數(shù)值分析第五版_李慶揚_王能超_易大義主編課后習題答案（numerical analysis of the fifth edition of li qingyang wang nengchao _ _ _ yi editor of afterschool rcise answers）

1、數(shù)值分析第五版_李慶揚_王能超_易大義主編課后習題答案（Numerical analysis of the fifth edition of Li Qingyang Wang Nengchao _ _ _ Yi editor of after-school exercise answers）The first chapter is introduction1., the relative error is the error.Solution: the relative error of the approximation isAnd the error isAnd then there is

2、The relative error of the 2. set is 2%, and the relative error is obtained.Solution: the condition number of a function isAgain,alsoAnd 23. the following numbers are approximate numbers obtained by four "Five", that is, the error limit is not more than half of the last one. Try to point ou

3、t how many effective digits they are:,Solution: five significant digits;Is the two significant digit;Is the four significant digit;Is the five significant digit;Is the two significant digit.4. use the formula (2.3) to find the error limits of the following approximations: (1), (2), (3)All of them ar

4、e given by third questions.Solution:5 to calculate the ball volume, the relative error is limited to 1. What is the relative error limit allowed when the radius of measurement is R?Solution: the volume of a sphere isWhat are the conditions of the function?alsoTherefore, the relative error limit is a

5、llowable when the radius of measurement is R6. set, according to recursive formula (n=1,2,.)Calculate to. If you take (5 significant digits), how much will the calculation be?Solution:.Followed by substitutionNamely,If taken,The error limit is.7. seek the two roots of an equation so that it has at l

6、east 4 bits of significant digits.Solution:,So the root of the equation should besoWith 5 significant digitsWith 5 significant digits8. when N is full, how do we ask for it?solutionSet up.be9., the square side length of about 100cm, how to measure to make the area error is not more than?Solution: th

7、e area function of a square is.At that time, if,beTherefore, when the length error limit is not more than 0.005cm, the error of the area is not exceeded10., suppose that G is accurate, and that the measurement of T has a second error, which proves that when t increases, the absolute error of S incre

8、ases, while the relative error decreases.Solution:When added, the absolute error increasesThe relative error decreases as the increase remains constant.11. sequence satisfies recurrence relation (n=1,2,. ),If (three significant digits), what is the time error? Is the calculation process stable?Solut

9、ion:alsoalsoThe time to arrival error is calculated, and the calculation process is unstable.12. calculate, take, use the following equation to calculate, which one obtains the result best?,.Solution: set up,If so, then.If the y value is calculated, thenIf the y value is calculated, thenIf the y val

10、ue is calculated, thenThe best results are obtained by calculation.13., the value of the request. If the square is opened with 6 bit function table, how much is the error when we ask for the logarithm? Use another equivalent formula.How much is the error in calculating the logarithm?solution,set upb

11、esoIf equivalent formula is usedbeRight now,The second chapter interpolation method1. at that time, the two interpolation polynomial.Solution:Then the two Lagrange interpolation polynomial isTable of values given in 2.X 0.4, 0.5, 0.6, 0.7, 0.8LNX, -0.916291, -0.693147, -0.510826, -0.356675, -0.22314

12、4Approximate values are computed by linear interpolation and two interpolations.Solution: by form,If the linear interpolation method is used, that is,beIf the two interpolation method is used,3. give the entire function table, step size, if the function table has 5 significant digits, we study the t

13、otal error bound of the approximate value by linear interpolation.Solution: the approximate value, the error can be divided into two parts, on the one hand, X is approximate, with 5 digits, have certain error propagation in the subsequent calculation process; on the other hand, the approximate value

14、 function by interpolation, linear interpolation method using the interpolation is not more than 0, there will be some errors. Therefore, the calculation of the total error bound should be integrated with the above two factors.Then,ordertakeorderbeAt that time, the linear interpolation polynomial wa

15、sInterpolation remainderWhen building a function table, the data in the table has 5 bits of significant digits, and there is an error propagation process in the calculation.The total error bound is4. set for each other node, verify:(1)(2)Prove(1) OrderIf the interpolation node is, then the quadratic

16、 interpolation polynomial of the function is.Interpolation remainderalsoFrom the above question, we can see that.Acquired evidence.5 set and verify:Solution: as the interpolation node, the linear interpolation polynomial is=Interpolation remainder6. on the given equidistant node function table, if t

17、he approximate value of two interpolation is used, if the truncation error is not exceeded, how long should we use the function table's step size H?Solution: if the interpolation nodes are sum, then the interpolation remainder of the piecewise polynomial interpolation is twoSet the step size to

18、h, i.e.If the truncation error is not exceeded, then7. if,The solution is based on the definition of forward difference operator and central difference operator.8. if it is a m degree polynomial, remember that the k order difference is a sub polynomial, and (is a positive integer).Solution: the expa

19、nsion of a function isamongPolynomial againOrder polynomialOrder polynomialThus, the process is recursive polynomialIs constantWhen a positive integer,9. proofProveTo permit10. proofProof: it can be seen from the above conclusionAcquired evidence.11. proofProveAcquired evidence.12., if there is a di

20、fferent root,Prove：Proof: there is a distinct solid rootAndorderbeandorderbealsoAcquired evidence.The 13. order difference has the following properties to prove:(1) if, then(2) if, thenProve：(1)Acquired evidence.+Acquired evidence.14. seeking.Solution:ifbe15. two point three times more than that of

21、Hermite interpolation isSolution:If the interpolation polynomial satisfies the conditionInterpolation remainderThe interpolation condition shows thatAndCan be writtenWhich is about the undetermined function,Now consider as a fixed point as a functionAccording to the nature of the remainder, there ar

22、eBy Rolle's theorem, existence andThat is, there are four separate zeros in the upper.According to Rolle's theorem, there is at least one zero point between the two zeros,So there are at least three separate zeros,And so on, there's at least one zero point.Remember to makealsoWhich depen

23、ds onSection three Hermite interpolation, if the node is set up, step, i.e.On the cell16., a number of times not more than 4 times the polynomial P (x), so that it satisfiesSolution: the use of Hermite interpolation can get the number of not more than 4 polynomialset upAmong them, A is the undetermi

24、ned constantthus17., in the upper, the piecewise linear interpolation function is calculated according to the equidistant nodes, and the midpoint and the value of each node are calculated, and the error is estimated.Solution:ifThen step lengthOn the cell, the piecewise linear interpolation function

25、isThe values at the midpoint between the nodes areThen,Then,Then,Then,Then,erroralsoorderThe stationary points are as follows18. seek the piecewise linear interpolation function and estimate the error.Solution:On the interval,The function is piecewise linear interpolation functions on cellsError is1

26、9. for in Hermite interpolation, and estimation error.Solution:On the interval,orderThe piecewise Hermite interpolation function in the interval forError isalso20. the given data table is as follows:Xj 0.25, 0.30, 0.39, 0.45, 0.53Yj 0.5000, 0.5477, 0.6245, 0.6708, 0.7280Try to find the three spline

27、interpolation and satisfy the condition:Solution:The resulting equations in the form of matrices are21 M02 M12 M22 M312 M4The solution of this equation is obtainedThe three spline expression isWill be replaced byThe resulting equations for the start of the matrix areThe solution of this equation is

28、obtainedThe three spline expression isWill be replaced by21. if the three spline function is proved:If the formula is an interpolation node, thenProve：Thus thereThe third chapter, function approximation and curve fitting1,The Bernstein polynomials are given.Solution:Bernstein polynomialamongThen,The

29、n,2. at that time, the confirmationProve：If so, then3. prove that the function is linearly independentProve：ifThe inner product of the upper right and the upper ends of the upper two ends are obtained respectivelyThe coefficient matrix of this equation is Hilbert matrix, and the symmetric positive d

30、efinite is nonsingular,Only zero solution a=0.Function linearly independent.4. Calculate the following functions:M and N are positive integers,Solution:If so, thenMonotonically increasingIf so, thenIf M and N are positive integersThen,Then,Monotonically decreasingThen,Monotonically decreasing.ifThen

31、,Monotonically decreasing.5. ProveProve：6. Yes, definitionAsk if they are inner products.Solution:Order (C is constant, and)beandThis is when and only then, contradictoryAn inner product that cannot be constructed.If so, thenThen,If so, thenAnd,That is, and only thenTherefore, the inner product can

32、be constructed.7. The test proof is an orthogonal polynomial with upper weight.Solution:If so, thenOrder, and thenChebyshev polynomials are weighted orthogonal in the interval, andThe orthogonal polynomials are weighted on the upper right.also8. For the weight function, the orthogonal polynomial of

33、the first term coefficient is 1Solution:If so, the inner product on the interval isDefinitionamong9. It is proved that the class of Chebyshev polynomial family given by the teaching material formula is an orthogonal polynomial with second weights.Prove：ifOrders are availableThen,Then,Again, soAcquir

34、ed evidence.10. It is proved that Chebyshev polynomials satisfy differential equationsProve：Chebyshev polynomialThus thereAcquired evidence.11. Suppose that in the upper continuous, the zeroth best uniform approximation polynomial is obtainedSolution:Continuous on closed intervalExist, maketakeThe s

35、um is the deviation of the 2 from "positive" and "negative".By the Chebyshev theoremZero order best uniform approximation polynomial of P.12. Select the constant to make it minimal, and ask if the solution is uniqueSolution:orderThe upper is odd functionThe highest term coefficie

36、nt is 1, and the polynomial is 3 times.The minimum deviation is 0.Thus there13. Seek the best approximation polynomial at the upper one and estimate the error.Solution:Thus, the best approximation polynomial at one time isThat isError limit is14. The best approximation polynomial at the upper.Soluti

37、on:Thus, the best approximation polynomial at one time is15. Finding the best uniform approximation polynomial of three times on the interval.Solution:OrderAndOrderIf the interval is the best three approximation, the polynomial should be satisfiedWhenWhen,The deviation between the polynomial and the

38、 zero is minimumThen, the three best uniform approximation polynomial is the three best uniform approximation polynomial16. On the best square approximation polynomial.Solution:ifAnd thenThe normal equations areSolutionSo the best square approximation polynomial is.17. Finding the best approximation

39、 polynomial for a function on a given interval:Solution:ifAnd there areThe normal equations areThereby solvingSo the best square approximation polynomial is.ifAnd there areThe normal equations areThereby solvingSo the best square approximation polynomial is.ifAnd there areThe normal equations areThe

40、reby solvingSo the best square approximation polynomial is.ifAnd there areThe normal equations areThereby solvingSo the best square approximation polynomial is18. On the basis of the Legendre polynomial expansion, the three best square approximation polynomial is obtained.Solution:According to Legen

41、dre polynomial expansionbeThus the three best square approximation polynomial is19. To observe the rectilinear motion of an object, the following data are obtained:Time t (s) 0.9, 1.9, 3, 3.9, 5Distance s (m) 010305080110Equation of motion.Solution:The motion distance of the observed object is appro

42、ximately linear with the motion time, and the linear equation is chosenorderbeThe normal equations areThereby solvingSo the equation of motion of an object is20. The experimental data are as follows:192531384419, 32.3, 49, 73.3, 97.8Using the least square method to obtain the empirical formula of th

43、e shape, and calculate the mean square error.Solution:If so, thenbeThe normal equations areThereby solvingsoThe mean square error is21. In a temple in the reaction by the solution concentration and the relationship between the scores of experiment time are as follows:time05101520253035, 40455055conc

44、entration1.27, 2.16, 2.86, 3.44, 3.87, 4.15, 4.37, 4.51, 4.58, 4.62, 4.64Least square method.Solution:Observe the characteristics of the data given; adopt equationsBoth sides take the logarithm at the same timetakebeThe normal equations areThereby solvingtherefore22. Given a record, the discrete spe

45、ctrum is calculated by FFT algorithm.Solution:be01234567 43210123 44440484 04801600023, the division will be continued fractions.solution24. In order for the Pade approximation.Solution:From Taylor at the sitehave tothusThat isThereby solvingalsobeso25. In order for the Pade approximation.Solution:F

46、rom Taylor at the sitehave tothusThat isSolutionalsobesoThe fourth chapter is numerical integration and numerical differentiation1. determine the specific parameters in the following quadrature formulas, make their algebraic accuracy as high as possible, and show the algebraic accuracy of the constr

47、ucted quadrature formula:Solution:To solve the algebraic precision of quadrature formula, according to the definition of algebraic precision, namely quadrature formula for the polynomial times of less than m can be accurately established, but for the m+1 polynomial is not accurate, to verify the sol

48、ution.(1) ifOrderOrderOrderThereby solvingOrderSo founded.OrderSo when,soHaving 3 algebraic accuracy.(2) ifOrderOrderOrderThereby solvingOrderSo founded.OrderSo when,Therefore,Having 3 algebraic accuracy.(3) ifOrderOrderOrderThereby solvingorOrderIt does not hold water.Therefore, the original quadra

49、ture formula has 2 algebraic accuracy.(4) ifOrderOrderOrderIt isOrderOrderSo when,Therefore,Having 3 algebraic accuracy.2. the following integrals are calculated by using the trapezoidal formula and the Simpson formula, respectively:Solution:The complex trapezoid formula isThe complex Simpson formul

50、a isThe complex trapezoid formula isThe complex Simpson formula isThe complex trapezoid formula isThe complex Simpson formula isThe complex trapezoid formula isThe complex Simpson formula is3. Direct verification of Cortes's textbook formula (2. 4) has 5 intersection algebraic precision.Prove：Co

51、tes formulaOrderOrderOrderOrderOrderOrderOrderTherefore, the Cotes formulas with algebraic accuracy of 5 degrees.4. Using Simpson formula to find the integral and estimate the error.Solution:Simpson formulaRight now,Thus thereError is5. Derive the following three rectangular quadrature formulas:Prov

52、e：The points are scored on both sides at the same timeThat isThe points are scored on both sides at the same timeThat isTwo points on the same side at the same timeThat is6. If the integral trapezoid formula is used to calculate the integral and the number of segments should be equal, the truncation

53、 error can not be exceeded If the Simpson formula is used to achieve the same accuracy, how many equal fractions should be equal?Solution:When the complex trapezoid formula is adopted, the remainder term isalsosoIf so, thenWhen the interval is equal,It isTherefore, when the interval 213 is equal, the error requirement can be metIn the use of the complex Simpson formula, the rema