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Math Help - Using numerical methods to solve problems

  1. #1
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    Using numerical methods to solve problems

    Question One.
    Use an appropriate method to find the root of x + x + 1 = 0 correct to 2 decimal places. Start with initial interval -0.8 and -0.6. Show all working.

    Question Two.
    Human blood pressure P (in mm of mercury) varies periodically with time (t in seconds). The first time during the first second of a cycle when the pressure P is 90 mm can be expressed by:
    80 + 300t - 2700t = 90
    or:
    f(t) = 2700t - 300t + 10 = 0

    Take an initial value of t = 0.05 seconds and use a method different to Question One to find a solution to f(t) = 0 accurate to 4 decimal places.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Kiwigirl
    Question One.
    Use an appropriate method to find the root of x + x + 1 = 0 correct to 2 decimal places. Start with initial interval -0.8 and -0.6. Show all working.
    Let:

    f(x)=x^3+x+1.

    We wish to find a zero of f in the interval [-0.8,-0.6]. We know it has a zero
    in this interval as it changes sign between -0.8 and -0.6.


    Here we are going to use binary chop, In the following tableau we have
    in each row a x_1 lower limit of the interval containing the zeros, f_1 the
    value of the function at the lower limit, x_2 the upper limit of the interval
    containing the zeros, f_2 the value of the function at the upper limit, x_{mid}
    the mid-point of the interval and f_{mid} the value of the function at the mid=point.

    To obtain the following row from a row the value of the upper or lower limit
    of the interval is replaced by the mid-point according to the sign of the
    function at the mid-point. This process is continued until the upper and
    lower limits of the interval are equal to two significant digits.


    \begin{array}{|c|c|c|c|c|c|}<br />
\hline x_1 & f_1 & x_2 & f_2 & x_{mid} & f_{mid}\\ \hline<br />
-0.8 & -0.312 & -0.6 & 0.184 & -0.7 & -0.043\\<br />
-0.7 & -0.043 & -0.6 & 0.184 & -0.65 & 0.074\\<br />
-0.7 & -0.043 & -0.65 & 0.074 & -0.675 & 0.175\\<br />
-0.7 & -0.043 & -0.675 & 0.175 & -0.6875 & -0.0124\\<br />
-0.6875 & -0.0124 &-0.675 & 0.175&-0.681&0.0026\\<br />
-0.6875 & -0.0124 &-0.681&0.0026&-0.684&-0.0046\\<br />
-0.684&-0.0046&-0.681&0.0026&\ & \ \\ \hline<br />
\end{array}

    So the zero we seek is x\approx -0.68.

    RonL
    Last edited by CaptainBlack; May 5th 2006 at 11:53 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Kiwigirl
    Question Two.
    Human blood pressure P (in mm of mercury) varies periodically with time (t in seconds). The first time during the first second of a cycle when the pressure P is 90 mm can be expressed by:
    80 + 300t - 2700t = 90
    or:
    f(t) = 2700t - 300t + 10 = 0

    Take an initial value of t = 0.05 seconds and use a method different to Question One to find a solution to f(t) = 0 accurate to 4 decimal places.
    Here we will use Newton-Raphson itteration. This generates a new estimate
    for the root from an old estimate using the relation:

    <br />
t_{new}=t_{old}-\frac{f(t_{old})}{f'(t_{old})}<br />


    <br />
\begin{array}{|c|c|c|c|}\hline <br />
t_{old} & f(t_{old}) & f'(t_{old}) & t_{new} \\ \hline<br />
0.05 & -4.66 & -279.8 & 0.033333\\<br />
0.03333 & 0.10 & -291.0 & 0.033677\\<br />
0.033677 & 0.000025&-290.8 & 0.033677\\ \hline\end{array}<br />

    So to four significant digits 0.033677 is the required root.

    RonL
    Last edited by CaptainBlack; May 5th 2006 at 12:34 PM.
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