Mathematics - 2011-12-14 (page 3 of 3)

@FreakEnum first of all, 3 does not divide 98, second of all we wanted numbers that are divisible by all three numbers and 3 is way too small to be divisible by even one of them.

FreakEnum

@robjohn then 48, 98, 105 is factors of n

@robjohn I must say I'd got you Q wrong (chat.stackexchange.com/transcript/message/2715696#2715696)

robjohn

9:16 PM

what is the smallest number divisible by 48, 98, and 105?

that is, all 3 numbers.

FreakEnum

@robjohn LCM of those number , 11760

robjohn

and that is...

FreakEnum

11760

robjohn

yes; so what must be true of every number that is divisible by 48, 98, and 105?

FreakEnum

@robjohn must be lcm of those three numbers :)

anon

9:20 PM

(or a multiple thereof)

robjohn

yes, so can you count how many non-negative numbers less than or equal to 123456 are divisible by 11760?

:2715906 ah, but anon is correct.

FreakEnum

@robjohn no :(

@anon sorry , I hope you didn't mind that

robjohn

how many non-negative numbers less than or equal to 10 are divisible by 3?

0,3,6,9

FreakEnum

aah

robjohn

$\left\lfloor\frac{10}{3}+1\right\rfloor$

FreakEnum

9:25 PM

@robjohn what is that | bar?

N3buchadnezzar

function [ Y ] = Delighet(N,a,b,c )
p = 1:N
list1 = p((mod([1:N],a)==0))
list2 = p((mod([1:N],b)==0))
list3 = p((mod([1:N],c)==0))

G = [list1 list2 list3];
H = unique(G);
Y = numel(H);
end

heh, finally

robjohn

$\lfloor x\rfloor$ is the greatest number not greater than $x$

FreakEnum

@robjohn aah , got it

robjohn

$\lfloor x\rfloor$ is called the floor function of $x$

@N3buchadnezzar so, was I right?

FreakEnum

@robjohn you added 1 , for 0?

robjohn

9:27 PM

yes

N3buchadnezzar

Delighet(1234567,98,48,105)

ans =

48081

robjohn

try 123456, not 1234567

FreakEnum

@robjohn but I don't know how to find floor of some value ? should I search google?

N3buchadnezzar

4808

robjohn

@N3buchadnezzar did you count 0?

anon

9:29 PM

floor it by inspection

N3buchadnezzar

As you see, start at 1 not zero

@robjohn why would you start at zero ?

robjohn

@N3buchadnezzar So 4809 was correct :-)

N3buchadnezzar

Indeed

FreakEnum

@anon what does that mean? :(

N3buchadnezzar

But, now I can calculate it for any number, with any three numbers ^^

robjohn

9:30 PM

@FreakEnum look here chat.stackexchange.com/transcript/message/2716001#2716001

@N3buchadnezzar the hard way :-p

N3buchadnezzar

lets try

heh

if you and me were asked to calcuate this for 200 different numbers, I would be done before you

FreakEnum

@robjohn Idiot me , sorry for pestering for nothing

robjohn

@N3buchadnezzar not if I wrote a program to do it my way :-)

N3buchadnezzar

But robjohn ?

Would it not be easier to calculate the opposite?

That the number is not divisible by any of the three numbers

robjohn

@FreakEnum $\lfloor3.99\rfloor=3$

N3buchadnezzar

9:32 PM

the remaining numbers would be divisible by atleast one of the three numbers.

FreakEnum

@robjohn then |3.50| = 3? and |3.20| = 3 ?

robjohn

@N3buchadnezzar now you're confusing me...

@FreakEnum yes, but you need to use the floor, not the absolute values.

N3buchadnezzar

Assume we have found all the numbers below N that is not divisble by a, b or c.

FreakEnum

@robjohn I don't know how to write in TEX :(

anon

he just didn't want to bother writing something for floor signs

N3buchadnezzar

9:34 PM

We call this set for P

robjohn

$\lfloor x\rfloor$ is the floor of $x$, $|x|$ is the absolute value of $x$

N3buchadnezzar

Then all the numbers divisible by atleast, a b or c. Would be N - P

robjohn

@FreakEnum okay, just so you know the difference.

@N3buchadnezzar pardon? what are N and P?

FreakEnum

@robjohn Thanks a lot for such a nice explanation :)

robjohn

@FreakEnum glad that I could help.

N3buchadnezzar

9:37 PM

1234567 - floor( 1234567/lcm(48,98,105) )

robjohn

@N3buchadnezzar Ack, no!

N3buchadnezzar

I figured so too

robjohn

how would that give you how many were divisible by at least one?

N3buchadnezzar

But you do agree, if we had found all numbers not divisble, by 48,98,105 equal to or below 123456

It would be easy to figure out the atleast one?

anon

"the atleast one"? If you mean the least one, that would be 1, but what's the point? And if you mean at least one, then of course, but again what's the point?

robjohn

9:46 PM

The question we are looking at is to find out how many integers in 1...123456 are divisible by at least one of 48, 98, or 105.

I will leave out 0 so that your program works :-)

FreakEnum

@robjohn answer comes 11 > ideone.com/V3mGG

robjohn

@FreakEnum which is $\left\lfloor\frac{123456}{11760}+1\right\rfloor$

they are including $0$. :-)

anon

in order to count the number of #s divisble by at least one of three numbers a, b, c less than a given n, you need to define S(m) to be the set of all multiples of m less than the given number n, then compute |S(a)|+|S(b)|+|S(c)|-|S(gcd(a,b))|-|S(gcd(a,c))|-|S(gcd(b,c))|+|S(gcd(a,b,c))|, or something roughly like that I believe...

FreakEnum

@robjohn aah , thanks

robjohn

@anon look above at chat.stackexchange.com/transcript/message/2715326#2715326 and following.

$\left\lfloor\frac{123456}{48}\right\rfloor+ \left\lfloor\frac{123456}{98}\right\rfloor+ \left\lfloor\frac{123456}{105}\right\rfloor- \left\lfloor\frac{123456}{2352}\right\rfloor- \left\lfloor\frac{123456}{1680}\right\rfloor- \left\lfloor\frac{123456}{1470}\right\rfloor+ \left\lfloor\frac{123456}{11760}\right\rfloor
$

without 0

FreakEnum

9:50 PM

@robjohn is that relates to my Q also?

anon

oh, I'm late

robjohn

or $\left\lfloor\frac{123456}{48}+1\right\rfloor+ \left\lfloor\frac{123456}{98}+1\right\rfloor+ \left\lfloor\frac{123456}{105}+1\right\rfloor- \left\lfloor\frac{123456}{2352}+1\right\rfloor- \left\lfloor\frac{123456}{1680}+1\right\rfloor- \left\lfloor\frac{123456}{1470}+1\right\rfloor+ \left\lfloor\frac{123456}{11760}+1\right\rfloor$ counting $0$

N3buchadnezzar

FFs

robjohn

I had mentioned inclusion-exclusion.

that is a diagram of the 3 set case of inclusion-exclusion.

N3buchadnezzar

And I claim it is easier to calculate Red - Green

than to do the 6 sets

robjohn

9:53 PM

how would you compute the green?

N3buchadnezzar

If 1234567 is divisible by 48 , 105 and 98

ah

robjohn

you were computing everything outside of the innermost almost triangular area, weren't you?

which is not the same.

N3buchadnezzar

No, in the code. I found all numbers divisible by 48, 105 and 98

These formed 3 sets of numbers

robjohn

do you mean "or"?

N3buchadnezzar

I then made a new set H consisting of set1,2 and 3. The answer is the number of unique elements in H.

set1 = all numbers divisible by 48 less than 1234567

robjohn

9:58 PM

@FreakEnum no, that is the answer to how many numbers from 1 to 123456 are divisible by 48, 98, or 105.

@FreakEnum your question was how many numbers from 1 to 123456 are divisible by 48, 98, and 105.

N3buchadnezzar

Make a list of all the numbers below N divisible by a.

Make a list out of all the numbers in the new list divisible by b

make a third and final list of all the numbers in the previous list divisible by c

robjohn

That will do it, sort of like a sieve.

N3buchadnezzar

Rather cumbersome by hand though

robjohn

Floor[123456/48] + Floor[123456/98] + Floor[123456/105] -
Floor[123456/LCM[48, 98]] - Floor[123456/LCM[48, 105]] -
Floor[123456/LCM[98, 105]] + Floor[123456/LCM[48, 98, 105]]

N3buchadnezzar

floor( floor( floor(123456/48)/98 )/105 )

robjohn

10:05 PM

@N3buchadnezzar what is that?

N3buchadnezzar

Good question

robjohn

That would be something less than 11.

N3buchadnezzar

I thought floor(123456/48)

would give the amount of numbers divisible by 48?

robjohn

@N3buchadnezzar yes

N3buchadnezzar

Oh I see

I forgot about the Venn diagram again

Although it should infact be called an Euler diagram

robjohn

10:08 PM

Floor[123456/LCM[48,98,105]] would give you the count of numbers 1...123456 that are divisible by 48, 98, and 105

N3buchadnezzar

I have a short question, hopefully

let

Simplify ( phi^2 - 1 )/phi )^phi

Where phi is the golden ratio

robjohn

1

N3buchadnezzar

It should be 1

^^

robjohn

I was right the first time :-)

$\phi-1/\phi=1$

N3buchadnezzar

You could argue that the inside almost looks like n^2 -1 - n ?

Jonas Teuwen

10:12 PM

Phew :-). My lecture on the uniqueness of representing measures is postponed until January :-).

robjohn

what is n?

@JonasTeuwen so now you can worry about it all over break? :-)

N3buchadnezzar

Is not phi one of the solutions to n^2-1-n ?

Jonas Teuwen

@robjohn Yes.

robjohn

@N3buchadnezzar it is.

N3buchadnezzar

then

n^2 - 1 - n = 0

robjohn

10:13 PM

so $\phi^2-1=\phi$

N3buchadnezzar

n^2 - n = 1

robjohn

and $(\phi^2-1)/\phi=1$

N3buchadnezzar

Yeah

robjohn

and $1^\phi=1$

N3buchadnezzar

I felt my argument was rather weak, but its good to see my thinkin was correct.

QED

10:17 PM

I am a bit bored.

N3buchadnezzar

undress, put on a monocle, and a pirate hat. Place butter in your bottom. and dance like you do not care

Cheers me up everytime =)

robjohn

@QED last time you say that around here, I bet...

hah

Abba? Queen!

10:27 PM

@JonasTeuwen concerned parent.

N3buchadnezzar

ABBA =)

Jonas Teuwen

It has a very catchy tune that song.

QED

I was listening to Rocket Man actually

N3buchadnezzar

10:48 PM

►

QED

yeah she's nice

that's only 3 years ago?

Jonas Teuwen

►

N3buchadnezzar

I will try to restrain from spamming, but this one is amazing.

►

QED

This Tik Tak thing is a bit complicated

Dylan Moreland

What has happened here

YouTubes all over

:)

Jonas Teuwen

10:55 PM

Mine is brilliant.

Dylan Moreland

Am I supposed to guess what it is???

Jonas Teuwen

JDH always pops up with very strange mathematics.

Dylan Moreland

Is my first version here so ambiguous?

It somehow bothers me that someone somewhere in the world thinks that I don't know that the group of unipotent 2x2 matrices is abelian.

N3buchadnezzar

You do know that?!

Dylan Moreland

One of my many secrets.

QED

11:00 PM

I don't understand how standard models are so hard to pin down if you need choice to build ultrafilters?

N3buchadnezzar

Like clark kent

Tries to find Dylans cryptonite

Matt

Hi.

anon

hey Matt

Jonas Teuwen

@DylanMoreland Well, four people.

Matt

In this question, how is $[0,1]^{[0,1]}$ a product space? I thought this denoted all functions from $[0,1]$ to $[0,1]$.

(I was weak today: friend is over again and we've been drinking again : ( )

anon

11:09 PM

Well, it's like [0,1]x[0,1]x... except uncountably many in the product. Instead of making the component index an element of N make it an element of [0,1] and viola

Jonas Teuwen

Yes, give it the product topology.

Matt

So it's not all functions from $[0,1]$ to $[0,1]$?

anon

Is [0,1]x[0,1]x[0,1]x... the set of all functions N->[0,1]?

Matt

No?

anon

Okay, then someone with topology background get in here and make this clear

QED

11:15 PM

why did you say "No?"

Jonas Teuwen

@Matt A function of $\mathbf N \to [0, 1]$ is just a sequence consisting out of $0$'s and $1$'s right?

So what is an element of the first set?

Matt

What's the first set?

Jonas Teuwen

$\prod_{i = 1}^\infty [0, 1]^i$.

Stuff of the form $(0, 1, 1, 1, 1, 0, \ldots)$.

Matt

Why is it sequences of zeros and ones? If anything it should be sequences of reals in $[0,1]$.

@JonasTeuwen Why to the power of $i$?

Jonas Teuwen

Oh, sorry, I'm sleepy 8-).

I've read $\{0, 1\}$.

Matt

11:20 PM

phew

We were both very confused there : )

Jonas Teuwen

Yes 8-). Sorry for that.

QED

Hello?

Matt

Sorry, my computer just switched off. I hope this doesn't mean my motherboard problem is back.

It's late here, maybe we should sleep over it.

Dylan Moreland

Dupespotting..

Woops, wrong link :p

This one.

Matt

Done.

QED

11:27 PM

I wish people wouldn't ignore me when I ask questions

Beginners tend to do that the most when they are trying to get answers rather than learning something.

anon

heh heh

Matt

@QED You asked me why I said "No?", sorry. I thought it was a rhetorical question. I said that because I think the answer should be "no" but I'm not sure enough to say it without a question mark.

QED

thanks

Dylan Moreland

@QED That's always bad. "Instead of clarifying with you I'll wait for someone who can read my mind and provide a full write-up."

Matt

@DylanMoreland No, I was actually thinking in the meantime.

anon

11:30 PM

to be fair, he said he was drinking again shrugs

(directed at QED's remark)

Dylan Moreland

@Matt Oh I didn't know that he was talking about you

N3buchadnezzar

Its better to think and derive, than to drink and drive

Dylan Moreland

I thought he was talking about question-askers on the site. My apologies!

QED

Well when I said "beginners ..." that doesn't apply to Matt (he's more advanced than me at math), I was just reflecting on the general situation

anon

awkward turtle

QED

11:31 PM

Not just questions on this but also other sites and chat in the past

Matt

Oh, sorry to both of you. My mistake : )

@anon : D

Given the situation I think the right decision is to go to bed.

I can still figure it out tomorrow : )

QED

I don't think so

Matt

Good night folks!

QED

Well I was sort of hoping to say something but I guess you are going then

Matt

And thanks : )

Jonas Teuwen

11:33 PM

@Matt Bye.

Dylan Moreland

@Jonas Hmph, thanks.

Jonas Teuwen

:).

QED

After figuring out what that "No?" meant I was going to say about the isomorphism between pairs of X, triples of X, .. and 2 -> X, 3 -> X, ...

But he ignored me until he was just about to leave

Dylan Moreland

@QED What's going on?

QED

What do you mean? You've been here the whole time

Dylan Moreland

11:36 PM

I'll scroll up.

Aha.

The cold shoulder.

QED

I think it happens a lot to me because I say things in weird ways or ask questions people don't want to consider.

or I maybe just come across as some kind of really cruel horrible person

Dylan Moreland

Something about that Gravatar.

anon

no, I see it as a general pattern in discussions

QED

oh, Maybe people think I'm not so knowledgeable which is actually quite accurate

Dylan Moreland

I wouldn't read too much into it :)

anon

11:40 PM

I feel the temptation sometimes to ignore questions I don't want to consider myself - I am merely pseudonymous here anyway, after all

QED

Yeah, I guess I am thinking out loud here and nobody really needs to hear any of this.

N3buchadnezzar

@QED You are a very skilled and knowledgeble person

(Also a cruel and horrible) =)

anon

I find it interesting that the idea of infinity isn't absolute. I don't understand the set/model theory involved but it's an interesting paradox.

QED

I was freaked out enough by the lowenheim-skolem thing

> I don't understand how standard models are so hard to pin down if you need choice to build ultrafilters?

that was my question before

anon

I don't even know what an ultrafilter is, or what you mean by pining down models, so it kind of flew over my head :)

QED

11:46 PM

I am referencing this math.stackexchange.com/a/91603/16697

This is what I have in mind when thinking about ultrafilters:

* You can construct ultrafilters using the choice axiom

* You can construct nonstandard arithmetics and analysis with them

t.b.

So, I was looking forward to seeing an answer to this thread... *disappointed*

anon

goddamnit Sasha, answering a low-hanging one before I get a chance to even see it >:(

Asaf Karagila

Just before I hit the hay, the hammer calls.

anon

hammer? it's already closed, yes?

oh, t.b. came in some seconds ago

N3buchadnezzar

This will sound stupd

anon

11:53 PM

^ that reminds me

N3buchadnezzar

Could anyone explain to me in a logical fasion why

log_a(b) * log_b(a) = 1

Without rewriting it to ln ?

QED

why not?

anon

a full page ad in a recent SciAm begins, first sentence: "Artificial Intelligence (AI) is the quest to achieve comuters [sic] that equal or exceed human performance on complex intellectual tasks." I thought that was ironic.

Dylan Moreland

Heh.

anon

@N3bu: a=b^c <==> a^1/c=b

Dylan Moreland

11:57 PM

$\log_a(b) \log_b(a) = \log_b(a^{\log_a(b)}) = \log_b b$, right?

t.b.

right.

Dylan Moreland

Hurray.

N3buchadnezzar