66 lines
1.7 KiB
Python
66 lines
1.7 KiB
Python
import math
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#抛物线法解方程
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def MullerSolve(fx,x0,x1,x2,err1,err2,N):
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count = 0
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f0 = fx(x0)
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f1 = fx(x1)
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f2 = fx(x2)
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q = (x2 - x1) / (x1 - x0)
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p = 0
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a = 0
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b = 0
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c = 0
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while True:
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p = (x2 - x0) / (x1 - x0)
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a = q**2 * f0 - q*p*f1 + q*f2
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b = q**2 *f0 - p**2 *f1 + (p + q)*f2
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c = p*f2
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h1 = 0
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if b < 0:
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h1 = -2 * c / (b - (b**2 - 4*a*c)**0.5)
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else:
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h1 = -2 * c / (b + (b**2 - 4*a*c)**0.5)
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x3 = x2 + h1 * (x2 - x1)
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f3 = fx(x3)
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print(f"k={count}: x{count}={x0:.5}, f(x{count})={f0:.5}; x{count+1}={x1:.5}, f(x{count+1})={f1:.5}; x{count+2}={x2:.5}, f(x{count+2})={f2:.5}; x{count+3}={x3:.5}, f(x{count+3})={f3:.5}")
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print(f"p={p:.5}, q={q:.5}, a={a:.5}, b={b:.5}, c={c:.5}, h={h1:.5}")
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k = err1 + 1
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if abs(f3) < 1:
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k = abs(x3 - x2)
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else:
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k = abs(x3 - x2) / abs(f3)
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if abs(f3) < err2 or k < err1:
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return x3, 1
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count += 1
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if count > N:
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return None, 0
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x0 = x1
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x1 = x2
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x2 = x3
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f0 = f1
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f1 = f2
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f2 = f3
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q = h1
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if __name__ == "__main__":
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err1 = 1e-5 # 精度要求 P152
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err2 = 1e-5 # 精度要求 P152
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N = 100 # 最大迭代次数
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x0 = 0.3 # 初始值1
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x1 = 0.5 # 初始值2
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x2 = 0.4 # 初始值3
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fx = lambda x: 8*x**4 - 8*x**2 + 1 #原函数
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result, status = MullerSolve(fx, x0, x1, x2, err1, err2, N)
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if status == 1:
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print(f"fx收敛 解为: {result}")
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else:
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print("fx不收敛")
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