How to derive Lobachevsky's formula for the angle of parallelism
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I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:
$$Pi(x)=2tan^{-1}left(e^{-x}right)$$
Any help? Thanks
P.S. I couldn't find anything by google search either.
differential-geometry hyperbolic-geometry noneuclidean-geometry
edited Dec 24 '18 at 9:38
Blue
49.4k870157
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
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add a comment |
$begingroup$
I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:
$$Pi(x)=2tan^{-1}left(e^{-x}right)$$
Any help? Thanks
P.S. I couldn't find anything by google search either.
differential-geometry hyperbolic-geometry noneuclidean-geometry
edited Dec 24 '18 at 9:38
Blue
49.4k870157
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
$endgroup$
$begingroup$
have a look at en.wikipedia.org/wiki/Angle_of_parallelism, I doubt there is a proof it is more an axiom defining what a curvature of -1 means
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– Willemien
Jul 28 '16 at 5:52
add a comment |
$begingroup$
I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:
$$Pi(x)=2tan^{-1}left(e^{-x}right)$$
Any help? Thanks
P.S. I couldn't find anything by google search either.
differential-geometry hyperbolic-geometry noneuclidean-geometry
edited Dec 24 '18 at 9:38
Blue
49.4k870157
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
$endgroup$
I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:
$$Pi(x)=2tan^{-1}left(e^{-x}right)$$
Any help? Thanks
P.S. I couldn't find anything by google search either.
differential-geometry hyperbolic-geometry noneuclidean-geometry
differential-geometry hyperbolic-geometry noneuclidean-geometry
edited Dec 24 '18 at 9:38
Blue
49.4k870157
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
edited Dec 24 '18 at 9:38
Blue
49.4k870157
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
edited Dec 24 '18 at 9:38
Blue
49.4k870157
edited Dec 24 '18 at 9:38
Blue
49.4k870157
edited Dec 24 '18 at 9:38
Blue
49.4k870157
49.4k870157
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
asked Jul 21 '16 at 12:37
Marek KurczynskiMarek Kurczynski
115210
115210
$begingroup$
have a look at en.wikipedia.org/wiki/Angle_of_parallelism, I doubt there is a proof it is more an axiom defining what a curvature of -1 means
$endgroup$
– Willemien
Jul 28 '16 at 5:52
add a comment |
$begingroup$
have a look at en.wikipedia.org/wiki/Angle_of_parallelism, I doubt there is a proof it is more an axiom defining what a curvature of -1 means
$endgroup$
– Willemien
Jul 28 '16 at 5:52
$begingroup$
have a look at en.wikipedia.org/wiki/Angle_of_parallelism, I doubt there is a proof it is more an axiom defining what a curvature of -1 means
$endgroup$
– Willemien
Jul 28 '16 at 5:52
$begingroup$
have a look at en.wikipedia.org/wiki/Angle_of_parallelism, I doubt there is a proof it is more an axiom defining what a curvature of -1 means
$endgroup$
– Willemien
Jul 28 '16 at 5:52
add a comment |
1 Answer
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Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable
$$arctan(z) = frac{1}{2}ileft[ln(1-iz) -ln(1+iz)right]$$
Maybe it can help you somehow.
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
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1 Answer
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1 Answer
1
active
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active
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active
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votes
$begingroup$
Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable
$$arctan(z) = frac{1}{2}ileft[ln(1-iz) -ln(1+iz)right]$$
Maybe it can help you somehow.
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
$endgroup$
add a comment |
$begingroup$
Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable
$$arctan(z) = frac{1}{2}ileft[ln(1-iz) -ln(1+iz)right]$$
Maybe it can help you somehow.
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
$endgroup$
add a comment |
$begingroup$
Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable
$$arctan(z) = frac{1}{2}ileft[ln(1-iz) -ln(1+iz)right]$$
Maybe it can help you somehow.
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
$endgroup$
Hint An interesting relation between arcus tangent and exponential function are made clear in one complex variable
$$arctan(z) = frac{1}{2}ileft[ln(1-iz) -ln(1+iz)right]$$
Maybe it can help you somehow.
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
answered Jul 30 '16 at 21:22
mathreadlermathreadler
15.4k72263
15.4k72263
add a comment |
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$begingroup$
have a look at en.wikipedia.org/wiki/Angle_of_parallelism, I doubt there is a proof it is more an axiom defining what a curvature of -1 means
$endgroup$
– Willemien
Jul 28 '16 at 5:52