interpolation, extrapolation & polynomial approximation

Download Interpolation, Extrapolation & Polynomial Approximation

Post on 13-Jan-2017

213 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • Interpolation, Extrapolation&

    Polynomial Approximation

    November 21, 2017

    Interpolation, Extrapolation & Polynomial Approximation

  • Introduction

    In many cases we know the values of a function f (x) at a set of pointsx1, x2, ..., xN , but we dont have the analytic expression of the functionthat lets us calculate its value at an arbitrary point. We will try toestimate f (x) for arbitrary x by drawing a curve through the xi andsometimes beyond them.

    The procedure of estimating the value of f (x) for x [x1, xN ] is calledinterpolation while if the value is for points x / [x1, xN ] extrapolation.The form of the function that approximates the set of points should be aconvenient one and should be applicable to a general class of problems.

    Interpolation, Extrapolation & Polynomial Approximation

  • Polynomial Approximations

    Polynomial functions are the most common ones while rational andtrigonometric functions are used quite frequently.

    We will study the following methods for polynomial approximations:

    Lagranges Polynomial

    Hermite Polynomial

    Taylor Polynomial

    Cubic Splines

    Interpolation, Extrapolation & Polynomial Approximation

  • Lagrange Polynomial

    Lets assume the following set of data:

    x0 x1 x2 x3x 3.2 2.7 1.0 4.8f (x) 22.0 17.8 14.2 38.3

    f0 f1 f2 f3

    Then the interpolating polynomial will be of 4th order i.e.ax3 + bx2 + cx + d = P(x). This leads to 4 equations for the 4 unknowncoefficients and by solving this system we get a = 0.5275, b = 6.4952,c = 16.117 and d = 24.3499 and the polynomial is:

    P(x) = 0.5275x3 + 6.4952x2 16.117x + 24.3499

    It is obvious that this procedure is quite laboureus and Lagrangedeveloped a direct way to find the polynomial

    Pn(x) = f0L0(x) + f1L1(x) + ....+ fnLn(x) =n

    i=0

    fiLi (x) (1)

    where Li (x) are the Lagrange coefficient polynomials

    Interpolation, Extrapolation & Polynomial Approximation

  • Lj(x) =(x x0)(x x1)...(x xj1)(x xj+1)...(x xn)

    (xj x0)(xj x1)...(xj xj1)(xj xj+1)...(xj xn)(2)

    and obviously:

    Lj(xk) = jk =

    {0 if j 6= k1 if j = k

    where jk is Kroneckers symbol.

    ERRORSThe error when the Lagrange polynomial is used to approximate acontinuous function f (x) is :

    E (x) = (x x0)(x x1)....(x xn)f (n+1)()

    (n + 1)!where [x0, xN ] (3)

    NOTE Lagrange polynomial applies for evenly and unevenly spaced points. If the points are evenly spaced then it reduces to a much simpler form.

    Interpolation, Extrapolation & Polynomial Approximation

  • Lagrange Polynomial : Example

    EXAMPLE: Find the Lagrange polynomialthat approximates the function y = cos(x).We create the table

    xi 0 0.5 1fi 1 0.0 -1

    The Lagrange coeffiecient polynomials are:

    L1(x) =(x x2)(x x3)

    (x1 x2)(x1 x3)=

    (x 0.5)(x 1)(0 0.5)(0 1)

    = 2x2 3x + 1,

    L2(x) =(x x1) (x x3)

    (x2 x1)(x2 x3)=

    (x 0)(x 1)(0.5 0)(0.5 1)

    = 4x(x 1)

    L3(x) =(x x1)(x x2)

    (x3 x1)(x3 x2)=

    (x 0)(x 0.5)(1 0)(1 0.5)

    = 2x2 x

    thus

    P(x) = 1 (2x2 3x + 1) 0 4x(x 1) + (1) (2x2 x) = 2x + 1

    Interpolation, Extrapolation & Polynomial Approximation

  • The error will be:

    E (x) = x (x 0.5) (x 1)3 sin()

    3!

    e.g. for x = 0.25 is E (0.25) 0.24.

    Interpolation, Extrapolation & Polynomial Approximation

  • Newton Polynomial

    Forward Newton-Gregory

    Pn(xs) = f0 + s f0 +s(s 1)

    2!2f0 +

    s(s 1)(s 2)3!

    3f0 + ...

    = f0 +

    (s1

    )f0 +

    (s2

    )2f0 +

    (s3

    )3f0 + ...

    =n

    i=0

    (si

    )i f0 (4)

    where xs = x0 + s h and

    fi = fi+1 fi (5)2fi = fi+2 2fi+1 + fi (6)3fi = fi+3 3fi+2 + 3fi+1 fi (7)

    nfi = fi+n nfi+n1 +n(n 1)

    2!fi+n2

    n(n 1)(n 2)3!

    fi+n3 + (8)

    Interpolation, Extrapolation & Polynomial Approximation

  • Newton Polynomial

    ERROR The error is the same with the Lagrange polynomial :

    E (x) = (x x0)(x x1)....(x xn)f (n+1)()

    (n + 1)!where [x0, xN ] (9)

    x f (x) f (x) 2f (x) 3f (x) 4f (x) 5f (x)0.0 0.000

    0.2030.2 0.203 0.017

    0.220 0.0240.4 0.423 0.041 0.020

    0.261 0.044 0.0320.6 0.684 0.085 0.052

    0.246 0.0960.8 1.030 0.181

    0.5271.0 1.557

    Table: Example of a difference matrix

    Backward Newton-Gregory

    Pn(x) = f0+

    (s1

    )f1+

    (s + 1

    2

    )2f2+...+

    (s + n 1

    n

    )nfn

    (10)

    Interpolation, Extrapolation & Polynomial Approximation

  • x F (x) f (x) 2f (x) 3f (x) 4f (x) 5f (x)0.2 1.06894

    0.112420.5 1.18136 0.01183

    0.12425 0.001230.8 1.30561 0.01306 0.00015

    0.13731 0.00138 0.0000011.1 1.44292 0.01444 0.00014

    0.15175 0.01521.4 1.59467 0.01596

    0.167711.7 1.76238

    Newton-Gregory forward with x0 = 0.5:

    P3(x) = 1.18136 + 0.12425 s + 0.01306

    (s2

    )+ 0.00138

    (s3

    )= 1.18136 + 0.12425 s + 0.01306 s (s 1)/2 + 0.00138 s (s 1) (s 2)/6= 0.9996 + 0.3354 x + 0.052 x2 + 0.085 x3

    Newton-Gregory backwards with x0 = 1.1:

    P3(x) = 1.44292 + 0.13731 s + 0.01306

    (s + 1

    2

    )+ 0.00123

    (s + 2

    3

    )= 1.44292 + 0.13731 s + 0.01306 s (s + 1)/2 + 0.00123 s (s + 1) (s + 2)/6= 0.99996 + 0.33374 x + 0.05433 x2 + 0.007593 x3

    Interpolation, Extrapolation & Polynomial Approximation

  • Hermite Polynomial

    This applies when we have information not only for the values of f (x)but also on its derivative f (x)

    P2n1(x) =n

    i=1

    Ai (x)fi +n

    i=1

    Bi (x)fi (11)

    where

    Ai (x) = [1 2(x xi )Li (xi )] [Li (x)]2 (12)

    Bi (x) = (x xi ) [Li (x)]2 (13)

    and Li (x) are the Lagrange coefficients.

    ERROR: the accuracy is similar to that of Lagrange polynomial of order2n!

    y(x) p(x) = y(2n+1)()

    (2n + 1)![(x x1)(x x2) . . . (x xn)]2 (14)

    Interpolation, Extrapolation & Polynomial Approximation

  • Hermite Polynomial : Example

    Fit a Hermite polynomial to thedata of the table:

    k xk yk yk

    0 0 0 01 4 2 0

    The Lagrange coefficients are:

    L0(x) =x x1x0 x1

    =x 40 4

    = x 44

    L1(x) =x x0x1 x0

    =x

    4

    L0(x) =1

    x0 x1= 1

    4L1(x) =

    1

    x1 x0=

    1

    4

    Thus

    A0(x) =[

    1 2 L0(x x0)] L20 =

    [1 2

    (

    1

    4

    )(x 0)

    ](

    x 44

    )2A1(x) =

    [1 2 L0(x x1)

    ] L21 =

    [1 2

    1

    4(x 4)

    ](

    x

    4

    )2=

    (3

    x

    2

    )(

    x

    4

    )2B0(x) = (x 0)

    (x 4

    4

    )2= x

    (x 4

    4

    )2B1(x) = (x 4)

    (x

    4

    )2And the Hermite polynomial is:

    P(x) = (6 x)x2

    16.

    Interpolation, Extrapolation & Polynomial Approximation

  • Taylor Polynomial

    It is an alternative way of approximating functions with polynomials. Inthe previous two cases we found the polynomial P(x) that gets the samevalue with a function f (x) at N points or the polynomial that agrees witha function and its derivative at N points. Taylor polynomial has the samevalue x0 with the function but agrees also up to the Nth derivative withthe given function. That is:

    P(i)(x0) = f(i)(x0) and i = 0, 1, ..., n

    and the Taylor polynomial has the well known form from Calculus

    P(x) =Ni=0

    f (i)(x)

    i !(x x0)i (15)

    ERROR: was also estimated in calculus

    EN(x) = (x x0)N+1f (N+1)()

    (N + 1)!(16)

    Interpolation, Extrapolation & Polynomial Approximation

  • Taylor Polynomial : Example

    We will show that for the calculation of e = 2.718281828459... with13-digit approximation we need 15 terms of the Taylor expansion.All the derivatives at x = 1 are:

    y0 = y(1)0 = y

    (2)0 = ... = y

    (n)0 = 1

    thus

    p(x) =n

    i=1

    1

    i !xn = 1 + x +

    x2

    2+ ...+

    1

    n!xn

    and the error will be

    |En| = xn+1e

    (n + 1)!=

    e

    16!

Recommended

View more >