CE 30125 - Lecture 2

LECTURE 2

INTRODUCTION TO INTERPOLATION

• Interpolation function: a function that passes exactly through a set of data points.

• Interpolating functions to interpolate values in tables

x sin(x) 0.0 0.000000 0.5 0.479426 1.0 0.841471 1.5 0.997495 2.0 0.909297 2.5 0.598472

• In tables, the function is only specified at a limited number or discrete set of indepen- dent variable values (as opposed to a continuum function). • We can use interpolation to find functional values at other values of the independent variable, e.g. sin(0.63253)

p. 2.1 CE 30125 - Lecture 2

• In numerical methods, like tables, the values of the function are only specified at a discrete number of points! Using interpolation, we can describe or at least approximate the function at every point in space.

• For numerical methods, we use interpolation to • Interpolate values from computations • Develop numerical integration schemes • Develop numerical differentiation schemes • Develop finite element methods

• Interpolation is typically not used to obtain a functional description of measured data since errors in the data may lead to a poor representation. • Curve fitting to data is handled with a separate set of techniques

p. 2.2 CE 30125 - Lecture 2

Linear Interpolation • Linear interpolation is obtained by passing a straight line between 2 data points

y g(x)

f(x) f(x1)

f(x0)

x x0 x1

fx = the exact function for which values are known only at a discrete set of data points

gx = the interpolated approximation to fx

x0 , x1 = the data points (also referred to as interpolation points or nodes)

• In tabular form:

xo fxo x gx

x1 fx1

p. 2.3 CE 30125 - Lecture 2

• If gx is a linear function then

gx= Ax+ B (1)

where A and B are unknown coefficients

• To pass through points xo fxo and x1 fx1 we must have:

gxo = fxo  Axo + B = fxo (2)

gx1 = fx1  Ax1 + B = fx1 (3)

• 2 unknowns and 2 equations solve for AB

• Using (2)

Bfx= o – Axo Substituting into (3)

Ax1 + fxo – Axo = fx1

p. 2.4 CE 30125 - Lecture 2

fx– fx A = ------1 o- x1 – xo

fxx – fxx B = ------o 1 1 o- x1 – xo

• Substituting for A and B into equation (1)

x1 – x xx– o gx= fxo ------+ fx1 ------x1 – xo x1 – xo

This is the formula for linear interpolation

p. 2.5 CE 30125 - Lecture 2

Example 1

• Use values at xo and x1 to get an interpolated value at x = 0.632 using linear interpola- tion

Table 1: x fx= sinx

xo = 0.5 fxo = 0.47942554

0.632 g0.632 = ?

x1 = 1.0 fx1 = 0.84147099

1.0– 0.632 0.632– 0.5 g0.632 = 0.479425------+ 0 . 8 4147099------1.0– 0.5 1.0– 0.5

g0.632 = 0.57500

p. 2.6 CE 30125 - Lecture 2

Error for Linear Interpolating Functions • Error is defined as:

ex fx– gx

•ex represents the difference between the exact function fx and the interpolating or approximating function gx .

• We note that at the interpolating points xo and x1

exo = 0

ex1 = 0

• This is because at the interpolating point we have by definition

gxo = fxo

gx1 = fx1

p. 2.7 CE 30125 - Lecture 2

Derivation of e(x)

f(x) f(x) g(x) f(x1) f(x0)

x x0 x1

ex fx– gx

Step 1

• Expand fx in Taylor Series (T.S.) about xo

2 2 df xx– o d f fx= fx++xx– ------where x  x (4) o o dx 2! 2 o xx= dx o x = 

• The third term is the actual remainder term and represents all other terms in the series since it is evaluated at x =  !

p. 2.8 CE 30125 - Lecture 2

Step 2 df • Express ------in terms of fx and fx dx o 1 xx= o

• We can accomplish this by simply evaluating the T.S. in (4) at xx= 1 .

2 2 df x1 – xo d f fx= fx++x – x ------(5) 1 o 1 o dx 2! 2 xx= dx o x = 



2 2 df fx1 fxo x1 – xo 1 d f ------= ------– ------– ------ ------(6) dx x – x x – x 2! x – x 2 xx= 1 o 1 o 1 o dx o x = 



2 df fx1 fxo x1 – xo d f ------= ------– ------– ------(7) dx x – x x – x 2 2 xx= 1 o 1 o dx o x = 

• We note that this is a discrete approximation to the first derivative (a F.D. Formula)

p. 2.9 CE 30125 - Lecture 2

Step 3

• Substitute Equation 7 into T.S. form of fx , Equation (4).

• This gives us an expression for fx in terms of the discrete values fxo and fx1 .

2 2 2 fx1 fxo x1 – xo d f xx– o d f fx = fx + xx– ------– ------– ------+ ------(8)  o o 2 2 x1 – xo x1 – xo 2 dx 2 dx x =  x =  

2 2 xx– o xx– o xx– o –x1 + xo xx– o d f fx = fx ++------fx – ------fx ------+ ------(9)  o 1 o 2 x1 – xo x1 – xo 2 2 dx x =  

2 fxo fx1 xx– o d f fx = x – x – x + x ------++xx– ------– x ++x xx– ------(10)  1 o o o 1 o o 2 x1 – xo x1 – xo 2 dx x = 



2 x1 – x xx– o xx– o xx– 1 d f fx = fx ------++fx ------(11)  o 1 2 x1 – xo x1 – xo 2 dx x = 

p. 2.10 CE 30125 - Lecture 2

• The first part of Equation (11) is simply the linear interpolation formula. The second part is in fact the error. Thus:

ex fx– gx



x – x xx– xx– xx– d2f ex fx ------1 - ++fx ------o------o 1 ------ o 1 2 x1 – xo x1 – xo 2 dx x = 

x1 – x xx– o –fxo ------– fx1 ------x1 – xo x1 – xo 

2 xx– o xx– 1 d f ex= ------xo  x1 2 dx2 x = 

• If we assume that the interval xo x1 is small, then the second derivative won’t change dramatically in the interval! 2 2 2 2 d f d f d f d f xo + x1 ------ ------where xm  ------dx2 dx2 dx2 dx2 2 x =  xx= o xx= 1 xx= m

p. 2.11 CE 30125 - Lecture 2

• Thus we typically evaluate the derivative term in the error expression using the midpoint in the interval 1 d2f ex ---xx– o xx– 1 ------2 dx2 xx= m • Another problem is that we typically don’t know the second derivative at the midpoint of the interval, xm • However using finite differencing formulae we can approximate this derivative knowing the functional values at the interpolating points

• Maximum error occurs at the midpoint for linear interpolation (where xx– o xx– 1 is the largest) 1 d2f max exx xx ---xm – xo xm – x1 ------0 1 2 dx2 xx= m • However

hx 1 – xo and h h --- = x – x and --- = x – x 2 m o 2 1 m

p. 2.12 CE 30125 - Lecture 2

• Thus h2 d2f max exx xx= ------0 1 8 dx2 xm

• Notes on Error for linear interpolation • The error expression has a polynomial and a derivative portion.

• Maximum error occurs approximately at the midpoint between xo and x1

• Error increases as the interval h increases

2 2 • Error increases as f x increases. Again note that f x can be approximated with finite difference (F.D.) formulae if at least 3 surrounding functional values are available. (We will discuss F.D. formulae later.)

p. 2.13 CE 30125 - Lecture 2

Example 2

• Compute an error estimate for the problem in Example 1.

• Recall we found that

g0.632 = 0.57500

• Error is estimated as: 2 1 d f ex ---xx– o xx– 1 ------2 dx2 xx= m

• Since x = 0.750 is the midpoint at the interval 0.5 1.0 , we have

2 1 d f e0.632  --- 0.632– 0.5 0.632– 1.0 ------2 dx2 x = 0.750 

d2f e 0.632 –0.024288------ 2 dx x = 0.75

p. 2.14 CE 30125 - Lecture 2

• Since we have not yet extensively discussed approximating derivatives using discrete 2 d f values, we will compute ------using analytical methods: dx2 x = 0.75

2 d f ------==–sin0.750 – 0.68164 dx2 x = 0.75

2 d f • Substituting in the value for ------, we obtain an estimate for the error: dx2 x = 0.75

e0.632  –0.024288 –0.68164 = 0.016555

• Computing the actual error (the actual solution - the estimated error):

Ex= sinx – gx

E0.632 ==sin0.632 – 0.57500 0.01576

• The estimated error, ex is a good approximation of the actual error Ex!

p. 2.15