[PDF] LECTURE 2 INTRODUCTION TO INTERPOLATION

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LECTURE 2 INTRODUCTION TO INTERPOLATION

CE 30125 - Lecture 2 p 2 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

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CE 30125 - Lecture 2

p. 2.1

LECTURE 2INTRODUCTION TO INTERPOLATION• Interpolation function: a function that passes exactly through a set of data points

• Interpolating functions to interpolate values in tables • 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) xsin(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

CE 30125 - Lecture 2

p. 2.2 • 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

CE 30125 - Lecture 2

p. 2.3

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

points = the exact function for which values are known only at a discrete set of data points = the interpolated approximation to the data points (also referred to as interpolation points or nodes) • In tabular form: y f(x 1 f(x 0 x 0 x 1 f(x) xg(x) fxgx fx x 0 , x 1 x o fx o x gx x 1 fx 1

CE 30125 - Lecture 2

p. 2.4 • If is a linear function then (1) where and are unknown coefficients • To pass through points and we must have: (2) (3) • 2 unknowns and 2 equations solve for • Using (2)

Substituting into (3)

gx gxAx B+= A B x o fx o x 1 fx 1 gx o fx o Ax o B+fx o gx 1 fx 1 Ax 1 B+fx 1 AB Bfx o Ax o Ax 1 fx o Ax o -+fx 1

CE 30125 - Lecture 2

p. 2.5 • Substituting for and into equation (1)

This is the formula for linear

interpolation Afx 1 fx o x 1 x o Bfx o x 1 fx 1 x o x 1 x o A B gxfx o x 1 x- x 1 x o ----------------------fx 1 xx o x 1 x o

CE 30125 - Lecture 2

p. 2.6 Example 1• Use values at and to get an interpolated value at using linear interpola- tion

Table 1:

x o x 1 x0.632= x fxxsin= x o 0.5= fx o

0.47942554=

0.632 g0.632?= x 1 1.0= fx 1

0.84147099=

g0.6320.4794251.0 0.632-

1.0 0.5-------------------------------- 0.841470990.632 0.5-

1.0 0.5--------------------------------+=

g0.6320.57500=

CE 30125 - Lecture 2

p. 2.7 Error for Linear Interpolating Functions• Error is defined as: • represents the difference between the exact function and the interpolating or approximating function . • We note that at the interpolating points and • This is because at the interpolating point we have by definition exfxgx- ex fx gx x o x 1 ex o 0= ex 1 0= gx o fx o gx 1 fx 1

CE 30125 - Lecture 2

p. 2.8

Derivation of

e(x) Step 1• Expand in Taylor Series (T.S.) about where (4) • The third term is the actual remainder term and represents all other terms in the series since it is evaluated at ! xf(x 1 f(x 0 x 0 x 1 f(x) g(x) f(x) exfxgx- fx x o fxfx o xx o -df dx------ xx o xx o 2

2!---------------------d

2 f dx2 x= x o x x=

CE 30125 - Lecture 2

p. 2.9

Step 2• Express in terms of and

• We can accomplish this by simply evaluating the T.S. in (4) at . (5) (6) (7) • We note that this is a discrete approximation to the first derivative (a F.D. Formula) df dx------ xx o fx o fx 1 xx 1 fx 1 fx o x 1 x o -df dx------ xx o x 1 x o 2

2!-----------------------d

2 f dx2 x= df dx------ xx oquotesdbs_dbs16.pdfusesText_22