Mathematics - 2016-12-02 (page 6 of 11)

Axoren

07:00

And this is a result of the limit, which may be more than I can explain right now.

Jasch1

Ok

Axoren

Their density will actually be $\frac 1 m$ if $m$ is a natural number.

Jasch1

I get that

and if b is not very large right?

or even if it is?

Axoren

To be large enough to matter, $b$ cannot be a natural number.

Jasch1

k

Axoren

07:01

Because there exists no natural number large enough to matter.

Jasch1

You sure about that?

It seems odd

Axoren

Pretty sure. It's how limits end up working in this case.

Brody

$mx+b$ describes the multiples of $m$ shifted up by $b$. The numbers we're looking at still appear so-many times with or without the shift

Jasch1

So it wont affect it?

Balarka Sen

I heard Danse Macabre a few weeks ago. Liked it.

Jasch1

07:03

Is there any way to do this with a quadratic or higher rule that can be factored

Brody

Compare {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,...} against {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,...}

Jasch1

I understand why b wont affect it

how do i write expts in latex?

Axoren

\$e^x\$

If you only have every $m$th number after $b$ numbers, then your "found" equation after having checked $c$ numbers is $f(c) = \frac{\frac c m- \frac b m}{c}$

Please tell me if I messed that up, @Brody

I don't feel comfortable with it.

Jasch1

$x^2$ + 5x + 6

Axoren

@Jasch1 Generally, you should write the full equation between \$ \$

Jasch1

07:06

Ok

Axoren

Like $x^2 + 5x + 6$

Jasch1

Will do that next time

but eqautions that can be factored

Axoren

It's a good habit to get into if you're writing latex.

Brody

I'm personally not sure about those @Jasch1. That's an interesting one to investigate.

Jasch1

I'm supposed to do some of my own work so that may be it

Axoren

07:08

@Jasch1 Instead of defining your linear rule that way, instead define your "found" function.

Jasch1

Nobody else did their own work becuase its fucking hard to come up with something that hasnt been discovered yet

Axoren

$f(c) = x^2 + 5x + 6$

Brody

I'm not sure what you're talking about @Axoren, sorry :/

Axoren

If you have that, your density is ratios are $\frac{f(c)}{c}$

Jasch1

Ok thats easy

But would be and easier way to reach the desnity

Axoren

07:09

A good approximation for the density is to pick $c$ to be REALLY REALLY LARGE

Like, $10^{10^{10}}$

Jasch1

I would think that there would be some way to do it with factoring

Because it seems pretty easy with binomials

Axoren

Or just pick something like one million

So, be careful.

Kasmir Khaan

hi guys what is the idea of surface integral ?

and flux

I dont understand what am calculating

Axoren

If you've checked $c$ numbers so far, you can't have found $c^2$ of those numbers in your subset.

Kasmir Khaan

please help anyone

Axoren

07:11

That means you've found more numbers than you've checked, @Jasch1

Jasch1

brb for two min @Axoren

Axoren

@KasmirKhaan Do you know what a surface is?

Jasch1

need to talk to a friend real quick

Im back

Kasmir Khaan

@Axoren yes sir

Axoren

You're measuring the area of the associated region on that surface

$user image$

Kasmir Khaan

07:15

but why in the formula we take dot product with normal vector

Axoren

Which formula are you using?

Specifically for the flux?

Jasch1

@Axoren is there an easy way to get density with wolfram alpha?

Kasmir Khaan

yes flux and all

and we need orientation of surfaces

am all comfused

Jasch1

@Axoren all I need to know is if there's a way to do natural density of wolfram alpha

Balarka Sen

@KasmirKhaan Given a vector field on the surface, you want to compute how much stuff "flows out of your surface".

Axoren

07:19

@Jasch1 Well, for nice $f(c)$ you can just say $\lim_{c\to\infty} \frac{f(c)}{c}$ which is written in Wolfram as wolframalpha.com/input/…

The example I gave there is $f(c) = \frac c m + b$

Balarka Sen

So if you have a vector field $F$, you dot with the normal $n$ to look at the magnitude of the component of $F$ on that direction.

Jasch1

so linear

Axoren

@Jasch1 Yeah, that would be linear.

Balarka Sen

That's precisely the amount which "flows out normal to the surface"

Axoren

But you can replace the above with any $f(c)$

Balarka Sen

07:20

Integrating that is precisely the flux

Jasch1

So I'm gonna mess around until I get something for quadratic

Kasmir Khaan

@BalarkaSen thanks mate , but the orientation why is it important ?

Axoren

@BalarkaSen Is that how people normally integrate over surfaces? I normally just sum over the arc lengths of all horizontal slices.

Null

@KajHansen hi

Brody

@Axoren Interesting, but this formulation is a bit confusing. No?

Balarka Sen

07:21

@KasmirKhaan Look at the moebius strip. Without orientation the normal field on the surface can be "twisty"

Kaj Hansen

Hey

Axoren

@Brody It's technically the same one as the wikipedia.

Fargle

Hello @Kaj

Axoren

It's the function $a(n)$ associated with $A(n)$

Kaj Hansen

Hey there

Kasmir Khaan

07:22

@BalarkaSen yes my teacher gave that exemple as well , well i have to read the book again and ill post few question here

Axoren

I thought that would be easier for someone who's not a mathematician

Balarka Sen

@Axoren You're thinking of iterated integrals. That works, but the usual definition is Riemann sum.

Kasmir Khaan

thanks @Axoren @BalarkaSen

Balarka Sen

Sum over stuff lying over small squares below

Under integrability assumptions, these two are equivalent

Axoren

@BalarkaSen So people normally integrate over the bivectors?

Balarka Sen

07:23

I don't know what that means

Axoren

I thought you were more versed in differential geometry than I.

Jasch1

@Axoren Where would I put the rule

Balarka Sen

I have never in my life have heard of the term "bivector"

Axoren

@Jasch1 wolframalpha.com/input/?i=lim+(here)%2Fc+as+c+goes+to+infinity

Jasch1

whenever I put a rule where c is in the numerator i get infitite

Brody

07:24

@Axoren $a(n)$ is just the number of times stuff from $A$ appears in the first $n$ positive integers. That's all

Axoren

@Brody That's the same definition of $f(c)$

The the letters are associated with English words that have meaning.

Brody

Remind me how you define $f(c)$ @Axoren

Jasch1

And use c as the domain

right?

Axoren

@BalarkaSen a bivector is an area element (a square in a vector space would be one)

A bivector is like a parallelogram-shaped region defined by two vectors in the vector space

Jasch1

@Axoren whenever I put in anything with a degree above 1 it returns infinity

Balarka Sen

07:26

Eh, sure. The right term for that is a 2-tensor.

Axoren

@Brody How many numbers you've found in your subset out of all the numbers you checked.

@Jasch1 You need to ensure that $f(c) \le c$

Brody

@Axoren Isn't that precisely $a(n)$ with different symbols?

Axoren

Think about it: How can you find more numbers in the subset than numbers that you've checked.

Jasch1

@Axoren under all values of c?

Brody

Hold on, trying to understand what you mean by checking numbers

Axoren

07:27

@Brody I don't know how I could have misspoken. That's exactly what I said earlier, isn't it?

@Jasch1 $f(c)$ is how many numbers you've found so far after checking $c$ numbers. If you only checked $2$ numbers, could you have found $3$?

Brody

So for perfect squares, among the first 20 positive integers there are 4. Is that your $f(20)$?

Jasch1

Nope

So then it wont work for quaratic

Axoren

@Brody Yes

@Jasch1 Instead, use $f(c) = \sqrt c + u$

Brody

@Axoren Ah, I see now

Jasch1

@Axoren Now its always 0

Axoren

07:31

@Jasch1 That's because the density of all perfect squares is so small, that that set is practically empty.

Jasch1

@Axoren seems like it only works with degree one expressions

Axoren

@Jasch1 No, it still works. It's just that that's the actual density.

Jasch1

k

so anything with degree < 1 will be 0

Axoren

Yes.

Any degree higher than 1 will also be nonsensical.

Jasch1

and anything with degree > 1 will be $\infty$ if the leading coefficient is positive

Axoren

07:33

Because you'll have found more numbers than you've checked.

Jasch1

Whats infinity in latex?

Axoren

\$\infty\$

@Jasch1 Yeah, because that'll make $f(c) > c$

Which isn't possible. It's nonsensical.

Jasch1

any if the elading coeffieceint is negative, you get - $\infty$

makes sense, just writing it all so i dont forget

Axoren

$0 \le f(c) \le c$ should be enough to define any valid function.

Jasch1

Under any circumstances

Null

07:36

. math.stackexchange.com/questions/115822/… look at the related links board. hilarious

Axoren

If you want a function that's strictly positive, you just need to make sure it follows a linear rule $\frac c m$

Kaj Hansen

"Find the limit", "Evaluate the limit", "How do I solve the limit"

Jasch1

Is there a latex wiki?

Axoren

"Push it to the _______."

Brody

32 links. Goodness

Null

07:38

@Axoren ouch. bad jokes incoming :D

Axoren

@Jasch1 sharelatex.com/learn

On the left, there's a section for mathematics

Brody

@Axoren What I gather so far is that we "invert the rule" that describes our subset and choose that as our $f(c)$

Jasch1

\ f(c) \

Steve

How do I show that $3x3$ matrices are linearly independent?

Jasch1

Why is that not working?

Axoren

07:40

@Brody Likely, that's a sufficient argument. How how do you invert $ax^2 + bx + c$?

@Jasch1 use \$ f(c) \$

Brody

@Axoren Apply the quadratic formula

Axoren

@Brody That grants you the roots. How does that help you find an inverse?

Brody

Instead of $0$ you have $y$, so the discriminant looks like $b^2+4a(y-c)$

Jasch1

@Axoren What if \$ f(c) \$ = c + 8?

Why is that wrong?

Brody

Granted, quadratics are not invertible, but this will give you the $x$ that map to $y$

Axoren

07:42

@Jasch1 While you do get a density of $1$, it doesn't make sense for that to have been how many natural numbers you found so far.

You found 8 more natural numbers than you've seen so far.

Jasch1

lets say f(c) = c + 8

{0, 8, 16, 24...}

Axoren

@Brody Any parabola fails the horizontal line test.

Jasch1

@Axoren What would that return

Axoren

@Jasch1 Specifically what would what return?

Brody

@Axoren Exactly, hence why I'm cautious to say it's inverted, but you have an equality that defines the inverse relation

Jasch1

07:43

The density

O yea nvm

Axoren

@Jasch1 The equation can be used to give you a number. Because you applied it to a nonsensical situation, it doesn't mean anything sensical.

Jasch1

lmao im retarded

from f(c) = c + 8 I got that set

Brody

$$x=\dfrac{-b\pm\sqrt{b^2+4a(y-c)}}{2a}$$

Axoren

It's like if we applied Newton's First Law of Motion to your shoe size and age.

Those are numbers, but they don't match up with physical quantities those equations were originally designed to work with

Jasch1

So f(c) = mc - b ?

where b must be positive?

Alessandro

07:45

@steve I don't understand the question, are you working in a space of matrices?

Axoren

@Jasch1 If $m \le 1$, it's sensical.

Brody

I'm still curious as to how taking the limit of $f(c)/c$ provides the density, not that I've thought too hard about it so far either

Axoren

Also, sensical isn't a word, I don't think. But I'm defining it right now as the opposite of nonsensical.

Steve

@Alessandro vector space, $\mathbb{R}^{3\times 3}$

Brody

sensical

Adjective: sensical ‎(comparative more sensical, superlative most sensical)

(neologism) That makes sense; showing internal logic; sensible.
1986, Fred D'Agostino, Chomsky's System of Ideas, Clarendon Press, p. 189:
A nonsensical sentence, then, is one which is inconsistent with S, while a sensical sentence is one which is consistent with S.
1998, William Storm, After Dionysus: a theory of the tragic, Cornell University Press, p. 41
It contains […] no intrinsic propositions concerning whether its effects are sensical or not.

(4 more not shown…)

Axoren

07:46

@Brody I'm assuming that you're counting all the natural numbers in order, but the benefit of that is that no matter what order you count all the natural numbers in, you'll get the same result.

Null

@Axoren how did you define nonsensical?

Axoren

@Brody Spellcheck disagrees.

Jasch1

@Brody I'm taking his word for it, i dont understand precalc

Axoren

@Null Not making sense.

Brody

Spellcheck is awfully narrow-minded.

Alessandro

07:47

@steve so you have 2 matrices and want to show that they are linearly independent?

Null

so, not not making sense-> making sense for sensical? @Axoren

Axoren

It still thinks my last name should be Amoral.

Steve

@Alessandro i have 4 but yeah

Brody

lol @Axoren

Alessandro

@steve that's just $\mathbb{R}^9$ by the way

Axoren

07:48

@Null English is one of the few languages for which double negatives become positives.

I'm totally going to wave my hand at this one.

Brody

@Axoren At least in standard English, and standard English doesn't even exist!

Axoren

Brody confirmed that it exists "somewhere"

So someone agrees with me

Jasch1

@Axoren Even if m ≤ 1, it doesnt return infinity

It return m

Axoren

@Jasch1 It should return $\frac 1 m$

Brody

Hey @Axoren, let's be the two original co-founders of the World Royal Academy of the English Language. Right now

Jasch1

07:49

No, just m

wolframalpha.com/input/…

Steve

@Alessandro I mean matrices/vectors like these: $$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
$$

Alessandro

@steve I know

Axoren

@Jasch1 $\frac{f(c)}{c} = \frac{\frac{c}{m}+b}{c}$

Secret

abc.net.au/news/2016-12-01/…

Axoren

You're found function is the reverse of what you're using.

Alessandro

07:51

The vector space of $n\times n$ matrices over a field $K$ is isomorphic (as vector space) to $K^{n^2}$ @steve

Axoren

Wait, nevermind

Jasch1

@Axoren that would mimic a set built with mc+ b right?

Axoren

Yeah, I'm dumb

You're using it right because $m \le 1$

Jasch1

Can you fix the precalc and they way you would test it

Axoren

So $m = \frac 1 2$ if you're picking every second number

Jasch1

07:52

because I was getting some weird answers

Steve

@Alessandro I'm not quite sure how I show this when the vectors have dimensions like $3\times 3$

Axoren

@Jasch1 Explain?

Jasch1

nvm I think I get it

Alessandro

Do you agree that the space of $3\times 3$ matrices over R is a $9$ dimensional vector space?

Jasch1

so if the rule is 2m + b, our function rule with the limits should be (1/2m) + b

right?

Axoren

07:53

@Brody In regards to that Academy. I say we demand the Brits relinquish their erroneous u's.

Jasch1

m should be c

Axoren

@Jasch1 Yeah. In reality, you should just drop the $b$

Jasch1

k

Axoren

Because again, in the actual function, it doesn't matter

Alessandro

I mean, scalar multiplication is defined componentwise, and so is vector addition, it doesn't matter if you write them in a line or a square

Brody

07:54

@Axoren I'm wondering if your argument hinges on whether $f^{-1}(n)\approx |\{1,2,3,\ldots,n\}\cap\{f(m)\le n : m\in\mathbb{N}\}|$ for large $n$

Axoren

So, you get $mc + b \to \frac c m$

Alessandro

Matrix multiplication is another matter, but that's not part of the vector space structure

Brody

@Axoren That's what I was thinking!

Jasch1

@Axoren if we use a quadratic the results is still the same that we got before right?

Brody

Also, establishing the standard pronunciations for caramel, crayon, and others

Axoren

07:55

@Jasch1 Quadratic rules don't make sense. They're too big.

No matter what quadratic rule you pick, you will eventually have found more natural numbers in your subset than exist.

Jasch1

and anything with degree > 1 will be $\infty$ if the leading coefficient is positive

and anything with degree > 1 will be - $\infty$ if the leading coefficient is negative

for even degree

or both

Brody

Yes, for even degree @Jasch1

Axoren

Also, any degree higher than 2 doesn't make sense.

Brody

$+\infty$ for either positive or negative leading coefficients, with odd degrees

Axoren

Makes even less so than how much sense degree 2 functions make

Because at some point, a cubic curve can change direction

You can start finding LESS numbers than you've found so far as you look at more numbers.

Jasch1

07:59

But we understand how the evens work

Brody

The notion of the limit makes sense, but the quotient no longer represents a natural density

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