Mathematics

Ted Shifrin

12:00 AM

exhaust manifolds are more tiring, though, than intake manifolds.

Balarka Sen

(Just like numbers are interesting, to give an example.)

Meow Mix

They're more exhausting?

Ted Shifrin

Zach, that makes the humor totally unsubtle and much less clever.

@MATHASKER: I'm about to leave. I assume you'll have it all figured out.

MATH ASKER

@TedShifrin if iknow that the angle is 30 degrees how do i convert it to a bearing angle?

Ted Shifrin

What angle is 30º?

MATH ASKER

12:02 AM

the third side was a 30 degree

Ted Shifrin

That makes no sense.

Physics Guy

EXACTLY.

Just sayin'

MATH ASKER

i found the third side to be 627.7 miles long and then to find the third angle as i had 3 sides i did law of cosines

Ted Shifrin

You don't need law of cosines because you have a right triangle.

Just use the angles in the picture to figure out the angle between the third side (at the right vertex) and due north.

It's just like what you've been doing.

You know all three angles in the triangle.

MATH ASKER

but the 150 degrees its slanted, can i just turn it into like a straight line

I just know the middle one

Ted Shifrin

12:04 AM

One is 90º-60º10'. One is 90º. So the last is ...

You know every angle in the picture if you just use the kind of reasoning we've already used.

MATH ASKER

nvm i forgot about 60

Ted Shifrin

Remember you have to use all the pieces of information!

OK, I'm outta here.

MATH ASKER

ok

thanks for helping :D

was it supposed to be 60 degrees 10 mins?

Fruitful Approach

0

Q: Can you extend this Cayley graph of prime relationships $f(a,b,c) = d$?

Here is the Cayley graph of an associated polynomial $f = xy + z$. As you can see the graph is based at any one of its elements and has generating set red arrow = $(2,3,5)(11,13,17)$, and blue arrows change and are shown in the image. See, I can't seem to fit the green arrows into the group be...

Prime relationships form a group is all I'm sayin

Physics Guy

12:46 AM

Fawad

1:10 AM

@PhysicsGuy .

Akiva Weinberger

Interesting read.

Shared by the AoPS Facebook page.

The TL;DR is that a retired professor solved a classical conjecture in probability theory (something that some people spent decades of their lives on), using apparently not-too-complicated ideas.

Title: "A Retiree Discovers an Elusive Math Proof—And Nobody Notices"

(The TL;DR part of that is a combination of people being skeptical of proposed proofs of big conjectures, and him being bad at publishing the proof)

> Known as the Gaussian correlation inequality (GCI), the conjecture originated in the 1950s, was posed in its most elegant form in 1972 and has held mathematicians in its thrall ever since. “I know of people who worked on it for 40 years,” said Donald Richards, a statistician at Pennsylvania State University. “I myself worked on it for 30 years.”

The statement of the conjecture itself looks to be pretty simple, as well.

Fawad

A Long-Sought Proof, Found and Almost Lost. @AkivaWeinberger same article. So that proof is with people or lost now?

2

Akiva Weinberger

> Original story reprinted with permission from Quanta Magazine, an editorially independent publication of the Simons Foundation whose mission is to enhance public understanding of science by covering research developments and trends in mathematics and the physical and life sciences. - Wired

@Fawad Almost lost, but with people

Mary Star

8:29 AM

Suppose we have the following graph of a function:

We see that the function is concave, right?

Secret

9:13 AM

[Superrandom]

Northern cardinal

The northern cardinal (Cardinalis cardinalis) is a North American bird in the genus Cardinalis; it is also known colloquially as the redbird or common cardinal. It can be found in southern Canada, through the eastern United States from Maine to Texas and south through Mexico. It is found in woodlands, gardens, shrublands, and swamps. The northern cardinal is a mid-sized songbird with a body length of 21–23 cm (8.3–9.1 in). It has a distinctive crest on the head and a mask on the face which is black in the male and gray in the female. The male is a vibrant red, while the female is a dull reddish...

Martin Sleziak

9:25 AM

@LeilaHatami Sorry for pinging you about completely unrelated issue, but I thought that it might be reasonable to let you know that there was a discussion on meta about a tag-synonym you have recently suggested: Do we really want (discrete-mathematics) $\to$ (combinatorics) synonym?

2

If there is some further discussion needed about this you can comment (or answer) on meta or we can discuss this in the tagging chat room.

Wow, I am a bit surprised how quickly the above message got starred.

Anonymous

10:07 AM

Is there a list of mathematical topic from easiest to hardest?

valentin

10:33 AM

How does (1-x)^2 > 0 follow from 2(1+x^2)>(1+x)^2?

I see it in 2.16 of paradigmshiftnow.net/fundamental_reality/QUANTUM%20MECHANICS/…

Leila Hatami

@MartinSleziak Thank you dear Martin. I will read it.

valentin

Ok, I see. It is simple.

Martin Sleziak

11:03 AM

@LeilaHatami Thanks for the response. If needed, we can discuss this further in the Tagging chat room. In fact, I have already left there a few messages about direction of synonym. (Making two tags synonymous is not a symmetric operation.)

Leila Hatami

11:42 AM

@MartinSleziak Dear martin, it's not necessary, I have already convinced.

Martin Sleziak

@LeilaHatami Ok. Still, in case you plan to suggest some other synonyms in the future, I guess both things are useful to know: 1) It is not symmetric, so at least in some cases the decision which of them will be master tag matters. 2) It might be useful to post on meta first before suggesting the synonym - at least in the cases of large tags where the creation of the synonym would influence many questions.

Alucard

12:29 PM

@SimplyBeautifulArt hi, how are you

MATH ASKER

12:41 PM

If i have 98 miles per 1 hr = w * 10 in can i divide the 10 out and get 98 miles / 10 inc.?

BAYMAX

Hello@MartinSleziak , you byusy now a days?

Martin Sleziak

Yes.

BAYMAX

ok,I see without you the rooms are empty

:)

Martin Sleziak

lol

Without some proper context, that sounds quite funny. However, I am glad to give some other people an opportunity to talk in various chat rooms. (And I noticed that calculus chat room was quite active even without me.)

BAYMAX

yes that was!

ok,all the best for your work.

If you will be available sometimes we can discuss ring theory ?

Martin Sleziak

12:47 PM

I'd guess that there are probably quite a few romantic ballads about empty room without you.

@BAYMAX Why not simply posting your question here on in algebra chat room (or both) and see whether somebody will respond?

You know, I am not the only person on this site...

MATH ASKER

The question was As Zach speeds away from the police, who are in hot pursuit of him, he calculates his speed to be 98 mph. If he is
sporting 20-inch (diameter) wheels on his ride, find the speed of the wheel in rpm as he eludes the po-po.

So 98 mph is the v rigjt?

BAYMAX

Yes Will post when I start reading , lol :) , ok.

@MartinSleziak

skill patrol

@MATHASKER yes

BAYMAX

BTW nice background music @MartinSleziak

wait how do we do this link thing like "empty room with you " , instead of pasting the link ?

skill patrol

[empty room with you](linkURL)

BAYMAX

12:56 PM

wow wow

Martin Sleziak

@BAYMAX For any message in chat, you can view the source: chat.stackexchange.com/messages/36512419/history (Just go to transcript, click on the little arrow and then click on "history")

BAYMAX

thanks @MartinSleziak @skillpatrol

skill patrol

np pal :-)

MATH ASKER

@skillpatrol how would i convert the inches to minutes i know i can multiply 1 rev/2pi but idk what to do after that

skill patrol

@MATHASKER Start by making a sketch.

MATH ASKER

1:08 PM

I did draw but I'm confused how would i turn the inches to minutes

Maybe convert the miles to inches, cancel the inches out

skill patrol

do you know what rpm means?

Leila Hatami

1:34 PM

Can someone suggest something similar to these stunning results?

4

A: Visually stunning math concepts which are easy to explain

I recently find some stunning visualizations... I prefer to share them all: $5)$ Mean inequalities [from Proof without words] $4)$ Streographic projection [by H.Segerman] $3)$ Farey-Ford Tessellation in non-euclidean geometry [by F.Bonahon] $2)$ Steiner Porism [by Wikipedia] $1)$ P...

Simply Beautiful Art

1:46 PM

@Alucard as I always say, I am doing largely fine :-)

@LeilaHatami I will have to ask you to be more specific to answer your question.

skill patrol

@MATHASKER have a look at this

user8469759

2:17 PM

Hi guys
out of curiosity
if a positive sequence a_n -> 1 (strictly positive) does this happen if and only if log(a_n) -> 0?
I've tried to prove it by myself using the epsilon delta definitions
but I can't work out all the details
for each epsilon there an m such that -epsilon < a_n - 1 < epsilon for all n >= m
which is equivalent to
1 - espilon < a_n < 1 + epsilon iff log(1 - epsilon) < log(a_n) < log(1+epsilon)
assuming natural logs I have log(1 + epsilon) < epsilon, but I don't have log(1 - epsilon) > -epsilon

Alucard

2:29 PM

@SimplyBeautifulArt that is very good to hear. I wish you that your luck never depletes :D

samjoe

3:22 PM

Hello everyone, I am a highschool student who can handle basic algebra and trigonometry, geometry, calculus pretty well but find it hard to do discrete mathematics

Mats Granvik

@samjoe Eat chocolate, it will help you with the discrete mathematics.

4

samjoe

discrete mathematics, mostly combinatorics, binomial theorem application

I eat a lot of chocolate already

Alucard

@samjoe can you explain a little bit further where yur problem is? also, follow Mats advice

samjoe

:)

The problem is that I find it hard to solve multinomial theorem problems, and binomial problems (not the basic ones), also I have hard time handling a complete new series

Simply Beautiful Art

@Alucard I meant large, as in FAIL

Akiva Weinberger

3:27 PM

Interesting: something of negative mass will feel an upwards force due to gravity, but still accelerate downwards.

Taking $g\approx10\rm\frac ms$, the force $F$ will be $-mg\hat\imath$, which would be upwards. The acceleration will be $\frac Fm=-g\hat\imath$, which would be downwards.

Alucard

@AkivaWeinberger but how do you explain negative mass? couldn't this be just another force in the opposite direction of gravity?

Akiva Weinberger

The force would be upwards, yes. But the acceleration would be downwards.

It would also have negative kinetic energy.

Pretty sure that such an object is very impossible.

Alucard

If one is the object, probably not very good for the stomach...

Mary Star

4:04 PM

Hello!!

We have the following relations:

If we double i then s will also be doubled.
If d gets halved, then n be multiplied by 4.
If s gets multiplied by 4, then d will be doubled.

Can we get from these relations a formula?

user8469759

You can get a system from those descriptions

Mary Star

What system? @user8469759

Could you give me a hint?

Alucard

Bad Boys Theme Song [HD] (HD Audio)

►

:D

user8469759

I'm thinking

You have

s = f(i) s.t. 2s = f(2i)
n = g(d) s.t. 4n = g(d/4)
d = h(s) s.t. 2d = h(4s)

sorry

4n = g(d/2)

you probably have to use those relationshion to extrapolates the formula

Mary Star

And how could we solve that system without knowing the functions f, g, h? @user8469759

user8469759

4:17 PM

well as a guess it seems to me that s is a linear function of i, n is somekind of reciprocal

because

s = f(i) and 2s = f(2i) implies s = f(2i)/2 = f(i) then 2f(i) = f(2i)

so I would just say s = i...

sorry s = constant*i

while for n = g(d) and 4n = g(d/4) implies n = g(d/4)/4 = g(d) then 4g(d) = g(d/4)

let me do it again

4n = g(d/2) => n = g(d/2)/4 = g(d) => 4g(d) = g(d/2)

you can solve these functional equations

do you have any conditions on i,d,n,s ?

are they integers?

Mary Star

No. But I want to check also if the following are correct:

- d*i^2= s^2*n
-d^2*i=s*n
-d^2*i^2=s^2/n
- d^2*i^2=s*n
- d^2*i=s/n

How could we check these conditions?

user8469759

start with the first one

as I said you have f(i) = f(2i)/2

you can iterate and you have

f(i) = f(2i)/2 = f(2*2*i)/2*2 = f(2*2*2*i)/2*2*2 = ... = f(2^k*i)/2^k

but anyway if you have those five

you'd probably better just veryfing the condition

or maybe not...

let me think

Physics Guy

arxiv.org/pdf/1603.04246.pdf What do you think about this?

Alucard

Holy sh.. :D

arctic tern

I could have sworn I saw that paper months ago

ah, it was submitted a year ago. it just got revised.

Sha Vuklia

4:49 PM

Hi guys. Say we have $o(\vert t\vert)$ as $t\to0$. Now my teacher says that $\begin{aligned}\lim_{t\to0}\frac{o(\vert t\vert)}{t}=0
\end{aligned}$. But first of all, we know that $\begin{aligned}\lim_{t\to0}\frac{o(\vert t\vert)}{\vert t\vert}=0
\end{aligned}$. How do we know for sure that we can get rid of the absolute values? What if the function indeed goes to zero in the right-hand limit, but doesn't go to zero in the left-hand limit?

this is where my question comes from^

never mind, I asked on the main site!

arctic tern

@ShaVuklia do you know what $o(|t|)$ means?

means it represents an $f$ such that $f/|t|\to0$

surely if $f/|t|\to0$ then $f/t\to0$

Sha Vuklia

oh right, let me see

arctic tern

$f/|t|\to0$ implies $|f/|t||\to0$ i.e. $|f/t|\to0$ implies $f/t\to0$

Sha Vuklia

oh of course!!

but that means that $o(\vert t\vert)=o(t)$ right?

arctic tern

yes

Sha Vuklia

5:01 PM

cool, thanks!

BlueMoon93

Hello friends

I just posted about an article

I quoted relevant information

But I'm not sure if I should lnik the article or not

0

Q: Partial derivative of multi-variate probability

I'm looking at an algorithm and its implementation. In the article: $$ \Delta_t(a) = \frac{\partial V(\pi)}{\partial \pi(a)} . \eta $$ where $V$ is a standard value function using a function $Q$ to represent the expected value of each action $a$, and a probability $\pi$ for playing each action ...

algorithms partial-derivative

Alucard

@BlueMoon93 Why you have doubt that this might not be good to link the article?

What sayss your gut-instinct?

BlueMoon93

5:17 PM

About relevance, I guess

I dont hang out often in MathExchange

But i linked it already

s.harp

5:33 PM

Just found out next semester the topology seminar is about fibre bundles, looking forward to it :)

Sha Vuklia

Why is it true that $o(\Vert t\vec u\Vert)=o(\vert t\vert)\cdot\Vert\vec u\Vert$? As far as I can rewrite, we have:
$$
\lim_{t\to0}\frac{o(\Vert t\vec u\Vert)}{\Vert t\vec u\Vert}=\lim_{t\to0}\frac{1}{\Vert\vec u\Vert}\cdot\frac{o(\Vert t\vec u\Vert)}{\vert t\vert}=0,
$$
so why don't we say $\begin{align}o(\Vert t\vec u\Vert)=\frac{1}{\Vert\vec u\Vert} o(\vert t\vert)\end{align}$?
I know it doesn't matter if we multiply by a constant $\Vert\vec u\Vert$, but I'm just confused why they multiply and don't divide by the constant. In the end it doesn't matter of course, because the constant disa…

Alucard