Mathematical Methods For Computing(11BS203)
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-1Chapter-1NUMERICAL METHODSThe limitations of analytical methods have led the engineers and scientists to evolvenumerical methods. Numerical methods are techniques by which mathematical problems areformulated so that they can be solved with arithmetic and logic al operations. Because digitalcomputers excel at performing such operations, numerical methods are sometimes referred to ascomputer mathematics. With the advantage of high speed digital computers and increasingdemand for numerical answers to various problems, numerical techniques have becomeindispensible tool in the hands of engineers.Why should we study nume rical methods?There are several reasons to study numerical methods:Numerical methods greatly expand the types of problems you can address. They arecapable of handling large systems of equations, nonlinearities, and complicatedgeometries that are not uncommon in engineering and science and that are oftenimpossible to solve analytically with standard calculus. As such, they greatly enhanceyour problem-solving skills.Numerical methods allow you to use canned software with insight. During your career,you will invariably have occasion to use commercially available prepackaged computerprograms that involve numerical methods. The intelligent use of these programs is greatlyenhanced by an understanding of the basic theory underlying the methods. In the absenceof such understanding, you will be left to treat such packages as black boxes with littlecritical insight into their inner workings or the validity of the results they produce.Many problems cannot be approached using canned programs. If you are conversant withnumerical methods, and are adept at computer programming, you can design your ownprograms to solve problems without having to buy or commission expensive software.Numerical methods are an efficient vehicle for learning to use computers. Becausenumerical methods are expressly designed for computer implementation, they are idealfor illustrating the computers powers and limitations. When you successfully implementnumerical methods on a computer, and then apply them to solve otherwise intractableproblems, you will be provided with a dramatic demonstration of how computers canserve your professional development. At the same time, you will also learn toacknowledge and control the errors of approximation that are part and parcel of largescale numerical calculations.Numerical methods provide a vehicle for you to reinforce your understanding ofmathematics. Because one function of numerical methods is to reduce highermathematics to basic arithmetic operations, they get at the .nuts and bolts. of someotherwise obscure topics. Enhanced understanding and insight can result from thisalternative perspective.Drgsk@KLUniversityPage 1
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-1What is the nature of the nume rical methods?Numerical methods are often, of a repetitive nature. These consist in repeated executionof the same process where at each step the result of the preceding is used. This is known asiteration process and is repeated till the result is obtained to a desired degree of accuracy.Mathematical Model: A mathematical model can be defined as a formulation or equation thatexpresses the essential features of a physical system or process in mathematical terms. In a verygeneral sense, it can be represented as a functional relationship of the formDependent variable f (independent variables, parameters, forcing functions)where the dependent variable is a characteristic that typically reflects the behavior or state of thesystem; the independent variables are usually dimensions, such as time and space, along whichthe systems behavior is being determined; the parameters are reflective of the systems propertiesor composition; and the forcing functions are external influences acting upon it.A simple mathematical model: (Bungee-Jumper problem)(free falling body)Let m be the mass of the bungee jumper and v(t) be the velocityat time t after jumping. From the Newton second law of motion(time rate of change of momentum of a body is equal to theresulting force acting on it).F ma,where F is the net force acting on the jumper, a is its acceleration(m/s2 ). The net force is composed of two opposite forces: thedownward pull of gravity FD and the upward force of airresistance FU i.e.F F D F U.Again from the Newton second law of motion the downwardforce FD mg due to gravity, where g is the acceleration due togravity (9.81 m/s2 ).Air resistance can be formulated in a variety of ways. From fluid mechanics a good firstapproximation would be to assume that it is proportional to the square of the velocity,FU – Cd v2 ,Where Cd is proportionality constant called the drag coefficient (kg/m). Thus, thegreater the fall velocity, the greater the upward force due to air resistance. The parameterCd accounts for properties of the falling object, such as shape or surface roughness, that affectair resistance. For the present case, Cd might be a function of the type of clothing or theorientation used by the jumper during free fall. Therefore,F mg – Cd v2ma mg – Cd v2Drgsk@KLUniversityPage 2
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)mdvdtChapter-1mg C d v 2Cd 2dvgvdtmIs a model that relates the acceleration of a falling object to the forces acting on it. It is adifferential equation. If the jumper is initially at rest i.e. v 0 at t 0, then the solution of theinitial value problem isgC dgmv(t)tanht .CdmErrors : Engineers and scientists constantly find themselves having to accomplish objectivesbased on uncertain information. Although perfection is a laudable goal, it is rarely if everattained. For example, despite the fact that the model developed from Newton’s second law is anexcellent approximation, it would never in practice exactly predict the jumpers fall. A variety offactors such as winds and slight variations in air resistance would result in deviations from theprediction. If these deviations are systematically high or low, then we might need to develop anew model. However, if they are randomly distributed and tightly grouped around the prediction,then the deviations might be considered negligible and the model deemed adequate. Numericalapproximations also introduce similar discrepancies into the analysis.The errors associated with both calculations and measurements can be characterized withregard to their accuracy and precision. Accuracy refers to how closely a computed or measuredvalue agrees with the true value. Precision refers to how closely individual computed ormeasured values agree with each other.Numerical errors arise from the use of approximations to represent exact mathematicaloperations and quantities. For such errors, the relationship between the exact, or true, resultand the approximation can be formulated asTrue value approximation error.ThereforeTrue error true value – approximation.(1)For numerical methods, the true value will only be known when we deal with functions thatcan be solved analytically. Such will typically be the case when we investigate the theoreticalbehavior of a particular technique for simple systems. However, in real-world applications,we will obviously not know the true answer a priori. For these situations, an alternative is tonormalize the error using the best available estimate of the true value that is, to theapproximation itself, as inapproximat e errorεa100%(2)approximat ionwhere the subscript ‘a’ signifies that the error is normalized to an approximate value. Note alsothat for real- world applications, Eq(1) cannot be used to calculate the error term in the numeratorof Eq. (2). One of the challenges of numerical methods is to determine error estimates in theabsence of knowledge regarding the true value. For example, certain numerical methods useiteration to compute answers. In such cases, a present approximation is made on the basis of aDrgsk@KLUniversityPage 3
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-1previous approximation. This process is performed repeatedly, or iteratively, to successivelycompute (hopefully) better and better approximations. For such cases, the error is often estimatedas the difference between the previous and present approximations. Thus, percent relative error isdetermined according topresent approximat ion - previous approximat ionsεa100% .present approximat ionSolution of algebraic and transcendental equations:A problem of great importance in applied mathematics and engineering is that ofdetermining the solution (root) of an equationf (x) 0,Where the function f may be given explicitly, for examplef ( x) a0 x n a1 x n 1 an 1 x an , a0 0a polynomial of degree ‘n’ in ‘x’, or f(x) may be known only implicitly as a transcendentalfunction. A number ‘c’ is a solution of f(x) 0 if f(c) 0. Such a solution is a root of f(x) 0.Roots in Engineering and Science:Although they arise in other problem contexts, roots of equations frequently occur in thearea of design. Table-1 lists a number of fundamental principles that are routinely used indesign work. Mathematical equations or models derived from these principles are employed topredict dependent variables as a function of independent variables, forcing functions, andparameters. Note that in each case, the dependent variables reflect the state or performance of thesystem, whereas the parameters represent its properties or composition.Table – 1 (Fundamental principles eTime andpositionMass balanceConcentration orTime andquantity of masspositionForce balanceMagnitude andTime anddirection of forcepositionEnergy balanceChanges inTime andkinetic andpositionpotential energyNewton’s laws ofAcceleration,Time andmotionvelocity orpositionlocationKirchhoff’s lawsCurrents andTimevoltagesFundamentalPrincipleHeat balanceDrgsk@KLUniversityin design problems)ParametersThermal properties of material,system geometry.Chemical behavior of material, masstransfer, system geometry.Strength of material, structuralproperties, system geometry.Thermal properties, mass of material,system geometry.Mass of material, system geometry,dissipative parametersElectrical properties (resistance,capacitance, inductance)Page 4
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-1Medical studies have established that a bungeejumpers chances of sustaining a significant vertebraeinjury increase significantly if the free- fall velocityexceeds 36 m/s after 4 s of free fall. Suppose, we want todetermine the mass at which this criterion is exceededgiven a drag coefficient of 0.25 kg/m.We know that the analytical solution can be used topredict fall velocity as a function of time:v(t)gmtanhCdgC dt .mAn alternative way of looking at the problem involvessubtracting v(t) from both sides to give a new function:f(m)gmtanhCdgC dtmv(t).Now we can see that the answer to the problem is the value of m that makes the functionequal to zero. Hence, we call this a roots problem.Bisection method:It is a simplest method of bracketing the roots of a function and requires an initial interval whichis guaranteed to contain a root -- if a and b are the endpoints of the interval then f(a) must differin sign from f(b). This ensures that the function crosses zero at least once in the interval. If avalid initial interval is used then these algorithm cannot fail, provided the function is wellbehaved.On each iteration, the interval is bisectedand the value of the function at themidpoint is calculated. The sign of thisvalue is used to determine which half ofthe interval does not contain a root. Thathalf is discarded to give a new, smallerinterval containing the root. This methodcan be continued indefinitely until theinterval is sufficiently small. At any time,the current estimate of the root is taken asthe midpoint of the interval.Bisection method has linear convergence. Linear convergence means that successive significantfigures are won linearly with computational effort.When an interval contains more than one root, the bisection method can find one of them. Whenan interval contains a singularity, the bisection method converges to that singularity.Drgsk@KLUniversityPage 5
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-1Working Procedure1. Given algebraic or transcendental equation put in the form of f(x) 0.2. Choose ‘a’ and ‘b’ such that f(a) and f(b) are opposite sign or f(a)f(b) 0.a b3. The first approximate root of the equation in (a, b) is x1.24. If f(x1 ) 0, then x1 is the root of the given equation. Otherwise, bisect the interval (a, b)at mid point x1 . Now, the root lies in (a, x1 ) or (x1 , b) according to f(x1 ) positive ornegative.5. Again bisect the interval as before and continue the process until the root is found indesired accuracy.Example#1 Find a root of the equation x3 – 4x – 9 0, using bisection method in four stages.Sol. Let f(x) x3 – 4x – 9, then the equation become f(x) 0.f(2) (2)3 – 4(2) – 9 -9 0f(3) (3)3 – 4(3) – 9 6 0.Since f(2) and f(3) are opposite sign, the root lies between 2 and 3.2 3 5Therefore, the first approximate root x12.5.22f(x1 ) f(2.5) (2.5)3 – 4(2.5) – 9 -3.375 0.Since f(2.5) and f(3) are opposite sign, the root lies between 2.5 and 3.2.5 3The second approximate root x 22.75.2f(x2 ) f(2.75) (2.75)3 – 4(2.75) – 9 0.7969 0.Since f(x1 ) and f(x2 ) are opposite sign, the root lies between 2.5 and 2.75.2.5 2.752.625.2f(x3 ) f(2.625) (2.625)3 – 4(2.625) – 9 -1.4121 0.Since f(x3 ) and f(x2 ) are opposite sign, the root lies between 2.625 and 2.75.The third approximate root x32.625 2.752.6875.2Hence the approximate root of the given equation in fourth stage by bisectionmethod is 2.6875.The fourth approximate root x4Example#2 By using the bisection method, find an approximate root of the equation sin x 1/x,that lies between x 1 and x 1.5 (measured in radians). Carry out computations upto the 7 thstage.Sol. Let f(x) x sin x – 1, then the equation become f(x) 0.f(1) 1 sin(1) – 1 -0.15849 0f(1.5) 1.5 sin(1.5) – 1 0.49625 0.Since f(1) and f(1.5) are opposite sign, the root lies between 1 and 1.5.Drgsk@KLUniversityPage 6
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-11 1.5 2.51.25. Then22f(x1 ) f(1.25) 1.25 sin(1.25) – 1 0.18627 0.Therefore, root lies between 1 and x1 1.25.1 1.25 2.25Thus, the second approximate to the root is x21.125. Then22f(x2 ) f(1.125) 1.125 sin(1.125) – 1 0.01509 0.Therefore, root lies between 1 and x2 1.125.1 1.125 2.125Thus, the third approximate to the root is x31.0625. Then22f(x3 ) f(1.0625) 1.0625 sin(1.0625) – 1 – 0.0718 0.Therefore, root lies between x3 1.0625 and x2 1.125.1.0625 1.125Thus, the fourth approximate to the root is x41.09375. Then2f(x4 ) f(1.09375) 1.09375 sin(1.09375) – 1 – 0.02836 0.Therefore, root lies between x4 1.09375 and x2 1.125.1.09375 1.125Thus, the fifth approximate to the root is x41.10937. Then2f(x5 ) f(1.10937) 1.10937 sin(1.10937) – 1 – 0.00664 0.Therefore, root lies between x5 1.10937 and x2 1.125.1.10937 1.125Thus, the sixth approximate to the root is x61.11719. Then2f(x6 ) f(1.11719) 1.11719 sin(1.11719) – 1 0.00421 0.Therefore, root lies between x5 1.10937 and x6 1.11719.1.10937 1.11719Thus, the seventh approximate to the root is x71.11328.2Hence the desired approximate to the root is 1.11328.Therefore, the first approximate root x1Example#3. Use bisection to determine the drag coefficient needed so that an 80-kg bungeejumper has a velocity of 36 m/s after 4 s of free fall. The acceleration of gravity is 9.81 m/s2.Start with initial guesses of x l 0.1 and x u 0.2 and iterate until the approximate relative errorfalls below 2%.Sol. We know that the velocity of the bungee jumper at any time t isgC dgmv(t)tanht .CdmHere mass of the jumper m 80 kg, velocity of the jumper at t 4 sec, v(4) 36m/s and g 9.81 m/s2 . Substituting the values, we have9.81Cd(9.81)8036tanh(4) .Cd80Rewriting the equationDrgsk@KLUniversityPage 7
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)781.8tanh 4 0.11375C dCd781.8tanh 4 0.11375C dCdf(Cd) 0 using bisection method.Let f(Cd )36Chapter-10.36 . Now, we have to find the root of the equationIteration Method: Consider an equation f(x) 0 which can take in the form x φ(x), whereφ(x) satisfies the following conditions;(i)For two real numbers a and b, a x b implies a φ (x) b and(ii)For all x1, x2 lying in the interval (a, b), we have φ (x1 ) – φ (x2 ) m x1 – x2 , where m isa constant such that 0 m 1.Then the equation x φ (x) has a unique root in the interval (a, b). To find the approximate rootof this equation, we use the following iterative formula;The nth approximate root is xn φ (xn-1 ), n 1, 2, 3, . We start with an initial approximationx0 and continue iterations until the two successive approximations are identical.The following theorem states the sufficient condition for the convergence of the iterativeformula xn φ (xn-1 ).Theorem#1. Let I (a, b) an interval containing a root of the equation x φ (x). If φ'(x) 1, forall x in I, then for any value of x0 in I, the iteration xn φ (xn-1 ) is convergent for all n 1.Example#1. Determine the approximate root of the equation x3 – 2x – 5 0 near to x 2, usingfixed point iterative method.Sol. Given equation x3 – 2x – 5 0 can be written as x3 2x 5 and x (2x 5)1/3 . This is of2the form x φ (x), where φ (x) (2x 5)1/3 . Also (x)and φ '(x) 1, for 2 x 3.233(2x 5)Hence the iteration xn φ (xn-1 ), n 1, 2, 3, near x 2 is convergent. Let us take the initialapproximation x0 2.The first approximation to the root is x1 φ (x0 ) φ (2) [2(2) 5]1/3 91/3 2.08008.The second approximation to the root is x2 φ (x1 ) φ (2.08008) [2(2.08008) 5]1/3 2.09235.The third approximation to the root is x3 φ (x2 ) φ (2.09235) [2(2.09235) 5]1/3 2.09422.The fourth approximation to the root is x4 φ (x3 ) φ (2.09422) [2(2.09422) 5]1/3 2.09450.The fifth approximation to the root is x5 φ (x4 ) φ (2.09450) [2(2.09450) 5]1/3 2.09454.The sixth approximation to the root is x6 φ (x5 ) φ (2.09454) [2(2.09454) 5]1/3 2.09455.Since x5 and x6 are identical up to 4 decimal places, we take x6 2.09455 as the required root ofthe given equation.Example#2. Determine the approximate root of the equation tan-1 x – x 1 0, using fixed pointiterative method.Drgsk@KLUniversityPage 8
II/IV B.Tech CSE & ECM Mathematical Methods for Computing (11BS203)Chapter-1Sol. The given equation can be written as x φ (x), where φ (x) 1 tan-1 x. Clearly,1and φ '(x) 1 for 0 x 1. Hence the iteration xn φ (xn-1 ), n 1, 2, 3, is(x)1 x2convergent. Let us take the initial approximation x0 1.The first approximation to the root is x1 φ (x0 ) φ (1) 1 tan-1 (1) 1.7854.The first approximation to the root is x2 φ (x1 ) φ (1.7854) 1 tan-1 (1.7854) 2.0602.Similarly the successive iterations are x3 2.1189, x4 2.1318, x5 2.1322, x6 2.13231,x7 2.13132.Since x6 and x7 are identical up to 4 decimal places, we take x7 2.13132 as the required root ofthe given equation.Newton-Raphson Method:Newton-Raphson method, is a root-finding algorithm that uses the first few terms of the Taylorseries of a function f(x) in the vicinity of a suspected root.h2f(x h) f(x) hf (x)f (x) 2!Newton algorithm begins with aninitial guess for the location of theroot. On each iteration, a line tangentto the function f is drawn at thatposition. The point where this linecrosses the x-axis becomes the newguess. Newton method convergesquadratically for single roots, andlinearly for mul
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