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Thread: Iterative methods

  1. #1
    Feb 2008

    Iterative methods


    I am not a math major but an ocean scientist. I am trying to solve an equation for wave motion under water and I need help.

    I would like to use Matlab to solve for a variable using an iterative approach but I have to understand what that really means before I do that. The equation is:

    Where z, v, and U are known integers, * = multiply, / = divide

    I would like to find u_star,

    U = (5.5/u_star) + (2.5*(ln((u_star*z)/v)))/u_star

    Do you have any suggestions about which iterative method would work best?
    Do you have any suggestions about where to go to find a SIMPLE explanation about the application of an iterative method online?
    Should I not do this because it is really complicated?

    Thank you for reading my post and thank you for any help/suggestions you have to offer, they are greatly appreciated.

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  2. #2
    Junior Member pinion's Avatar
    Jan 2008
    i'm not sure what matlab is, but iterative means you check many values for a variable (your u\_star). it's a guess and check thing. the difficulty is that u\_star is not necessarily an integer .

    say you had the equation 2^y = 3^x, and you knew what x was. say in this case x = 5. a simple iterative method of finding the value of y would be to replace y with 0, then 1, then 2, and so on, until the value of 2^y is greater than 3^5 ( 243). in this case, we would find y to be between 7 and 8.

    knowing that 7 < y < 8 is not much of an answer, right? so now is when it gets a little harder.

    at this point, you could do a binary search - which means you would guess 7.5, because that's right between 7 and 8. so, 2^{7.5} = 181.blahblah, which is less than 3^5 ( 243). now you would guess y right between 7.5 and 8, so guess y = 7.75. 2^{7.75} is still less than 243, so you would guess between 7.75 and 8, which is 7.875.... you would keep doing this until the error between 2^y and 3^5 was minimal. you get to choose how much error is acceptable.

    also, in the case of 2^y = 3^5, you can get an answer algebraically, which is y=\frac{ln(3^5)}{ln(2)}, however, in the case of your equation, i don't see an immediate algebraic solution. if you do find one, i would love to see it.

    if this is confusing or if you would like more explanation about a binary search feel free to ask questions. a binary search is just one iterative method you might use, i'm sure you could find a better (faster) way.
    Last edited by pinion; Feb 11th 2008 at 12:11 PM.
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