Revival-Math Q&A

v8archie

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I'd have thought that (1,0) represents the real number 1, rather than zero.

You are correct that we are on the unit circle, but we won't ever get back to 1 or -1, because $\pi$ is irrational and thus for $n,m \in \mathbb{Z}$, $n\pi = m \; \Rightarrow \; n = m = 0$, there being no other solutions. In fact, we never get to -1 at all!

So what we have is two infinite sets consisting of points on the unit circle, both of which fill in more and more of the circle as $|n| \to \infty$. But the two sets are disjoint (they share no members). There's perhaps nothing too surprising about that. In the number $e^{i\theta}$, $\theta$ is a real number and we are comfortable with the odd integers and the even integers being a pair of disjoint sets within the reals. But I fine it challenging to think of these two sets on the unit circle being disjoint, because my mind wants to identify them, as $|n| \to \infty$, with the unit circle. Of course, this is a bad idea, because that would suggest that in some sense $1 = -1$ in the complex plane.

Anyway, over to you Hoempa.
 
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Hoempa

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v8archie said:
I'd have thought that (1,0) represents the real number 1, rather than zero.
O pitty I wrote it's 0:eek:

v8archie said:
You are correct that we are on the unit circle, but we won't ever get back to 1 or -1, because \(\displaystyle \pi\) is irrational and thus for \(\displaystyle n,m \in Z, n\pi=m\implies n=m=0\), there being no other solutions. In fact, we never get to -1 at all!
Would that mean that \(\displaystyle e^{i\pi} \ne -1\)?

Here's my question.
1.) Some positive integer partitions of 2014 are 2013+1, 1007+1007 etc. We can define the product of a partition as the product of the terms. For example: 2013+1 would give 2013*1 and 1007+1007 would give 1007*1007.
For what partition would we get the highest such product?

2.) If we allow terms of partitions to be non-integers, but still positive, for example 0.5 + 2013.5, (which has as the product 0.5*2013.5) but we'd still require an integer amount of terms (in case someone could think of a noninteger amount) what partition would give the highest such product?
 
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v8archie

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1) For integer $n = 2k$ the largest product comes form $k + K$. For $n=2k+1$, the largest is $k + (k+1)$.

2) $\frac{n}{2} + \frac{n}{2}$ for partitions into two as per the examples. Are we allowed more pieces? Not that I think it makes any difference.
 
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Hoempa

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Hm, by partitions, I mean any integer amount of terms, so 800+800+414 counts as well. Maybe my examples having only two terms are a bit off.
 

v8archie

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It makes no difference. A 3-partition a + b + c iis a 2-partition (a + b) + c. And similarly for any n-partition.

Actually, the product in this case is different (but smaller).
 
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Hoempa

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Okay, so a 2-partition is also a 1 partition since any 2-partition a + b is also (a + b).

But 800 + 800 + 414 gives a higher product, 800^2 * 414 than (800 + 800) + 414; 1600 * 414. It's about the highest product we can find under the given rules

Alternatively (equivalently?), we could write the question as 2014 = \(\displaystyle \sum_{i=1}^n a_i \) where a_i > 0 and n in N and for 1.) also a_i in N. Find the set of elements \(\displaystyle a_i\) such that \(\displaystyle \prod_{i=1}^n a_i \) is maximal.

2.) \(\displaystyle a_i\) in R. What would be the highest product now?
 

Hoempa

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v8archie said:
Actually, the product in this case is different (but smaller).
Ah, yes. How large can it get?
 

v8archie

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You got me. I shouldn't try these in my head.

The best you can do with an $n$-partition of $a$ is $$\left(\frac{a}{n}\right)^n$$ (I'd need to prove this for $n\gt 2$, although replacing any two numbers $b,b$ by $b+c,b-c$ changes their product from $b^2 - c^2$, so that ought to do it.

So, does the product above have a maximum? Yes, because when $\frac{a}{n} \lt 1$ the product is less than 1.

The product $$\left(\frac{a}{x}\right)^x = e^{x\log a - x\log x}$$
is maximised when the derivative is zero. That is
$$\log a - \log x - 1 =0 \qquad \Longrightarrow \qquad x = \frac{a}{e}$$

So the maximum value of the product is
$$e^\frac{a}{e}$$
or a little less if we allow only integers. For 2014 my phone chokes on the number.
 
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Hoempa

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v8archie said:
You got me.
:cool::spin:

You aren't there yet. For 1.) A prime factorisation can be given.
For 2.), e^(2014/e) isn't allowed; it hasn't got an integer amount of terms.
 

v8archie

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Well I left unspoken the assumtion that we'd test the integers either side of $\frac{a}{e}$. In some cases the best partition is not unique, e.g. $a = 10$, but they. Probably will be once $\lfloor\frac{a}{n}\rfloor \gt 2$.

Ah! I've just realised that $n \approx \frac{a}{e} \; \Rightarrow \; \frac{a}{n} \approx e$. So $\lfloor\frac{a}{n}\rfloor = 2$.

To get an $n$-partition, we set each $a_i$ in the partition to 2 and then add 1 to $a - 2n \lt n$ of them.

So the $n$-partition is $$a_i = \begin{cases} 3 & 1 \le i \le a - 2n \\ 2 & a - 2n \lt i \le a \\ \end{cases}$$
and $n = \lfloor\frac{a}{n}\rfloor$ or $n=\lfloor\frac{a}{n}\rfloor+1$.

Note, we can always turn a pair of 2s in the partition into a single 4 without changing either the sum or the product.

So, for 2014 we have $740 \lt \lfloor \frac{2014}{e} = 740.9 \lt 741$.

Taking $n=740$ we get a partition $$a_i = \begin{cases} 3 & 1 \le i \le 534 \\ 2 & 535 \le i \le 740 \end{cases}$$
and $534 \times 3 + 206 \times 2 = 2014$ as required.

For $n=741$ the partition is $$a_i = \begin{cases} 3 & 1 \le i \le 532 \\ 2 & 533 \le i \le 741 \end{cases}$$
We have replaced two 3s with three 2s so the product has reduced since $3^2 \gt 2^3$. Thus we always want the lower partition.

And the product for that is $3^534 \times 2^206 = $
62363305591989832933404109356460339797199221914017331977460606784302737365473042633893490038331148725065928912469563054482062618158578602766995927555252411964471771848167796259962976851374479135205372385884930891176916836822196624263278847347475738643808545926399912446880330504323947268781746284352041340292527292416
 
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