
Algorithms for Factoring Square-Free Polynomials over
... of a(x), say ai (x), divides v(x) − s. Then ai (x) also divides m(v(x)) since m(v(x)) ≡ 0 mod a(x). From m(x) = q(x) · (x − s) + r, we have m(v(x)) = q(v(x)) · (v(x) − s) + r, which implies that ai (x) | r, since ai (x) | m(v(x)) and ai (x) | (v(x) − s). But r is a constant, so this is a contradicti ...
... of a(x), say ai (x), divides v(x) − s. Then ai (x) also divides m(v(x)) since m(v(x)) ≡ 0 mod a(x). From m(x) = q(x) · (x − s) + r, we have m(v(x)) = q(v(x)) · (v(x) − s) + r, which implies that ai (x) | r, since ai (x) | m(v(x)) and ai (x) | (v(x) − s). But r is a constant, so this is a contradicti ...
MATH NEWS
... Degree of a Monomial: The degree of a non-zero monomial is the sum of the exponents of the variable symbols that appear in the monomial. Standard Form of a Polynomial Expression in One Variable: A polynomial expression with one variable symbol is in standard form if it is expressed as ...
... Degree of a Monomial: The degree of a non-zero monomial is the sum of the exponents of the variable symbols that appear in the monomial. Standard Form of a Polynomial Expression in One Variable: A polynomial expression with one variable symbol is in standard form if it is expressed as ...
Sample homework solutions for 2.2 Jim Brown
... 2 | c2 , which implies 2 | c. So we can write c = 2k for some integer k. Our equation becomes 2k 2 = 3d2 . This shows that 2|3d2 , but since 2 - 3, we must have 2|d2 by Proposition 2.5 of Chapter 1. Thus √ 2|d,√a contradiction to the fact that gcd(c, d) = 1. Thus it must be that 2 + 3 is irrational. ...
... 2 | c2 , which implies 2 | c. So we can write c = 2k for some integer k. Our equation becomes 2k 2 = 3d2 . This shows that 2|3d2 , but since 2 - 3, we must have 2|d2 by Proposition 2.5 of Chapter 1. Thus √ 2|d,√a contradiction to the fact that gcd(c, d) = 1. Thus it must be that 2 + 3 is irrational. ...
The Multivariate Resultant is NP-hard in any Characteristic
... B. Grenet, P. Koiran, N. Portier polynomials, so that the square case is an important one. This result is sometimes known as Bertini’s theorem (as explained toward the end of this section, we will use an effective version of this result in one of our NP-hardness proofs). In this paper, we focus on t ...
... B. Grenet, P. Koiran, N. Portier polynomials, so that the square case is an important one. This result is sometimes known as Bertini’s theorem (as explained toward the end of this section, we will use an effective version of this result in one of our NP-hardness proofs). In this paper, we focus on t ...
Test 2 Working with Polynomials
... Donkey Kong is competing in a shot-put challenge at the Olympics. His throw can be modeled by the function h(t) = -5t2 + 8.5t + 1.8, where h is the height, in metres, of a shot-put t seconds after it is thrown. Determine the remainder when h(t) is divided by (t – 1.4). What does this value represent ...
... Donkey Kong is competing in a shot-put challenge at the Olympics. His throw can be modeled by the function h(t) = -5t2 + 8.5t + 1.8, where h is the height, in metres, of a shot-put t seconds after it is thrown. Determine the remainder when h(t) is divided by (t – 1.4). What does this value represent ...
Math/Stat 2300 Smoothing (4.3): Low
... Math/Stat 2300 Smoothing (4.3): Low-Order Polynomial Models from text A First Course in Mathematical Modeling, Giordano, Fox, Horton, Weir, 2009. We want to find methods that retain the advantages of the higher-order polynomials without the disadvantages. One technique is to choose a low-order polyn ...
... Math/Stat 2300 Smoothing (4.3): Low-Order Polynomial Models from text A First Course in Mathematical Modeling, Giordano, Fox, Horton, Weir, 2009. We want to find methods that retain the advantages of the higher-order polynomials without the disadvantages. One technique is to choose a low-order polyn ...
TRANSCENDENCE BASES AND N
... term a constant in k, which is just as good as monic. We then get two integral extensions k[z1, …, zs-1] k[z1, …zs-1, ys] = k[y1, …, ys] k[x1, …, xn], so by transitivity of integral extensions we contradict the minimality of s. First assume k is infinite. Set zj = yj- ujys for j < s, where uj ...
... term a constant in k, which is just as good as monic. We then get two integral extensions k[z1, …, zs-1] k[z1, …zs-1, ys] = k[y1, …, ys] k[x1, …, xn], so by transitivity of integral extensions we contradict the minimality of s. First assume k is infinite. Set zj = yj- ujys for j < s, where uj ...
LHF - Maths, NUS
... Construct M M L M1 where x, M k ( x ) SO(n 1) satisfies g(t , x)M (x) g(t , x) and yM (x) y if k k k k -1 yg(t , x) yg(t , x) 0.M is continuous and completes P. k k -1 ...
... Construct M M L M1 where x, M k ( x ) SO(n 1) satisfies g(t , x)M (x) g(t , x) and yM (x) y if k k k k -1 yg(t , x) yg(t , x) 0.M is continuous and completes P. k k -1 ...
Approximating the volume of a convex body
... neighbor not in G, choose again. Note that H has degree 2n, but exponentially many vertices, so we need to show that the walk mixes very quickly. This is plausible though, and is easily seen when G is just a cube with side length ≤ n/δ. Since the path Pn/δ mixes in time polynomial in n/δ, and the cu ...
... neighbor not in G, choose again. Note that H has degree 2n, but exponentially many vertices, so we need to show that the walk mixes very quickly. This is plausible though, and is easily seen when G is just a cube with side length ≤ n/δ. Since the path Pn/δ mixes in time polynomial in n/δ, and the cu ...
Algebra 1 Unit 3: Systems of Equations
... A box with no top is to be made from an 8 inch by 6 inch piece of metal by cutting identical squares from each corner and turning up the sides. The volume of the box is modeled by the polynomial 4x3 – 28x2 + 48x. Factor the polynomial completely. Then use the dimensions given on the box and show tha ...
... A box with no top is to be made from an 8 inch by 6 inch piece of metal by cutting identical squares from each corner and turning up the sides. The volume of the box is modeled by the polynomial 4x3 – 28x2 + 48x. Factor the polynomial completely. Then use the dimensions given on the box and show tha ...