What Is the Smallest Possible Value of the Product in This Mathematical Problem?

[remaan]

Finding the least value - Help Please !

Homework Statement

In this problem, we are required to find the smallest possible value for the product :
∏ (i=1,n) (1-xi)/(1+xi)

Given that : x1, x2, ...,xn are non-negative Real numbers and their sum is 1/2.


. The attempt at a solution

I tried doing the following :
given that their sum is 1/2 , we can benefit from that finding the n from n(n+1)/2
from that we have 2 values of n which are 1 and 0.

If n = 0 we will get 0
and if n = 1 we get 1

so 0 is the least possible value -
What do you think ?
AM I in the right direction ?

 

[lanedance]

I don't follow your argument? and i would read it as requiring n>=1

at a quick look, I would try the following, (not tested)
- see if you can expand the products to get any further info / use the constarint
- consider the case x_i = 1/2n for insight on the impact of n
- consider optimisation using differentiation (probably with Lagrange multiplier to capture constraint)

 

[remaan]

Yes, I will try these suggestions -

 

[remaan]

Sorry, but I am not sure I get what you mean by :
"consider the case x_i = 1/2n for insight on the impact of n"

In other words, how did you conclude that x_i = 1/2n ?

 

[lanedance]

just guessed as start, its the symmetric solution (xi=xj) of
x1 + x2 + .. + xn = 1/2

its not necessarily the solution, but may give insight to the problem & the effect of n

 

[lanedance]

is n variable?

 

[remaan]

I don't think that n is a varible,
But rather n is the number of fractions that we multiply together each time.

so if we started with n =1 we'll get some thing like :

(1/1+x) - ( x/1+x)

So if n = 2

we will have the same quantity but multiplied by itself one more time

and if n = 3 it's multiplied by itself 3 times and so on.

How does using the Given 1/2 relates to finding the smallest possible value in this case ?

In other words, the pattern that I could identfy is :
n = 1 (1/1+x - x/1+x) ^1
n = 2 (1/1+x - x/1+x) ^2

so on,
so the general formula for that could be :
(1/1+x - x/1+x) ^n

Any Extra hints ?

 

[remaan]

** It gets a little complicated when I tried to extract it -

 

[Dick]

Sorry, but I am not sure I get what you mean by :
"consider the case x_i = 1/2n for insight on the impact of n"

In other words, how did you conclude that x_i = 1/2n ?

lanedance may have guessed it, but you can show it's true using a Lagrange multiplier form to express the constraint, as lanedance already suggested but didn't follow up on. Haven't you learned that? If not you probably shouldn't have been given the question. Hint: maximize log of the product instead of the product.

 

[lanedance]

minimise, but log is a great idea

 

[Dick]

minimise, but log is a great idea

Uh, right, minimize. It is 3AM here.

 

[remaan]

Thanks all for your great effort trying to make me progress with this problem.

The last thing I got after visiting my prof. is that the least value is 1/3

and I have to show that this is true.

** Do you think using using a counter example method would be a good choice ?

 

[lanedance]

probably not, i would try the minimisation previuosly suggested

 

[Dick]

Thanks all for your great effort trying to make me progress with this problem.

The last thing I got after visiting my prof. is that the least value is 1/3

and I have to show that this is true.

** Do you think using using a counter example method would be a good choice ?

1/3 is the n=1 case. Apparently your prof wants you to minimize over n as well. You need to solve it for each fixed n first. Use a Lagrange multiplier approach.

 

[remaan]

For the method you suggested,
I think that I will use some thing like :

∇f(P) = λ ∇g(P).

However, I am not sure about the f and g.

should f be 1-xi/1+xi

and what about g ?

should I choose λ to be 1/3 ?

** I can't really see how does n comes into this formula ??

??

 

[lanedance]

g is the constraint, don't choose lambda see what comes orm working it through

you can treat n as a variable and minimise with respect to it, however i would probably start by looking at the case for a given n