How to calculate the inverse agm(1,x)?
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If we define the following function $$agm(1,x) =y$$ and its inverse, $$agm^{-1}(y)=(1,x)$$ is it possible to calculate $agm^-1(y)$. Where $agm(x)$ is the sequence $$a_{n+1}=frac{a_n+b_n}{2}$$ and $$b_{n+1}=sqrt(a_nb_n)$$I know how to reverse the iterations of $agm(x)$ using $$a_{n-1}=a_n-sqrt(a_n^2-b_n^2)$$ and that $$b_{n-1}=a_n+sqrt(a_n^2-b_n^2)$$ However for those to work I would need two values, how can I calculate $x$ or $(1,x)$ using $y$.
calculus inverse-function
asked Dec 24 '18 at 3:59
user10560552user10560552
355
$endgroup$
add a comment |
$begingroup$
If we define the following function $$agm(1,x) =y$$ and its inverse, $$agm^{-1}(y)=(1,x)$$ is it possible to calculate $agm^-1(y)$. Where $agm(x)$ is the sequence $$a_{n+1}=frac{a_n+b_n}{2}$$ and $$b_{n+1}=sqrt(a_nb_n)$$I know how to reverse the iterations of $agm(x)$ using $$a_{n-1}=a_n-sqrt(a_n^2-b_n^2)$$ and that $$b_{n-1}=a_n+sqrt(a_n^2-b_n^2)$$ However for those to work I would need two values, how can I calculate $x$ or $(1,x)$ using $y$.
calculus inverse-function
asked Dec 24 '18 at 3:59
user10560552user10560552
355
$endgroup$
$begingroup$
It is not clear to me what you want to do. Do you want to find $agm^{-1}$ or to reverse the agm iteration? Can you give a simple example?
$endgroup$
– Somos
Dec 24 '18 at 4:04
$begingroup$
Of course, the $agm(1,2) = 1.4567910310469068692...$, what i want to know is how to calculate $agm^{-1}(1.4567910310469068692)=(1,2)$ or $agm^{-1}(1.4567910310469068692)=2$ only for the special case of $agm(1,x)$
$endgroup$
– user10560552
Dec 24 '18 at 4:16
add a comment |
$begingroup$
If we define the following function $$agm(1,x) =y$$ and its inverse, $$agm^{-1}(y)=(1,x)$$ is it possible to calculate $agm^-1(y)$. Where $agm(x)$ is the sequence $$a_{n+1}=frac{a_n+b_n}{2}$$ and $$b_{n+1}=sqrt(a_nb_n)$$I know how to reverse the iterations of $agm(x)$ using $$a_{n-1}=a_n-sqrt(a_n^2-b_n^2)$$ and that $$b_{n-1}=a_n+sqrt(a_n^2-b_n^2)$$ However for those to work I would need two values, how can I calculate $x$ or $(1,x)$ using $y$.
calculus inverse-function
asked Dec 24 '18 at 3:59
user10560552user10560552
355
$endgroup$
If we define the following function $$agm(1,x) =y$$ and its inverse, $$agm^{-1}(y)=(1,x)$$ is it possible to calculate $agm^-1(y)$. Where $agm(x)$ is the sequence $$a_{n+1}=frac{a_n+b_n}{2}$$ and $$b_{n+1}=sqrt(a_nb_n)$$I know how to reverse the iterations of $agm(x)$ using $$a_{n-1}=a_n-sqrt(a_n^2-b_n^2)$$ and that $$b_{n-1}=a_n+sqrt(a_n^2-b_n^2)$$ However for those to work I would need two values, how can I calculate $x$ or $(1,x)$ using $y$.
calculus inverse-function
calculus inverse-function
asked Dec 24 '18 at 3:59
user10560552user10560552
355
asked Dec 24 '18 at 3:59
user10560552user10560552
355
asked Dec 24 '18 at 3:59
user10560552user10560552
355
asked Dec 24 '18 at 3:59
user10560552user10560552
355
asked Dec 24 '18 at 3:59
user10560552user10560552
355
355
$begingroup$
It is not clear to me what you want to do. Do you want to find $agm^{-1}$ or to reverse the agm iteration? Can you give a simple example?
$endgroup$
– Somos
Dec 24 '18 at 4:04
$begingroup$
Of course, the $agm(1,2) = 1.4567910310469068692...$, what i want to know is how to calculate $agm^{-1}(1.4567910310469068692)=(1,2)$ or $agm^{-1}(1.4567910310469068692)=2$ only for the special case of $agm(1,x)$
$endgroup$
– user10560552
Dec 24 '18 at 4:16
add a comment |
$begingroup$
It is not clear to me what you want to do. Do you want to find $agm^{-1}$ or to reverse the agm iteration? Can you give a simple example?
$endgroup$
– Somos
Dec 24 '18 at 4:04
$begingroup$
Of course, the $agm(1,2) = 1.4567910310469068692...$, what i want to know is how to calculate $agm^{-1}(1.4567910310469068692)=(1,2)$ or $agm^{-1}(1.4567910310469068692)=2$ only for the special case of $agm(1,x)$
$endgroup$
– user10560552
Dec 24 '18 at 4:16
$begingroup$
It is not clear to me what you want to do. Do you want to find $agm^{-1}$ or to reverse the agm iteration? Can you give a simple example?
$endgroup$
– Somos
Dec 24 '18 at 4:04
$begingroup$
It is not clear to me what you want to do. Do you want to find $agm^{-1}$ or to reverse the agm iteration? Can you give a simple example?
$endgroup$
– Somos
Dec 24 '18 at 4:04
$begingroup$
Of course, the $agm(1,2) = 1.4567910310469068692...$, what i want to know is how to calculate $agm^{-1}(1.4567910310469068692)=(1,2)$ or $agm^{-1}(1.4567910310469068692)=2$ only for the special case of $agm(1,x)$
$endgroup$
– user10560552
Dec 24 '18 at 4:16
$begingroup$
Of course, the $agm(1,2) = 1.4567910310469068692...$, what i want to know is how to calculate $agm^{-1}(1.4567910310469068692)=(1,2)$ or $agm^{-1}(1.4567910310469068692)=2$ only for the special case of $agm(1,x)$
$endgroup$
– user10560552
Dec 24 '18 at 4:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have
$$text{agm}(1,x)=frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}$$ Suing series expansions for large values of $x$, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{pi x}{4log (2)+2 log left({x}right)}+frac{1-pi log
left({4 x}right)}{8 x log ^2left({4
x}right)}+Oleft(frac{1}{x^3}right)$$ So, if we consider the first term only and solve for $x$ we have, as an approximation,
$$x=-frac{2,k }{pi },W_{-1}left(-frac{pi }{8,k}right)qquad text{where} qquad k=text{agm}(1,x)tag 1$$ where appears the second branch of Lambert function.
To check how good or bad is this approximation, give $x$ a value, compute $text{agm}(1,x)$ and apply $(1)$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
5 & 2.60401 & 4.94933 \
10 & 4.25041 & 9.97492 \
15 & 5.74991 & 14.9833 \
20 & 7.16581 & 19.9875 \
25 & 8.52468 & 24.9900 \
30 & 9.84096 & 29.9917 \
35 & 11.1236 & 34.9929 \
40 & 12.3787 & 39.9937 \
45 & 13.6105 & 44.9944 \
50 & 14.8223 & 49.9950 \
55 & 16.0167 & 54.9955 \
60 & 17.1955 & 59.9958 \
65 & 18.3605 & 64.9962 \
70 & 19.5129 & 69.9964 \
75 & 20.6539 & 74.9967 \
80 & 21.7844 & 79.9969 \
85 & 22.9053 & 84.9971 \
90 & 24.0173 & 89.9972 \
95 & 25.1209 & 94.9974 \
100 & 26.2167 & 99.9975
end{array}
right)$$
If $x$ is mall, using simple Padé approximants, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{1+frac{29 }{24}(x-1)+frac{61}{192} (x-1)^2}{1+frac{17
}{24} (x-1)+frac{5}{192} (x-1)^2}$$ leading to
$$x=frac{55-63k-4 sqrt{229 k^2-194 k+109}}{5 k-61}qquad text{where} qquad k=text{agm}(1,x)tag 2$$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
1.0 & 1.00000 & 1.00000 \
1.5 & 1.23734 & 1.50004 \
2.0 & 1.45679 & 2.00079 \
2.5 & 1.66450 & 2.50377 \
3.0 & 1.86362 & 3.01071 \
3.5 & 2.05604 & 3.52325 \
4.0 & 2.24303 & 4.04282 \
4.5 & 2.42546 & 4.57073 \
5.0 & 2.60401 & 5.10818
end{array}
right)$$
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
add a comment |
$begingroup$
From standard facts, such as in the Wikipedia article Arithmetic-geometric mean, $, textrm{agm}(1,k) = frac{pi/2}{K(k')},$ where $,k,$ is the modulus, $,k'=sqrt{1-k^2},$ is the complementary modulus, and $,K(k),$ is the complete elliptic integral of the first kind. In order to find $,textrm{agm}^{-1},$ you need to find the inverse of $,K().,$ This involves the Jacobi amplitude function described in the Jacobi elliptic functions article.
As a practical matter, I suggest restricting to $,0<k<1,$ although there is analytic continuation, but you may run into branch cuts or worse.
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
$endgroup$
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
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Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
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
$begingroup$
We have
$$text{agm}(1,x)=frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}$$ Suing series expansions for large values of $x$, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{pi x}{4log (2)+2 log left({x}right)}+frac{1-pi log
left({4 x}right)}{8 x log ^2left({4
x}right)}+Oleft(frac{1}{x^3}right)$$ So, if we consider the first term only and solve for $x$ we have, as an approximation,
$$x=-frac{2,k }{pi },W_{-1}left(-frac{pi }{8,k}right)qquad text{where} qquad k=text{agm}(1,x)tag 1$$ where appears the second branch of Lambert function.
To check how good or bad is this approximation, give $x$ a value, compute $text{agm}(1,x)$ and apply $(1)$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
5 & 2.60401 & 4.94933 \
10 & 4.25041 & 9.97492 \
15 & 5.74991 & 14.9833 \
20 & 7.16581 & 19.9875 \
25 & 8.52468 & 24.9900 \
30 & 9.84096 & 29.9917 \
35 & 11.1236 & 34.9929 \
40 & 12.3787 & 39.9937 \
45 & 13.6105 & 44.9944 \
50 & 14.8223 & 49.9950 \
55 & 16.0167 & 54.9955 \
60 & 17.1955 & 59.9958 \
65 & 18.3605 & 64.9962 \
70 & 19.5129 & 69.9964 \
75 & 20.6539 & 74.9967 \
80 & 21.7844 & 79.9969 \
85 & 22.9053 & 84.9971 \
90 & 24.0173 & 89.9972 \
95 & 25.1209 & 94.9974 \
100 & 26.2167 & 99.9975
end{array}
right)$$
If $x$ is mall, using simple Padé approximants, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{1+frac{29 }{24}(x-1)+frac{61}{192} (x-1)^2}{1+frac{17
}{24} (x-1)+frac{5}{192} (x-1)^2}$$ leading to
$$x=frac{55-63k-4 sqrt{229 k^2-194 k+109}}{5 k-61}qquad text{where} qquad k=text{agm}(1,x)tag 2$$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
1.0 & 1.00000 & 1.00000 \
1.5 & 1.23734 & 1.50004 \
2.0 & 1.45679 & 2.00079 \
2.5 & 1.66450 & 2.50377 \
3.0 & 1.86362 & 3.01071 \
3.5 & 2.05604 & 3.52325 \
4.0 & 2.24303 & 4.04282 \
4.5 & 2.42546 & 4.57073 \
5.0 & 2.60401 & 5.10818
end{array}
right)$$
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
add a comment |
$begingroup$
We have
$$text{agm}(1,x)=frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}$$ Suing series expansions for large values of $x$, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{pi x}{4log (2)+2 log left({x}right)}+frac{1-pi log
left({4 x}right)}{8 x log ^2left({4
x}right)}+Oleft(frac{1}{x^3}right)$$ So, if we consider the first term only and solve for $x$ we have, as an approximation,
$$x=-frac{2,k }{pi },W_{-1}left(-frac{pi }{8,k}right)qquad text{where} qquad k=text{agm}(1,x)tag 1$$ where appears the second branch of Lambert function.
To check how good or bad is this approximation, give $x$ a value, compute $text{agm}(1,x)$ and apply $(1)$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
5 & 2.60401 & 4.94933 \
10 & 4.25041 & 9.97492 \
15 & 5.74991 & 14.9833 \
20 & 7.16581 & 19.9875 \
25 & 8.52468 & 24.9900 \
30 & 9.84096 & 29.9917 \
35 & 11.1236 & 34.9929 \
40 & 12.3787 & 39.9937 \
45 & 13.6105 & 44.9944 \
50 & 14.8223 & 49.9950 \
55 & 16.0167 & 54.9955 \
60 & 17.1955 & 59.9958 \
65 & 18.3605 & 64.9962 \
70 & 19.5129 & 69.9964 \
75 & 20.6539 & 74.9967 \
80 & 21.7844 & 79.9969 \
85 & 22.9053 & 84.9971 \
90 & 24.0173 & 89.9972 \
95 & 25.1209 & 94.9974 \
100 & 26.2167 & 99.9975
end{array}
right)$$
If $x$ is mall, using simple Padé approximants, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{1+frac{29 }{24}(x-1)+frac{61}{192} (x-1)^2}{1+frac{17
}{24} (x-1)+frac{5}{192} (x-1)^2}$$ leading to
$$x=frac{55-63k-4 sqrt{229 k^2-194 k+109}}{5 k-61}qquad text{where} qquad k=text{agm}(1,x)tag 2$$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
1.0 & 1.00000 & 1.00000 \
1.5 & 1.23734 & 1.50004 \
2.0 & 1.45679 & 2.00079 \
2.5 & 1.66450 & 2.50377 \
3.0 & 1.86362 & 3.01071 \
3.5 & 2.05604 & 3.52325 \
4.0 & 2.24303 & 4.04282 \
4.5 & 2.42546 & 4.57073 \
5.0 & 2.60401 & 5.10818
end{array}
right)$$
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
add a comment |
$begingroup$
We have
$$text{agm}(1,x)=frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}$$ Suing series expansions for large values of $x$, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{pi x}{4log (2)+2 log left({x}right)}+frac{1-pi log
left({4 x}right)}{8 x log ^2left({4
x}right)}+Oleft(frac{1}{x^3}right)$$ So, if we consider the first term only and solve for $x$ we have, as an approximation,
$$x=-frac{2,k }{pi },W_{-1}left(-frac{pi }{8,k}right)qquad text{where} qquad k=text{agm}(1,x)tag 1$$ where appears the second branch of Lambert function.
To check how good or bad is this approximation, give $x$ a value, compute $text{agm}(1,x)$ and apply $(1)$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
5 & 2.60401 & 4.94933 \
10 & 4.25041 & 9.97492 \
15 & 5.74991 & 14.9833 \
20 & 7.16581 & 19.9875 \
25 & 8.52468 & 24.9900 \
30 & 9.84096 & 29.9917 \
35 & 11.1236 & 34.9929 \
40 & 12.3787 & 39.9937 \
45 & 13.6105 & 44.9944 \
50 & 14.8223 & 49.9950 \
55 & 16.0167 & 54.9955 \
60 & 17.1955 & 59.9958 \
65 & 18.3605 & 64.9962 \
70 & 19.5129 & 69.9964 \
75 & 20.6539 & 74.9967 \
80 & 21.7844 & 79.9969 \
85 & 22.9053 & 84.9971 \
90 & 24.0173 & 89.9972 \
95 & 25.1209 & 94.9974 \
100 & 26.2167 & 99.9975
end{array}
right)$$
If $x$ is mall, using simple Padé approximants, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{1+frac{29 }{24}(x-1)+frac{61}{192} (x-1)^2}{1+frac{17
}{24} (x-1)+frac{5}{192} (x-1)^2}$$ leading to
$$x=frac{55-63k-4 sqrt{229 k^2-194 k+109}}{5 k-61}qquad text{where} qquad k=text{agm}(1,x)tag 2$$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
1.0 & 1.00000 & 1.00000 \
1.5 & 1.23734 & 1.50004 \
2.0 & 1.45679 & 2.00079 \
2.5 & 1.66450 & 2.50377 \
3.0 & 1.86362 & 3.01071 \
3.5 & 2.05604 & 3.52325 \
4.0 & 2.24303 & 4.04282 \
4.5 & 2.42546 & 4.57073 \
5.0 & 2.60401 & 5.10818
end{array}
right)$$
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
We have
$$text{agm}(1,x)=frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}$$ Suing series expansions for large values of $x$, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{pi x}{4log (2)+2 log left({x}right)}+frac{1-pi log
left({4 x}right)}{8 x log ^2left({4
x}right)}+Oleft(frac{1}{x^3}right)$$ So, if we consider the first term only and solve for $x$ we have, as an approximation,
$$x=-frac{2,k }{pi },W_{-1}left(-frac{pi }{8,k}right)qquad text{where} qquad k=text{agm}(1,x)tag 1$$ where appears the second branch of Lambert function.
To check how good or bad is this approximation, give $x$ a value, compute $text{agm}(1,x)$ and apply $(1)$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
5 & 2.60401 & 4.94933 \
10 & 4.25041 & 9.97492 \
15 & 5.74991 & 14.9833 \
20 & 7.16581 & 19.9875 \
25 & 8.52468 & 24.9900 \
30 & 9.84096 & 29.9917 \
35 & 11.1236 & 34.9929 \
40 & 12.3787 & 39.9937 \
45 & 13.6105 & 44.9944 \
50 & 14.8223 & 49.9950 \
55 & 16.0167 & 54.9955 \
60 & 17.1955 & 59.9958 \
65 & 18.3605 & 64.9962 \
70 & 19.5129 & 69.9964 \
75 & 20.6539 & 74.9967 \
80 & 21.7844 & 79.9969 \
85 & 22.9053 & 84.9971 \
90 & 24.0173 & 89.9972 \
95 & 25.1209 & 94.9974 \
100 & 26.2167 & 99.9975
end{array}
right)$$
If $x$ is mall, using simple Padé approximants, we have
$$frac{pi (1+x)}{4 Kleft(left(frac{1-x}{1+x}right)^2right)}=frac{1+frac{29 }{24}(x-1)+frac{61}{192} (x-1)^2}{1+frac{17
}{24} (x-1)+frac{5}{192} (x-1)^2}$$ leading to
$$x=frac{55-63k-4 sqrt{229 k^2-194 k+109}}{5 k-61}qquad text{where} qquad k=text{agm}(1,x)tag 2$$
$$left(
begin{array}{ccc}
x_{given} & text{agm}(1,x) & x_{calc} \
1.0 & 1.00000 & 1.00000 \
1.5 & 1.23734 & 1.50004 \
2.0 & 1.45679 & 2.00079 \
2.5 & 1.66450 & 2.50377 \
3.0 & 1.86362 & 3.01071 \
3.5 & 2.05604 & 3.52325 \
4.0 & 2.24303 & 4.04282 \
4.5 & 2.42546 & 4.57073 \
5.0 & 2.60401 & 5.10818
end{array}
right)$$
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
edited Dec 24 '18 at 6:37
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
answered Dec 24 '18 at 6:02
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
add a comment |
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
Thank you this is what i was looking for, will this method also work for complex numbers? And is it possible to get more precise results using your formulas.
$endgroup$
– user10560552
Dec 24 '18 at 16:22
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
I'm assuming you would get more precise results by inputing more precise results.
$endgroup$
– user10560552
Dec 24 '18 at 16:41
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
@user10560552. I am not sure to understand what you mean. I did not input anything. Do you want something better ? I can build an empirical model. Let me know the range to cover. Cheers & Merry Xmas
$endgroup$
– Claude Leibovici
Dec 24 '18 at 17:21
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
You don't have to go through that much trouble. Sorry my comment was not clear. What I meant is if i use your second equation and $k=1.4567910310469068692$ instead of $k=1.45679$ would i get a more precise result.
$endgroup$
– user10560552
Dec 24 '18 at 17:41
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
$begingroup$
Let me clarify : for building the tablle, for a given $x$, $text{agm}(1,x)$ is exactly computed with illimited precision; so when table contains $k=1.45679$ is is the shorcut writing of $k=1.4567910310469068692$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 2:42
add a comment |
$begingroup$
From standard facts, such as in the Wikipedia article Arithmetic-geometric mean, $, textrm{agm}(1,k) = frac{pi/2}{K(k')},$ where $,k,$ is the modulus, $,k'=sqrt{1-k^2},$ is the complementary modulus, and $,K(k),$ is the complete elliptic integral of the first kind. In order to find $,textrm{agm}^{-1},$ you need to find the inverse of $,K().,$ This involves the Jacobi amplitude function described in the Jacobi elliptic functions article.
As a practical matter, I suggest restricting to $,0<k<1,$ although there is analytic continuation, but you may run into branch cuts or worse.
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
$endgroup$
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
$begingroup$
Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
add a comment |
$begingroup$
From standard facts, such as in the Wikipedia article Arithmetic-geometric mean, $, textrm{agm}(1,k) = frac{pi/2}{K(k')},$ where $,k,$ is the modulus, $,k'=sqrt{1-k^2},$ is the complementary modulus, and $,K(k),$ is the complete elliptic integral of the first kind. In order to find $,textrm{agm}^{-1},$ you need to find the inverse of $,K().,$ This involves the Jacobi amplitude function described in the Jacobi elliptic functions article.
As a practical matter, I suggest restricting to $,0<k<1,$ although there is analytic continuation, but you may run into branch cuts or worse.
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
$endgroup$
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
$begingroup$
Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
add a comment |
$begingroup$
From standard facts, such as in the Wikipedia article Arithmetic-geometric mean, $, textrm{agm}(1,k) = frac{pi/2}{K(k')},$ where $,k,$ is the modulus, $,k'=sqrt{1-k^2},$ is the complementary modulus, and $,K(k),$ is the complete elliptic integral of the first kind. In order to find $,textrm{agm}^{-1},$ you need to find the inverse of $,K().,$ This involves the Jacobi amplitude function described in the Jacobi elliptic functions article.
As a practical matter, I suggest restricting to $,0<k<1,$ although there is analytic continuation, but you may run into branch cuts or worse.
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
$endgroup$
From standard facts, such as in the Wikipedia article Arithmetic-geometric mean, $, textrm{agm}(1,k) = frac{pi/2}{K(k')},$ where $,k,$ is the modulus, $,k'=sqrt{1-k^2},$ is the complementary modulus, and $,K(k),$ is the complete elliptic integral of the first kind. In order to find $,textrm{agm}^{-1},$ you need to find the inverse of $,K().,$ This involves the Jacobi amplitude function described in the Jacobi elliptic functions article.
As a practical matter, I suggest restricting to $,0<k<1,$ although there is analytic continuation, but you may run into branch cuts or worse.
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
edited Dec 24 '18 at 6:03
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
answered Dec 24 '18 at 4:42
SomosSomos
14.4k11336
14.4k11336
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
$begingroup$
Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
add a comment |
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
$begingroup$
Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
$begingroup$
Thanks for the quick and helpful response.
$endgroup$
– user10560552
Dec 24 '18 at 5:06
$begingroup$
Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
$begingroup$
Just a question, is this formula still viable with complex numbers.
$endgroup$
– user10560552
Dec 24 '18 at 5:21
add a comment |
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$begingroup$
It is not clear to me what you want to do. Do you want to find $agm^{-1}$ or to reverse the agm iteration? Can you give a simple example?
$endgroup$
– Somos
Dec 24 '18 at 4:04
$begingroup$
Of course, the $agm(1,2) = 1.4567910310469068692...$, what i want to know is how to calculate $agm^{-1}(1.4567910310469068692)=(1,2)$ or $agm^{-1}(1.4567910310469068692)=2$ only for the special case of $agm(1,x)$
$endgroup$
– user10560552
Dec 24 '18 at 4:16