Interpolation with unequal interval

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

Interpolation
To ﬁnd the value of y for an x between diﬀerent x - values
x0, x1, . . . , xn is called problem of interpolation.

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.

Interpolation
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. ”

Interpolation
Let us assign polynomial Pn of degree n (or less) that assumes
the given data values

Interpolation
Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn

Interpolation
Pn(x0) = y0, Pn(x1) = y1, . . ., Pn(xn) = yn
This polynomial Pn is called interpolation polynomial.

Interpolation
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).

Interpolation with unequal intervals
Lagrange’s interpolation formula with unequal intervals:
Let y = f(x) be continuous and diﬀerentiable in the interval
(a, b).

(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.

(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.

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

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 + . . .

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

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.

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

formula becomes
x = g(y) ≈ Pn(y) =
(y − y1)(y − y2) . . . (y − yn)
(y0 − y1)(y0 − y2) . . . (y0 − yn)
x0

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 + . . .

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

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.

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.

Example
Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

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

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,

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
f(x) ≈ Pn(x) =
(x − x1)(x − x2)(x − x3)
(x0 − x1)(x0 − x2)(x0 − x3)
y0

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
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

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
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
+
(x − x0)(x − x1)(x − x3)
(x2 − x0)(x2 − x1)(x2 − x3)
y2

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
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
+
(x − x0)(x − x1)(x − x3)
(x2 − x0)(x2 − x1)(x2 − x3)
y2
+
(x − x0)(x − x1)(x − x2)
(x3 − x0)(x3 − x1)(x3 − x2)
y3

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
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
+
(x − x0)(x − x1)(x − x3)
(x2 − x0)(x2 − x1)(x2 − x3)
y2
+
(x − x0)(x − x1)(x − x2)
(x3 − x0)(x3 − x1)(x3 − x2)
y3
for x = 155
∴ f(155) =

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

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

Example
Ex. Using Lagrange’s formula, ﬁnd x for y = 7 for the following data:
x 1 3 4
y 4 12 19

Example
Ex. Using Lagrange’s formula, express the function
3x2 + x + 1
(x − 1)(x − 2)(x − 3)

Example
Sol.: Let us evaluate y = 3x2 + x + 1 for x = 1, x = 2 and x = 3

Example
These values are x0 = 1, x1 = 2 and x2 = 3 and

Example
y0 = 5,

Example
y0 = 5, y1 = 15 and y2 = 31

Example
y0 = 5, y1 = 15 and y2 = 31
y =
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
y0 +
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
y1

Example
y0 = 5, y1 = 15 and y2 = 31
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

Example
y0 = 5, y1 = 15 and y2 = 31
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)

Example
Thus
3x2 + x + 1
(x − 1)(x − 2)(x − 3)

Example
Thus
3x2 + x + 1
(x − 1)(x − 2)(x − 3)
=
2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2)
(x − 1)(x − 2)(x − 3)
=
2.5
(x − 1)
-
15
(x − 2)
+
15.5
(x − 3)

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

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)

truncation error as
E(f, x) = f(x) − Pn(x) =
(x − x0)(x − x1) . . . (x − xn)
(n + 1)!
f(n+1)(ξ)
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

truncation error as
E(f, x) = f(x) − Pn(x) =
(x − x0)(x − x1) . . . (x − xn)
(n + 1)!
f(n+1)(ξ)
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)|

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.

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.

Disadvantages:
To increase the degree the addition of another interpolation point
leads to re-computation.
i.e. no previous work is useful.

Disadvantages:
E.g: In calculating Pk(x), no obvious advantage can be taken of
the fact that one already has calculated Pk−1(x).

Disadvantages:
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.

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.

Divided Difference
Then the first divided difference of f for the arguments
x0, x1, . . . , xn are defined by ,

Divided Difference
f(x0, x1) =
f(x1) − f(x0)
x1 − x0

Divided Difference
f(x0, x1) =
f(x1) − f(x0)
x1 − x0
f(x1, x2) =
f(x2) − f(x1)
x2 − x1

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

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

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.

Divided Difference
Properties:
The divided diﬀerence operator is linear.

Divided Difference
Properties:
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.

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.

adding new terms.
Newton’s general interpolation formula is one such formula and
terms in it are called divided diﬀerences.

adding new terms.
By the deﬁnition of divided diﬀerence,

adding new terms.
f(x, x0) =
f(x) − f(x0)
x − x0

adding new terms.
f(x, x0) =
f(x) − f(x0)
x − x0
∴
f(x) = f(x0) + (x − x0)f(x, x0) − −(1)

Further
f(x, x0, x1) =
f(x, x0) − f(x0, x1)
x − x1

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)

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)

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)

multiplying equation (2) by (x − x0)

multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)
and so on,

and so on, and ﬁnally the last term (4) by
(x − x0) (x − x1) ... (x − xn−1) and

(x − x0) (x − x1) ... (x − xn−1) and adding (1), (2) , (3) up to (4)
we obtain

(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.

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

Example
Ex. Obtain the divided diﬀerence table for the data:
x -1 0 2 3
y -8 3 1 12

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

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

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

Example
The Newtons divided diﬀerence formula gives:

Example
f(x) = f(x0) +

Example
f(x) = f(x0) + (x − x0)f[x0, x1] + (x − x0)(x − x1)f[x0, x1, x2]

Example
+

Example
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]

Example
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
+

Example
+ (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]

Example
+ (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]
= ...
= 3x4 − 5x3 + 6x2 − 14x + 5

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.

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

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) +

Example

Example
+

Example
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]

Example
+ (x − x0)(x − x1)(x − x2)f[x0, x1, x2, x3]
= ...
= x3 + 17

Example
Ex. Find the missing term in the following table:
x 0 1 2 3 4
y 1 3 9 - 81

Example
Sol. Divided diﬀerence table:

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) + ...

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.

f(x) =
(xi+1 − x)3Mi
6h
+

f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h

f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +

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

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),

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.

Example
Ex. Obtain cubic spline for the following data:
x 0 1 2 3
y 2 -6 -8 2

Example
Sol. Since points are equispaced with h = 1 and n = 3,

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

Example
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0

Example
Mi−1 + 4Mi + Mi+1 =
6
h2
(yi−1 − 2yi + yi+1), i = 1, 2
also M0 = 0 , M3 = 0
∴ for i = 1,

Example
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)

Example
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,

Example
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)

Example
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)

Example
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

Example
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 =

Example
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 =

Example
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

Example
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)
Now the cubic spline in (xi ≤ x ≤ xi+1) is

Example
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)
f(x) =
(xi+1 − x)3Mi
6h
+

Example
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)
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h

Example
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)
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +

Example
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)
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)

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)

Example
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)

Example
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,

Example
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)

Example
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)

Example
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

Example
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 =

Example
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 =

Example
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

Example
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)

Example
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)
f(x) =
(xi+1 − x)3Mi
6h
+

Example
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)
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h

Example
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)
f(x) =
(xi+1 − x)3Mi
6h
+
(x − xi)3Mi+1
6h
+
(xi+1 − x)
h
yi −
h2
6
Mi +

Example
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)
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)

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

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)

Example
1. We have
lim
x→0+
f(x) = 0

Example
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)

Example
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1

Example
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x)

Example
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
Now
f (x) = 15x2 − 6x, −1 ≤ x ≤ 0
= −15x2 − 6x, 0 ≤ x ≤ 1
we have,
lim
x→0+
f (x) = 0

Example
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
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)

Example
1. We have
lim
x→0+
f(x) = 0 = lim
x→0−
f(x)
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

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