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back to boardGCD (P(x), P'(x)) = const ? As I understand, this conditions means P(x) has only simple roots. I tried various solutions to find roots and have got an impression this is not true. Re: GCD (P(x), P'(x)) = const ? I think you have problem with precision. It is very hard to solve this problem with numeric solver, try to calculate exact solution Re: GCD (P(x), P'(x)) = const ? True. Reimplementing same thing for 64-bit precision (in Pascal) works okay. Re: GCD (P(x), P'(x)) = const ? Posted by svr 27 Sep 2008 15:30 I succeed with numeric Barstow method and avoid monkey Ferrari formulas. Because evil Olympiad precision 1.e-09 I was forced to implement Barstow above BigGecimal in Java. Who want good tests use MathCad. It's integration in case rather accurate. For those who inerested I write Barstow method from my prog: shorten- also my method for cutting off long BigDecimals public static BigDecimal Barstow(BigDecimal a[], int N, BigDecimal u0, BigDecimal v0, BigDecimal eps, int MaxIter, int MaxLen)[]
{ BigDecimal u, v; BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String str = null; int k, n,faza; u = u0; v = v0; BigDecimal[] b = new BigDecimal[N+1]; BigDecimal[] c = new BigDecimal[N+1]; BigDecimal rez[] = new BigDecimal[N+3]; BigDecimal d, s, q,h; BigDecimal Z = new BigDecimal("0.0"); BigDecimal bb1, bb2; h = new BigDecimal("0.0000000000000000000000000000000000000001"); faza = 0; for (k = 0; k <= MaxIter; k++) {
for (n = 0; n <= N; n++) { if (n < 1) bb1 = Z; else bb1 = b[n - 1]; if (n < 2) bb2 = Z; else bb2 = b[n - 2]; b[n] = shorten(a[n].add(bb1.multiply(u)).add(bb2.multiply(v)), MaxLen); } for (n = 0; n <= N - 1; n++) { if (n < 1) bb1 = Z; else bb1 = c[n - 1]; if (n < 2) bb2 = Z; else bb2 = c[n - 2]; c[n] = shorten((b[n].add(bb1.multiply(u))).add(bb2.multiply(v)), MaxLen); } d = shorten(((b[N].multiply(c[N - 3])).subtract(b[N - 1].multiply(c[N - 2]))), MaxLen); q = shorten((c[N - 2].multiply(c[N - 2])).subtract(c[N - 1].multiply(c[N - 3])), MaxLen); d = shorten(d.divide(q, BigDecimal.ROUND_FLOOR), MaxLen); s = shorten(((b[N - 1].multiply(c[N - 1])).subtract(b[N].multiply(c[N - 2]))), MaxLen); s = shorten(s.divide(q, BigDecimal.ROUND_FLOOR), MaxLen); if ((d.abs().compareTo(eps) < 0) && (s.abs().compareTo(eps) < 0)) { if (faza == 0) { faza = 1; MaxLen = MaxLen * 4; eps = eps.multiply(h); } else break; } u = u.add(d); v = v.add(s); k = k; } rez[0] = u; rez[1] = v; for (n = 0; n <= N; n++) rez[n + 2] = b[n]; return rez; }
Edited by author 27.09.2008 15:32 Re: GCD (P(x), P'(x)) = const ? I implemented Barstow in C++ as svr suggested but wa11 ... Is there any possibility that good initial guesses obtain a result up to desired precision using Barstow without long arithmetics ? |
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