第一章
一、问题
设 S N = ∑ j = 2 N 1 j 2 ? 1 S_N = \sum_{j=2}^N \frac{1}{j^2 - 1} SN=∑j=2Nj2?11，其精确值为 1 2 ( 3 2 ? 1 N ? 1 N 1 ) \frac{1}{2}(\frac{3}{2} - \frac{1}{N} - \frac{1}{N 1}) 21(23−N1−N+11)
- 编制按从大到小的顺序 S N = 1 2 2 − 1 + 1 3 2 − 1 + ⋯ + 1 N 2 − 1 S_N = \frac{1}{2^2 - 1} + \frac{1}{3^2 - 1} + \cdots + \frac{1}{N^2 - 1} SN=22−11+32−11+⋯+N2−11，计算 S N S_N SN 的通用程序
- 编制按从小到大的顺序 S N = 1 N 2 − 1 + 1 ( N − 1 ) 2 − 1 + ⋯ + 1 N 2 − 1 S_N = \frac{1}{N^2 - 1} + \frac{1}{(N-1)^2 - 1} + \cdots + \frac{1}{N^2 - 1} SN=N2−11+(N−1)2−11+⋯+N2−11，计算 S N S_N SN 的通用程序
- 按两种顺序分别计算 S 1 0 2 S_{10^2} S102， S 1 0 4 S_{10^4} S104， S 1 0 6 S_{10^6} S106，并指出有效位数
- 通过本上机题你明白了什么
二、通用程序
def my_fun(n):
sn, sn1, sn2 = 1/2 * (3/2 - 1/n - 1/(n+1)), 0, 0
for i in range(2, n+1):
sn1 += 1/(i**2 - 1)
for i in range(n, 1, -1):
sn2 += 1/(i**2 - 1)
print('The N we use in calculate Sn:', n, '\nThe accurate sum: Sn =', sn,
'\nFrom large to small sum: Sn1 =', sn1, '\nFrom small to large sum: Sn2 =', sn2)
print('abs(Sn-Sn1) =', abs(sn-sn1), '\nabs(Sn-Sn2) =', abs(sn-sn2))
print('-' * 55)
if __name__ == '__main__':
N = [100, 10000, 1000000]
for num in N:
my_fun(num)
三、程序结果
The N we use in calculate Sn: 100
The accurate sum: Sn = 0.740049504950495
From large to small sum: Sn1 = 0.7400495049504949
From small to large sum: Sn2 = 0.7400495049504949
abs(Sn-Sn1) = 1.1102230246251565e-16
abs(Sn-Sn2) = 1.1102230246251565e-16
-------------------------------------------------------
The N we use in calculate Sn: 10000
The accurate sum: Sn = 0.7499000049995
From large to small sum: Sn1 = 0.7499000049995057
From small to large sum: Sn2 = 0.7499000049995
abs(Sn-Sn1) = 5.662137425588298e-15
abs(Sn-Sn2) = 0.0
-------------------------------------------------------
The N we use in calculate Sn: 1000000
The accurate sum: Sn = 0.7499990000005
From large to small sum: Sn1 = 0.7499990000005217
From small to large sum: Sn2 = 0.7499990000004999
abs(Sn-Sn1) = 2.1649348980190553e-14
abs(Sn-Sn2) = 1.1102230246251565e-16
-------------------------------------------------------
四、有效位分析
| n==10^2 | n==10^4 | n==10^6 | |
|---|---|---|---|
| 从小到大 | 15 | ? | 15 |
| 从大到小 | 15 | 13 | 13 |
五、结果分析
- 从有效位可以看出，程序算法对实验的误差具有一定的影响，一个好的算法可以使结果更加精确
- 以该程序为例，从大到小会出现“大数吃小数”的现象，导致其精度随轮次的增加而相应减少
- 从小到大顺序可能因为机器问题，导致其误差为 0
第二章
一、问题
- 给定初值 x 0 x_0 x0 及容许误差 ϵ \epsilon ϵ，编制 Newton 法解方程 f ( x ) = 0 f(x) = 0 f(x)=0 根的通用程序
- 给定方程 f ( x ) = x 3 3 − x = 0 f(x) = \frac{x^3}{3} - x = 0 f(x)=3x3−x=0，易知其有三个根 x 1 ∗ = − 3 x_1^* = -\sqrt{3} x1∗=−3
， x 2 ∗ = 0 x_2^* = 0 x2∗=0， x 3 ∗ = 3 x_3^* = \sqrt{3} x3∗=3
- 由 Newton 方法的局部收敛性可知存在 δ > 0 \delta \gt 0 δ>0，当 x 0 ∈ ( − δ , δ ) x_0 \in (-\delta, \delta) x0∈(−δ,δ) 时 Newton 迭代序列收敛于根 x 2 ∗ x_2^* x2∗，试确定尽可能大的 δ \delta δ
- 试取若干初始值，观察当 x 0 ∈ ( − ∞ , − 1 ) x_0 \in (-\infty, -1) x0∈(−∞,−1)， ( − 1 , − δ ) (-1, -\delta) (−1,−δ)， ( − δ , δ ) (-\delta, \delta) (−δ,δ)， ( δ , 1 ) (\delta, 1) (δ,1)， ( 1 , + ∞ ) (1, +\infty) (1,+∞) 时 Newton 序列是否收敛以及收敛于哪一根
- 通过本上机题，你明白了什么
二、通用程序
from sympy import * # 牛顿法通用程序 def newton_fun(): x = Symbol('x') epson = eval(input('Enter the absolute error: ')) # 误差 x0 = eval(input('Enter the initial value: ')) # 初值 fun = eval(input('Enter the function: ')) # 函数 derivative = diff(fun, x, 1) # 一阶导 x1 = x0 - fun.subs('x', x0) / derivative.subs('x', x0) while abs(x1 - x0) > epson: # Newton 迭代格式 x0, x1 = x1, x1 - fun.subs('x', x1) / derivative.subs('x', x1) print(x0, x1) # 返回收敛值 def discriminant(left, right, x0=0): x = Symbol('x') epson, fun = 1e-13, x ** 3 / 3 - x x0 = (left + right) / 2 if x0 == 0 else x0 derivative = diff(fun, x, 1) x1 = x0 - fun.subs('x', x0) / derivative.subs('x', x0) while abs(x1 - x0) > epson: x0, x1 = x1, x1 - fun.subs('x', x1) / derivative.subs('x', x1) return x1 # 二分法搜索 def search(): l, r, delta, e = 0.0, 1.0, 0.0, 1e-13 while abs(delta - (l + r) / 2) > e: delta = (l + r) / 2 res = discriminant(l, r) if res == 0: l = (l + r) / 2 # 收敛于 0 则将 l 设为中点 else: r = (l + r) / 2 # 不收敛于 0（收敛于根号3） 则将 r 设为中点 return l # delta 值可能使函数不收敛于0，故返回 l if __name__ == '__main__': newton_fun() print('-' * 50) delta = search() print('delta =', delta) print('-' * 50 标签：abs轮速传感器s102端盖