Hello all,

I have a complex function

http://www.mathhelpforum.com/math-he...4574d9bd53.png

How can I prove that it has a minimum at certain point.

Can anyone give me some suggestions for finding the minimum.

Thanks

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- Jul 20th 2010, 04:29 AMbjkrishnaFinding minima of a complex function
Hello all,

I have a complex function

http://www.mathhelpforum.com/math-he...4574d9bd53.png

How can I prove that it has a minimum at certain point.

Can anyone give me some suggestions for finding the minimum.

Thanks - Jul 20th 2010, 04:38 AMAckbeet
I have a number of questions:

1. Have you tried the usual Calc I method?

2. What is D(s)?

3. What is d(s)?

4. What are any constraints on any of the parameters, especially the domain for x, and the allowed values of m? (We can see at a glance that 1 is not in the domain.)

5. What does it mean to sum over D(s)? Do you mean there's a dummy variable running over the values in D(s)? - Jul 21st 2010, 08:38 PMbjkrishna
1.I have differentiated the equation.

But i need not get the extreme point.I need to prove that there exists a global minimum for the function.

2.D(s) is just a set representation.there are no variables in D(s)

3.d(s) for particular set of node.

4.m > 0 and $\displaystyle P_w > 0 $

5.the value $\displaystyle \frac{1}{1-P_w}$ sums for all the nodes in D(s)

Thanks - Jul 22nd 2010, 02:00 AMAckbeet
Your answers are a bit mystifying.

1. In the OP, you asked how you can prove the function has a minimum at a certain point (this means you're talking about a local minimum). In your latest post in this thread, you're asking how you can show that the function has a global minimum. Which is it?

2. Are you saying that the summation variable for $\displaystyle D(s)$ does not show up in your expression anywhere? If it doesn't, then you could simply remove the summation symbol and replace it with multiplication by the cardinality of $\displaystyle D(s)$. The same thing goes for the product over $\displaystyle d(s)$: if nothing in the expression depends on $\displaystyle r$ explicitly, then you can replace the product with exponentiation to the power of the cardinality of $\displaystyle d(s)$ thus:

$\displaystyle \displaystyle{F(x)=\frac{(a^{x}-1)^{m}}{2a/m}|D(s)|\left(\frac{1}{1-x}\right)^{\!\!m}\left(\frac{m\,2^{2x^{2}}-1}{2^{n}-1}\right)^{\!|d(s)|}}.$

Is it your opinion that this expression is equivalent to your original expression? Here, I've used $\displaystyle |D(s)|$ for the cardinality of $\displaystyle D(s)$, and $\displaystyle |d(s)|$ for the cardinality of $\displaystyle d(s)$.

3. What are "nodes" in this context?

4. What is $\displaystyle P_{w}$? There is no appearance of that symbol in the original expression. How does it relate to your original expression?

5. What's the difference, if any, between $\displaystyle D(s)$ and $\displaystyle d(s)$?

6. Is this function given to you, or have you derived it as part of a bigger problem? If the latter, could you please type up that problem, so I can have more context? Thanks!

7. A comment: the $\displaystyle 1/(1-x)$ factors are very troubling for proving there is a global minimum. If $\displaystyle m$ is allowed to be odd, then it seems to me that

$\displaystyle \displaystyle{\lim_{x\to 1^{+}}F(x)=-\infty},$

which would prove that there is no global minimum, unless $\displaystyle -\infty$ is an acceptable answer. On the other hand, if $\displaystyle m$ is only allowed to be even, then you still have a chance of proving that there is a finite global minimum.

8. What is $\displaystyle a$?