Given a table of data poins of an unknown or complicated function f : we want to find a (simpler) function p s.t. px (

Transcription

1 Iterpolato 1

2 Iterpolato Gve a table of data pos of a ukow or complcated fucto f : y y y y y we wat to fd a (smpler) fucto p s.t. p ( ) = y for = 0... p s sad to terpolate the table or terpolate f at 0,. p s called a terpolat. The pots are called odes. For betwee odes, we ca estmate f( ) by p( ). 2

3 Data Pots 3

4 Pecewse Lear Iterpolato 4

5 Polyomal Iterpolato Iterpolate the table wth a polyomal. 5

6 Sple Iterpolato Iterpolate the table wth a sple. 6

7 Polyomal Sple 7

8 Polyomal Iterpolato Gve a table of + 1 data pots: y y y y y there s a uque polyomal p of degree that terpolates the table. 8

9 Smplest Case: = 0 Gve a sgle data pot: y y 0 0 fd a polyomal p of degree 0 such that p( ) = y. 0 0 Soluto: p( ) = y 0 9

10 Lear Iterpolato Gve two data pots: y y 0 1 y 0 1 fd a lear fucto p such that p ( ) = y ad p ( ) = y Soluto: p ( ) y y y = y y p ( ) = y+ ( )

11 Polyomal Iterpolato: a smple case y Soluto (kow as cardal polyomals) : L( ) = ( 0) ( 1)( + 1) ( ) ( ) ( )( ) ( ) = j ( j) ( j) = j= 0 j= 0 j= 0 j j j j Verfcato: L( ) j 0 f = δj = 1 f = j j 11

12 Now we wat a polyomal that terpolates ths table: y y 0 0 Soluto: y L( ) 12

13 Polyomal Iterpolato: Lagrage Form Fd a polyomal of degree that terpolates ths table: y y y y y Soluto: For each, the degree polyomal L( ) y terpolates y y 0 0 = So, p ( ) = L( y ) (havg degree ) terpolates = 0 the gve + 1 data pots. 13

14 Lagrage Iterpolato: eample Fd a polyomal of degree 2 that terpolates ths table: y Soluto: = p ( ) = L( y ) = 0 ( 2)( 3) ( 1)( 3) ( 1)( 2) = (1 2 )(1 3) (2 1)(2 3) (3 1)(3 2) = 2( 2)( 3) 7( 1)( 3) + 3( 1)( 2) 14

15 Polyomal Iterpolato: Newto Form Fd a polyomal of degree that terpolates ths table: y = f ( ) y y y y Newto form: p ( ) = a+ a( ) + a( )( ) + a ( )( )( ) a( )( ) ( )

16 Dervg Newto Iterpolatg Polyomal Suppose p k ( ) terpolates y 0 1 y y 0 1 k k Fd p k+ 1 ( ) that terpolates y 0 1 k k+ 1 y y y 0 1 k k+ 1 Let p ( ) p ( ) + q( ), where q( ) has zeros at,, k+ 1 = k 0 k ad q ( ) = y p( ). k+ 1 k+ 1 k k+ 1 That s, q( ) terpolates y 0 1 k k y p ( ) k+ 1 k k+ 1 16

17 Dervg Newto Iterpolatg Polyomal (cot'ed) q ( ) terpolates 0 1 k k+ 1 y y p ( ) k+ 1 k k+ 1 ( ) So, q ( ) = y p( ) k+ 1 k k+ 1 ( )( ) ( ) 0 1 ( )( ) ( ) k+ 1 0 k+ 1 1 k+ 1 k k = a ( )( ) ( ) for some costat a. k k k+ 1 So, p ( ) = p ( ) + a ( )( ) ( ). k+ 1 k k k 17

18 Polyomal Iterpolato: Newto Form p ( ) = a 0 0 p ( ) = a + a ( ) p ( ) = a + a ( ) + a ( )( ) p ( ) = a + a ( ) + a ( )( ) a ( )( )( ) a( 0)( 1) ( 1) 18

19 Computg the Coeffcets a k The polyomal of degree that terpolates a fucto at odes,,, Newto's form s 0 1 f p ( ) = a + a ( ) + a ( )( ) a ( )( )( ) a ( )( ) ( ) a = f[,,, ] k k Ths quatty s called the dvded dfferece of order k for f. 19

20 That s, the polyomal of degree that terpolates a fucto f at odes,,, Newto's form s: p ( ) = f[ ] f[, ]( ) f[,, 0 1 ]( )( ) f[,,, ]( )( )( ) f[,,, ]( )( ) ( ) f[,,, ] s called the dvded dfferece of order k for f. k 20

21 Dvded Dffereces Formula f[ ] = f( ) f[,,, ] Or + 1 j f[,,, ] + 1 j = = f [,,, ] f [,,, ] + 1 j j f [,,, ] f [,,, ] j + 1 j 1 j j 21

22 Dvded Dffereces Table f[ ] = f( ) f[,] f[,,] f[,,,] f[ ] = f( ) f[, ] f[ ] = f( ) f[,, ] f [, ] f [,,, ] f[ ] = f( ) f[,, ] f[ ] = f( ) f[, ]

23 Newto Iterpolato: eample Fd a polyomal of degree 2 that terpolates ths table: y Soluto: f[ ] f[, ] f [,, ] p ( ) = 4+ 3( 1) 2 ( 1)( 2) 23

24 Eample: f ( ) f[ ] f[,] f[,,] f[,,,] 0 = = /6 2 2 = = Iterpolatg polyomal Newto form: p ( ) = 3+ ( 1) + ( 1)( ) 2 ( 1) ( )( 0)

25 Newto Iterpolato: eample f( )

26 Polyomal Iterpolato usg Matrces Table: y Let p( ) = a + b + c + d terpolate the table. The p( 3) = 0, p( 2) = 3, p(0) = 3, p(2) = 15, ad we have a system of lear equatos ukows a, b, c, d: a b c d 3 2 ( 3) + ( 3) + ( 3) + = 0 a b c d 3 2 ( 2) + ( 2) + ( 2) + = 3 a + b + c + d = 3 2 (0) (0) (0) 3 a + b + c + d = 3 2 (2) (2) (2) 15 26

27 I matr form: 3 2 ( 3) ( 3) 3 1 a ( 2) ( 2) 2 1 b 3 = 3 2 (0) (0) 0 1 c (2) (2) 2 1 d 15 Reduced row-echelo form: a b c d 1 a = 1 3 b = 3 = 1 c = 1 3 d = 3 = Thus, p ( )

28 Augmeted matr for the system: 3 2 ( 3) ( 3) ( 2) ( 2) (0) (0) (2) (2) Reduced row-echelo form:

29 Usg Matlab Augmeted matr for the system: M = 3 2 ( 3) ( 3) ( 2) ( 2) (0) (0) (2) (2) Eter the matr to Matlab: >> >> = [-3, -2, 0, 2]'; y = [ 0, -3, 3, -15]'; M = [.^3,.^2,, oes(sze()), y]; 29

30 Use the Matlab commad rref(m) to covert the matr to reduced row-echelo form: rref( M) a = b = c = d = 3 >> MR = rref(m); >> p = MR(:, ed)'; The terpolatg polyomal s: p 3 2 ( ) =

31 >> = [-3, -2, 0, 2]'; y = [0, -3, 3, -15]'; >> M = [.^3,.^2,, oes(sze()), y]; >> MR = rref(m); >> p = MR(:, ed)'; >> = -3:0.01:2; >> y = polyval(p, ); >> plot(,y,'r*',,y) 31

32 Aother eample >> = [0, 1, 2, 3, 4, 5, 6]'; >> y = [4, 1, 1, 2, 0, 0, 8]'; >> m = [.^6,.^5,.^4,.^3,.^2,,oes(sze()),y]; >> mr = rref(m); >> p = mr(:, ed); >> = 0:0.01:6; >> y = polyval(p, ); >> plot(, y, 'ro',, y) 32

33 Chebyshev Nodes Equally spaced odes are ofte ot good. A better choce s the set of + 1 Chebyshev odes: O the terval [ 1, 1] = cos π For ay terval [ a, b], map [ 1, 1] to [ a, b] learly: = ( a+ b) + ( b a)cos π

34 Ruge fucto: f ( ) 1 = 1 + 2

35 Degree-10 polyomal terpolatg f( ) at 11 equally spaced odes. 1 = 1 + 2

36 1 Iterpolatg f( ) = usg 11 Chebyshev odes

37 Sple Iterpolato Iterpolate a set of data pots wth a sple. 37

38 Sple of Degree 1 Iterval: [ a, b] Kots: pots t such that a= t < t < t < < t = b Sple of degree 1: pecewse lear fucto. 38

39 Sple of Degree 2 Iterval: [ a, b] Kots: pots t such that a= t < t < t < < t = b Sple of degree 2: pecewse quadratc fucto wth cotuous dervatv e. 39

40 Cubc Sple (Sple of Degree 3) Iterval: [ a, b] Kots: pots t such that a= t < t < t < < t = b Cubc sple: a fucto S such that 1. S, S, S are cotuous o [ a, b] 2. S s a cubc polyomal o each subterval [ t, t ]. (So, S = pecewse cubc fucto wth cotuous S ad S.) + 1 Natural cubc sple: addto, S ( a) = S ( b) = 0. 40

41 Kots: a= t < t < t < < t = b Cubc sple: S0( ) [ t0, t1] S1( ) [ t1, t2] S ( ) = S 1( ) [ t 1, t] Each S s a cubc polyomal. S -1 t S For each teral t, S ( t ) = S ( t ) = y, S ( t ) = S ( t 1 1 ), ad S ( t ) = S ( t ). S ( t ) = y, S ( t ) = y Natural cubc sple: addto, S ( a) = S ( b) = 0. 41

42 Cubc Sple: eample y S ( ) A = a + b + c+ d = B = e + f + g+ h 3 2 ( ) for [ 1, 0] 3 2 ( ) for [0, 1] Iterpolato: S( 1) = 1 A( 1) = 1 a+ b c+ d = 1 A(0) = 2 d = 2 S(0) = 2 B(0) = 2 h= 2 S(1) = 1 B(1) = 1 e+ f + g+ h= 1 42

43 A = a + b + c S ( ) = B = e + f+ g 2 ( ) 3 2 for [ 1, 0] 2 ( ) 3 2 for [0, 1] A ( ) = 6a+ 2b for [ 1, 0] S ( ) = B ( ) = 6e+ 2f for [0, 1] Cotuty of S ad S at 0: A (0) = B (0) c= A (0) = B (0) b= f Natural cubc sple: A ( 1) = 0 6a+ 2b= 0 B (1) = 0 6e+ 2 f = 0 g 43

44 We have 8 ukows ( a, b, c, d, e, f, g, h) ad 8 equatos: a+ b c+ d = 1 a= 1 d = 2 b= 3 h= 2 c= 1 e+ f + g+ h= 1 d = 2 c= g e= 1 b= f f = 3 6a+ 2b= 0 g = 1 6e+ 2f = 0 h= 2 S( ) = for [ 1, 0] B ( ) = e + f + g+ h for [0, 1] 3 2 A ( ) a b c d =

45 Sple Iterploato wth Matlab: eample >> = [0, 1, 2, 3, 4, 5, 6]; >> y = [4, 1, 1, 2, 0, 0, 8]; >> = 0:0.01:6; >> y = sple(, y, ); >> plot(, y, 'ro',, y) 45

46 1 Sple Iterploato of Ruge fucto: f( ) =

47 A applcato of polyomal terpolato Secret sharg: There s a secret s, whch s a postve teger. We wat to dvde s to shares s, each gve to a user u, 1, such that ay t or more users together ca recover s; t 1 or fewer users ca ot recover s; where 0 < t. A soluto for ths problem s called a ( t, )-threshold secret sharg scheme.

48 Basc Idea What ca be uquely determed by t peces of formato? Notes: All computatos are doe modulo p for some large prme umber p. That s, all computatos are tegers wthout roudg errors.

49 Shamr's secret sharg scheme 1. Costruct a ( t 1)-degree polyomal p( ) = a, where a,, a are radomly chose, ad let a = s. 1 t 1 2. Choose dstct values,,. j J j J\{ } j 1 3. Let share s = (, y ), where y = p( ), 1. Gve t shares (, y ), J, J = t, p( ) ad s ca be t 1 recovered usg the Lagrage terpolato formula: p ( ) = y j ad s = a0 = p(0) = y. = 0 0 J j J\{ } j

50 Shamr's secret sharg scheme: eample Suppose s = 541, = 7, t = 4. Let 3 2 p ( ) = Seve shares (, y ), each gve to a user: y Ay 4 users, usg ther shares, ca recostruct p ( ) ad compute the secret s = p(0) = 541.