Outline. Numerical Analysis Basics. Some Higher-Level Languages. Programming Languages. BASIC and VBA MATLAB

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umerical Analysis Basics Larry Caretto Mechanical Engineering 39 umerical Analysis of Engineering Systems February 5, 214 Outline Programming Languages Binary umbers and Data Types Limits on the size and accuracy of numbers stored on computers Errors in computing Character and date representations Introduction to numerical solution of algebraic equations 2 Programming Languages Instructions to manipulate information stored in computer memory Machine language that actually does wor is all in binary numbers Assembly language, used in some applications, has commands that are similar to actual computer operation Higher-level languages are easier for humans to understand and use 3 Some Higher-Level Languages Fortran earliest higher-level language COBOL original business language PASCAL used for instruction in 197s ADA developed by US Department of Defense in 197s C developed at Bell labs to be applied as higher and lower level language C++ variant of C that uses objectoriented programming 4 BASIC and VBA MATLAB Beginners All-purpose Symbolic Instruction Code developed for instructional use at Dartmouth in 1964 Used by various microcomputers in 197s (Apple, Microsoft, etc.) Visual Basic for Applications is objectoriented version used to support many programs (Office, LabView, SolidWors, etc. 5 MATrix LABoratory initially developed as interface for students to Fortran linear algebra programs Widely used in control system analysis Provides simple interface for quic answers to complex problems MATLAB coding language available for advanced problem solving 6 ME 39 umerical Analysis of Engineering Systems 1

Common Language Features Data types: integer, real, string, Replacement statements: assign a value to a variable Expressions: mathematical, logical, etc. Choice statements: if, case Looping statements Count controlled loops Conditional loops Arrays 7 A Binary Machine What are the next numbers in this sequence? Computers use binary numbers, 1, 1, 11, 1, 11, 11, 111, 1, This places limits on accuracy 11, 11, 111, 11, 111, 111 Data types translate zeros and ones in computer hardware into data (numbers and text) that humans recognize The number of binary bits (a binary digit, or 1) used to represent a number affects its maximum size and accuracy 1111 1 11 11 111 8 umbers and Bases We want to display binary numbers and show their equivalence to normal (baseten) numbers We will also mention octal (base eight) and hexadecimal numbers (base 16) To discuss this we will develop a general format starting with base-ten numbers using an example to get a general format 9 Decimal umbers What does the decimal number 132 mean? It s 2 times 1 + 3 times 1 + 1 times 1 It s 2 times 1 + 3 times 1 1 + 1 times 1 2 How can we generalize this? Define d as the digit in position of a number, counting from right to left, with d being the rightmost digit In the example of 132, d = 2, d 1 = 3 and d 2 = 1 What are d d = 1, d 1 = 9, 1 values for 7491? d 2 = 4 and d 3 = 7 General Decimal umbers Start with example from previous chart We wrote 132 as 2 times 1 + 3 times 1 1 + 1 times 1 2 So 132 = 2(1 ) + 3(1 1 ) + 1(1 2 ) We also said that for 132, d = 2, d 1 = 3 and d 2 = 1 So, for the general three-digit number we have d 2 d 1 d = d 2 (1 2 ) + d 1 (1 1 ) + d (1 ) How can we write an -digit number? 1 2 1 d d d d d (1 ) d (1 ) d (1 ) (1 ) 1 2 1 1 2 1 d d 1 1 d 2 d (1 ) 11 Generalizing the Base Rewrite general base-1 equation d 1 1 d 2 d (1 ) to apply to any base, b d 1 1 d 2 d ( b ) d 1d 2 b Can use base, b, as subscript for clarity Common bases for computing: 2, 8, 16 Base 16 digits use a 16 = 1 1, b 16 = 11 1, c 16 = 12 1, d 16 = 13 1, e 16 = 14 1, and f 16 = 15 1 What is abc 16 in base ten? (abc) 16 = 12(16 ) + 11(16 1 ) + 1(16 2 ) = 2748 1 c 16 = 12 1 b 16 = 11 1 a 16 = 1 1 12 ME 39 umerical Analysis of Engineering Systems 2

Different Bases Decimal 1 2 3 4 5 Binary 1 1 11 1 11 Octal 1 2 3 4 5 Hex 1 2 3 4 5 Decimal 6 7 8 9 1 11 Binary 11 111 1 11 11 111 Octal 6 7 1 11 12 13 Hex 6 7 8 9 a b One octal digit ranges over all possible combinations of 3 binary digits ( to 111) One hex digit ranges from to 1111 13 Integer umbers on Computer Integer data types fill memory available with zeros and ones (one bit required) What is maximum number that can be stored in 2 bytes (16 bits)? 15 d15d14 d (2 ) so that all d 1 gives 15 15 14 1 1111111111111111 1(2 ) 2 2 2 2 65 535 Can show maximum size for bits = 2 1 14 Integer umbers on Computer II 16-bit integer numbers can range from to 65,535 1 Use offset to have range from 32,768 1 to 32,767 1 Long data type with 4 bytes (32 bits) has range of 2 32 1 = 4,294,967,295 With offset range is 2,147,483,648 1 to 2,147,483,647 1 15 Real umbers on Computer Use power notation for number. For example,.123456x1 4 In this example 123456 is called the significand and 4 is called the exponent One bit is required for the sign (+ or ) IEEE standard 754 for floating-point numbers started 1985, revised 28 Single data type uses 4 bytes (32 bits) with one sign bit, 23 bits for the significand and 8 bits for the exponent 16 Real umbers on Computer Use power notation for number. For example,.123456x1 4 In this example 123456 is called the significand and 4 is called the exponent One bit is required for the sign (+ or ) IEEE standard 754 for floating-point numbers started 1985, revised 28 Single data type uses 4 bytes (32 bits) with one sign bit, 23 bits for the significand and 8 bits for the exponent 17 Real umbers on Computer II A binary significand will have the form.1dddd (d s are remaining digits, or 1) The lead 1 is not stored so the 23 bits for the significand are effectively 24 bits The maximum size of the 24 -bit significand is 2 24 1 = 16,277,214; log 1 (16,277,214) = 7.22 1 significant figures Eight-bit exponent size = 2 8 1 = 255 1 Scaled from 128 to 127 giving maximum power of 2 127 = 1 38.23 18 ME 39 umerical Analysis of Engineering Systems 3

Real umbers on Computer III Type double uses 8 bytes (64 bits) 52 bits for significand; 11 for exponent Maximum exponent is 1 38 Gives about 16 significant figures VBA does not have quadruple type that uses 16 bytes (128 bits) 112 bits for significand, 15 for exponent Maximum exponent is 1 4931 and accuracy is 34 significant figures 19 umerical Analysis Errors Truncation error caused by converting calculus equations into algebraic equations by truncating Taylor series Roundoff error caused by inexact conversion of decimal numbers into binary representation Error propagation is growth in errors as calculations proceed due to different error types 2 Engineering Errors Modeling Errors arise when a simplified model of a complex process is used Done when some parts of process do not have complete mathematical description Also done to reduce computer time and/or storage for detailed processes Human error, sometimes called blunders Character Codes ASCII uses 1 byte ( to 377 8 = 255 1 ) See char() function on Excel worsheet = 65 to 9 is A to Z = 97 to 122 is a to z Unicode used to accommodate multiple character sets for different languages (Asian, Arabic, Hebrew, etc.) Range is to 1FFFF 16 = 1114111 1 Initial range same as ASCII 21 22 Date Variables Really date and time Excel and VBA use 8 bytes for storage Integer part is number of days since reference date Date range is 1/1/1 to 12/31/9999 Fractional part is time in fraction of a day (9 am =.375, noon =.5, etc.) Use format function to convert this fraction to conventional hours, minutes, seconds 23 Solution of Equations First topic in numerical analysis Some equations lie 3x + 15 = 5(4x+2) or.2 = sin(x) can be solved for x Get x = 5 and x = sin -1 (.2) for examples Some equations cannot be solved directly, e.g.: 3x 2 = cos(x) Want general methods for solving such equations 24 ME 39 umerical Analysis of Engineering Systems 4

Example Problem Plot to solve ax = sin(x) What is solution of ax = sin(x) Could use graphing calculator Algorithms for finding roots of equations used as parts of larger computer programs that requires such solutions Graphing equation to be solved is used here to show how roots (solutions) can be visualized ote: there may be more than one root 25 26 Methods for Finding Roots Two classes of methods One initial guess Two initial guesses that bracet root (i.e. root lies between the initial guesses) Some require more than two initial guesses When location of root cannot be estimated a search procedure is required to bracet root Methods that bracet root are usually slower but surer 27 General Algorithms Any equation in one unnown cay be written in the form f(x) = The equations shown previously, ax = sin(x) (a = 1.1 or.5), can be written as ax sin(x) = or sin(x) ax = Almost all algorithms based on this form Root means that f(x) = As f(x) decreases we are getting closer to a root 28 Braceting a Root Plot shows two roots, f(x) = and values of f(x) on opposite sides of root Basic idea: when a root, f(x) =, is braceted the values of f(x) on opposite sides of the root have opposite signs If f(x)f(x+dx) < there is a root, f(x) =, between x and x+dx 29 How to Bracet a Root Often some physical information about a problem will let you now that there is a root at some approximate location x* Pic a value of Dx and find f(x* + Dx) and f(x* Dx) If the product of these two f values is negative, the root is braceted If product is positive increase Dx and retry If Dx is too large you may have braceted two roots and have f(x* + Dx)f(x* Dx) > 3 ME 39 umerical Analysis of Engineering Systems 5

Process and otation In solving f(x) = we use iteration We mae an initial estimate, called x, of the correct root (where f(x) = ) Sometimes we mae two initial estimates, x and x 1 (or even more) We have an iteration procedure that uses previous estimates, x, x -1, etc. to find a new value x +1 Iterations continues until x value is found that gives desired accuracy 31 Secant Method Starts with two initial guesses, x and x 1, which need not bracet root Each iteration uses values of two successive guesses, x and x -1 to find new guess, x +1 Approximate behavior of f(x) near guess as straight line: f f(x ) = m(x x ) Slope m found from two latest guesses m = f x f x 1 x x 1 32 Secant Method II Substitute equation for slope, m, into linear approximation: f f(x ) = m(x x ) f f x = f x f x 1 x x Tae new guess, x +1 as value of x that sets f = in linear approximation f x = f x f x 1 x +1 = x f x x +1 x f x f x 1 33 Secant Method III Start with two initial guesses x 1 and x Repeat the following steps to get a new guess, x +1, from the guesses x and x -1 Compute x +1 from the equation x +1 = x f x f x f x 1 Compute f(x +1 ) Repeat the process until two successive values are close enough 34 Secant Example Solve f(x) =.5x sin(x) = Pic x = 2 and x 1 = 2.5 Initial guesses do not have to bracet root f(x ) = f(2) =.5(2) sin(2) =.89 f(x 1 ) = f(2.5) =.5(2.5) sin(2.5) =.473 Apply x +1 = x f x f x f x 1 general x 1 x equation x 2 = x 1 f x 1 f x 1 f x to first iteration x 2 = 2.5.473 2.5 2.473.89 = 3.2 f x 2 =.5 3.2 sin 3.21 =.224 35 Secant Example II ote: All significant figures of calculator or spreadsheet used in calculations Rounding shown here to save space x +1 = x f x Apply f x f x 1 general x 2 x 1 x 3 = x 2 f x 2 equation f x 2 f x 1 to second x iteration 3 = 3.2.224 3.2 2.5.224.473 = 2.98 f x 3 =.5 2.98 sin 2.98 =.131 Can you find x4 and f(x 4 )? 36 ME 39 umerical Analysis of Engineering Systems 6

Secant Example III x f(x ) x - x -1 2-8.9E-1 1 2.5-4.73E-1 5.E-1 2 3.249382516827 2.24E-1 7.E-1 3 2.978849281731-1.31E-2 2.3E-1 4 2.9913497389318-1.11E-4 1.3E-2 5 2.9914565334953 1.4E-7 1.1E-4 6 2.99145643339981-7.95E-13 1.E-7 7 2.991456433458.E+ 7.7E-13 8 2.991456433458.E+.E+ 37 Quiz One solutions I Function myint(a as Double, b as _ Double, n as Double) as Double If n <> -1 Then myint = (b^(n+1) a^(n+1)) _ /(n+1) ElseIf a*b > Then myint = log(b/a) Else myint = Undefined End If End Function 38 Quiz Solutions II Write the following Excel formula in cell B4 and copy the equation to cells C4:E4 to get the results for all cells: =myint(a1, B1, C1,) For part 3 the values in the cells give [2 (1+1) 1 (1+1) ] / (1 + 1) = 1.5 [(-2) (1+1) 2 (1+1) ] / (1 + 1) = [(-1) (4+1) (-1) (4+1) ] / (4 + 1) = 39 ME 39 umerical Analysis of Engineering Systems 7