Loading in 5 sec....

Computers in Civil Engineering 53:081 Spring 2003PowerPoint Presentation

Computers in Civil Engineering 53:081 Spring 2003

- 103 Views
- Uploaded on
- Presentation posted in: General

Computers in Civil Engineering 53:081 Spring 2003

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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

Quad

Cities

Dresden

LaSalle

Braidwood

Quad-Cities Nuke Station Diffuser Curve

Fist-order (linear)

Third-order (cubic)

Second-order (quadratic)

- General comments
- Linear Interpolation
- Quadratic Interpolation
- General Form

The notation: means the first order interpolatingpolynomial

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:

Set in (1) to find

Substitute in (1) and evaluate at to find:

Substitute in (1) and evaluate at to find:

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:

do i=0,n-1

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

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.

Example 18.5 in text

Determine ln(2) using the following table

MATLAB function interp1 is very useful for this

Tuesday 15 April