## Calculate the discriminant and the conductor of an elliptical curve using magma

To calculate the minimum discrimination and the conductor of an elliptical curve using magma for example this elliptic curve, we use this command

E: = EllipticCurve ((0,8,0,48,0));

E;

Elliptical curve defined by y ^ 2 = x ^ 3 + 8 * x ^ 2 + 48 * x on the rational field

F: = MinimalModel (E);

F;

Elliptical curve defined by y ^ 2 = x ^ 3 – x ^ 2 + 2 * x – 2 on the rational field

D: = Discriminant (F);

N: = conductor (E);

My quastion is how to calculate the discriminant and the conductor when the curve has a variable coefficient for example this elliptical curve

$$y ^ 2 = x (x-a) (x-D ^ {p} zeta ^ {k})$$
Or $$a, D$$ are integers and $$zeta ^ {k}$$ is the k-th power of unity

## Undecomposable vector bundles on elliptical curves

Let $$X$$ be a complex and smooth projective curve of genus 3,
$$E$$ an elliptical curve, and $$f: X to E$$ a finished card of
degree 2. Either $$L$$ to be a line group on $$X$$, and $$R ^ 0f _ * (L)$$
its direct image. The question is $$R ^ 0f _ * (L)$$ indecomposable?

## rational points – Two other generators necessary for an elliptical curve Z / 6

We are looking for elliptic curves of rank 8 with the Z / 6 torsion subgroup using recently discovered families similar to those of Kihara (the Kihara family is described in https://arxiv.org/pdf/1503.03667.pdf ).

Today we came across a curve

$$(0.8169768624655967629114128598.0, -451787550647310420612086468536366715869054405951830599,0)$$

Magma Calculator (http://magma.maths.usyd.edu.au/calc/) and mwrank with $$-b12$$ return 6 generators for this curve.
Magma V2.20-10 (STUDENT) runs out of memory when running the following code:

SetClassGroupBounds("GRH");
E := EllipticCurve((0,8169768624655967629114128598,0,-451787550647310420612086468536366715869054405951830599,0));
MordellWeilShaInformation(E);

Sagemath returns $$8$$ for the upper limit of the analytical rank, even for max_Delta =$$3.3$$ (we always test more than max_Delta):

E = EllipticCurve((0,8169768624655967629114128598,0,-451787550647310420612086468536366715869054405951830599,0))
E.analytic_rank_upper_bound(max_Delta=3.3,root_number="compute")

Is there a way to find two other generators?

A similar question for the $$6$$ <= Rank (E) <= $$7$$ the situation was successfully resolved by Jeremy Rouse (One more generator needed for a Z / 6 elliptical curve) but our software chokes when we try to follow its instructions.

We are ready to award a bonus $$100$$ (all my current reputation, how can I transfer it?) for the two generators. In addition, your name (with ours) will be published at the bottom of the page here: https://web.math.pmf.unizg.hr/~duje/tors/z6.html

Max

## private key – How to subtract two points on an elliptical curve?

Here is the sample code for subtracting two points.

# -*- coding: utf-8 -*-

def OnCurve(x,y): # Check if the point is on the curve
A = (y*y)%P
B = (x*x*x)%P
C = False
if A == (B + 7):
C = True
return C

m = ((yq-yp) * modinv(xq-xp,P))%P
xr = (m*m-xp-xq)%P
yr = (m*(xp-xr)-yp)%P
return (xr,yr)

def legendre_symbol(a,p):
ls = pow(a, (p - 1) / 2, p)
return -1 if ls == p - 1 else ls

def modsqrt(a,p):
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return p
elif p % 4 == 3:
return pow(a, (p + 1) / 4, p)
s = p - 1
e = 0
while s % 2 == 0:
s /= 2
e += 1
n = 2
while legendre_symbol(n, p) != -1:
n += 1
x = pow(a, (s + 1) / 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in xrange(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m

def modinv(a,n): # Extended Euclidean Algorithm in elliptic curves
lm, hm = 1,0
low, high = a%n,n
while low > 1:
ratio = high/low
nm = hm - lm * ratio
new = high - low * ratio
hm = lm
high = low
lm = nm
low = new
return lm % n

m = ((yq-yp) * modinv(xq-xp,P))%P
xr = (m*m-xp-xq)%P
yr = (m*(xp-xr)-yp)%P
return (xr,yr)

def ECsub(xp,yp,xq,yq): # EC point subtraction
X = (((yp+yq)*modinv(xq-xp,P))**2-xp-xq)%P
A = (xp + X + xq)%P
B = modsqrt(A,P)
B1 = P - B
Y = yq - (xq - X) * B
X = X % P
Y = Y % P
if not OnCurve(X,Y):
Y = yq - (xq - X) * B1
Y = Y % P
return X,Y

P = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1

Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424

print Gx,Gy

print Ax,Ay

Bx,By = ECsub(Ax,Ay,Gx,Gy)
print Bx,By

It works perfectly in Python 2.7.13.

In the earlier version of Python, there are OverflowError: (34, 'Result too large').

## Fourier analysis – Reference (fundamental soil and grad estimation, etc.): a particular elliptical PDE

in $$mathbb {R} ^ d$$, consider the following equation
$$Delta u -x cdot nabla u = f$$
or $$f$$ may be $$C ^ infty$$ and exponentially rapid decay.

I would like to know the fundamental soil. to this equation or to a gradient estimate, preferably punctually. In particular I am interested in the following quantity
begin {align *} int _ { mathbb {R} ^ d} | nabla u | ^ 2e ^ {- frac {| x | ^ 2} {2}} dx. end {align *}

Thank you!

## ca. classical analysis and odes – an identity between two elliptical integrals

I want a direct change of variable proof of identity

$$int_0 ^ ( arctan frac {sqrt {2}} { sqrt { sqrt {3}}) 1 / sqrt {1- frac {2+ sqrt {3}} {4} sin ^ 2 phi d phi} = int_0 ^ arctan frac {1} { sqrt { sqrt {3}} 1 / sqrt {1- frac {2+ sqrt {3}} {4} sin ^ 22 phi d phi} = ,.$$

I need it as part of an article on Legendre's proof of the "third singular module".

## elliptical pde – A classic problem of uniqueness in a problem of constraint minimization

Consider the following constraint minimization problem

$$inf _ { | u | _p = 1} int _ { mathbb {R} ^ N} | nabla u | ^ 2 + V (x) u ^ 2 , dx$$
or $$| cdot | _p$$ is the $$L ^ p$$ standard, $$2 and $$V (x)$$ is not negative with $$lim_ {| x | to infty} V (x) = infty$$

This is a classic approach to build a weak solution for the next semi-linear EDP
$$– Delta u + V (x) u = | u | ^ {p-2} u quad text {in} ; mathbb {R} ^ N,$$
which benefits from the homogeneity of the non-linear term.

The existence of the minimizer is a classic and relatively easy result since the condition of $$V$$ indicates the uniform decay of the minimization sequence and thus recovers the compactness of any minimization sequence.

I am curious to know if the result is unique or not unique for this minimization problem. I'm trying to find something related to Google to inspire me, but that does not work well so far.

For a particular case $$V (x) = | x | ^ 2$$ which is radial, I wonder if the minimizer must be radial? If this is done, then I think the corresponding Euler-Lagrange EDE can help us show off its singularity.

Thank you very much for any discussion or reference.

## elliptical pde – Coupled (solid-fluid) heat transfer problem in a heat sink

I'm trying to solve a heat transfer problem coupled between a solid and a fluid (I've subtended the main equations and labeled them). Eqn. (3) is the partial-integral differential equation that I arrive after pairing the two. I've added my attempt with an ansatz but this approach gives me trivial solutions.

Anyone can suggest another way to solve Eqn. (3) or report any gaps in the current approach?

$$alpha, beta, gamma$$ are constants
$$underbrace { frac { partial T_f} { partial x} + alpha (T_f – T (x, y)) = 0} _ {FLUID} Rightarrow T_f = e ^ {- alpha x} int e ^ { alpha x} T mathrm {d} x \ Rightarrow T_f = alpha e ^ {- alpha x} Bigg ( int_0 ^ xe ^ { alpha s} T (s, y) mathrm {d} s + frac {T_ {fi}} { alpha} Bigg) tag 1$$
$$T_f (x = 0) = T_ {fi}$$ is a known quantity.
$$underbrace { Bigg ( frac { partial ^ 2} { partial x ^ 2} + frac { partial ^ 2} { partial y ^ 2} Bigg) T- beta (T-T_f) = 0} _ {SOLID} tag 2$$
Replacing with (1) in (2):
$$nabla ^ 2 T – beta T + beta Bigg ( alpha e ^ {- alpha x} Bigg ( int_0 ^ xe ^ { alpha s} T (s, y) mathrm {d} s frac {T_ {fi}} { alpha} Bigg) Bigg) = 0 tag 3$$
(3) is dictated by the following limiting conditions:
$$frac { partial T} { partial x} green_ {x = 0} = frac { partial T} { partial x} green_ {x = L} = frac { partial T} { partial y} green_ {y = d} = 0, frac { T partial} { partial y} green_ {y = 0} = gamma$$

Attempt
Using the ansatz:
$$T (x, y) = sum_ {k = 0} ^ { infty} f_k (y) cos ( frac {n pi x} {L}) tag 4$$
and substituting in (3) and then performing the integrations and using orthogonality, we come to
$$f_k & # 39; (y) = Big (( tfac {n pi} {L}) ^ 2+ beta Big) f_k (y) tag 5$$
When I try to apply the $$y$$boundary-conditions on (5), it results in a trivial solution.

## co.combinatorics – Weierstrass elliptical function as Laurent series

Could someone help me understand how
$$f_0 (z) = wp ( logz; i pi, log rho)$$
or $$wp$$ denotes the elliptical function of Weierstrass and $$i pi$$, $$log rho$$ are his half-periods.
$$f_0 (z) = sum_ {n = 1} ^ { infty} frac {n rho ^ {2n} z ^ {- n}} {1- rho ^ {2n}} + sum_ { n = 1} ^ { infty} frac {nz ^ {n}} {1- rho ^ {2n}}.$$
This was obtained on page 214 in Z. Nehari and B. Schwarz, on the coefficients of the Laurent univalent series, Proc. Bitter. Math. Soc., 1954, 212-217.
I learned that it was a Fourier series $$z$$ and a series of Laurent in $$e ^ {2 pi i z}$$
$$f_0 (z) = sum_ {n = – infty} ^ { infty} a_n e ^ {2 pi i z}$$
then
$$wp left ( frac { log z} {2 pi i} right) = sum_ {n = – infty} ^ { infty} a_n z ^ n,$$
how to find the Fourier coefficients using
$$sum_ {m = – infty} ^ { infty} frac {1} {(z + m) ^ 2} = frac {d} {d} {d}} frac {1} {1- e ^ {2 pi iz}} = frac {2 pi ie ^ {2 pi iz}} {{1 -e ^ {2 pi iz}) ^ 2} = sum_ {k geq 0} (2 pi ik) e ^ {2 pi iz}.$$

I've also found that
$$wp (z; tau) = frac {1} {z ^ 2} + sum _ {(n, m) neq (0,0)} left ( frac {1} {(zm tau + n) ^ 2} – frac {1} {(m tau + n) ^ 2} right)$$
and
$$frac {1} {(2 pi i) ^ 2} wp (z; tau) = frac {1} {12} + sum_ {m = – infty} ^ { infty} frac {q ^ m _ { tau} q_z} {(1 – q ^ m _ { tau} q_z) ^ 2} – 2 sum_ {n = 1} ^ { infty} frac {nq ^ n _ { tau}} {1-q ^ n _ { tau}}$$

## Check if a point is on an elliptical curve

I have the point (2.1) and I would like to check if it's on my elliptical curve.

The elliptic curve is defined by y ^ 2 = x ^ 3 + 3x + 1

My solution to check if the point is on the curve was to replace it with the equation:

(1) ^ 2 = (2) ^ 3 + 3 (1) + 1

At the inspection, it can be seen that the left side is not equal to the right side and that the point is not on the curve.

However, I suspect that it is inaccurate. I was just looking for directions. Thank you very much.