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Interpolation polynomial * Means finding an approximate value to a - - PowerPoint PPT Presentation
Interpolation polynomial * Means finding an approximate value to a - - PowerPoint PPT Presentation
Interpolation polynomial * Means finding an approximate value to a function as a polynomial of degree n using n+1 points. Methods of solution 1- Quadratic Lagrange x x x x 1 2 0 f x y y y 0 1 2 P
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* Means finding an approximate value to a function as a polynomial of degree n using n+1 points.
Interpolation polynomial Methods of solution
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1- Quadratic Lagrange
2 1 1 2 2
P x L y L y L y
x
x
1
x
2
x y
1
y
2
y
1 2 1 2
x x x x L x x x x
2 1 1 1 2
x x x x L x x x x
1 2 2 2 1
x x x x L x x x x
f x
SLIDE 5
Example
x
1 2 f(x) 1.2 4.5 3 3.8
* Interpolate the following data:
1 2 1 2
x x x x L x x x x
2 3 1 2 1 3 x x
2
1 5 6 2 x x
2 1 1 1 2
x x x x L x x x x
1 3 2 1 2 3 x x
2
4 3 x x
1 2 2 2 1
x x x x L x x x x
1 2 3 1 3 2 x x
2
1 3 2 2 x x
2 1 1 2 2
P x L y L y L y
1 2
1.2 4.5 3.8 L L L
2
2 9.3 6.1 x x
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Example
x
9 9.5 f(x) 2.19 2.25 11 2.39
* Given that:
1 2 1 2
x x x x L x x x x
9.5 11 9 9.5 9 11 x x
2
20.5 104.5 x x
2 1 1 1 2
x x x x L x x x x
9 11 9.5 9 9.5 11 x x
2
4 20 99 3 x x
1 2 2 2 1
x x x x L x x x x
9 9.5 11 9 11 9.5 x x
2
1 18.5 85.5 3 x x
f x Ln x
approximate . If , evaluate the error.
2 9.2
P
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2 1 1 2 2
P x L y L y L y
1 2
2.19 2.25 2.39 L L L
2
1.58 29.11 136.2 x x
2(9.2)
2.1192 P
Error = Exact value - Approximate value
2
(9.2) 9.2 f P
Exact value (9.2) Ln
2.2192
2.2192-2.1192 0.1
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1 2 1 1 1
+
n n n
P x a a x x a x x x x a x x x x x x
x
x
1
x
2
x y
1
y
2
y
0,
a y
1 1 1
, y y a x x ( ) f x
2- Newton divided difference
1 n
x
1 n
y
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Example
x
9 9.5 f(x) 2.19 2.25 11 2.39
* Given that:
f x Ln x
approximate . If , evaluate the error.
2 9.2
P
x y 9 2.19 9.5 2.25 11 2.39 First Stage
2.25 2.19 9.5 9 0.12
Second Stage
2.39 2.25 11 9.5 0.093 0.093 0.12 11 9 0.0135
a
1
a
2
a
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2 1 2 1
P x a a x x a x x x x
2.19 0.12 9 0.0135 9 9.5 x x x
2
0.0135 +0.36975 0.04425 x x
2(9.2)
2.21481 P
Error = Exact value - Approximate value
2
(9.2) 9.2 f P
Exact value (9.2) Ln
2.2192
2.2192-2.21481 0.00439
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Example
x
1 f(x) 1 3 2 6
* Given that: approximate .
3 1.8
P
x y 1 1 3 2 6 4 8 First Stage
3 1 1 2
Second Stage
6 3 2 1 3 3 2 2 1 2
a
1
a
2
a
4 8
8 6 4 2 1 1 3 4 1 2 3
Third Stage
2 1 3 2 4 7 24
3
a
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3 1 2 1 3 1 2
+ P x a a x x a x x x x a x x x x x x
1 7 1 2 1 1 2 2 24 x x x x x x
3 2
7 1.375 0.916667 1 24 x x x
3(1.8)
5.404 P
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