Chapter 5 - Interpolation(newton's diff)

Engineering Mathematics 4
AQ090-3-2 and VE1
Interpolation

AQ090-3-2-EMT4 & Engineering Mathematics 4 Interpolation
Topic & Structure of The Lesson
 Introduction
 Newton’s Forward Difference
 Newton’s Backward Difference
 Newton’s Divided-difference Interpolating polynomials
 Lagrange Interpolation
 Inverse Lagrange Interpolation

• At the end of this topic, You should be able to
Learning Outcomes
Understand the difference operators and the use
of interpolation.

• If you have mastered this topic, you should be able to use the
following terms correctly in your assignments and exams:
Key Terms You Must Be Able To Use
 Interpolation
 First Difference
 Second Difference
 Forward Difference
 Backward Difference
 Lagrange

Interpolation
: 𝑥 0 … . 𝑥 1 … .. 𝑥 2 … . 𝑥 𝑛
: 𝑦0 … . 𝑦 1 … .. 𝑦 2 …. 𝑦 𝑛
Arguments
Entries
• The process of finding the values of function f(x) (f(x) is unknown) for
any intermediate value of the argument ‘x’ between and is called
interpolation. The process of finding the value of a function outside the
interval of argument ‘x’ is called extrapolation.
• The process of computing the value of the unknown function, with
given discrete set of points. To find the value of the function, we first
determine an interpolating polynomial and then compute the
approximate function at the given point.
• The resulting polynomial will always be the same!

The interval difference (h)
is same for all sequence
of values.

Newton’s Forward Difference
• This formula is particularly useful for interpolating the
values of f(x) when x is near to
• If we have the data from the unknown function f(x):
x …
y …

Values of x Values of y First Difference (y) Second Difference (y) Third Difference (y)
Newton’s Forward Difference Table

Newton’s Forward Difference
𝑃=
𝑥 − 𝑥0
h
h = equal interval between x

Example 1
Compute the value of f(x), when x=3 by applying the appropriate
interpolation formula from the polynomial function, correct to the value of
4 significant figure.
x 0 5 10 15 20
f(x) 1.0 1.6 3.8 8.2 15.4
x=3 is near to , so Newton forward difference

Values of x Values of y () () () ()
Forward difference table

𝑃=
𝑥 − 𝑥0
h
=
3 −0
5 −0
=
3
5
The value of f(x), when x=3 is 1.2016

Example 2
Temperature x °C 140 150 160 170 180
Pressure y 3.685 4.854 6.302 8.076 10.225
Calculate the pressure at temperature using the appropriate interpolation
method.
x=142 is near to , so Newton forward difference

𝑃=
𝑥 − 𝑥0
h
=
142 −140
150 −140
=
2
10
=
1
5
The pressure at temperature is 3.8987

Newton’s Backward Difference
• This formula is particularly useful for interpolating the
values of f(x) when x is near to , i.e. when x is at the end of the
table.
x …
y …

Newton’s Backward Difference Table
Values of x Values of y First Difference (y) Second Difference (y) Third Difference (y)

Newton’s Backward Difference
𝑃=
𝑥 − 𝑥𝑛
h
h = equal interval between x

Example 3
Compute the value of f(x), when x=17 by applying the appropriate
interpolation formula from the polynomial function, correct to the value of
4 significant figure.
x 0 5 10 15 20
f(x) 1.0 1.6 3.8 8.2 15.4
x=17 is near to , so Newton backward difference

Values of x Values of y () () () ()
Backward difference table

The value of f(x), when x=17 is 10.7104
𝑃=
𝑥 − 𝑥𝑛
h
=
17 −20
20 −15
=
−3
5

Example 4
Using Newton’s backward interpolation formula, calculate f(7.5) from below
x 1 2 3 4 5 6 7 8
y 1 8 27 64 125 216 343 512
[Answer: 421.875

x y () () () () () () ()
216
343
Backward difference table

.5)=512+(−0 .5 )(169 )+
(− 0.5) (−0.5 +1)
2 !
( 42)+
(− 0.5) (− 0.5+1) (−0.5+ 2
3 !
The value of f(x), when x=7.5 is 421.875
𝑃=
𝑥 − 𝑥𝑛
h
=
7.5 −8
8 −7
=− 0.5

The interval difference
(h) is NOT same for all
sequence of values.

Newton’s Divided Difference
• Newton’s divided difference interpolation formula is an
interpolation technique used when the interval difference is not same
for all sequence of values.
x …
y …

Value
s of x
Values
of y
First Divided
Difference
Second Divided
Difference
Third Divided
Difference

Example 5
Compute f(8) and f(9) for the data
using Newton’s divided difference formula.
x 4 5 7 10 11 13
f(x) 48 100 294 900 1210 2028

Divided difference
table
x y
4 48
5 100
7 294 0
0
10 900 0
11 1210
13 2028

Now applying the Newton’s divided difference formula
¿48+(𝑥−4)(52)+(𝑥−4)(𝑥−5)(15)+(𝑥−4)(𝑥−5)(𝑥−7)(1)+(𝑥−4)(𝑥−5)(𝑥−7)(𝑥−10)(0)
Subs. x=8

Subs. x=9

ANS: v(16)=392.0767 / 392.0737 / m/s

Lagrange’s Method
• If the intervals are unequal (hsame)
• Lagrange’s Method will form Lagrange Interpolating
Polynomial (Continuous Polynomial of N – 1 degree that passes
through a given set of N data points).

Example 5
Using appropriate formula, to determine y corresponding to x = 10
x: 5 6 9 11
y: 12 13 14 16
Since the argument spacing is unequal (6 – 5 ≠ 9 – 6) we use Lagrange’s
interpolation formula to solve for f(10) at x = 10.

Using appropriate formula, to determine y corresponding to x = 10
x: 5 6 9 11
y: 12 13 14 16

Inverse Lagrange’s Method
• If the intervals are unequal (hsame)
• The technique of finding an estimate of a value of an independent
variable x corresponding to a given value of the dependent variable y
within the range of the observed values of y.

Example 7
Using appropriate interpolation formula, find x corresponding to y =
0.3
x: 0.4 0.6 0.8
y: 0.3638 0.3332 0.2897
[Answer: 0.7616]

Notes:
Newton divide difference and Lagrange
interpolation can use for BOTH of h equal
and h is not equal.
Newton forward and backward
interpolation only can solve for h equal.

Recall
Summary of Main Teaching
Points
 Interpolation
 Newton’s Forward Difference
 Newton’s Backward Difference
 Newton’s Divided Difference
 Lagrange’s Interpolation
 Inverse Lagrange’s Interpolation

Question and Answer Session
Q & A

What we will cover next
 Numerical Differentiation

Chapter 5 - Interpolation(newton's diff)

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