1. Numerical Methods - Interpolation
Unequal Intervals
Dr. N. B. Vyas
Department of Mathematics,
Atmiya Institute of Tech. and Science, Rajkot (Guj.)
niravbvyas@gmail.com
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
2. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
3. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
To find the value of y for an x which falls outside the range of x
(x < x0 or x > xn) is called the problem of extrapolation.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
4. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
To find the value of y for an x which falls outside the range of x
(x < x0 or x > xn) is called the problem of extrapolation.
Theorem by Weierstrass in 1885, “Every continuous
function in an interval (a,b) can be represented in that
interval to any desired accuracy by a polynomial. ”
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
5. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
To find the value of y for an x which falls outside the range of x
(x < x0 or x > xn) is called the problem of extrapolation.
Theorem by Weierstrass in 1885, “Every continuous
function in an interval (a,b) can be represented in that
interval to any desired accuracy by a polynomial. ”
Let us assign polynomial Pn of degree n (or less) that assumes
the given data values
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
6. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
To find the value of y for an x which falls outside the range of x
(x < x0 or x > xn) is called the problem of extrapolation.
Theorem by Weierstrass in 1885, “Every continuous
function in an interval (a,b) can be represented in that
interval to any desired accuracy by a polynomial. ”
Let us assign polynomial Pn of degree n (or less) that assumes
the given data values
Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
7. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
To find the value of y for an x which falls outside the range of x
(x < x0 or x > xn) is called the problem of extrapolation.
Theorem by Weierstrass in 1885, “Every continuous
function in an interval (a,b) can be represented in that
interval to any desired accuracy by a polynomial. ”
Let us assign polynomial Pn of degree n (or less) that assumes
the given data values
Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn
This polynomial Pn is called interpolation polynomial.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
8. Interpolation
To find the value of y for an x between different x - values
x0, x1, . . . , xn is called problem of interpolation.
To find the value of y for an x which falls outside the range of x
(x < x0 or x > xn) is called the problem of extrapolation.
Theorem by Weierstrass in 1885, “Every continuous
function in an interval (a,b) can be represented in that
interval to any desired accuracy by a polynomial. ”
Let us assign polynomial Pn of degree n (or less) that assumes
the given data values
Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn
This polynomial Pn is called interpolation polynomial.
x0, x1, . . . , xn is called the nodes ( tabular points, pivotal
points or arguments).
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
9. Interpolation with unequal intervals
Lagrange’s interpolation formula with unequal intervals:
Let y = f(x) be continuous and differentiable in the interval
(a, b).
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
10. Interpolation with unequal intervals
Lagrange’s interpolation formula with unequal intervals:
Let y = f(x) be continuous and differentiable in the interval
(a, b).
Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of x
and y, where the values of x need not necessarily be equally
spaced.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
11. Interpolation with unequal intervals
Lagrange’s interpolation formula with unequal intervals:
Let y = f(x) be continuous and differentiable in the interval
(a, b).
Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of x
and y, where the values of x need not necessarily be equally
spaced.
It is required to find Pn(x), a polynomial of degree n such that y
and Pn(x) agree at the tabulated points.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
12. Interpolation with unequal intervals
Lagrange’s interpolation formula with unequal intervals:
Let y = f(x) be continuous and differentiable in the interval
(a, b).
Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of x
and y, where the values of x need not necessarily be equally
spaced.
It is required to find Pn(x), a polynomial of degree n such that y
and Pn(x) agree at the tabulated points.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
14. Lagrange’s Interpolation
This polynomial is given by the following formula:
y = f(x) ≈ Pn(x) =
(x − x1)(x − x2) . . . (x − xn)
(x0 − x1)(x0 − x2) . . . (x0 − xn)
y0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
15. Lagrange’s Interpolation
This polynomial is given by the following formula:
y = f(x) ≈ Pn(x) =
(x − x1)(x − x2) . . . (x − xn)
(x0 − x1)(x0 − x2) . . . (x0 − xn)
y0
+
(x − x0)(x − x2) . . . (x − xn)
(x1 − x0)(x1 − x2) . . . (x1 − xn)
y1 + . . .
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
16. Lagrange’s Interpolation
This polynomial is given by the following formula:
y = f(x) ≈ Pn(x) =
(x − x1)(x − x2) . . . (x − xn)
(x0 − x1)(x0 − x2) . . . (x0 − xn)
y0
+
(x − x0)(x − x2) . . . (x − xn)
(x1 − x0)(x1 − x2) . . . (x1 − xn)
y1 + . . .
+
(x − x0)(x − x1) . . . (x − xn−1)
(xn − x0)(xn − x1) . . . (xn − xn−1)
yn
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
17. Lagrange’s Interpolation
This polynomial is given by the following formula:
y = f(x) ≈ Pn(x) =
(x − x1)(x − x2) . . . (x − xn)
(x0 − x1)(x0 − x2) . . . (x0 − xn)
y0
+
(x − x0)(x − x2) . . . (x − xn)
(x1 − x0)(x1 − x2) . . . (x1 − xn)
y1 + . . .
+
(x − x0)(x − x1) . . . (x − xn−1)
(xn − x0)(xn − x1) . . . (xn − xn−1)
yn
NOTE:
The above formula can be used irrespective of whether the values
x0, x1, . . . , xn are equally spaced or not.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
18. Lagrange’s Inverse Interpolation
In the Lagrange’s interpolation formula y is treated as dependent
variable and expressed as function of independent variable x.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
19. Lagrange’s Inverse Interpolation
In the Lagrange’s interpolation formula y is treated as dependent
variable and expressed as function of independent variable x.
Instead if x is treated as dependent variable and expressed as the
function of independent variable y, then Lagrange’s interpolation
formula becomes
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
20. Lagrange’s Inverse Interpolation
In the Lagrange’s interpolation formula y is treated as dependent
variable and expressed as function of independent variable x.
Instead if x is treated as dependent variable and expressed as the
function of independent variable y, then Lagrange’s interpolation
formula becomes
x = g(y) ≈ Pn(y) =
(y − y1)(y − y2) . . . (y − yn)
(y0 − y1)(y0 − y2) . . . (y0 − yn)
x0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
21. Lagrange’s Inverse Interpolation
In the Lagrange’s interpolation formula y is treated as dependent
variable and expressed as function of independent variable x.
Instead if x is treated as dependent variable and expressed as the
function of independent variable y, then Lagrange’s interpolation
formula becomes
x = g(y) ≈ Pn(y) =
(y − y1)(y − y2) . . . (y − yn)
(y0 − y1)(y0 − y2) . . . (y0 − yn)
x0
+
(y − y0)(y − y2) . . . (y − yn)
(y1 − y0)(y1 − y2) . . . (y1 − yn)
x1 + . . .
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
22. Lagrange’s Inverse Interpolation
In the Lagrange’s interpolation formula y is treated as dependent
variable and expressed as function of independent variable x.
Instead if x is treated as dependent variable and expressed as the
function of independent variable y, then Lagrange’s interpolation
formula becomes
x = g(y) ≈ Pn(y) =
(y − y1)(y − y2) . . . (y − yn)
(y0 − y1)(y0 − y2) . . . (y0 − yn)
x0
+
(y − y0)(y − y2) . . . (y − yn)
(y1 − y0)(y1 − y2) . . . (y1 − yn)
x1 + . . .
+
(y − y0)(y − y1) . . . (y − yn−1)
(yn − y0)(yn − y1) . . . (yn − yn−1)
xn
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
23. Lagrange’s Inverse Interpolation
In the Lagrange’s interpolation formula y is treated as dependent
variable and expressed as function of independent variable x.
Instead if x is treated as dependent variable and expressed as the
function of independent variable y, then Lagrange’s interpolation
formula becomes
x = g(y) ≈ Pn(y) =
(y − y1)(y − y2) . . . (y − yn)
(y0 − y1)(y0 − y2) . . . (y0 − yn)
x0
+
(y − y0)(y − y2) . . . (y − yn)
(y1 − y0)(y1 − y2) . . . (y1 − yn)
x1 + . . .
+
(y − y0)(y − y1) . . . (y − yn−1)
(yn − y0)(yn − y1) . . . (yn − yn−1)
xn
This relation is referred as Lagrange’s inverse interpolation
formula.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
24. Example
Ex. Given the table of values:
x 150 152 154 156
y =
√
x 12.247 12.329 12.410 12.490
Evaluate
√
155 using Lagrange’s interpolation formula.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
25. Example
Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
26. Example
Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156
y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
27. Example
Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156
y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490
By Lagrange’s interpolation formula,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
28. Example
Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156
y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490
By Lagrange’s interpolation formula,
f(x) ≈ Pn(x) =
(x − x1)(x − x2)(x − x3)
(x0 − x1)(x0 − x2)(x0 − x3)
y0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
29. Example
Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156
y0 = 12.247, y1 = 12.329, y2 = 12.410 and y3 = 12.490
By Lagrange’s interpolation formula,
f(x) ≈ Pn(x) =
(x − x1)(x − x2)(x − x3)
(x0 − x1)(x0 − x2)(x0 − x3)
y0
+
(x − x0)(x − x2)(x − x3)
(x1 − x0)(x1 − x2)(x1 − x3)
y1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
33. Example
Ex. Compute f(0.4) for the table below by the Lagrange’s
interpolation:
x 0.3 0.5 0.6
f(x) 0.61 0.69 0.72
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
34. Example
Ex. Using Lagrange’s formula, find the form of f(x) for the following
data:
x 0 1 2 5
f(x) 2 3 12 147
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
35. Example
Ex. Using Lagrange’s formula, find x for y = 7 for the following data:
x 1 3 4
y 4 12 19
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
36. Example
Ex. Using Lagrange’s formula, express the function
3x2 + x + 1
(x − 1)(x − 2)(x − 3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
37. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
38. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
39. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
y0 = 5,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
40. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
y0 = 5, y1 = 15 and y2 = 31
By Lagrange’s interpolation formula,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
41. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
y0 = 5, y1 = 15 and y2 = 31
By Lagrange’s interpolation formula,
y =
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
y0 +
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
y1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
42. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
y0 = 5, y1 = 15 and y2 = 31
By Lagrange’s interpolation formula,
y =
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
y0 +
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
y1
+
(x − x0)(x − x1)
(x2 − x0)(x2 − x1)
y2
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
43. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
y0 = 5, y1 = 15 and y2 = 31
By Lagrange’s interpolation formula,
y =
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
y0 +
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
y1
+
(x − x0)(x − x1)
(x2 − x0)(x2 − x1)
y2
substituting above values, we get
y = 2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
44. Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3
These values are x0 = 1, x1 = 2 and x2 = 3 and
y0 = 5, y1 = 15 and y2 = 31
By Lagrange’s interpolation formula,
y =
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
y0 +
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
y1
+
(x − x0)(x − x1)
(x2 − x0)(x2 − x1)
y2
substituting above values, we get
y = 2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
45. Example
Thus
3x2 + x + 1
(x − 1)(x − 2)(x − 3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
47. Error in Interpolation
Error in Interpolation:
We assume that f(x) has continuous derivatives of order upto
n + 1 for all x ∈ (a, b). Since, f(x) is approximated by Pn(x), the
results contains errors. We define the error of interpolation or
truncation error as
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
48. Error in Interpolation
Error in Interpolation:
We assume that f(x) has continuous derivatives of order upto
n + 1 for all x ∈ (a, b). Since, f(x) is approximated by Pn(x), the
results contains errors. We define the error of interpolation or
truncation error as
E(f, x) = f(x) − Pn(x) =
(x − x0)(x − x1) . . . (x − xn)
(n + 1)!
f(n+1)(ξ)
where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
49. Error in Interpolation
Error in Interpolation:
We assume that f(x) has continuous derivatives of order upto
n + 1 for all x ∈ (a, b). Since, f(x) is approximated by Pn(x), the
results contains errors. We define the error of interpolation or
truncation error as
E(f, x) = f(x) − Pn(x) =
(x − x0)(x − x1) . . . (x − xn)
(n + 1)!
f(n+1)(ξ)
where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x)
since, ξ is an unknown, it is difficult to find the value of error.
However, we can find a bound of the error. The bound of the
error is obtained as
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
50. Error in Interpolation
Error in Interpolation:
We assume that f(x) has continuous derivatives of order upto
n + 1 for all x ∈ (a, b). Since, f(x) is approximated by Pn(x), the
results contains errors. We define the error of interpolation or
truncation error as
E(f, x) = f(x) − Pn(x) =
(x − x0)(x − x1) . . . (x − xn)
(n + 1)!
f(n+1)(ξ)
where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x)
since, ξ is an unknown, it is difficult to find the value of error.
However, we can find a bound of the error. The bound of the
error is obtained as
|E(f, x)| ≤
|(x − x0)(x − x1) . . . (x − xn)|
(n + 1)!
max
a≤ξ≤b
|f(n+1)(ξ)|
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
51. Example
Ex. Using the data sin(0.1) = 0.09983 and sin(0.2) = 0.19867, find
an approximate value of sin(0.15) by Lagrange interpolation.
Obtain a bound on the error at x = 0.15.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
52. Lagrange’s Interpolation
Disadvantages:
In practice, we often do not know the degree of the interpolation
polynomial that will give the required accuracy, so we should be
prepared to increase the degree if necessary.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
53. Lagrange’s Interpolation
Disadvantages:
In practice, we often do not know the degree of the interpolation
polynomial that will give the required accuracy, so we should be
prepared to increase the degree if necessary.
To increase the degree the addition of another interpolation point
leads to re-computation.
i.e. no previous work is useful.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
54. Lagrange’s Interpolation
Disadvantages:
In practice, we often do not know the degree of the interpolation
polynomial that will give the required accuracy, so we should be
prepared to increase the degree if necessary.
To increase the degree the addition of another interpolation point
leads to re-computation.
i.e. no previous work is useful.
E.g: In calculating Pk(x), no obvious advantage can be taken of
the fact that one already has calculated Pk−1(x).
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
55. Lagrange’s Interpolation
Disadvantages:
In practice, we often do not know the degree of the interpolation
polynomial that will give the required accuracy, so we should be
prepared to increase the degree if necessary.
To increase the degree the addition of another interpolation point
leads to re-computation.
i.e. no previous work is useful.
E.g: In calculating Pk(x), no obvious advantage can be taken of
the fact that one already has calculated Pk−1(x).
That means we need to calculate entirely new polynomial.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
56. Divided Difference
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
57. Divided Difference
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
Then the first divided difference of f for the arguments
x0, x1, . . . , xn are defined by ,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
58. Divided Difference
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
Then the first divided difference of f for the arguments
x0, x1, . . . , xn are defined by ,
f(x0, x1) =
f(x1) − f(x0)
x1 − x0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
59. Divided Difference
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
Then the first divided difference of f for the arguments
x0, x1, . . . , xn are defined by ,
f(x0, x1) =
f(x1) − f(x0)
x1 − x0
f(x1, x2) =
f(x2) − f(x1)
x2 − x1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
60. Divided Difference
The second divided difference of f for three arguments
x0, x1, x2 is defined by
f(x0, x1, x2) =
f(x1, x2) − f(x0, x1)
x2 − x0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
61. Divided Difference
The second divided difference of f for three arguments
x0, x1, x2 is defined by
f(x0, x1, x2) =
f(x1, x2) − f(x0, x1)
x2 − x0
and similarly the divided difference of order n is defined by
f(x0, x1, . . . , xn) =
f(x1, x2, . . . , xn) − f(x0, x1, . . . , xn−1)
xn − x0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
62. Divided Difference
Properties:
The divided differences are symmetrical in all their arguments;
that is, the value of any divided difference is independent of the
order of the arguments.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
63. Divided Difference
Properties:
The divided differences are symmetrical in all their arguments;
that is, the value of any divided difference is independent of the
order of the arguments.
The divided difference operator is linear.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
64. Divided Difference
Properties:
The divided differences are symmetrical in all their arguments;
that is, the value of any divided difference is independent of the
order of the arguments.
The divided difference operator is linear.
The nth order divided differences of a polynomial of degree n are
constant, equal to the coefficient of xn.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
65. Newton’s Divided Difference Interpolation
An interpolation formula which has the property that a
polynomial of higher degree may be derived from it by simply
adding new terms.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
66. Newton’s Divided Difference Interpolation
An interpolation formula which has the property that a
polynomial of higher degree may be derived from it by simply
adding new terms.
Newton’s general interpolation formula is one such formula and
terms in it are called divided differences.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
67. Newton’s Divided Difference Interpolation
An interpolation formula which has the property that a
polynomial of higher degree may be derived from it by simply
adding new terms.
Newton’s general interpolation formula is one such formula and
terms in it are called divided differences.
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
By the definition of divided difference,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
68. Newton’s Divided Difference Interpolation
An interpolation formula which has the property that a
polynomial of higher degree may be derived from it by simply
adding new terms.
Newton’s general interpolation formula is one such formula and
terms in it are called divided differences.
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
By the definition of divided difference,
f(x, x0) =
f(x) − f(x0)
x − x0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
69. Newton’s Divided Difference Interpolation
An interpolation formula which has the property that a
polynomial of higher degree may be derived from it by simply
adding new terms.
Newton’s general interpolation formula is one such formula and
terms in it are called divided differences.
Let f(x0), f(x1), . . . , f(xn) be the values of a function f
corresponding to the arguments x0, x1, . . . , xn where the intervals
x1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.
By the definition of divided difference,
f(x, x0) =
f(x) − f(x0)
x − x0
∴
f(x) = f(x0) + (x − x0)f(x, x0) − −(1)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
70. Newton’s Divided Difference Interpolation
Further
f(x, x0, x1) =
f(x, x0) − f(x0, x1)
x − x1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
71. Newton’s Divided Difference Interpolation
Further
f(x, x0, x1) =
f(x, x0) − f(x0, x1)
x − x1
which yields
f(x, x0) = f(x0, x1) + (x − x1)f(x, x0, x1) − −(2)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
72. Newton’s Divided Difference Interpolation
Further
f(x, x0, x1) =
f(x, x0) − f(x0, x1)
x − x1
which yields
f(x, x0) = f(x0, x1) + (x − x1)f(x, x0, x1) − −(2)
Similarly
f(x, x0, x1) = f(x0, x1, x2) + (x − x2)f(x, x0, x1, x2) − −(3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
73. Newton’s Divided Difference Interpolation
Further
f(x, x0, x1) =
f(x, x0) − f(x0, x1)
x − x1
which yields
f(x, x0) = f(x0, x1) + (x − x1)f(x, x0, x1) − −(2)
Similarly
f(x, x0, x1) = f(x0, x1, x2) + (x − x2)f(x, x0, x1, x2) − −(3)
and in general
f(x, x0, ..., xn−1) = f(x0, x1, ..., xn) + (x − xn)f(x, x0, x1, ..., xn) − −(4)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
74. Newton’s Divided Difference Interpolation
multiplying equation (2) by (x − x0)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
75. Newton’s Divided Difference Interpolation
multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)
and so on,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
76. Newton’s Divided Difference Interpolation
multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)
and so on, and finally the last term (4) by
(x − x0) (x − x1) ... (x − xn−1) and
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
77. Newton’s Divided Difference Interpolation
multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)
and so on, and finally the last term (4) by
(x − x0) (x − x1) ... (x − xn−1) and adding (1), (2) , (3) up to (4)
we obtain
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
78. Newton’s Divided Difference Interpolation
multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)
and so on, and finally the last term (4) by
(x − x0) (x − x1) ... (x − xn−1) and adding (1), (2) , (3) up to (4)
we obtain
f(x) =
f (x0) + (x − x0) f (x0, x1) + (x − x0) (x − x1) f (x0, x1, x2) + ... +
(x − x0) (x − x1) ... (x − xn−1) f (x0, x1, ..., xn)
This formula is called Newton’s divided difference formula.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
79. Newton’s Divided Difference Interpolation
The divided difference upto third order
x y 1stdiv.diff. 2nddiv.diff. 3rddiv.diff.
x0 y0
[x0, x1]
x1 y1 [x0, x1, x2]
[x1, x2] [x0, x1, x2, x3]
x2 y2 [x1, x2, x3]
[x2, x3]
x3 y3
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
80. Example
Ex. Obtain the divided difference table for the data:
x -1 0 2 3
y -8 3 1 12
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
81. Example
Sol. We have the following divided difference table for the data:
x y First d.d Second d.d Third d.d
-1 -8
3 + 8
0 + 1
= 11
0 3
−1 − 11
2 + 1
= −4
1 − 3
2 − 0
= −1
4 + 4
3 + 1
= 2
2 1
11 + 1
3 − 0
= 4
12 − 1
3 − 2
= 11
3 12
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
82. Example
Ex. Find f(x) as a polynomial in x for the following data by Newtons
divided difference formula:
x -4 -1 0 2 5
f(x) 1245 33 5 9 1335
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
83. Example
Sol. We have the following divided difference table for the data:
x y 1st d.d 2nd d.d 3rd d.d 4th d.d
-4 1245
−404
-1 33 94
−28 −14
0 5 10 3
2 13
2 9 88
442
5 1335
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
84. Example
The Newtons divided difference formula gives:
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
85. Example
The Newtons divided difference formula gives:
f(x) = f(x0) +
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
86. Example
The Newtons divided difference formula gives:
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
87. Example
The Newtons divided difference formula gives:
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
88. Example
The Newtons divided difference formula gives:
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
89. Example
The Newtons divided difference formula gives:
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
+
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
90. Example
The Newtons divided difference formula gives:
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
+ (x − x0)(x − x1)(x − x2)(x − x3)f[x0, x1, x2, x3, x4]
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
92. Example
Ex. Find f(x) as a polynomial in x for the following data by Newtons
divided difference formula:
x -2 -1 0 1 3 4
f(x) 9 16 17 18 44 81
Hence, interpolate at x = 0.5 and x = 3.1.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
93. Example
Sol. We form the divided difference table for the given data.
x f(x) 1st d.d 2nd d.d 3rd d.d 4th d.d
−2 9
7
−1 16 −3
1 1
0 17 0 0
1 1
1 18 4 0
13 1
3 44 8
37
4 81
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
94. Example
Since, the fourth order differences are zeros, the data represents a
third degree polynomial. Newtons divided difference formula
gives the polynomial as
f(x) = f(x0) +
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
95. Example
Since, the fourth order differences are zeros, the data represents a
third degree polynomial. Newtons divided difference formula
gives the polynomial as
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
96. Example
Since, the fourth order differences are zeros, the data represents a
third degree polynomial. Newtons divided difference formula
gives the polynomial as
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
97. Example
Since, the fourth order differences are zeros, the data represents a
third degree polynomial. Newtons divided difference formula
gives the polynomial as
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
98. Example
Since, the fourth order differences are zeros, the data represents a
third degree polynomial. Newtons divided difference formula
gives the polynomial as
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
= ...
= x3 + 17
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
99. Example
Ex. Find the missing term in the following table:
x 0 1 2 3 4
y 1 3 9 - 81
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
101. Example
Sol. Divided difference table:
By Newton’s divided difference formula
f(x) =
f (x0) + (x − x0) f (x0, x1) + (x − x0) (x − x1) f (x0, x1, x2) + ...
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
102. Spline Interpolation
Spline interpolation is a form of interpolation where the
interpolant is a special type of piecewise polynomial called a
spline
Consider the problem of interpolating between the data points
(x0, y0), (x1, y1), . . . , (xn, yn) by means of spline fitting.
Then the cubic spline f(x) is such that
(i) f(x) is a linear polynomial outside the interval (x0, xn)
(ii) f(x) is a cubic polynomial in each of the subintervals,
(iii) f (x) and f (x) are continuous at each point.
Since f(x) is cubic in each of the subintervals f (x) shall be
linear.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
105. Spline Interpolation
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
106. Spline Interpolation
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
107. Spline Interpolation
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1
where Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1),
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
108. Spline Interpolation
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1
where Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1),
i = 1, 2, 3, ..., (n − 1)
and M0 = 0, Mn = 0, xi+1 − xi = h.
which gives n + 1 equations in n + 1 unknowns Mi(i = 0, 1, ..., n)
which can be solved. Substituting the value of Mi gives the
concerned cubic spline.
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
109. Example
Ex. Obtain cubic spline for the following data:
x 0 1 2 3
y 2 -6 -8 2
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
110. Example
Sol. Since points are equispaced with h = 1 and n = 3,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
111. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
112. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
113. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
114. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
115. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
116. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
117. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
118. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
119. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
120. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
121. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
122. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Now the cubic spline in (xi ≤ x ≤ xi+1) is
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
123. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
124. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
125. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
126. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1 —(3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
127. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 36; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 72 —(2)
solving these, we get M1 =4.8 and M2 =16.8
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1 —(3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
128. Example
Ex. The following values of x and y are given:
x 1 2 3 4
y 1 2 5 11
Find the cubic splines and evaluate y(1.5) and y (3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
129. Example
Sol. Since points are equispaced with h = 1 and n = 3,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
130. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
131. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
132. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
133. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
134. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2,
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
135. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
136. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
137. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
138. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
139. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
140. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
141. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Now the cubic spline in (xi ≤ x ≤ xi+1) is
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
142. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
143. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
144. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
145. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1 —(3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
146. Example
Sol. Since points are equispaced with h = 1 and n = 3, the cubic
spline can be determined from
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1, M0 + 4M1 + M2 = 6(y0 − 2y1 + y2)
therefore, 4M1 + M2 = 12; —(1)
for i = 2, M1 + 4M2 + M3 = 6(y1 − 2y2 + y3)
M1 + 4M2 = 18 —(2)
solving these, we get M1 =2 and M2 =4
Now the cubic spline in (xi ≤ x ≤ xi+1) is
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +
(x − xi)
h
yi+1 −
h2
6
Mi+1 —(3)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
147. Example
Ex. Find whether the following functions are cubic splines ?
1.
f(x) = 5x3 − 3x2, −1 ≤ x ≤ 0
= −5x3 − 3x2, 0 ≤ x ≤ 1
2.
f(x) = −2x3 − x2, −1 ≤ x ≤ 0
= 2x3 + 3x2, 0 ≤ x ≤ 1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
148. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
149. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
150. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
151. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
152. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
153. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x) = 0
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
154. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x) = 0 = lim
x→0−
f (x)
therefore, the function f (x) is continuous on (−1, 1).
f (x)
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
155. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x) = 0 = lim
x→0−
f (x)
therefore, the function f (x) is continuous on (−1, 1).
f (x) = 30x−6, −1 ≤ x ≤ 0
= −30x − 6, 0 ≤ x ≤ 1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals
156. Example
Sol. In both the examples, f(x) is a cubic polynomial in both
intervals (1, 0) and (0, 1).
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
therefore, given function f(x) is continuous on (−1, 1).
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x) = 0 = lim
x→0−
f (x)
therefore, the function f (x) is continuous on (−1, 1).
f (x) = 30x−6, −1 ≤ x ≤ 0
= −30x − 6, 0 ≤ x ≤ 1
Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals