I have scoured the newsgroups and web and can't seem to find an anwser
to this.

I need to find the max of an equation. where the equation is in this
form 1/(1-e^(z1-x))*(1-1/(1+e^(z2-x))*.........*(1-1/(1+e^(zn-x)).

Where there are n number of z values that are know. Meaning that z is
not a variable but different constants.

I can calculate the slope at a specific value of x but want to solve
for x where F'(x) = 0.

Any help?

Thanks,
John

You can use bisection method to approximate the root of
the equation. It is easy to use excel to do.


-----Original Message-----
I have scoured the newsgroups and web and can't seem to
find an anwser
to this.

I need to find the max of an equation. where the
equation is in this
form 1/(1-e^(z1-x))*(1-1/(1+e^(z2-x))*.........*(1-1/
(1+e^(zn-x)).

Where there are n number of z values that are know.
Meaning that z is
not a variable but different constants.

I can calculate the slope at a specific value of x but
want to solve
for x where F'(x) = 0.

Any help?

Thanks,
John
.

If you can't find a mathematical solution you might
consider running the equation in a loop progressively
incrementing the x variable and extract the maximum result:

For x = 0.1 To 100 Step 0.1
F = (Your equation here)
Peek = Application.Max(Peek, F)
Next
MsgBox Peek

Regards,
Greg

-----Original Message-----
I have scoured the newsgroups and web and can't seem to
find an anwser
to this.

I need to find the max of an equation. where the
equation is in this
form 1/(1-e^(z1-x))*(1-1/(1+e^(z2-x))*.........*(1-1/(1+e^
(zn-x)).

Where there are n number of z values that are know.
Meaning that z is
not a variable but different constants.

I can calculate the slope at a specific value of x but
want to solve
for x where F'(x) = 0.

Any help?

Thanks,
John
.

I can't tell if your equation is listed correctly, but here is just a guess.
I most likely am wrong here, but it looks like your equation is a decreasing
function that really does not have a Maximum value, except at points of
discontinuity when you have 1/0.

Your first term is:
1/(1-E^(z1-x))

This is a decreasing function as x tends towards infinity. When x = z1, you
have 1/0 which is a discontinuity (value tends to infinity). Just above x,
the value is a large number. The limit as x goes to infinity is 1.
Each of the other terms (1/(1-E^(z2-x))) have a limit of 0 as x tends
towards infinity. (With jumps as x approaches z2). So, my guess is that it
is hard to tell what you mean by "Maximum" value with the given equation.

--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


"John Hall" wrote in message
m...
I have scoured the newsgroups and web and can't seem to find an anwser
to this.

I need to find the max of an equation. where the equation is in this
form 1/(1-e^(z1-x))*(1-1/(1+e^(z2-x))*.........*(1-1/(1+e^(zn-x)).

Where there are n number of z values that are know. Meaning that z is
not a variable but different constants.

I can calculate the slope at a specific value of x but want to solve
for x where F'(x) = 0.

Any help?

Thanks,
John

Your conclusion about the overall function is correct in that there is
a pole at Z1. But, at no other value of x.

Check the function again, Dana.

1/(1-e^(z1-x))*(1-1/(1+e^(z2-x))*.........*(1-1/(1+e^(zn-x))

The Z2..Zn components are different from the Z1 element. Each of the
former are bounded +1 (large negative x) to 0 (large postitive x).
Like a S-shaped innovation adoption curve seen through a vertical
mirror. At x=Zi, the function has a value of 1/2.

The first function, as you pointed out is unbounded at x=Z1.
Basically, the function is monotonically decreasing. For x<Z1, it goes
from -0 (large negative x) to -infinity (x-Z1 from below) and from
+infinity (x=Z1+) to 1 (large positive x).

But, beyond the existence of the pole at Z1, I don't see any way to
easily characterize the product of the individual components.

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article ,
says...
I can't tell if your equation is listed correctly, but here is just a guess.
I most likely am wrong here, but it looks like your equation is a decreasing
function that really does not have a Maximum value, except at points of
discontinuity when you have 1/0.

Your first term is:
1/(1-E^(z1-x))

This is a decreasing function as x tends towards infinity. When x = z1, you
have 1/0 which is a discontinuity (value tends to infinity). Just above x,
the value is a large number. The limit as x goes to infinity is 1.
Each of the other terms (1/(1-E^(z2-x))) have a limit of 0 as x tends
towards infinity. (With jumps as x approaches z2). So, my guess is that it
is hard to tell what you mean by "Maximum" value with the given equation.

Your conclusion about the overall function is correct in that there is
a pole at Z1. But, at no other value of x.

Oops! You're right. I copied a 1-e^.. for the remaining terms(like the
first term) when it should have been a 1+e^.. Ahh. :)

--
Dana DeLouis
Using Windows XP & Office XP

<snip