Can you solve it?

>n can be any positive integer; you can pick as many terms A as you want as long as they sum to p
Spose all A should be positive as well.

so n is a variable too? the way it was written, i would have though n is an arbitrary constant.

>and so on...

What? Is there supposed to be a pattern there?

In what book did you find this?

>In what book did you find this?
I didn't, just thought it up..

Sorry for the lack of clarity

in that case, my guess would be to have a1=a2=...=an, and have n be the largest positive integer such that a1,a2, etc are greater than two.

>I didn't, just thought it up..

I was assuming this much.

This is probably not solvable, as the set of As are partitions of p, which tend to be many and random for big p.

Then comes the fact that the way you order the partition will also affect the value of x.

If I was you I would first try to solve the easier questions:

If [math] A_1 + ... + A_n = p [/math] then find the biggest possible value for [math] x = A_1* ... *A_n [/math]

And

Find an upper bound for [math] A_1^{A_2^{...^{A_n}}} [/math] in terms of [math] p [/math]

The first one is easier because the commutativity of multiplication allows us to forget about reordering and consider only the "biggest" partition of p.

The second one is easier because instead of finding the actual biggest value, you can simply find an upper bound, prove it is an upper bound and then maybe start working to decrease it.

A trivial upper bound is obviously p to the p to the p to the p, etc. n times.

Now start decreasing it.

And if you care about solving it then maybe read more about partitions.

>And if you care about solving it then maybe read more about partitions.
Ehh not really, I'm a pleb and am not studying math
Just thought someone here might enjoy it//was moderately curious if there was a simple solution and thus how much of a retard I am.

>I didn't, just thought it up..

Please state the problem more concise because as it is stated now it is a mess.

Could you use Lagrange multipliers here?

if n did not vary that's what i'd try. but since n is a parameter, it becomes a discrete optimization problem.

I mean I don't think it's that confusing really...
Anyway I made this picture which more or less illustrates how the problem evolved in my mind when I was thinking of it.

If we can have [math] A_i < 0 [/math] then it is pretty easy to let [math] x [/math] be as big as you want. Just take [math] x = (-k)^{k+p} [/math] for big [math] k [/math] such that [math] k + p [/math] is an even integer.

>If we can have [math] A_i < 0 [/math] then it is pretty easy to let [math] x [/math] be as big as you want. Just take [math] x = (-k)^{k+p} [/math] for big [math] k [/math] such that [math] k + p [/math] is an even integer.
Yeah I realised that up here

>if there was a simple solution

There is probably not a simple solution. With things like these what number theorists do is instead find formulas for upper bounds that get more accurate as you put in bigger outputs.

Finding closed form exact formulas was something mathematicians gave up a long time ago.

No, because there is no "maximum" of a set of complex numbers; e.g. if p = -1/2 we can have x = -1^(1/2). You'll need to impose more conditions.

See
for a version with appropriate restrictions

Regardless of what p is, the maximum value of x would be:

$\x = A_1^{A_2^{...^A_n}} where $\A_1 < A_2 < ... < A_{n-1} < A_n

Made a little script to test random combinations for different values of p.
It seems to try to stick to each of the terms being equal and as close to ~2.7 as possible +- say .5.
When each of the terms gets above 3 it adds another one to keep the individual values down.
No idea if it holds for big p though because lol can't go much the maximum grows quite rapidly with p.

ah, then it's probably e.

should have known...

>should have known...
Indeed, you should have.

oh, (A_1, ..., A_n) is an integer partition of p?

Infinity.

Taking exponents isn't associative, so without brackets this question is poorly defined. All the more reason why everyone should use Polish notation.

Sounds about right

i did, somewhere in the back of my head, there was a voice telling me "there's nothing special about 2 you moron, it's not 2"

as a warm-up let's solve this for n=2

and i'll assume all the A's are >=0

in this case, we want to maximize the function

x^(p-x) over the interval [0,p]

i'll instead consider the function f(x)=(p-x)ln(x), which clearly has maxima at the same points

the derivative is given by

f'(x)=-1-ln(x)+p/x

note this is a strictly decreasing function of x

moreover, it is clear the f'(x) tends to infinity as x tends to 0

consider the case p=0

since the derivative is a decreasing function of x, it follows that it is positive on the entire interval [0,p] (except at the endpoint if p=1)

therefore f is an increasing function on [0,p], so the maximum occurs at the right endpoint p=x. the value is f(p)=0

the p>1 case is more complicated

in this case, f'(p)=-ln(p)1; in particular pe>1, and it follows that W(pe)>1 (since W is increasing)

so f(x0)>0, which means it is the extremum on the interval [0,p]

the n=3 case looks a lot messier; maybe someone else sees how to proceed?

the answer is infinity

I think you're over-complicating your [math]n=2[/math] solution; determining
[eqn] \max\{A_1^{A_2},A_2^{A_1}\} [/eqn]
comes down to comparing the two ratios
[eqn] \frac{A_1}{A_2},\frac{\log A_1}{\log A_2} [/eqn]
If the first is larger, then
[eqn]
A_2^{A_1}
[/eqn]
is the max; otherwise, the second one is the max.
From here, it is clear how one would go about determining
[eqn]
\max_{(\sigma_1, \cdots, \sigma_n) \in S_n}\{A_{\sigma_1}^{\cdots^{A_{\sigma_n}}}\}
[/eqn]
It requires a lot of computation. I cannot think of an easier solution.
(This renders correctly in Veeky Forums's TeX preview; sorry if it doesn't as an actual post)

maybe i mis-interpreted the problem, but i assumed the Ai's were real numbers, not integers

if you require the Ai's to be integers, then yes you can just brute force it

By the way, this post only addresses OP's problem as he intended to put it (seen in ), where A_1, ..., A_n formed a partition of p, meaning there are only finitely many tuples to compare.

But see this though for the 'real' constraints of the problem

Assuming that A_n can be negative, the answer is infinity for all n > 2.
100000^(.1^(-100000.1+p))

x^p ?
or infinity?

if A_n can be negative, then there is no "maximum":

and now the A's have to be integers? OP STATE YOUR PROBLEM CORRECTLY YOUR"RE MAKING ME LOOK LIKE A FOOL HERE

A_n doesn't have to be a natural number, so I can create an infinite sum using negatives or rationals that adds up to p and creates a product power towards an absolute value of -1/12, the largest possible number.

isn't this just 2^2^...^2^3?

Bump

Good question

0

In reality, the entire Universe is a zero-sum game of "bubbles" waiting to be popped back into a value of zero by the infinite decay of entropy. Mathematics is designed as an abstract language to explain that, therefor, all values in math should tend towards their maximum of zero as time progresses to infinity.