Mathematics - 2017-06-21 (page 2 of 11)

Akiva Weinberger

1:00 AM

I am most definitely confused.

Dodsy

yeah it's hard.

maybe I'll find an easier one

Typhon

@Zee are you ok?

Akiva Weinberger

P: This is not a semiprime.
S: I knew it's not a semiprime, because my number is not the sum of two primes. (Read: It's 2+composite)
P: Now that I know the sum is 2 more than a composite, I can solve it.

S: So can I.

Simply Beautiful Art

@AkivaWeinberger so I posted the problem I had: math.stackexchange.com/questions/2330515/prove-the-limit-exists

Dodsy

huh.

Akiva Weinberger

1:03 AM

That doesn't necessarily mean that 2 is one of the numbers.

Annelise Toft

Can someone check my proof please? I did it myself and I think there's something wrong on it math.stackexchange.com/questions/2330469/…

Semiclassical

Akiva Weinberger

Which are we deleting, the light ones of the dark ones

Semiclassical

Light ones, I guess.

Typhon

oh geez!

i got it!

Dodsy

1:09 AM

P: My number has more than two eligible factors.
S: My number must be the sum of two numbers that produce a number with more than two factors.

is what I see it as

What is it ty.

Semiclassical

Now I just need to figure out a way to check conductivity...hrm.

Akiva Weinberger

Wrap it up into a cylinder and look at homology

Semiclassical

What I'd love is an option like: Pick two sets of vertices, and figure out whether or not they're connected.

Akiva Weinberger

(Con't) Wait no that's a stupid idea don't do that

Semiclassical

In mathematica, I mean.

I think I can make Mathematica split the graph into connected components, and then check if any of the graphs have both a vertex from the left edge and from the right edge

Typhon

1:14 AM

@Semiclassical the probability of there being a connection from left to right is the probability of there not being a connection from top to bottom including a piece on the top and bottom.

Akiva Weinberger

There could be a connection from left to right and a connection from top to bottom

Typhon

uuh

see the image

the red is a path from top to bottom

blocking paths from left to right?

Dodsy

Ty

Akiva Weinberger

Consider a single straight-line red path from top to bottom

Dodsy

Right.

Typhon

1:15 AM

oh geez

XD

sorry

Akiva Weinberger

On the other hand, maybe you're on to something...

Dodsy

What's interesting about a line from top to bottom

Akiva Weinberger

@Dodsy You could also have a straight-line light blue path from left to right

They could cross.

Unless you make the conditions for "connecting" even stronger for the light blue stuff

Dodsy

is that the chances are equal to $1-(1-(0.5^n)^n$ and the chance of it going left to right is $1-(1-(0.5^{n+1})^{n+1}$

Akiva Weinberger

Like, prohibit the use of any vertices in right edges

Dodsy

1:17 AM

meaning that

the chances of it going top to bottom is always one step behind it going left to right

Typhon

hmmm idk

Dodsy

in pure horizontal or vertical lines

Typhon

i honestly am kind of sick of this problem

Dodsy

of course, they both approach zero as $n\to\infty$

Semiclassical

I'm staying not sick of it by focusing on the more practical problem of "how do I simulate it in Mathematica"

Typhon

1:18 AM

meh

Dodsy

Here's an easy math problem

Semiclassical

Especially since I have a fairly easy way of making the probability not 1/2.

Dodsy

Write 271 as the sum of positive real numbers so as to maximize their product.

Typhon

im considering the possibility of differentiation as an operation upon a vector

and therefore, integration is the inverse operation

Semiclassical

AM-GM inequality is decisive there.

Typhon

1:19 AM

i'm using postfix notation for function

so as to represent a function as a vector of numerical codes

Dodsy

The answer is 135 + 136

right?

Semiclassical

Doubtful. @dodsy

Oh. Do you mean just two numbers?

Typhon

@Semiclassical me?

Dodsy

oh

I didn't think of that.

I thought it had to be two numbers.

but that makes the problem more interesting.

Semiclassical

If it's just two numbers it'll be 135.5+135.5

Dodsy

1:21 AM

oh right, I thought they had to be whole numbers :P

Semiclassical

lol

Typhon

271 = 20 + 20 + 231

Dodsy

okay so a possibly infinite amount of numbers?

that aren't whole numbers.

Semiclassical

I think if you allow infinite you might break it. Lemme check.

Typhon

20*20 = 400 > 135

136 < 231

just keep pulling out 20's

that's my strategy

Semiclassical

1:22 AM

If I've got n real numbers, then the arithmetic mean is going to be 271/n

and the geometric mean will be product^(1/n).

Typhon

271 = 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 11

Dodsy

hm

jesus christ hahaa

Typhon

20^13 * 11 = 901120000000000000

Semiclassical

The former bounds the latter , so $271/n\geq \sqrt[n]{\text{product}}$

Dodsy

split those 20's into 10''s

Typhon

1:24 AM

@Dodsy fair point

Semiclassical

hence the product is bounded above by $(271/n)^{n}$

Dodsy

1100000000000000000000000000

if you split into 10's

how about 5's?

Semiclassical

with equality only if I divide it up evenly.

Typhon

10^27 * 1 = 1000000000000000000000000000

darnit

Dodsy

5.5511151231257827021181583404541e+37

Semiclassical

1:25 AM

which...goes to zero. huh.

Dodsy

if you do 5's

Semiclassical

I am error.

Dodsy

9.495567745759798747473242269562e+42

if you do 2.5's

Semiclassical

To get a sense of where I'm going with this, suppose you divide 271 into 100 equal chunks.

then you've got 2.71^100 ~ 2 x 10^43

PVAL-inactive

The way I usually teach that is looking at x(271-x)=y

Typhon

1:26 AM

2^135 * 1 = 4.3556142965880123323311949751266e+40

@Dodsy somewhere between 2 and 5 is the peak

Dodsy

right.

PVAL-inactive

which is downwards opening parabola

Semiclassical

Oh, dur, I see what I'm doing wrong

PVAL-inactive

and you want the integer closest to the vertex

if you want an integer.

Dodsy

hm

its any real number.

Akiva Weinberger

1:29 AM

Obviously the way to do the grid problem is to provide a bijection between the success events and the failure events

The ones where there is a connection and the ones where there isn't

Typhon

3^90 * 1 = 8.7279635680877124258913974794767e+42

Dodsy

Someone mentioned a bijection before

Akiva Weinberger

And clearly the $n\times(n+1)$ condition on the dimensions comes into it.

Typhon

so if it is one particular number...

PVAL-inactive

Yeah I looked at both sets in the 1 by 2 case and I couldn't come up with one.

Semiclassical

1:30 AM

Ahah, I figured out the number thing!

Typhon

than it is between 3 and 2

Semiclassical

Here's an observation: Suppose I break 271 into 271 equal pieces. What's the product then?

PVAL-inactive

I think there's some trick where you reduce the thing to smaller things

Dodsy

271/3 is 90$\frac{1}{3}$

Akiva Weinberger

So maybe you somehow you squish it horizontally and stretch it vertically a certain way, rotate it 90 degrees, and then flip colors

Typhon

1:30 AM

@Semiclassical 1?

Akiva Weinberger

Or something

Semiclassical

Right.

So increasing the number of pieces eventually makes things worse not better.

Dodsy

1.2587901705733172032541887968498e+43

Typhon

@Semiclassical we know that

2 went down from 2.5

Semiclassical

well, I didn't :P

Typhon

1:31 AM

3 is less than 2.5

Dodsy

no it isn't

Typhon

it's just a matter of finagling the numbers down the chain

Dodsy

you didn't do the math right.

Typhon

@Dodsy fractional powers are not allowed...

Dodsy

why not?

Typhon

1:32 AM

because

we split it into a sum of numbers

Semiclassical

Eh, he said it as "real numbers" earlier.

Typhon

that means an integer number of coefficients

Semiclassical

But you do need an integer number of terms. So if that's what you're saying I agree with it.

Dodsy

hm

Typhon

^^^

Dodsy

1:33 AM

I see.

Typhon

so that means integer powers

@Semiclassical exactly what i meant

PVAL-inactive

So are you guys trying to find the number of terms you want to divide it up so the product is maximal?

Typhon

@Dodsy if you have a 3^1/3 multiple then where did that term appear in the sum?

Semiclassical

If I've got it right, though, the max would occur (if you could have non-integer number of terms) when $n=271/e$ :)

Simply Beautiful Art

@Typhon why such small numbers you are working with?

Typhon

1:34 AM

i mean, you could divide by 3^(1/3) to get a weird sum... but that isn't what we had or what you were doing.

Semiclassical

And $271/3\approx 99.6$, so either $n=99$ or $n=100$.

Dodsy

We basically need a number such that $(\frac{271}{x})^x$ is the maximum.

Typhon

@SimplyBeautifulArt "Write 271 as the sum of positive real numbers so as to maximize their product."

Simply Beautiful Art

Ah, that...

Semiclassical

Yeah, that's how I just got $271/e$.

Typhon

1:34 AM

we found that really small numbers with really big powers work better

Of course!

e

Akiva Weinberger

Here's a nice puzzle: Given $s$ ($s > 0$) points in the plane such that every three of them are contained in a disk of radius $1$. Prove that all $s$ points are contained in a disk of radius $1$.

Typhon

im such a dumbass

Semiclassical

It helps to differentiate the log of that function instead of the function itself. (Log is an increasing function so it doesn't change the location of the critical point)

PVAL-inactive

so you're looking at x^(271/x) and maximizing that

Simply Beautiful Art

@Typhon You didn't know that? Think I did a problem of that sort a month ago or so.

PVAL-inactive

1:35 AM

that's not so hard

Dodsy

uhhhh

yeah I guess so

I may have written it wrong above.

Simply Beautiful Art

:|

Semiclassical

(271/x)^x, not the other way around @PVAL-inactive

Dodsy

right

so let's differentiate it.

and find the vertex.

Akiva Weinberger

What are you trying to do?

PVAL-inactive

1:36 AM

@semi it shouldn't matter if x is the size of the partition the number of partitions is 271/x

and vice versa

Semiclassical

uh

a^b != b^a

Simply Beautiful Art

How do you guys feel about the closure of this question?

7

Q: The mystik spiral(challenge)

Still one of my favorite problems. Feel free to attempt to solve it. Start at a 0,0 Travel along the x axis in the positive direction a distance of 1 Turn 90 degrees counterclockwise go forward 1/2 of the distance of the previous step Repeat steps 3 and 4 forever. how far from the origin to you...

recreational-mathematics

Semiclassical

and you definitely don't have 271/x = x

Typhon

e^99 * 1.8900989825545216993315403360864 = 9.889030319346946770560030967138e+42 * 1.8900989825545216993315403360864 = 1.8691246145048480893269096030194e+43

Dodsy

$y' =(271/x)^x(ln 271 − ln x − 1)$

PVAL-inactive

1:37 AM

I'm using x as the size of the partition

Semiclassical

@SimplyBeautifulArt Given that I know a duplicate of it, I'd want it closed

PVAL-inactive

your using it as the number of partitions

Dodsy

now solve for y' = 0

Semiclassical

ahhh

Typhon

@Dodsy e was the best number

it's because of analysis stuff

Semiclassical

1:38 AM

@Dodsy yeah, that gives $x=271/e$ as the ideal number of terms

Typhon

the nature of the product integral

which has a pretty large significance here for this stuff

Simply Beautiful Art

@Semiclassical Hm... well, anything closed as a dupe would be good, but not closed as off-topic. At least, I don't think that one should be.

Akiva Weinberger

$e^x\ge x+1\iff e^{1/e}\ge x^{1/x}$

Dodsy

Oh you guys already solved it?

Typhon

i should have noticed we were narrowing around e and picked up on that

@Dodsy 271 when split into one number is best done with the constant e with a straggler term.

Dodsy

1:39 AM

or did my differentiation suggestion help

Semiclassical

(I notice that, by the way, because my answer to the earlier verison of that is my highest-voted answer :P)

Dodsy

So you guys solved this with no help from me!?

Typhon

nah. @Semiclassical mentioned e and I realized that was it

Dodsy

LIES.

Semiclassical

40

Q: Fun Geometric Series Puzzle

I recently was reminded of a puzzle I solved in college and thought I'd give it a shot again. However, being distanced from college math, I am having a harder time remembering how I arrived at the solution. The problem is as follows: Imagine you are standing in the middle of an open field. You w...

calculus puzzle

weee

Akiva Weinberger

1:40 AM

Proof: Substitute $x\mapsto\frac xe-1$ in $e^x\ge x+1$.

Typhon

@Dodsy it's possible something other than e might perform better. It's the best so far.

Dodsy

yes but I solved it best.

Akiva Weinberger

We get $e^{x/e-1}\ge\frac xe$

Dodsy

Well,the answer is e.

Typhon

wait...

Dodsy

1:40 AM

clearly.

Typhon

welll...

Akiva Weinberger

which means $e^{x/e}\ge x$

Dodsy

but you guys didn't prove it.

Typhon

this is assuming it is all the same term

Akiva Weinberger

which means $e^{1/e}\ge x^{1/x}$

Dodsy

1:40 AM

I did, by finding the derivative.

Typhon

and not a mixture

PVAL-inactive

I know how to differentiate functions like that

I've taught it for seven or eight years

Typhon

@Dodsy you have to compare the integral of a function to it's product integral

over a range of some piecewise constants

Dodsy

@PVAL-inactive of course.

no, we just had to find the vertex.

Typhon

e just seems obvious from that as a guess

Dodsy

1:41 AM

as PVAL said.

Typhon

????

what are we talking about here

im referring to the 271 sum game

Semiclassical

anyways, n=100 gives 1.98e43 as the highest result.

Dodsy

yes.

Semiclassical

What I find cute is that it's big but not infinite i.e. there is a point where it starts to go down.

Typhon

i only know enough calculus to do basic stuff

Dodsy

1:43 AM

Well, the first derivative, if we set it equal to zero

Semiclassical

(The fact that they picked 271 as to be close to 100e is icing on the cake.)

Dodsy

we can find the vertex.

which is what we were basically looking for.

Semiclassical

Right.

Typhon

what if have the terms are ideally one number with half a different number?

Semiclassical

The way I'd prove the entire thing, to be clear.

Dodsy

1:44 AM

better than churning numbers.

Semiclassical

First, show that if you've got $n$ terms then the product is maximized when you pick all the numbers the same.

That's most easily done with the arithmetic mean - geometric mean inequality.

Dodsy

Don't know what that is, but agree.

Semiclassical

It's cute: The arithmetic mean of a sequence of positive reals, $\frac{1}{n}(x_1+x_2+\cdots+x_n)$, is at least as big as the geometric mean $(x_1x_2\cdots x_n)^{1/n}$.

PVAL-inactive

I think if you look at (M-x_1-x_2-...-x_n)(x_1)(x_2)...(x_n) you'll get a function with a unique maximum as well

Dodsy

hm.

PVAL-inactive

1:46 AM

so you can do calculus to show that as well.

Semiclassical

And furthermore the only way to get equality is when $x_1=x_2=\cdots=x_n$.

Typhon

@Semiclassical the problem is really equivalent to "find a number such that the only combination of summed terms which gives itself or higher is itself by itself"

Dodsy

huh that's interesting.

Typhon

that number is e

Semiclassical

hmm, an interesting way to put it

Anyways, you're given that $x_1+x_2+\cdots +x_n =271.$

Typhon

1:47 AM

i cannot split e into two numbers a and b such that a*b >= e and a+b <= e

Dodsy

I guess that would be the intuitive way to see it.

Typhon

proving that e is that number is beyond me

probably real analysis stuff

Semiclassical

So you've got the bound $x_1 x_2\cdots x_n\leq (271/n)^{n}$, with equality only when $x_1=x_2=\cdots = x_n$.

Typhon

but it's a funky number and if I had to bet money, I'd bet it on e.

Balarka Sen

@AkivaWeinberger Say $p_1, p_2, \cdots, p_s$ be those points, and let $S_i$ be the union of all the radius 1 disks containing $p_i$ (that's a disk of radius 2 centered at $p_i$, I guess). Every triplet $S_i, S_j, S_k$ intersect nontrivially and $S_i \cap S_j \cap S_k$ contains a disk of radius $1$ containing $p_i, p_j, p_k$. By Helly's theorem $\bigcap S_i$ is nonempty, and is contained in $S_i \cap S_j \cap S_k$... hmm, if only I can fit that inside the radius 1 disk it contains.

Semiclassical

1:48 AM

Now you just need to show that the largest possible value of the RHS is achieved when $n$ is as close to $271/e$ as possible

Dodsy

@Typhon no it's just calculus.

Like I said

finding the vertex.

by finding the first derivative and setting it to zero.

Typhon

@Dodsy what vertex?

Semiclassical

Well, there's more than one way to approach it.

Dodsy

There is a vertex here.

@Semiclassical no I mean, it'st not real analysis

Semiclassical

But once you've got it down to $f(x)=(271/x)^x$, then yeah.

Dodsy

1:49 AM

and it's not that difficult.

Typhon

and I meant proving there do not exist any two real numbers a and b such that a*b >= e and a+b <= e?

Dodsy

Oh, sorry.

Typhon

you used one attempt

Akiva Weinberger

I like how "$a$ is in a disk of radius 1 centered at $b$" is equivalent to "$b$ is in a disk of radius 1 centered at $a$"

Dodsy

I thought you meant you'd have a difficult time proving that $e$ is the answer for the question.

Akiva Weinberger

1:50 AM

It's a nice duality

Typhon

mine however shows that e is the best one to use... in general

Akiva Weinberger

They both mean $d(a,b)\le1$

Dodsy

no..

Typhon

@Dodsy me?

Semiclassical

I think @typhon may be on a good track here, actually.

Balarka Sen

1:51 AM

@Akiva Basically symmetry of the metric

Dodsy

hm.

Akiva Weinberger

@BalarkaSen Yeah

Dodsy

I'll think, one moment.

Semiclassical

Suppose you've got a number N and you split it up into n parts. You want to maximize the product.

Typhon

im just guessing that proving that property of e shows it is the best possible number or (at the very least) that nothing below e works better.

Dodsy

1:52 AM

Well, see I was looking at it from finding the vertex, but let me think here.

Typhon

@Semiclassical split N into \lfloor N/e \rfloor e's and a straggler term of N - \lfloor N/e \rfloor

Semiclassical

@Typhon's guess, if I take him right, is that you'll do best when you can take all the pieces as close to $e$ as possible.

Typhon

^^^^

Semiclassical

And that's a rather interesting characterization.

Typhon

my hypothesis is that if you have e as N

Semiclassical

1:53 AM

I think it may be right, though I don't immediately see it.

Typhon

then the best answer is e

and everything higher stems from that conjecture

and just splitting into more e's

Dodsy

What if the number was 400.

Semiclassical

$400/e\approx 147.152$

So take $147$ pieces, each of size 2.72109. (That's only 1% different than e itself).

Typhon

@Dodsy 147 e's with a straggler term of 0.41257121652035040203774171115861

Dodsy

right.

Typhon

1:54 AM

ok dayum

maybe if the last term is less than one you put it into the last e

Dodsy

is that the maximum?

Typhon

if only cause we don't want to lessen the product

Semiclassical

I wonder if we're missing something, but if so I'm not seeing it.

Typhon

@Dodsy 146 e's with a straggler term will do better obviously

Dodsy

alright.

Semiclassical

1:55 AM

okay, back later

Typhon

if only because 0.41257121652035040203774171115861 is < 1

200 2's is 1.6069380442589902755419620923412e+60

e^146 = 2.5526681395254551047668755808654e+63

3^133 = 2.8650148523904757106795721053232e+63

Balarka Sen

I wonder if I should go to sleep or keep awake

Typhon

forgot the straggler term

derp

3.1308530449793956373980291825113 * e^146 = 7.9920288174551598698208857155677e+63

ummm

the hypothesis seems to hold true

e is the best value still

@Dodsy

@PVAL-inactive hmm?

Dodsy

wait what is this ty.

oh I see.

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