By Antonio Machì

This e-book bargains with numerous subject matters in algebra helpful for machine technological know-how purposes and the symbolic remedy of algebraic difficulties, mentioning and discussing their algorithmic nature. the themes lined diversity from classical effects resembling 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 services, to arrive the matter of the polynomial factorisation, specially through Berlekamp’s process, and the discrete Fourier remodel. simple algebra strategies are revised in a sort fitted to implementation on a working laptop or computer algebra method.

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Mathematik für Wirtschaftswissenschaftler 2: Lineare Algebra, Funktionen mehrerer Variablen

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Extra resources for Algebra for Symbolic Computation (UNITEXT)

Example text

Remark. The ring A is also a vector space over K, as is easily seen; so it is an algebra. Example. Let m0 (x) = x − 1, m1 (x) = x2 + 1, m2 (x) = x2 − 2, and m(x) = m0 (x)m1 (x)m2 (x) = x5 − x4 − x3 + x2 − 2x + 2. Compute L0 (x). With m0 (x) = x − 1 and m1 (x)m2 (x) = x4 − x2 − 2, we get −x3 − x2 1 · (x − 1) − (x4 − x2 − 2) = 1, 2 2 and hence L0 (x) = − 12 (x4 − x2 − 2). For L1 (x), with m1 (x) = x2 + 1, m0 (x)m2 (x) = x3 − x2 − 2x + 2 we ﬁnd 4 − x2 x+1 3 · (x2 + 1) + (x − x2 − 2x + 2) = 1, 6 6 and L1 (x) = x+1 3 6 (x − x2 − 2x + 2) = 16 (x4 − 3x2 + 2).

Determine the p-adic expansion of 1 . 1 we have: 1 · (1 − p) + 1 · p = 1, and hence: 1 1 =1+ p, 1−p 1−p and c0 = 1. From this follows 1 1 1 2 = 1 + (1 + p)p = 1 + p + p , 1−p 1−p 1−p and c1 = 1. Going on like this, or by induction, we ﬁnd: 1 = 1 + p + p2 + · · · . 1−p 2. More generally, let us expand 1 , 1 − pn this time by running the algorithm. We have: 1 ≡ (1 − pn )c0 = c0 − pn c0 mod p, and hence c0 = 1. Moreover, 1 ≡ (1 − pn )(1 + c1 p) = 1 − pn + c1 p − c1 pn−1 mod p2 . If n ≥ 2, we have c1 ≡ 0 mod p.

3 Newton’s method 55 Now, f (e1 ) ≡ 0 mod p, so c solves the previous congruence mod p2 if and only if it solves f (e1 ) cf (e1 ) ≡ − mod p. p But the latter admits the solution c ≡ −( f (e1 ) mod p)p. f (e1 )p So we have the new approximation e2 = e1 − ( f (e1 ) mod p)p. f (e1 )p In general, in order to pass from en to en+1 , consider the congruence: f (x + cpn ) ≡ f (x) + cpn f (x) mod pn+1 , which yields the value c≡− f (en ) mod p, f (en )pn and the new approximation en+1 = en − ( f (en ) mod p)pn .