Iterative methods

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 (Crying).

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. (Headbang)

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.