endgroup$
add a comment |
$begingroup$
Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
begin{equation*}
prod_{s=1}^{2n}sum_{k=1}^{2n}(-i)^ksinfrac{skpi}{2n+1}=(-1)^nfrac{2n+1}{2^n},
end{equation*}
where $i=sqrt{-1}$.
Is it a known equality? If it is true, would you please give me some insights on how to derive this equality?
ca.classical-analysis-and-odes sequences-and-series
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
$endgroup$
$begingroup$
Maybe it helps if you use $sin(x)=(e^{ix}-e^{-ix})/2i$. Then it seems to be related with discrete Fourier transform.
$endgroup$
– user35593
23 hours ago
$begingroup$
In possible connection with the matrix of Discrete Sine Transform Version DST-I (en.wikipedia.org/wiki/Discrete_sine_transform).
$endgroup$
– Jean Marie Becker
28 mins ago
add a comment |
$begingroup$
Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
begin{equation*}
prod_{s=1}^{2n}sum_{k=1}^{2n}(-i)^ksinfrac{skpi}{2n+1}=(-1)^nfrac{2n+1}{2^n},
end{equation*}
where $i=sqrt{-1}$.
Is it a known equality? If it is true, would you please give me some insights on how to derive this equality?
ca.classical-analysis-and-odes sequences-and-series
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
$endgroup$
Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
begin{equation*}
prod_{s=1}^{2n}sum_{k=1}^{2n}(-i)^ksinfrac{skpi}{2n+1}=(-1)^nfrac{2n+1}{2^n},
end{equation*}
where $i=sqrt{-1}$.
Is it a known equality? If it is true, would you please give me some insights on how to derive this equality?
ca.classical-analysis-and-odes sequences-and-series
ca.classical-analysis-and-odes sequences-and-series
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
edited 8 hours ago
GH from MO
62.4k5 gold badges158 silver badges238 bronze badges
62.4k5 gold badges158 silver badges238 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
asked yesterday
W. WangW. Wang
483 bronze badges
483 bronze badges
$begingroup$
Maybe it helps if you use $sin(x)=(e^{ix}-e^{-ix})/2i$. Then it seems to be related with discrete Fourier transform.
$endgroup$
– user35593
23 hours ago
$begingroup$
In possible connection with the matrix of Discrete Sine Transform Version DST-I (en.wikipedia.org/wiki/Discrete_sine_transform).
$endgroup$
– Jean Marie Becker
28 mins ago
add a comment |
$begingroup$
Maybe it helps if you use $sin(x)=(e^{ix}-e^{-ix})/2i$. Then it seems to be related with discrete Fourier transform.
$endgroup$
– user35593
23 hours ago
$begingroup$
In possible connection with the matrix of Discrete Sine Transform Version DST-I (en.wikipedia.org/wiki/Discrete_sine_transform).
$endgroup$
– Jean Marie Becker
28 mins ago
$begingroup$
Maybe it helps if you use $sin(x)=(e^{ix}-e^{-ix})/2i$. Then it seems to be related with discrete Fourier transform.
$endgroup$
– user35593
23 hours ago
$begingroup$
Maybe it helps if you use $sin(x)=(e^{ix}-e^{-ix})/2i$. Then it seems to be related with discrete Fourier transform.
$endgroup$
– user35593
23 hours ago
$begingroup$
In possible connection with the matrix of Discrete Sine Transform Version DST-I (en.wikipedia.org/wiki/Discrete_sine_transform).
$endgroup$
– Jean Marie Becker
28 mins ago
$begingroup$
In possible connection with the matrix of Discrete Sine Transform Version DST-I (en.wikipedia.org/wiki/Discrete_sine_transform).
$endgroup$
– Jean Marie Becker
28 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have
$$
sum_{k=0}^{2n}(-i)^ksinfrac{skpi}{2n+1}=frac{h(s)-h(-s)}{2i},quadtext{where}\
h(s)=sum_{k=0}^{2n}e^{i(-pi/2+frac{pi s}{2n+1})k}=frac{1-e^{-ipi(2n+1)/2+ipi s}}{1-e^{i(-pi/2+frac{pi s}{2n+1})}}=frac{1+i(-1)^{n+s}}{1+ie^{ifrac{pi s}{2n+1}}}.
$$
The numerators for $s$ and $-s$ are the same, and
$$
frac1{1+ie^{itheta}}-
frac1{1+ie^{-itheta}}=frac{2sintheta}{2icos theta}=-itantheta,
$$
so the product reads as
$$
2^{-2n}prod_{s=1}^{2n} (1+i(-1)^{n+s})tan frac{spi}{2n+1}.
$$
The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that
$$
prod_{s=1}^{2n}tan frac{spi}{2n+1}=(-1)^n(2n+1).
$$
This should be well known, and in any case it is standard:
using the formula
$$
itan theta=frac{e^{2itheta}-1}{e^{2itheta}+1}
$$
we get
$$
(-1)^nprod_{s=1}^{2n}tan frac{spi}{2n+1}=prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1},quadtext{where}, omega=e^{2pi i/(2n+1)}.
$$
We have $prod_{s=1}^{2n}(z-omega^s)=1+z+ldots+z^{2n}=:P(z)$, therefore $$prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1}=frac
{P(1)}{P(-1)}=2n+1$$
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
$endgroup$
1
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have
$$
sum_{k=0}^{2n}(-i)^ksinfrac{skpi}{2n+1}=frac{h(s)-h(-s)}{2i},quadtext{where}\
h(s)=sum_{k=0}^{2n}e^{i(-pi/2+frac{pi s}{2n+1})k}=frac{1-e^{-ipi(2n+1)/2+ipi s}}{1-e^{i(-pi/2+frac{pi s}{2n+1})}}=frac{1+i(-1)^{n+s}}{1+ie^{ifrac{pi s}{2n+1}}}.
$$
The numerators for $s$ and $-s$ are the same, and
$$
frac1{1+ie^{itheta}}-
frac1{1+ie^{-itheta}}=frac{2sintheta}{2icos theta}=-itantheta,
$$
so the product reads as
$$
2^{-2n}prod_{s=1}^{2n} (1+i(-1)^{n+s})tan frac{spi}{2n+1}.
$$
The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that
$$
prod_{s=1}^{2n}tan frac{spi}{2n+1}=(-1)^n(2n+1).
$$
This should be well known, and in any case it is standard:
using the formula
$$
itan theta=frac{e^{2itheta}-1}{e^{2itheta}+1}
$$
we get
$$
(-1)^nprod_{s=1}^{2n}tan frac{spi}{2n+1}=prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1},quadtext{where}, omega=e^{2pi i/(2n+1)}.
$$
We have $prod_{s=1}^{2n}(z-omega^s)=1+z+ldots+z^{2n}=:P(z)$, therefore $$prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1}=frac
{P(1)}{P(-1)}=2n+1$$
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
$endgroup$
1
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
add a comment |
$begingroup$
We have
$$
sum_{k=0}^{2n}(-i)^ksinfrac{skpi}{2n+1}=frac{h(s)-h(-s)}{2i},quadtext{where}\
h(s)=sum_{k=0}^{2n}e^{i(-pi/2+frac{pi s}{2n+1})k}=frac{1-e^{-ipi(2n+1)/2+ipi s}}{1-e^{i(-pi/2+frac{pi s}{2n+1})}}=frac{1+i(-1)^{n+s}}{1+ie^{ifrac{pi s}{2n+1}}}.
$$
The numerators for $s$ and $-s$ are the same, and
$$
frac1{1+ie^{itheta}}-
frac1{1+ie^{-itheta}}=frac{2sintheta}{2icos theta}=-itantheta,
$$
so the product reads as
$$
2^{-2n}prod_{s=1}^{2n} (1+i(-1)^{n+s})tan frac{spi}{2n+1}.
$$
The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that
$$
prod_{s=1}^{2n}tan frac{spi}{2n+1}=(-1)^n(2n+1).
$$
This should be well known, and in any case it is standard:
using the formula
$$
itan theta=frac{e^{2itheta}-1}{e^{2itheta}+1}
$$
we get
$$
(-1)^nprod_{s=1}^{2n}tan frac{spi}{2n+1}=prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1},quadtext{where}, omega=e^{2pi i/(2n+1)}.
$$
We have $prod_{s=1}^{2n}(z-omega^s)=1+z+ldots+z^{2n}=:P(z)$, therefore $$prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1}=frac
{P(1)}{P(-1)}=2n+1$$
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
$endgroup$
1
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
add a comment |
$begingroup$
We have
$$
sum_{k=0}^{2n}(-i)^ksinfrac{skpi}{2n+1}=frac{h(s)-h(-s)}{2i},quadtext{where}\
h(s)=sum_{k=0}^{2n}e^{i(-pi/2+frac{pi s}{2n+1})k}=frac{1-e^{-ipi(2n+1)/2+ipi s}}{1-e^{i(-pi/2+frac{pi s}{2n+1})}}=frac{1+i(-1)^{n+s}}{1+ie^{ifrac{pi s}{2n+1}}}.
$$
The numerators for $s$ and $-s$ are the same, and
$$
frac1{1+ie^{itheta}}-
frac1{1+ie^{-itheta}}=frac{2sintheta}{2icos theta}=-itantheta,
$$
so the product reads as
$$
2^{-2n}prod_{s=1}^{2n} (1+i(-1)^{n+s})tan frac{spi}{2n+1}.
$$
The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that
$$
prod_{s=1}^{2n}tan frac{spi}{2n+1}=(-1)^n(2n+1).
$$
This should be well known, and in any case it is standard:
using the formula
$$
itan theta=frac{e^{2itheta}-1}{e^{2itheta}+1}
$$
we get
$$
(-1)^nprod_{s=1}^{2n}tan frac{spi}{2n+1}=prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1},quadtext{where}, omega=e^{2pi i/(2n+1)}.
$$
We have $prod_{s=1}^{2n}(z-omega^s)=1+z+ldots+z^{2n}=:P(z)$, therefore $$prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1}=frac
{P(1)}{P(-1)}=2n+1$$
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
$endgroup$
We have
$$
sum_{k=0}^{2n}(-i)^ksinfrac{skpi}{2n+1}=frac{h(s)-h(-s)}{2i},quadtext{where}\
h(s)=sum_{k=0}^{2n}e^{i(-pi/2+frac{pi s}{2n+1})k}=frac{1-e^{-ipi(2n+1)/2+ipi s}}{1-e^{i(-pi/2+frac{pi s}{2n+1})}}=frac{1+i(-1)^{n+s}}{1+ie^{ifrac{pi s}{2n+1}}}.
$$
The numerators for $s$ and $-s$ are the same, and
$$
frac1{1+ie^{itheta}}-
frac1{1+ie^{-itheta}}=frac{2sintheta}{2icos theta}=-itantheta,
$$
so the product reads as
$$
2^{-2n}prod_{s=1}^{2n} (1+i(-1)^{n+s})tan frac{spi}{2n+1}.
$$
The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that
$$
prod_{s=1}^{2n}tan frac{spi}{2n+1}=(-1)^n(2n+1).
$$
This should be well known, and in any case it is standard:
using the formula
$$
itan theta=frac{e^{2itheta}-1}{e^{2itheta}+1}
$$
we get
$$
(-1)^nprod_{s=1}^{2n}tan frac{spi}{2n+1}=prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1},quadtext{where}, omega=e^{2pi i/(2n+1)}.
$$
We have $prod_{s=1}^{2n}(z-omega^s)=1+z+ldots+z^{2n}=:P(z)$, therefore $$prod_{s=1}^{2n}
frac{omega^s-1}{omega^s+1}=frac
{P(1)}{P(-1)}=2n+1$$
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
edited 9 hours ago
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
answered 22 hours ago
Fedor PetrovFedor Petrov
55.4k6 gold badges132 silver badges252 bronze badges
55.4k6 gold badges132 silver badges252 bronze badges
1
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
add a comment |
1
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
1
1
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
$begingroup$
Thank you very much for your quick and detailed reply. The proof is extremely beautiful.
$endgroup$
– W. Wang
15 hours ago
add a comment |
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$begingroup$
Maybe it helps if you use $sin(x)=(e^{ix}-e^{-ix})/2i$. Then it seems to be related with discrete Fourier transform.
$endgroup$
– user35593
23 hours ago
$begingroup$
In possible connection with the matrix of Discrete Sine Transform Version DST-I (en.wikipedia.org/wiki/Discrete_sine_transform).
$endgroup$
– Jean Marie Becker
28 mins ago