@meer2kat We're substituting $x^2$ for $a$ and $y^3$ for $b$. You'll get the same result.
You might try a test case: what happens when you write down $(x^2+y^3)^3$, ansd what do you get when you do $(a+b)^3$ and then substitute $a=x^2, b=y^3$?
okay, @mike can you help me out one more time? what if you had say $x^6y^12$ again and you want to find the coefficient for something like $((x/y)^2+y^6)^{10}$ ?
The moral of the story is :A physicist and a mathematician are sitting in a faculty lounge. Suddenly, the coffee machine catches on fire. The physicist grabs a bucket and leap towards the sink, filled the bucket with water and puts out the fire. Second day, the same two sit in the same lounge. Again, the coffee machine catches on fire. This time, the mathematician stands up, got a bucket, hands the bucket to the physicist, thus reducing the problem to a previously solved one. @meer2kat
1) I wasn't able to ask any question on there, since I didn't want to make people look down on me for asking dumb questions, e.g. I was emotionally attached to my account
So I started this one and asked two really dumb questions that I already knew the answer to(now both deleted I believe), just so I wouldn't feel bad for asking any other dumb questions.
2) Many of my questions were really trivial, and if friends at university found my account they would probably judge me for them. Whereas the two ones at the start of this would both be outliers, and I could just palm them off as trolls
@ᴇʏᴇs It's just sitting there
I have asked four questions on it this year and over 80 in total
So there are two interpretations I suppose, my interpretation: You learn math so poorly that you honestly think you had once solved something like that, since it looks familiar, and you surface learnt math to such an extent that you can't see why it is meaningless, and the second interpretation, Engineers are generally extremely arrogant
I am not to familiar, but the octions have a certain operation, and I don't think it is the one inherited from just looking at the products of $\mathbb{C}$
@Committingtoachallenge I am upset because I did a number of things these few days that made me worse, unintentionally of course. Now I need to try to fix it.
About that thing I wrote (htmlpublish.com/newTestDocStorage/DocStorage/…), that's a very early draft, which is why it looks unfinished. In any case, now you have puzzles to try to solve!
@meer2kat Sometimes, I need to do something to resolve my OCD issues. But then instead of thinking clearly and making the right decisions and getting some help from others, I sometimes make the wrong decisions and end up getting people to do some things which don't help. When I realise it later, I will need their help again to fix it and I am afraid I would upset them. I try to minimise upsetting people, but sometimes I make mistakes and need more help than if I had thought clearly.
@meer2kat So, I just hope that I won't upset them, and I hope that they would still help me in the end. Something like that. And then I feel guilty that I had ruined it all again with the mistake I just made.
A square matrix with nonzero entries only along its main diagonal we call a "diagonal matrix". What of a matrix with all nonzero entries the diagonal specifically above that? A super diagonal matrix? Generalizing further, what name should be used for a matrix with nonzero entries only along the $k^{th}$ diagonal to the up-right from the main diagonal?
well, yes., though I'm trying to prove something about jordan forms @ commitin, but even more generally I'm trying to prove a result about powers of jordan blocks
and that would only work for the superdiagonal (one offset) case, and i'm trying to use an appropriate name for the third, fourth, ... kth powers of a jordan block
@JasperLoy no i mean my job doesn't quite fulfill the requirements of bills so everytime i think i'm good to go a bill comes up and ruins my mask of financial security. aka, college.
Oh, I'm pretty sure I've got the proof worked out, however its frustrating to write. Letting $M$ be an $n\times n$ matrix with complex entries, prove that if $M^n = I$ that $M$ is diagonalizable.
Me: AHA! It's 6 choose 3! No it's not. Me: But I can't even solve this. It's not possible without a fraction. And fractions aren't allowed. Are you sure because I think they are. Me: Hits head on table.
Galois Theory, this is just the first step in proving Kummer's Theorem
Let $n$ be a positive integer and let $F$ be a subfield of $\mathbb{C}$ containing $\mu_n$, the $n^{th}$ roots of unity. Let $K/F$ be a Galois extension with cyclic Galois group $G$ of order $n$. Then $K=F(\sqrt[n]{\beta})$ for some $\beta\in F$.
Come to think of it... I don't think I've ever used my d30's yet... still fun to have them though. Maybe once in order to randomly determine the day of the month for a scenario,
Please be gentle as I do not have any degree in maths.
By using a compass/straighedge method to construct Metatron's cube, a regular dodecahedron can be inferred from intersecting points. I'm looking for the ratio between the lengths of the edges (blue) of the dodecahedron and the radius of the ...
@Rubito Let $X_i$ be the random variable for how many calls the $i^{th}$ callcenter employee can handle. Assuming the employee's stats are independent, then $E[\sum\limits_{i=1}^k X_i] = k\cdot 100$ and $Var[\sum\limits_{i=1}^k X_i] = k\cdot 15$ (unless I made a mistake somewhere)
i still need some steps to calculate intersecting points to make the dodecahedron, but a lot of plumbing is done and i can rotate the entire object nicely
@rubito I had made a mistake. $Var[\sum\limits_{i=1}^k X_i] = $ since they are assumed independent $ = k Var(X_i) = k 15^2$, so stddev is $\sqrt{k}15$. We are looking for the least $k$ such that $Pr(\sum X_i \leq 30000) = .99$
You would have as mentioned previously that $Pr(Z\leq 2.33)=.99$, so we try and get $X=\sum X_i$ such that $X\leq E[X] + 2.33 \sqrt{Var[X]} = 30000$
So, try and solve $E[X] + 2.33\sqrt{Var[X]} = 30000$, gives you $k\cdot 300 + 2.33\cdot \sqrt{k}\cdot 15 = 30000$
this is quadratic in $\sqrt{k}$,
I see now that I made a transcription error. $E[X] = 100\cdot k$, not $300\cdot k$ $100(\sqrt{k})^2 + 34.95 (\sqrt{k}) - 30000 = 0$, by quadratic formula gives me $\sqrt{k} = 17.14...$
so I feel as though I'm still making an error calculating somewhere, as that gives me $k\approx 294$
yea, that'll change things around a bit. so $X\geq E[X]-2.33\sqrt{Var[X]}=30000$, giving us $k\cdot 100 - 2.33\cdot \sqrt{k}\cdot 15 = 30000$, and the quadratic equation $100(\sqrt{k})^2 - 34.95(\sqrt{k})-30000=0$
since if we lower the number of people a bit, our probability will drop below 99%, and we can't have a fraction number of people., we'll use 307 employees then
It seems in the lectureslides, they are using $X$ to be the random variable for an individual employee's performance (whereas I labeled it as $X_i$), and they are using $T$ to be the random variable for the total combined performance of all employees (whereas I used $X=\sum X_i$)
$E[T]$ is the expected value of $T$, i.e. the number of calls that the employees field in total.
or are capable of fielding rather
in the lecture slides, they are using $n$ for the number of employees (whereas I had used $k$)
MSE pedagogy question: How do people deal with situations like these. I've almost given the answer away :/ http://math.stackexchange.com/questions/1203763/prove-fx-0-for-all-x-in-a-b/1203776#1203776
Great! Thanks a lot man, I was mainly confused as to why the Z score was negative and I wasn't sure my answer was right. But, we seem to have the same answer now.
@JMoravitz, I admit that I tend to give away more than I probably should in general. I'm actively trying to work on this though because mathematical pedagogy interests me.
@kaj I wouldn't respond or just tell them at this point they should think about it more, and that you explaining the situation won't help them understand. Plus the guy is a math grad, should be able to do some independent thinking.
Well that is probably really rude so I will delete that
@KajHansen It interests me aswell, and MSE is an amazing platform for improving teaching abilities. I look forward to the day when I can answer peoples questions atleast once a day(that are of an interesting level)
That what I ended up doing earlier today in chat, I don't necessarily think they were fishing for answers but it seemed like the best way to help after you practically gave them the answer but they are still not getting it. I think they just become relient on the leading and stop thinking about the problem themselves @kaj
Another hint that the person is to use to people leading him to the answer is that they end everything with a question mark to see if the people doing the leading thinks it is right instead of trying to work through it. They are basically guessing and not thinking through it. @jmoravitz
@PaulPlummer do you agree with me if i say that : $]-\varepsilon,\varepsilon[\cap \overline{A}=(]-\varepsilon,-\frac1x[\cap \overline{A})\cup (]-\frac1x, \varepsilon[\cap \overline{A})$ for $0<\frac1x<\varepsilon$ or this
