Mathematics - 2017-01-06 (page 5 of 6)

Semiclassical

21:01

The other unfortunate truth is that a lot of things are more easily proven using complex numbers than trig identities. so to some extent you're stuck proving things in a more involved way than is really necessary.

Mahmoud

@Semiclassical Fortunately, I don't have any Calculus/Complex Analysis to worry about for now, but those identities seem to be generating randomly, to the point that any substitution of one in another creates more and more stuff to memorize.

Semiclassical

Yeah, it's hard to avoid that impression.

I guess if I were to dedicate a few things to memory...

One would be the sum-to-product identities, e.g. $\sin(A+B)=\sin A\cos B+\cos A\sin B$.

the way I remember them is that $\sin(A+B)$ and $\cos(A+B)$ should both be expressible in terms of $\sin A$ and $\cos A$ using certain coefficients.

But if $\sin(A+B)=a\sin A+b\cos A$, then when $A=0$ you have $\sin B=a(0)+b(1)=b$ so $b=\sin B$.

Mahmoud

My teacher promised us that he'll make the whole class recite randomly any given identity he happens to ask about any student, therefore a mistake (like confusing - for +) will result in a punishment : writing the formula $100000+$ times.

Semiclassical

and when $A=90^\circ$ you have $\sin(B+90^\circ)=\cos B=a(1)+b(0)=a$

that's dumb.

Admittedly, I do have the signs memorized at this point: for sine, they add. for cosine, they subtract.

but still dumb.

Mahmoud

We said : Isn't $100000$ a lot ? Answer : No, not really, it's just doing $10000$ $10$ times.

Semiclassical

21:11

Best response: Have the class increase their volume by 10 decibels.

That's just doing the same volume 10 times at once.

Mahmoud

LOL.

Semiclassical

The frustration of recitation does not vary on a log scale.

(i.e. the rise in frustration when going from 1 recitation to 10 recitations is not the same as the rise when going from 10000 to 100000 recitations)

Mahmoud

The words Mathematics and Recitation don't go well together.

Semiclassical

They really don't.

Mike Miller

@BalarkaSen I'm not 100% sure that's going to do what you want it to do. Remember that while $M(G,n)$ is unique up to homotopy equivalence (when it's simply connected), but maps from $M(G,n)$ do not have the same classification properties. So I don't think your pinching map will be well-defined up to homotopy.

Mahmoud

21:15

@Semiclassical $\sin^2(x)=\frac{1+\cos(2x)}{2}$ But memorizing something like this, is .. tricky.

Semiclassical

I tend to remember the form $\cos 2x = \cos^2 x-\sin^2 x$, due to the similarity with $\cos^2 x+\sin^2 x=1$.

Mahmoud

If only I could find some geometric intuition behind all of them, I know there should be

Mike Miller

In general it's unclear why you're mapping out of $M(G, n)$. You should really get groups $pi_X$ as maps out of $\Sigma X$. Then we have $\pi_X(Y) = [S^1 \wedge X, Y] = \pi_1 \text{Map}(X,Y)$.

Semiclassical

If I then use that second identity to eliminate $\cos^2 x$ from the first, I get $\cos 2x=1-2\sin^2 x$.

And that rearranges to the identity you gave above.

Mike Miller

But this is only defined when I have chosen a homeomorphism of the domain with $S^1 \wedge X$.

Mahmoud

21:18

Does this have something to do with $\frac{d}{dx} \sin(x)=cos(x)$ ?

Semiclassical

Sure, you can understand it as a consequence of $\frac{d}{dx}\sin x=\cos x$.

If you differentiate $\sin^2 x$ using the product rule, you get $2\sin x\frac{d}{dx}\cos x=2\sin x\cos x$.

Null

@Mahmoud (arbitary formular)$\cdot 10^5$

Semiclassical

Now, that second expression is familiar to me: it's the double-angle identity for sine, $\sin 2x=2\sin x\cos x$.

So $\frac{d}{dx}\sin^2 x=\sin 2x$.

Mahmoud

Now to the integration part ?

Semiclassical

Yeah.

If I take indefinite integrals on both sides, I get back $\sin^2 x$ on the left and on the right I get $-\frac{1}{2}\cos 2x+C$ where $C$ is a (momentarily) unknown constant.

To figure out what $C$ is, I evaluate both sides for some simple value---say, $x=0$.

The left is just $0^2=0$, whereas the right is $-1/2+C$.

For those to agree, I need $C=1/2$. So $\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x$ as you said earlier.

Animesh Ashish

21:23

Maybe we should use limits and find if it's defined or not.

Semiclassical

Actually, though, I just realized there's a simpler approach.

Animesh Ashish

And what would that be?

Semiclassical

If you start from $\sin 2x=2\sin x\cos x$ and differentiate both sides, you get

Animesh Ashish

Got you!

Semiclassical

$2\cos 2x=2(-\sin x)(\sin x)+2(\cos x)(\cos x)$

dividing both sides by 2 and simplifying gives $\cos 2x=\cos^2 x-\sin^2 x$, which is the form I gave earlier.

Animesh Ashish

21:25

Semiclassical, are you a graduate?

Semiclassical

Grad student in physics, yes.

Animesh Ashish

Year?

Semiclassical

Long enough, I'll say that much.

Animesh Ashish

College?

Semiclassical

University of Minnesota.

Animesh Ashish

21:26

Did you apply for SAT?

Semiclassical

SAT?

Animesh Ashish

Scholastic Aptitude Test?

Jorge Fernández Hidalgo

@Semiclassical can you help me understand this phrase from a song?

Semiclassical

You don't exactly apply for that. You take that as a matter of course in order to apply for college.

Animesh Ashish

Wait, I need help first

Jorge Fernández Hidalgo

21:27

"I can hold you down, like I'm givin' lessons in physics (Right)" @Semiclassical

Semiclassical

I should point out, though, that the SAT is relevant for getting into college/university as an undergraduate

Mahmoud

I found this beautiful diagram from YouTube :

Jorge Fernández Hidalgo

it's from a famous song by Iggy Azzalea but I don't understand how physics poffesors hold you down.

Animesh Ashish

I am an Undergrad and I need help.

Mahmoud

Semiclassical

21:29

You have to have graduated from college in order to get into grad school, and for that the relevant aptitude tests are different.

@Mahmoud Figure out which angles in there are $\theta$ and that diagram becomes pretty straightforward.

especially given that the white line has length 1.

@JorgeFernándezHidalgo Could not tell you.

Animesh Ashish

I can appply for SAT , get a good SAT score and be in a good University such a Stony Brooke/Ivy League?

Null

If you want some quick points: math.stackexchange.com/questions/2086746/…

Mahmoud

Isn't it the same thing you showed me earlier ?

Semiclassical

No idea, frankly. Undergraduate admissions are a process into which I have very little insight.

Mahmoud

And yes the circle is of radius $1$

Semiclassical

21:30

Talk to an academic counselor.

@Mahmoud I showed you one of the other angles that's the same as $\theta$.

But there's a few right triangles in there.

Null

Oh, if you gave me upvotes thanks, tho, i don't think my questiuon is that good^^

Semiclassical

Though I think that one we discussed earlier may be enough, as I look at that diagram.

Animesh Ashish

Weren't you an undergrad once?

Semiclassical

For instance, you can use that angle to get the $\cot\theta$ relation.

Animesh Ashish

@semic

Semiclassical

21:33

Yes, and all old men were once boys. Doesn't mean they know how to use twitter.

Animesh Ashish

@Semiclassical : Help my out in taking decisions of my life. The most cruel and important ones.

Jorge Fernández Hidalgo

@Semiclassical dropping some wisdom there m8.

Animesh Ashish

How do I upload a picture here? I have a question to ask. Also, Where is the chatroom in the app?

Semiclassical

I'm not going to pretend to have insight into a process which I took part in over a decade ago, and which may well have changed in the intervening years.

Animesh Ashish

I understand. Sorry to forecast a stubborn attitude by insisting.

Semiclassical

21:35

Plus, it's not just your SAT scores. There are a ton of things which affect college admissions: extracurricular activities, for instance.

I really have no idea what an Ivy League school is looking for.

Mahmoud

The other angles are $\pi /_2 - \theta$ In both the Pink and Blue triangles.

Semiclassical

@Mahmoud Yeah.

Mahmoud

But what does that tell us ?

Semiclassical

well, for one.

Mahmoud

$\sin(\frac{\pi}{2}-\phi)=\cos(\phi)$ ? This way ?

Semiclassical

21:37

Yeah, that's where I was going.

and similarly $\cot(\phi)=\tan \theta$ if $\phi=\pi/2-\theta$.

What this diagram won't help you with is angle addition, alas.

Mahmoud

That's a good one,

Semiclassical

There's only two angles in this diagram, and the second one is determined by the first. So we don't have any way to get stuff like $\cos(A+B)$ from here.

you can get that from a diagram like this: goo.gl/images/c6XnWn

but I'm not sure how easy that is to remember.

Jorge Fernández Hidalgo

@Semiclassical noice

Semiclassical

honestly, the angle addition formulas for cos/sin are something I do have memorized at this point. they really are useful to solidly know.

Mahmoud

Check this article,

betterexplained.com/articles/intuitive-trigonometry

sine and cosine are percentages ?

Semiclassical

21:44

I wouldn't put it quite like that, but it does capture something correct.

Mahmoud

It makes sense because $-1\le \sin(x)\le 1$

Semiclassical

Suppose I draw a right triangle. there will be some base angle, and that angle determines the proportions of the right triangle.

Mahmoud

Well if you replace $1$ with $100%$

Semiclassical

each of the legs of said triangle will be some fraction of the hypotenuse. if I make that triangle bigger, those lengths will change but the fractions stay the same.

and those fractions are just the cosine and sine of the base angle.

So in that sense it is fair to call them percentages. the point where I don't like that statement is that those percentages, sine and cosine, are not independent quantities.

Mahmoud

And hypotenuse will be always $1$ so this is why I never recognized the fraction.

Semiclassical

21:49

if I know that the sine is 1/sqrt(2), then I know that the cosine had better be 1/sqrt(2) as well. (or -1/sqrt(2), depending on where I am on the unit circle)

So they're percentages but not independently so.

They have to be consistent with the Pythagorean theorem.

Also, you'll note that they don't touch at all on the various identities that come with the trig functions.

so there's limits to how useful this actually is.

Mahmoud

Anyway, this is better than saying SOH-CAH-TOA.

Semiclassical

eh, yes and no. the trouble is, you still have to remember which fraction goes with sine and which fraction goes with cosine.

At the end of the day, we refer to sine and cosine because that's what people have done for a long long time now.

Mahmoud

And because of $e^{\pi i}=\cos(x)+i\sin(x)$

Semiclassical

You can use intuition to explain why you'd want to have fractions like sine and cosine. But you can't use it to explain why cosine is the ratio of the adjacent length to the hypotenuse.

Evinda

@JorgeFernándezHidalgo Do you know a game which has an interesting algorithm in order to win ?

Semiclassical

21:54

eh, not really. I could've chosen to call those functions Rp(x) and Ip(x), for real part and imaginary part.

They'd still have the same properties as cosine and sine, but not the same names.

Since they do have those names, we have to remember them. And for that, ye olde SOH-CAH-TOA is simple enough as a mnemonic.

Jorge Fernández Hidalgo

@Evinda do you know the nim game?

Semiclassical

So there is a certain 'hard core' of stuff that you want to know without having to think about it.

Jorge Fernández Hidalgo

Or the game in which two players repeteadly remove $1,2,3$ or $4$ coins froma stack with $100$ coins.

Mahmoud

@Semiclassical This diagram contradicts the other one, or does it ?

TheGreatDuck

woah woah woah.

Semiclassical

21:57

It's a different triangle.

They've got the wall distance being 1, not the hypotenuse.

TheGreatDuck

tangent function is the red's slope

Semiclassical

If I rescale it so that the hypotenuse is 1, then the construction given in the other diagram indeed gives tan(theta).

TheGreatDuck

"tangent = wall height"

nope

Mahmoud

But the blue line will get longer, right ?

Evinda

@JorgeFernándezHidalgo I read about it right now. Do you think that it would be a good presentation topic?

Semiclassical

21:59

In this particular framework, it works. But that's precisely because they chose the wall distance to be 1.

TheGreatDuck

since when is the height of any right triangle tangent?

tangent is the slope of the hypotenuse!

Semiclassical

$\tan\theta = \frac{\tan\theta}{1}.$

TheGreatDuck

oh

you scaled the base to 1.

Mahmoud

@TheGreatDuck Slope$\neq$Height

TheGreatDuck

-_-

Semiclassical

22:00

Hence what I said about their choice of wall distance. It's special to that.

TheGreatDuck

@Mahmoud tangent = slope. height is not tangent!

oh

well the image is deceptive mildly

Semiclassical

well, rise over run = rise if the run is exactly 1 :p

TheGreatDuck

it appears to indicate a general relationship

:p

anyway

Semiclassical

Yeah. Within the context of that article, it's fine.

TheGreatDuck

just popping by

cya

Semiclassical

22:01

But out of context it's definitely misleading.

TheGreatDuck

^^^

I'm flagging that chat post as needing closure for lack of context. jk

Semiclassical

lol

Mahmoud

シ

Jorge Fernández Hidalgo

@Evinda it's one of the most important games out there, mainly because of the sprague grundy theorem.

Semiclassical

A better description would be: Wall height = wall distance * tangent, ladder length = wall distance * secant

I think what makes that article reasonable is that it connects trig with various applications.

That's the trouble with it as well, though. A lot of the applications of trig identities are not as simple as this.

TheGreatDuck

22:04

^^^

all trig articles involving real life things should use irrational numbers for their things if only because it enforces usage of all the identities in a truly correct nontrivial way.

Semiclassical

:/

Mahmoud

It's just trying to give a funny approach,

We have some fancy new vocab terms. Imagine seeing the Vitruvian “TAN GENTleman” projected on the wall. You climb the ladder, making sure you can “SEE, CAN’T you?”. (Yeah, he’s [A word was here, use your imagination]… won’t forget the analogy now, will you?)

LOL

mercio

what

TheGreatDuck

...

Semiclassical

@Mahmoud As sources of mnemonics, sure.

Mahmoud

22:06

@mercio betterexplained.com/articles/intuitive-trigonometry

TheGreatDuck

NSWF

Semiclassical

But eh. At some point, you need to pick a mnemonic and stick with it.

mercio

what @ great duck

Semiclassical

For me, SOH-CAH-TOA does that just fine.

TheGreatDuck

not safe for work

Semiclassical

22:07

My preferred example of a trig identity statement that is very useful in algebra and fairly useless in terms of geometry:

TheGreatDuck

i prefer to just remember polar coordinates

remember the recipricals

Semiclassical

$\cos n \theta$ and $\frac{\sin{(n+1)\theta}}{\sin \theta}$ are both polynomials in $\cos \theta$ for any nonnegative integer $n$.

TheGreatDuck

and just use everything else when i need it

irl one rarely needs the complicated rules for trig

good for algebra manipulation

but rarely useful for trig

Semiclassical

Good for certain problems.

Rarely relevant for actual geometric problems.

TheGreatDuck

@Semiclassical i just mean that knowing every rule is excessive.

Semiclassical

22:10

Sure, sure.

TheGreatDuck

usually the basic rules and law of sines gets you through everything decently

in the real world if you actually find a spot where those don't work you're either going to know it by habit or you'll look it up cause it is so rare.

imo

granted

the half-angle formulae are essential for integration

but im only referring to trig and triangles so that's not relevant

Mahmoud

Here is a well animated video,

Tattoos on Math

►

mercio

just use exp(it) = cos(t) + i sin(t) and forget everything else

TheGreatDuck

technically i could also bring up spherical triangles in spherical geometry

Semiclassical

Well, the polynomial observation is useful as a rule of thumb. It suggests that if you see a complicated trig identity, you usually can rearrange it to a polynomial identity in $x=\cos\theta$.

Mahmoud

22:12

The title is a little misleading.

TheGreatDuck

@mercio that is a really bad idea

mercio

also yeah the half angle formula are good because it parametrizes the circle with a single variable

Semiclassical

@mercio Weierstrass substitution is surprisingly handy, yes.

TheGreatDuck

@Semiclassical im thinking in the sense of you have a flat geometric figure (either in space or the plane) with angles and you wish to deduce all the angles and lengths if possible. some things become less relevant.

important sure

but less relevant

Semiclassical

Agreed.

Evinda

22:14

@JorgeFernándezHidalgo Interesting. Can you suggest me a good source about it?

TheGreatDuck

sin and cos have other applications, yes but I am 100% purely referring to geometric purposes.

Semiclassical

@TheGreatDuck Where that line becomes a bit blurry is with integration in, say, spherical coordinates.

TheGreatDuck

if other identities have other uses than you learn the identities more fluently that you use in your area.

Semiclassical

But good old Euclidean geometry? Standard trig is all you should need.

I feel about obscure trig identities about how I feel about pi:

smbc-comics.com/?id=1777

Mahmoud

How is $\tan$ related to $\sec$ ?

TheGreatDuck

22:17

@Semiclassical well as I was going to say, most people do not learn spherical geometry. In fact, our advanced geometry class didn't really teach it (basically a two week blurb), and I just know a lot about it cause I did an extra project regarding it.

Semiclassical

Take a look at that ladder diagram you had earlier, and use the Pythagorean theorem.

mercio

i don't even know what sec is

TheGreatDuck

@Mahmoud sec is 1/cos

Semiclassical

sec = 1/cos

csc is 1/sin

TheGreatDuck

of course

...

you're not stupid

if anything

they meant me

Semiclassical

22:18

okay, time to go wait for the light rail in the cold.

Jorge Fernández Hidalgo

@Evinda john conways book on games and numbers

Mahmoud

@Semiclassical Hope it comes in time $:)$

Evinda

@JorgeFernándezHidalgo Nice... I will look for it

Dair

@Mahmoud $\tan(x) = \frac{\sin(x)}{\cos(x)}$ so $\frac{\tan(x)}{\sin(x)} = \sec(x)$ or $\sec(x)\sin(x) = \tan(x)$

Mahmoud

Never mind : $\frac{d}{dx}\tan x=\sec ^2x$

And similarly : $\frac{d}{dx}\cot x=-\csc ^2x$

Dair

22:26

@Mahmoud Do you know how to derive the equalities?

Simple Art

@Dair Yes, I do :D

Mahmoud

Yes, derivative of sine and cosine + product and quotient rules.

Simple Art

Say

This is related

1

Q: The $n$th derivative of the hyperbolic cotangent

While working on a problem, I came to this: What is the $n$th derivative of the hyperbolic cotangent? For simplicity, let $c=\coth(x)$. $c=c$ $c'=-c^2+1$ $c''=2c^3-2c$ Etc. It appears to be representable as a polynomial of $c$. Any ideas on what the coefficients are? Update: It appear...

Since you know that

$$1+\tan^2x=\sec^2x$$

We can say that $\frac d{dx}\tan x=1+\tan^2x$

And what is the derivative of that, in terms of $\tan$?

What is, for example, the 100th derivative of $\tan$ as a polynomial degree 101 of $\tan$.

Think about that for a bit :D

And today I finally got lowered to second place in the ranking

but that's cuz I was sleeping and such. I'm back in first :D

Mahmoud

Congratz :D

Simple Art

Thanks

Mike Miller

22:31

@BalarkaSen Do you know a better name for $\langle a, b \mid [a,b]^2\rangle$?

Simple Art

And it appears ye who surpasses me shall gain little more rep today

So I don't have competition

By the way, there's this one chick I found on MSE

and I think she's beautiful

what to do? XD

FYI, I am age 17

Mahmoud

How can you find .. On MSE ?

Dair

Let's ask the magic conch:

"Nothing"

Simple Art

XD

@Mahmoud with my daily activity, I've seen most of the users who participate

:P

1

A: An integral of a rational function on the real line

If you take the counter-clockwise contour of a closed semi-circle with radius $R$ in the complex plane with a contour along the real line from $-R$ to $+R$, we have $$I_0=\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx=P.V.\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx$$ $$\lim_{R\to\infty}\oint_{\gamma...

I finid it quite funny that I can post this answer quite nicely

Yet I haven't the slightest clue on basic real analysis

@MikeMiller What do you think of me?

Dair

Complex analysis makes soooo many things easier lol

Simple Art

22:36

@Dair The best of things :D

mercio

omg it's dair nooooooooooooooooooo

Simple Art

@mercio LMAO

Dair

haha

Mahmoud

Well, unless I misunderstood you (misconceptions happen) : Go for it (>‘o’)> @SimpleArt

Dair

hi @mercio

Simple Art

22:37

XD Maybe, I'll try...

mercio

hi ;w;

Simple Art

Hi

Mahmoud

Hi

Simple Art

Why the face?

Mahmoud

To look more natural.

Like ヽ(´▽｀)ノ

Simple Art

22:38

Ah, I see. To fit in with the rest of us

Mahmoud

(>‘o’)> @SimpleArt <('o'<)

Simple Art

@Mahmoud despair.com/collections/demotivators/products/…

tilper

To fit in with the rest of us? Bah, conformity! I'm so angry I could overturn this table.

(╯°□°）╯︵ ┻━┻

mercio

D :

Simple Art

despair.com/collections/demotivators/products/…

mercio

22:41

tables are sacred

don't hurt them

Simple Art

Guys, seriously, check the links

I'll send a pic so you don't have to if you don't want to. One moment please.

cdn.shopify.com/s/files/1/0535/6917/products/…

Null

what does a scalar(?) do to a matrix? like $c\cdot A$

tilper

Scales it

mercio

it multiplies every entry of the matrix by $c$

Simple Art

Yeah, literally

Dair

22:42

@Null It multiplies each entry by $c$...

Null

yeah, but which entries

tilper

I like how the dare to be different poster is sold out.

Simple Art

XD

Dair

all the entries

tilper

All of 'em

Null

22:42

@Dair ah ok

Simple Art

@tilper HAHAHAHA

I love that site

despair.com/collections/demotivators/products/downsizing

SteamyRoot

Sweet, finally 44 gold medals in Euclid: The Game

Mahmoud

@SimpleArt Maybe,

cdn.shopify.com/s/files/1/0535/6917/products/…

Simple Art

:D

Mahmoud

That was supposed to be D:

Dair

22:45

geese brilliant.org adding group theory now...

Simple Art

D:

despair.com/collections/demotivators/products/…

Mike Miller

@SimpleArt I don't

Simple Art

=D Ok that's fine

Mary Star

Let $p$ be a prime, $n\in \mathbb{N}$ and $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ irreducible. We have that $a\in \overline{\mathbb{F}_p}$ is a root of $f$.
We have that $\mathbb{F}_p(a)$ is a finite extension of $\mathbb{F}_p$. How can we show that the extension $\mathbb{F}_p(a)/\mathbb{F}_p$ is normal?

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$Mathematics$ Mathematics