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lecture 10 to be used for the purpose of studying
Typology: Lecture notes
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Newton- Cotes
Finite difference methods
Spline interpolation
Newton’s polynomials
Errors in FD methods
Simpson’s rules
-5 1 2 3 4 5 6 7 8
0
5
10
15
20
x
y
7th order polynomial
x = [1 2 3 4 5 6 7 8]; y = [4 7 5 3 8 0 3 2 ];
Example: 7th^ order polynomial fit to 8 data points
f x a 1 x^2 a 2 xa 3 Careful – descending powers!
polyfit returns the coefficients p 1 , p 2 ,… in vector “p”. Function polyval takes
these coefficients along with a set of x values (xi) and calculates the polynomial f (x)
xi = [1:0.1:8]; yii = polyval(p,xi);
figure; plot(x,y,'ro',xi,yii); xlabel('x'); ylabel('y'); title('7th order polynomial');
p = polyfit(x,y,7);
20000 2002 2004 2006 2008 2010 2012
20
40
60
80
100
x
f(x)
Shifted fit
20000 2002 2004 2006 2008 2010 2012
20
40
60
80
100
x
f(x)
Raw fit - Large x-values
figure; plot(x,y,'ro',xi,yi); xlim([2000 2012]); ylim([0 100]); xlabel('x'); ylabel('f(x)'); title('Raw fit - Large x-values');
x = [2001 2002 2003 2004 2005 2006 2007 2009 2010 2011]; xi = x-2006; y = [18 28 23 28 26 24 50 28 36 39];
Shifting Data for Polynomial Fitting
(See note on Canvas > Resources > Supplementary Materials for an explanation of why polyfit can experience problems fitting high order polynomials to large x-values)
x = [2001 2002 2003 2004 2005 2006 2007 2009 2010 2011]; y = [18 28 23 28 26 24 50 28 36 39]; p = polyfit(x,y,9); xi = [2001:0.1:2011]; yi = polyval(p,xi); figure; plot(x,y,'ro',xi+2006,yi); xlim([2000 2012]); ylim([0 100]); xlabel('x'); ylabel('f(x)'); title('Shifted fit');
p = polyfit(xi,y,9); xi = [-5:0.1:5]; yi = polyval(p,xi);
Equivalent to interp1(x,Y,xi,’spline’)
-5 0 2 4 6 8
0
5
10
15
20
x
y
Cubic Spline
-5 0 2 4 6 8
0
5
10
15
20
x
y
7th order polynomial
Newton- Cotes
Finite difference methods
Spline interpolation
Newton’s polynomials
Errors in FD methods
Simpson’s rules
x
f (x)
a (^) b
n
h ba
1 2 3 … (^) n
h
b
a
n
b
a
I f x f x f (^) n x a 0 a 1 xa 2 x^2 anx^ n
Replace the original integrand with an nth^ order polynomial function:
x
f (x)
a (^) b Zero-order polynomial (constant)
x
f (x)
a (^) b 1st-order polynomial (linear)
x
f (x)
a (^) b 2nd-order polynomial (quadratic)
^
b
a
b
a
I f x f 1 x f 1 x a 0 a 1 x
Recall Lecture 9:
x 0 x x 1
f (x)
f (x 1 )
f 1 (x)
f (x 0 )
1 0
1 0 (^1 0) x x x x
f x f x f x f x
The 1st^ order polynomial that passes through [x 0 , f (x 0 )] and [x 1 , f (x 1 )] is:
x
f (x)
a (^) b 1st-order (linear)
b a
f x f a f b f a
1
We replace the true f ( x ) (black curve) with a 1st^ order polynomial f 1 ( x ) (red line) as an approximation for the purpose of integration
x
f (x)
a b 1st-order (linear)
b
a
b
a
I f x f 1 x
b a
f b f a I f a
b
a
(^)
b a
f b f a f x f a
1
2
I ba f a f^ b
The trapezoid rule:
f (a)
f (b)
from a = 0 to b = 0.
(^00) 0.2 0.4 0.6 0.8 1
1
2
3
4
x
y = f(x)
(^00) 0.2 0.4 0.6 0.
1
2
3
4
x
y = f(x)
1.640533 (exact)
… a true relative error of 89.5%
2
I h f a f b
(^00) 0.2 0.4 0.6 0.8 1
1
2
3
4
x
y = f(x)
Using the Trapezoid rule gives
Trapezoid rule:
3 2
Oh f a f b I h The error associated with using the Trapezoid rule is O(h^3 ):
(^00) 0.2 0.4 0.6 0.8 1
1
2
3
4
x
y = f(x) h = 0.
(^00) 0.2 0.4 0.6 0.8 1
1
2
3
4
x
y = f(x)
h = 0.
We can break down the interval from a to b into multiple smaller sections, each with smaller h, applying the Trapezoid rule to each section and summing the individual areas:
(^00) 0.2 0.4 0.6 0.8 1
1
2
3
4
x
y = f(x)^ h^ = 0.^
This is the “multiple application” or Composite Trapezoid Rule
Can we develop more efficient numerical integration algorithms with errors that reduce more quickly as the number of segments (n) increases?
^
b
a
n
b
a
I f x f x f (^) n x a 0 a 1 xa 2 x^2 anx^ n Use higher order polynomial functions to approximate f (x):
x
f (x)
a b
1st-order (linear)
Trapezoid rule
x
f (x)
a (^) b
2nd-order (quadratic)
Simpson’s 1/3 rule
x
f (x)
a (^) b
3rd-order (cubic)
Simpson’s 3/8 rule
b
a
b
a
I f x f 2 x f 2 x a 0 a 1 xa 2 x^2
^0 ^4 ^1 ^ ^2 3
f x f x f x h I 2
b a h
This can be integrated and simplified to: (^) where
“1/3 rule”
If we know the value of f (x) at x = a, b, and midway between a and b, ( data points) we can use Simpson’s 1/3 rule
x
f (x)
a = x 0 b = x 2
2 2
0 0
2 0 1 0 2 0 1
x x
x x
I (^) f x (^) b b x x b x x x x dx
x 1