Cryptography Reference
In-Depth Information
Figure 2.1(a) on page 10. The points don't appear on the graph
because y is imaginary. For the curve in Figure 2.1(b) on page 10, x
moves to the left along the x -axis, from the point on the x -axis back
to the point at infinity, corresponding to the fact that ω 1 = 2 ω 2 + it
for appropriate t (see Exercise 9.5).
9.7 Define the elliptic integral of the second kind to be
E ( k )= 1
0
1
k 2 x 2
1 − x 2
dx,
1 <k< 1 .
(a) Show that
E ( k )= π/ 2
0
(1 − k 2 sin 2 θ ) 1 / 2 dθ.
(b) Show that the arc length of the ellipse
x 2
a 2
+ y 2
b 2
=1
with b ≥ a> 0equals4 bE ( 1 ( a/b ) 2 ).
This connection with ellipses is the origin of the name “elliptic inte-
gral.” The relation between elliptic integrals and elliptic curves, as in
Section 9.4, is the origin of the name “elliptic curve.”
For more on
elliptic integrals, see [78].
9.8 Let E be the elliptic curve y 2 =4 x 3
4 x . Show that
ω 2 =
1
1
dx
x ( x 2
1) = 1
t 3 / 4 (1
t ) 1 / 2 dt = β (1 / 4 , 1 / 2) ,
2
0
where β ( p, q )= 0
t p− 1 (1
t ) q− 1 dt is the beta function . A classical
result says that
β ( p, q )= Γ( p )Γ( q )
Γ( p + q ) .
Therefore,
ω 2 = 1
Γ(1 / 4)Γ(1 / 2)
Γ(3 / 4)
.
2
 
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