Chapter 1 Numerical approximation of data : interpolation, least squares method
|
|
- Karen Whitehead
- 5 years ago
- Views:
Transcription
1 Chapter 1 Numerical approximation of data : interpolation, least squares method
2 I. Motivation 1 Approximation of functions
3 Evaluation of a function Which functions (f : R R) can be effectively evaluated in any point?
4 Evaluation of a function Which functions (f : R R) can be effectively evaluated in any point? the power functions : f(x) = x m, m N the polynomial functions : f(x) = a 0 + a 1 x + a 2 x a m x m
5 Evaluation of a function Which functions (f : R R) can be effectively evaluated in any point? the power functions : f(x) = x m, m N the polynomial functions : f(x) = a 0 + a 1 x + a 2 x a m x m How can we evaluate other functions in a given point? for instance : f(x) = cos(x), f(x) = sin(x) exp(x),...
6 Evaluation of a function Which functions (f : R R) can be effectively evaluated in any point? the power functions : f(x) = x m, m N the polynomial functions : f(x) = a 0 + a 1 x + a 2 x a m x m How can we evaluate other functions in a given point? for instance : f(x) = cos(x), f(x) = sin(x) exp(x),... approximation by a polynomial function : using a Taylor series about the given point, searching a polynomial having the same values as the function in some close points Lagrange interpolation
7 Principles of Lagrange interpolation f(x) = sin( πx 2 )(x2 + 3)
8 Principles of Lagrange interpolation f(x) = sin( πx 2 )(x2 + 3) 4 points on the curve : ( 1, 4), (1, 4), (2, 0), (3, 12)
9 Principles of Lagrange interpolation f(x) = sin( πx 2 )(x2 + 3) Lagrange interpolating polynomial 4 points on the curve : P polynomial of degree 3 satisfying ( 1, 4), P ( 1) = 4 (1, 4), P (1) = 4 (2, 0), P (2) = 0 (3, 12) P (3) = 12
10 Principles of Lagrange interpolation, with 6 points f(x) = sin( πx 2 )(x2 + 3) Lagrange interpolating polynomial 6 points on the curve : P polynomial of degree 5 satisfying (x i, y i ) 0 i 5, P (x i ) = y i for 0 i 5 Remark : ouside the interval defined by the (x i ), the Lagrange interpolating polynomial has nothing to do with f.
11 I. Motivation 1 Approximation of functions 2 Curve approximation
12 Piecewise interpolation f(x) = sin( πx 2 )(x2 + 3)
13 Piecewise interpolation f(x) = sin( πx 2 )(x2 + 3) piecewise affine approximation
14 Piecewise interpolation f(x) = sin( πx 2 )(x2 + 3) piecewise affine approximation
15 Piecewise interpolation f(x) = sin( πx 2 )(x2 + 3) piecewise affine approximation Applications : Calculation of an approximate value of the length of the curve the area under the curve (here see chapter f(x)dx)
16 Cubic spline f(x) = sin( πx 2 )(x2 + 3) s cubic spline Principle : between two consecutive points, s is a cubic polynomial s(x i ) = f(x i ) s C 2 + two conditions on the boundary points
17 Cubic spline f(x) = sin( πx 2 )(x2 + 3) s cubic spline Principle : between two consecutive points, s is a cubic polynomial s(x i ) = f(x i ) s C 2 + two conditions on the boundary points
18 Bézier curves : principle P 0, P 1,, P n, are n + 1 given control points
19 Bézier curves : principle P 0, P 1,, P n, are n + 1 given control points The corresponding Bézier curve is defined by n M(t) = Bn(t)P i i, 0 t 1 i=0 where B i n are Bernstein polynomial defined by B i n = C i nx i (1 X) n i, with C i n = n! i!(n i)!.
20 Bézier curves : with more points
21 Application of the Bézier curves
22 I. Motivation 1 Approximation of functions 2 Curve approximation 3 Fitting of statistical data
23 Study of statistical data Some experimental measurements X : noise level in the factory (in db), Y : time used to do a definite work (in minutes) X Y A physical measure always contains some noise. Can we find a law linking Y and X (Y = f(x))? Can we predict the value of Y for X = 66dB?
24 Linear regression Principle We are looking for a and b such that d(a, b) = (y i (ax i + b)) 2 is minimal. The straight line y = ax + b is the linear regression line. The polynomial P = ax + b is the least squares fitting polynomial of the cloud of points.
25 II. Lagrange interpolating polynomial : theoretical study 1 Study of an example
26 Lagrange interpolation in 1 point f(x) = sin( πx 2 )(x2 + 3) 1 point on the curve : Search for P 0, such that (x 0, y 0 ) = (1, 4) deg P 0 0 and P 0 (x 0 ) = y 0
27 Lagrange interpolation in 1 point f(x) = sin( πx 2 )(x2 + 3) 1 point on the curve : Search for P 0, such that (x 0, y 0 ) = (1, 4) deg P 0 0 and P 0 (x 0 ) = y 0
28 Lagrange interpolation in 1 point f(x) = sin( πx 2 )(x2 + 3) 1 point on the curve : Search for P 0, such that (x 0, y 0 ) = (1, 4) deg P 0 0 and P 0 (x 0 ) = y 0 P 0 = 4
29 Lagrange interpolation in 2 points f(x) = sin( πx 2 )(x2 + 3) 2 points on the curve : Search for P 1, such that (x 0, y 0 ) = (1, 4) deg P 1 1 and (x 1, y 1 ) = ( 1, 4) P 1 (x 0 ) = y 0 and P 1 (x 1 ) = y 1
30 Lagrange interpolation in 2 points f(x) = sin( πx 2 )(x2 + 3) 2 points on the curve : Search for P 1, such that (x 0, y 0 ) = (1, 4) deg P 1 1 and (x 1, y 1 ) = ( 1, 4) P 1 (x 0 ) = y 0 and P 1 (x 1 ) = y 1
31 Lagrange interpolation in 2 points f(x) = sin( πx 2 )(x2 + 3) 2 points on the curve : Search for P 1, such that (x 0, y 0 ) = (1, 4) deg P 1 1 and (x 1, y 1 ) = ( 1, 4) P 1 (x 0 ) = y 0 and P 1 (x 1 ) = y 1 P 1 = 4X
32 Lagrange interpolation in 3 points f(x) = sin( πx 2 )(x2 + 3) 3 points on the curve : Search for P 2, such that (x 0, y 0 ) = (1, 4) deg P 2 2 and (x 1, y 1 ) = ( 1, 4) P 2 (x 0 ) = y 0, P 2 (x 1 ) = y 1 (x 2, y 2 ) = (3, 12) and P 2 (x 2 ) = y 2
33 Lagrange interpolation in 3 points f(x) = sin( πx 2 )(x2 + 3) 3 points on the curve : Search for P 2, such that (x 0, y 0 ) = (1, 4) deg P 2 2 and (x 1, y 1 ) = ( 1, 4) P 2 (x 0 ) = y 0, P 2 (x 1 ) = y 1 (x 2, y 2 ) = (3, 12) and P 2 (x 2 ) = y 2
34 Lagrange interpolation in 3 points Idea Search for L 0, such that deg L 0 = 2 and L 0 (x 1 ) = L 0 (x 2 ) = 0 and L 0 (x 0 ) = 1. Search for L 1, such that deg L 1 = 2 and L 1 (x 0 ) = L 1 (x 2 ) = 0 and L 1 (x 1 ) = 1. Search for L 2, such that deg L 0 = 2 and L 2 (x 0 ) = L 2 (x 1 ) = 0 and L 2 (x 2 ) = 1. Prove that P 2 = y 0 L 0 + y 1 L 1 + y 2 L 2 is a solution of the pb. Give the expression of P.
35 Lagrange interpolation in 3 points Idea Search for L 0, such that deg L 0 = 2 and L 0 (x 1 ) = L 0 (x 2 ) = 0 and L 0 (x 0 ) = 1. Search for L 1, such that deg L 1 = 2 and L 1 (x 0 ) = L 1 (x 2 ) = 0 and L 1 (x 1 ) = 1. Search for L 2, such that deg L 0 = 2 and L 2 (x 0 ) = L 2 (x 1 ) = 0 and L 2 (x 2 ) = 1. Prove that P 2 = y 0 L 0 + y 1 L 1 + y 2 L 2 is a solution of the pb. Give the expression of P. P 2 = 3X 2 + 4X + 3
36 Lagrange interpolation in 3 points Other idea P 1 = 4X satisfies P 1 (1) = 4, P 1 ( 1) = 4 and deg P 1 = 1. Q = (X + 1)(X 1) satisfies Q(1) = Q( 1) = 0 and deg Q = 2 (in fact Q = L 2 ) Search for P 2 under the form P 2 = P 1 + αq = 4X + α(x + 1)(X 1).
37 Lagrange interpolation in 3 points Other idea P 1 = 4X satisfies P 1 (1) = 4, P 1 ( 1) = 4 and deg P 1 = 1. Q = (X + 1)(X 1) satisfies Q(1) = Q( 1) = 0 and deg Q = 2 (in fact Q = L 2 ) Search for P 2 under the form P 2 = P 1 + αq = 4X + α(x + 1)(X 1). P 2 (3) = 12 α = 3 and P 2 = 4X 3(X + 1)(X 1) = 3X 2 + 4X + 3
38 II. Lagrange interpolating polynomial : theoretical study 1 Study of an example 2 Existence and uniqueness of the Lagrange interpolating polynomial
39 The mathematical problem Formulation Let (n + 1) points be given : (x i, y i ) 0 i n with (x i, y i ) R 2 for all i, x i x j for all i j. Is it possible to find a polynomial P with real coefficients satisfying P (x i ) = y i 0 i n? Degree of P? number of equations : n + 1 number of unknowns (coefficients (a i )) less that n + 1 deg P n
40 The Lagrange basis For 0 j n, let us define It satisfies : L j = n i=0,i j X x i x j x i. L j (x j ) = 1 and L j (x i ) = 0 for all i j L j (x i ) = δ i,j. and deg L j = n 0 j n.
41 Solution of the problem + uniqueness Existence of a solution n Let P = y j L j, we have j=0 deg P n, n P (x i ) = y j L j (x i ) = y i for all 0 i n. j=0 P is a solution of the Lagrange interpolation problem. Uniqueness Can we find another solution to the problem, Q? If Q exists, deg Q n and deg(p Q) n, P (x i ) Q(x i ) = (P Q)(x i ) = 0 for 0 i n. P Q = 0 and the solution is unique.
42 Main theorem Theorem Hypotheses : Let us consider (n + 1) points of R 2 : (x i, y i ) 0 i n, x i x j for all i j Then, there exists a unique polynomial P R n [X] satisfying P (x i ) = y i 0 i n. P is the Lagrange interpolating polynomial that passes through the (n + 1) points (x i, y i ) 0 i n. In the case where y i = f(x i ) for all 0 i n (with f a given function), P is the Lagrange interpolating polynomial of f in the points (x i ) 0 i n.
43 II. Lagrange interpolating polynomial : theoretical study 1 Study of an example 2 Existence and uniqueness of the Lagrange interpolating polynomial 3 Interpolation error result
44 Presentation of the problem Comparison of f and P (4 points) E(x) = f(x) P (x)
45 Presentation of the problem Comparison of f and P (4 points) E(x) = f(x) P (x) with a zoom around the points
46 Presentation of the problem Comparison of f and P (6 points) E(x) = f(x) P (x)
47 Presentation of the problem Comparison of f and P (6 points) E(x) = f(x) P (x) with a zoom around the points What can be said about E(x)? Can it be bounded?...
48 Interpolation error Theorem Hypotheses : f : [a, b] R, f C n+1 ([a, b]), (x i ) 0 i n, n + 1 distinct real numbers of [a, b]. P n : Lagrange interpolating polynomial of f in the points (x i ) 0 i n. Then, for all x [a, b], there exists ξ x [a, b] such that with Π n = f(x) P n (x) = n (X x i ). i=0 1 (n + 1)! Π n(x)f (n+1) (ξ x ),
49 Consequence As a consequence, we get : with x [a, b] f(x) P n (x) M n+1 = max ξ [a,b] f (n+1) (ξ). 1 (n + 1)! M n+1 Π n (x), It does not imply the convergence of P n (x) towards f(x). It is not necessary interesting to increase n.
50 III. Lagrange interpolating polynomial : practical computation 1 Cost of the computation of the interpolating polynomial
51 With the Lagrange basis L j (x) = n (x x i ) i=0,i j n i=0,i j (x j x i ) Cost of the computation :, ( ) 2(n 1) mult. + 1 div.
52 With the Lagrange basis L j (x) = n i=0,i j n i=0,i j (x x i ) (x j x i ), for j = 0 to n Cost of the computation : (n + 1) ( ) 2(n 1) mult. + 1 div.
53 With the Lagrange basis L j (x) = n i=0,i j n i=0,i j (x x i ) (x j x i ), for j = 0 to n ( ) Cost of the computation : (n + 1) 2(n 1) mult. + 1 div. n P (x) = y j L j (x) final cost 2n 2. j=0
54 With the Lagrange basis L j (x) = n i=0,i j n i=0,i j (x x i ) (x j x i ), for j = 0 to n ( ) Cost of the computation : (n + 1) 2(n 1) mult. + 1 div. n P (x) = y j L j (x) final cost 2n 2. j=0 Other main drawback of the method : what happens if we finally want to add one more point (x n+1, y n+1 )? All must be started again from zero.
55 With an other basis Idea Write the polynomial in the basis n 1 1, (X x 0 ), (X x 0 )(X x 1 ),, (X x j ). P j=0 n 1 = α 0 + α 1 (X x 0 ) + + α n (X x j ). j=0
56 With an other basis Idea Write the polynomial in the basis n 1 1, (X x 0 ), (X x 0 )(X x 1 ),, (X x j ). j=0 n 1 P n = α 0 + α 1 (X x 0 ) + + α n (X x j ). j=0
57 With an other basis Idea Write the polynomial in the basis n 1 1, (X x 0 ), (X x 0 )(X x 1 ),, (X x j ). j=0 n 1 P n = α 0 + α 1 (X x 0 ) + + α n (X x j ). j=0 Now, if we add one point (x n+1, y n+1 ), we have : P n+1 = P n + α n+1 n j=0 (X x j ) we just need to calculate α n+1 to get P n+1.
58 Cost of the computation of P n (x) n 1 P n (x) = α 0 + α 1 (x x 0 ) + + α n (x x j ) ( j=0 ( = α 0 + (x x 0 ) α 1 + (x x 1 ) α 2 + ( ))). Hörner s algorithm for the computation of p = P n (x) Cost p α n for k from n 1 to 0 end p α k + (x x k )p n additions + n multiplications
59 Cost of the computation of P n (x) n 1 P n (x) = α 0 + α 1 (x x 0 ) + + α n (x x j ) ( j=0 ( = α 0 + (x x 0 ) α 1 + (x x 1 ) α 2 + ( ))). Hörner s algorithm for the computation of p = P n (x) Cost p α n for k from n 1 to 0 end p α k + (x x k )p n additions + n multiplications and the computation of the coefficients α i?
60 III. Lagrange interpolating polynomial : practical computation 1 Cost of the computation of the interpolating polynomial 2 The divided difference method
61 Calculation of the first α i n = 0, 1 first point (x 0, y 0 ), P 0 = α 0 : P 0 (x 0 ) = y 0 = α 0 = y 0 n = 1, + (x 1, y 1 ), P 1 = y 0 + α 1 (X x 0 ) : n = 2, + (x 2, y 2 ), P 1 (x 1 ) = y 1 = α 1 = y 1 y 0 x 1 x 0 P 2 = y 0 + y 1 y 0 x 1 x 0 (X x 0 ) + α 2 (X x 0 )(X x 1 ) P 2 (x 2 ) = y 2 = α 2 = y 2 y 1 x 2 x 1 y 1 y 0 x 1 x 0 x 2 x 0
62 Recurrence formula Assume we have : Interpolating points : (x 0, y 0 ) (x n 1, y n 1 ), n 1 P n 1 = α 0 + j=1 j 1 α j k=0 Interpolating points : (x 1, y 1 ) (x n, y n ), (X x k ), (α j ) 0 j n 1 known Then, n 1 Q n 1 = β 0 + j=1 β j j (X x k ), (β j ) 0 j n 1 known k=1 X x 0 x n x 0 Q n 1 + x n X x n x 0 P n 1
63 Recurrence formula Assume we have : Interpolating points : (x 0, y 0 ) (x n 1, y n 1 ), n 1 P n 1 = α 0 + j=1 j 1 α j k=0 Interpolating points : (x 1, y 1 ) (x n, y n ), (X x k ), (α j ) 0 j n 1 known Then, n 1 Q n 1 = β 0 + j=1 β j j (X x k ), (β j ) 0 j n 1 known k=1 X x 0 x n x 0 Q n 1 + x n X x n x 0 P n 1 = P n
64 Recurrence formula Assume we have : Interpolating points : (x 0, y 0 ) (x n 1, y n 1 ), n 1 P n 1 = α 0 + j=1 j 1 α j k=0 Interpolating points : (x 1, y 1 ) (x n, y n ), (X x k ), (α j ) 0 j n 1 known Then, and n 1 Q n 1 = β 0 + α n = j=1 β j j (X x k ), (β j ) 0 j n 1 known k=1 X x 0 x n x 0 Q n 1 + x n X x n x 0 P n 1 = P n 1 1 β n 1 α n 1 = β n 1 α n 1. x n x 0 x n x 0 x n x 0
65 The divided differences x 0 f(x 0 ) x 1 f(x 1 ) x 2 f(x 2 ).. x n 1 f(x n 1 ) x n f(x n )
66 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] x 2 f[x 2 ].. x n 1 f[x n 1 ] x n f[x n ]
67 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] x 2 f[x 2 ] f[x 1 ] f[x 0 ] x 1 x 0 f[x 2 ] f[x 1 ] x 2 x 1... x n 1 f[x n 1 ] x n f[x n ] f[x n 1 ] f[x n 2 ] x n 1 x n 2 f[x n ] f[x n 1 ] x n x n 1
68 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ]... x n 1 f[x n 1 ] f[x n 2, x n 1 ] x n f[x n ] f[x n 1, x n ]
69 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ] f[x 1, x 2 ] f[x 0, x 1 ] x 2 x x n 1 f[x n 1 ] f[x n 2, x n 1 ] x n f[x n ] f[x n 1, x n ] f[x n 1, x n ] f[x n 2, x n 1 ] x n x n 2 f[x n 1, x n ] f[x n 2, x n 1 ] x n x n 2
70 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ] f[x 0, x 1, x 2 ].... x n 1 f[x n 1 ] f[x n 2, x n 1 ] f[x n 3, x n 2, x n 1 ] x n f[x n ] f[x n 1, x n ] f[x n 2, x n 1, x n ]
71 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ] f[x 0, x 1, x 2 ] x n 1 f[x n 1 ] f[x n 2, x n 1 ] f[x n 3, x n 2, x n 1 ]... x n f[x n ] f[x n 1, x n ] f[x n 2, x n 1, x n ] f[x 0,, x n ] with f[x 0,, x n ] = f[x 1,, x n ] f[x 0,, x n 1 ] x n x 0.
72 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ] f[x 0, x 1, x 2 ] x n 1 f[x n 1 ] f[x n 2, x n 1 ] f[x n 3, x n 2, x n 1 ]... x n f[x n ] f[x n 1, x n ] f[x n 2, x n 1, x n ] f[x 0,, x n ] P n = f[x 0 ]+f[x 0, x 1 ](X x 0 ) +f[x 0, x 1, x 2 ](X x 0 )(X x 1 ) n f[x 0,, x n ] (X x j ). j=0
73 The divided differences x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ] f[x 0, x 1, x 2 ] x n 1 f[x n 1 ] f[x n 2, x n 1 ] f[x n 3, x n 2, x n 1 ]... x n f[x n ] f[x n 1, x n ] f[x n 2, x n 1, x n ] f[x 0,, x n ] Cost of the computation n2 2 div. and n2 sub.
74 Result Theorem The Lagrange interpolating polynomial of f in the points (x i ) 0 i n reads n j 1 P n = f[x 0 ] + f[x 0,, x j ] (X x k ), j=1 k=0 where f[ ] denotes the divided difference of f defined by induction f[x i ] = f(x i ) for 0 i n f[x i,, x i+k ] = f[x i+1,, x i+k ] f[x i,, x i+k 1 ] x i+k x i for 0 i n k, 1 k n.
75 IV. A few words about Hermite interpolation 1 Presentation of the problem
76 One example f(x) = sin( πx 2 )(x2 + 3) Consider two points : (x 0, y 0 ) = ( 1, 4) (x 1, y 1 ) = (3, 12)
77 One example f(x) = sin( πx 2 )(x2 + 3) Consider two points : (x 0, y 0 ) = ( 1, 4) (x 1, y 1 ) = (3, 12) Search P such that P (x 0 ) = f(x 0 ), P (x 1 ) = f(x 1 )
78 One example f(x) = sin( πx 2 )(x2 + 3) Consider two points : (x 0, y 0 ) = ( 1, 4) (x 1, y 1 ) = (3, 12) Search P such that P (x 0 ) = f(x 0 ), P (x 1 ) = f(x 1 ) P = 4 2(X + 1) = 6 2X
79 One example f(x) = sin( πx 2 )(x2 + 3) Consider two points : (x 0, y 0 ) = ( 1, 4) (x 1, y 1 ) = (3, 12) Search Q such that Q(x 0 ) = f(x 0 ), Q(x 1 ) = f(x 1 ) Q (x 0 ) = f (x 0 ), Q (x 1 ) = f (x 1 ) P = 4 2(X + 1) = 6 2X
80 One example f(x) = sin( πx 2 )(x2 + 3) Consider two points : (x 0, y 0 ) = ( 1, 4) (x 1, y 1 ) = (3, 12) Search Q such that Q(x 0 ) = f(x 0 ), Q(x 1 ) = f(x 1 ) Q (x 0 ) = f (x 0 ), Q (x 1 ) = f (x 1 ) P = 4 2(X + 1) = 6 2X Q = P + (X + 1)(X 3)(αX + β)
81 One example f(x) = sin( πx 2 )(x2 + 3) Consider two points : (x 0, y 0 ) = ( 1, 4) (x 1, y 1 ) = (3, 12) Search Q such that Q(x 0 ) = f(x 0 ), Q(x 1 ) = f(x 1 ) Q (x 0 ) = f (x 0 ), Q (x 1 ) = f (x 1 ) P = 4 2(X + 1) = 6 2X Q = P + (X + 1)(X 3)(αX + β) = P + (X + 1)(X 3)( 1) = X 2 3
82 With 2 other points f(x) = sin( πx 2 )(x2 + 3) P = 0 Q = π 2 X3 + π 4 X2 + 3π 2 X
83 The mathematical problem Generalities The Hermite interpolation takes into account the values of the function in some points (x i ) 0 i k, the values of the successive derivatives of the function until order α i in x i. Formulation f is a sufficiently smooth function defined on [a, b], x 0,..., x k are (k + 1) given points of [a, b], α 0,..., α k are (k + 1) integers. Is it possible to find P satisfying 0 i k, P (j) (x i ) = f (j) (x i ), 0 j α i?
84 IV. A few words about Hermite interpolation 1 Presentation of the problem 2 Main results
85 Analysis of the problem Degree of P Number of equations : Degree : n = k + Definition k (α i + 1) = k i=0 k α i. i=0 k α i. i=0 P is the Hermite interpolating polynomial of f in the points (x i ) 0 i k with the orders (α i ) 0 i k
86 Theorem Theorem : existence and uniqueness + interpolation error Hypotheses : (x i ) 0 i k, (k + 1) points in [a, b], (α i ) 0 i k, (k + 1) integers, f : [a, b] R, f C n+1 ([a, b]), n = k + Then, there exists a unique polynomial P n R n [X] such that k i=0 α i 0 i k, P (j) n (x i ) = f (j) (x i ), 0 j α i. Furthermore, for all x [a, b], there exists ξ x [a, b] such that with Ω n = f(x) P n (x) = k (X x i ) αi+1. i=0 1 (n + 1)! Ω n(x)f (n+1) (ξ x ),
87 V. Least squares method 1 The case of linear regression
88 Linear regression We are looking for a 0 and b 0 such that the following distance is minimal : d(a, b) = n (y i (ax i + b)) 2. i=1 Necessary condition d a (a 0, b 0 ) = 0 and d b (a 0, b 0 ) = 0 n n n a 0 x 2 i + b 0 x i = x i y i i=1 i=1 i=1 n n a 0 x i + b 0 n = y i i=1 i=1
89 Existence of a unique candidate (a 0, b 0 ) Matrix of the linear system n n x 2 i A = i=1 i=1 n n Invertibility det A = n 2 ( 1 n i=1 x i x i = n n x 2 i X 2 ) = n i=1 1 n n i=1 x 2 i X X 1 n (x i X) 2 = n 2 V(X). Conclusion As soon as two x i are different, det A 0 and there exists a unique (a 0, b 0 ) susceptible to be a minimum of d : a 0 = Cov(X, Y ) V(X) i=1 and b 0 = Ȳ a 0 X.
90 (a 0, b 0 ) is a minimizer of d After some computations, we prove that d(a, b) d(a 0, b 0 ) = n ((a 0 a)x i + b 0 b) 2 i=1 It yields (a, b) R 2 d(a, b) d(a 0, b 0 ). Remark on the matrix A n n x 2 i x i A = i=1 i=1 n n i=1 x i = BT B with B = x 1 1 x x n 1
91 V. Least squares method 1 The case of linear regression 2 Generalization
92 Presentation of the problem Given points The cloud of points is still given by (x i ) 1 i n and (y i ) 1 i n. A space of functions For some independent functions (ϕ 1,, ϕ m ), let us define U = {ϕ; ϕ = m u i ϕ i } i=1 Search for a minimizer We are looking for ϕ U such that n i=0 y i ϕ (x i ) 2 = min ϕ U n y i ϕ(x i ) 2. i=0
93 Main result Theorem As soon as two x i are different, the least squares problem admits a unique solution m ϕ = u i ϕ i. i=1 Futhermore, the vector u = (u 1,, u m) is the unique solution of the linear system B T Bu = B T y, with B = ϕ 1 (x 1 ) ϕ m (x 1 ).. ϕ 1 (x n ) ϕ m (x n ) Remark In the linear case, m = 2 and ϕ 1 (x) = x, ϕ 2 (x) = 1.
3.1 Interpolation and the Lagrange Polynomial
MATH 4073 Chapter 3 Interpolation and Polynomial Approximation Fall 2003 1 Consider a sample x x 0 x 1 x n y y 0 y 1 y n. Can we get a function out of discrete data above that gives a reasonable estimate
More informationINTERPOLATION. and y i = cos x i, i = 0, 1, 2 This gives us the three points. Now find a quadratic polynomial. p(x) = a 0 + a 1 x + a 2 x 2.
INTERPOLATION Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). As an example, consider defining and x 0 = 0, x 1 = π/4, x
More informationInterpolation Theory
Numerical Analysis Massoud Malek Interpolation Theory The concept of interpolation is to select a function P (x) from a given class of functions in such a way that the graph of y P (x) passes through the
More informationWe consider the problem of finding a polynomial that interpolates a given set of values:
Chapter 5 Interpolation 5. Polynomial Interpolation We consider the problem of finding a polynomial that interpolates a given set of values: x x 0 x... x n y y 0 y... y n where the x i are all distinct.
More informationCubic Splines MATH 375. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Cubic Splines
Cubic Splines MATH 375 J. Robert Buchanan Department of Mathematics Fall 2006 Introduction Given data {(x 0, f(x 0 )), (x 1, f(x 1 )),...,(x n, f(x n ))} which we wish to interpolate using a polynomial...
More informationLecture 10 Polynomial interpolation
Lecture 10 Polynomial interpolation Weinan E 1,2 and Tiejun Li 2 1 Department of Mathematics, Princeton University, weinan@princeton.edu 2 School of Mathematical Sciences, Peking University, tieli@pku.edu.cn
More informationLecture Note 3: Interpolation and Polynomial Approximation. Xiaoqun Zhang Shanghai Jiao Tong University
Lecture Note 3: Interpolation and Polynomial Approximation Xiaoqun Zhang Shanghai Jiao Tong University Last updated: October 10, 2015 2 Contents 1.1 Introduction................................ 3 1.1.1
More informationScientific Computing
2301678 Scientific Computing Chapter 2 Interpolation and Approximation Paisan Nakmahachalasint Paisan.N@chula.ac.th Chapter 2 Interpolation and Approximation p. 1/66 Contents 1. Polynomial interpolation
More informationExam 2. Average: 85.6 Median: 87.0 Maximum: Minimum: 55.0 Standard Deviation: Numerical Methods Fall 2011 Lecture 20
Exam 2 Average: 85.6 Median: 87.0 Maximum: 100.0 Minimum: 55.0 Standard Deviation: 10.42 Fall 2011 1 Today s class Multiple Variable Linear Regression Polynomial Interpolation Lagrange Interpolation Newton
More informationChapter 4: Interpolation and Approximation. October 28, 2005
Chapter 4: Interpolation and Approximation October 28, 2005 Outline 1 2.4 Linear Interpolation 2 4.1 Lagrange Interpolation 3 4.2 Newton Interpolation and Divided Differences 4 4.3 Interpolation Error
More informationInterpolation. Escuela de Ingeniería Informática de Oviedo. (Dpto. de Matemáticas-UniOvi) Numerical Computation Interpolation 1 / 34
Interpolation Escuela de Ingeniería Informática de Oviedo (Dpto. de Matemáticas-UniOvi) Numerical Computation Interpolation 1 / 34 Outline 1 Introduction 2 Lagrange interpolation 3 Piecewise polynomial
More informationLecture Note 3: Polynomial Interpolation. Xiaoqun Zhang Shanghai Jiao Tong University
Lecture Note 3: Polynomial Interpolation Xiaoqun Zhang Shanghai Jiao Tong University Last updated: October 24, 2013 1.1 Introduction We first look at some examples. Lookup table for f(x) = 2 π x 0 e x2
More informationApplied Numerical Analysis Quiz #2
Applied Numerical Analysis Quiz #2 Modules 3 and 4 Name: Student number: DO NOT OPEN UNTIL ASKED Instructions: Make sure you have a machine-readable answer form. Write your name and student number in the
More informationComputational Physics
Interpolation, Extrapolation & Polynomial Approximation Lectures based on course notes by Pablo Laguna and Kostas Kokkotas revamped by Deirdre Shoemaker Spring 2014 Introduction In many cases, a function
More informationJim Lambers MAT 460/560 Fall Semester Practice Final Exam
Jim Lambers MAT 460/560 Fall Semester 2009-10 Practice Final Exam 1. Let f(x) = sin 2x + cos 2x. (a) Write down the 2nd Taylor polynomial P 2 (x) of f(x) centered around x 0 = 0. (b) Write down the corresponding
More informationGeometric Lagrange Interpolation by Planar Cubic Pythagorean-hodograph Curves
Geometric Lagrange Interpolation by Planar Cubic Pythagorean-hodograph Curves Gašper Jaklič a,c, Jernej Kozak a,b, Marjeta Krajnc b, Vito Vitrih c, Emil Žagar a,b, a FMF, University of Ljubljana, Jadranska
More informationCubic Splines; Bézier Curves
Cubic Splines; Bézier Curves 1 Cubic Splines piecewise approximation with cubic polynomials conditions on the coefficients of the splines 2 Bézier Curves computer-aided design and manufacturing MCS 471
More informationArsène Pérard-Gayot (Slides by Piotr Danilewski)
Computer Graphics - Splines - Arsène Pérard-Gayot (Slides by Piotr Danilewski) CURVES Curves Explicit y = f x f: R R γ = x, f x y = 1 x 2 Implicit F x, y = 0 F: R 2 R γ = x, y : F x, y = 0 x 2 + y 2 =
More informationInterpolation. Chapter Interpolation. 7.2 Existence, Uniqueness and conditioning
76 Chapter 7 Interpolation 7.1 Interpolation Definition 7.1.1. Interpolation of a given function f defined on an interval [a,b] by a polynomial p: Given a set of specified points {(t i,y i } n with {t
More informationM2R IVR, October 12th Mathematical tools 1 - Session 2
Mathematical tools 1 Session 2 Franck HÉTROY M2R IVR, October 12th 2006 First session reminder Basic definitions Motivation: interpolate or approximate an ordered list of 2D points P i n Definition: spline
More informationLectures 9-10: Polynomial and piecewise polynomial interpolation
Lectures 9-1: Polynomial and piecewise polynomial interpolation Let f be a function, which is only known at the nodes x 1, x,, x n, ie, all we know about the function f are its values y j = f(x j ), j
More information8.7 MacLaurin Polynomials
8.7 maclaurin polynomials 67 8.7 MacLaurin Polynomials In this chapter you have learned to find antiderivatives of a wide variety of elementary functions, but many more such functions fail to have an antiderivative
More informationInterpolation and extrapolation
Interpolation and extrapolation Alexander Khanov PHYS6260: Experimental Methods is HEP Oklahoma State University October 30, 207 Interpolation/extrapolation vs fitting Formulation of the problem: there
More informationSome notes on Chapter 8: Polynomial and Piecewise-polynomial Interpolation
Some notes on Chapter 8: Polynomial and Piecewise-polynomial Interpolation See your notes. 1. Lagrange Interpolation (8.2) 1 2. Newton Interpolation (8.3) different form of the same polynomial as Lagrange
More informationFinite Elements. Colin Cotter. January 18, Colin Cotter FEM
Finite Elements January 18, 2019 The finite element Given a triangulation T of a domain Ω, finite element spaces are defined according to 1. the form the functions take (usually polynomial) when restricted
More informationPolynomial Review Problems
Polynomial Review Problems 1. Find polynomial function formulas that could fit each of these graphs. Remember that you will need to determine the value of the leading coefficient. The point (0,-3) is on
More informationIntroduction to Computer Graphics. Modeling (1) April 13, 2017 Kenshi Takayama
Introduction to Computer Graphics Modeling (1) April 13, 2017 Kenshi Takayama Parametric curves X & Y coordinates defined by parameter t ( time) Example: Cycloid x t = t sin t y t = 1 cos t Tangent (aka.
More informationQ 0 x if x 0 x x 1. S 1 x if x 1 x x 2. i 0,1,...,n 1, and L x L n 1 x if x n 1 x x n
. - Piecewise Linear-Quadratic Interpolation Piecewise-polynomial Approximation: Problem: Givenn pairs of data points x i, y i, i,,...,n, find a piecewise-polynomial Sx S x if x x x Sx S x if x x x 2 :
More informationLagrange Interpolation and Neville s Algorithm. Ron Goldman Department of Computer Science Rice University
Lagrange Interpolation and Neville s Algorithm Ron Goldman Department of Computer Science Rice University Tension between Mathematics and Engineering 1. How do Mathematicians actually represent curves
More informationChapter 3 Interpolation and Polynomial Approximation
Chapter 3 Interpolation and Polynomial Approximation Per-Olof Persson persson@berkeley.edu Department of Mathematics University of California, Berkeley Math 128A Numerical Analysis Polynomial Interpolation
More informationPolynomials. p n (x) = a n x n + a n 1 x n 1 + a 1 x + a 0, where
Polynomials Polynomials Evaluation of polynomials involve only arithmetic operations, which can be done on today s digital computers. We consider polynomials with real coefficients and real variable. p
More informationTwo hours. To be provided by Examinations Office: Mathematical Formula Tables. THE UNIVERSITY OF MANCHESTER. 29 May :45 11:45
Two hours MATH20602 To be provided by Examinations Office: Mathematical Formula Tables. THE UNIVERSITY OF MANCHESTER NUMERICAL ANALYSIS 1 29 May 2015 9:45 11:45 Answer THREE of the FOUR questions. If more
More informationChapter 8: Taylor s theorem and L Hospital s rule
Chapter 8: Taylor s theorem and L Hospital s rule Theorem: [Inverse Mapping Theorem] Suppose that a < b and f : [a, b] R. Given that f (x) > 0 for all x (a, b) then f 1 is differentiable on (f(a), f(b))
More information(0, 0), (1, ), (2, ), (3, ), (4, ), (5, ), (6, ).
1 Interpolation: The method of constructing new data points within the range of a finite set of known data points That is if (x i, y i ), i = 1, N are known, with y i the dependent variable and x i [x
More informationNumerical Analysis: Interpolation Part 1
Numerical Analysis: Interpolation Part 1 Computer Science, Ben-Gurion University (slides based mostly on Prof. Ben-Shahar s notes) 2018/2019, Fall Semester BGU CS Interpolation (ver. 1.00) AY 2018/2019,
More informationOutline. 1 Interpolation. 2 Polynomial Interpolation. 3 Piecewise Polynomial Interpolation
Outline Interpolation 1 Interpolation 2 3 Michael T. Heath Scientific Computing 2 / 56 Interpolation Motivation Choosing Interpolant Existence and Uniqueness Basic interpolation problem: for given data
More informationNeville s Method. MATH 375 Numerical Analysis. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Neville s Method
Neville s Method MATH 375 Numerical Analysis J. Robert Buchanan Department of Mathematics Fall 2013 Motivation We have learned how to approximate a function using Lagrange polynomials and how to estimate
More informationMathematics for Engineers. Numerical mathematics
Mathematics for Engineers Numerical mathematics Integers Determine the largest representable integer with the intmax command. intmax ans = int32 2147483647 2147483647+1 ans = 2.1475e+09 Remark The set
More informationHermite Interpolation
Jim Lambers MAT 77 Fall Semester 010-11 Lecture Notes These notes correspond to Sections 4 and 5 in the text Hermite Interpolation Suppose that the interpolation points are perturbed so that two neighboring
More informationKeyframing. CS 448D: Character Animation Prof. Vladlen Koltun Stanford University
Keyframing CS 448D: Character Animation Prof. Vladlen Koltun Stanford University Keyframing in traditional animation Master animator draws key frames Apprentice fills in the in-between frames Keyframing
More informationExample 1 Which of these functions are polynomials in x? In the case(s) where f is a polynomial,
1. Polynomials A polynomial in x is a function of the form p(x) = a 0 + a 1 x + a 2 x 2 +... a n x n (a n 0, n a non-negative integer) where a 0, a 1, a 2,..., a n are constants. We say that this polynomial
More informationLECTURE 7, WEDNESDAY
LECTURE 7, WEDNESDAY 25.02.04 FRANZ LEMMERMEYER 1. Singular Weierstrass Curves Consider cubic curves in Weierstraß form (1) E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6, the coefficients a i
More informationAnalysis II: Basic knowledge of real analysis: Part V, Power Series, Differentiation, and Taylor Series
.... Analysis II: Basic knowledge of real analysis: Part V, Power Series, Differentiation, and Taylor Series Kenichi Maruno Department of Mathematics, The University of Texas - Pan American March 4, 20
More informationMA2501 Numerical Methods Spring 2015
Norwegian University of Science and Technology Department of Mathematics MA5 Numerical Methods Spring 5 Solutions to exercise set 9 Find approximate values of the following integrals using the adaptive
More informationInterpolation APPLIED PROBLEMS. Reading Between the Lines FLY ROCKET FLY, FLY ROCKET FLY WHAT IS INTERPOLATION? Figure Interpolation of discrete data.
WHAT IS INTERPOLATION? Given (x 0,y 0 ), (x,y ), (x n,y n ), find the value of y at a value of x that is not given. Interpolation Reading Between the Lines Figure Interpolation of discrete data. FLY ROCKET
More informationReview for Exam 2. Review for Exam 2.
Review for Exam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation
More informationNovember 20, Interpolation, Extrapolation & Polynomial Approximation
Interpolation, Extrapolation & Polynomial Approximation November 20, 2016 Introduction In many cases we know the values of a function f (x) at a set of points x 1, x 2,..., x N, but we don t have the analytic
More informationInput: A set (x i -yy i ) data. Output: Function value at arbitrary point x. What for x = 1.2?
Applied Numerical Analysis Interpolation Lecturer: Emad Fatemizadeh Interpolation Input: A set (x i -yy i ) data. Output: Function value at arbitrary point x. 0 1 4 1-3 3 9 What for x = 1.? Interpolation
More informationApplied Numerical Analysis Homework #3
Applied Numerical Analysis Homework #3 Interpolation: Splines, Multiple dimensions, Radial Bases, Least-Squares Splines Question Consider a cubic spline interpolation of a set of data points, and derivatives
More informationSPLINE INTERPOLATION
Spline Background SPLINE INTERPOLATION Problem: high degree interpolating polynomials often have extra oscillations. Example: Runge function f(x = 1 1+4x 2, x [ 1, 1]. 1 1/(1+4x 2 and P 8 (x and P 16 (x
More information1 Piecewise Cubic Interpolation
Piecewise Cubic Interpolation Typically the problem with piecewise linear interpolation is the interpolant is not differentiable as the interpolation points (it has a kinks at every interpolation point)
More informationApplied Numerical Analysis (AE2220-I) R. Klees and R.P. Dwight
Applied Numerical Analysis (AE0-I) R. Klees and R.P. Dwight February 018 Contents 1 Preliminaries: Motivation, Computer arithmetic, Taylor series 1 1.1 Numerical Analysis Motivation..........................
More informationNumerical Methods I: Interpolation (cont ed)
1/20 Numerical Methods I: Interpolation (cont ed) Georg Stadler Courant Institute, NYU stadler@cims.nyu.edu November 30, 2017 Interpolation Things you should know 2/20 I Lagrange vs. Hermite interpolation
More informationNumerical Methods of Approximation
Contents 31 Numerical Methods of Approximation 31.1 Polynomial Approximations 2 31.2 Numerical Integration 28 31.3 Numerical Differentiation 58 31.4 Nonlinear Equations 67 Learning outcomes In this Workbook
More informationCurve Fitting and Interpolation
Chapter 5 Curve Fitting and Interpolation 5.1 Basic Concepts Consider a set of (x, y) data pairs (points) collected during an experiment, Curve fitting: is a procedure to develop or evaluate mathematical
More informationReview of Section 1.1. Mathematical Models. Review of Section 1.1. Review of Section 1.1. Functions. Domain and range. Piecewise functions
Review of Section 1.1 Functions Mathematical Models Domain and range Piecewise functions January 19, 2017 Even and odd functions Increasing and decreasing functions Mathematical Models January 19, 2017
More informationCHAPTER 3 Further properties of splines and B-splines
CHAPTER 3 Further properties of splines and B-splines In Chapter 2 we established some of the most elementary properties of B-splines. In this chapter our focus is on the question What kind of functions
More informationUniversity of Houston, Department of Mathematics Numerical Analysis, Fall 2005
4 Interpolation 4.1 Polynomial interpolation Problem: LetP n (I), n ln, I := [a,b] lr, be the linear space of polynomials of degree n on I, P n (I) := { p n : I lr p n (x) = n i=0 a i x i, a i lr, 0 i
More informationScientific Computing: An Introductory Survey
Scientific Computing: An Introductory Survey Chapter 7 Interpolation Prof. Michael T. Heath Department of Computer Science University of Illinois at Urbana-Champaign Copyright c 2002. Reproduction permitted
More informationMATH ASSIGNMENT 07 SOLUTIONS. 8.1 Following is census data showing the population of the US between 1900 and 2000:
MATH4414.01 ASSIGNMENT 07 SOLUTIONS 8.1 Following is census data showing the population of the US between 1900 and 2000: Years after 1900 Population in millions 0 76.0 20 105.7 40 131.7 60 179.3 80 226.5
More informationEmpirical Models Interpolation Polynomial Models
Mathematical Modeling Lia Vas Empirical Models Interpolation Polynomial Models Lagrange Polynomial. Recall that two points (x 1, y 1 ) and (x 2, y 2 ) determine a unique line y = ax + b passing them (obtained
More informationMath 651 Introduction to Numerical Analysis I Fall SOLUTIONS: Homework Set 1
ath 651 Introduction to Numerical Analysis I Fall 2010 SOLUTIONS: Homework Set 1 1. Consider the polynomial f(x) = x 2 x 2. (a) Find P 1 (x), P 2 (x) and P 3 (x) for f(x) about x 0 = 0. What is the relation
More informationGeometric Interpolation by Planar Cubic Polynomials
1 / 20 Geometric Interpolation by Planar Cubic Polynomials Jernej Kozak, Marjeta Krajnc Faculty of Mathematics and Physics University of Ljubljana Institute of Mathematics, Physics and Mechanics Avignon,
More informationNUMERICAL METHODS. x n+1 = 2x n x 2 n. In particular: which of them gives faster convergence, and why? [Work to four decimal places.
NUMERICAL METHODS 1. Rearranging the equation x 3 =.5 gives the iterative formula x n+1 = g(x n ), where g(x) = (2x 2 ) 1. (a) Starting with x = 1, compute the x n up to n = 6, and describe what is happening.
More information(f(x) P 3 (x)) dx. (a) The Lagrange formula for the error is given by
1. QUESTION (a) Given a nth degree Taylor polynomial P n (x) of a function f(x), expanded about x = x 0, write down the Lagrange formula for the truncation error, carefully defining all its elements. How
More informationCHAPTER 4. Interpolation
CHAPTER 4 Interpolation 4.1. Introduction We will cover sections 4.1 through 4.12 in the book. Read section 4.1 in the book on your own. The basic problem of one-dimensional interpolation is this: Given
More informationAdditional exercises with Numerieke Analyse
Additional exercises with Numerieke Analyse March 10, 017 1. (a) Given different points x 0, x 1, x [a, b] and scalars y 0, y 1, y, z 1, show that there exists at most one polynomial p P 3 with p(x i )
More informationInterpolation and Polynomial Approximation I
Interpolation and Polynomial Approximation I If f (n) (x), n are available, Taylor polynomial is an approximation: f (x) = f (x 0 )+f (x 0 )(x x 0 )+ 1 2! f (x 0 )(x x 0 ) 2 + Example: e x = 1 + x 1! +
More informationNumerical Methods-Lecture VIII: Interpolation
Numerical Methods-Lecture VIII: Interpolation (See Judd Chapter 6) Trevor Gallen Fall, 2015 1 / 113 Motivation Most solutions are functions Many functions are (potentially) high-dimensional Want a way
More informationInterpolation and polynomial approximation Interpolation
Outline Interpolation and polynomial approximation Interpolation Lagrange Cubic Splines Approximation B-Splines 1 Outline Approximation B-Splines We still focus on curves for the moment. 2 3 Pierre Bézier
More informationCS 323: Numerical Analysis and Computing
CS 323: Numerical Analysis and Computing MIDTERM #2 Instructions: This is an open notes exam, i.e., you are allowed to consult any textbook, your class notes, homeworks, or any of the handouts from us.
More informationTaylor Series and Maclaurin Series
Taylor Series and Maclaurin Series Definition (Taylor Series) Suppose the function f is infinitely di erentiable at a. The Taylor series of f about a (or at a or centered at a) isthepowerseries f (n) (a)
More informationMCE693/793: Analysis and Control of Nonlinear Systems
MCE693/793: Analysis and Control of Nonlinear Systems Input-Output and Input-State Linearization Zero Dynamics of Nonlinear Systems Hanz Richter Mechanical Engineering Department Cleveland State University
More informationMath 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C
Math 6 Fall 4 Exam. October 3, 4. The following questions have to do with the integral (a) Evaluate dx. Use integration by parts (x 3 dx = ) ( dx = ) x3 x dx = x x () dx = x + x x dx = x + x 3 dx dx =
More informationMa 530 Power Series II
Ma 530 Power Series II Please note that there is material on power series at Visual Calculus. Some of this material was used as part of the presentation of the topics that follow. Operations on Power Series
More informationEngg. Math. II (Unit-IV) Numerical Analysis
Dr. Satish Shukla of 33 Engg. Math. II (Unit-IV) Numerical Analysis Syllabus. Interpolation and Curve Fitting: Introduction to Interpolation; Calculus of Finite Differences; Finite Difference and Divided
More informationLECTURE NOTES ELEMENTARY NUMERICAL METHODS. Eusebius Doedel
LECTURE NOTES on ELEMENTARY NUMERICAL METHODS Eusebius Doedel TABLE OF CONTENTS Vector and Matrix Norms 1 Banach Lemma 20 The Numerical Solution of Linear Systems 25 Gauss Elimination 25 Operation Count
More informationExamples of the Fourier Theorem (Sect. 10.3). The Fourier Theorem: Continuous case.
s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem. The Fourier Theorem:
More informationLecture 1: Interpolation and approximation
ecture notes on Variational and Approximate Methods in Applied Mathematics - A Peirce UBC ecture : Interpolation and approximation (Compiled 6 August 207 In this lecture we introduce the concept of approximation
More informationTaylor Series. Math114. March 1, Department of Mathematics, University of Kentucky. Math114 Lecture 18 1/ 13
Taylor Series Math114 Department of Mathematics, University of Kentucky March 1, 2017 Math114 Lecture 18 1/ 13 Given a function, can we find a power series representation? Math114 Lecture 18 2/ 13 Given
More informationLECTURE 16 GAUSS QUADRATURE In general for Newton-Cotes (equispaced interpolation points/ data points/ integration points/ nodes).
CE 025 - Lecture 6 LECTURE 6 GAUSS QUADRATURE In general for ewton-cotes (equispaced interpolation points/ data points/ integration points/ nodes). x E x S fx dx hw' o f o + w' f + + w' f + E 84 f 0 f
More information1. Let A be a 2 2 nonzero real matrix. Which of the following is true?
1. Let A be a 2 2 nonzero real matrix. Which of the following is true? (A) A has a nonzero eigenvalue. (B) A 2 has at least one positive entry. (C) trace (A 2 ) is positive. (D) All entries of A 2 cannot
More informationDepartment of Applied Mathematics and Theoretical Physics. AMA 204 Numerical analysis. Exam Winter 2004
Department of Applied Mathematics and Theoretical Physics AMA 204 Numerical analysis Exam Winter 2004 The best six answers will be credited All questions carry equal marks Answer all parts of each question
More informationyou expect to encounter difficulties when trying to solve A x = b? 4. A composite quadrature rule has error associated with it in the following form
Qualifying exam for numerical analysis (Spring 2017) Show your work for full credit. If you are unable to solve some part, attempt the subsequent parts. 1. Consider the following finite difference: f (0)
More informationMathematics for Control Theory
Mathematics for Control Theory Geometric Concepts in Control Involutivity and Frobenius Theorem Exact Linearization Hanz Richter Mechanical Engineering Department Cleveland State University Reading materials
More informationMathematical Olympiad Training Polynomials
Mathematical Olympiad Training Polynomials Definition A polynomial over a ring R(Z, Q, R, C) in x is an expression of the form p(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, a i R, for 0 i n. If a n 0,
More informationMTAEA Differentiation
School of Economics, Australian National University February 5, 2010 Basic Properties of the Derivative. Secant Tangent Applet l 3 l 2 l 1 a a 3 a 2 a 1 Figure: The derivative of f at a is the limiting
More informationNumerical Analysis: Approximation of Functions
Numerical Analysis: Approximation of Functions Mirko Navara http://cmp.felk.cvut.cz/ navara/ Center for Machine Perception, Department of Cybernetics, FEE, CTU Karlovo náměstí, building G, office 104a
More information1 + lim. n n+1. f(x) = x + 1, x 1. and we check that f is increasing, instead. Using the quotient rule, we easily find that. 1 (x + 1) 1 x (x + 1) 2 =
Chapter 5 Sequences and series 5. Sequences Definition 5. (Sequence). A sequence is a function which is defined on the set N of natural numbers. Since such a function is uniquely determined by its values
More informationNumerical Methods I: Polynomial Interpolation
1/31 Numerical Methods I: Polynomial Interpolation Georg Stadler Courant Institute, NYU stadler@cims.nyu.edu November 16, 2017 lassical polynomial interpolation Given f i := f(t i ), i =0,...,n, we would
More informationTaylor Series and Numerical Approximations
Taylor Series and Numerical Approximations Hilary Weller h.weller@reading.ac.uk August 7, 05 An introduction to the concept of a Taylor series and how these are used in numerical analysis to find numerical
More informationExamination paper for TMA4215 Numerical Mathematics
Department of Mathematical Sciences Examination paper for TMA425 Numerical Mathematics Academic contact during examination: Trond Kvamsdal Phone: 93058702 Examination date: 6th of December 207 Examination
More informationPolynomial interpolation over finite fields and applications to list decoding of Reed-Solomon codes
Polynomial interpolation over finite fields and applications to list decoding of Reed-Solomon codes Roberta Barbi December 17, 2015 Roberta Barbi List decoding December 17, 2015 1 / 13 Codes Let F q be
More informationCHALLENGE! (0) = 5. Construct a polynomial with the following behavior at x = 0:
TAYLOR SERIES Construct a polynomial with the following behavior at x = 0: CHALLENGE! P( x) = a + ax+ ax + ax + ax 2 3 4 0 1 2 3 4 P(0) = 1 P (0) = 2 P (0) = 3 P (0) = 4 P (4) (0) = 5 Sounds hard right?
More information[2] (a) Develop and describe the piecewise linear Galerkin finite element approximation of,
269 C, Vese Practice problems [1] Write the differential equation u + u = f(x, y), (x, y) Ω u = 1 (x, y) Ω 1 n + u = x (x, y) Ω 2, Ω = {(x, y) x 2 + y 2 < 1}, Ω 1 = {(x, y) x 2 + y 2 = 1, x 0}, Ω 2 = {(x,
More informationWe have to prove now that (3.38) defines an orthonormal wavelet. It belongs to W 0 by Lemma and (3.55) with j = 1. We can write any f W 1 as
88 CHAPTER 3. WAVELETS AND APPLICATIONS We have to prove now that (3.38) defines an orthonormal wavelet. It belongs to W 0 by Lemma 3..7 and (3.55) with j =. We can write any f W as (3.58) f(ξ) = p(2ξ)ν(2ξ)
More informationMT804 Analysis Homework II
MT804 Analysis Homework II Eudoxus October 6, 2008 p. 135 4.5.1, 4.5.2 p. 136 4.5.3 part a only) p. 140 4.6.1 Exercise 4.5.1 Use the Intermediate Value Theorem to prove that every polynomial of with real
More informationData Analysis-I. Interpolation. Soon-Hyung Yook. December 4, Soon-Hyung Yook Data Analysis-I December 4, / 1
Data Analysis-I Interpolation Soon-Hyung Yook December 4, 2015 Soon-Hyung Yook Data Analysis-I December 4, 2015 1 / 1 Table of Contents Soon-Hyung Yook Data Analysis-I December 4, 2015 2 / 1 Introduction
More informationIntro Polynomial Piecewise Cubic Spline Software Summary. Interpolation. Sanzheng Qiao. Department of Computing and Software McMaster University
Interpolation Sanzheng Qiao Department of Computing and Software McMaster University January, 2014 Outline 1 Introduction 2 Polynomial Interpolation 3 Piecewise Polynomial Interpolation 4 Natural Cubic
More informationChapter 2 Interpolation
Chapter 2 Interpolation Experiments usually produce a discrete set of data points (x i, f i ) which represent the value of a function f (x) for a finite set of arguments {x 0...x n }. If additional data
More information