Download Algebra for Symbolic Computation by Antonio Machì (auth.) PDF

By Antonio Machì (auth.)

This publication offers with a number of issues in algebra helpful for laptop technology functions and the symbolic remedy of algebraic difficulties, mentioning and discussing their algorithmic nature. the themes lined variety from classical effects reminiscent of the Euclidean set of rules, the chinese language the rest theorem, and polynomial interpolation, to p-adic expansions of rational and algebraic numbers and rational capabilities, to arrive the matter of the polynomial factorisation, in particular through Berlekamp’s technique, and the discrete Fourier rework. uncomplicated algebra strategies are revised in a kind suited to implementation on a working laptop or computer algebra system.

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In other words, it allows us to find the p-adic expansion of a root of an arbitrary polynomial f (x) with integer coefficients (within the mentioned limits). Indeed, let: m a k xk ; f (x) = k=0 note that, for all integer c, k (x + cp)k = i=o k k−i x (cp)i ≡ xk + kcpxk−1 mod p2 , i since, for i > 1, (cp)i = ci pi ≡ 0 mod p2 . From this follows: m m ak (x + cp)k ≡ f (x + cp) = k=0 m ak kxk−1 mod p2 , ak xk + cp k=0 k=0 and hence: f (x + cp) ≡ f (x) + cpf (x) mod p2 . Now, if e1 is a root of f (x) mod p such that f (e1 ) ≡ 0 mod p, for a root e2 ≡ e1 + cp mod p2 the following must hold: 0 ≡ f (e1 + cp) = f (e1 ) + cpf (e1 ) mod p2 .

Xn . For a monomial xi we have: xi = xi0 L0 (x) + xi1 L1 (x) + . . + xin Ln (x). 4 Polynomial interpolation 29 From this it follows that the change of basis matrix from the basis (Lk (x)) to the basis (xi ) is the (n + 1) × (n + 1) Vandermonde matrix: ⎞ ⎛ 1 1 ... 1 ⎜ x0 x1 . . xn ⎟ ⎟ ⎜ 2 2 2 ⎟ ⎜ V = ⎜ x0 x1 . . xn ⎟ . ⎜ .. .. ⎟ .. ⎝ . . ⎠ n n x0 x1 . . xnn (Since this is a change of basis matrix, V is non-singular. ) The inverse matrix V −1 of V is the change of basis matrix from (xi ) to (Lk (x)).

N−1, the value of Lk (x) and Lk (x) at x = xk is 1, while (n) at x = xj , j = k, it is 0. Ln is zero at the same points. From this follows that the difference u(n) (x)−u(n−1) (x) is divisible by (x−x0 )(x−x1 ) · · · (x−xn−1 ), so: u(n) (x) = u(n−1) (x) + an (x − x0 )(x − x1 ) · · · (x − xn−1 ). This argument leads to the Newton interpolation formula: u(x) = u(n) (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )(x − x1 ) + · · · + an (x − x0 )(x − x1 ) · · · (x − xn−1 ). 3) is clear. 8) does not depend on xk+1 , .

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