Complete Elliptic Integral of the First Kind Identity The 2019 Stack Overflow Developer Survey Results Are InIntegral involving Complete Elliptic Integral of the First Kind K(k)How to show that the Complete Elliptic Integral of the First Kind increases in m?Derivative of the elliptic integral of the first kindAsymptotic expansion of the complete elliptic integral of the first kindAsymptotic behavior of elliptic integral (first kind)Identity for the real part of the elliptic integral of the first kind: referencesAsymptotic expansion of complete elliptic integral of third kindAn identity for the complete elliptic integral of the first kindSolving for argument of complete elliptic integral of first kindAsymptotic expression for the complete Elliptic integral of the first kind
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Complete Elliptic Integral of the First Kind Identity
The 2019 Stack Overflow Developer Survey Results Are InIntegral involving Complete Elliptic Integral of the First Kind K(k)How to show that the Complete Elliptic Integral of the First Kind increases in m?Derivative of the elliptic integral of the first kindAsymptotic expansion of the complete elliptic integral of the first kindAsymptotic behavior of elliptic integral (first kind)Identity for the real part of the elliptic integral of the first kind: referencesAsymptotic expansion of complete elliptic integral of third kindAn identity for the complete elliptic integral of the first kindSolving for argument of complete elliptic integral of first kindAsymptotic expression for the complete Elliptic integral of the first kind
$begingroup$
Is there an identity for $fracK'(k)K(k)=?$ where $K(k)=int_0^fracpi2frac1sqrt1-k^2sin^2(x)dx=int_0^1frac1sqrt(1-t^2)(1-k^2t^2)dt$ is the Complete Elliptic Integral of the First Kind and $K'(k)=K(sqrt1-k^2)$ ? The closest I got was showing that $$1.):fracK'(k)K(frac1-k1+k)=frac21+k$$ and from the book "Pi and the AGM" i got an identity showing $$2.):fracK'(k)K(k)=2fracK'(frac2sqrtk1+k)K(frac2sqrtk1+k)$$ however this identity isn't particularly useful for me. Is there a identity that is similar to the identity I derived for $1.)$
calculus definite-integrals rational-functions elliptic-integrals
asked Mar 23 at 1:06
user10560552user10560552
536
$endgroup$
add a comment |
$begingroup$
Is there an identity for $fracK'(k)K(k)=?$ where $K(k)=int_0^fracpi2frac1sqrt1-k^2sin^2(x)dx=int_0^1frac1sqrt(1-t^2)(1-k^2t^2)dt$ is the Complete Elliptic Integral of the First Kind and $K'(k)=K(sqrt1-k^2)$ ? The closest I got was showing that $$1.):fracK'(k)K(frac1-k1+k)=frac21+k$$ and from the book "Pi and the AGM" i got an identity showing $$2.):fracK'(k)K(k)=2fracK'(frac2sqrtk1+k)K(frac2sqrtk1+k)$$ however this identity isn't particularly useful for me. Is there a identity that is similar to the identity I derived for $1.)$
calculus definite-integrals rational-functions elliptic-integrals
asked Mar 23 at 1:06
user10560552user10560552
536
$endgroup$
$begingroup$
The identity you suggested isn't useful for me because it simplifies to $$fracK'(k)K(k)=fracK'(k)K(k)$$ I was hoping for a rational expression in terms of k or a series similar to example 1.) that I gave in my question.
$endgroup$
– user10560552
Mar 23 at 4:46
$begingroup$
$$ K(k)=fracpi2theta_3^2(q),quad q= e^-pi K'/K $$ and special values for $K,K'$ are related to modular equations.
$endgroup$
– Jack D'Aurizio
Mar 23 at 9:52
add a comment |
$begingroup$
Is there an identity for $fracK'(k)K(k)=?$ where $K(k)=int_0^fracpi2frac1sqrt1-k^2sin^2(x)dx=int_0^1frac1sqrt(1-t^2)(1-k^2t^2)dt$ is the Complete Elliptic Integral of the First Kind and $K'(k)=K(sqrt1-k^2)$ ? The closest I got was showing that $$1.):fracK'(k)K(frac1-k1+k)=frac21+k$$ and from the book "Pi and the AGM" i got an identity showing $$2.):fracK'(k)K(k)=2fracK'(frac2sqrtk1+k)K(frac2sqrtk1+k)$$ however this identity isn't particularly useful for me. Is there a identity that is similar to the identity I derived for $1.)$
calculus definite-integrals rational-functions elliptic-integrals
asked Mar 23 at 1:06
user10560552user10560552
536
$endgroup$
Is there an identity for $fracK'(k)K(k)=?$ where $K(k)=int_0^fracpi2frac1sqrt1-k^2sin^2(x)dx=int_0^1frac1sqrt(1-t^2)(1-k^2t^2)dt$ is the Complete Elliptic Integral of the First Kind and $K'(k)=K(sqrt1-k^2)$ ? The closest I got was showing that $$1.):fracK'(k)K(frac1-k1+k)=frac21+k$$ and from the book "Pi and the AGM" i got an identity showing $$2.):fracK'(k)K(k)=2fracK'(frac2sqrtk1+k)K(frac2sqrtk1+k)$$ however this identity isn't particularly useful for me. Is there a identity that is similar to the identity I derived for $1.)$
calculus definite-integrals rational-functions elliptic-integrals
calculus definite-integrals rational-functions elliptic-integrals
asked Mar 23 at 1:06
user10560552user10560552
536
asked Mar 23 at 1:06
user10560552user10560552
536
asked Mar 23 at 1:06
user10560552user10560552
536
asked Mar 23 at 1:06
user10560552user10560552
536
asked Mar 23 at 1:06
user10560552user10560552
536
536
$begingroup$
The identity you suggested isn't useful for me because it simplifies to $$fracK'(k)K(k)=fracK'(k)K(k)$$ I was hoping for a rational expression in terms of k or a series similar to example 1.) that I gave in my question.
$endgroup$
– user10560552
Mar 23 at 4:46
$begingroup$
$$ K(k)=fracpi2theta_3^2(q),quad q= e^-pi K'/K $$ and special values for $K,K'$ are related to modular equations.
$endgroup$
– Jack D'Aurizio
Mar 23 at 9:52
add a comment |
$begingroup$
The identity you suggested isn't useful for me because it simplifies to $$fracK'(k)K(k)=fracK'(k)K(k)$$ I was hoping for a rational expression in terms of k or a series similar to example 1.) that I gave in my question.
$endgroup$
– user10560552
Mar 23 at 4:46
$begingroup$
$$ K(k)=fracpi2theta_3^2(q),quad q= e^-pi K'/K $$ and special values for $K,K'$ are related to modular equations.
$endgroup$
– Jack D'Aurizio
Mar 23 at 9:52
$begingroup$
The identity you suggested isn't useful for me because it simplifies to $$fracK'(k)K(k)=fracK'(k)K(k)$$ I was hoping for a rational expression in terms of k or a series similar to example 1.) that I gave in my question.
$endgroup$
– user10560552
Mar 23 at 4:46
$begingroup$
The identity you suggested isn't useful for me because it simplifies to $$fracK'(k)K(k)=fracK'(k)K(k)$$ I was hoping for a rational expression in terms of k or a series similar to example 1.) that I gave in my question.
$endgroup$
– user10560552
Mar 23 at 4:46
$begingroup$
$$ K(k)=fracpi2theta_3^2(q),quad q= e^-pi K'/K $$ and special values for $K,K'$ are related to modular equations.
$endgroup$
– Jack D'Aurizio
Mar 23 at 9:52
$begingroup$
$$ K(k)=fracpi2theta_3^2(q),quad q= e^-pi K'/K $$ and special values for $K,K'$ are related to modular equations.
$endgroup$
– Jack D'Aurizio
Mar 23 at 9:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't know if you're asking for something like this: http://mathworld.wolfram.com/ModularEquation.html
Or this: https://en.m.wikipedia.org/wiki/Elliptic_integral (look under complete elliptic integral of the first kind under "relation to Jacobi theta function")
The relation you derived can be used in the identity for the theta function to find the value of it when the imaginary part is equal to 2, and using some relationships between the other theta functions you can evaluate values of all the theta functions for all powers of 2, which allows you to calculate values of the j invariant for all powers of 2 as well (just thought that was interesting)
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
$endgroup$
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
add a comment |
$begingroup$
The OEIS sequence A227503 may be what you are looking for. By definition the
Jacobi nome q is $, q = exp(-pi K'(k)/K(k)),$ which implies
$,K'(k)/K(k) = -log(q)/pi.,$ This can be expanded as a power series in $,k.,$ The sequence A227503 is closely related to that series expansion.
$$ fracK'(k)K(k) = -frac1pileft(log(x)+8left(x + 13, x^2/2 + 184, x^3/3+dotsright)right) $$
where $, x := k^2/16.,$ and the coefficients are from OEIS A227503.
There are several other methods depending on your needs and you can choose among them.Read the DLMF Chapter 19 on Elliptic integrals for much more details. In particular, equations 19.5.5 and 19.5.6.
P.S. In your question you mention an identity from "Pi and the AGM".
Note that in the context of $ q = exp(-pi K'(k)/K(k)) $ that the map $ k to 2sqrtk/(1+k) $ corresponds to $ q to sqrtq $ and its inverse mapping
$ k to (1-k')/(1+k') $ corresponds to $ q to q^2. $ These are known as ascending and descending transformations.
answered Mar 23 at 5:08
SomosSomos
14.8k11337
$endgroup$
$begingroup$
Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
$endgroup$
– user10560552
Mar 23 at 6:16
$begingroup$
Okay thanks a lot this is very helpful.
$endgroup$
– user10560552
Mar 23 at 6:23
$begingroup$
for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
$endgroup$
– user10560552
Mar 23 at 23:17
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
$endgroup$
– user10560552
Mar 24 at 0:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't know if you're asking for something like this: http://mathworld.wolfram.com/ModularEquation.html
Or this: https://en.m.wikipedia.org/wiki/Elliptic_integral (look under complete elliptic integral of the first kind under "relation to Jacobi theta function")
The relation you derived can be used in the identity for the theta function to find the value of it when the imaginary part is equal to 2, and using some relationships between the other theta functions you can evaluate values of all the theta functions for all powers of 2, which allows you to calculate values of the j invariant for all powers of 2 as well (just thought that was interesting)
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
$endgroup$
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
add a comment |
$begingroup$
I don't know if you're asking for something like this: http://mathworld.wolfram.com/ModularEquation.html
Or this: https://en.m.wikipedia.org/wiki/Elliptic_integral (look under complete elliptic integral of the first kind under "relation to Jacobi theta function")
The relation you derived can be used in the identity for the theta function to find the value of it when the imaginary part is equal to 2, and using some relationships between the other theta functions you can evaluate values of all the theta functions for all powers of 2, which allows you to calculate values of the j invariant for all powers of 2 as well (just thought that was interesting)
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
$endgroup$
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
add a comment |
$begingroup$
I don't know if you're asking for something like this: http://mathworld.wolfram.com/ModularEquation.html
Or this: https://en.m.wikipedia.org/wiki/Elliptic_integral (look under complete elliptic integral of the first kind under "relation to Jacobi theta function")
The relation you derived can be used in the identity for the theta function to find the value of it when the imaginary part is equal to 2, and using some relationships between the other theta functions you can evaluate values of all the theta functions for all powers of 2, which allows you to calculate values of the j invariant for all powers of 2 as well (just thought that was interesting)
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
$endgroup$
I don't know if you're asking for something like this: http://mathworld.wolfram.com/ModularEquation.html
Or this: https://en.m.wikipedia.org/wiki/Elliptic_integral (look under complete elliptic integral of the first kind under "relation to Jacobi theta function")
The relation you derived can be used in the identity for the theta function to find the value of it when the imaginary part is equal to 2, and using some relationships between the other theta functions you can evaluate values of all the theta functions for all powers of 2, which allows you to calculate values of the j invariant for all powers of 2 as well (just thought that was interesting)
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
answered Mar 23 at 6:17
uhhhhidkuhhhhidk
1266
1266
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
add a comment |
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
$begingroup$
What I wanted was a relation such as example 1.) in my question. Something of a rational expression. I don't know if $l=sqrt1-k^2$ satisfies the modular form.
$endgroup$
– user10560552
Mar 23 at 8:13
add a comment |
$begingroup$
The OEIS sequence A227503 may be what you are looking for. By definition the
Jacobi nome q is $, q = exp(-pi K'(k)/K(k)),$ which implies
$,K'(k)/K(k) = -log(q)/pi.,$ This can be expanded as a power series in $,k.,$ The sequence A227503 is closely related to that series expansion.
$$ fracK'(k)K(k) = -frac1pileft(log(x)+8left(x + 13, x^2/2 + 184, x^3/3+dotsright)right) $$
where $, x := k^2/16.,$ and the coefficients are from OEIS A227503.
There are several other methods depending on your needs and you can choose among them.Read the DLMF Chapter 19 on Elliptic integrals for much more details. In particular, equations 19.5.5 and 19.5.6.
P.S. In your question you mention an identity from "Pi and the AGM".
Note that in the context of $ q = exp(-pi K'(k)/K(k)) $ that the map $ k to 2sqrtk/(1+k) $ corresponds to $ q to sqrtq $ and its inverse mapping
$ k to (1-k')/(1+k') $ corresponds to $ q to q^2. $ These are known as ascending and descending transformations.
answered Mar 23 at 5:08
SomosSomos
14.8k11337
$endgroup$
$begingroup$
Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
$endgroup$
– user10560552
Mar 23 at 6:16
$begingroup$
Okay thanks a lot this is very helpful.
$endgroup$
– user10560552
Mar 23 at 6:23
$begingroup$
for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
$endgroup$
– user10560552
Mar 23 at 23:17
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
$endgroup$
– user10560552
Mar 24 at 0:31
add a comment |
$begingroup$
The OEIS sequence A227503 may be what you are looking for. By definition the
Jacobi nome q is $, q = exp(-pi K'(k)/K(k)),$ which implies
$,K'(k)/K(k) = -log(q)/pi.,$ This can be expanded as a power series in $,k.,$ The sequence A227503 is closely related to that series expansion.
$$ fracK'(k)K(k) = -frac1pileft(log(x)+8left(x + 13, x^2/2 + 184, x^3/3+dotsright)right) $$
where $, x := k^2/16.,$ and the coefficients are from OEIS A227503.
There are several other methods depending on your needs and you can choose among them.Read the DLMF Chapter 19 on Elliptic integrals for much more details. In particular, equations 19.5.5 and 19.5.6.
P.S. In your question you mention an identity from "Pi and the AGM".
Note that in the context of $ q = exp(-pi K'(k)/K(k)) $ that the map $ k to 2sqrtk/(1+k) $ corresponds to $ q to sqrtq $ and its inverse mapping
$ k to (1-k')/(1+k') $ corresponds to $ q to q^2. $ These are known as ascending and descending transformations.
answered Mar 23 at 5:08
SomosSomos
14.8k11337
$endgroup$
$begingroup$
Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
$endgroup$
– user10560552
Mar 23 at 6:16
$begingroup$
Okay thanks a lot this is very helpful.
$endgroup$
– user10560552
Mar 23 at 6:23
$begingroup$
for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
$endgroup$
– user10560552
Mar 23 at 23:17
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
$endgroup$
– user10560552
Mar 24 at 0:31
add a comment |
$begingroup$
The OEIS sequence A227503 may be what you are looking for. By definition the
Jacobi nome q is $, q = exp(-pi K'(k)/K(k)),$ which implies
$,K'(k)/K(k) = -log(q)/pi.,$ This can be expanded as a power series in $,k.,$ The sequence A227503 is closely related to that series expansion.
$$ fracK'(k)K(k) = -frac1pileft(log(x)+8left(x + 13, x^2/2 + 184, x^3/3+dotsright)right) $$
where $, x := k^2/16.,$ and the coefficients are from OEIS A227503.
There are several other methods depending on your needs and you can choose among them.Read the DLMF Chapter 19 on Elliptic integrals for much more details. In particular, equations 19.5.5 and 19.5.6.
P.S. In your question you mention an identity from "Pi and the AGM".
Note that in the context of $ q = exp(-pi K'(k)/K(k)) $ that the map $ k to 2sqrtk/(1+k) $ corresponds to $ q to sqrtq $ and its inverse mapping
$ k to (1-k')/(1+k') $ corresponds to $ q to q^2. $ These are known as ascending and descending transformations.
answered Mar 23 at 5:08
SomosSomos
14.8k11337
$endgroup$
The OEIS sequence A227503 may be what you are looking for. By definition the
Jacobi nome q is $, q = exp(-pi K'(k)/K(k)),$ which implies
$,K'(k)/K(k) = -log(q)/pi.,$ This can be expanded as a power series in $,k.,$ The sequence A227503 is closely related to that series expansion.
$$ fracK'(k)K(k) = -frac1pileft(log(x)+8left(x + 13, x^2/2 + 184, x^3/3+dotsright)right) $$
where $, x := k^2/16.,$ and the coefficients are from OEIS A227503.
There are several other methods depending on your needs and you can choose among them.Read the DLMF Chapter 19 on Elliptic integrals for much more details. In particular, equations 19.5.5 and 19.5.6.
P.S. In your question you mention an identity from "Pi and the AGM".
Note that in the context of $ q = exp(-pi K'(k)/K(k)) $ that the map $ k to 2sqrtk/(1+k) $ corresponds to $ q to sqrtq $ and its inverse mapping
$ k to (1-k')/(1+k') $ corresponds to $ q to q^2. $ These are known as ascending and descending transformations.
answered Mar 23 at 5:08
SomosSomos
14.8k11337
edited Mar 24 at 1:15
answered Mar 23 at 5:08
SomosSomos
14.8k11337
answered Mar 23 at 5:08
SomosSomos
14.8k11337
answered Mar 23 at 5:08
SomosSomos
14.8k11337
14.8k11337
$begingroup$
Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
$endgroup$
– user10560552
Mar 23 at 6:16
$begingroup$
Okay thanks a lot this is very helpful.
$endgroup$
– user10560552
Mar 23 at 6:23
$begingroup$
for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
$endgroup$
– user10560552
Mar 23 at 23:17
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
$endgroup$
– user10560552
Mar 24 at 0:31
add a comment |
$begingroup$
Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
$endgroup$
– user10560552
Mar 23 at 6:16
$begingroup$
Okay thanks a lot this is very helpful.
$endgroup$
– user10560552
Mar 23 at 6:23
$begingroup$
for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
$endgroup$
– user10560552
Mar 23 at 23:17
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
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– user10560552
Mar 24 at 0:31
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Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
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– user10560552
Mar 23 at 6:16
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Thank you this is very helpful and what I wanted. However is it possible it can be expressed as a rational expression?
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– user10560552
Mar 23 at 6:16
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Okay thanks a lot this is very helpful.
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– user10560552
Mar 23 at 6:23
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Okay thanks a lot this is very helpful.
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– user10560552
Mar 23 at 6:23
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for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
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– user10560552
Mar 23 at 23:17
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for my purposes I need more accuracy, is there a way to employ argument reduction for $fracK'(k)K(k)$ similar to how $ln(fracx2^n)=ln(x)-n*ln(2)$?
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– user10560552
Mar 23 at 23:17
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
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– user10560552
Mar 24 at 0:31
$begingroup$
Sorry if i wasn't clear with my question I did read it and understand that $j=frac1-k'1+k'$ $$K(k)=(1+j)(K(j))$$ however since the series starts to diverge for $k$ around $1$ is there was a way to employ an argument reduction for $fracK'(frack2)K(frack2)$
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– user10560552
Mar 24 at 0:31
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The identity you suggested isn't useful for me because it simplifies to $$fracK'(k)K(k)=fracK'(k)K(k)$$ I was hoping for a rational expression in terms of k or a series similar to example 1.) that I gave in my question.
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– user10560552
Mar 23 at 4:46
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$$ K(k)=fracpi2theta_3^2(q),quad q= e^-pi K'/K $$ and special values for $K,K'$ are related to modular equations.
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– Jack D'Aurizio
Mar 23 at 9:52