用牛顿法求一个数的根
原文:https://www . geesforgeks . org/find-number-of-use-newtons-method/
给定一个整数 N 和一个公差等级 L ，任务是使用牛顿法求该数的平方根。 举例:
输入: N = 16，L = 0.0001 输出: 4 4 2 = 16 输入: N = 327，L = 0.00001 输出: 18.0831
牛顿法: 设 N 为任意数，则 N 的平方根可由公式 给出
根= 0.5 * (X + (N / X))其中 X 是可以假设为 N 或 1 的任何猜测。
- 在上式中， X 是 N 的任意假定平方根，根是 N 的正确平方根。
- 公差极限是 X 与允许根之间的最大差值。
方法:可以按照以下步骤计算答案:
- 将 X 分配给 N 本身。
- 现在，开始一个循环，继续计算根，它肯定会向 N 的正确平方根移动。
- 检查假设的 X 和计算的根之间的差异，如果还不在公差范围内，则更新根并继续。
- 如果计算出的根在允许的公差范围内，则脱离循环。
- 打印根。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the square root of
// a number using Newtons method
double squareRoot(double n, float l)
{
// Assuming the sqrt of n as n only
double x = n;
// The closed guess will be stored in the root
double root;
// To count the number of iterations
int count = 0;
while (1) {
count++;
// Calculate more closed x
root = 0.5 * (x + (n / x));
// Check for closeness
if (abs(root - x) < l)
break;
// Update root
x = root;
}
return root;
}
// Driver code
int main()
{
double n = 327;
float l = 0.00001;
cout << squareRoot(n, l);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the square root of
// a number using Newtons method
static double squareRoot(double n, double l)
{
// Assuming the sqrt of n as n only
double x = n;
// The closed guess will be stored in the root
double root;
// To count the number of iterations
int count = 0;
while (true)
{
count++;
// Calculate more closed x
root = 0.5 * (x + (n / x));
// Check for closeness
if (Math.abs(root - x) < l)
break;
// Update root
x = root;
}
return root;
}
// Driver code
public static void main (String[] args)
{
double n = 327;
double l = 0.00001;
System.out.println(squareRoot(n, l));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to return the square root of
# a number using Newtons method
def squareRoot(n, l) :
# Assuming the sqrt of n as n only
x = n
# To count the number of iterations
count = 0
while (1) :
count += 1
# Calculate more closed x
root = 0.5 * (x + (n / x))
# Check for closeness
if (abs(root - x) < l) :
break
# Update root
x = root
return root
# Driver code
if __name__ == "__main__" :
n = 327
l = 0.00001
print(squareRoot(n, l))
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the square root of
// a number using Newtons method
static double squareRoot(double n, double l)
{
// Assuming the sqrt of n as n only
double x = n;
// The closed guess will be stored in the root
double root;
// To count the number of iterations
int count = 0;
while (true)
{
count++;
// Calculate more closed x
root = 0.5 * (x + (n / x));
// Check for closeness
if (Math.Abs(root - x) < l)
break;
// Update root
x = root;
}
return root;
}
// Driver code
public static void Main()
{
double n = 327;
double l = 0.00001;
Console.WriteLine(squareRoot(n, l));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the square root of
// a number using Newtons method
function squareRoot(n, l)
{
// Assuming the sqrt of n as n only
let x = n;
// The closed guess will be stored in the root
let root;
// To count the number of iterations
let count = 0;
while (true)
{
count++;
// Calculate more closed x
root = 0.5 * (x + (n / x));
// Check for closeness
if (Math.abs(root - x) < l)
break;
// Update root
x = root;
}
return root.toFixed(4);
}
let n = 327;
let l = 0.00001;
document.write(squareRoot(n, l));
// This code is contributed by divyesh072019.
</script>
Output:
18.0831
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