# interpolation, extrapolation & polynomial approximation

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• 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!

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