# Revival-Math Q&A

#### v8archie

Math Team
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

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

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?

1 person

#### v8archie

Math Team
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

Math Team
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

Math Team
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

Math Team
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

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

#### v8archie

Math Team
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

Math Team
v8archie said:
You got me.
: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

Math Team
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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