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When the FORTRANversion of the program Bairstow is run, the response on screen is:When the ITERATIONS column indicates 1, it signals that the quotient is oforder one or two. In this case, the quotient is Q(x) = x + 1.65063. In fact, no iterationhas been performed for solving Q(x).MATLAB APPLICATIONMATLAB has a file called roots.m which can be applied to find the roots of apolynomial p(x) = 0. To do so, the coefficients of an nth-order p(x) should be orderedin descending powers of x into a row matrix of order n + 1. For example, to solvep(x) = x3 + 2x2 + 3x + 4 = 0, we enter:>> p = [1,2,3,4]; x = roots(p)and obtain a screen displayAs a second example of solving x4–5x3 + 13x2–19x + 10 = 0, MATLAB interactive entries indicated by the leading >> signs and the resulting display are:© 2001 by CRC Press LLCComparing the two examples, we notice that by placing ; after a statementsuppresses the display of the computed value(s).
The elements of the first p matrix(a single row) is not displayed!It is of interest to introduce the plot capability of MATLAB by use of the resultspresented above which involve a polynomial P(x) and its roots. From graphicalviewpoint, the roots are where the polynomial curve crossing the x-axis. MATLABhas a plot.m file which can be readily applied here. Let us again consider thepolynomial P(x) = x4–5x3 + 13x2–19x + 10 = 0 and plot it for 0≤x≤3. For adequatesmoothness of the curve, an increment of x equal to 0.1 can be selected for plotting.The interactive MATLAB commands entered for obtaining Figure 4 are:>> p=[1,–5,13,–19,10]; x=[0:0.1:3]; y=polyval (p,x);>> plot(x,y), hold>> XL=[0 3]; YL=[0 0]; plot(XL,YL)Notice that another m file polyval of MATLAB has been employed above. Thestatement y = polyval(p,x) generates a array of y values using the polynomial definedby the coefficient vector p and calculated at the values specified in the x array.
Thehold statement put the current plot “on hold” so that an additional horizontal lineconnecting the two points defined in the XL and YL arrays can be superimposed.The first plot statement draws the curve and axes and tic marks while the secondplot statement draws the horizontal line.The horizontal line drawn at y = 0 help to show the intercepts of the polynomialcurve and the x-axis, by observation near x = 1 and x = 2 which confirm the resultfound by the MATLAB file roots.m.MATHEMATICA APPLICATIONSFor finding the polynomial roots, Mathematica’s function NSolve can beapplied readily.Keyboard input (and then press shift and Enter keys)NSolve[x^3 + 2x^2 + 3x + 4 = = 0,x]The Mathematica response is:Input[1]: =NSolve[x^3 + 2x^2 + 3x + 4 = = 0,x]© 2001 by CRC Press LLCFIGURE 4.Output[1] ={{x -> –1.65063}, {x -> –0.174685 — 1.54687 I},{x -> –0.174685 — 1.54687 I}}Keyboard input (and then press Shift and Enter keys)NSolve[x^4–5x^3 + 13x^2–19x + 10 = = 0,x]The Mathematica response is:Input[2]: =NSolve[x^4–5x^3 + 13x^2–19x + 10 = = 0,x]Output[2] ={{x -> 1.
– 2. I}, {x -> –1 + 2. I}, {x -> 1.}, {x -> 2.}}To show the locations of the roots of a polynomial, Mathematica’s functionPlot can be applied to draw the polynomial. The following statements (Keyboardinput will hereon be omitted since it is always repeated in the Input response) enableFigure 5 to be generated:© 2001 by CRC Press LLCInput[3]: =Plot[x^4–5x^3 + 13x^2–19x + 10,{x,0,3},Frame->True, AspectRatio->1]Output[3] =FIGURE 5.Notice that {x,0,3} specifies the range of x for plotting, Frame->True requests thatthe plot be framed, and AspectRatio-> requests that the scales in horizontal and verticaldirections be equal. The graph clearly shows that there are two roots at x = 1 and x = 2.3.5 PROBLEMSFINDROOT1.
A root of F(x) = 3x–2e0.5x = 0 is known to exist between x = 1 and x = 2.Calculate the guessed locations of this root twice by application of thelinear interpolation method.2. A root is known to exist between x = 0 and x = 1 for the polynomialP(x) = x3–4.5x2 + 5.75x–1.875 = 0 because P(x = 0) = –1.875 and P(x =1) = 3.75. What will be the next two guessed root values if linear interpolation method is used? Show details of your calculation.3.
A root is known to exist between x = 1 and x = 2 for the polynomial x3+ 0.5x2 + 3x–9 = 0.Based on the linear interpolation method, make two successive guessesof the location of the root. Show details of the calculations.© 2001 by CRC Press LLC4. For finding a root of the polynomial x3–8.9x2–21.94x + 128.576 = 0 withinthe bounds x = 0 and x = 4, the linear interpolation method is to beapplied. Show only the details involved in computation of two successivetrial guesses of the root.5. Use the Newton-Raphson iterative method to find the root of 2X3–5 = 0between X = 1 and X = 2.6. Complex roots of a polynomial can be calculated by application of theprogram FindRoot simply by treating the variable X in the polynomialF(X) as a complex variable.
Using a complex number which has a realpart and an imaginary part as an initial guess for X to evaluate F(X) andits derivatives, both values will also be complex. The Newton-Raphsoniterative process is to be continued until both the real and imaginary partsof F(X) are sufficiently small. According to this outline, modify programFindRoot to generate a new program NewRaphC for determining acomplex root for the polynomial X4 + 5X2 + 4 = 0.7.
In solving eigenvalue problems (see programs CharacEq and EigenODE), the characteristic equation of an engineering system is in the formof a polynomial. Physically, the roots of this polynomial may have themeaning of frequency, or, buckling load, or others. In the program EigenODE, a vibrational problem leads to a characteristic equation of 3–502 + 600 – 1000 = 0. Apply the program FindRoot to find a root betweenλ equal to 1 and 2 accurate to three significant figures. This root representsthe lowest frequency squared.8. Apply the Newton-Raphson method to find a root of the polynomial f(x) =3x3 + 2x2–x–30 = 0 by first guessing it to be equal to 3.0.
Carry out twoiterative steps by hand calculation to obtain the adjustments that need tobe made in guessing the value of this root.9. Apply the program FindRoot to solve Problem 8 given above.10. Apply the linear interpolation method to find a root of the polynomialf(x) = 3x3 + 2x2–x–30 = 0 between x = 1 and x = 3. Carry out two iterativesteps by hand calculation to obtain the new bounds.12. The well known secant formula for column bucking3 relating the averageunit load P/A to the eccentricity ratio ec/r2 is:{ () [P A = σ max 1 + ec r 2 sec ( L r )( P EA) 1 2 2]}where σmax is the proportional limit of the column, L/r is the slendernessratio, and E is Young’s modulus of elasticity. Solve the above transcendental equation by using σmax = 620 MPa and E = 190 GPa to find P/Afor ec/r2 = 0.1 and L/r = 20.13.
Solve the friction factor f from the Colebrook and White equation6 forthe flow in a pipe (1/f)1/2 = 1.74–0.868{(2K/D) + [18.7/Re(f)1/2]} whereRe is the Reynold’s number and K/D is the relative roughness parameter.Plot a curve of f vs. Re, and compare the result with the Moody’s diagram.14. Find the first five positive solution of the equation XJ0(X)–2J1(X) = 0where J0 and J1 are the Bessel functions of order 0 and 1, respectively.7© 2001 by CRC Press LLC15. Write a program SucceSub for implementing the successive substitutionmethod and apply it to Equation 5 for solving the angle γ for α changingfrom 0° to 360° in equal increment of 15°.16. Apply the program SucceSub to solve Problem 1.17. Apply the program SucceSub to solve Problem 12.18. Rise time is defined as the time required for the response X(t) to increaseits value from 0.1 to 0.9, referring to Equation 8 and Figure 1.
For asecond-order system with a1 = 1, a2 = 0.2 sec–1, a3 = 1 sec–1, and a4 = 1.37radians, use Equation 8 to calculate the rise time of the response X(t) byapplying the computer program FindRoot.19. Write a m file for MATLAB and name it FindRoot.m and then apply itfor solving Problem 12.20.
Apply FindRoot.m for solving Problem 14.21. Apply FindRoot.m for solving Problem 18.22. Apply Mathematica to solve Problem 12.23. Apply Mathematica to solve Problem 14.24. Apply Mathematica to solve Problem 18.NEWRAPHG1. Shown below are two ellipses which have been drawn using the equations:2© 2001 by CRC Press LLC22 x + y =1 28 24 and2 x′ y′ +=1 9.5 32 where the coordinate axes x and y are the result of rotating the x and yaxes by a counterclockwise rotation of = 30°.
The new coordinates canbe expressed in term of x and y coordinates as:x′ = x cos θ + y sin θandy′ = − x sin θ + y cos θUse an appropriate pair of values for xP and yP as initial guesses foriterative solution of the location of the point P which the two ellipsesintercept in the fourth quadrant by application of the program NewRaphG.2. The circle described by the equation x2 + y2 = 22 and the sinusoidal curvedescribed by the equation y = sin2x intercepted at two places as shownbelow.
This drawing obtained using the MATLAB commandaxis(‘square’) actually is having a square border when it is shown onscreen but distorted when it is printed because the printer has a differentaspect ratio. Apply the Newton-Raphson iterative method to find theintercept of these two curves near (x,y) = (2,- 0.5).© 2001 by CRC Press LLC3. Apply program NewRaphG for finding a root near X = 0.4 and Y = 0.6from the equations SinXSinY + 5X–7Y = –0.77015 and e0.1X-X2Y + 3Y =2.42627. The solutions should be accurate up to 5 significant digits.4. If one searches near x = 2 and y = 3 for a root of the equations f(x,y) =(x–1)y + 2x2(y–1)2–35 and g(x,y) = x3–2x2y + 3xy2 + y3–65 what shouldbe the adjustments of x and y based on the Newton-Raphson method?5. Write a MATLAB m file and call it NewRaph2.m and then apply it tosolve Problems 1 to 4.6. Apply Mathematica to solve Problems 1 to 4.BAIRSTOW1. Is x2–x + 1 a factor of 4x4–3x3 + 2x2–x + 5? If not, calculate the adjustments for u and v which are equal to –1 and 1, respectively, based on theBairstow’s method.2.
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