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a conjectured continued fraction for $tanleft(frac{zpi}{4z+2n}right)$
Continued fraction for $tan(nx)$How does $cos(2pi/257)$ look like in real radicals?conjectured general continued fraction for the quotient of gamma functionsContinued fraction for $tan(nx)$Continued fraction and double series.a conjectured continued-fraction for $displaystylecotleft(frac{zpi}{4z+2n}right)$ that leads to a new limit for $pi$conjectured general continued fraction for the quotient of gamma functionsRamanujan's continued fraction for ratio of gamma valuesFound an recursive identity (involving a continued fraction) for which some simplification is needed.Continued Fraction Identity ProblemA conjectured continued fraction for $phi^phi$conjectured continued fraction related to certain polynomialsConjectured continued fraction for the Jacobi theta function $vartheta_{4}(q)$
$begingroup$
Given a complex number $begin{aligned}frac{z}{n}=x+iyend{aligned}$ and a gamma function $Gamma(z)$ with $xgt0$, it is conjectured that the following continued fraction for $displaystyletanleft(frac{zpi}{4z+2n}right)$ is true
$$begin{split}displaystyletanleft(frac{zpi}{4z+2n}right)&=frac{displaystyleGammaleft(frac{z+n}{4z+2n}right)Gammaleft(frac{3z+n}{4z+2n}right)}{displaystyleGammaleft(frac{z}{4z+2n}right)Gammaleft(frac{3z+2n}{4z+2n}right)}\&=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}end{split}$$
Corollaries:
By taking the limit(which follows after abel's theorem)
$$
begin{aligned}lim_{zto0}frac{displaystyletanleft(frac{zpi}{4z+2}right)}{2z}=frac{pi}{4}end{aligned},
$$
we recover the well known continued fraction for $pi$
$$begin{aligned}cfrac{4}{1+cfrac{1^2}{3+cfrac{2^2}{5+cfrac{3^2}{7+ddots}}}}=piend{aligned}$$
If we let $z=1$ and $n=2$,then we have the square root of $2$ $$begin{aligned}{1+cfrac{1}{2+cfrac{1} {2+cfrac{1}{2+cfrac{1}{2+ddots}}}}}=sqrt{2}end{aligned}$$
Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $xgt0$?
Update:I initially defined the continued fraction $displaystyletanleft(frac{zpi}{4z+2}right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.
number-theory special-functions gamma-function continued-fractions conjectures
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
$endgroup$
add a comment |
$begingroup$
Given a complex number $begin{aligned}frac{z}{n}=x+iyend{aligned}$ and a gamma function $Gamma(z)$ with $xgt0$, it is conjectured that the following continued fraction for $displaystyletanleft(frac{zpi}{4z+2n}right)$ is true
$$begin{split}displaystyletanleft(frac{zpi}{4z+2n}right)&=frac{displaystyleGammaleft(frac{z+n}{4z+2n}right)Gammaleft(frac{3z+n}{4z+2n}right)}{displaystyleGammaleft(frac{z}{4z+2n}right)Gammaleft(frac{3z+2n}{4z+2n}right)}\&=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}end{split}$$
Corollaries:
By taking the limit(which follows after abel's theorem)
$$
begin{aligned}lim_{zto0}frac{displaystyletanleft(frac{zpi}{4z+2}right)}{2z}=frac{pi}{4}end{aligned},
$$
we recover the well known continued fraction for $pi$
$$begin{aligned}cfrac{4}{1+cfrac{1^2}{3+cfrac{2^2}{5+cfrac{3^2}{7+ddots}}}}=piend{aligned}$$
If we let $z=1$ and $n=2$,then we have the square root of $2$ $$begin{aligned}{1+cfrac{1}{2+cfrac{1} {2+cfrac{1}{2+cfrac{1}{2+ddots}}}}}=sqrt{2}end{aligned}$$
Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $xgt0$?
Update:I initially defined the continued fraction $displaystyletanleft(frac{zpi}{4z+2}right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.
number-theory special-functions gamma-function continued-fractions conjectures
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
$endgroup$
2
$begingroup$
That is impressive. How did you come up with it?
$endgroup$
– marty cohen
Oct 13 '15 at 16:20
2
$begingroup$
Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question.
$endgroup$
– Yuriy S
Mar 13 '16 at 19:01
$begingroup$
@Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far
$endgroup$
– Yuriy S
Apr 7 '16 at 17:04
$begingroup$
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 12 '18 at 17:02
add a comment |
$begingroup$
Given a complex number $begin{aligned}frac{z}{n}=x+iyend{aligned}$ and a gamma function $Gamma(z)$ with $xgt0$, it is conjectured that the following continued fraction for $displaystyletanleft(frac{zpi}{4z+2n}right)$ is true
$$begin{split}displaystyletanleft(frac{zpi}{4z+2n}right)&=frac{displaystyleGammaleft(frac{z+n}{4z+2n}right)Gammaleft(frac{3z+n}{4z+2n}right)}{displaystyleGammaleft(frac{z}{4z+2n}right)Gammaleft(frac{3z+2n}{4z+2n}right)}\&=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}end{split}$$
Corollaries:
By taking the limit(which follows after abel's theorem)
$$
begin{aligned}lim_{zto0}frac{displaystyletanleft(frac{zpi}{4z+2}right)}{2z}=frac{pi}{4}end{aligned},
$$
we recover the well known continued fraction for $pi$
$$begin{aligned}cfrac{4}{1+cfrac{1^2}{3+cfrac{2^2}{5+cfrac{3^2}{7+ddots}}}}=piend{aligned}$$
If we let $z=1$ and $n=2$,then we have the square root of $2$ $$begin{aligned}{1+cfrac{1}{2+cfrac{1} {2+cfrac{1}{2+cfrac{1}{2+ddots}}}}}=sqrt{2}end{aligned}$$
Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $xgt0$?
Update:I initially defined the continued fraction $displaystyletanleft(frac{zpi}{4z+2}right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.
number-theory special-functions gamma-function continued-fractions conjectures
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
$endgroup$
Given a complex number $begin{aligned}frac{z}{n}=x+iyend{aligned}$ and a gamma function $Gamma(z)$ with $xgt0$, it is conjectured that the following continued fraction for $displaystyletanleft(frac{zpi}{4z+2n}right)$ is true
$$begin{split}displaystyletanleft(frac{zpi}{4z+2n}right)&=frac{displaystyleGammaleft(frac{z+n}{4z+2n}right)Gammaleft(frac{3z+n}{4z+2n}right)}{displaystyleGammaleft(frac{z}{4z+2n}right)Gammaleft(frac{3z+2n}{4z+2n}right)}\&=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}end{split}$$
Corollaries:
By taking the limit(which follows after abel's theorem)
$$
begin{aligned}lim_{zto0}frac{displaystyletanleft(frac{zpi}{4z+2}right)}{2z}=frac{pi}{4}end{aligned},
$$
we recover the well known continued fraction for $pi$
$$begin{aligned}cfrac{4}{1+cfrac{1^2}{3+cfrac{2^2}{5+cfrac{3^2}{7+ddots}}}}=piend{aligned}$$
If we let $z=1$ and $n=2$,then we have the square root of $2$ $$begin{aligned}{1+cfrac{1}{2+cfrac{1} {2+cfrac{1}{2+cfrac{1}{2+ddots}}}}}=sqrt{2}end{aligned}$$
Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $xgt0$?
Update:I initially defined the continued fraction $displaystyletanleft(frac{zpi}{4z+2}right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.
number-theory special-functions gamma-function continued-fractions conjectures
number-theory special-functions gamma-function continued-fractions conjectures
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
edited Mar 20 at 21:51
Daniele Tampieri
2,66221022
2,66221022
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
asked Sep 22 '15 at 8:00
NiccoNicco
1,064827
1,064827
2
$begingroup$
That is impressive. How did you come up with it?
$endgroup$
– marty cohen
Oct 13 '15 at 16:20
2
$begingroup$
Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question.
$endgroup$
– Yuriy S
Mar 13 '16 at 19:01
$begingroup$
@Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far
$endgroup$
– Yuriy S
Apr 7 '16 at 17:04
$begingroup$
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 12 '18 at 17:02
add a comment |
2
$begingroup$
That is impressive. How did you come up with it?
$endgroup$
– marty cohen
Oct 13 '15 at 16:20
2
$begingroup$
Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question.
$endgroup$
– Yuriy S
Mar 13 '16 at 19:01
$begingroup$
@Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far
$endgroup$
– Yuriy S
Apr 7 '16 at 17:04
$begingroup$
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 12 '18 at 17:02
2
2
$begingroup$
That is impressive. How did you come up with it?
$endgroup$
– marty cohen
Oct 13 '15 at 16:20
$begingroup$
That is impressive. How did you come up with it?
$endgroup$
– marty cohen
Oct 13 '15 at 16:20
2
2
$begingroup$
Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question.
$endgroup$
– Yuriy S
Mar 13 '16 at 19:01
$begingroup$
Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question.
$endgroup$
– Yuriy S
Mar 13 '16 at 19:01
$begingroup$
@Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far
$endgroup$
– Yuriy S
Apr 7 '16 at 17:04
$begingroup$
@Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far
$endgroup$
– Yuriy S
Apr 7 '16 at 17:04
$begingroup$
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 12 '18 at 17:02
$begingroup$
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 12 '18 at 17:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The proposed continued fraction
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}
end{equation}
can be written as
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z/left( 2z+n right)}{1+cfrac{(n)/left( 2z+n right)cdot(4z+n)/left( 2z+n right)} {3+cfrac{(2z+2n)/left( 2z+n right)cdot(6z+2n)/left( 2z+n right)}{5+cfrac{(4z+3n)/left( 2z+n right)cdot(8z+3n)/left( 2z+n right)}{7+ddots}}}}
end{equation}
Denoting $u=cfrac{z}{4z+2n}$, the factors of the numerators are
begin{equation}
frac{n}{2z+n}=1-4u,;quadfrac{4z+n}{2z+n}=1+4u,;quadfrac{2z+2n}{2z+n}=2-4u,;quadfrac{6z+2n}{2z+n}=2+4u,;cdots
end{equation}
Then, the fraction can be simplified as
begin{equation}
displaystyletanleft(pi uright)=cfrac{4u}{1+cfrac{cfrac{1-16u^2}{1cdot3}} {1+cfrac{cfrac{4-16u^2}{3cdot5}}{1+cfrac{cfrac{9-16u^2}{5cdot7}}{1+ddots}}}}
end{equation}
It is thus a special case of the continued fraction found in this answer:
begin{equation}
tanleft(alphatan^{-1}zright)=cfrac{alpha z}{1+cfrac{frac{(1^2-alpha^2)z^2}{1cdot 3}} {1+cfrac{frac{(2^2-alpha^2)z^2}{3cdot 5}}{1+cfrac{frac{(3^2-alpha^2)z^2}{5cdot 7}}{1+ddots}}}}
end{equation}
here $z=1$ and $alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
$endgroup$
add a comment |
$begingroup$
The ratio
$$tandfrac{pi z}{4z+2n}
= dfrac{Gammaleft(dfrac{z+n}{4z+2n}right)Gammaleft(dfrac{3z+n}{4z+2n}right)}{Gammaleft(dfrac{z}{4z+2n}right)Gammaleft(dfrac{3z+2n}{4z+2n}right)}hspace{100mu}tag1$$
can be obtained, applying "real" identity
$$sinpi x = dfracpi{Gamma(x)Gamma(1-x)}hspace{100mu}tag2$$
to the expression
$$tandfracpi2dfrac z{2z+n}
= dfrac{sinpidfrac z{4z+2n}}{sinpidfrac{z+n}{4z+2n}},$$
so it looks nice and quite correct.
Continued fraction can be obtained, using known continued fraction of the tangent function in the form of
$$tan dfrac{pi x}4 = cfrac x{1+operatorname{
Large K}hspace{-27mu}phantom{Big|}_{k=1}^{large ^{,infty}}cfrac{(2k-1)^2-x^2}2}hspace{100mu}tag3$$
with
$$x=dfrac{2z}{2z+n}.$$
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
The proposed continued fraction
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}
end{equation}
can be written as
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z/left( 2z+n right)}{1+cfrac{(n)/left( 2z+n right)cdot(4z+n)/left( 2z+n right)} {3+cfrac{(2z+2n)/left( 2z+n right)cdot(6z+2n)/left( 2z+n right)}{5+cfrac{(4z+3n)/left( 2z+n right)cdot(8z+3n)/left( 2z+n right)}{7+ddots}}}}
end{equation}
Denoting $u=cfrac{z}{4z+2n}$, the factors of the numerators are
begin{equation}
frac{n}{2z+n}=1-4u,;quadfrac{4z+n}{2z+n}=1+4u,;quadfrac{2z+2n}{2z+n}=2-4u,;quadfrac{6z+2n}{2z+n}=2+4u,;cdots
end{equation}
Then, the fraction can be simplified as
begin{equation}
displaystyletanleft(pi uright)=cfrac{4u}{1+cfrac{cfrac{1-16u^2}{1cdot3}} {1+cfrac{cfrac{4-16u^2}{3cdot5}}{1+cfrac{cfrac{9-16u^2}{5cdot7}}{1+ddots}}}}
end{equation}
It is thus a special case of the continued fraction found in this answer:
begin{equation}
tanleft(alphatan^{-1}zright)=cfrac{alpha z}{1+cfrac{frac{(1^2-alpha^2)z^2}{1cdot 3}} {1+cfrac{frac{(2^2-alpha^2)z^2}{3cdot 5}}{1+cfrac{frac{(3^2-alpha^2)z^2}{5cdot 7}}{1+ddots}}}}
end{equation}
here $z=1$ and $alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
$endgroup$
add a comment |
$begingroup$
The proposed continued fraction
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}
end{equation}
can be written as
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z/left( 2z+n right)}{1+cfrac{(n)/left( 2z+n right)cdot(4z+n)/left( 2z+n right)} {3+cfrac{(2z+2n)/left( 2z+n right)cdot(6z+2n)/left( 2z+n right)}{5+cfrac{(4z+3n)/left( 2z+n right)cdot(8z+3n)/left( 2z+n right)}{7+ddots}}}}
end{equation}
Denoting $u=cfrac{z}{4z+2n}$, the factors of the numerators are
begin{equation}
frac{n}{2z+n}=1-4u,;quadfrac{4z+n}{2z+n}=1+4u,;quadfrac{2z+2n}{2z+n}=2-4u,;quadfrac{6z+2n}{2z+n}=2+4u,;cdots
end{equation}
Then, the fraction can be simplified as
begin{equation}
displaystyletanleft(pi uright)=cfrac{4u}{1+cfrac{cfrac{1-16u^2}{1cdot3}} {1+cfrac{cfrac{4-16u^2}{3cdot5}}{1+cfrac{cfrac{9-16u^2}{5cdot7}}{1+ddots}}}}
end{equation}
It is thus a special case of the continued fraction found in this answer:
begin{equation}
tanleft(alphatan^{-1}zright)=cfrac{alpha z}{1+cfrac{frac{(1^2-alpha^2)z^2}{1cdot 3}} {1+cfrac{frac{(2^2-alpha^2)z^2}{3cdot 5}}{1+cfrac{frac{(3^2-alpha^2)z^2}{5cdot 7}}{1+ddots}}}}
end{equation}
here $z=1$ and $alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
$endgroup$
add a comment |
$begingroup$
The proposed continued fraction
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}
end{equation}
can be written as
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z/left( 2z+n right)}{1+cfrac{(n)/left( 2z+n right)cdot(4z+n)/left( 2z+n right)} {3+cfrac{(2z+2n)/left( 2z+n right)cdot(6z+2n)/left( 2z+n right)}{5+cfrac{(4z+3n)/left( 2z+n right)cdot(8z+3n)/left( 2z+n right)}{7+ddots}}}}
end{equation}
Denoting $u=cfrac{z}{4z+2n}$, the factors of the numerators are
begin{equation}
frac{n}{2z+n}=1-4u,;quadfrac{4z+n}{2z+n}=1+4u,;quadfrac{2z+2n}{2z+n}=2-4u,;quadfrac{6z+2n}{2z+n}=2+4u,;cdots
end{equation}
Then, the fraction can be simplified as
begin{equation}
displaystyletanleft(pi uright)=cfrac{4u}{1+cfrac{cfrac{1-16u^2}{1cdot3}} {1+cfrac{cfrac{4-16u^2}{3cdot5}}{1+cfrac{cfrac{9-16u^2}{5cdot7}}{1+ddots}}}}
end{equation}
It is thus a special case of the continued fraction found in this answer:
begin{equation}
tanleft(alphatan^{-1}zright)=cfrac{alpha z}{1+cfrac{frac{(1^2-alpha^2)z^2}{1cdot 3}} {1+cfrac{frac{(2^2-alpha^2)z^2}{3cdot 5}}{1+cfrac{frac{(3^2-alpha^2)z^2}{5cdot 7}}{1+ddots}}}}
end{equation}
here $z=1$ and $alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
$endgroup$
The proposed continued fraction
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z}{2z+n+cfrac{(n)(4z+n)} {3(2z+n)+cfrac{(2z+2n)(6z+2n)}{5(2z+n)+cfrac{(4z+3n)(8z+3n)}{7(2z+n)+ddots}}}}
end{equation}
can be written as
begin{equation}
displaystyletanleft(frac{zpi}{4z+2n}right)=cfrac{2z/left( 2z+n right)}{1+cfrac{(n)/left( 2z+n right)cdot(4z+n)/left( 2z+n right)} {3+cfrac{(2z+2n)/left( 2z+n right)cdot(6z+2n)/left( 2z+n right)}{5+cfrac{(4z+3n)/left( 2z+n right)cdot(8z+3n)/left( 2z+n right)}{7+ddots}}}}
end{equation}
Denoting $u=cfrac{z}{4z+2n}$, the factors of the numerators are
begin{equation}
frac{n}{2z+n}=1-4u,;quadfrac{4z+n}{2z+n}=1+4u,;quadfrac{2z+2n}{2z+n}=2-4u,;quadfrac{6z+2n}{2z+n}=2+4u,;cdots
end{equation}
Then, the fraction can be simplified as
begin{equation}
displaystyletanleft(pi uright)=cfrac{4u}{1+cfrac{cfrac{1-16u^2}{1cdot3}} {1+cfrac{cfrac{4-16u^2}{3cdot5}}{1+cfrac{cfrac{9-16u^2}{5cdot7}}{1+ddots}}}}
end{equation}
It is thus a special case of the continued fraction found in this answer:
begin{equation}
tanleft(alphatan^{-1}zright)=cfrac{alpha z}{1+cfrac{frac{(1^2-alpha^2)z^2}{1cdot 3}} {1+cfrac{frac{(2^2-alpha^2)z^2}{3cdot 5}}{1+cfrac{frac{(3^2-alpha^2)z^2}{5cdot 7}}{1+ddots}}}}
end{equation}
here $z=1$ and $alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
answered Mar 31 at 18:42
Paul EntaPaul Enta
5,45611435
5,45611435
add a comment |
add a comment |
$begingroup$
The ratio
$$tandfrac{pi z}{4z+2n}
= dfrac{Gammaleft(dfrac{z+n}{4z+2n}right)Gammaleft(dfrac{3z+n}{4z+2n}right)}{Gammaleft(dfrac{z}{4z+2n}right)Gammaleft(dfrac{3z+2n}{4z+2n}right)}hspace{100mu}tag1$$
can be obtained, applying "real" identity
$$sinpi x = dfracpi{Gamma(x)Gamma(1-x)}hspace{100mu}tag2$$
to the expression
$$tandfracpi2dfrac z{2z+n}
= dfrac{sinpidfrac z{4z+2n}}{sinpidfrac{z+n}{4z+2n}},$$
so it looks nice and quite correct.
Continued fraction can be obtained, using known continued fraction of the tangent function in the form of
$$tan dfrac{pi x}4 = cfrac x{1+operatorname{
Large K}hspace{-27mu}phantom{Big|}_{k=1}^{large ^{,infty}}cfrac{(2k-1)^2-x^2}2}hspace{100mu}tag3$$
with
$$x=dfrac{2z}{2z+n}.$$
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
$endgroup$
add a comment |
$begingroup$
The ratio
$$tandfrac{pi z}{4z+2n}
= dfrac{Gammaleft(dfrac{z+n}{4z+2n}right)Gammaleft(dfrac{3z+n}{4z+2n}right)}{Gammaleft(dfrac{z}{4z+2n}right)Gammaleft(dfrac{3z+2n}{4z+2n}right)}hspace{100mu}tag1$$
can be obtained, applying "real" identity
$$sinpi x = dfracpi{Gamma(x)Gamma(1-x)}hspace{100mu}tag2$$
to the expression
$$tandfracpi2dfrac z{2z+n}
= dfrac{sinpidfrac z{4z+2n}}{sinpidfrac{z+n}{4z+2n}},$$
so it looks nice and quite correct.
Continued fraction can be obtained, using known continued fraction of the tangent function in the form of
$$tan dfrac{pi x}4 = cfrac x{1+operatorname{
Large K}hspace{-27mu}phantom{Big|}_{k=1}^{large ^{,infty}}cfrac{(2k-1)^2-x^2}2}hspace{100mu}tag3$$
with
$$x=dfrac{2z}{2z+n}.$$
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
$endgroup$
add a comment |
$begingroup$
The ratio
$$tandfrac{pi z}{4z+2n}
= dfrac{Gammaleft(dfrac{z+n}{4z+2n}right)Gammaleft(dfrac{3z+n}{4z+2n}right)}{Gammaleft(dfrac{z}{4z+2n}right)Gammaleft(dfrac{3z+2n}{4z+2n}right)}hspace{100mu}tag1$$
can be obtained, applying "real" identity
$$sinpi x = dfracpi{Gamma(x)Gamma(1-x)}hspace{100mu}tag2$$
to the expression
$$tandfracpi2dfrac z{2z+n}
= dfrac{sinpidfrac z{4z+2n}}{sinpidfrac{z+n}{4z+2n}},$$
so it looks nice and quite correct.
Continued fraction can be obtained, using known continued fraction of the tangent function in the form of
$$tan dfrac{pi x}4 = cfrac x{1+operatorname{
Large K}hspace{-27mu}phantom{Big|}_{k=1}^{large ^{,infty}}cfrac{(2k-1)^2-x^2}2}hspace{100mu}tag3$$
with
$$x=dfrac{2z}{2z+n}.$$
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
$endgroup$
The ratio
$$tandfrac{pi z}{4z+2n}
= dfrac{Gammaleft(dfrac{z+n}{4z+2n}right)Gammaleft(dfrac{3z+n}{4z+2n}right)}{Gammaleft(dfrac{z}{4z+2n}right)Gammaleft(dfrac{3z+2n}{4z+2n}right)}hspace{100mu}tag1$$
can be obtained, applying "real" identity
$$sinpi x = dfracpi{Gamma(x)Gamma(1-x)}hspace{100mu}tag2$$
to the expression
$$tandfracpi2dfrac z{2z+n}
= dfrac{sinpidfrac z{4z+2n}}{sinpidfrac{z+n}{4z+2n}},$$
so it looks nice and quite correct.
Continued fraction can be obtained, using known continued fraction of the tangent function in the form of
$$tan dfrac{pi x}4 = cfrac x{1+operatorname{
Large K}hspace{-27mu}phantom{Big|}_{k=1}^{large ^{,infty}}cfrac{(2k-1)^2-x^2}2}hspace{100mu}tag3$$
with
$$x=dfrac{2z}{2z+n}.$$
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
answered 2 days ago
Yuri NegometyanovYuri Negometyanov
12.5k1729
12.5k1729
add a comment |
add a comment |
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That is impressive. How did you come up with it?
$endgroup$
– marty cohen
Oct 13 '15 at 16:20
2
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Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question.
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– Yuriy S
Mar 13 '16 at 19:01
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@Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far
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– Yuriy S
Apr 7 '16 at 17:04
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I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance.
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– GNUSupporter 8964民主女神 地下教會
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