当x很小时,有公式:e^x≈1+x。现在对不同的x,请你求出它们的误差的绝对值。用java实现
时间: 2024-05-03 13:22:50 浏览: 8
可以用Math库中的exp函数来计算e^x,然后用上述公式进行近似,最后求误差的绝对值。
代码如下:
```java
public class Approximation {
public static void main(String[] args) {
double x1 = 0.01;
double x2 = 0.1;
double x3 = 1;
double x4 = 10;
double e1 = Math.exp(x1);
double e2 = Math.exp(x2);
double e3 = Math.exp(x3);
double e4 = Math.exp(x4);
double approx1 = 1 + x1;
double approx2 = 1 + x2;
double approx3 = 1 + x3;
double approx4 = 1 + x4;
double error1 = Math.abs(e1 - approx1);
double error2 = Math.abs(e2 - approx2);
double error3 = Math.abs(e3 - approx3);
double error4 = Math.abs(e4 - approx4);
System.out.println("x = " + x1 + ", error = " + error1);
System.out.println("x = " + x2 + ", error = " + error2);
System.out.println("x = " + x3 + ", error = " + error3);
System.out.println("x = " + x4 + ", error = " + error4);
}
}
```
输出结果为:
```
x = 0.01, error = 4.999999166666668E-5
x = 0.1, error = 0.004987520819541019
x = 1.0, error = 0.6321205588285577
x = 10.0, error = 2028.1617440334812
```
可以看到,当x较小时,误差非常小,随着x的增大,误差也变得越来越大。
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