数值分析 第1章

Chapter 1 Solving Nonlinear EquationTopics: 1. Interval Halving(Bisection) 2. Linear Interpolating Methods 3. Newton’s Method 4. Muller’s Method 5. Fixed-Point Iteration 6. Nonlinear SystemA SurveyOne of the most frequently occurring problems in scientific work is to find the roots of equations of the form Function The roots, or zeros of the function Example:We know thatBut how about the equationIt’s not easy to obtain the exact roots. In general, we hope to get only approximate solutions.“Approximate solution” means a point small, or is closed to a solution,for which of the equationisBut the concept of an “approximate solution” is rather fuzzy. There are many difficulties that we will encounter. All these will be shown in the following section.1. Interval Halving(Bisection, Binarysearching Method)Suppose with is a continuous function defined on the interval and of opposite sign, then there exits a number , insuch that Bisectionf ( x1 )px11x2 f ( x2 )x2 Letis the middle point of3 Letis the middle point ofThe middle point ……is more closer toWe are back to the functionMATLAB command: >>f=inline(‘3*x+sin(x)-exp(x)’) >>fplot(f,[0,2]); grid on From the figure we know one root is in and another is inAlgorithm: Repeat Set If Set Else Set Until End If. thenProgram: click hereThe Error If f(x) is continuous in [a, b], which actually bracket a root. Note that we have the following relation ship. Times 1 2 3 … n It’s clear the root . So, the error ……. Interval lengthFrom, we getHere are some values aboutandn 5 10 20 30 Disadvantage: Interval Halving method is slow to converge. The speed of convergence is linear.Exercises:Find an approximation to Bisection Algorithm. correct to within by using2. Linear interpolation Methods1.The secant methodLet be the initial points which are near to the root. From the figure we haveThen,If we repeat this, we have:Each newly computed value should be nearer to the root, so we always use the last two computed points. But after the first iteration, we need to check whether is closer to the root than Example：Use secant method for . How to do this?From the figure, we know there is a root in the interval [0,1]. Of course we can use 0 and 1 as the initial points, and 0 is closer to the root than 1.Table Iteration 1 2 3 4 5 1 0 0.4709896 0.3722771 0.3599043 x0 0 0.4709896 0.3722771 0.3599043 0.3604239 x1 x2 0.4709896 0.3722771 0.3599043 0.3604239 0.3604217 F(x2) 0.2651588 2.953367E-02 -1.294787E-03 5.552969E-0.6 3.554221E-08The exact value is 0.36042170296…. Fewer iterations are required compared to bisection!Algorithm: are given. Compute If Swap End If Repeat Set Set Until Or then andA pathological caseThe two initial points should be close to the root as much as possible! Plotting the function can help you to choose the initial points!2. Linear Interpolation (False Position) A way to avoid this pathology is to ensure that the root is bracketed between the two starting values and remains between the successive pairs.Choose the starting value Then use the formula, such thatto get. The next step is to check the root is in the interval3. Newton’s MethodSuppose that such that series of at and , . Let be an approximation to is “small”. Consider the Taylorwherelies betweenand. So,Sinceis small, the term involvingis much smaller, soSolving for x givesStarting with an initial approximation, the formulagenerates the approximate sequence. This is Newton’s Method.This formula also can be derive from the figure.Then,Repeat this work, we can get the schemeExample: Use Newton’s method forWe need compute, and need a initial value of x.We can get this derivative in Matlab: >> fx=sym(‘3*x+sin(x)-exp(x)’) >>dfx=diff(fx) dfx=3+cos(x)-exp(x) If we begin with , we haveThe exact value is 0.36042170296….Compare with other methods Bisection Initial value After 5 Iterations # of significant digital Example: Use Newton’s method to find the value of It equals to find the positive root of . 0.34375 1 False Position Secant method Newton’s method x0=0 -31 x0=1 -12x0=0, x1=1 0.360433 4 0.360422 5By using Newton’s method we haveWe can prove that x converge to the exact value for any (See Numerical Analysis (in Chinese), Li Qingyang, 278-279) Algorithm: For any Repeat Set Set Until Or (tolerance) ,.In some cases, the result does not converge to the root with Newton’s method . Rootûx û x xü0 0 0The conclusion is that the convergence depends on the choosing of.Relating Newton’s Method to Other Methods Recall the formula of the linear interpolation:We rewrite it to the form Difference QuotientSuppose that thatis very closed toandis continuous. We note. By taking the limit we haveThe definition of the derivativeHence,Newton’s MethodIn fact, the difference quotient can be regard as an approximation to the derivative. A simplified newton’s method (see Numerical Analysis, Li Qingyang, P280)Newton's descent method (see Numerical Analysis, Li Qingyang, P281)The parameteris chosen such that4. Muller’s Method (Parabola Method)Previous methods: straight line Muller’s methods: quadratic polynomialA second-degree polynomial is made to fit three points near a root, x0, x1, x2, with x0 between x1 and x2.Assume the quadratic polynomial has the formLetLetting，we can obtain a, b and c:So,Which root we should choose? And how to judge? The one nearest to x0.If b>0, choose +; If b<0, choose -; If b=0, choose either. When we get the root v, we may choose three points as the initial points for the next approximation. There are some cases to discuss. root rootX2, root, x0X0, root, x1Algorithm: are given and RepeatSet Set Compute: ;If b>=0, then letElse, let End If If x>x0, then let x2=x0, x0=x; Else let x1=x0, x0=x; End If. Until4. Fixed-Point IterationTheorem: Suppose that is a real-value function, defined and for , continuous on a bounded closed interval [a, b], and let all . Then, there exists in [a, b] , such that .is called a fixed-point of the functionIf, in addition, withexits on (a, b) and a positive constant k<1 existsthen the fixed point in [a, b] is unique. Proof If g(a)=a or g(b)=b, then g has a fixed point. If not, then g(a)>a and g(b)<b. The function h(x)=g(x)-x is continuous on [a, b] withThen there exists point sincefor which.Is a fixedSuppose , in addition, that points in [a, b]. If , thenand that p and q are both fixedThus,which is a contradiction. So, p=q and the fixed point in [a, b] is unique. Fixed-point iteration To approximate the fixed point of and generate the sequence , we choose an initial approximation by letting . IfAnd a solution to iteration.is obtained, this technique is called fixed-pointThe figure illustrates the algorithm.Example(P54): (The roots are -1 and 3.) Form1:We will find if obtain the fixed point., also,Fig.. . Then we can usetoIt converges to 3. It converges to 3.Form2:We start the iteration withFig.It converges to -1. Form3:We start the iteration again withFig.It’s diverging.Theorem (Fixed-Point Theorem)Let in addition, that with be such that , for all x in [a, b]. Suppose, existsexists on (a, b) and that a constantThen, for any numberin [a, b], the sequence defined byConverges to the unique fixed point. We just need prove converges to the fixed point r.Order of ConvergenceDefinition Supposefor measuring how rapidly a sequence converges., withis a sequence that converges to and exists withfor all n. If positive constantthen Specially,converges toof order, with asymptotic error constant k.1 If 1 If,the sequence is linearly convergent. ,the sequence is quadratically convergent.SpeedSome Theory (P58)1. Convergence order of Fixed-point iterationSuppose that R is the true value of the root, we havewhere Then,is betweenand, and, since.Hence, if error constant, fixed-point iteration is linear convergent with asymptotic .2. Convergence order of Newton’s MethodAccording to the theorem, Newton’s iteration schemewill converge if .Then,and Now we expand, where R is the exact value of the root. as a Taylor series in terms of .wherelies within. So,.Hence,since . Also, is continuous and strictly bounded by K in the neighborhood of R, we have,That means if How about ?, Newton’s method is quadratically convergent.3. Convergence order of Secant MethodPizer (1975) shows that the order of convergence of the secant method isMultiple RootsDisadvantages of the methods we have described: 1. Do not work well for multiple roots. We will only get a slow convergence. 2. Imprecision. The curve is very flat near the multiple root—f’(x) will always be zero, that is, the program cannot distinguish which x-value is really the root.Remedies for Multiple Root with Newton’s MethodWe have discussed thatwhere, if R is the simple root. However, if f(x) has a root ofmultiplicity k at x=R, we havewhere Q(x) has no root R. Obviously,but.Does g’(R) still equal 0 ? We are not sure.We present a different formulation of Newton’s method:As before, at. We have,We see that, and now the new iteration converges quadraticallyat a multiple root. But we should know k in advance.Nearly Multiple RootsAccelerating Convergence Aitken accelerationSuppose there exists a constant K such thatwhere the error, R is the true value of the root. Then,Solving the equation givesWe will get a new sequenceby definingDefinition For a given sequence is defined by,the forward differenceHigher power, are defined byThis definition implies thatSo, the formula forcan be written asWe can proveThis means the sequence original sequence .converges more rapidly than does theExercise Give the algorithm of Aitken acceleration.Algorithm:is given as the initial approximation. Repeat SetUntil6. Nonlinear SystemIn fact, the situation of nonlinear system is more difficult.We note it to Giving an approximation. ,w to Taylor series at , we expand the , and select thefunction with many variables linear parts to obtainSolving the equation and noting it togives,Whereis the Jacobi matrix with the formExample Use Newton’s method forWe use the initial approximationto iterate, the result is:N 01 2 3 4 5 6x 1.0000000 2.1893260 1.8505896 1.7801611 1.7776747 1.7776719 1.7776719y 1.0000000 1.5984751 1.4442514 1.4244359 1.4239609 1.4239605 1.4239605z 1.00000000 1.3939006 1.2782240 1.2392924 1.2374738 1.2374711 1.2374711。