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Engineering Mathematics Test 11

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Engineering Mathematics Test 11
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  • Question 1
    1 / -0

    The following equation needs to be numerically solved by using Newton – Raphson method.

    x log10 x – 1.2 = 0

    The iterative equation for this purpose is (n indicates the iteration level)
    Solution

    Newton – Raphson Method:

    xn+1=xnf(xn)f(xn)

    Here f(x) = x log10 x – 1.2

    ddx(log10x)=1xlog10e

    f(x)=x.(1xlog10e)+log10x=log10e+log10x

    f’(x) = 0.43429 + log10 x

    xn+1=xnf(xn)f(xn)=xnxnlog10xn1.20.43429+log10xn

    xn+1=0.43429xn+xnlog10xnxnlog10xn+1.20.43429+log10xn

    xn+1=0.43429xn+1.20.43429+log10xn
  • Question 2
    1 / -0
    what is the type of convergence of Secant method
    Solution

    order of convergence is linear for bisection method,successive approximation method and Regula falsi method.

    super linear for secant method

    quadratic for newton rapson method

  • Question 3
    1 / -0
    In case of Newton-Raphson algorithm, the type of convergence is
    Solution

    xn+1=xnf(xn)f(xn)

    Suppose xn differs from the root α by a small quantity ϵn so that x0 = α + ϵn and xn+1 = α + ϵn+1  

    Now the above equation will become

    α+εn+1=α+εnf(α+εn)f(α+εn)εn+1=εnf(α+εn)f(α+εn)

    By Taylor’s expansion

    =nf(α)+εnf(α)+12!εn2fα+f(α)+εnf(α)+f(α)=0,=εnεnf(α)+12!εn2fα+f(α)+εnf(α)+

     Neglecting third and higher powers of ϵn

    =εn2f(α)2[f(α)+εnf(α)]εn22.f(α)f(α)

    This shows that the subsequent error at each step is proportional to the square of the previous error and as such the convergence is quadratic.
  • Question 4
    1 / -0
    Newton-Raphson method is to be used to find the root of equation 4x – e2x + sin x = 0. If the initial trial value for the root is taken as 0.5, the next approximation for the root would be _______ (corrected up to three decimal places)
    Solution

    Concept:

    According to Newton-Raphson Method

    Xn+1=Xnf(Xn)f(Xn)

    Calculation:

    f(x) = 4x – e2x + sin x = 0

    f’(x) = 4 – 2e2x + cos x

    f(x0) = - 0.23885

    f’(x0) = -0.55898

    x1=x0f(x0)f(x0) 

    =0.5(0.238850.55898) 

    = 0.0727

  • Question 5
    1 / -0
    If the differential equation dydx=x+2y,y(0)=0 is solved using the Euler’s method with step size h = 0.25, then y2 is equal to ______ (up to three decimal places).
    Solution

    Concept:

    To use Euler's Method to generate a numerical solution to an initial value problem of the form:

    y=f(x,y),y(x0)=y0

    We decide upon what interval, starting at the initial condition, we desire to find the solution. We chop this interval into small subdivisions of length h. Then, using the initial condition as our starting point, we generate the rest of the solution by using the iterative formulas:

    xn+1=xn+h

    yn+1=yn+hf(xn,yn)

    Application:

    Given differential equation is,

    dydx=x+2y 

    y(0) = 0

    y0 = 0, x0 = 0

    h = 0.25

    First iteration:

    y0 = 0, x0 = 0, h = 0.25

    y’(x0, y0) = 0

    y(0.25) = y1 = y0 + h y’(x0, y0) = 0

    x1 = x0 + h = 0 + 0.25 = 0.25

    Second iteration:

    y1 = 0, x1 = 0.25, h = 0.25

    y’(x1, y1) = 0.25

    y2 = y1 + hy’(x1, y1)

    = 0 + 0.25 (0.25) = 0.0625

    x2 = x1 + h = 0.5

    y2 = 0.0625

  • Question 6
    1 / -0
    The Newton-Raphson iteration Xn+1=13(2xn+NXn2) can be used to compute
    Solution

    Concept:

    According to Newton-Raphson Method

    Xn+1=Xnf(Xn)f(Xn)

    Calculation:

    The given Newton-Raphson iterative formula is,

    Xn+1=13(2xn+NXn2)

    Xn+1=13(2Xn3+NXn2) 

    Let us suppose the formula converges to the root after n iterations

    Then Xn = Xn+1 = x (root)

    The given iterative formula becomes

    x=13(2x3+Nx2) 

    3x3 = 2x3 + N

    x=N3 

    Therefore, the given iterative formula is used to find N3

  • Question 7
    1 / -0
    The second approximation to a real root of the equation x3 – 2x – 5 = 0 by method of false position between 2 and 3 is ______
    Solution

    f(x) = x3 – 2x – 5 = 0

    a = 2, b = 3

    f(a) = f(2) = -1

    f(b) = f)3) = 16

    The root lies between 2 and 3.

    First iteration:

    x1=af(b)bf(a)f(b)f(a)=(2)(16)(3)(1)16(1) 

    = 2.0588

    f(x1) = -0.39 < 0

    The root lies between x1 = 2.0588 and b = 3

    Second iteration:

    a = 2.0588     b = 3

    x2=(2.0588)(16)3(0.39)16(0.39) 

    = 2.081

  • Question 8
    1 / -0
    Only one of the real roots of f(x) lies in the interval 2 ≤ x ≤ 4 and bisection method is used to find its value. The minimum number of iterations required to achieve an accuracy of 0.2% is ______
    Solution

    Concept:

    In Bisection method, for f(x) = 0 in the interval (a, b), the interval width is reduced by a factor of one one-half at each step, at the end of the nth step, the new interval will be [an, bn] of length |ba|2n.

    The error, ε|ba|2n 

    Calculation:

    In the given question,

    a = 2, b = 4

    ε = 0.2% = 0.002

    0.002|42|2n 

    0.00222n 

    2n ≥ 1000

    n ≥ 10

    The minimum number of iterations required are, n = 10

  • Question 9
    1 / -0
    An equation sinx=1x is required to be solved by the Bisection method where x lies between 1 and 1.5 (x is in radian). The approximate root of the fourth iteration will be: (Correct upto four decimal places)
    Solution

    f(x) = x sin x – 1

    f(1) = -0.158529 < 0

    f(1.5) = 0.496242 > 0

    So root will lie between 1 & 1.5

    x1=12(1+1.5)=1.25

    f(x1) = 0.18623 > 0

    so root lies between 1 & 1.25

    x2=12(1+1.25)=1.125

    f(x2 ) = 0.01505 > 0

    So root lies between 1 & 1.125

    x3=1+1.252=1.0625

    f(x3) = -0.0718 < 0

    f(x2) > 0 & f(x3) < 0

    So root lies between x2 & x3

    x4=x2+x32=1.125+1.06252=1.09375
  • Question 10
    1 / -0
    Consider the first order initial value problem y’ + y = 0, y(0) = 1 for x = 0.1, the solution obtained using a single iteration  of the third order Runge Kutta method with step-size h = 0.1 is _________.
    Solution

    Concept:

    dydx=f(x,y),y(x0)=y0

    Runge-Kutta Method of first order:

    k1 = h f(x0, y0)

    y1 = y0 + k where k = k1 = h f (x0, y0)

    y2 = y1 + k where k = h f (x1, y1)

    Runge-Kutta Method of Second order:

    k1 = h f(x0, y0)

    k2 = h f (x0 + h, y0 + k1)

    k=k1+k22

    y1 = y0 + k

    Runge-Kutta Method of Third order:

    k1 = h f (x0, y0)

    k2=hf(x0+h2,y0+k12)k=hf(x0+h,y0+k1)  

    k3=hf(x0+h,y0+k)k=16(k1+4k2+k3)  

    y1 = y0 + k

    Runge-Kutta Method of fourth order:

    k1 = h f (x0, y0)

    k2=hf(x0+h2,y0+k12)k3=hf(x0+h2,y0+k22)  

    k4=hf(x0+h,y0+k3)k=16(k1+2k2+2k3+k4)  

    y1 = y0 + k

    Calculation:

    Question is about third order Runge Kutta method with step-size h = 0.1.

    dydx=y=f(x,y);x0=0,y0=1

    f(x0, y0 ) = f(0, 1) = -1

    k1 = h f (x0, y0) = 0.1 × (-1) = -0.1

    k2=hf(x0+h2,y0+k12)=hf(0.05,0.95)

    = 0.1 [-0.95] = -0.095

    k’ = h f (x0 + h, y0 + k1) = h f (0.1, 0.9) = =0.9 h

    k’ = -0.09

    k3 = h f (x0 + h, y0 + k’) = h f (0.1, 0.91) = -0.91 h

    k3 = -0.091

    k=16(k1+4k2+k3)=16(0.1+4(0.095)+0.091)=0.095167  

    y1 = y0 + k = 1 - 0.095167 = 0.9
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