UNIT-2
Finite Differences
and
Interpolation
with
Equal Intervals
Differences:
denoted as ∆f.
If for given arguments are
and its functional values are
The forward difference is respectively define as:
Calculation of Higher order Difference: (Second order)
In this way one can calculate----------
In this way one can obtain-------
Exm. Construct difference table for following data.
x | 0 | 1 | 2 | 3 | 4 | 5 |
F(x) | 12 | 15 | 20 | 27 | 39 | 52 |
(2) Back ward Difference: The Back ward difference is denoted as f.
If for given arguments are
and its functional values are
The Backward difference is respectively define as:
Calculation of Higher order Difference: (Second order)
In this way one can obtain-------
x | 0 | 1 | 2 | 3 | 4 |
F(x) | 2 | 3 | 12 | 35 | 78 |
Exm. Construct difference table for following data.
In this way one can obtain-------
x | 0 | 1 | 2 | 3 | 4 |
F(x) | 2 | 3 | 12 | 35 | 78 |
Exm. Construct difference table for following data.
The Central difference
Define as:
The Shift Operator (E)
Define as:
The Mean or Average Operator
Define as:
In usual notation prove:
Exa. Find the missing term of the following data.
f(1)=8, f (2)=12, f(3)=19, f(4)=29
Sol.: First of all we construct difference table for the given data.
x | F(x)=y | | | |
1 | 8 | | | |
| | 4 | | |
2 | 12 | | 3 | |
| | 7 | | 0 |
3 | 19 | | 3 | |
| | 10 | | |
4 | 29 | | | |
We have to find out f(6):
First of all we note the following formula
From the given data the difference value of x means
h=1, initial value x0=1 we have to given four values,
So we can obtain expansion up to ∆3----
In calculation of f(6)=f(x5)
In usual notation prove that
Let y=f(x) be a continuous function and the value of x are equally space .
Expression of Polynomial
Let f(x) be a given polynomial of degree n which is to be expressed in terms of factorial notation.
Let
Expression of Polynomial
Let f(x) be a given polynomial of degree n which is to be expressed in terms of factorial notation.
Let
Substituting x=0, in all the above equations, we have
Sub. Ai’s we get
Exa. Express in factorial notation and hence find its differences in factorial notation, the interval of differencing being unity.
Sol. Here given polynomial of degree four is given.
So, we need in our difference table up to fourth difference.
x | f(x) | | | | |
0 | 5 | | | | |
| | -20 | | | |
1 | -15 | | -10 | | |
| | -30 | | 12 | |
2 | -45 | | 2 | | 72 |
| | -28 | | 84 | |
3 | -73 | | 86 | | |
| | 58 | | | |
4 | -15 | | | | |
We have
Find the polynomial f(x), which satisfy the following
data and hence find the value of f(1.5).
x | 1 | 2 | 3 | 4 | 5 |
y | 4 | 13 | 34 | 73 | 136 |
Sol. We have to construct a difference table for
Given data.
x | f(x) | | | |
1 | 4 | | | |
| | 9 | | |
2 | 13 | | 12 | |
| | 21 | | 6 |
3 | 34 | | 18 | |
| | 39 | | 6 |
4 | 73 | | 24 | |
| | 63 | | |
5 | 136 | | | |
As the third difference is constant.
So, we have a polynomial of degree three.
We know that
Self-Practice example: Write down the polynomial which satisfy the following data.
x | -1 | 0 | 1 | 2 |
f(x) | 1 | 1 | 1 | -5 |
Interpolation:
The method of finding the value of f(x) for a given
Value of x is known as interpolation.
NEWTON’S FORWARD DIFFERENCE INTERPOLATION FORMULA:
Let y=f(x) be a function of x. Given the set of n+1
value
Where x is independent and y dependent variable.
The value of x is at equally spaced.
So, y=f(x) is a polynomial of x degree n.
Such that the value of y agree at the given point
can be written as
(1)
Where are constant to be determine.
Put in equation (1)
Again, put in equ.(1)
Put in equation (1)
Similarly putting
Now
Similarly, we can find remaining terms, using all the above values in equation (1), we get
Exa.: The population of town were as under
Year | 1891 | 1901 | 1911 | 1921 | 1931 |
Population (In thousand) | 46 | 66 | 81 | 93 | 101 |
When Finding is starting of the table than we can apply Newton’s forward formula
Estimate the population for the year 1895.
Solution: We have to find x=1895 which is starting of the table, so we can apply Newton’s forward
interpolation formula.
NOW, NEWTON’S FORWARD DIFFERENCE INTERPOLATION FORMULA:
Difference table:
x | y | | | | |
1891 | 46 | | | | |
| | 20 | | | |
1901 | 66 | | -5 | | |
| | 15 | | 2 | |
1911 | 81 | | -3 | | -3 |
| | 12 | | -1 | |
1921 | 93 | | -4 | | |
| | 8 | | | |
1931 | 101 | | | | |
For given data:
Sub. All these values in N.F..I.F.
Self-Practice Example:
Find f(1.02) having given:
x | 1.00 | 1.10 | 1.20 | 1.30 |
f(x) | 0.8415 | 0.8912 | 0.9320 | 0.9636 |
Stirling ‘s Interpolation Formula
Let y=f(x) is a function of x.
Given set of n+1 value
Of x & y. The value of x is equally spaced.
To find a polynomial of a x of degree n we consider, the following diff. table in which central co-ordinate as , corresponding to
By Gauss’s forward formula, we have
By Gauss’s Backward formula, we have
Adding equation (1) and (2) gives the formula
To find a polynomial of a x of degree n we consider, the following diff. table in which central co-ordinate as , corresponding to
Bessel’s Interpolation Formula
Difference table:
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The Bessel’s formula is of the form
Where are constant to be determine.
We know that
Sub. these values of (a), (b),(c) and (d) in equation (1)
Comparing the coefficient of from
Equ. (2) and (3)
Using all these values in equation (1), we get
Bessel’s Formula
To find a polynomial of a x of degree n we consider, the following diff. table in which central co-ordinate as , corresponding to
Everett’s Interpolation Formula
Let y=f(x) is a function of x. Given a set of n+1
values of x and y. The value of x are at equally spaced.
Difference table:
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For this formula we can use even order term from
The table.
The Everett’s formula is of the form
Where are constant to be determine.
We know that
Sub. these values of (a), (b),(c) and in equation (1)
Comparing the coefficient of equ. (2) and (3)
Consider
Using the values of In equation (1)
Everett’s Formula