Evaluate $int_0^fracpi2 frac1sqrta^2 cos^2 theta+b^2 sin^2 theta d theta$
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up vote
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$int_0^fracpi2 frac1sqrta^2 cos^2 theta+b^2 sin^2 theta d theta$
$ = int_0^fracpi2 frac1asec theta frac1 sqrt1+(b/a)^2 tan^2 theta d theta$
But i know $d(tan theta) = sec^2 theta$. But this is not in that form. Can you tell me how to proceed further?
integration definite-integrals elliptic-integrals
asked Aug 12 at 7:40
Magneto
845213
add a comment |Â
up vote
3
down vote
favorite
$int_0^fracpi2 frac1sqrta^2 cos^2 theta+b^2 sin^2 theta d theta$
$ = int_0^fracpi2 frac1asec theta frac1 sqrt1+(b/a)^2 tan^2 theta d theta$
But i know $d(tan theta) = sec^2 theta$. But this is not in that form. Can you tell me how to proceed further?
integration definite-integrals elliptic-integrals
asked Aug 12 at 7:40
Magneto
845213
2
This integral leads to an elliptic one.
– Dr. Sonnhard Graubner
Aug 12 at 7:58
@mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral.
– Magneto
Aug 12 at 8:40
@Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind.
– Jack D'Aurizio♦
Aug 13 at 5:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$int_0^fracpi2 frac1sqrta^2 cos^2 theta+b^2 sin^2 theta d theta$
$ = int_0^fracpi2 frac1asec theta frac1 sqrt1+(b/a)^2 tan^2 theta d theta$
But i know $d(tan theta) = sec^2 theta$. But this is not in that form. Can you tell me how to proceed further?
integration definite-integrals elliptic-integrals
asked Aug 12 at 7:40
Magneto
845213
$int_0^fracpi2 frac1sqrta^2 cos^2 theta+b^2 sin^2 theta d theta$
$ = int_0^fracpi2 frac1asec theta frac1 sqrt1+(b/a)^2 tan^2 theta d theta$
But i know $d(tan theta) = sec^2 theta$. But this is not in that form. Can you tell me how to proceed further?
integration definite-integrals elliptic-integrals
asked Aug 12 at 7:40
Magneto
845213
asked Aug 12 at 7:40
Magneto
845213
asked Aug 12 at 7:40
Magneto
845213
asked Aug 12 at 7:40
Magneto
845213
845213
2
This integral leads to an elliptic one.
– Dr. Sonnhard Graubner
Aug 12 at 7:58
@mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral.
– Magneto
Aug 12 at 8:40
@Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind.
– Jack D'Aurizio♦
Aug 13 at 5:36
add a comment |Â
2
This integral leads to an elliptic one.
– Dr. Sonnhard Graubner
Aug 12 at 7:58
@mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral.
– Magneto
Aug 12 at 8:40
@Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind.
– Jack D'Aurizio♦
Aug 13 at 5:36
2
2
This integral leads to an elliptic one.
– Dr. Sonnhard Graubner
Aug 12 at 7:58
This integral leads to an elliptic one.
– Dr. Sonnhard Graubner
Aug 12 at 7:58
@mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral.
– Magneto
Aug 12 at 8:40
@mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral.
– Magneto
Aug 12 at 8:40
@Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind.
– Jack D'Aurizio♦
Aug 13 at 5:36
@Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind.
– Jack D'Aurizio♦
Aug 13 at 5:36
add a comment |Â
1 Answer
1
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oldest
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up vote
6
down vote
accepted
$$I(a, b) =int_0^fracpi2 frac1sqrta^2 sin^2 t + b^2 cos^2 tdt$$
$$ = int_0^fracpi2 frac1sqrtb^2-(b^2-a^2)sin^2 tdt=frac1b int_0^fracpi2 frac1sqrt1-left(1-fraca^2b^2right)sin^2 tdt=frac1b Kleft(1-fraca^2b^2right)$$ Or this can be written as: $I(a, b)=fracpi2M(a,b),$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=lim_ntoinftya_n=lim_ntoinftyb_n$ where
$$a_0=a, b_0=b, a_n+1=fraca_n+b_n2, b_n+1=sqrta_nb_n$$
answered Aug 12 at 8:11
Zacky
2,2741327
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
1
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
2
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
1
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
$$I(a, b) =int_0^fracpi2 frac1sqrta^2 sin^2 t + b^2 cos^2 tdt$$
$$ = int_0^fracpi2 frac1sqrtb^2-(b^2-a^2)sin^2 tdt=frac1b int_0^fracpi2 frac1sqrt1-left(1-fraca^2b^2right)sin^2 tdt=frac1b Kleft(1-fraca^2b^2right)$$ Or this can be written as: $I(a, b)=fracpi2M(a,b),$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=lim_ntoinftya_n=lim_ntoinftyb_n$ where
$$a_0=a, b_0=b, a_n+1=fraca_n+b_n2, b_n+1=sqrta_nb_n$$
answered Aug 12 at 8:11
Zacky
2,2741327
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
1
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
2
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
1
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
 |Â
show 1 more comment
up vote
6
down vote
accepted
$$I(a, b) =int_0^fracpi2 frac1sqrta^2 sin^2 t + b^2 cos^2 tdt$$
$$ = int_0^fracpi2 frac1sqrtb^2-(b^2-a^2)sin^2 tdt=frac1b int_0^fracpi2 frac1sqrt1-left(1-fraca^2b^2right)sin^2 tdt=frac1b Kleft(1-fraca^2b^2right)$$ Or this can be written as: $I(a, b)=fracpi2M(a,b),$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=lim_ntoinftya_n=lim_ntoinftyb_n$ where
$$a_0=a, b_0=b, a_n+1=fraca_n+b_n2, b_n+1=sqrta_nb_n$$
answered Aug 12 at 8:11
Zacky
2,2741327
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
1
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
2
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
1
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
 |Â
show 1 more comment
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$$I(a, b) =int_0^fracpi2 frac1sqrta^2 sin^2 t + b^2 cos^2 tdt$$
$$ = int_0^fracpi2 frac1sqrtb^2-(b^2-a^2)sin^2 tdt=frac1b int_0^fracpi2 frac1sqrt1-left(1-fraca^2b^2right)sin^2 tdt=frac1b Kleft(1-fraca^2b^2right)$$ Or this can be written as: $I(a, b)=fracpi2M(a,b),$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=lim_ntoinftya_n=lim_ntoinftyb_n$ where
$$a_0=a, b_0=b, a_n+1=fraca_n+b_n2, b_n+1=sqrta_nb_n$$
answered Aug 12 at 8:11
Zacky
2,2741327
$$I(a, b) =int_0^fracpi2 frac1sqrta^2 sin^2 t + b^2 cos^2 tdt$$
$$ = int_0^fracpi2 frac1sqrtb^2-(b^2-a^2)sin^2 tdt=frac1b int_0^fracpi2 frac1sqrt1-left(1-fraca^2b^2right)sin^2 tdt=frac1b Kleft(1-fraca^2b^2right)$$ Or this can be written as: $I(a, b)=fracpi2M(a,b),$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=lim_ntoinftya_n=lim_ntoinftyb_n$ where
$$a_0=a, b_0=b, a_n+1=fraca_n+b_n2, b_n+1=sqrta_nb_n$$
answered Aug 12 at 8:11
Zacky
2,2741327
answered Aug 12 at 8:11
Zacky
2,2741327
answered Aug 12 at 8:11
Zacky
2,2741327
answered Aug 12 at 8:11
Zacky
2,2741327
2,2741327
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
1
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
2
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
1
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
 |Â
show 1 more comment
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
1
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
2
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
1
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
I have already seen this in wiki. But can u tell me what is value of this M(a,b)?
– Magneto
Aug 12 at 8:41
1
1
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
@Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $cos(x)$ or $e^x$.
– mr_e_man
Aug 12 at 8:43
2
2
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=fracpi2bfrac1Kleft(1-fraca^2b^2right)$$ As said above, you would need to learn new properties, see: en.wikipedia.org/wiki/…
– Zacky
Aug 12 at 8:47
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
@Zacky How about $$ int_0^fracpi2 fraccos(A cos(t))a^2 sin^2 t + b^2 cos^2 tdt?$$
– Dinesh Shankar
Aug 16 at 20:16
1
1
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
For the moment I have no idea, I guess its related to Bessel function, maybe we can rewrite the numerator with euler formula then take the real part out? You may want to post this as a new question.
– Zacky
Aug 16 at 21:30
 |Â
show 1 more comment
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2
This integral leads to an elliptic one.
– Dr. Sonnhard Graubner
Aug 12 at 7:58
@mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral.
– Magneto
Aug 12 at 8:40
@Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind.
– Jack D'Aurizio♦
Aug 13 at 5:36