Newton’s forward/backward difference:
Newton Forward Difference:
If the x value of a data set are uniformly spaced then let us define for the data set.
$$ S= \frac {x-x_0}{\Delta x} $$
This variable is called interpolating variable. i.e. s is linear in ‘x’.
Now the unique nth degree polynomial can also be represented as,
$$ P_n(x)=f_0+s\Delta f_0+ \frac {s(s-1)}{2!}\Delta^2 f_0 + \frac {s(s-1)(s-2)}{3!}\Delta^3 f_0+ \cdots + \frac {s(s-1)(s-2)\cdots (s-(n-1))}{n!}\Delta^n f_0 $$
This is Newton’s forward difference polynomial.
You can also represent this forward difference polynomial using binomial coefficient representation
$$ \text{i.e.} \binom{s}{i}= \frac {s(s-1)(s-2)\cdots (s-(i-1))}{i!} $$
$$ \therefore P_n(x)= f_0+ \binom{s}{1}\Delta f_0+ \binom{s}{2}\Delta^2 f_0+\cdots $$
Example:
Find Solution using Newton’s Forward Difference formula(x = 1895)
| x |
1891 |
1901 |
1911 |
1921 |
1931 |
| y |
46 |
66 |
81 |
93 |
101 |
Solution:
Newton’s forward difference table is
| x |
y |
Δy |
Δ2y |
Δ3y |
Δ4y |
| 1891 |
46 |
|
|
|
|
|
|
20 |
|
|
|
| 1901 |
66 |
|
-5 |
|
|
|
|
15 |
|
2 |
|
| 1911 |
81 |
|
-3 |
|
-3 |
|
|
12 |
|
-1 |
|
| 1921 |
93 |
|
-4 |
|
|
|
|
8 |
|
|
|
| 1931 |
101 |
|
|
|
|
$$ Δx=x_1-x_0=1901-1891=10 $$
$$ S= \frac {x-x_0}{\Delta x} $$
$$ S= \frac {1895-1891}{10} $$
$$ S= 0.4 $$
Newton’s forward difference interpolation formula is
$$ P_n(x)=f_0+s\Delta f_0+ \frac {s(s-1)}{2!}\Delta^2 f_0 + \frac {s(s-1)(s-2)}{3!}\Delta^3 f_0+ \cdots + \frac {s(s-1)(s-2)\cdots (s-(n-1))}{n!}\Delta^n f_0 $$
$$ =46+0.4\times 20+\frac {0.4(0.4-1)}{2}\times(-5)+\frac {0.4(0.4-1)(0.4-2)}{6}\times(2)\frac {0.4(0.4-1)(0.4-2)(0.4-3)}{24}\times(-3) $$
$$ =46+8+0.6+0.128+0.1248 $$
$$ =54.8528 $$
Newton Backward Difference Polynomial:
As explained for Newton’s forward difference polynomials, we can also use backward difference formulas to develop polynomials.
We can use backward differences
For a uniformly spaced data, if xn is the base in an (n+1) data (x0,f0),(x1,f1),..,(xn,fn)
$$ \Delta f_n-f_n-f_{n-1} $$
$$ S= \frac {x-x_n}{\Delta x} $$
We have the unique nth degree polynomial in the power series as:
$$ P_n(x)=f_n+s\Delta f_n+ \frac {s(s+1)}{2!}\Delta^2 f_n + \frac {s(s+1)(s+2)}{3!}\Delta^3 f_n+ \cdots + \frac {s(s+1)(s+2)\cdots (s+(n-1))}{n!}\Delta^n f_n $$
Example:
Find Solution using Newton’s Backward Difference formula(x = 1925)
| x |
1891 |
1901 |
1911 |
1921 |
1931 |
| y |
46 |
66 |
81 |
93 |
101 |
Solution:
Newton’s backward difference table is
| x |
y |
Δy |
Δ2y |
Δ3y |
Δ4y |
| 1891 |
46 |
|
|
|
|
|
|
20 |
|
|
|
| 1901 |
66 |
|
-5 |
|
|
|
|
15 |
|
2 |
|
| 1911 |
81 |
|
-3 |
|
-3 |
|
|
12 |
|
-1 |
|
| 1921 |
93 |
|
-4 |
|
|
|
|
8 |
|
|
|
| 1931 |
101 |
|
|
|
|
$$ Δx=x_1-x_0=1901-1891=10 $$
$$ S= \frac {x-x_n}{\Delta x} $$
$$ S= \frac {1925-1931}{10} $$
$$ S= -0.6 $$
Newton’s backward difference interpolation formula is
$$ P_n(x)=f_n+s\Delta f_n+ \frac {s(s+1)}{2!}\Delta^2 f_n + \frac {s(s+1)(s+2)}{3!}\Delta^3 f_n+ \cdots + \frac {s(s+1)(s+2)\cdots (s+(n-1))}{n!}\Delta^n f_n $$
$$ =101+(-0.6)\times 8+\frac {-0.6(-0.6+1)}{2}\times(-4)+\frac {-0.6(-0.6+1)(-0.6+2)}{6}\times(-1)\frac {-0.6(-0.6+1)(-0.6+2)(-0.6+3)}{24}\times(-3) $$
$$ =101-4.8+0.48-0.056+0.1008 $$
$$ =96.8368 $$
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