12:32 AM
Why do we call $\iff$ "if and only if"? It translates so terribly in english
eh nvm im just overanalyzing

what's the wrong with "if and only if"?

12:53 AM
@Obliv I don't understand the complaint.

1:05 AM
I think the connotation of "only" makes it seem like we're hyper-fixated on a single relation

@Obliv I don't understand.

like it sounded as though if and only if coupled two statements and no other statements were related to them idk
hence why I said I'm just overanalyzing
Like $(A \iff B) \implies \nexists C$ s.t. $(C \implies A) \lor (C \implies B)$ and $\nexists C$ s.t. $A \implies C$ nor $B \implies C$ or something lol

1:37 AM
what's the significance of a normal subgroup? The center of a ring/group is the set of elements that commute with the rest of the structure but I can't grasp the significance of equal left/right cosets
let $G$ be a group, $N$ be a normal subgroup, and $G/N$ be the quotient group. Is the significance that the cosets are reflexive&symmetric
nvm
"If $N$ is a normal subgroup of a group $G$ and if $x^2 \in N$ for every $x \in G$, prove that every nonidentity element of the quotient group $G/N$ has order 2."
this problem doesn't explicitly mention if $|G|,|N|$ are finite or not
does this matter?

2:31 AM
it does not

2:55 AM
can I get a hint for this problem @thorgott
Let $N$ be a normal subgroup of a group $G$ and let $x^2 \in N$ for every $x \in G$.
I can immediately say that the quotient group $G/N$ exists right
and that the element are of the form $Nx$ for $x \in G$
I was going to argue the order of $N$ must divide $G$ and the order of the elements must divide the order of the group/subgroup but idk if that argument works for possibly infinite $|G|$

sure, N being normal in G guarantees you those things (among many others)
yeah, don't argue anything about the order of N
maybe fix x and try to compute the order of the element Nx in G/N

Well, $(Nx)^2=(Nx)(Nx)=Nx^2$

i agree with that (as rewritten)

Does $x^2 \in N$ imply something here ... :P

it does

3:04 AM
normal sg has the property if $x \in G$, then $xNx^{-1} = N$. soo...

i don't always recommend doing this, but, what would x^2 being in N have to imply if the result you're being asked to prove is true

that $e \in N$

i guess also lurking in the bacgkround here is the general issue of when two cosets represented as Nx and Ny (for potentially different x, y in G) are equal to one another

1 sec let me get my worksheet
I know either $xNx^{-1} = N$ or $N \cap xNx^{-1} = \{e\}$
idk if that's even relevant

it's related (all of this stuff is related) but maybe not the best thing. do you know under what circumstances a coset Ng is equal to N?
where here g is just some element of G

3:10 AM
I should :D
i mean intuitively it's when $g \in N$
but idk

well, yes, that's right, although, getting from "idk" to "of course, for reasons i understand" is maybe 99% of this exercise
anyway, that's how x^2 being in N tells you about what Nx^2 is

I'm gonna ruminate
lol

3:27 AM
i've ruminated
it makes sense

@Pizza -1/3 is not the limit of that expression

for additive group $(\mathbb{Z}\times\mathbb{Z})/\langle(5,5)\rangle$ an element $(1,1)$ for example, has order $5$ right since $(1,1)+(1,1)+(1,1)+(1,1)+(1,1) = (0,0)$?
trying to find an element of infinite order
it's cool because $(a,b)$ have to be out of sync
like for no $k \in \mathbb{Z}$ do we have $a^k \equiv b^k \pmod 5$

obliv: someone nitpicky might not like your notation in that first equality (which is written an equality of ordered pairs, when you mean something like the elements of the quotient group represented by those ordered pairs) but yes

yeah sorry i'm sloppy with my notation. I'm sure I should be using $\oplus$ and stuff too

i don't have any opinion about what you should be using (there are any number of OK choices), most such abuses of notation are perfectly fine in contexts other than this exact one (i.e. exercises directed at understanding quotient groups). it is definitely important to be aware of when one is abusing notation. i don't know that it is as important to always use unambiguous, non-abused forms of notation

3:36 AM
all notation/language is abusive to what really matters anyway maaaan..
how is munchkin these days.. ok that sounds horrible when not said by Ted

i don't think it's a coincidence that basically the only time you see \oplus and other funny operation symbols for the binary operation on a group is in those sections in a group theory book
she's doing great

yeah I don't think it caught on very much. Even the book doesn't really use it and it says "i probably should be using this notation but whatever"
"for our purposes its fine"
etc

3:54 AM
hmm so in finding order of an element of a quotient group $G/N$ we have $a \in G$ has order $k$ if $Na^k = N$ so for the example above we need $(5,5)(a,b)^k=(5,5)$
allegedly $(1,0)$ has infinite order
but I don't see how. Isn't $(1,0)^5 = e$
oh well ill figure this out tomorrow

4:47 AM
"(1,0)^5" (maybe also a bad idea to mix additive and multiplicative notation?) is (5,0). this is not in the subgroup generated by (5,5) because [why? why would it be?]
reading over the above, you seem to observe (by implication) that if (x,y) is in the subgroup generated by (5,5), then x and y are congruent mod 5. this is certainly true, but the converse is not. this is not an if-and-only-if test for membership in the subgroup generated by (5,5)
it would be wise to care slightly more about how you are using notation, at least for now. try to make clear in notation whether a calculation refers to elements of the "non-quotiented" object ZxZ, or to elements of the quotient object (if a single calculation refers to both, make sure that it's clear which is which). it doesn't matter which one you pick, but it's important to pick one, and bad practice to do a half-and-half mix that leaves a reader to decide where it might be taking place

5:19 AM
@YourLordJoyBoy Then write those things in some notebook.

5:46 AM
one of the message options is 'permalink' which gives a persistent URL that you can store locally (e.g. in a list of bookmarks)

2 hours later…
7:46 AM
A supposedly very dumb question. If n is a positive integer, then if (1/2)^n <= k (0 < k <= 1), then is it true that 1/2 <= nth root of k ?

yes. this is true even without assuming k <= 1

Thank you. I've spent four days in figuring out this problem: mathb.in/78273 (I have also proven the upper bound, too), and I tend to disbelieve even the basic laws of functions. Inequalities are evil sometimes.
On the other hand, I've learned a ton about AM-GM-HM inequality.
So yeah, a book of David A. Santos is a good exercise source

8:04 AM
"I tend to disbelieve even the basic laws of functions"
I assume this only means you are skeptical about them rather than straight up rejecting them on belief that they're wrong

I'm like, 'hey, I've seen some ideas that made me doubt I understand the basics properly'. Of course I do accept them, I'm just asking myself 'maybe THAT is the exception'
Happy to prove myself wrong

if you look at almost any word for too long, it begins to look misspelled

This is exactly what happens. I'm happy I know some new tricks, though! :)

8:37 AM
@PM2Ring Yes, I also checked on wolfram

2

Show that $(x_1+1)(x_2+1)...(x_n+1)=1+x_1 +x_2+...x_n+$ "other terms". How can the following be expanded ? $$(x_1+1)(x_2+1)...(x_n+1)$$ I have a problem where I have to expand and show that I'll have a form of $1+x_1 +x_2+...x_n+$all the other terms. I don't have a math background so I'm...

Is there something similar to this but for (1-x_1)(1-x_2)...(1-x_n)?

@Rusurano $1-x_1-x_2-...-x_n + ...$ ?

Is (some positive trash)-(x_1+x_2+...+x_n) possible?

positive?
$x_i$ are not assumed to be positive or negative, they don't even have to be real numbers

Oh, wait. I am solving a problem that involves (x_i) being a sequence of n nonnegative reals
I also know x_1+x_2+...+x_n <= 1/2. I have to prove (1-x_1)(1-x_2)...(1-x_n) >= 1/2, and I am doing just that
So far I have shown only that (1/2)^n <= (1-x_1)(1-x_2)...(1-x_n)

8:49 AM
Sure. Bernoulli inequality seems like a way to go

1 hour later…
10:07 AM
how to compute the integral in a) ?

10:24 AM
@Koro It is a contour integral: en.wikipedia.org/wiki/Contour_integration

11:05 AM
Let there be a partition $P=\{x_1,\ldots,x_n\}$ of the domain $[a,b]$ of a Riemann integrable function $f$. Let $m_i(f)=\inf\limits_{I_i}f$ and $M_i(f)=\sup\limits_{I_i}f$, where $I_i=[x_{i-1},x_i]$. Define the oscillation of $f$ at $x$ to be $$\mathrm{osc}(f;x) = \lim_{h \to 0^+} \sup \{ |f(x')-f(x'')| : x', x'' \in (x-h,x+h) \cap [a,b] \}.$$
I'm reading proof of the Lebesgue criterion and I'm stuck on a basic claim. We want to prove that if $f$ is bounded on $[a,b]$ and Riemann integrable, then the set of discontinuities has measure zero. If a function is discontinuous at $x$, then $\mathrm{osc}(f;x)>0$ and so we construct the sets $$A_k=\left\{x\in[a,b]:\mathrm{osc}(f;x)\geq\frac{1}{k}\right\}.\tag1$$
It is then claimed that for a given $i$, if $x\in A_k$ and $x\in (x_{i-1},x_i)$, then $$(M_i(f)-m_i(f))\Delta x_i\geq\frac{1}{k}\Delta x_i \quad\text{by } (1).$$
I don't understand this inequality. The oscillation is not $M_i(f)-m_i(f)$. How does it follow?
I'm reading this is here, page 7 in the pdf.

11:35 AM
I think I found a useful clue, namely that $\sup_A f-\inf_A f=\sup \{f(x)-f(y): x,y \in A\}$.

11:58 AM
@psie yes
$M_i(f)-m_i(f)\geq \text{osc}(f; x)$

makes sense

12:31 PM
Prove the continuum hypothesis is independent from the axiom of dependent choice

1:21 PM
@SoumikMukherjee Going forward I may just post what's said to me in a notepad. Or paste them in a document.

1:42 PM
Anyways..... kinda considering you guys like math experts to a degree since you're like way ahead of me when it comes to the subject.
So I must once again ask for some help if it's alright with you. Multiply polynomials

(10x + 6)(10x - 5)

@XanderHenderson The problem is (10x + 6)(10x - 5)

$$100x^2-50x+60x-30$$
can you take it from there?

@YourLordJoyBoy That is a problem, not a question.

$$100x^2 +10x-30$$

1:50 PM
@JohnZimmerman Just handing the answer to someone doesn't really help them to learn the material that they are trying to learn. "Teach a man to fish..."

@JohnZimmerman You fell into a trap

Truthfully I am trying to learn the material. I don't want it just answered TBH.

especially the easier the question is, the more pressure there should be for the author to solve the question themselves, in my opinion

I'm honestly trying to do these on my own and have done many of them alone. The ones I've been asking about are the ones I've struggled to figure out alone.

look up the F.O.I.L. method

1:54 PM
@JohnZimmerman Oh, how I hate FOIL. I can't tell you how many students I have seen mess up $(a+b+c)(d+e)$ because they learned "FOIL".

@YourLordJoyBoy what you want to understand is how to multiply terms like $(a+b)(c+d)$

@Jakobian Once more, you are correct.

this boils down to identities that any real numbers satisfy like distributivity: $a(c+d) = ac+ad$
So now if you have a product of sums of terms $(a_1+...+a_n)(b_1+...+b_m)$ then you should try to think on how to multiply such an expression out and what would it lead you to

@XanderHenderson but that is a different problem ;)

@JohnZimmerman No. The problem is that students learn "FOIL", and they try to use it everywhere. FOIL is terrible.
Just learn to distribute.

1:57 PM
I barely even use FOIL meself.

and to do this you should use distributive property twice: $$(a_1+...+a_n)(b_1+...+b_n) = a_1(b_1+...+b_m)+...+a_n(b_1+...+b_m) = \ ?$$

So you're saying, use the distributive property

yes, twice

yeah @XanderHenderson I agree with you

I don't know what FOIL is

1:59 PM
I think we all can agree FOIL SUCKS
FOIL: First Outer Inner Last :(

well anyway. Can you tell me what $(a_1+...+a_n)(b_1+...+b_m)$ would be?
@Jakobian here it should be $b_m$ by the way, not $b_n$

foil doesn't suck. I stand by foil
for multiply 2 binomials

Yeah. I'm wearing my foil hat right now too

helps with you now... cosmic radiation

2:04 PM
I don't think foil should be used everywhere, That's the thing about it.

I agree

@Jakobian Why am I confused?

@YourLordJoyBoy How can I know

foil would have you the answer on a silver platter by now

idk about this FOIL but it doesn't seem like an actual explanation for why things are how they are

2:07 PM
@JohnZimmerman How many students have you taught how have learned FOIL?

I could just tell you what $(a_1+...+a_n)(b_1+...+b_m)$ is and that would probably be enough but that wouldn't give an explanation on how I arrived at the result

gotta draw those foil arrows tho

The biggest thing is ensuring I'm doing the math correctly.

@YourLordJoyBoy Again, the key idea is that multiplication distributes over addition: $$a (b+c) = ab+ ac \qquad\text{and}\qquad (a+b)c = ac + bc.$$

@XanderHenderson Yes, agreed.

2:11 PM
right, and maybe this is not a completely clear, but from those distribution laws above we also have more general distribution laws $$a(b_1+b_2+...+b_m) = ab_1+...+ab_m,\ (a_1+a_2+...+a_n)b = a_1b+...+a_nb$$
think of above distrubtional law applied over and over again to arrive at above formulas

Man I wish I could get chatjax working, it'd make it easier to read the math stuff I think.

do you have bookmark bar enabled?

I didn't :o
But I do now

that was probably the problem
do you have "start ChatJax" as one of your bookmarks

I do, yes

2:15 PM
click it

Clicked
WAIT! ITS WORKING!
There we go, I see what you were saying now.
I think I can do this now! So you're saying (a+b)c = ac + bc?

Yes.

Now I see it
That means I can do THIS
Perform the indicated operations.
(14t^4 - 4t^2 + 8t) + (4t^4 - 8t + 27) - (8t^4 + t2 + 14)
Wait WHA

@Jakobian @YourLordJoyBoy can you answer my question here based on what I wrote above that?

@YourLordJoyBoy Use $to enclose 2:20 PM Perform the indicated operations. (14t^4 - 4t^2 + 8t) + (4t^4 - 8t + 27) - (8t^4 + t2 + 14)$
uGH STILL NOT WORKING
But part of it has been fixed

Both sides

Perform the indicated operations.
$(14t^4 - 4t^2 + 8t) + (4t^4 - 8t + 27) - (8t^4 + t2 + 14)$
I SEE IT NOW!
@Jakobian Let me try it.
@JohnZimmerman Ah I see what you were saying there now.

Also watch this video and then try.

Video?

@SoumikMukherjee oh okay. If you want to help them instead then be my guest

2:25 PM
@Jakobian Would that be $am$ instead of $an$?
Okay that did not come out correctly.

_
underscore

@Jakobian ?

it should be $a_1, a_2, ..., a_n$, thats how I numerated those
@YourLordJoyBoy a_n

Ah, you're saying $a_n$
BAM I GET IT NOW!

$n$ terms labelled $a_1, a_2, ..., a_n$ respectively

2:28 PM
If I'm a bit energetic, I apologize. I'm just so glad this is finally working. I love this coding stuff.
@Jakobian I think that makes sense.

anyway first application of distributive law gets you $$(a_1+...+a_n)(b_1+...+b_m) = a_1(b_1+...+b_m)+...+a_n(b_1+...+b_m)$$

@Jakobian Actually he was struggling on whether -(a+b)=-a-b or a+b yesterday. So I assume it would be hard for him to answer what you wrote. I hope I didn't made you angry by directing him to a YouTube video

can you apply it one more time and tell me what it is
@SoumikMukherjee I didn't want them to get into specific examples before we discuss this abstractly

Sorry, the biggest thing is that while not a short one I am kind of on a time limit for studying. A week from tomorrow is the final, and the study guide is due a week from yesterday. And this may very well be the last time I tackle algebra, at least for a grade. If I go into it again I'm planning on doing it as a personal challenge.

I mean, how would you apply distributive law to $a_i(b_1+...+b_m)$ for $i = 1, ..., n$?

2:32 PM
@Jakobian Oh okay, sorry for the intervention.

@JakobianIt would depend on which numbers are in place of $a_i(b_1+...+b_m)$ for $1,...,n?$ I did this correctly right?
YES IT WORKED!

@YourLordJoyBoy what do you mean?
just apply the distributive law
@Jakobian I even wrote it here above, apply it for $a = a_i$

@Jakobian I get it now! I guess what I'm trying to say is what part of the equation replaces $a_i$ and so on.

doesn't matter what $a_i$ and $b_j$ will be for now
just treat them as some real numbers we don't want to talk about in detail
you're trying to derive what could you possibly do if those were, say, real numbers
after you know the answer you won't have to repeat it anymore and apply the result directly

@Jakobian Was trying to put my answer into a worded explanation.

2:38 PM
I will show you how to apply this to your multiplication of polynomials problem

@Jakobian THANK YOU

okay well do you agree that $$(a_1+...+a_n)(b_1+...+b_m) = a_1b_1+...+a_1b_m + ... + a_nb_1+....+a_nb_m$$
after application of second distributive law
$a_1b_1+a_1b_2+...+a_1b_m + ... + a_nb_1+a_nb_2+...+a_nb_m$

@Jakobian First off, I keep wanting to replace the a in your username with o for some reason. But second and most importantly, I agree, yes.

okay but whats the most important in a sum? Its not the order in which we sum but what we sum and here we are summing all the terms $a_ib_j$ for $i = 1, ..., n$ and $j = 1, ..., m$
so the conclusion is that $(a_1+...+a_n)(b_1+...+b_m)$ is a sum over all those products $a_ib_j$
and this is actually what we mean with multiplication term by term

@Jakobian Agreed. I think if I've learned anything it is that the order is irrelevant in these cases.

2:43 PM
you want to take $a_i$ from first sum, $b_j$ from second sum and multiply them, and sum them for all such choices

@Jakobian Yes, I see.

and this is the rule you want to always be applying with those multiplications

@Jakobian Alright, I think I'm getting that.

@YourLordJoyBoy so now your problem is to multiply $(10x+6)(10x-5)$

@Jakobian Correct

2:45 PM
so now lets set $a_1 = 10x$, $a_2 = 6$ and $b_1 = 10x$ and $b_2 = -5$
$b_2 = -5$ since $10x-5 = 10x+(-5)$
it doesn't matter that we have a minus there as long as we make it a plus
so now you want to obtain $(a_1+a_2)(b_1+b_2)$

@Jakobian Yes! I think mentally I wanted to jump to that. But I shouldn't have. My tutor was telling me the same thing, to slow down.

so what do you do, you take the sum $a_1b_1+a_1b_2+a_2b_1+a_2b_2$
well this is just to show you how this really works
but once you know how it works you should start applying this in your head
without any $a_i$ or $b_j$
just memorize this rule without any variables

@Jakobian And I believe I'm starting to. I want to keep it in my head so next week I cam apply it where it counts.
can not cam

by without any variables I mean something like, okay we have $(10x)(10x)+6(10x)+(-5)(10x)+6\cdot (-5)$
you don't even try to name them as $a_i$ and $b_j$

@Jakobian Yes, I believe I'm beginning to see that.

2:49 PM
just treat each of those terms as their own black box
okay and now what? you simplify of course
so you have $100x^2+60x-50x-30$
and now you further simplify

@Jakobian Exactly! I believe I'm understanding it!

so you take the terms with $x$ together because its the only thing you can do, and you obtain $100x^2+10x-30$
and now you can't really do anything so you leave it like this

@Jakobian Yes, it makes sense!

try some example like, lets say $(1+x+x^2)(1-x)$
could you try to multiply it out

And I went ahead and snipped the conversation

2:53 PM
remember you don't have to introduce any $a_i$ or $b_j$, just remember to multiply everything in the first batch with everything in the second batch and add all of the terms together

@Jakobian Will do, and I'll get to answering that problem once I return from dogsitting for our neighbore. And lunch. I should be back close to 12, brb. And seriously man, the help is appreciated.

3:20 PM
I've attempted this 5 times but am not getting the right answer.
The correct answer is $sec^{-1}(2x-7)$.
Can someone please take a look at this and tell me what am I doing wrong? Thanks :)

0

If a real analytic function $f$ is involutive i.e. $f(f(x))=x$ and its Mellin transform can be taken, does this imply that $\tilde f$ inherits a functional equation as well? I'm reminded by Poisson summation involving the Jacbobi theta function and its functional equation corresponding to a new f...

thoughts on this "question?"

@Swan when u get $4(t^2+1)$ it should be $4(t^2+\frac{1}{4})$
I don't know if that's the only mistake
also should be $\int \frac{2dt}{(2x-7)^2}$
where are you getting the answer from because when I plug it into a calculator I get the same answer as you @swan
with the modification of the factor 4 and 2 I mentioned

3:49 PM
@Obliv thanks a lot for the help. The answer is from the time when my teacher had solved it in class. Maybe both the answers are different representation of the same thing

@Jakobian $10(2x-1)(5x+3)$

oops
your handwriting irks me :) the x's look like u's. Yeah that's also a valid strategy.. maybe the solutions are equivalent ðŸ¤¡

10(2x - 1)(5x + 3)

10 * 2x - 10 * 1
10 * 1 - 10 * x

20x - 10

10 - 10x?
@Jakobian Trying to think if I should distribute or foil.

you can do either one
in either order
an important concept is that ordinary multiplication is commutative which means xy = yx
and associative
meaning x*(yz)=(xy)*z
just recite the holy word PEMDAS it will lead you into the light

4:15 PM
@Obliv they are equivalent but as I am not great at handling inverse trigo ratios so it took me some time to derive it. And yeah, I'm not fond of my handwriting either

Nice, I should memorize those trig identities/relations.. they come up all the time in physics

@Obliv Sweetness Obliv :D I swear, you and @Jakobian even @JohnZimmerman, you've all been a big help to me.

I wouldn't mind it as much if it were consistent.. sometimes you draw it like a u and sometimes like an x :P

Lets not forget @ronjon
Why do I want to dub @Swan the Black Swan?

4:38 PM
Hope Ted is ok.
@psie you pinged me on a ticket (volume of rectangles) and then disappeared.

I don't think I've spoken with Ted. But the situation is oddly familiar to whats happening with a missing student from my class.
After spring break, he never showed up for class again so I don't know what became of him.

@copper.hat I mailed him 3 weeks ago and he said that he is doing fine, taking a long break from the chatroom

@SoumikMukherjee That's a relief to hear. I wish I could hear something from the classmate I mentioned.

I hope your classmate is doing fine

@SoumikMukherjee Same. We messaged him as my teacher said to do but nothing.