Stranger Things

THIS is not math anymore.

@Mahmoud Out of curiousity, what other sort of weird problems showed up in the HW?

Many Many @Semiclassical

heh

were they mostly of that existential/universal quantifier sort?

$a$ and $b$ and $c$ are odd numbers.

Prove that $$ax^2+bx+c=0$$ has no solution in $\mathbb Q$

ah.

And yes they were @Semiclassical

Some with the summation.

If you've got access to the rational root theorem, that's immediate

actually, though

maybe not immediate

a brute-force approach would be to say that the part of the quadratic formula that leads to an irrational answer is the square-root part, i.e. $b^2-4ac$

so you want to show that this is never a perfect square.

Amazing ! :D

The password at this coffee shop is wedonthavewifi. Only later when the owner asked me if I was able to connect did I realize what she meant.

eh, only amazing if I can figure out why it's never a perfect square :/

LOL, @MikeMiller

though, if $b^2-4ac=m^2$ for some integer $m$

and find contradiction ?

then $4ac=m^2-b^2=(m-b)(m+b)$

yeah, somehow. not that I see it at the moment

since $b$ is odd, $m-b$ and $m+b$ are either both even or both odd

if it's odd, the LHS doesn't work

so you'd need $m-b$ and $m+b$ to both be even, i.e. $m$ odd

don't think that's enough, though

I guess I'm not seeing it, right now

$$(\forall n\in \mathbb N^*) \sum_{k=1}^n (-1)^{k+1} k^2=\frac {(-1)^n+1 n(n+1)}{2}$$

@MikeMiller We do the opposite at my home. There's a shop opposite to where I live, let's call it XYZ Shop. We named our wifi XYZ_Shop_Guest_WiFi

ah, the formula for an alternating sum of perfect squares

That's awful

@Mahmoud probably want to use induction for that

@Krijn why would you do that

It's a bad shop that attracts quite a lot of vandalism to the neighborhood

So I don't mind trolling them

also, still need to change that to (-1)^{n+1}

Yes I tried but got stuck as always

gotcha.

@Krijn what the hell do they sell

but yeah, nothing terribly interesting about that one. they give you the formula, so use induction

The $k+1$ part

the harder problem would be to find said formula if they didn't actually give it to you

@MikeMiller Drugs, mostly

Do you know how you'd do induction here?

@Krijn Well I'm all ofr that

Yes but I don't know how to play with the Summation

@MikeMiller Weed is allowed here, and that's all fine and sunshine

okay. Just to check: What's the first step in any proof by induction?

So that I can use my assumption as usual

Verify for $n_0$

the n=0 base case, yes

Then Assume it's true for some $m\in\mathbb S$

This is more of a situation where they are clearly selling drugs behind the counter but they masquerade as another shop

you wouldn't want to do induction on $k$ here, though. you're already using that for the summation index

Oh yes

yeah

I don't know the English idiom for that unfortunately

$m$ ?

that's fine, yeah

And then prove for $m+1$

@Krijn the english word I know is 'front'

right, @Mahmoud

Usually that'll include putting the assumption into the formula for $m+1$

so you assume $$\sum_{k=1}^{m} (-1)^{k+1} k^2=\frac {(-1)^{m+1} m(m+1)}{2}$$

And then some clever $$\text{Algebra}$$

heh, yes

Which is the part where I mostly fail horribly. :(

Well, induction is usuaully simple for summation formulas for one simple reason: How does the summation change when you go from $m$ to $m+1$?

@Mike I realized today that I'm not particularly fond of Algebra

You write down the summation for $m$

It's number theory

and then add the whole thing for $m+1$

right. so all that happens is you add the $(m+1)$th term

I add it computed ?

right

That was very helpful :D

more symbolically, $$\sum_{k=1}^{m+1} (-1)^{k+1} k^2=(-1)^{m+2} (m+1)^2+\sum_{k=1}^{m} (-1)^{k+1} k^2$$

but what do you know about the second part of that?

@Krijn As it happens, you still end up using a lot of algebra

One more problem to go if you don't mind :I

heh. first, do you see what you'd do next?

I replace the second term in RHS with the assumption :)

@MikeMiller Very much, but algebra is vast and used in everything

I used to think I'd be able to get away without much of it

right. so it becomes $(-1)^{m+2} (m+1)^2+\frac{1}{2}(-1)^{m+1}m(m+1)$

@MikeMiller I remember a student asking me (I was a TA for a course on group theory) "Why do I need this course if I want to do mathematical physics"?

oof

Factor by $(-1)^{m+2}m(m+1)$

something like that

probably just $(m+1)$, since $m$ doesn't appear in the first term

but at that point it's just some tedious manipulation

it's straightforward and boring and therefore not worth lingering on too much

Hello

what's the other problem? @Mahmoud

I have some questions about domain, for several problems I'm working through.

$$|x|\le1 \land|a|\le1 \Rightarrow |ax^2+x-a|\le \frac 54$$

Oh, wauw, I just wrote the best $\xi$ ever

\implies works for implies

yeah.

not going to lie, that's the kind of problem my brain just sort of goes 'nope' at.

not because it's that hard, I doubt it is

Nope at ?

but just can't get myself excited about :/

Yeah Absolute value inequality, not much going on :P

yeah.

I'd probably default to using calculus out of laziness

But solving it is a real frustration for me.

I have the function $V(x) = \sqrt{1-(\frac{x}{2})^2}$

And I think the domain would be $(-\infty, \infty)$ but that is probably wrong

I still didn't learn about calculus officially in class, but I know how to compute Derivatives, thanks to KhanAcademy :)

Oh, wait, no that's definitely wrong.

nice

Graph it !

@Mahmoud, I'd recommend The Cartoon Guide to Calculus, it's what I'm working through right now.

Okay thanks :)

rip

brother was in car accident

@user3502615, oh, I'm so sorry (I know that sounds superficial, but it is sincere).

hes not seriously injured, luckily

his friend died though

who was in the car with him

It's only defined on $[-2,2]$ @heather

@Mahmoud, okay, so you'd say $\sqrt{1-(\frac{x}{2})^2} < 0$, and then solve to figure out the domain?

Interesting function :)

No, $1-\sqrt{\frac {x}{2}^2}$

I think you need a comma between the x and 2 in your \frac command

mhm

You'd want \frac{x}{2}

and a backslash before sqrt

@Mahmoud, but that doesn't make sense...why would the $1-$ be outside the square root?

Just ignore that

I'm still learning MathJax

But my equation must be wrong, because solving it, unless I solved wrong, gave $x>2$

which is exact opposite of what the upper bound should be.

The square root over the reals is defined if and only if whatever is under the root, is positive

$1-(x/2)^2$?

@SteamyRoot Or zero.

Yes ! Oh meh.

Oh, right. Non-Belgian people defined positive as >0 >.<

It must be greater than Or equal to zero

then comes computation @heather

@SteamyRoot, right, but solving that gave the wrong answer. I probably solved wrong.

@SteamyRoot Flemish or Walloon?

Anyway, @Mahmoud, put the ^2 outside the brackets. And if you want brackets to stretch, like when around a fraction, use \left( ... \right)

Flemish :p

k.

@mahmoud those curves of $y=ax^2-x-a$ plotted from $-1$ to $1$ and for $a$ varied from $-1$ to $1$

@Mahmoud, yeah, that's what I did wrong. I solved for <0, not greater than or equal to 0

Ah, ik heb toevallig in Leuven gestudeerd het afgelopen semester.

How to send images ?

@Semiclassical Thanks for graphing those :D

@Mahmoud, okay, I got the less than or equal to 2 part, but how do you get greater than or equal to -2?

Ahzo - welke vakken/proffen?

Mhm

Eigenlijk alleen interessant Alg. Geom. van Nero Budur

You forgot to set the absolute value when taking the square roof function @heather

Sorry, still bad at MathJax

Hmm... heb ik ook gevolgd, toen hij voor het eerst lesgaf. Vak was wel interessant maar zijn lesgeven was toen afschuwelijk

I can't write it :/

Oh, duh! It's $\pm$ whatever I'm doing, and of course multiplying by a negative switches the sign...

\pm is the mathjax notation

so then the domain is [-2, 2]

victory!

Yay !

@SteamyRoot Ja, vrij accuraat.

@Mahmoud, that is, until I start the next problem =P

@heather :)

Now how to send images in chat ? >:I

Well, you can click upload right next to the send button and upload a file.

I don't see it.

There is only one button.

Also, if there's a mathematical symbol you don't know the LaTeX/MathJax for: try detexify.kirelabs.org/classify.html

@SteamyRoot Can you tell me how to send images ? please

Might need rep. for that

rep. ?

Okay, let me see if I did the next problem right: if I have the function $g(\theta) = \frac{\tan \theta}{\theta^2 - \frac{\pi}{9}}$ then I have the domain $\{\theta\in\mathbb{R}|\theta \neq \sqrt{\frac{\pi}{9}}\}$, right?

Reputation, like on MSE

@heather Is $\tan \theta$ defined for all $\theta \in \mathbb{R}$?

Plus or minus again @heather

@SteamyRoot, I have no idea.

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

When is $\cos \theta$ zero?

@Mahmoud, curses, that thing keeps coming back to haunt me. But seriously, though, isn't that implied in the square root?

@SteamyRoot, when the x coordinate is 0?

It happens to me all the time

Oh, gosh.

The angle @heather not x coordinate

@Mahmoud, but the x coordinate of the intersection of the angle and the unit circle is equivalent the the cosine. That's what I meant.

Think about the unit circle

You know the $x$ intersection, what's left is the angle $\theta$

so if $\sin \theta = 0$, then $\tan \theta$ is undefined, and then problem.

At what $\theta$ ?

uh, any $\theta$?

I'm not quite sure what you mean.

You mean $cos(\theta)=0 \forall x$

When does the cosine equal zero

When $\theta$ is $\pm 90$?

:D

Oh, lightbulb moment.

the domain would really be $\{\theta \in \mathbb{R}|\theta \neq \sqrt{\frac{\pi}{9}}, \pm 90\}$, then?

$\frac {\pi}{2}+k\pi$

Yes

sweet!

But don't write it in degrees

Thanks for all your help. Sorry I take so long to get these things; dunno how I'm going to make it through calculus at this rate. =)

No degrees, okay.

It's okay

I took longer when first learning this stuff :P

And note that it repeats to be negative every $k\pi$

So then $\pi/2$ and $3\pi/2$ instead of $\pm 90$

Every $k\pi$? What does that mean?

pi/2+kpi

Imagine sitting on the top of the unit circle

at pi/2 the cosine is 0

right

if you go half way around it'll be negative too

the cosine??

adding another half you get to 5pi/2 which makes the cosine equal zero again

And so on and so forth.

oh, okay (i think)

You think ? What's the issue ?

tan=sin/cos right ?

yeah

So what's the problem ?

Nothing, I've got it =) I just was still thinking about it

:D

=) thanks again for all your help!

Next problem, hopefully I'll get it on my own =P

Ok I'll go to bed now, exam tomorrow.

G'night.

Next problem, hopefully I'll get it on my own =P

Yes, just focus.

@Mahmoud, good night, and good luck on the exam!

Thanks :)

Bye.

=)

$e^{2u}=1$...how to solve for $u$?

Hmm, could you square root both sides to get $e^u=1$?

And then...oh, wait, could you do $u = \ln 1$?

And then you get $u = 0$! It is $\pm$, because of the square root, but it doesn't matter because it is $0$!

Okay, so then the domain for $A(x) = (1 - e^{2x})^{-1}$ is $\{x \in \mathbb{R}|x\neq 0\}$, right?

@heather are you only interested in real number solutions?

@arctictern, yes.

then yes u=0 is the only time exp(2u)=1

@arctictern, thank you!

mmhmm

So then for the function $T(u) = (1 - e^{2u})^{-\frac{1}{2}}$ would you have the domain $\{u\in\mathbb{R}|u<0\}$?

yes

@JessyCat Given any divisor $d\mid n$ (let's say $n=dk$), the subgroup of $\Bbb Z/n\Bbb Z$ generated by $d$ is given by $\{0,d,2d,\cdots,(k-1)d\}$. This is cyclic of order $k$. There are a set of coset representatives given by $\{0,1,2,\cdots,d-1\}$ (all integers mod $n$ of course). Given two representatives, to compute the sum in the quotient group you just add the representatives and discard excess multiples of $d$. In this way, the quotient group is cyclic of order $d$.

@AndrewThompson I figured out what they meant by that one thing where they said they wrote down a map that was "induced" by a map $[0,1]^2 \to [0,1]$.

@heather Ah, the same problem here than there in PSE ?

What's up, mathematicians ?

im ok

I'm happy about this circumstance.

I'm ok, too.

just tired

Yeah, me, too. It's 2 AM here.

i didnt have any sleep last night

@phy

oops

Why ?

@PhysicsGuy

where do you live?

i was at the hospital

Germany.

ah, guten tag?

What did you do ?

something happened

Ja, guten Tag.

@user3502615 accident ?

yeah i guess

dont want to talk about it

some people in the #math freenode channel

Ok.

told me to e-mail a professor

Do you want to talk about maths ?

because in my current circumstance, i cant receive guidance from my math teachers

(they haven't taken the courses)

Okay

is that a good idea?

What ?

I don't understand.

i'm currently self-learning maths, but i have no one to "guide" me

because my teachers havent taken the courses

Oh, I understand.

What are you learning right now ?

currently reviewing abstract algebra

Oh yeah, and what are going to do after that one ?

commutative algebra

Ok.

im interested in algebraic topology

That's very interestng, yes.

i've already taken the following undergraduate courses: calc sequence, diff eq, linear algebra, real analysis, complex analysis

I understand.