La interpolación es la técnica de estimar el valor de una función para cualquier valor intermedio de la variable independiente, mientras que el proceso de calcular el valor de la función fuera del rango dado se llama extrapolación .
Diferencias directas : Las diferencias y1 – y0, y2 – y1, y3 – y2, ……, yn – yn–1 cuando se denotan por dy0, dy1, dy2, ……, dyn–1 se denominan respectivamente primeras diferencias directas. Así, las primeras diferencias directas son: 
FÓRMULA DE INTERPOLACIÓN HACIA ADELANTE DE GREGORY DE NEWTON : 
Esta fórmula es particularmente útil para interpolar los valores de f(x) cerca del comienzo del conjunto de valores dado. h se llama el intervalo de diferencia y u = ( x – a ) / h , Aquí a es el primer término.
Ejemplo :
Input : Value of Sin 52
Output :
Value at Sin 52 is 0.788003
A continuación se muestra la implementación del método de interpolación directa de Newton.
C++
// CPP Program to interpolate using
// newton forward interpolation
#include <bits/stdc++.h>
using namespace std;
// calculating u mentioned in the formula
float u_cal(float u, int n)
{
float temp = u;
for (int i = 1; i < n; i++)
temp = temp * (u - i);
return temp;
}
// calculating factorial of given number n
int fact(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
int main()
{
// Number of values given
int n = 4;
float x[] = { 45, 50, 55, 60 };
// y[][] is used for difference table
// with y[][0] used for input
float y[n][n];
y[0][0] = 0.7071;
y[1][0] = 0.7660;
y[2][0] = 0.8192;
y[3][0] = 0.8660;
// Calculating the forward difference
// table
for (int i = 1; i < n; i++) {
for (int j = 0; j < n - i; j++)
y[j][i] = y[j + 1][i - 1] - y[j][i - 1];
}
// Displaying the forward difference table
for (int i = 0; i < n; i++) {
cout << setw(4) << x[i]
<< "\t";
for (int j = 0; j < n - i; j++)
cout << setw(4) << y[i][j]
<< "\t";
cout << endl;
}
// Value to interpolate at
float value = 52;
// initializing u and sum
float sum = y[0][0];
float u = (value - x[0]) / (x[1] - x[0]);
for (int i = 1; i < n; i++) {
sum = sum + (u_cal(u, i) * y[0][i]) /
fact(i);
}
cout << "\n Value at " << value << " is "
<< sum << endl;
return 0;
}
Java
// Java Program to interpolate using
// newton forward interpolation
class GFG{
// calculating u mentioned in the formula
static double u_cal(double u, int n)
{
double temp = u;
for (int i = 1; i < n; i++)
temp = temp * (u - i);
return temp;
}
// calculating factorial of given number n
static int fact(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
public static void main(String[] args)
{
// Number of values given
int n = 4;
double x[] = { 45, 50, 55, 60 };
// y[][] is used for difference table
// with y[][0] used for input
double y[][]=new double[n][n];
y[0][0] = 0.7071;
y[1][0] = 0.7660;
y[2][0] = 0.8192;
y[3][0] = 0.8660;
// Calculating the forward difference
// table
for (int i = 1; i < n; i++) {
for (int j = 0; j < n - i; j++)
y[j][i] = y[j + 1][i - 1] - y[j][i - 1];
}
// Displaying the forward difference table
for (int i = 0; i < n; i++) {
System.out.print(x[i]+"\t");
for (int j = 0; j < n - i; j++)
System.out.print(y[i][j]+"\t");
System.out.println();
}
// Value to interpolate at
double value = 52;
// initializing u and sum
double sum = y[0][0];
double u = (value - x[0]) / (x[1] - x[0]);
for (int i = 1; i < n; i++) {
sum = sum + (u_cal(u, i) * y[0][i]) /
fact(i);
}
System.out.println("\n Value at "+value+" is "+String.format("%.6g%n",sum));
}
}
// This code is contributed by mits
Python3
# Python3 Program to interpolate using
# newton forward interpolation
# calculating u mentioned in the formula
def u_cal(u, n):
temp = u;
for i in range(1, n):
temp = temp * (u - i);
return temp;
# calculating factorial of given number n
def fact(n):
f = 1;
for i in range(2, n + 1):
f *= i;
return f;
# Driver Code
# Number of values given
n = 4;
x = [ 45, 50, 55, 60 ];
# y[][] is used for difference table
# with y[][0] used for input
y = [[0 for i in range(n)]
for j in range(n)];
y[0][0] = 0.7071;
y[1][0] = 0.7660;
y[2][0] = 0.8192;
y[3][0] = 0.8660;
# Calculating the forward difference
# table
for i in range(1, n):
for j in range(n - i):
y[j][i] = y[j + 1][i - 1] - y[j][i - 1];
# Displaying the forward difference table
for i in range(n):
print(x[i], end = "\t");
for j in range(n - i):
print(y[i][j], end = "\t");
print("");
# Value to interpolate at
value = 52;
# initializing u and sum
sum = y[0][0];
u = (value - x[0]) / (x[1] - x[0]);
for i in range(1,n):
sum = sum + (u_cal(u, i) * y[0][i]) / fact(i);
print("\nValue at", value,
"is", round(sum, 6));
# This code is contributed by mits
C#
// C# Program to interpolate using
// newton forward interpolation
using System;
class GFG
{
// calculating u mentioned in the formula
static double u_cal(double u, int n)
{
double temp = u;
for (int i = 1; i < n; i++)
temp = temp * (u - i);
return temp;
}
// calculating factorial of given number n
static int fact(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
// Driver code
public static void Main()
{
// Number of values given
int n = 4;
double[] x = { 45, 50, 55, 60 };
// y[,] is used for difference table
// with y[,0] used for input
double[,] y=new double[n,n];
y[0,0] = 0.7071;
y[1,0] = 0.7660;
y[2,0] = 0.8192;
y[3,0] = 0.8660;
// Calculating the forward difference
// table
for (int i = 1; i < n; i++) {
for (int j = 0; j < n - i; j++)
y[j,i] = y[j + 1,i - 1] - y[j,i - 1];
}
// Displaying the forward difference table
for (int i = 0; i < n; i++) {
Console.Write(x[i]+"\t");
for (int j = 0; j < n - i; j++)
Console.Write(y[i,j]+"\t");
Console.WriteLine();
}
// Value to interpolate at
double value = 52;
// initializing u and sum
double sum = y[0,0];
double u = (value - x[0]) / (x[1] - x[0]);
for (int i = 1; i < n; i++) {
sum = sum + (u_cal(u, i) * y[0,i]) /
fact(i);
}
Console.WriteLine("\n Value at "+value+" is "+Math.Round(sum,6));
}
}
// This code is contributed by mits
PHP
<?php
// PHP Program to interpolate using
// newton forward interpolation
// calculating u mentioned in the formula
function u_cal($u, $n)
{
$temp = $u;
for ($i = 1; $i < $n; $i++)
$temp = $temp * ($u - $i);
return $temp;
}
// calculating factorial of given number n
function fact($n)
{
$f = 1;
for ($i = 2; $i <= $n; $i++)
$f *= $i;
return $f;
}
// Driver Code
// Number of values given
$n = 4;
$x = array( 45, 50, 55, 60 );
// y[,] is used for difference table
// with y[,0] used for input
$y = array_fill(0, $n,
array_fill(0, $n, 0));
$y[0][0] = 0.7071;
$y[1][0] = 0.7660;
$y[2][0] = 0.8192;
$y[3][0] = 0.8660;
// Calculating the forward difference
// table
for ($i = 1; $i < $n; $i++)
{
for ($j = 0; $j < $n - $i; $j++)
$y[$j][$i] = $y[$j + 1][$i - 1] -
$y[$j][$i - 1];
}
// Displaying the forward difference table
for ($i = 0; $i < $n; $i++)
{
print($x[$i] . "\t");
for ($j = 0; $j < $n - $i; $j++)
print($y[$i][$j] . "\t");
print("\n");
}
// Value to interpolate at
$value = 52;
// initializing u and sum
$sum = $y[0][0];
$u = ($value - $x[0]) / ($x[1] - $x[0]);
for ($i = 1; $i < $n; $i++)
{
$sum = $sum + (u_cal($u, $i) * $y[0][$i]) /
fact($i);
}
print("\nValue at " . $value .
" is " . round($sum, 6));
// This code is contributed by mits
Javascript
<script>
// javascript Program to interpolate using
// newton forward interpolation
// calculating u mentioned in the formula
function u_cal(u , n)
{
var temp = u;
for (var i = 1; i < n; i++)
temp = temp * (u - i);
return temp;
}
// calculating factorial of given number n
function fact(n)
{
var f = 1;
for (var i = 2; i <= n; i++)
f *= i;
return f;
}
// Number of values given
var n = 4;
var x = [ 45, 50, 55, 60 ];
// y is used for difference table
// with y[0] used for input
var y=Array(n).fill(0.0).map(x => Array(n).fill(0.0));
y[0][0] = 0.7071;
y[1][0] = 0.7660;
y[2][0] = 0.8192;
y[3][0] = 0.8660;
// Calculating the forward difference
// table
for (var i = 1; i < n; i++) {
for (var j = 0; j < n - i; j++)
y[j][i] = y[j + 1][i - 1] - y[j][i - 1];
}
// Displaying the forward difference table
for (var i = 0; i < n; i++) {
document.write(x[i].toFixed(6)+" ");
for (var j = 0; j < n - i; j++)
document.write(y[i][j].toFixed(6)+" ");
document.write('<br>');
}
// Value to interpolate at
var value = 52;
// initializing u and sum
var sum = y[0][0];
var u = (value - x[0]) / (x[1] - x[0]);
for (var i = 1; i < n; i++) {
sum = sum + (u_cal(u, i) * y[0][i]) /
fact(i);
}
document.write("\n Value at "+value.toFixed(6)+" is "+sum.toFixed(6));
// This code is contributed by 29AjayKumar
</script>
Producción:
45 0.7071 0.0589 -0.00569999 -0.000699997 50 0.766 0.0532 -0.00639999 55 0.8192 0.0468 60 0.866 Value at 52 is 0.788003
Diferencias hacia atrás : Las diferencias y1 – y0, y2 – y1, ……, yn – yn–1 cuando se denotan por dy1, dy2, ……, dyn, respectivamente, se denominan primeras diferencias hacia atrás. Así, las primeras diferencias hacia atrás son: 
FÓRMULA DE INTERPOLACIÓN HACIA ATRÁS DE GREGORY DE NEWTON : 
Esta fórmula es útil cuando se requiere el valor de f(x) cerca del final de la tabla. h se llama intervalo de diferencia y u = ( x – an ) / h , aquí an es el último término.
Ejemplo :
Input : Population in 1925
Output :
Value in 1925 is 96.8368
A continuación se muestra la implementación del método de interpolación hacia atrás de Newton.
C++
// CPP Program to interpolate using
// newton backward interpolation
#include <bits/stdc++.h>
using namespace std;
// Calculation of u mentioned in formula
float u_cal(float u, int n)
{
float temp = u;
for (int i = 1; i < n; i++)
temp = temp * (u + i);
return temp;
}
// Calculating factorial of given n
int fact(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
int main()
{
// number of values given
int n = 5;
float x[] = { 1891, 1901, 1911,
1921, 1931 };
// y[][] is used for difference
// table and y[][0] used for input
float y[n][n];
y[0][0] = 46;
y[1][0] = 66;
y[2][0] = 81;
y[3][0] = 93;
y[4][0] = 101;
// Calculating the backward difference table
for (int i = 1; i < n; i++) {
for (int j = n - 1; j >= i; j--)
y[j][i] = y[j][i - 1] - y[j - 1][i - 1];
}
// Displaying the backward difference table
for (int i = 0; i < n; i++) {
for (int j = 0; j <= i; j++)
cout << setw(4) << y[i][j]
<< "\t";
cout << endl;
}
// Value to interpolate at
float value = 1925;
// Initializing u and sum
float sum = y[n - 1][0];
float u = (value - x[n - 1]) / (x[1] - x[0]);
for (int i = 1; i < n; i++) {
sum = sum + (u_cal(u, i) * y[n - 1][i]) /
fact(i);
}
cout << "\n Value at " << value << " is "
<< sum << endl;
return 0;
}
Java
// Java Program to interpolate using
// newton backward interpolation
class GFG
{
// Calculation of u mentioned in formula
static double u_cal(double u, int n)
{
double temp = u;
for (int i = 1; i < n; i++)
temp = temp * (u + i);
return temp;
}
// Calculating factorial of given n
static int fact(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
// Driver code
public static void main(String[] args)
{
// number of values given
int n = 5;
double x[] = { 1891, 1901, 1911,
1921, 1931 };
// y[][] is used for difference
// table and y[][0] used for input
double[][] y = new double[n][n];
y[0][0] = 46;
y[1][0] = 66;
y[2][0] = 81;
y[3][0] = 93;
y[4][0] = 101;
// Calculating the backward difference table
for (int i = 1; i < n; i++)
{
for (int j = n - 1; j >= i; j--)
y[j][i] = y[j][i - 1] - y[j - 1][i - 1];
}
// Displaying the backward difference table
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= i; j++)
System.out.print(y[i][j] + "\t");
System.out.println("");;
}
// Value to interpolate at
double value = 1925;
// Initializing u and sum
double sum = y[n - 1][0];
double u = (value - x[n - 1]) / (x[1] - x[0]);
for (int i = 1; i < n; i++)
{
sum = sum + (u_cal(u, i) * y[n - 1][i]) /
fact(i);
}
System.out.println("\n Value at " + value +
" is " + String.format("%.6g%n",sum));
}
}
// This code is contributed by mits
Python3
# Python3 Program to interpolate using
# newton backward interpolation
# Calculation of u mentioned in formula
def u_cal(u, n):
temp = u
for i in range(n):
temp = temp * (u + i)
return temp
# Calculating factorial of given n
def fact(n):
f = 1
for i in range(2, n + 1):
f *= i
return f
# Driver code
# number of values given
n = 5
x = [1891, 1901, 1911, 1921, 1931]
# y is used for difference
# table and y[0] used for input
y = [[0.0 for _ in range(n)] for __ in range(n)]
y[0][0] = 46
y[1][0] = 66
y[2][0] = 81
y[3][0] = 93
y[4][0] = 101
# Calculating the backward difference table
for i in range(1, n):
for j in range(n - 1, i - 1, -1):
y[j][i] = y[j][i - 1] - y[j - 1][i - 1]
# Displaying the backward difference table
for i in range(n):
for j in range(i + 1):
print(y[i][j], end="\t")
print()
# Value to interpolate at
value = 1925
# Initializing u and sum
sum = y[n - 1][0]
u = (value - x[n - 1]) / (x[1] - x[0])
for i in range(1, n):
sum = sum + (u_cal(u, i) * y[n - 1][i]) / fact(i)
print("\n Value at", value, "is", sum)
# This code is contributed by phasing17
C#
// C# Program to interpolate using
// newton backward interpolation
using System;
class GFG
{
// Calculation of u mentioned in formula
static double u_cal(double u, int n)
{
double temp = u;
for (int i = 1; i < n; i++)
temp = temp * (u + i);
return temp;
}
// Calculating factorial of given n
static int fact(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
// Driver code
static void Main()
{
// number of values given
int n = 5;
double[] x = { 1891, 1901, 1911,
1921, 1931 };
// y[][] is used for difference
// table and y[][0] used for input
double[,] y = new double[n,n];
y[0,0] = 46;
y[1,0] = 66;
y[2,0] = 81;
y[3,0] = 93;
y[4,0] = 101;
// Calculating the backward difference table
for (int i = 1; i < n; i++)
{
for (int j = n - 1; j >= i; j--)
y[j,i] = y[j,i - 1] - y[j - 1,i - 1];
}
// Displaying the backward difference table
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= i; j++)
Console.Write(y[i,j]+"\t");
Console.WriteLine("");;
}
// Value to interpolate at
double value = 1925;
// Initializing u and sum
double sum = y[n - 1,0];
double u = (value - x[n - 1]) / (x[1] - x[0]);
for (int i = 1; i < n; i++)
{
sum = sum + (u_cal(u, i) * y[n - 1,i]) /
fact(i);
}
Console.WriteLine("\n Value at "+value+" is "+Math.Round(sum,4));
}
}
// This code is contributed by mits
PHP
<?php
// PHP Program to interpolate using
// newton backward interpolation
// Calculation of u mentioned in formula
function u_cal($u, $n)
{
$temp = $u;
for ($i = 1; $i < $n; $i++)
$temp = $temp * ($u + $i);
return $temp;
}
// Calculating factorial of given n
function fact($n)
{
$f = 1;
for ($i = 2; $i <= $n; $i++)
$f *= $i;
return $f;
}
// Driver Code
// number of values given
$n = 5;
$x = array(1891, 1901, 1911,
1921, 1931);
// y[][] is used for difference
// table and y[][0] used for input
$y = array_fill(0, $n, array_fill(0, $n, 0));
$y[0][0] = 46;
$y[1][0] = 66;
$y[2][0] = 81;
$y[3][0] = 93;
$y[4][0] = 101;
// Calculating the backward difference table
for ($i = 1; $i < $n; $i++)
{
for ($j = $n - 1; $j >= $i; $j--)
$y[$j][$i] = $y[$j][$i - 1] -
$y[$j - 1][$i - 1];
}
// Displaying the backward difference table
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j <= $i; $j++)
print($y[$i][$j] . "\t");
print("\n");
}
// Value to interpolate at
$value = 1925;
// Initializing u and sum
$sum = $y[$n - 1][0];
$u = ($value - $x[$n - 1]) / ($x[1] - $x[0]);
for ($i = 1; $i < $n; $i++)
{
$sum = $sum + (u_cal($u, $i) *
$y[$n - 1][$i]) / fact($i);
}
print("\n Value at " . $value .
" is " . round($sum, 4));
// This code is contributed by chandan_jnu
?>
Javascript
<script>
// javascript Program to interpolate using
// newton backward interpolation
// Calculation of u mentioned in formula
function u_cal(u , n)
{
var temp = u;
for (var i = 1; i < n; i++)
temp = temp * (u + i);
return temp;
}
// Calculating factorial of given n
function fact(n)
{
var f = 1;
for (var i = 2; i <= n; i++)
f *= i;
return f;
}
// Driver code
// number of values given
var n = 5;
var x = [ 1891, 1901, 1911,
1921, 1931 ];
// y is used for difference
// table and y[0] used for input
var y = Array(n).fill(0.0).map(x => Array(n).fill(0.0));
y[0][0] = 46;
y[1][0] = 66;
y[2][0] = 81;
y[3][0] = 93;
y[4][0] = 101;
// Calculating the backward difference table
for (var i = 1; i < n; i++)
{
for (var j = n - 1; j >= i; j--)
y[j][i] = y[j][i - 1] - y[j - 1][i - 1];
}
// Displaying the backward difference table
for (var i = 0; i < n; i++)
{
for (var j = 0; j <= i; j++)
document.write(y[i][j] + "\t");
document.write('<br>');;
}
// Value to interpolate at
var value = 1925;
// Initializing u and sum
var sum = y[n - 1][0];
var u = (value - x[n - 1]) / (x[1] - x[0]);
for (var i = 1; i < n; i++)
{
sum = sum + (u_cal(u, i) * y[n - 1][i]) /
fact(i);
}
document.write("\n Value at " + value +
" is " +sum);
// This code is contributed by 29AjayKumar
</script>
Producción:
46 66 20 81 15 -5 93 12 -3 2 101 8 -4 -1 -3 Value at 1925 is 96.8368
Complejidad de tiempo: O (N * N), N es el número de filas.
Complejidad espacial: O (N * N), para almacenar elementos en la tabla de diferencias.
Este artículo es una contribución de Shubham Rana . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA