How to Maximize a Nonlinear Function with Limited Variables?

[ericdavid]

Hi to everyone,

I'm optimizing a nonlinear function but I'm struggling to achieve it. The function is the following:

X and i are relationed so i doesn't go to infinite. Do you have any idea how to maximize this function?

Thanks in advance,

Eric

 

[mfb]

What do you mean by "related"? i is a summation index, not a free variable. The starting value is missing.

The sum has an explicit formula, this should be easy to simplify.

 

[Mark44]

Hi to everyone,

I'm optimizing a nonlinear function but I'm struggling to achieve it. The function is the following:
View attachment 107800

X and i are relationed so i doesn't go to infinite.

What does this mean? Your summation appears to be from i = <something> to $\infty$.

Do you have any idea how to maximize this function?

The summation in the formula for F(x) is unclear. Is this what you have in mind for the summation?
$$\sum_{i = 0}^{\infty}e^{-i(x + 2)}$$

 

[ericdavid]

Yes, it is from i to ∞, but if I want to optimize it I guess that I will have to set a bound.
And yes, the summation corresponds to what you have posted. Let's say that i is related to x as follows: i=T/x, T is a time. Therefore, x determines i.

 

[mathman]

How can the summation index (i) start at i?

 

[Mark44]

Yes, it is from i to ∞,

This makes no sense -- "from i to ∞". i is merely the index. You have to give a starting value, such as 0 or 1 or whatever, and an ending value, such as 10 or N or ∞.

but if I want to optimize it I guess that I will have to set a bound.
And yes, the summation corresponds to what you have posted. Let's say that i is related to x as follows: i=T/x, T is a time. Therefore, x determines i.

This also makes no sense. x is presumably a real number, and a summation index is usually an integer.

The summation that I wrote can be expanded like so:
$1 + e^{-(x + 2)} + e^{-2(x + 2)} + e^{-3(x + 2)} + \dots$

It's not at all clear to me or the other people replying in this thread what you're trying to do.

 

[ericdavid]

Apologies, I wanted to say that it goes from i=1 to inifinite. I will reformulate my question so you can understand it better.
I want to maximize this function:

And x is related to i as follows; i=T/x, where T is a constant. Due to the nature of the problem, i is an integer.

Thanks for your patience.

Eric

 

[Mark44]

And x is related to i as follows; i=T/x, where T is a constant.

This still doesn't make sense. i is an index of the summation, and x occurs outside the summation (as well as being part of the things being summed).
i takes on an infinite number of values: 1, 2, 3, ... in the summation, but the x that multiplies the summation can't change with the change in index values.

Based on what you said, you can write the summation as $\sum_{i = 1}^{\infty}e^{-i(T/i + 2)}$, but you can't replace either x outside the summation by T/i.

It seems that you're trying to come up with a formula for a function that involves a summation, without understanding how a summation works.

 

[ericdavid]

If we forget about the relationship between i and T, how would you optimize it?

 

[Ssnow]

I'm optimizing a nonlinear function

Your question is related to find the max of this function "in general" or with constraints ?
I don't understand well how this function is defined, $i$ is the integer index in the sum but what about $x$?, is real or integer? Has this function a domain?
If it is a real function try with the derivative...

Ssnow

 

[mfb]

View attachment 107822

If that is your function, it is identical to (just a different notation)
$$F(x)=\frac{1}{x+x \left( e^{-1(x+2)} + e^{-2(x+2)} + e^{-3(x+2)} + ... \right)}$$
You see how i disappears just by rewriting it? i cannot depend on anything.

You can use$$\sum_{i=1}^\infty e^{-i(x+2)} = \sum_{i=1}^\infty \left(e^{-(x+2)}\right)^i$$
The sum on the right (a sum over qⁱ for some q) can be evaluated with a well-known formula.

Afterwards, you can use that a maximum of your function (which does not occur at 0) is a minimum of the inverse, 1/F(x). From there it should not be hard to look for minima.

 

[ericdavid]

Your question is related to find the max of this function "in general" or with constraints ?
I don't understand well how this function is defined, $i$ is the integer index in the sum but what about $x$?, is real or integer? Has this function a domain?
If it is a real function try with the derivative...

Ssnow

Hi Ssnow,

x is a real, I'm working on a domain and i do not have any constrains more.

 

[Ssnow]

Hi, ok thanks for the clarification, I suggest you to follow the suggestions of @mfb.

 

[ericdavid]

If that is your function, it is identical to (just a different notation)
$$F(x)=\frac{1}{x+x \left( e^{-1(x+2)} + e^{-2(x+2)} + e^{-3(x+2)} + ... \right)}$$
You see how i disappears just by rewriting it? i cannot depend on anything.

You can use$$\sum_{i=1}^\infty e^{-i(x+2)} = \sum_{i=1}^\infty \left(e^{-(x+2)}\right)^i$$
The sum on the right (a sum over qⁱ for some q) can be evaluated with a well-known formula.

Afterwards, you can use that a maximum of your function (which does not occur at 0) is a minimum of the inverse, 1/F(x). From there it should not be hard to look for minima.

Thanks for your advice, so far the best I've received.

Can I use this formula?

By the way, how do you introduce formulas in this forum?

 

[mfb]

You can use this formula, to take the limit n->infinity you have to check if z is in the correct range for that.

You can use LaTeX for formulas. The quote in your post has two examples.

 

[ericdavid]

In this formula, the summaton begins at i=0, while in my case it's i=1. How can I solve that?

Thanks

 

[mfb]

You can replace i by i+1 everywhere (!) in your expression, then simplify.

There is a German Wikipedia article about - no English version, but the formulas are international.

 

[ericdavid]

In fact, my function is this one:

I posted a simplified one before, can I still express this one with a well-known formula?

Thanks

 

[mfb]

No.

Where was the point of the other function if that is not what you actually want to solve?