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Computers in Civil Engineering 53:081 Spring 2003

Computers in Civil Engineering 53:081 Spring 2003. Lecture #6. Roots of Equations (Chapter 5 of 3rd Edition of C 2 ). Solve for x please …. 4x=8 (x-1)(x+1)=0 e -x =x cos(x)cosh(x)+1=0. Problem Formulation. Find roots of equations in form: f(x)=0.

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Computers in Civil Engineering 53:081 Spring 2003

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  1. Computers in Civil Engineering53:081 Spring 2003 Lecture #6 Roots of Equations(Chapter 5 of 3rd Edition of C2)

  2. Solve for x please … • 4x=8 • (x-1)(x+1)=0 • e-x=x • cos(x)cosh(x)+1=0

  3. Problem Formulation Find roots of equations in form:f(x)=0 (We can put any of our one-equation engineering problems into this form)Question: What do we mean by roots? Question: Why do we need to find the roots of such equations? Answer: Example of uniform flow in open channel

  4. Uniform Flow Colebrook-White Formula T(y)=surface width y A(y)=area P(y) = wetted perimeter

  5. Continue... In many instances Q, So, etc. are known, and one wants to solve for y. In other words: given Q and channel geometry, solve: f(y) = 0 Problem: Explicit solution does not exist! (especially since A(y) and P(y) are also functions) Solution:Numerically find root of equation, e.g. find value of depth y such that the equation is satisfied. (Note that you can move Q to the right side and say 0=f(y). What value of y causes f(y) to be zero?)

  6. Outline By definition f(xr) = 0 • Single roots • Multiple roots • Systems of equations fi(x1,x2,...,xn)=0 for i = 1,...,n

  7. Graphical Method • Plot function and observe zero crossing- ALWAYS do this first! • Useful for finding initial conditions of systematic methods f(x) x

  8. Example Consider the function: i.e. or

  9. The Graphical Approach 1.0 0.5 f(x) 0.0 0.2 0.4 0.6 0.8 -0.5 -1.0 x

  10. General Discussion • Bracketing Methods • Require two initial guesses that bracket the true root • Open Methods • Require only one initial guess • Special case: multiple roots. For example: Two roots at 1

  11. Bracketing Methods • Bisection Method • False-Position Method (Regula-Falsi)

  12. Possible Situations f(x) f(x) f(xu) f(xl) x x f(xl)f(xu) < 0 Odd number of roots in bracket f(xl)f(xu) > 0 No root in bracket f(x) f(x) x x f(xl)f(xu) > 0 Even number of roots in bracket f(xl)f(xu) < 0 Odd number of roots in bracket

  13. Difficult Situations (f(xl)f(xu)<0 but even number of roots f(x) f(x) f(xlu) x x f(xl)

  14. The Bisection Method 1. Choose lower xl and upper xu bounds. 2. Check if f(xl) f(xu) < 0. If not, go to 1. 3. Estimate root by xr = 0.5(xl+xu) 4. Check which interval contains the root. If f(xl) f(xr) < 0 set xu = xr and go to 5. If f(xl) f(xr) > 0 set xl = xr and go to 5. If f(xl) f(xr) = 0 stop. (is this possible?) 5. Is the required accuracy met? If yes: Stop. If not: Go to 3.

  15. Example: Bisection Method f(x) 4 1 0 x 2 3 xl xu

  16. Termination Criteria Iteratively estimate error: In the present context: Where is the prescribed stopping criterion.

  17. Example: f(x) = e-x -x 1.0 By definition f(xr) = 0 0.5 f(x) 0.0 0.2 0.4 0.6 0.8 -0.5 -1.0 xu xl x

  18. Termination Criteria Iteration xr|t|% |a|% 1 0.5 11.8 2 0.75 32.2 33.3 3 0.625 10.2 20.0 4 0.5625 0.819 11.1 5 0.59375 4.69 5.3 (The root is approximately at 0.567)

  19. Roots of Equations … to be continued(Chapter 5 of 3rd Edition of C2)

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