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Mathematical Background
Published in Alfred J. Menezes, Paul C. van Oorschot, Scott A. Vanstone, Handbook of Applied Cryptography, 2018
Alfred J. Menezes, Paul C. van Oorschot, Scott A. Vanstone
2.192 Example (polynomial division) Consider the polynomials g(x) = x6+x5+x3+x2+x+1 and h(x) = x4 + x3 + 1 in ℤ2[x]. Polynomial long division of g(x) by h(x) yields g(x)=x2h(x)+(x3+x+1). Hence g(x) mod h(x) = x3 + x + 1 and g(x) div h(x)= x2.
Optimization in Product Design – Synthesis
Published in S. Ratnajeevan H. Hoole, Yovahn Yesuraiyan R. Hoole, Finite Elements-based Optimization, 2019
S. Ratnajeevan H. Hoole, Yovahn Yesuraiyan R. Hoole
The minima must be the roots of df/dx = 0. It is readiy evident that at x = 1, df/dx is zero so x = 1 is a root where there lies a minimum of f(x) and (x–1) is a factor of df/dx. Dividing df/dx by (x–1) by polynomial long division we get 4x2+x–1 as the quotient with zero remainder. Solving 4x2+x−1=0
Combinatorics
Published in Erchin Serpedin, Thomas Chen, Dinesh Rajan, Mathematical Foundations for SIGNAL PROCESSING, COMMUNICATIONS, AND NETWORKING, 2012
It may be shown that polynomial long division, as in an earlier example, produces the Fibonacci numbers. But there is a more useful and fascinating consequence of our calculation of f(x). Recall that for the earlier case of the sequence an = 3n − 2n, the generating function was also a rational function, and that the bases of the exponents were reciprocals of the roots of the denominator. Alternatively, we may use partial fractions or the characteristic equation.
Why does trigonometric substitution work?
Published in International Journal of Mathematical Education in Science and Technology, 2018
Solution. Let be defined by . We want to use a substitution that will transform the integrand into one that is more accessible. The terms and appear to be the problem. If we selected the substitution x = u6, then the denominator of the integrand would become a polynomial in u. So, let be defined by g(u) = u6. The function g has domain (0, ∞), is differentiable, satisfies g′(u) ≠ 0 for all u in (0, ∞) and it has range (0, ∞) which is also the domain of the function f. Using the substitution x = g(u), we obtain Thus, (via polynomial long division). The anti-derivative, G(u), of f (g(u))g′(u) is Since , we conclude from Theorem 2.1 that Thus, F′(x) = f(x) for all x in (0, ∞).