概説微分積分の問題3.19の解答。
exp(0.5)、cos(1)、log(1.1)を小数第二位まで求める問題。ラグランジュの剰余項Rn(x)の絶対値を評価する必要があるのだけど、評価方法を経験則的にやってるので一般化できてない。
import math def trunc(x): if x > 0: return math.floor(x) return math.ceil(x) def solve(x, taylor_term, eval_R): n = 0 while True: # calc f(x) result = 0. for i in range(0, n + 1): result += taylor_term(x, i) # calc |Rn(x)| r = eval_R(x, n + 1) # Check the range of f(x) min_fx = trunc((result + r) * 100) / 100 max_fx = trunc((result - r) * 100) / 100 print "f(x) = %.4f, |Rn(x)| = %.4f" % (result, r) if min_fx == max_fx: return min_fx, n # run next loop n += 1 def factor(n): factor_n = 1 for i in range(2, n + 1): factor_n *= i return factor_n def taylor_term1(x, n): return float(x ** n) / factor(n) def R1(x, n): # since c = 1/2 < 2 return 2. * (x ** n) / factor(n) def taylor_term2(x, n): if n % 2 == 1: return 0 return (float(x) ** n) * ((-1) ** (n / 2)) / factor(n) def R2(x, n): # since sin(or cos) <= 1 return (float(x) ** n) / factor(n) def taylor_term3(x, n): if n == 0: return 0 return float(x ** n) * ((-1) ** (n - 1)) / n def R3(x, n): # since 1 + c > 1 return float(x ** n) / float(n) print "f(x) = %.2f, n = %d" % solve(1. / 2, taylor_term1, R1) print "f(x) = %.2f, n = %d" % solve(1., taylor_term2, R2) print "f(x) = %.2f, n = %d" % solve(.1, taylor_term3, R3)
f(x) = 1.0000, |Rn(x)| = 1.0000 f(x) = 1.5000, |Rn(x)| = 0.2500 f(x) = 1.6250, |Rn(x)| = 0.0417 f(x) = 1.6458, |Rn(x)| = 0.0052 f(x) = 1.6484, |Rn(x)| = 0.0005 f(x) = 1.64, n = 4 f(x) = 1.0000, |Rn(x)| = 1.0000 f(x) = 1.0000, |Rn(x)| = 0.5000 f(x) = 0.5000, |Rn(x)| = 0.1667 f(x) = 0.5000, |Rn(x)| = 0.0417 f(x) = 0.5417, |Rn(x)| = 0.0083 f(x) = 0.5417, |Rn(x)| = 0.0014 f(x) = 0.54, n = 5 f(x) = 0.0000, |Rn(x)| = 0.1000 f(x) = 0.1000, |Rn(x)| = 0.0050 f(x) = 0.0950, |Rn(x)| = 0.0003 f(x) = 0.09, n = 2
