求出e=1+1/1!+1/2!+1/3!+……+1/n!+……的近似值的java applet程序

//求出e=1+1/1!+1/2!+1/3!+……+1/n!+……的近似值,要求误差小于0.0001

import java.applet.*;

import java.awt.*;

import java.awt.event.*;

public class AT1_1 extends Applet implements ActionListener

{

TextField text1;

Button Button1;

public void init()

{

text1 = new TextField("0",10);

Button1 = new Button("清除");

add(text1);

add(Button1);

text1.addActionListener(this);

Button1.addActionListener(this);

}

public void start(){}

public void stop(){}

public void destory(){}

public void paint(Graphics g)

{

g.drawString("在文本区输入数字n后回车",10,100);

g.drawString("文本区显示1+1/1!+1/2!+1/3!+……+1/n!+……的近似值",10,120);

}

public void actionPerformed(ActionEvent e)

{

if(e.getSource()==text1)

{

double sum=1,a=1;

int i=1;

int n=0;

try

{

n = Integer.valueOf(text1.getText()).intValue();

while(i<=n)

{

a = a*(1.0/i);

sum = sum + a;

i=i+1;

}

sum=sum*10000;

sum=Math.round(sum);

sum=sum/10000;

text1.setText(""+sum);

}

catch(NumberFormatException Event)

{

text1.setText("请输入数字字符");

}

}

else if(e.getSource()==Button1)

{

text1.setText("0");

}

}

}