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12:00 AM
usefulness would mean no use? what means full of use?
ah no lol, i thought of uselessness haha
 
I'm investigating the shape of the p.m.f. for random variable $X$ which is hypergeometrically distributed. The probability is given by $$P(x=k)=\dfrac{\binom{K}{k}\binom{J}{j}}{\binom{N}{n}}.$$We have a population size of $N$ with $K$ successes and $J$ failures. The sample counts follow the same notation in lower-case letters.
Keeping all else constant, the only varying component in the p.m.f. is $\dfrac{1}{k!(K-k)!}$. So I'm wondering what sort of shape this gives us. $\dfrac{1}{k!}$ decreases very fast as $k$ increases, but $\dfrac{1}{(K-k)!}$ gets larger?
 
if K is constant, surely K-k gets smaller as k increases. therefore 1/(K-k) increases, and so does 1/(K-k)!
@Brody
 
@Null Right. Didn't think to consider $1/(K-k)$.
 
hi @KajHansen
 
Would I apply the formula $\frac{1}{n}\sum_i x_i P(x_i)$ to derive the mean?
 
12:15 AM
@AkivaWeinberger so given any countable subset of the real numbers, you can construct a number not in it
that's cool
 
@Brody (K-k)! has to be properly defined tho for all values i think. do you have such a definition?
 
it has been long time since i quit maths
 
@Brody Have you checked to see that the sum of the probabilities is $1$?
 
wonder if i can be back
 
@Null It's just discrete, and I require $0\le k \le K$.
 
12:16 AM
@Brody ah ok
 
@robjohn No, but I will do that.
 
@Brody If $k$ is the only variable, then $2^{-K}\binom{K}{k}$ is a probability distribution
I don't think that this is true for the function you cite.
 
Well, the way I formulated the p.m.f., $k+j=n$. So I suppose I cannot simply ignore the other binomial coefficient :/
 
@Brody Ah... that is important to mention.
 
Since it's really $\binom{K}{k}\binom{N-K}{n-k} \div \binom{N}{n}$. More complicated than I earlier thought.
 
12:19 AM
@Brody and is it true that $K+J=N$?
 
yes @robjohn
 
@Null, it's easier because the "b" would be fixed, but the $n$ varies with the $n$th root
We have $\sqrt[n]{n} \rightarrow c \iff \ln( \sqrt[n]{n} ) \rightarrow \ln(c)$
 
@Brody Then the Vandermonde Identity says it is a probability density.
 
Hey is "Robjohn" short for Robert John or Robot John
 
$\ln( \sqrt[n]{n} ) = \frac{\ln(n)}{n}$
 
12:21 AM
@AkivaWeinberger :p
 
We can investigate the limit of $\frac{\ln(n)}{n}$ with L'hopital, e.g.
 
@robjohn Interesting. Thanks.
 
And find that it converges to zero. Therefore, $\sqrt[n]{n} \rightarrow e^0 = 1$.
 
What hypotheses do I need to do a change of coordinates on a contour integral?
 
Will include Vandermonde's identity (at least a mention of it) in my notes on the distribution
 
12:22 AM
Cauchy-Goursat?
 
@KajHansen the only disadvantage I see with b, is that one has to consider b<1 and $b\geq 1$
 
You'd have to take cases I think, yeah
 
So I computed the second Chern class of a complete intersection and it's not very insightful. I guess that's the intended effect @TedShifrin? :P
 
Man, one of my questions is 12 views away from getting me a "Popular Question" medal :P
 
@KajHansen care to share the link? :d
 
12:25 AM
Could you link it @KajHansen? :p
lol
 
@Brody 2 dumbs one thought haha
 
http://math.stackexchange.com/questions/729611/is-a-finite-group-uniquely-determined-by-the-orders-of-its-elements

I feel dirty linking it and thus using chat for such selfish purposes.
 
@KajHansen or substitute $n=e^x$
 
@KajHansen well, i could have searched your profile... and then I would feel dirty lol
 
@Sophie, doh, I always forget the "substitution" route for these things. I bet that would've saved me a lot of combined time.
I'd have my own stalker :D
 
12:27 AM
Meaning, I want to take a contour integral, move the contour (homotopically) to another contour, and do an integration by substitution
when am I allowed to do that?
 
@KajHansen what you think? "
if i want to show an equality about spans. I have to show an equality about sets or? (sets of vectors to be precise) "
 
@Kaj you look so confused in your profile pic :P
not that i look any better; i don't
 
If you guys are visiting that question, do make sure to upvote the two answers. They were downright fantastic.
 
I just look like some triangles and diamonds.
In real life, too.
 
@KajHansen i will, if i understood them ;)
 
12:29 AM
As @Null said.
 
Yeah, spans are sets @Null
 
@KajHansen so the classic $x\in bla\implies ...$ and the backwards direction?
 
@meow-mix, I've thought the same thing, haha. It's a screenshot YouTube made for previews or something. It made some better ones, but they aren't well-suited for the 150x150 pixel format, or w/e it is
 
@KajHansen In the words of Ted, my profile pictures look like "mugshots"
 
Wow, missing two online homework assignments really wrecked my stats grade. :/ Ah well, lesson learned the hard way.
 
12:34 AM
I've gotten that on an old pic of mine as well @meow-mix !
 
@Brody learn to hack the grades? lesson learned the f'ed up way. (FOOLed :P)
 
Yeah @Null. In particular, if the elements that are spanning are linearly independent in both cases, say $Span(a, b) = Span(c, d)$, it suffices to show $c, d \in Span(a, b)$ or $a, b \in Span(c, d)$
Its somewhat scenario-specific
 
@Null not worth the risks associated with that, haha
Not super tech-savvy anyhows
 
@Brody it would surely proof a point or 2 tho.
 
@KajHansen Do I pronounce your name /kaı/ or /kædÊ’/
(Rhyming with "eye" or "badge")
 
12:37 AM
@Akiva, the former
 
I've been wrong this whole time Kadge D:
 
"j" is like that in Norse languages. See "fjord", e.g.
LOL @Brody
 
@KajHansen or "jajaja"
 
No worries, it happens a lot
That's spanish!
 
12:38 AM
@meow-mix …no
Unless you're a very excited German
 
yeah
"yayaya"
 
In Spanish, it's pronounced more like "hahaha" and means laughter.
 
Adverb: ja ‎(not comparable)
  1. (chiefly South Africa, informal) yes
  2. ja
  3. yes
  4. ja
  5. already, (in negative sentences) any more
(33 more not shown…)
Pronoun: ja
  1. I (first-person singular subjective)
  2. ja
  3. I
  4. ja f
  5. (third-person singular) instrumental form of ji.
(19 more not shown…)
Interjection: ja
  1. yes
  2. ja
  3. yes!
  4. "Ja!" riep hij luid toen er een doelpunt viel.
  5. Yes! he screamed loudly when they scored a goal.
(18 more not shown…)
Noun: ja n (singular definite jaet, plural indefinite jaer)
  1. yes
  2. foot
  3. ja m, n ‎(plural ja's, diminutive jaatje n)
  4. yes
(12 more not shown…)
Conjunction: ja
  1. and
  2. ja
  3. (coordinating) and
  4. ja
  5. if
(7 more not shown…)
Adjective: ja
  1. heavy
Verb: -ja ‎(infinitive kuja)
  1. to come
  2. go ja (past jelê)
  3. to eat
Article: ja
  1. the
oops sorry for spamming chat, dont know why it did that
 
That's longer than expected, lol
 
Sometimes I wish chat would just post the link and not a preview of the link
It does that to me a lot, and I hate it when it happens
 
12:40 AM
Type "[https://en.wiktionary.org/wiki/ja](https://en.wiktionary.org/wiki/ja)"
 
any field is a vectorspace over itself?
 
I tend to always prefer the previews shown
 
Yeah @Null, and every ring is a module over itself
 
@Null Indeed. All the axioms for a vector space follow from the field axioms.
 
12:41 AM
i will be able to change my name to a less silly name in a week or so
 
@meow-mix What was it before this?
 
I hate the name-change time restrictions
I suppose I understand why they exist, but they should relax it down to 2 weeks IMO
 
user<numbers>
 
What inspired meow-mix? What will be the new name?
 
12:42 AM
Hello all,
 
Hey there
 
the new name will be my irl name :)
 
Ah cool
 
if only i had a cool name
 
We're a rare breed on here it seems, lol
 
12:43 AM
On the subject: @Aksel'sRose, what does your username mean?
@meow-mix What's your irl name
 
Zach Hauk
 
@AkivaWeinberger Aksel is my son and everyone always asks if its like Axle Rose (its not, at all) kinda became a joke in the family
 
boring and german
 
@meow-mix For some reason I read Hulk the first time
 
12:44 AM
Hey folks, question from SE: stackoverflow.com/questions/41003611/…
 
The Incredible Hauk
 
Wondering if my interpretation is correct
 
Haukeye
 
hi @Brody @meow @Kaj @Aksel @DogAteMy (collapses with exhaustion)
 
Let $A=\{1\}$,$B=\{2\}$. Then $A\cap B=\emptyset$. $\langle\emptyset\rangle=\vec{0}$. Now let furthermore $K=\mathbb{R}$ and $V=\mathbb{R}$. Then $\langle A\rangle=\mathbb{R}$ and $\langle B\rangle=\mathbb{R}$. $\mathbb{R}\cap \mathbb{R}=\mathbb{R}$. Therefore $\langle A\cap B\rangle =\langle A\rangle \cap \langle B\rangle$ is false. can someone look over this?
 
12:44 AM
@Brody many people i know mispronounce it as "hawk"
 
beautiful @Akiva
 
@meow-mix Is it Howk, then?
 
Hey @Ted
 
so much, to the point which i am conditioned to pronounce it that way
yeah, @Akiva
 
Yello 'dere @Ted
 
12:46 AM
Hi @TedShifrin
 
alrighty, im on the last topic of my studying (Bilinear transformations). I've played around with it for a couple days, and this particular question is driving me nuts. I feel like it should be simple, so im likely overthinking. Can anyone help with it?
 
Is it like sauerkraut @meow-mix?
 
@Brody yes, like a sauerhauk
 
The Incredible Sauerhaukeye
 
lol nice
 
12:47 AM
0
Q: Bilinear proof of transformations

Aksel'sRoseI realize this question has been posted but it is on hold. It piqued my interest and now I cant figure it out so Im looking for some help. Question Let V and W be vector spaces over a field F and let $T \in Hom (V,W)$. For each $g \in Bil(W\times W)$, define $g(T): V \times V \to F$ by setting ...

 
Be back soon
 
@Ted now that i've read some more content, time to do more exercises! i'm thrilled
 
Is $ \sum_{n=1}^{\infty} (n^2 + 2n)^{-1} $ a geometric series ??
 
@Maks no
you beat me to it, @Akiva
 
12:48 AM
Yes @meow-mix
 
Yes or no ?
 
And how do I know it converges to 3/4 ?
??
 
My guess is it's a telescoping series somehow. Try partial fractions
Yeah
 
@meow: that sounded like a sarcastic thrilled.
 
12:51 AM
I will try
 
@TedShifrin maybe a little
thus far the exercises haven't been so bad, though
 
I told you there are much more interesting ones.
 
@TedShifrin Got some exercises for me?
 
Can someone say if this is right or wrong? "Let $F$ be a field, $V$ a vectorspace over $F$ and $A,B\subset V$, then the following holds: $\langle A\cup B\rangle _F=\langle A\rangle _F + \langle B\rangle _F$"
 
Maximum of a function like 7.9*a1 + 8.1*a2 + 0.4*a3 + ........ + 2.6*a50 where a_n is between 0 an d10, inclusively, and sum(a) = 100... Is it trivial, say that the 10 largest coeffs should all get a value of 10? stackoverflow.com/q/41003611/483620
 
12:52 AM
LOL, you want more, DogAteMy?
 
hi chat
 
OK, here are two, DogAteMy: (1) Suppose $a_n\to 0$ and the partial sums $s_n$ are bounded ($|s_n|\le M$ for all $n$). Must $\sum a_n$ converge? (2) If $\sum a_n$ converges absolutely, then does $\sum a_n^3$ converge? What if the original series just converges conditionally?
 
who is DogAteMy? lol
 
12:55 AM
Me
 
Of course, one-word answers do not suffice ... :D
 
For (1), let the first two be $\frac12$, the next three be $-\frac13$, the next four be $\frac14$, etc
 
ah, have you seen this question before, DogAteMy?
 
Alternatively, $\sin(\sqrt{x+1})-\sin(\sqrt x)$ probably
@TedShifrin Probably, or some variant, I don't know
(2) For converging absolutely, yes I think?
Because, for $n$ large enough, $|a_n|<1$
and so $\sum-|a_n|<\sum a_n^3<\sum|a_n|$ if we start the sum from late enough
so it converges
 
(Just work with $|a_n^3|$, of course, DogAteMy.)
 
12:59 AM
That too. So it converges absolutely.
 
I'm gonna have to give you harder ones ... Or differential geometry. But you're not quite done yet.
 
If the original series just converges conditionally…
Then it should converge also?
But I'm not sure
 
hi chat
 
I'll let you ponder, DogAteMy.
 
1:01 AM
Now I'm less sure…
 
Do you know what curvature of a curve is, DogAteMy?
 
@akiva Got any other things like that hex thing you'd be interested in seeing?
 
2
Q: Surface integral (divergens theorem) parameterization

Jacksoja$\int \int \bar{F}\bar{N}dS$ $\bar{F} =(4x^{3}+y^{2}-z , x^{2}+y^{3}+z , x^{3}+z+1 )$ $\gamma : z=4-4x^{2}-y^{2}$ $z\geq 0$ , Normal is pointing upward. My attempt : $r(s,t)= (s,t,4-4s^{2}-t^{2})$ but when I took the cross product of the partials dot normal vector it did not simplify. Is th...

what is dogatemy ?
 
@Jacksoja: So maybe you should use the Divergence Theorem and not compute the surface integral directly.
 
@Semiclassical I've been thinking about the question of how many loops a random thing should have. I wonder if, for the square thing at least, there's a relationship between the number of loops in $A$ and $A'$ (where $A$ is a thing based on an $n\times n$ grid of squares, and $A'$ is the same as $A$ but with all of the squares switched with the other "coloring")
@Jacksoja Me
 
1:03 AM
Damn, my nickname is causing so much difficulty.
 
But I haven't looked into it yet
 
konichiwa, Dogatemi
 
glares @Brody
 
<.<
 
1:05 AM
@Brody >.>
 
itadogatimasu!
 
@Danu: I mean, Chern classes tell you subtle things. I don't know what you expect for insightful. You can do the top Chern class of the complete intersection and get its Euler characteristic.
 
@Semiclassical Idea: To get rid of "small loops" in the square version (i.e. little squares), we can divide the squares into pairs and require that the things in each pair have the same "color" as each other. Specifically, for each even row, pair up squares with $x$-coordinate $2n$ and $2n+1$; in each odd row, pair up $2n-1$ and $2n$.
Sorry for the long description, I'm not sure how to say it in words
 
Hello @Ted.
 
Btw, Friday's my last final @Ted (ironically, the linear algebra exam). And then I'll have plenty more time to work on the books and not freak over this semester's end. :)
 
1:07 AM
But you end up with a certain "bricklaying" tiling made of pairs of squares
 
Hi @Fargle. Good luck, @Brody.
 
i'm still pondering the general form of that question, @Akiva
 
and if each square in each pair has the same "color" then we can't get any tiny squares
@meow-mix Which question?
 
later im going to read a research paper that i believes proves it for prime numbered cases
@AkivaWeinberger the $4$-cycles
 
1:08 AM
So, Brody, was the linear algebra course any good in the end?
 
@TedShifrin Are there any exercises in chapter 2 that you'd recommend I look at before forging ahead (which I may or may not have already done)? Assuming you have a copy of Rudin on hand, that is, and that you're not busy...>_>
 
Hey guys rumor has it that you guys were looking for a good contour integration question on the site so I did you all a favor and asked one
0
Q: Contour integration by substitution?

GFauxPasLet's say I have a contour integral on a non-closed contour with starting point $z_0$ and ending point $z_1$. Am I allowed to do a substitution like this? And under what assumptions? $\displaystyle \int_{z_0}^{z_1} f (z) \, \mathrm d z = \int_{u^{-1}(z_0)}^{u^{-1}(z_1)} f (u (z))u'(z) \, \mathr...

 
@TedShifrin Thanks. The structure is three 100 pt exams and a 150 pt final exam. I actually failed one from not doing anything during that time, but I still have a chance of getting an A in the course.
 
ah, @Akiva; this is the paper: arxiv.org/pdf/1608.04809v1.pdf
the only research paper i can comprehend :P
 
if $V$ is a vectorspace over F, and $A\subset V$, then $\langle A\rangle _F$ is definitly a subspace too. (even if A is empty, as 0 is always a subspace) right?
 
1:11 AM
There are some good ones, @Fargle, but still, maybe forging is better for learning more analysis. I never remember how to do 17, 18; 30 is baby Baire Category. 27 is significant. But moving on is OK.
 
what does $\langle A \rangle$ denote? span?
 
yep
 
@TedShifrin I'll keep 27 in the back of my head. Thanks.
 
what does the $F$ denote? in field $F$?
 
No, $0$ is not a substace; $\{ 0 \}$ is a subspace
 
1:12 AM
i.e. linear combinations with scalars in $F$
but yes, you are correct @Null
 
meow-mix the bottom half of your face is missing in the chat
but not when I click your picture
 
i know! i dont know why!
 
@GFauxPas polite applause for @GFauxPas
 
Thank you @TedShifrin
 
@meow: You're all there on my pages. I have a Mac. I wonder if it's OS related.
 
1:14 AM
Well, +60% counts as passing here (most of the time) so I didn't technically fail the exam.
 
@TedShifrin browser related?
you're using safari, correct?
 
no, I never use Safari. I'm on Chrome.
 
@TedShifrin oh; that's odd. maybe it is OS
it's doing that for linux also
 
maybe it's you, @meow :P
 
i'll try uploading another picture
should it be one with my cat(s)?
 
1:16 AM
Some of us are cat-people.
 
some of us are people-cats
 
Hello MSErs. Long time, no chat
 
@Ted The class wasn't really valuable. The professor just introduced new concepts and definitions and showed us how to do some computations. I know marginally more stuff but with no idea of connections or motivations or anything really insightful.
 
holy cow, it's Clarinetist.
 
well, i gotta go take a shower
 
1:17 AM
Got into an argument with a friend recently about whether pi is rational in base pi. I don't know how many times I can say "rationality is independent of base".
 
adios :]
 
I've never even seen bases other than positive integers
 
'Evening @Ted
 
Well, no, I disagree, @Fargle.
 
b'bye Haukeye
 
1:18 AM
@Fargle maybe he thought of rationality as "terminating"
 
@Brody: That sucks. But that's the case with most courses at most places, honestly.
 
@TedShifrin In what respect? Certainly $10_{\pi}$ is transcendental.
 
Is base-$\pi$ actually a thing used somewhere?
 
@GFauxPas Not as anything other than an example, I believe.
 
One of my longtime friends loves base Fibonacci.
 
1:18 AM
The coefficients of $\langle 1, \pi, \pi^2, \ldots \rangle$? Is that what base $\pi$ would mean?
 
Why, @Fargle? It has an integer representation.
 
@TedShifrin But it's still not of the form $S^n(0)/S^m(0)$.
 
WTF?
 
By $10_{\pi}$ I mean $\pi$ itself.
 
@TedShifrin Wouldn't doubt it too much, honestly. My former ODEs professor was very smart and pedagogical, and she teaches linear algebra. Should've taken her, though she was on our other campus
 
1:20 AM
But you're basing this on base-10 arithmetic, @Fargle, to decide what rational means.
 
@TedShifrin I'm not. Even in base $\pi$, $\pi$ is not a successor of zero, nor a ratio of two successors of zero.
 
I can do $\Bbb Z$ based on base $\pi$. It's a different universe.
 
Woudn't $10_\pi$ be what you get when you divide $10$ by $\pi^2$, and then take the reminder and divide it by $\pi$, and then take the remainder and divide it by $\pi^{-1}$, and so on?
 
And in fact successors of zero past what we call "3" would have non-terminating, non-repeating representations.
@GFauxPas $10_{\pi} = 1\cdot\pi^1 + 0\cdot\pi^0$
 
Oh right
I was thinking of $10_{10}$ converted to base $\pi$?
 
1:22 AM
I wasn't.
 
In that case, yes.
 
I was redefining $\Bbb Z$ and hence redefining $\Bbb Q$.
 
But then you'd need an infinite sequence to span every number in $\mathbb{Z}$, what's the point of that
 
@TedShifrin If you choose to define the integers as all things with integer representations in the base you're working in, that's fine, but you don't have closure. $2 + 3$ cannot be written as $m\pi + n$ for $m,n$ "normal" integers.
 
It's a different $\Bbb Z$. :)
 
1:24 AM
Is it just $\pi \mathbb{Z}$?
Wait no, it isnt
 
No, @GFauxPas, it's a subset of that.
I'm not quite sure what the digits will be :P
 
@TedShifrin But $\pi \Bbb Z$ doesn't contain 1...
 
So?
Now I'm not sure what base $\pi$ means at all. After all, what should the digits be?
 
@TedShifrin $0,1,2,3$.
 
Why? That's totally ad hoc.
 
1:26 AM
It would be the coefficients of the spanning set $\langle 1, \pi, \pi^2, \pi^3, \ldots \rangle$, of course
 
Why not continue mod $\pi$?
 
Base phi is a thing
 
No, that's not right.
 
@TedShifrin Because then you'd have uncountably many digits.
 
It isn't?
 
1:26 AM
Digits 0 and 1, with no two 1s next to each other
 
DogAteMy: That's what my friend's base Fibonacci amounts to.
 
"Phinary"
 
Right @Fargle :P
 
No, my definition would have countably many digits
 
@TedShifrin Doubt it
In phinary, integers require decimal points
They're related, though
 
1:27 AM
Anyhow, @Fargle: I declare this whole thing nonsense.
 
"decimal" points?
 
Oh yeah, you're right, DogAteMy.
 
Phinary points
 
radix points :P
 
1:27 AM
Hey guys
 
Man my question didn't even get any upboats :(
 
Not ${}^{\Large \cdot}$ dots or $\cdot$ dots but $.$ dots
 
I thought it was a good question :(
 
@TedShifrin If I can share how the discussion started, I was saying most properties of numbers hold regardless of base. Like, $\pi$ is transcendental, and it doesn't matter how we represent it.
 
need some help with matricies
 
1:28 AM
@AkivaWeinberger the generic term for the point delimiting between the integer part and the fractional part is "radix point"
 
I think this discussion just adds confusion.
 
Then he fired back with "well it's 10 in base pi and 10 is rational". But this ignores that what we mean by 10 normally is $1 + 1 + \cdots + 1$.
 
@TedShifrin Exercise: Express the even-indexed Lucas numbers $L_{2n}$ in phinary. Use this to write $1000$ in phinary.
 
I don't even know what Lucas numbers are. I quit.
 
He was using the definition of irrational as having a non-terminating non-repeating decimal representation; but the better definition of irrational is that it isn't in the set of rationals but it is in the closure of the rationals under etc
 
1:29 AM
@GFauxPas Right. That's what I'm getting at.
The property that rationals have repeating or terminating representations only holds in rational bases. So you can either throw out the concept of irrational bases entirely (which I'm okay with), or use the property of "ratio of two integers" and the definition of an integer as "some successor of zero, or its additive inverse" to divorce rationality from base entirely.
 
I don't "get" the dedekind cut definition; I understand the equivalence class of cauchy sequence definition, but I don't understand how a dedekind cut can define something that's "between" the cut
But my head is in contour integration for the time being
these past days
 
But I'll table this for the sake of @Ted's sanity.
 
I'm ignoring it. Proceed.
I'm leaving soon for dinner and bridge, anyhow.
 
I've said all I intend to anyway, haha. Now to real mathematics. opens Rudin
 
Bridge, as in, graph theory?
 
1:34 AM
If $\pi$ were rational then it would also be constructible. @Fargle
 
no, bridge as in card game.
 
@Brody I agree.
 
game theory and probability? nice
 
Tell him to square the circle. :)
 
@Brody: But our compass would be set to a different $1$.
 
1:35 AM
I have a straightedge of length $e$ units and I use it to square the circle if the circle is in base $e$
 
Not thinking that far @Ted, but right
 
I'm just sticking to my (perhaps wrong) POV. :P
 
im tryna using matrices for encryption of sorts. I have my message as $Matrix M = {{8,5,12,12,4}, {23,15,18,12,4}}$.
My encoder as $Matrix E = {{3,4},{-1,2}}$ I multiply M by E to give me matrix C $EM = C$ where $Matrix C = {{116, 75, 108, 84, 61},{38, 25, 24, 12, -7}}$.
When I try to decrypt $Matrix C$ with the inverse of $Matrix E$; $E'EM = E'C ; M = E'C$, I have my $Matrix E' = {{2/10, -4/10}, {1/10, 3/10}}$ so I do the following $E'C = {{2/10, -4/10}, {1/10, 3/10}} * {{116, 75, 108, 84, 61},{38, 25, 24, 12, -7}} = {{-18/5, -5/2, 6/5, 18/5, 89/10}, {23, 15, 18, 12, 4}}$
btw that was a total pain to type
and my syntax is wrong, whats the syntax for a matrix?
 
I think it's still unconstructible... not certain though oc
 
oh
How do I not do that?
 
1:37 AM
 
Either {{ ...},{...}} or [[ ...],[...]] ...
 
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
 
@TedShifrin thats what I did, didnt work out
rip me, okay let me try again
 
syntax where? here?
 
@TedShifrin yea i wrote the initial one using {{x,y},{x,y}}
 
1:39 AM
here it's begin{bmatrix} ..& .. & .. \\ .. & .. & end{bmatrix}
wait .. I didn't put $$'s ... and it still typeset.
 
@TedShifrin Put it in backticks , ` `
 
oh hell.
with \ in front of commands.
oh right
 
Because $F$ is a field, @Null, the only ideals are going to be $\{0\}$ and the whole field. If I'm not mistaken, $\langle A \rangle$ is the ideal generated by the elements of $A$.
 
hi @Kaj
 
Hey @Ted
 
1:42 AM
OK, time for me to disappear. Bye all.
 
Yeah - ideals of a ring $R$ are closed under addition and multiplication from outside elements. They're analogous to "spans" in vector spaces
 
Hello!!

Based on the definition of $+_k$, I want to give an inductive definition for the multiplication $\cdot_k$ in $\mathbb{Z}_k$, such that for all $x\in \mathbb{N}_0$ and $y\in \mathbb{N}_0$ it holds that $$x\cdot_k y=(x\cdot y)\mod k$$

What is an inductive definition?

Do we write $x\cdot y=y+y+\ldots +y$ ($n$-times) and we apply each time the addition $+_k$ ?
 
erugh
its a mess
 
:33955926 Put backslashes before "begin" and "end".
 
3 times a charm..
im tryna using matrices for encryption of sorts. I have my message as $Matrix M = \begin{bmatrix}8&5&12&12&4 \\ 23&15&18&12&4\end{bmatrix}$. My encoder as $Matrix E =\begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix}$
I multiply M by E to give me matrix C $EM = C$ where $Matrix C = \begin{bmatrix}116& 75& 108& 84& 61 \\ 38& 25& 24& 12& -7\end{bmatrix}$.
When I try to decrypt $Matrix C$ with the inverse of $Matrix E$; $E'EM = E'C ; M = E'C$, I have my $Matrix E' = \begin{bmatrix}2/10& -4/10 \\ 1/10& 3/10\end{bmatrix}$ so I do the following $\begin{bmatrix}2/10& -4/10 \\ 1/10& 3/10&\end{bmatrix} * \begin{bmatrix}116& 75& 108& 84& 61 \\ 38& 25& 24& 12& -7&\end{bmatrix}$ The problem is that the bottom line decrypts perfectly however the top doesn't, is it because the top one gets multiplied by a negative?
there we go
 
1:48 AM
There was a huge twist in the television show I was watching
 
The butler was the detective
@SylentNyte what does $*$ mean
 
@GFauxPas mutliplied
such as 2 * 3 = 6
 
oh, don't use $*$ for matrix multiplication
 
ah sorry, its too late to edit it unfortunately
 
Alas
 

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