Chapter 1 Numerical approximation of data : interpolation, least squares method

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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.

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