First Midterm Exam Solutions
... • Alternatively, find the line L which passes through P and is parallel to n. This line intersects the plane at R. Since the line passes through P (7, −2, −1) and is parallel to n = h2, −3, −1i, it can be described parametrically as h7, −2, −1i + th2, −3, −1i = h7 + 2t, −2 − 3t, −1 − ti. To find whe ...
... • Alternatively, find the line L which passes through P and is parallel to n. This line intersects the plane at R. Since the line passes through P (7, −2, −1) and is parallel to n = h2, −3, −1i, it can be described parametrically as h7, −2, −1i + th2, −3, −1i = h7 + 2t, −2 − 3t, −1 − ti. To find whe ...
Primality - Factorization
... Why does it work? If p and q are two prime divisors of n, then y 2 = x3 + ax + b (mod n) implies the same equation also modulo p and modulo q. These two smaller elliptic curves are now genuine groups. If these groups have Np and Nq elements, respectively, then for any point P on the original curve, ...
... Why does it work? If p and q are two prime divisors of n, then y 2 = x3 + ax + b (mod n) implies the same equation also modulo p and modulo q. These two smaller elliptic curves are now genuine groups. If these groups have Np and Nq elements, respectively, then for any point P on the original curve, ...
Moreover, if one passes to other groups, then there are σ Eisenstein
... Elliptic Modular Forms and Their Applications ...
... Elliptic Modular Forms and Their Applications ...
MATH 254A: RINGS OF INTEGERS AND DEDEKIND DOMAINS 1
... We are now ready to prove the following theorem: Theorem 2.1. A ring of integers OK is a Dedekind domain: i.e., it satisfies (i) OK is Noetherian; (ii) Every non-zero prime ideal of OK is maximal; (iii) OK is integrally closed in its field of fractions. Algebraic Geometry Remark 2.2. From an algebra ...
... We are now ready to prove the following theorem: Theorem 2.1. A ring of integers OK is a Dedekind domain: i.e., it satisfies (i) OK is Noetherian; (ii) Every non-zero prime ideal of OK is maximal; (iii) OK is integrally closed in its field of fractions. Algebraic Geometry Remark 2.2. From an algebra ...
A LOWER BOUND FOR AVERAGE VALUES OF DYNAMICAL
... z1 , . . . , zN belong to the filled Julia set of ϕ. Note that the case where ϕ is a polynomial is simpler than the general case, due to the fact that polynomial maps have a superattracting fixed point at infinity. When K is non-archimedean, the −CN log N term in (1.5) can be replaced by −CN , see [ ...
... z1 , . . . , zN belong to the filled Julia set of ϕ. Note that the case where ϕ is a polynomial is simpler than the general case, due to the fact that polynomial maps have a superattracting fixed point at infinity. When K is non-archimedean, the −CN log N term in (1.5) can be replaced by −CN , see [ ...
introduction to advanced mathematics, c1
... The construction and presentation of mathematical arguments through appropriate use of logical deduction and precise statements involving correct use of symbols and appropriate connecting language pervade the whole of mathematics at this level. These skills, and the Competence Statements below, are ...
... The construction and presentation of mathematical arguments through appropriate use of logical deduction and precise statements involving correct use of symbols and appropriate connecting language pervade the whole of mathematics at this level. These skills, and the Competence Statements below, are ...
Solutions - UConn Math
... If x 6= 0, then λ = 25 , while if y 6= 0 then λ = 100 and if z 6= 0, then λ = 41 . Thus, at least two of x, y, z must be 0, while all three cannot be 0 since the origin is not on the ellipsoid. The extreme values must therefore occur at the points where the ellipsoid intersects the coordinate axes. ...
... If x 6= 0, then λ = 25 , while if y 6= 0 then λ = 100 and if z 6= 0, then λ = 41 . Thus, at least two of x, y, z must be 0, while all three cannot be 0 since the origin is not on the ellipsoid. The extreme values must therefore occur at the points where the ellipsoid intersects the coordinate axes. ...
Graphs of Sine and Cosine Functions
... • We can have curves with a different period or amplitude. We need to add something to our equations. The general form of the equations of the sine and cosine curves now is: y = A sin bx ...
... • We can have curves with a different period or amplitude. We need to add something to our equations. The general form of the equations of the sine and cosine curves now is: y = A sin bx ...
Euclid Meets Bézout: Intersecting Algebraic Plane Curves with the
... The special case m = n = 1 of Bézout’s Theorem tells us that two (distinct) lines in the projective plane always intersect at a point (no parallel lines in K P2 !). But in general finding the intersection points, and especially their multiplicities, is a nontrivial business. It is this process whic ...
... The special case m = n = 1 of Bézout’s Theorem tells us that two (distinct) lines in the projective plane always intersect at a point (no parallel lines in K P2 !). But in general finding the intersection points, and especially their multiplicities, is a nontrivial business. It is this process whic ...
Presentation
... 1. Alice and Bob agree on an Elliptic Curve E (specified by the field F and parameters a, b) and a base point g on E. 2. (a) Alice secretly chooses integer x, computes X = xg and sends it to Bob. (b) Bob secretly chooses integer y, computes Y = yg and sends it to Alice. 3. (a) Alice computes: xY = x ...
... 1. Alice and Bob agree on an Elliptic Curve E (specified by the field F and parameters a, b) and a base point g on E. 2. (a) Alice secretly chooses integer x, computes X = xg and sends it to Bob. (b) Bob secretly chooses integer y, computes Y = yg and sends it to Alice. 3. (a) Alice computes: xY = x ...
Unit 2 Exercise 1 - Official Mathematics Revision Website
... Given that the centre of the above circle is C(4,3) and the radius of the circle is 5, find the coordinates of the intersection of the circle with the axis. Find the equation of the tangents at these three points and show two of these tangents are parallel. Find the fourth point on the circle at whi ...
... Given that the centre of the above circle is C(4,3) and the radius of the circle is 5, find the coordinates of the intersection of the circle with the axis. Find the equation of the tangents at these three points and show two of these tangents are parallel. Find the fourth point on the circle at whi ...
Topic 7 - Polynomials
... Find the coordinates of the points of intersection namely A, B and C where the line meets the graph . (NB the diagram is not drawn to scale) ...
... Find the coordinates of the points of intersection namely A, B and C where the line meets the graph . (NB the diagram is not drawn to scale) ...
The infinite fern of Galois representations of type U(3) Gaëtan
... A first important step in the proof of Theorem A is a result from the theory of p-adic families of automorphic forms for the definite unitary group U(3) ([2],[1]). Fix v a prime of E dividing p, so that Ev = Qp . Define a refined modular point as a pair (ρΠ , (ϕ1 /pk1 , ϕ2 /pk2 , ϕ3 /pk3 )) in X (ρ) ...
... A first important step in the proof of Theorem A is a result from the theory of p-adic families of automorphic forms for the definite unitary group U(3) ([2],[1]). Fix v a prime of E dividing p, so that Ev = Qp . Define a refined modular point as a pair (ρΠ , (ϕ1 /pk1 , ϕ2 /pk2 , ϕ3 /pk3 )) in X (ρ) ...
Add Maths Gym Sec 34.indb
... If the discriminant is negative, the straight line and the curve have no common point. If the discriminant is zero, the straight line touches the curve at one and only one point. It is a tangent to the curve. If the discriminant is positive, the straight line intersects the curve at two distinct poi ...
... If the discriminant is negative, the straight line and the curve have no common point. If the discriminant is zero, the straight line touches the curve at one and only one point. It is a tangent to the curve. If the discriminant is positive, the straight line intersects the curve at two distinct poi ...
arXiv:1705.08225v1 [math.NT] 23 May 2017
... Remark 2. According to the Langlands philosophy, the Galois representations associated to algebraic varieties are expected to be automorphic. In fact, this conjectural correspondence should be functorial: not only the Galois representations, but also the morphisms between them should have an automor ...
... Remark 2. According to the Langlands philosophy, the Galois representations associated to algebraic varieties are expected to be automorphic. In fact, this conjectural correspondence should be functorial: not only the Galois representations, but also the morphisms between them should have an automor ...
Solutions - SD308.org
... Consider y = x − 5x + 4. 5. Write the equation of the axis of symmetry. SOLUTION: To find the axis of symmetry, use the equation ...
... Consider y = x − 5x + 4. 5. Write the equation of the axis of symmetry. SOLUTION: To find the axis of symmetry, use the equation ...
Engr 6002: Ship Structures
... Statically Indeterminate Beams • Consider beam with fixed support at A and roller support at B. • From free-body diagram, note that there are four unknown reaction components. • Conditions for static equilibrium yield Fx 0 Fy 0 M A 0 ...
... Statically Indeterminate Beams • Consider beam with fixed support at A and roller support at B. • From free-body diagram, note that there are four unknown reaction components. • Conditions for static equilibrium yield Fx 0 Fy 0 M A 0 ...
Section 4.3-4.6 Section 4.4 Section 4.5 Section 4.6
... 14. A curve that passes through the point (0, 25) has the property that the slope at every point (x, y) is eight times the y coordinate. Find the equation of the curve. 15. A pie is taken from an oven, where the temperature is 450◦ , to a 75◦ room. After 15 minutes, the temperature of the pie reads ...
... 14. A curve that passes through the point (0, 25) has the property that the slope at every point (x, y) is eight times the y coordinate. Find the equation of the curve. 15. A pie is taken from an oven, where the temperature is 450◦ , to a 75◦ room. After 15 minutes, the temperature of the pie reads ...
Arithmetic and Hyperbolic Geometry
... 1. (Lang, 1965) Replace the error term shA(P) in the arithmetic variant of Theorem 2.2 with something sharper, e.g., (1 + s)loghA(P). 2. (Vojta, 1987) Prove a formula similar to (1) for rational points on varieties of higher dimension. In that case, however, one would first have to exclude a proper ...
... 1. (Lang, 1965) Replace the error term shA(P) in the arithmetic variant of Theorem 2.2 with something sharper, e.g., (1 + s)loghA(P). 2. (Vojta, 1987) Prove a formula similar to (1) for rational points on varieties of higher dimension. In that case, however, one would first have to exclude a proper ...
Math 113 Final Exam Solutions
... False: if Q were ha/bi, where a, b ∈ Z\{0}, then we would have that a/2b is not in Q. e) The field of quotients of Q[π] is isomorphic to Q(x), the field of rational functions. True, since π is transcendental over Q we have Q[π] ∼ = Q[x] and so its field of quotients is isomorphic to Q(x). f) A commu ...
... False: if Q were ha/bi, where a, b ∈ Z\{0}, then we would have that a/2b is not in Q. e) The field of quotients of Q[π] is isomorphic to Q(x), the field of rational functions. True, since π is transcendental over Q we have Q[π] ∼ = Q[x] and so its field of quotients is isomorphic to Q(x). f) A commu ...
x - Math TAMU
... 12. At what points on the curve x = t3 + 4t, y = 6t2 is the tangent parallel to the line with the equations x = −7t, y = 12t − 5? 13. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.9 ft/s, how fast is the angle between the ladd ...
... 12. At what points on the curve x = t3 + 4t, y = 6t2 is the tangent parallel to the line with the equations x = −7t, y = 12t − 5? 13. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.9 ft/s, how fast is the angle between the ladd ...
MATH 160 MIDTERM SOLUTIONS
... F, since in handwriting they are sometimes indistiguishable.) No explanations are required in this problem. (a) There exist integers x and y satisfying 709x + 100y = 4. TRUE, since gcd(709, 100) = 1, which divides 4. (The gcd can be computed using the Euclidean algorithm, or by observing that the on ...
... F, since in handwriting they are sometimes indistiguishable.) No explanations are required in this problem. (a) There exist integers x and y satisfying 709x + 100y = 4. TRUE, since gcd(709, 100) = 1, which divides 4. (The gcd can be computed using the Euclidean algorithm, or by observing that the on ...
June 2012
... Show that the four points (0, –1), (6, 7). ( – 2, 3) and (8, 3) are the vertices of a rectangle. Find the slope of a line which passes through the points (3, 2) and (-1, 5). Find the centre and radius of the given circle x2 + y2 – 4x + 6y=12. Find the equation of the parabola whose focus is ( –3, 2) ...
... Show that the four points (0, –1), (6, 7). ( – 2, 3) and (8, 3) are the vertices of a rectangle. Find the slope of a line which passes through the points (3, 2) and (-1, 5). Find the centre and radius of the given circle x2 + y2 – 4x + 6y=12. Find the equation of the parabola whose focus is ( –3, 2) ...
non-abelian classfields over function fields in special cases
... variable over finite fields (abbrev. function fields). An origin of this idea was the following question: is there any identity between the Riemann ^-function of an arithmetic field K and a " Selberg type ^-function " of a discrete group T, in some cases? Indeed, if we assume such an identity for a ...
... variable over finite fields (abbrev. function fields). An origin of this idea was the following question: is there any identity between the Riemann ^-function of an arithmetic field K and a " Selberg type ^-function " of a discrete group T, in some cases? Indeed, if we assume such an identity for a ...
Number Theory: Elliptic Curves, Problem Sheet 3
... line hits a k-point other than (0, 0) if and only if the quadratic x2 f3 (1, t) + xf2 (1, t) + f1 (1, t) has roots in k, which happens iff f2 (1, t)2 − 4f1 (1, t)f3 (1, t) is a square in k. Hence if we write s2 = f2 (1, t)2 − 4f1 (1, t)f3 (1, t) then the formula for the roots of a quadratic equatio ...
... line hits a k-point other than (0, 0) if and only if the quadratic x2 f3 (1, t) + xf2 (1, t) + f1 (1, t) has roots in k, which happens iff f2 (1, t)2 − 4f1 (1, t)f3 (1, t) is a square in k. Hence if we write s2 = f2 (1, t)2 − 4f1 (1, t)f3 (1, t) then the formula for the roots of a quadratic equatio ...