alpha, beta, gamma. it's wolfram alright

a new kind of scientist

Oh, huh, it didn't work

Mathematica gives me positive infinity for Limit[1/x, x -> 0], but it approaches from the right by default

@BalarkaSen Ah, they call it "Complex infinity". (So, when the magnitude goes to infinity but the angle doesn't converge, I guess. It'd correspond to the extra point on the Riemann sphere.)

$(v_i-m)^2=v_i^2-2mv_i+m^2$ @LegionMammal978

@FreeChoice Whoops, forgot the 2 :P

@AkivaWeinberger Ok, I see. So just the point at infinity

@BalarkaSen Ah, you can get it simply by typing in $1/0$ :P

still lame notation

there's a wolfram gamma?

There's a Wolfram Beta?

@FreeChoice So more like this?

I have a math question for you. Given the current bandwidth of the chat, how many more daily users of room 36 can we have on chat.stackexchange.com before there is too many messages being sent per second for anything to be able to keep track of a conversation or get help?

Eh, we're fiine

So far, at least

semivoluntary tongue click

thanks for accepting a causingunderflowseverywhere into your daily user stream

what's Maple SE?

Maple is a scientific programming language

kinda like Mathematica

yeah I heard of it okay

$\lim\limits_{\mathrm{users}\rightarrow\infty}\mathrm{Usefulness}_{36}(\mathrm{u‌​sers})=0$, though, so eventually we'll have a problem

@LegionMammal978 Yeah, up to line 9 it looks good.

mmmm I dont know about infinity.. a computer does not have infinite memory to allow an infinite people to be logged in, I guess we can use such a broad limit doesnt matter right

@FreeChoice And line 10?

@CausingUnderflowsEverywhere That's why the number of users only goes to infinity

@LegionMammal978 Oh I see, yeah it's good.

does mathmematical induction not work when whatever you're trying to prove isn't a function?

or does it work to prove that something is not a function

Example?

a circle

What?

...and the induction?

actually Im just trying to imagine when an induction can prove that the formula is incorrect

still lost

I tend to do that to people

Oh, a circle fails the vertical line test and so is not the graph of any function

Sounds like a potentially interesting question.

but I have no idea what that has to do with induction.

I mean, there was the whole Induction on Real Numbers question, but I doubt that's what you're talking about

Uhm, in what case does the induction step produce a L.S. =/= R.S.

Thanks for the reminder to find out what induction on real numbers is.

@LegionMammal978 Yeah, real induction is great

only in some calculation error?

Like like how the usual induction is equivalent to minimal counterexample, real induction is like infinum counterexample

So, given $L\ne R$, for parameter value $N$, we prove $L\ne R$ for parameter value $N+1$?

That sounds common enough.

But then what does this have to do with circles?

Going from $L=R$ to $L\ne R$ sounds interesting, though not very inductive.

That's a good question, well don't let my lack of understanding of induction let you get confused about what it has to do with circles and why I asked about the circle.

For example, what would a proof by induction look like for the statement for all even integers $2<n<10000000$ there exist prime numbers $p$ and $q$ such that $n=p+q$?

hmm

Actually, the induction step would be broken into cases.

$n<1000...$ and $n\ge 1000...$.

n = 1 + 3

@AkivaWeinberger so is that like the infinity that'd added to $\Bbb R^2$ to make it compact?

or, well to make the riemann sphere

what's wrong with 1000?

I didn't feel like writing out all of the zeros of my arbitrary upper bound.

This is a totally different topic, but I have some kind of abstract-algebraic-structure-thingy that I want to categorize. I know a variety of its properties (such as non-associative, right-distributive, there's two "types"(?) of elements, idempotent multiplication, etc.), and I want to know if something similar has been studied before and/or what the proper axioms are. I don't know if this is something worth asking on the main site. I can give more details if you want.

oh wait

n < 9999999 + 1

so 9999998 / 2 = 4999999 . therefore

I give up

if I'm given an cos value of x and it says that 3pi/2 < x < 2pi and it wants me to find sin(2x) how would I do this

I'm really stuck

do you know about inverse functions?

oh cool this sounds like something I'd understand for once

yes i think so

@MATHASKER and you know about the symmetries of $\cos$?

symmetries?

like the ids?

yes, specifically how it reflects around the Y axis

No i don't know about that i know that cos^2(x) = 1-sin^2(x) and the double and half ids

and also the sum and diff

cos of theta = 1/6

Since cosine gives the same output for multiple inputs, an inverse function for it can only get you so far here

I think but not sure

no like the questio gives u the value of cos (x)

What you'll have to do is "shift" and/or "flip" the result as necessary to find the specific inverse in the range you want

er

I just quite don't remember it, but lets say they give the value, how would I solve it

oh but what if its not in the unit circle

why dont we do the inverse of cos using 1/6 then see what that and the relating positions are?

For example you can always add $2\pi$ so the argument of cosine, without changing the value

So you can also add $2\pi$ to any inverse of cosine, and it's still valid

That's the periodic property

oh

You can similarly take advantage of it being an even function, because an argument to cosine can always be negated without changing the output

80.4 degrees does that make sense

or 1.4 rad

oh, ok then how would i find the sin (2x)

I only have the adj and hyp but not the opp

1.4 rad IS less than 2pi rad, but isnt 3pi /2 like close to 4.5 radians?

oh then may be it was 3/6 I don't remember the value

@MATHASKER the question as you stated above can be solved elegantly

so 360 - 80 = 280 degrees

im just throwing around guesses for now

I'd guess it's homework or something lol

how @MickLH

@CausingUnderflowsEverywhere ya when i drew it was somewhere after 270 degrees and before 360

with a pencil and paper

uhm a calculator?

@MATHASKER Just how I said above, use an inverse function to find a basis of your solution space, then find the solution inside your desired region, that'll be $x$. I'm sure you know what to do once you have $x$.

sin(2 whatever value close to 270 you got?

oh just multiply it by two

oh ok

540/2 = 270 so that thing close to 280 fits our range right

@MickLH im not misleading math asker am I with 2pi - theta being a related angle or something for cos?

Hint: $-1 < \sin(2x) < 0$

I think cos was different than sine, double check your answer math asker

oh ok it was an multiply question in today's test and i missed it, just wanted to know how i was going to do it

but like the answer was something like -3sqrt7/4

alright lets see

maybe not -3 but i know there was an sqrt7/4

Consider the domain of arc cosine for positive inputs

all x values

For all the positive values it's defined on, the output is bounded by $\left[0, \frac{\pi}{2}\right)$

oh

So if we negate it, then we're on $\left(-\frac{\pi}{2}, 0\right]$

And now if we add $2\pi$, we reach $\left(2\pi-\frac{\pi}{2}, 2\pi\right]$

oh because of that periodic thing right

First one is using the property that cosine is an even function, and the second one is using the property that it's $2\pi$ periodic, yes

Simplify that bound and you'll recognize it

oh ok

ill try

hi chat

hi semiclassical, how's things?

Hi

alright finally found the question

Given $cosϑ = 3/4 $ on the interval $3π/2<= ϑ <= 2π$$, find sin (2ϑ)$

Single dollar sign gives you inline math

do i still use the inverse thing?

after finding theta what do i do? add 2pi in this situation too?

I think it helps to put it on the unit circle.

Yes, re-do the procedure of "flipping" and/or "shifting"

but there is no cosine value of 3/4 in unit circle right

Sure there is.

It's not one of the labelled values, but it's still there.

The lack of retractions from the disk to its boundary is essentially why trampolines work.

Trampolines are disks attached to a fixed circle.

oh ok when i do that i know its like in between the 4th quadrant

If there were a way to move the main part of the trampoline entirely to its boundary without needing to rip or tear it, you could accidentally "break" a trampoline and leave a huge hole in the middle.

but then what, i remember getting till here then getting stuck

where is "here"

Well, it's in the fourth quadrant, so you know that the sine is negative.

oh ya true

Combine that with the pythagorean theorem and you can find what the sine in fact is.

i remembered that an then took out the negative options, i got 4 for denominator as hyp is 4 but no opposite

ohhh, I could have done that...I kept on thinking there was an ID or something

there's multiple identities that will get you there

after finding sin, did i just have to multiply it by 2? cause it asks for 2x

No.

Do you remember the double angle identity for sine?

ya divide the original thing by two and sqrt of 1+cos(x)/2

That's the half-angle identity.

hi @Semi

what's up

not much

oh nvm its the 2sinxcosx

right

you know what cos is, and we just concluded we could figure out what sine is.

tbh I think working with the angle directly gives more insight into what's actually going on, but $\sin(x)^2 + \cos(x)^2 = 1$ is probably more well known than $\sin(\arccos(x)) = \sqrt{1-x^2}$.

There's probably a nice geometric argument.

the former of those equations is just a consequence of pythagorean thm

Arguably, it just -is- the pythagorean theorem.

you get that

$o^2/h^2 + a^2 / h^2 = 1$

and so, $o^2 + a^2 = h^2$

@Semiclassical Wait. Are you saying trigonometry is related to triangles? Illuminati Confirmed.

which is obvious by pythagorean theorem

so i got for sin of theta = sqrt 7 / 4

You have 2 small issues with that solution

if its 2 * (sqrt7/4) * (3/4)

the 3 and 2 will be outside right

or can we multiply 3 and sqrt of 7

There's nothing wrong with the arithmetic, there's a subtle flaw in one of your logical steps though

what is it?

The sine was supposed to be negative

What did we conclude about sine based on the unit circle?

$\begin{array}{c|c}A&S\\\hline T&C\end{array}$

ohh ya so its - 3*sqrt7/8

cause of 4th qd

You used $\sqrt{x^2} = \pm x$ right? Didn't just negate it?

@AkivaWeinberger oooh pretty

$\begin{array}{c|c}S&A\\\hline T&C\end{array}$

I mean

I guess you just did the first one so I could mentally write the word AeSThetiC through it

The mnemonic I know is "All Students Take Calculus" (going through the quadrants counterclockwise)

The idea is, it's saying which of $\sin$, $\cos$, and $\tan$ are positive in each quadrant.

oh our teacher said somethig like all student talk constantly

lol

All three, only sine, only tangent, only cosine, respectively.

where did u get sqrt of x^2 from @MickLH ?

I'll take a guess: You took the square root of $\sin(x)^2$ at some point?

That seems like a fairly unnecessary mnemonic.

oh ya sqrt of x when i was trying to find the opposite

Sine is the y-coordinate on the unit circle and cos is the x-coordinate

I always fuck mnemonics up, if it's hard to remember I just have to learn how to derive it... or write it down.

So if you're above the x-axis sine is positive and below you're negative.

Still here?

oh

And similarly being right/left of the y-axis means cos is positive/negative.

@MATHASKER Ok so do you know that the square root actually has two results? Positive and negative?

if the equation gives me something like cos(2x) = 1, when i take the inverse of cos to cancel out the the cos in the left hand side, do i divide by two after taking the inverse or before?

And for tan, well---draw a line through the origin. If it passes through the first/third quadrant then it's a positive slope and tan is positive.

@MATHASKER I want to make sure you didn't just tack a minus sign onto your answer lol. That's not an ok manipulation to arbitrarily negate things. Using the negative output of square root however, is perfectly legal.

yes I knew about that, but i always forget to put it

oh

Well, you could reason as follows.

Since $x$ gives a point on the unit circle in the fourth quadrant, $-x$ would give a point in the first quadrant.

And then use the fact that $\sin$ is an odd function to write $\sin(2x)=-\sin(2(-x)).$

Then you can take positive square roots just fine.

Tic-tac-toe? $\begin{array}{c|c|c}\phantom O&&\phantom O\\\hline&X&\\\hline&&\end{array}$

ohh

O on the top right side

Akiva can you help me with an olympiad geometry problem

$\begin{array}{c|c|c}&&O\\\hline&X&\\\hline \color{Red}X&&\end{array}$

also, $\Large \times$ and $\bigcirc$ would probably look better

can you \big up the \times a little?

O on the top left side

$\begin{array}{c|c|c}O&\color{Red}X&O\\\hline&X&\\\hline X&&\end{array}$

@ZachHauk <3

O on the bottom middle

ok akiva it's \Large\times and \LARGE\circ: $\LARGE\times \; \LARGE\circ$

$\begin{array}{c|c|c}O&X&O\\\hline\color{Red}X &X&\\\hline X&O&\end{array}$

O on middle right

$\begin{array}{c|c|c}O&X&O\\\hline X &X&O\\\hline X&O&\color{Red}X\end{array}$

I think the \Large & \LARGE matched better, maybe vertical offset the \circ a px or two down?