Find Max of Curve Equation

[John Hall]

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

 

[Jessica]

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

 

[Greg Wilson]

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

 

[Dana DeLouis]

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.

 

[Tushar Mehta]

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

 

[Dana DeLouis]

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. :>)