Prove that $prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =frac {sqrt {2m+1}}{2^m}$
$begingroup$
Prove that $$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =frac {sqrt {2m+1}}{2^m}$$
My try:
$$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =prod_{r=1}^m left(frac {e^{frac {irpi}{2m+1}} -e^{frac {-irpi}{2m+1}}}{2i}right) =frac {1}{2^mi^m}left(prod_{r=1}^m e^{frac {-irpi}{2m+1}}right) prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$
But I can't get a clue to tackle $$prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$ . I tried to use it's relation with roots of unity but couldn't proceed much.
I also tried writing the $sin$ using Euler's reflection formula that for $zin (0,1)$ $$sin (pi z)=frac {pi}{Gamma(z)Gamma(1-z)}$$ where in our case $z=frac {r}{2m+1}$. But that too didn't last long.
Any help is greatly appreciated. Thanks.
calculus sequences-and-series trigonometry complex-numbers gamma-function
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
$endgroup$
add a comment |
$begingroup$
Prove that $$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =frac {sqrt {2m+1}}{2^m}$$
My try:
$$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =prod_{r=1}^m left(frac {e^{frac {irpi}{2m+1}} -e^{frac {-irpi}{2m+1}}}{2i}right) =frac {1}{2^mi^m}left(prod_{r=1}^m e^{frac {-irpi}{2m+1}}right) prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$
But I can't get a clue to tackle $$prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$ . I tried to use it's relation with roots of unity but couldn't proceed much.
I also tried writing the $sin$ using Euler's reflection formula that for $zin (0,1)$ $$sin (pi z)=frac {pi}{Gamma(z)Gamma(1-z)}$$ where in our case $z=frac {r}{2m+1}$. But that too didn't last long.
Any help is greatly appreciated. Thanks.
calculus sequences-and-series trigonometry complex-numbers gamma-function
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
$endgroup$
3
$begingroup$
Check out the answer here.
$endgroup$
– MisterRiemann
Dec 8 '18 at 12:03
add a comment |
$begingroup$
Prove that $$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =frac {sqrt {2m+1}}{2^m}$$
My try:
$$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =prod_{r=1}^m left(frac {e^{frac {irpi}{2m+1}} -e^{frac {-irpi}{2m+1}}}{2i}right) =frac {1}{2^mi^m}left(prod_{r=1}^m e^{frac {-irpi}{2m+1}}right) prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$
But I can't get a clue to tackle $$prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$ . I tried to use it's relation with roots of unity but couldn't proceed much.
I also tried writing the $sin$ using Euler's reflection formula that for $zin (0,1)$ $$sin (pi z)=frac {pi}{Gamma(z)Gamma(1-z)}$$ where in our case $z=frac {r}{2m+1}$. But that too didn't last long.
Any help is greatly appreciated. Thanks.
calculus sequences-and-series trigonometry complex-numbers gamma-function
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
$endgroup$
Prove that $$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =frac {sqrt {2m+1}}{2^m}$$
My try:
$$prod_{r=1}^m sin left( frac {rpi}{2m+1}right) =prod_{r=1}^m left(frac {e^{frac {irpi}{2m+1}} -e^{frac {-irpi}{2m+1}}}{2i}right) =frac {1}{2^mi^m}left(prod_{r=1}^m e^{frac {-irpi}{2m+1}}right) prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$
But I can't get a clue to tackle $$prod_{r=1}^m left(e^{frac {2irpi}{2m+1}} -1right)$$ . I tried to use it's relation with roots of unity but couldn't proceed much.
I also tried writing the $sin$ using Euler's reflection formula that for $zin (0,1)$ $$sin (pi z)=frac {pi}{Gamma(z)Gamma(1-z)}$$ where in our case $z=frac {r}{2m+1}$. But that too didn't last long.
Any help is greatly appreciated. Thanks.
calculus sequences-and-series trigonometry complex-numbers gamma-function
calculus sequences-and-series trigonometry complex-numbers gamma-function
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
asked Dec 8 '18 at 11:55
DarkraiDarkrai
6,4121442
6,4121442
3
$begingroup$
Check out the answer here.
$endgroup$
– MisterRiemann
Dec 8 '18 at 12:03
add a comment |
3
$begingroup$
Check out the answer here.
$endgroup$
– MisterRiemann
Dec 8 '18 at 12:03
3
3
$begingroup$
Check out the answer here.
$endgroup$
– MisterRiemann
Dec 8 '18 at 12:03
$begingroup$
Check out the answer here.
$endgroup$
– MisterRiemann
Dec 8 '18 at 12:03
add a comment |
1 Answer
1
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$begingroup$
Hint.
Calling
$$
P = prod_{r=1}^m sin left( frac {rpi}{2m+1}right)
$$ then
$$
P^2 = prod_{r=1}^{2m} sin left( frac {rpi}{2m+1}right)
$$
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
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1 Answer
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1 Answer
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$begingroup$
Hint.
Calling
$$
P = prod_{r=1}^m sin left( frac {rpi}{2m+1}right)
$$ then
$$
P^2 = prod_{r=1}^{2m} sin left( frac {rpi}{2m+1}right)
$$
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
$endgroup$
add a comment |
$begingroup$
Hint.
Calling
$$
P = prod_{r=1}^m sin left( frac {rpi}{2m+1}right)
$$ then
$$
P^2 = prod_{r=1}^{2m} sin left( frac {rpi}{2m+1}right)
$$
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
$endgroup$
add a comment |
$begingroup$
Hint.
Calling
$$
P = prod_{r=1}^m sin left( frac {rpi}{2m+1}right)
$$ then
$$
P^2 = prod_{r=1}^{2m} sin left( frac {rpi}{2m+1}right)
$$
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
$endgroup$
Hint.
Calling
$$
P = prod_{r=1}^m sin left( frac {rpi}{2m+1}right)
$$ then
$$
P^2 = prod_{r=1}^{2m} sin left( frac {rpi}{2m+1}right)
$$
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
answered Dec 8 '18 at 13:28
CesareoCesareo
9,3613517
9,3613517
add a comment |
add a comment |
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– MisterRiemann
Dec 8 '18 at 12:03