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Numerical Analysis for Engineers

Numerical Analysis for Engineers . Eng. Muhannad Al- Jabi. Analytical and Numerical Solutions :. Numerical methods are techniques by which mathematical problems are formulated so that they can be solved with arithmetic operations. Approximations and Round-Off Errors.

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Numerical Analysis for Engineers

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  1. Numerical Analysis for Engineers Eng. Muhannad Al-Jabi

  2. Analytical and Numerical Solutions: • Numerical methods are techniques by which mathematical problems are formulated so that they can be solved with arithmetic operations. Eng. Muhannad Al-Jabi

  3. Approximations and Round-Off Errors • For many engineering problems, we cannot obtain analytical solutions. • Numerical methods yield approximate results, results that are close to the exact analytical solution. We cannot exactly compute the errors associated with numerical methods. • Only rarely given data are exact, since they originate from measurements. Therefore there is probably error in the input information. • Algorithm itself usually introduces errors as well, e.g., unavoidable round-offs, etc … • The output information will then contain error from both of these sources. • How confident we are in our approximate result? • The question is “how much error is present in our calculation and is it tolerable?” Eng. Muhannad Al-Jabi

  4. Accuracy. How close is a computed or measured value to the true value • Precision (or reproducibility). How close is a computed or measured value to previously computed or measured values. • Inaccuracy(or bias). A systematic deviation from the actual value. • Imprecision(or uncertainty). Magnitude of scatter. Eng. Muhannad Al-Jabi

  5. Significant Figures • Number of significant figures indicates precision. Significant digits of a number are those that can be used with confidence, e.g.,the number of certain digits plus one estimated digit. 53,800 How many significant figures? 5.38 x 1043 5.380 x 1044 5.3800 x 1045 Zeros are sometimes used to locate the decimal point not significant figures. 0.00001753 4 0.0001753 4 0.001753 4 Eng. Muhannad Al-Jabi

  6. I can say the speed is between 48.5 and 48.9 km/h. Thus, we have a 3-significant figure reading. One might say the speed is between 48 and 49 km/h. Thus, we have a 2-significant figure reading. Eng. Muhannad Al-Jabi

  7. Rules regarding the zero: • Zeros within a number are always significant: 5001 and 50.01 have 4 significant figures • Zeros to the left of the first nonzero digit in a number are not significant: 0.00001845, 0.0001845, 0.001845 have 4 significant figures • Trailing zeros are significant: 6.00 has 3 sig. figures. • 45300 may have 3, 4, or 5 sig. figures, we can know if the number is written in the scientific notation: 4.53×104 has 3 sig. figures 4.530×104 has 4 sig. figures 4.5300×104 has 5 sig. figures Eng. Muhannad Al-Jabi

  8. Error Definitions True Value = Approximation + Error Et = True value – Approximation (+/-) True error Eng. Muhannad Al-Jabi

  9. For numerical methods, the true value will be known only when we deal with functions that can be solved analytically (simple systems). In real world applications, we usually not know the answer a priori. Then • Iterative approach, example Newton’s method Eng. Muhannad Al-Jabi

  10. Use absolute value. • Computations are repeated until stopping criterion is satisfied. • If the following criterion is met you can be sure that the result is correct to at least n significant figures. Pre-specified % tolerance based on the knowledge of your solution Eng. Muhannad Al-Jabi

  11. Round-off Errors • Numbers such as p, e, or cannot be expressed by a fixed number of significant figures. • Computers use a base-2 representation, they cannot precisely represent certain exact base-10 numbers. • Fractional quantities are typically represented in computer using “floating point” form, e.g., Integer part exponent mantissa Base of the number system used Eng. Muhannad Al-Jabi

  12. Eng. Muhannad Al-Jabi

  13. Word • How are numbers represented in computers? • Numbers are stored in what is called ‘word’. A word has a number of bits, each bit holds either 0 or 1. • For example, -173 is presented on a 16-bit computer as Eng. Muhannad Al-Jabi

  14. On a 16-bit computer, the range of numbers that can be represented is between -32,768 and 32,767. Eng. Muhannad Al-Jabi

  15. 156.78 (normal form) 0.15678×103(floating point form) Word Floating Point Representation Eng. Muhannad Al-Jabi

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  18. 156.78  0.15678x103 in a floating point base-10 system Suppose only 4 decimal places to be stored • Normalized to remove the leading zeroes. Multiply the mantissa by 10 and lower the exponent by 1 0.2941 x 10-1 Additional significant figure is retained Eng. Muhannad Al-Jabi

  19. Therefore for a base-10 system 0.1 ≤m<1 for a base-2 system 0.5 ≤m<1 • Floating point representation allows both fractions and very large numbers to be expressed on the computer. However, • Floating point numbers take up more room. • Take longer to process than integer numbers. • Round-off errors are introduced because mantissa holds only a finite number of significant figures. Eng. Muhannad Al-Jabi

  20. Rounding • Rounding up is to increase by one the digit before the part that will be discarded if the first digit of the discarded part is greater than 5. • If it is less than 5, the digit is rounded down. • If it is exactly 5, the digit is rounded up or down to reach the nearest even digit. • Round 1.14 to one decimal place: 1.1 • Round 1.15 to one decimal place: 2.2 • Round 21.857 to one decimal place: 21.8 Eng. Muhannad Al-Jabi

  21. Chopping Example: p=3.14159265358 to be stored on a base-10 system carrying 7 significant digits. p=3.141592 chopping error et=0.00000065 If rounded p=3.141593 et=0.00000035 • Some machines use chopping, becauserounding adds to the computational overhead. Since number of significant figures is large enough, resulting chopping error is negligible. Eng. Muhannad Al-Jabi

  22. Truncation Errors and the Taylor Series Eng. Muhannad Al-Jabi

  23. Taylor’s Theorem If the function f and its first n+1 derivatives are continuous on an interval containing a and x, then the value of the function at x is given by: where the remainder Rn is defined as Eng. Muhannad Al-Jabi

  24. Using Taylor’s Theorem, we can approximate any smooth function by a polynomial. • The zero-order approximation of the value of f(xi+1) is given f(xi+1)f(xi) • The first-order approximation is given by f(xi+1)f(xi) + f '(xi)(xi+1-xi) Eng. Muhannad Al-Jabi

  25. The complete Taylor series is given by The remainder term is where  is between xiand xi+1 Eng. Muhannad Al-Jabi

  26. Notes • We usually replace the difference (xi+1 - xi) by h. • A special case of Taylor series when xi = 0 is called Maclaurin series. Eng. Muhannad Al-Jabi

  27. Solution h = /3 - /4 = /12 Zero-order approximation: f(/3)  cos (/4) = 0.707106781 t = First-order approximation: f(/3)  cos (/4) – (/12) sin (/4) = 0.521986659 t = -4.4% See Table 4.1 Example Use Taylor series expansions with n = 0 to 6 to approximate f(x) = cos x near xi = /4 at xi+1 = /3. Eng. Muhannad Al-Jabi

  28. To get more accurate estimation of f(xi+1), we can do one or both of the following: • add more terms to the Taylor polynomial • reduce the value of h. Eng. Muhannad Al-Jabi

  29. Roots of equations All Iterative Eng. Muhannad Al-Jabi

  30. Bracketing Methods(Or, two point methods for finding roots)Chapter 5 • Two initial guesses for the root are required. These guesses must “bracket” or be on either side of the root. == > Fig. 5.1 • If one root of a real and continuous function, f(x)=0, is bounded by values x=xl, x=xu then f(xl) . f(xu) <0. (The function changes sign on opposite sides of the root) Eng. Muhannad Al-Jabi

  31. No answer (No root) Two roots( Might work for a while!!) Nice case (one root) Discontinuous function. Need special method Oops!! (two roots!!) Three roots ( Might work for a while!!) Figure 5.2 Figure 5.3 Eng. Muhannad Al-Jabi

  32. The Bisection Method For the arbitrary equation of one variable, f(x)=0 • Pick xl and xu such that they bound the root of interest, check if f(xl).f(xu) <0. • Estimate the root by evaluating (xl+xu)/2. • Find the pair • If f(xl). f[(xl+xu)/2]<0, root lies in the lower interval, then xu=(xl+xu)/2 and go to step 2. Eng. Muhannad Al-Jabi

  33. If f(xl). f[(xl+xu)/2]>0, root lies in the upper interval, then xl= [(xl+xu)/2, go to step 2. • If f(xl). f[(xl+xu)/2]=0, then root is (xl+xu)/2 and terminate. • Compare es with ea • If ea<es, stop. Otherwise repeat the process. Eng. Muhannad Al-Jabi

  34. Figure 5.6 Eng. Muhannad Al-Jabi

  35. Evaluation of Method Advantages • Easy • Always find root • Number of iterations required to attain an absolute error can be computed a priori. Disadvantages • Slow • Know a and b that bound root • Multiple roots • No account is taken of f(xl) and f(xu), if f(xl) is closer to zero, it is likely that root is closer to xl . Eng. Muhannad Al-Jabi

  36. How Many Iterations will It Take? • Length of the first Interval Lo=b-a • After 1 iteration L1=Lo/2 • After 2 iterations L2=Lo/4 • After k iterations Lk=Lo/2k Eng. Muhannad Al-Jabi

  37. If the absolute magnitude of the error is and Lo=2, how many iterations will you have to do to get the required accuracy in the solution? Eng. Muhannad Al-Jabi

  38. The False-Position Method(Regula-Falsi) • If a real root is bounded by xl and xu of f(x)=0, then we can approximate the solution by doing a linear interpolation between the points [xl, f(xl)] and [xu, f(xu)] to find the xr value such that l(xr)=0, l(x) is the linear approximation of f(x). == > Fig. 5.12 Eng. Muhannad Al-Jabi

  39. Procedure • Find a pair of values of x, xl and xu such that fl=f(xl) <0 and fu=f(xu) >0. • Estimate the value of the root from the following formula (Refer to Box 5.1) and evaluate f(xr). Eng. Muhannad Al-Jabi

  40. Use the new point to replace one of the original points, keeping the two points on opposite sides of the x axis. If f(xr)*f(xu)<0 then xl=xr If f(xr)*f(xl)<0 then xu=xr If f(xr)=0 then you have found the root and need go no further! Eng. Muhannad Al-Jabi

  41. See if the new xl and xu are close enough for convergence to be declared. If they are not go back to step 2. • Why this method? • Faster • Always converges for a single root. See Sec.5.3.1, Pitfalls of the False-Position Method Note: Always check by substituting estimated root in the original equation to determine whether f(xr) ≈ 0. Eng. Muhannad Al-Jabi

  42. example Determine the real roots of A)By using the bisection method B)By using the false position method Eng. Muhannad Al-Jabi

  43. Solution of part A Eng. Muhannad Al-Jabi

  44. Solution of part B Eng. Muhannad Al-Jabi

  45. Open Methods • Open methods start with a starting point or the initial guess. • Because the open methods do not bracketing the root, they may be converges or diverges. • On the other hand open methods are faster in finding the root. ( more efficient). • Open methods are:- simple fixed point iteration, Newton Raphson, Secant and multiple roots Eng. Muhannad Al-Jabi

  46. Simple fixed point iteration • Rearrange the function so that x is on the left side of the equation: • Fixed-point methods may sometime “diverge”, depending on the stating point (initial guess) and how the function behaves. Eng. Muhannad Al-Jabi

  47. Example: Eng. Muhannad Al-Jabi

  48. Convergence • x=g(x) can be expressed as a pair of equations: y1=x y2=g(x) (component equations) • Plot them separately. Eng. Muhannad Al-Jabi

  49. Fixed-Point Iteration Convergent Divergent Eng. Muhannad Al-Jabi

  50. Conclusion • Fixed-point iteration converges if • When the method converges, the error is roughly proportional to or less than the error of the previous step, therefore it is called “linearly convergent.” Eng. Muhannad Al-Jabi

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