Power point Presentation Numerical lecture7.ppt

1. 09/11/25
Lectures on Numerical Methods 1
Numerical Methods
Charudatt Kadolkar
Copyright 1996-98 © Dale Carnegie & Associates, Inc.

2. 09/11/25
Lectures on Numerical Methods 2
Interpolation Problem
– In general the function is known only for a few points or provided as a
table.
– Example: A log table only mentions following values.
x 1 1.5 2 3 3.5 4
Log(x) 0.0 0.17609 0.30103 0.47712 0.54407 0.60206
– The problem is to find log(2.5). We may use a straight line interpolation
between points 2 and 3.
– The straight line interpolation may be extended to a polynomial, since the
straight line uses only two datapoints, when we have several data points
available with us.
– We will find a polynomial which passes through all these points.

3. 09/11/25
Lectures on Numerical Methods 3
Polynomials
Forms of Polynomials
)
(
)
(
)
)(
(
)
(
)
(
)
(
)
(
1
2
1
2
1
1
0
2
2
1
0
2
2
1
0
n
n
n
n
n
n
c
x
c
x
a
c
x
c
x
a
c
x
a
a
P(x)
c
x
a
c
x
a
c
x
a
a
P(x)
x
a
x
a
x
a
a
P(x)



























form
s
Newton'
polynomial
a
form
power
Shifted
Polynomial
a
of
form
Power

4. 09/11/25
Lectures on Numerical Methods 4
Polynomials
• Nested Multiplication Algorithm
– Want to evaluate a polynomial P(x) given in Newton form, at x = z.
–
1. bn = an
2. For i = n – 1 to 0 step –1
3. bi = ai + ( z – ci+1 )* bi+1
4. Print the answer P(z) = b0.
–
)
(
)
(
)
)(
(
)
( 1
2
1
2
1
1
0 n
n c
x
c
x
a
c
x
c
x
a
c
x
a
a
P(x) 








 

)
(
)
(
)
)(
(
)
( 1
1
2
1
0 









 n
n c
x
z
x
b
c
x
z
x
b
z
x
b
b
P(x) 


5. 09/11/25
Lectures on Numerical Methods 5
Polynomials
• Nested Multiplication Algorithm Example
2
0
1
2
3
3
2
1
3
2
1
0
3
2
)
1
(
)
1
(
2
)
1
(
3
4
)
(
4
,
3
,
2
,
1
,
1
0
1
1
x
x
x
x
x
x
P
b
b
b
b
x
c
c
c
a
a
a
a
x
x
x
P(x)























and
get
We
for
algorithm
the
applying
After
And
Then

6. 09/11/25
Lectures on Numerical Methods 6
Polynomials
– To interpolate a function (unknown) whose values at n+1 distinct points
are known, we will use a polynomial of degree <= n.
• Uniqueness
– If two polynomials P(x) and Q(x) of degrees less than or equal to k agree
at k+1 distinct points, then P(x) and Q(x) are identical.
• Existence
– Lagranges polynomial given below is the proof of existence.
 
 
 



 



n
k
n
k
i
i
k
i
k
i
k
k
f
x
x
x
x
P(x)
x
f
f
0 0
Let

7. 09/11/25
Lectures on Numerical Methods 7
Lagrange’s Polynomial
• Algorithm
– The data is given as ( xi , fi ). To calculate p(x)
1. p = 0
2. For i = 1 to n
3. t = 1
4. For j = 0 to n
5. If j  i then
6. t = t * ( x – xi ) / ( xi – xj )
7. p = p + t * fi
8. P is the required value of p(x)

8. 09/11/25
Lectures on Numerical Methods 8
Method of Divided Differences
 
 
 
)
(
)
(
)
)(
(
)
(
.
.
,...,
1
,
0
,
1
0
,
0
1
0
2
,
0
0
1
,
0
0
,
0
,
1
,
,
1
,



















n
n
i
i
i
i
k
k
i
k
i
k
i
i
i
x
x
x
x
f
x
x
x
x
f
x
x
f
f
P(x)
f
f
i
k
x
x
f
f
f
n
i
f
x


is
then
polynomal,
ing
interpolat
The
and
Define
is
set
data
The

9. 09/11/25
Lectures on Numerical Methods 9
Method of Divided Differences
• Table
3
,
3
3
3
,
2
2
,
2
2
3
,
1
2
,
1
1
,
1
1
3
,
0
2
,
0
1
,
0
0
,
0
0
3
2
1
0
f
x
f
f
x
f
f
f
x
f
f
f
f
x
x

10. 09/11/25
Lectures on Numerical Methods 10
Method of Divided Differences
• Algorithm
– To create the table of divided differences.
1. For i = 1 to n
2. di,0 = fi
3. For i = 1 to n
4. For j = 1 to n-i
5. dj,i = (dj+1,i-1 – dj,i-1 ) / (xj+i – xj)
6. D would be required table