Mathematics - 2014-08-30 (page 4 of 5)

i'm struggling with $$\lim_{x \to 0^+} \ln x^2+2\ln x -3$$ - i know that $$\ln 0^+ = -\infty$$ but when i substitute that in, i get $$(-\infty)^2 + 2(-\infty) - 3$$ which evaluates to $$\infty - \infty -3$$, which is not much use. can anyone shed some light on this one?

Nick

@carpetjar: I miss the obvious sometimes.... All the time

@topper: Please check out a list of what are known as indeterminates in mathematics while someone explains.

carpet jar

@topper tried with $a\ln(x) = \ln(x^a)$?

topper

5:15 PM

@Nick Which of the several entries at en.wikipedia.org/wiki/Indeterminate? Anyway the given solution is $$\lim_{x \to 0^+} \ln x(\ln x+2) - 3 = (-\infty)(-\infty)-3 = \infty$$

@carpetjar No, what I tried was what I wrote. :) I find limits so frustrating, I never seem to be logical enough, no matter how hard I try.

Studentmath

Oh, it's $ln^2(x)$, not $ln(x^2)$

topper

@Studentmath Correct, sorry

Can't edit it now, argh

@Nick See @Studentmath's comment

Studentmath

But yeah, generally try to stuff things up. Also @Nick, that would be true if it would've been $ln(x^2)$ but it's $ln^2(x)$

topper

i'm struggling with $$\lim_{x \to 0^+} (\ln x)^2+2\ln x -3$$ - i know that $$\ln 0^+ = -\infty$$ but when i substitute that in, i get $$(-\infty)^2 + 2(-\infty) - 3$$ which evaluates to $$\infty - \infty -3$$, which is not much use. can anyone shed some light on this one?

correct version ^^

Nick

O_O I hat notations

I meant I hate them, but I could hat them too.

Studentmath

5:20 PM

@topper seeing that I would think first how to stuff the ln's up, which would've been either @Nick's way if it was $ln(x^2)$, or the way they suggest if it was yours. Also, note that you could 'view' $(-\infty)^2$ as $\infty^2$ more than as $\infty$.

Which then would give it more sense as to why it goes to $\infty$ even when you evaluated it the way you tried.

Nick

@topper: Yeah, just take the ln(x) common and do what studentmath says.

@BalarkaSen: btw, I forgot to thank you earlier for that hint that i totally didn't notice. I'm such a clutz.

topper

@Studentmath 1) Stuff the ln's up means it's just an idiom to factor them out? 2) Re the view of $(-\infty)^2$, can you explain? I have it quite clearly in my notes that $-\infty \cdot -\infty = \infty$

nabla blah

I forgot if the $Y$ in $f: X \rightarrow Y$ means the image or codomain of the function

Nick

@nablablah: Codomain.

nabla blah

Okay, thanks

topper

5:24 PM

I just worked out to use one dollar sign not two for latex...

Nick

@topper: Yeah, that uncrowds the chat a lot.

topper

Put it this way. I understand why the given solution works. I read it and it all makes sense. I don't understand why my way didn't work.

Studentmath

@topper 1. correct
2. Well, it's obviously true. But if you evaluate $ln(x)$ as $-\infty$, then $ln^2(x)$ is more 'like' $\infty^2$ in the sense of how *strong* it goes there. But then again, it's not really so, it's just to give you a sense of why evaluating it as you did still makes sense with the result of $\infty$

nabla blah

Is there a good webpage/online article (free) that explains the rigorous definition of a function (for real analysis class)

Studentmath

Now, I am not sure if it is legit to explain it that way (which is why factoring them out is safer), but still.

Anyone knows of a neat proof of the semi-perfection of the $G_{24}$ extended golay code?

Nick

5:27 PM

@nablablah: You just asked me which one the codomain was. Go to Khan Academy brush up on basics. Then, I'll give you as many links as you want.

@topper: Let me put it this way, why would we have limits if direct subsitution of a quantity can yield an answer?

I have officially become a badass with my replies!

topper

@Nick I thought that substitution was a valid way of finding a limit, unless it evaluates to $0 / 0$ or $\infty / \infty$

carpet jar

@topper no, it is not.

topper

Not sure what's more demanding, limits or latex. :P

Ah great, so I still don't understand even the basics

I'm not actually thick!

Nick

@topper: There are many more indeterminates than those. That's why I said in the beginning to look up a list

topper

@Nick Yup, and I asked you which from the list, because it wasn't clear

carpet jar

5:34 PM

lets define a function $f(x) = \frac{1}{x}$ for $x \in \mathbb{R}\backslash \{0\}$ and $f(0) = 0$

@topper now look what happens when you try to count the limit for 0...

Nick

@topper: $\infty - \infty$ is indeterminate. So substitution is not valid

topper

Penny dropped, thank you.

Nick

:D lol

topper

As an aside, it seems strange to me as a naive observer, that it's possible to get a different result by something like factorization, which doesn't change the value of the expression, only the syntax (as far as I can see)

Nick

Oh wait, you're referring to the last question.

Well, you need to research why we have limits for that.

topper

5:39 PM

Maybe I didn't express myself properly. I'm saying that before stuffing the $\ln x$s, I can't get to an answer. After stuffing, the expression has the same value, it's essentially the same thing, yet now I can get an answer.

Nick

Your basically asking why factoring works when direct substitution fails.

You're secretly killing the people that are causing a problem for you.

I hope that analogy works good.

@carpetjar: i don't know what else to say, can you help me out.

carpet jar

@Nick with what? I'm doing my linear algebra, no one wants to help me, so I am not reading all of it.

define the problem.

Nick

@carpetjar: Does it involve vector spaces and stuff like that or simple linear equation. I can do the latter, not the former.

topper

Yes and no. I mean, I see the expression as a black box that has an associated value (limit). It's not intuitive to me that rearranging the components in the black box would offer a different value. Anyway my PC is making that smell as if a capacitor is about to blow in the PSU, so I might be done soon for tonight. ;)

carpet jar

@Nick vector spaces and projections.

Nick

5:45 PM

@carpetjar: ... can't help. I haven't learnt them.

topper

Don't worry @nick, I'll just keep doing problems until it becomes intuitive enough that a casual observer would think it's intuitive.

Nick

@topper: What are you practicing from?

carpet jar

@Nick nvm, I will do it on my own, if noone is able to help.

topper

@Nick I'm doing questions on the tuition website Gool, which has material to cover my syllabus, then I have a question book from the university, and some past papers.

Nick

@carpetjar: Again, stress on the able, I would if I could, bud.

carpet jar

5:47 PM

@Nick yeah, I don't hold grudge to anyone.

thanks

Nick

@topper: Please consider finishing the Khan Academy Playlist at 1.5x or 2x speed on the topic of limits. The lectures will help. I promise.

topper

@Nick I'm happy to. But I think the problem is trying to retain the vast amount of stuff I've tried to learn in the last month. Thanks

Nick

@topper: there's a lot more in your path. Good Luck!

topper

And I've covered about ten topics of material. Limits was the first topic. Investigating functions is the tenth, so it's the first time I'm actually using the material in anger. Probably a normal part of learning

Nick

@topper: yes it is.

Good night :D

topper

5:55 PM

Goodnight!

Nick

@topper: One last thing, grab every resource you can find. You won't lose anything you've sincerely learnt.

topper

If anything, I think it's information overload, and not enough actual doing exercises. Trying to do as many as I can now. I have 6 days to blitz this... I also noticed from the past papers that the questions are somewhat easier than some of the exercise questions. So it's good to do relatively hard ones

And thanks so much for taking the time to help!

robjohn

@Chris'ssis When Lilly passed away about a year and a half ago, I was in a deep depression for a good month. I still have times where I blame myself, but I realize that it wasn't really my fault. Things happen. I am sure that you will learn to deal with the pain, but I know you will miss her deeply for a long time.

Nick

@topper: believe me, I have not helped you at all.

Ted Shifrin

@Nick: You got it straightened out?

Nick

6:01 PM

@TedShifrin: did you not get the thousand praises?

Ted Shifrin

LOL ... I am having issues logging into this chat from my iPad ... can't get it to work.

There is a lesson from what I was trying to communicate to you: Try to make things more concrete instead of totally general, and see if that helps you understand.

And, anyhow, I was too busy being very annoyed with Balarka.

Nick

@TedShifrin: Praise #972 stated thank you for not slapping me for not realizing it the moment you said it.

Ted Shifrin

LOL, oh, no slapping needed :)

well, not this time, anyhow

Nick

@TedShifrin: He's always like that but he's a prodigy. By the time he' my age, he'd have mastered the whole math.

Ted Shifrin

well, part of the point of this site is for people to get some good hints and understanding, not just "it's clear"

Nick

6:05 PM

@TedShifrin: I ignored his hint because I was focusing on yours, if it's any consolation.

Ted Shifrin

LOL ... I'm glad you got it sorted out. This is a standard argument that you will now never forget :)

Nick

@TedShifrin: speaking of the site, wt hppnd to crusade of answers.

Ted Shifrin

huh?

carpet jar

@TedShifrin could you help me with some linear algebra? I am not sure if I am doing it the right way.

Ted Shifrin

what's your question, @carpet?

Nick

6:07 PM

There are loads ofold easy unanswered questions on main. No one's doing them.

Ted Shifrin

I'm fed up with the homework-type users, @Nick, in general.

But there are usually plenty of people answering those.

Nick

@TedShifrin: ... should I feel guilty, now. I answer those questions a lot.

lol

carpet jar

We have a vector space of polynominals $a + bx + cx^2 + dx^3$, and scalar product $(f|g) =\int_0^1 f(x)g(x)dx$. Get an orthogonal projection of $u(x) = x^2$ on subspace $W = \mathbb{R}_1[\cdot ]$ and component of vector $u$ orthogonal to $W$

Balarka Sen

@TedShifrin Is it safe to unzip the lip now?

Ted Shifrin

hang on, @carpet ... I've got an emergency with one of my actual students to deal with.

carpet jar

6:10 PM

@TedShifrin fine, I will tell what have I done so far.

Nick

@BalarkaSen: Dude, glass of milk, helps to sleep.

carpet jar

@Nick not as good as two bottles of vodka

Nick

@BalarkaSen: i speak on behalf of Ted when I say, you can speak now.

@carpetjar: ... we're minors!

Balarka Sen

I hope @Ted's not ignoring me.

carpet jar

@Nick I see.

Balarka Sen

6:12 PM

I can be a jerk sometimes. I apologize.

Nick

@BalarkaSen: ... you are insightfully mature for you age and the only reason anyone treats you seriously is because you're almost always right. You're a kid, not a jerk.

Trust me, I know the difference.

carpet jar

@TedShifrin I think I have done a) part, i.e. I wanted to find $w = P_wu = a + bx$. It means that $(w - u|1) = 0$ and $(w - u|x) = 0$, because after we subtract the pararell part from $u$ we have only orthogonal, which must give $0$ after scalar producting it. This gives us $a = -\frac{1}{6}$ and $b=1$

Nick

@BalarkaSen: ... was there anything wrong in what I said?

Ted Shifrin

ok, @carpet, I'm back. What is $W$?

r9m

@robjohn that is a very neat proof of Hellmut's Identity ! :-) can I edit the post according to it ? :-) (when I say edit: I mean copy paste =P .. ) .. sorry for the late reply .. I fell asleep :|

carpet jar

6:17 PM

@TedShifrin $W=\mathbb{R}_1[\cdot ]$, i.e. all polynominals in the form $a + bx$, where $a, b\in\mathbb{R}$

Ted Shifrin

OK, I've never in my life seen notation like that.

carpet jar

@TedShifrin it seems that it's my facultie's personal notation

Ted Shifrin

Be careful. Are the scalars and $x$ already orthogonal?

You want an orthogonal basis, first of all, for $W$.

Nick

@carpetjar: My faculty has notations that actually spell out this is a notation

carpet jar

okay, so I assume $W$ has standard base. When I had no idea what I'm doing I tried to orthogonalize the standard base and got something very ugly.

Ted Shifrin

6:21 PM

You cannot project on the $xy$-plane in $\Bbb R^3$ by projecting independently on $(1,0)$ and $(1,2)$, say.

carpet jar

So how do I tackle this problem?

Ted Shifrin

No, no, you cannot assume that $\{1,x,x^2,x^3\}$ is an orthogonal basis for your vector space.

carpet jar

@TedShifrin right, because we have different scalar product.

Ted Shifrin

Use Gram-Schmidt.

Nick

Goodnight and don't forget $$1\cdot20\sqrt{e\cdot980}$$ All of you!

Ted Shifrin

6:22 PM

Night, @Nick.

carpet jar

night @Nick

Ted Shifrin

I guess the way you did part (a) was ok, actually, @carpet. You found an element $w$ of $W$ so that $w-u$ was orthogonal to some basis for $W$. You didn't project independently onto $1$ and $x$. So you're fine.

carpet jar

@TedShifrin then we receive very ugly base, like $\{1, x - 0.5, x^2 - \frac{1}{2\sqrt{3}}x - 1/3 - 1/\sqrt{3}, idontknowwhat\}$

Ted Shifrin

I haven't checked your work.

topper

I need to find the derivative $(\ln x)^2$. Please don't tell me the answer itself, but I'm planning to use the product rule $(\ln x)(\ln x)$. Right strategy? Is chain rule more / less relevant?

Ted Shifrin

6:23 PM

Well, to project on $W$, you only need an orthogonal basis for $W$ ... not for the entire space. But your approach should be fine, if you did the computation right.

Chain rule is easier, in general, @topper.

topper

@TedShifrin Okay, this is my chance to dive in and actually try to understand the thing...

Ted Shifrin

@topper: If I gave you $(\ln x)^5$, you certainly would not want to use product rule.

Nick

@topper: Differentiate $u^2$ , then multiply it with the derivative of $u$ where $u = \ln x$ . This is the Chain Rule! :D

carpet jar

@TedShifrin how comes, I didn't project independently on $1$ and $x$, when I was solving equations $(a + bx - x^2 | 1) = 0$ and $(a + bx - x^2|x) = 0$

Ted Shifrin

You just said that $u-w$ (for $w=a+bx\in W$) had to be orthogonal to the subspace $W$. You don't need any particular basis for $W$ to do that. However, if you want to add the individual projections onto individual basis guys, those guys need to be orthogonal.

topper

6:27 PM

@Nick I'm proud, I "smelt" that there was a Better Way (tm)

carpet jar

I see

Ted Shifrin

@carpet: If I compute the projection on $1$, I compute $\dfrac{\langle u,1\rangle}{\langle 1,1\rangle}1$. Then I would need to compute the projection onto $x-1/2$ and add the two.

OK, I'm going back to trying to get my iPad to get into chat. BBIAB.

carpet jar

$\langle a, b\rangle$ is a scalar product, isn't it?

@TedShifrin so I could as well compute the projection on $5$ and $x - 10$ and that would be still fine, because they span the same space as $1$ and $x - 1/2$, right?

nabla blah

Hi @Ted

topper

YES! Chain rule worked, now I'm not scared of it. I'm resigned to the fact that I can watch hours of videos and nod my head and "understand", but only by diving into exercises and hitting brick walls do I seem to retain the hard bits

Ted Shifrin

6:41 PM

Weird, I finally got back in ... @carpet: No. If you're going to project on basis vectors and add the projections, the basis must be orthogonal. Draw a picture in $\Bbb R^2$.

r9m

@robjohn is there an elementary way of proving the identity $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n}{n^2}\left(\sum\limits_{k=1}^{n}\frac{1}{k^2}\right) = \zeta(2)\zeta(3)+\zeta(5)$ ?

I mean a way that only manipulates the series ?

Ted Shifrin

@topper: Math is never a spectator sport. You have to do exercises, always!

6

carpet jar

@TedShifrin okay, I got it. what about finding a component of $u$ orthogonal to $W$?

Ted Shifrin

That's what $u-w$ is, @carpet!

Again, draw familiar pictures :)

carpet jar

How can I draw a picture of $12x^2$?

Ted Shifrin

6:47 PM

No, understand the pictures in $\Bbb R^2$ and $\Bbb R^3$ :)

You're just in a different version of $\Bbb R^4$ in this problem ...

carpet jar

thank you

Ted Shifrin

Most welcome ...

carpet jar

got time and energy for another one?

this piece I have no idea how to tackle

Studentmath

Rehi all

carpet jar

hiho @Studentmath

skullpatrol

7:04 PM

hi pal

carpet jar

@TedShifrin I will take it as a yes. Find own orthonormal basis and matrix in this basis of unitar operator $U$, which matrix in some orthonormal basis $B$ looks like $[U]_B := \frac{1}{9}((4 - 3i, 4i, -6 -2i), (-4i, 4 - 3i, -2 - 6i), (6 + 2i, -2 -6i, 1))$. Perform spectral decomposition of this operator and count $exp(U)$. I have no idea where to start.

Khallil

Heyoo!

carpet jar

@Khallil yo

Khallil

I've got a problem in proving why the derivative w.r.t. $x$ of $a^x$ is $\log(a) \cdot a^{x}$.

From first principles, I get $$ \dfrac{\text{d}}{\text{d}x} a^{x} = a^{x} \ \lim_{h \to 0} \dfrac{a^{h} - 1}{h} $$

Alex J Best

@khallil $a^x = e^{\log(a)x}$ then apply chain rule?

Khallil

7:17 PM

I wanted to do it via first principles, @Alex.

^_^

I'm having trouble evaluating the limit.

carpet jar

@Khallil so now use de l'Hospital and you are done!

Khallil

That's a circular argument.

carpet jar

I know

funny, isn't it?

Khallil

You're assuming the derivative of $a^x$ when doing L'Hop.

Nope, not at all.

😶

FractalHand

@robjohn Why do we have to say $h \neq 3$? chat.stackexchange.com/transcript/message/17421568#17421568

Alex J Best

7:31 PM

@Khallil ah, can't think how to do it in a nicer way than just following through the proof of the chain rule in this specific case, there could be some way though, I'm no good at this sort of thing! P.s. I read in the other chat you're going to warwick? That's pretty cool (I just graduated from there)!

Khallil

That's awesome! Did you do math there, @Alex?

Alex J Best

@Khallil I did discrete maths, it's what most uni's would call maths and computer science, but I ended up doing mostly maths I guess.

carpet jar

@AlexJBest discrete maths is the best maths!

Alex J Best

@carpetjar +1

Alexander Gruber

if stackexchange doesn't fix its pinging system I'm just going to have to find everyone whose username begins with alex and change it to something else.

skullpatrol

7:37 PM

@AlexanderGruber +0.999...

carpet jar

@AlexanderGruber ping

Khallil

It makes sense that you'd get pinged as well, @AlexanderGruber. Sorry. I'll ping the full username in future.

carpet jar

any kind person here to explain me some linear algebra? I've no idea where to start.

Khallil

Discrete math didn't look too appealing to me, @AlexJBest.
I was offered a place for it last year but I took a year out and reapplied for math on it's own.

Alexander Gruber

believe me, @Khalil, this is not the first time this has happened. :p

@carpetjar what is it

carpet jar

7:40 PM

Find own orthonormal basis and matrix in this basis of unitar operator $U$, which matrix in some orthonormal basis B looks like $[U]B:=19((4−3i,4i,−6−2i),(−4i,4−3i,−2−6i),(6+2i,−2−6i,1))$. Perform spectral decomposition of this operator and count exp(U). I have no idea where to start.

Alexander Gruber

do you know how to find orthonormal bases?

carpet jar

gram-schmidt algorithm - yeah.

Nick

@khalil: the proof by first principle is only 3 steps. Did you forget $$ \lim_{h \to 0} \frac{a^h -1}{h} = \ln a$$
Use it!!

Alexander Gruber

am i right that your matrix is $$B=19\left(\begin{array}{ccc} 4-3i & 4i & -6-2i\\ -4i & 4-3i & -2-6i \\ 6+2i &-2-6i &1 \\\end{array}\right)$$

carpet jar

the first parenthesis is first row, second is second row, etc

Nick

7:43 PM

Typing latex on a phone is so hard.

Alexander Gruber

better

carpet jar

@AlexanderGruber no, third column, second row it's $-2 -6i$

Khallil

Thanks @Nick, but that definition of the limit using L'Hop already assumes the derivative of $a^h$ which is what we are trying to find.

You said you had another proof for it, @Nick. What does it involve?

Nick

Hey, the sidebar with stars and pins isn't available in mobile chat

carpet jar

@AlexanderGruber that's more like it.

@Nick request desktop site then.

Alizter

7:47 PM

@AlexanderGruber What is your research?

carpet jar

@AlexanderGruber forgot about one thing. It's $1/9$ instead of $19$, but that's just a constant.

we won't do the calculations anyway

Nick

@khalil: it involves not using lhosp. Lhosp is a shortcut. Go read a proof for the identity I gave

Alexander Gruber

@Alizter in the past i've done research in the intersection of graph theory and finite solvable group theory, and experimental particle physics.

carpet jar

@AlexanderGruber what do we start with? I would use the gram-schmidt algorithm if I knew on what vectors should I apply it.

Alizter

@AlexanderGruber what are you doing now?

Nick

7:51 PM

@carpet: firstly, are you getting this ping. Second, that option isn't available.

carpet jar

@Nick get a browser which enables you this option. Puffin has it, but I don't like it.

Alexander Gruber

@Alizter i'm doing applied algebra. cryptography, HPC, stuff like that.

Alizter

@Nick $\displaystyle \lim_{h\to 0}\frac{a^h-1}{h}= \lim_{h\to 0}\frac{h\ln a + o(h^2)}{h}= \lim_{h\to 0}\ln a + o(h)=\ln a$

Alexander Gruber

i haven't published anything in that yet, though.

Alizter

@AlexanderGruber Are you postdoc?

Alexander Gruber

7:54 PM

@Alizter No, 2nd year grad student. i did most of my publishing during undergrad.

Alex J Best

@Khallil Ah, fair enough, I really enjoyed the course as a whole, but I like programming and that sort of thing anyway so I don't think my modules would have been very different if I had done the straight maths course.

Nick

@carpet: my typing speed has dropped from 60 wpm to 1 wpm on this phone.haha

carpet jar

@Nick and how many $\Latex$-per-minute?

Alizter

@AlexanderGruber That is interesting. I havn't heard of many undergrads publishing alot

well done

also you kinda look like Andy Samberg from the right angle

Nick

@alizter: I can't read that now. Phone can't use latex rendering. What was that?

Alizter

7:56 PM

@Nick I solved your limit using asymptotics :D

no need for Lh

Nick

@alizter: that's cool. Show it to Khalil.

Alexander Gruber

@Alizter yeah it's sort of rare, i chalk it up to having no gen eds, due to a previous degree, so i was able to concentrate completely and find my niche.

@Alizter I've been told my voice sounds like his too.

Alizter

@AlexanderGruber Unfortunately I can't hear pictures so I believe you.

Alexander Gruber

@Alizter thank god for that, this site would be a nightmare.

Nick

@carpet: 0 latex per minute. I died typing a tiny limit identity here.

Alizter

8:00 PM

@Nick When I first joined chat it was a good 2 or 3 days before anybody told me about ChatJax. I can't remember who but someone said when I asked "We are real men here so we read latex". I believed that and attepted to join

2

carpet jar

@Alizter I thought like it too for a minute and then looked at favourites.

Alizter

@carpetjar This was about a year ago so the people here were different.

carpet jar

@Alizter right, you might not have the favourite on the right then.

@AlexanderGruber could you please hint me with this algebra?

Nick

@alizter: I am unable to star what you said, so instead I say " +1"

Khallil

What does the $o(h^2)$ mean, @Alizter?

I personally don't enjoy programming, @AlexJBest. Or at least, I didn't enjoy the small amount I did in Visual Basic about 5 years ago.

blue

8:05 PM

@Khallil http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation

Alizter

@Khallil Basically some other stuff that is at least with a $h^2$ term

I don't really care what it is

It is basically a taylor expansion of $a^h$

Khallil

Doesn't that rely on the successive derivatives of $a^h$, thereby assuming the result we're trying to prove?

Nick

@khalil: bud, visual basic ain't programming. Try c++, python even JavaScript. That's where all the fun is.

Alizter

If you define $e^x$ as a limit you can manipulate it into that taylor series

Alex J Best

@Khallil Have fun with maths by computer if it's still running :) (I like programming and I hated that module!)

Alizter

8:08 PM

The taylor series of $e^x$ is special because it an be found without taylors theorem

Nick

@khalid: Were you not around when balarka explained little o?

Khallil

Nope, I missed @Balarka's little o explanation.

It's still running, @AlexJBest. :\

(It's a core module as well.)

@Alizter, the original question was:

55 mins ago, by Khallil

I've got a problem in proving why the derivative w.r.t. $x$ of $a^x$ is $\log(a) \cdot a^{x}$.

Alex J Best

@Khallil Ouch, it was pretty terrible in my year, I managed to drop it as it wasn't core for me! Oh well there's plenty of interesting stuff as well.

Nick

@khallil: I apologize for constantly misspelling your name. Auto correct and fat fingers do not mix well.

Khallil

That's fine! Don't worry about it. ^^

carpet jar

8:13 PM

@Nick +1. You will get +10 for some proper, non-trivial, compiling latex.

Nick

@carpet: everything I know is trivial. Please explain what is non trivial.

carpet jar

@Nick for example, a carpet jar is an example of non trivial jar.

Vladimir Putin

a

carpet jar

b

Vladimir Putin

I was testing something.

The periodic table of finite simple groups: math.uh.edu/~tomforde/Images/periodic-table-of-groups.pdf

@carpetjar Who are you?

Nick

8:17 PM

@carpet : $$ \text{this is non trivial} $$

carpet jar

$ whoami
as322527

@Nick +10

@Vÿska I'm a guy looking for someone who wants to help me with algebra.

linear algebra.

Vÿska

@carpetjar Have you made your question on the main site?

carpet jar

@Vÿska no, I'd like to talk about it

It's a question from my latest exam and I have no idea how to tackle it, so during solving this I will have some questions

I think my attitude to linear algebra is wrong, that's why I lack some basics

Nick

@carpet: as I grow weary and sllepy, me hath also unlocketh achievement! Good night for the $\infty^{\text{th}}$ time.

carpet jar

@Nick a lot of spelling mistakes you do. Good night. And +10 for one more time.

Khallil

8:22 PM

Good night, @Nick!

Nick

@carpet: those better be real points that you are wiring me. Otherwise you may have a 0.63% chance of regret. Before I sleep I wanna try latex again.

carpet jar

@Nick and how much does "may" multiply the chance by?

Nick

@carpet: it may multiply it just like any element $\in (0,1)$

carpet jar

@Nick randomness? That sounds like being a human...

night all;

Nick

@carpet: yes. You're right, humans are detrimental multiples.

... I'm poofed..

skullpatrol

8:30 PM

later pal

Nick

What if I never leave...

skullpatrol

never, say never :-)

Nick

What if I stuff that never into your.... Zzzzzz...

skullpatrol

:-O

Nick

... See what I did there?

skullpatrol

8:33 PM

no

Nick

Nah. C'mon it was a joke. I was sleepy so I fell asleep

skullpatrol

i know, i know

Nick

Maybe I shouldn't be chatting from my coffin.

I mean bed.

skullpatrol

sleep well, my friend

Nick

Why would I say I coffin?

I'm not a vampire.

skullpatrol

8:35 PM

we all must eventually sleep there...

Nick

... Or am I?

skullpatrol

Nah.

Alizter

@skullpatrol Are you part of a biker gang?

skullpatrol

maybe, why?

Nick

Skull, remember when I told you I got depressed. I cured it completely. Go out and find a ray of sunshine. That's the cure.

skullpatrol

8:38 PM

good job :)

Alizter

@skullpatrol What reason do you have for joining it?

Nick

@alizter: there's a great chance he is.

skullpatrol

:D

Yasser

HI, how can I evaluate the limit 1/\log(1+x)-1/x at $0$ ?

Khallil

Maybe you could try get everything under a common denominator?

Also, is it a left sided, right sided or a two-sided limit, @Yasser?

Yasser

8:45 PM

I tried @Khallil but I did not succeed, two-sided. I am trying with $\log(1+x)=x-x^2/2+o(x^2)$

I find $1/2$ with this method.

Khallil

L'Hop?

Ted Shifrin

Your approach is good, @Yasser. L'Hôpital is inferior :P

Yasser

@TedShifrin Thank you!

Ted Shifrin

But @Khallil's suggestion to use a common denominator is needed, too.

Khallil

I've not run into any o or O notation, so I've just got L'Hôp, @TedShifrin.
I managed to get one half as well, @Yasser.

Ted Shifrin

8:51 PM

$$\frac1{\log(1+x)}-\frac1x = \frac{x-\log(1+x)}{x\log(1+x)},$$ and now compute with 2nd degree Taylor polynomials. (The error terms go away in the limit.)

@Khallil: Do you know about Taylor polynomials?

Try doing $$\lim_{x\to 0}\frac1{\sin^2x}-\frac1{x^2}$$ by L'Hôpital, and you'll be converted to my viewpoint.

Khallil

Yes, and the special case of Taylor expansions called Maclaurin, @TedShifrin.

Just not the o or O notation. ^_^

Ted Shifrin

ok, @Khallil, then you know how to do it the way @Yasser's talking about. Go for it. And you can prove it works.

skullpatrol

Hi Professor @TedShifrin

Ted Shifrin

hi @skull

skullpatrol

we lost chris's sis :(

Balarka Sen

8:55 PM

@TedShifrin Are you ignoring me?

Yasser

@TedShifrin you are right, I thought this method just after trying to get everything under a common denominator ^^.

Ted Shifrin

@skull: I doubt permanently

Khallil

L'Hôp isn't getting me anywhere with that, @Ted!

Time to try out some series expansions.

Ted Shifrin

I rest my case, @Khallil. You'll have to do l'Hôpital 4 times, and it's very difficult to make zero mistakes.

But you should also use intelligence (and then try to prove why it works) to figure out, say, the degree $4$ Maclaurin polynomial for $\sin^2x$.

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$Mathematics$ Mathematics