Proving $int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$
using the following result
$int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $
Prove
$int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$
I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.
trigonometry binomial-theorem cauchy-integral-formula
edited Nov 24 at 15:25
Bernard
118k639112
asked Nov 24 at 13:17
frankfields
165
add a comment |
using the following result
$int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $
Prove
$int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$
I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.
trigonometry binomial-theorem cauchy-integral-formula
edited Nov 24 at 15:25
Bernard
118k639112
asked Nov 24 at 13:17
frankfields
165
add a comment |
using the following result
$int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $
Prove
$int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$
I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.
trigonometry binomial-theorem cauchy-integral-formula
edited Nov 24 at 15:25
Bernard
118k639112
asked Nov 24 at 13:17
frankfields
165
using the following result
$int _{gamma}(z+ frac{1}{z})^{2n}frac{dz}{z}= {2n choose {n}}2pi i $
Prove
$int ^{2pi} _{0} (cos(t)^{2n})={2n choose {n}}frac{2pi}{2^{2n}}$
I cant see how the second part can be put in this form is there any suggestions? Also used Cauchy integral formula and binomial expansion to prove first part.
trigonometry binomial-theorem cauchy-integral-formula
trigonometry binomial-theorem cauchy-integral-formula
edited Nov 24 at 15:25
Bernard
118k639112
asked Nov 24 at 13:17
frankfields
165
edited Nov 24 at 15:25
Bernard
118k639112
asked Nov 24 at 13:17
frankfields
165
edited Nov 24 at 15:25
Bernard
118k639112
edited Nov 24 at 15:25
Bernard
118k639112
edited Nov 24 at 15:25
Bernard
118k639112
118k639112
asked Nov 24 at 13:17
frankfields
165
asked Nov 24 at 13:17
frankfields
165
asked Nov 24 at 13:17
frankfields
165
165
add a comment |
add a comment |
1 Answer
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Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$
And the rest follows.
answered Nov 24 at 13:26
John_Wick
1,356111
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$
And the rest follows.
answered Nov 24 at 13:26
John_Wick
1,356111
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
add a comment |
Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$
And the rest follows.
answered Nov 24 at 13:26
John_Wick
1,356111
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
add a comment |
Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$
And the rest follows.
answered Nov 24 at 13:26
John_Wick
1,356111
Take $gamma={(x,y)|x^2+y^2=1}.$ Then $int_gamma (z+1/z)^{2n}(1/z)dz=int_0^{2pi}(2cos t)^{2n}(cos t-isin t)(-sin t+icos t)dt=iint_0^{2pi}(2cos t)^{2n}dt={2nchoose n}2pi i.$
And the rest follows.
answered Nov 24 at 13:26
John_Wick
1,356111
answered Nov 24 at 13:26
John_Wick
1,356111
answered Nov 24 at 13:26
John_Wick
1,356111
answered Nov 24 at 13:26
John_Wick
1,356111
1,356111
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
add a comment |
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
It's line integral, we are integrating over a curve where we have a parametric form of the curve $gamma={cos t+isin t : 0leq t<2pi}.$
– John_Wick
Nov 24 at 14:11
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
@John_Wick, what's the meaning of $gamma={(x,y)|(x^2+y^2=1)}$
– Dhamnekar Winod
Nov 24 at 14:13
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
I meant unit circle on complex plane centered at zero.
– John_Wick
Nov 24 at 14:16
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
services.math.duke.edu/education/ccp/materials/engin/cint1/… check here how to calculate line integral on the complex plane.
– John_Wick
Nov 24 at 14:47
add a comment |
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