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# Computers in Civil Engineering 53:081 Spring 2003 - PowerPoint PPT Presentation

Computers in Civil Engineering 53:081 Spring 2003. Lecture #14. Interpolation. Interpolation: Overview. Objective: estimate intermediate values between precise data points using simple functions Solutions Newton Polynomials Lagrange Polynomials Spline Interpolation. Interpolation.

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Computers in Civil Engineering53:081 Spring 2003

Lecture #14

Interpolation

• Objective: estimate intermediate values between precise data points using simple functions

• Solutions

• Newton Polynomials

• Lagrange Polynomials

• Spline Interpolation

Interpolation

Curve Fitting

Curve goes through data points

single value

Curve need not go through data points

multiple values

High-precision data points

Cities

Dresden

LaSalle

Braidwood

Fist-order (linear)

Third-order (cubic)

Second-order (quadratic)

• General comments

• Linear Interpolation

• Quadratic Interpolation

• General Form

The notation: means the first order interpolatingpolynomial

Linear Interpolation Formula

By similar triangles:

Rearrange:

Problem:

Estimate ln(2)(the true value is 0.69)

Solution:

We know that:

at x = 1 ln(x) =0

at x = e ln(x) =1 (e=2.718...)

Thus,

General form:

Equivalent form:

(f2(x) means second-order interpolating polynomial)

To solve for ,three points are needed:

Substitute in (1) and evaluate at to find:

Substitute in (1) and evaluate at to find:

Quadratic Interpolation

Note: this looks

like a second

derivative…

Problem

Estimate ln(2)(the true value is 0.69)

Solution

We know that:

at x = x0 = 1 ln(x) =0

at x = x1 = e ln(x) =1 (e=2.718...)

at x = x2= e2 ln(x) = 2

It would get pretty tedious to do this for third,

fourth, fifth, sixth, etc order polynominal

We need a plan:

Newton’s Interpolating Polynomials

To solve for , n+1 points are needed:

Solution

What does this [ ]

notation mean?

First finite divided difference:

Second finite divided difference:

nth finite divided difference:

Finite divided difference table, case n = 3:

fdd(i,1)=f(i)

enddo

do j=2,n

do i=1,n-j+1

fdd(i,j)=(fdd(i+1,j-1)-fdd(i,j-1))/

& (x(i+j-1)-x(i))

enddo

enddo

Divided Differences Pseudo Code

See the textbook!

• Data need not be equally spaced

• Arrangement of data does not have to be ascending or descending, but it does influence error of interpolation

• Best case is when the base points are close to the unknown value

• Estimate of relative error:

Error estimate for nth-order polynomial is the difference between the (n+1)th and nth-order prediction.

Relative Error As a Function of Order Polynomial

Example 18.5 in text

Determine ln(2) using the following table

MATLAB function interp1 is very useful for this

Midterm 2 Polynomial

Tuesday 15 April