# Math Help - Iterative methods

1. ## Iterative methods

Hello,

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.

Sophie

2. 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.