numerical_analysis

# numerical_analysis - Numerical Analysis You could say that...

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Unformatted text preview: Numerical Analysis You could say that some of the equations that you encounter in describing physical systems can’t be solved in terms of familiar functions and that they require numerical calculations to solve. It would be misleading to say this however, because the reality is quite the opposite. Most of the equations that describe the real world are sufficiently complex that your only hope of solving them is to use numerical methods. The simple equations that you find in introductory texts are there because they can be solved in terms of elementary calculations. When you start to add reality, you quickly reach a point at which no amount of clever analytical ability will get you a solution. That becomes the subject of this chapter. In all of the examples that I present I’m just showing you a taste of the subject, but I hope that you will see the essential ideas of how to extend the concepts. 11.1 Interpolation Given equally spaced tabulated data, the problem is to find a value between the tabulated points, and to estimate the error in doing so. As a first example, to find a value midway between given points use a linear interpolation: f ( x + h/ 2) ≈ 1 2 f ( x ) + f ( x + h ) This gives no hint of the error. To compute an error estimate, it is convenient to transform the variables so that this equation reads f (0) ≈ 1 2 f ( k ) + f (- k ) , where the interval between data points is now 2 k . Use a power series expansion of f to find an estimate of the error. f ( k ) = f (0) + kf (0) + 1 2 k 2 f 00 (0) + ··· f (- k ) = f (0)- kf (0) + 1 2 k 2 f 00 (0) + ··· Then 1 2 f ( k ) + f (- k ) ≈ f (0) + 1 2 k 2 f 00 (0) , (11 . 1) where the last term is your error estimate: Error = Estimate- Exact = + k 2 f 00 (0) / 2 = + h 2 f 00 (0) / 8 And the relative error is (Estimate- Exact)/Exact. As an example, interpolate the function f ( x ) = 2 x between 0 and 1. Here h = 1 . 2 1 / 2 ≈ 1 2 2 + 2 1 = 1 . 5 The error term is error ≈ (ln2) 2 2 x / 8 for x = . 5 = ( . 693) 2 (1 . 5) / 8 = . 090 , and of course the true error is 1 . 5- 1 . 414 = . 086 You can write a more general interpolation method for an arbitrary point between x and x + h . The solution is a simple extension of the above result. James Nearing, University of Miami 1 11—Numerical Analysis 2 The line passing through the two points of the graph is y- f = ( x- x )( f 1- f ) /h, x x 1 f = f ( x ) , f 1 = f ( x + h ) . At the point x = x + ph you have y = f + ( ph )( f 1- f ) /h = f (1- p ) + f 1 p As before, this approach doesn’t suggest the error, but again, the Taylor series allows you to work it out to be h 2 p (1- p ) f 00 ( x + ph ) / 2 . The use of only two points to do an interpolation ignores the data available in the rest of the table. By using more points, you can greatly improve the accuracy. The simplest example of this method is the 4-point interpolation to find the function halfway between the data points. Again, the independent variable has an increment h = 2 k , so the problem can be stated as one of finding the...
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## This note was uploaded on 01/30/2012 for the course PHYS 315 taught by Professor Nearing during the Fall '08 term at University of Miami.

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numerical_analysis - Numerical Analysis You could say that...

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