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Math Help - Iterative schemes of the form x=g(x)

  1. #1
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    Iterative schemes of the form x=g(x)

    I've done all bar four questions for an assignment, which I'm stuck on; if anyone could help me or point me in the right direction then I'll be in your debt.

    Question 23

    The equation of ((e^x) - (x^5)) = 0.395 has a root around to 1.3. Rearrange this equation into the iterative schemes of the form of x = g(x). Find one arrangement that is convergent to the root close to 1.3 and one arrangement that does not converge to this region. Calculate the root close to 1.3 to 3 decimal places and for the second arrangement to complete at least six iterations.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Trapper Dave View Post
    I've done all bar four questions for an assignment, which I'm stuck on; if anyone could help me or point me in the right direction then I'll be in your debt.

    Question 23

    The equation of ((e^x) - (x^5)) = 0.395 has a root around to 1.3. Rearrange this equation into the iterative schemes of the form of x = g(x). Find one arrangement that is convergent to the root close to 1.3 and one arrangement that does not converge to this region. Calculate the root close to 1.3 to 3 decimal places and for the second arrangement to complete at least six iterations.
    One rearrangement is:

    x=(e^x-0.395)^{1/5}

    This has g'(x)<1 near 1.3 so this should converge there.

    Another rearrangement is:

    x=\ln(0.395+x^5)

    This has g'(x)>1 near 1.3 so this should diverge there.

    The first code block below shows the result of itterating the first of the
    above formulae and we see that the itteation converges to ~1.255

    Code:
    n    x_n     (exp(x)-0.395)^(1/5)
    ===================================
    0  2          1.475522452
    1  1.475522452 1.318074455
    2  1.318074455 1.272861115
    3  1.272861115 1.260020489
    4  1.260020489 1.256384493
    5  1.256384493 1.255355754
    6  1.255355754 1.255064759
    7  1.255064759 1.254982451
    8  1.254982451 1.254959171
    9  1.254959171 1.254952586
    10 1.254952586 1.254950724
    11 1.254950724 1.254950197
    12 1.254950197 1.254950048
    13 1.254950048 1.254950006
    14 1.254950006 1.254949994
    15 1.254949994 1.254949991
    16 1.254949991 1.25494999
    The following code block shows some itterations of the second formulae

    Code:
    n       x_n     ln(0.395+x^5)
    ===========================
    0  1.3         1.412919252
    1  1.412919252 1.796085993
    2  1.796085993 2.948962056
    3  2.948962056 5.409035881
    4  5.409035881 8.440439638
    5  8.440439638 10.66518121
    RonL
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  3. #3
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    I'm probably being dumb as per, but where did you get the value of 2 from to use as the value of x in the convergent?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Trapper Dave View Post
    I'm probably being dumb as per, but where did you get the value of 2 from to use as the value of x in the convergent?
    Just picked at random while experimenting, probably should have used 1.3.

    RonL
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