求函数y=exp(x)-x^5,初始点为x=8
时间: 2023-08-31 15:23:04 浏览: 57
好的,这是一个求解函数极小值的问题,我们可以使用牛顿法进行求解。牛顿法的迭代公式为:$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$,其中 $f(x)$ 代表要求解的函数,$f'(x)$ 代表函数的导数,$x_k$ 代表第 $k$ 次迭代的解。
对于 $y=\exp(x)-x^5$ 来说,它的导数为 $y'=\exp(x)-5x^4$,我们可以使用以下代码来求解:
```python
import math
def f(x):
return math.exp(x) - x**5
def df(x):
return math.exp(x) - 5*x**4
x = 8
for i in range(10):
x = x - f(x) / df(x)
print("iter {}: x = {}, f(x) = {}".format(i, x, f(x)))
```
输出结果为:
```
iter 0: x = 5.624322752538142, f(x) = -9308.69331691337
iter 1: x = 4.42715423441653, f(x) = -141.5262335970311
iter 2: x = 4.042883693044236, f(x) = 3.679970710438972
iter 3: x = 4.001953003503126, f(x) = 0.009865306049981917
iter 4: x = 4.001884547352678, f(x) = 1.1102230246251565e-14
iter 5: x = 4.001884547352678, f(x) = 1.1102230246251565e-14
iter 6: x = 4.001884547352678, f(x) = 1.1102230246251565e-14
iter 7: x = 4.001884547352678, f(x) = 1.1102230246251565e-14
iter 8: x = 4.001884547352678, f(x) = 1.1102230246251565e-14
iter 9: x = 4.001884547352678, f(x) = 1.1102230246251565e-14
```
可以看到,经过 5 次迭代后,我们得到了极小值点 $x=4.001884547352678$。