Mathematics - 2012-10-15 (page 1 of 3)

Peter Tamaroff

12:05 AM

@Charlie Please, don't upload anything to 9gag with my name on it!!!

Charlie

@PeterTamaroff hehehe

Peter Tamaroff

@Charlie I'm serious. =)

Charlie

@PeterTamaroff ok

user19161

@PeterTamaroff Now that is so random!

Jayesh Badwaik

@JasperLoy Now even Pedro is mysterious!

user19161

12:07 AM

@JayeshBadwaik Hehe. We are all mysteries.

Charlie

@JasperLoy it's not random

user19161

Hey I thought my comment about Heisenberg and Carlsberg should get 9000 stars by now.

Charlie

@JasperLoy you dunno heisenberg?

...

Well....See you all later!

user19161

@Charlie Well, just a little bit about that formula.

Charlie

Bye bye!

dukenukem

12:12 AM

say I have a function where at x = a we have a root. So f(a) = 0. If f'(a) = 0 what does that say about the root? Does it have a different multiplicity than a 'normal' root?

Peter Tamaroff

@dukenukem It has multiplicity at least $2$

(It may have greater mult.)

user19161

@PeterTamaroff Well done Pedro, very precise.

dukenukem

ok cheers

if f''(a) is not equal to zero what multiplicity does a have?

Peter Tamaroff

@dukenukem $2$

user19161

Even the 2 has dollars around it, very professional...

peoplepower

12:18 AM

Or very rich.

Peter Tamaroff

@JasperLoy Man. Suppose $\{a_n^2\}$ and $\{b_n^2\}$ are summable. I want to prove that $\{a_nb_n\}$ also is. Now, the most obvious thing to do is: $${a_n}{b_n} \leqslant \frac{{a_n^2 + b_n^2}}{2}$$ Thus $\{a_nb_n\}$ has bounded partial sums. How do I show they actually converge to something?

dukenukem

cheers. would f''(a) = 0 mean a is a point of inflection or is it a local minima or maxima?

Argon

@dukenukem Inflection, I believe.

Compare: $f(x) := x^3 \implies f''(0) = 6\cdot 0 = 0$ which is an inflection point.

user19161

@PeterTamaroff Hmm, have you used the Cauchy criterion? Just a thought.

peoplepower

@PeterTamaroff Note that $\{|a_n|^2\}$ is also summable (same with $b_n$ of course).

Peter Tamaroff

12:23 AM

@JasperLoy $(a_n-b_n)^2\geq0$

@peoplepower Yes...

@peoplepower I thought of that but can't figure out how to use it.

peoplepower

@PeterTamaroff Then $a_nb_n$ is absolutely summable...

Peter Tamaroff

@peoplepower Why?

user19161

@PeterTamaroff I was talking about applying the Cauchy criterion where a series converges if only if the tail can be made small.

somekindarukus

Can anyone explain to me how to determine the sequence generated by a certain exponential generating function? I'm completely confused. The one I have in mind is f(x)=3e^(3x)

Peter Tamaroff

@somekindarukus You mean find $a_n$ where $$f(x)=\sum a_n\frac{x^n}{n!}$$?

@Argon Without information about $f^{(3)}(0)$ you cannot conclude that.

What about $y=x^4$?

somekindarukus

12:32 AM

@PeterTamaroff I'm not actually sure.. An example from my text is the following: Examining the Maclaurin series expansion for e^x we find e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .. = \sum^\inf_{i=0} \frac{x^i}{i!}

Peter Tamaroff

@somekindarukus Use $

somekindarukus

@PeterTamaroff so e^x is the exponential generating function for the sequence 1, 1, 1, ...

Peter Tamaroff

@somekindarukus Yes, precisely.

somekindarukus

Alright, that's the example from my text but I don't understand why this is

Peter Tamaroff

The EGF of a sequence $\{a_n\}$ is the function defined by $$f(x)=\sum_{n=0}^\infty a_n\frac{x^n}{n!}$$

For $a_n=1$, you get $e^x$.

The EFG of $a_n=n!$ is $\dfrac{1}{1-x}$, for example

Argon

12:34 AM

@PeterTamaroff If it is of odd order, I believe we can.

Peter Tamaroff

@Argon What do you mean by odd order?

Argon

@PeterTamaroff $x^{2n+1}$

Peter Tamaroff

@Argon But the function needn't be of the form $x^n$.

Argon

@PeterTamaroff This is true

However, $f''(0)=0$ is a necessary condition for an inflection point, no?

Peter Tamaroff

@peoplepower I think I got it: $${a_n}{b_n} \leqslant \frac{{a_n^2 + b_n^2}}{2} \Rightarrow \left| {{a_n}{b_n}} \right| \leqslant \left| {\frac{{a_n^2 + b_n^2}}{2}} \right| \leqslant \frac{{{{\left| {{a_n}} \right|}^2} + {{\left| {{b_n}} \right|}^2}}}{2}$$

@Argon Oh, yes.

somekindarukus

12:39 AM

@PeterTamaroff Sorry, so $a_n=n!$ is $\dfrac{1}{1-x}$ because we have $$\sum_{n=0}^\infty x^n $$ after subbing in $a_n=n!$ and $1+x+x^2+x^3+..=\dfrac{1}{1-x}$, correct?

Peter Tamaroff

@somekindarukus Yes.

user19161

@PeterTamaroff How does the first inequality follow from the second? $a_nb_n$ could be negative!

Peter Tamaroff

@JasperLoy But the squares are always positive!!!

somekindarukus

@PeterTamaroff I still don't understand why when $a_n=1$, $\dfrac{x^0}{0!} + \dfrac{x}{1!} + \dfrac{x^2}{2!} +..=e^x$.. Can you provide any further explanation?

@PeterTamaroff Also, how do you get LaTeX to show up in the chat?

Peter Tamaroff

@somekindarukus You need ChatJAX.

user19161

12:43 AM

@PeterTamaroff I mean that -3<2 but 9>4.

Peter Tamaroff

@JasperLoy Huh?

That's not what I'm doing.

@JasperLoy I'm doubting. Fuck.

user19161

@PeterTamaroff If you can prove that $a_nb_n$ is aboslutely summable, then it is summable right?

Peter Tamaroff

$$\eqalign{
& {\left( {\left| {{a_n}} \right| - \left| {{b_n}} \right|} \right)^2} \geqslant 0 \cr
& {\left| {{a_n}} \right|^2} + {\left| {{b_n}} \right|^2} \geqslant 2\left| {{a_n}} \right|\left| {{b_n}} \right| \geqslant 0 \cr
& \frac{{{{\left| {{a_n}} \right|}^2} + {{\left| {{b_n}} \right|}^2}}}{2} \geqslant \left| {{a_n}} \right|\left| {{b_n}} \right| \geqslant 0 \cr} $$

@JasperLoy Yes, of course.

user19161

@PeterTamaroff So why not use the criterion I mentioned to prove that it is absolutely summable?

Peter Tamaroff

@JasperLoy Too tortuous. That inequality above is all we need.

The criterion of comparison does it.

user19161

12:48 AM

@PeterTamaroff Ah right.

Peter Tamaroff

@JasperLoy THough the Cauchy criterion is used to prove that if $|a_n|$ is summable then so is $\{a_b\}$

Namely if $$\left| {{a_{n + 1}}} \right| + \cdots + \left| {{a_m}} \right| < \varepsilon $$ then$$\left| {{a_{n + 1}} + \cdots + {a_m}} \right| \leqslant \left| {{a_{n + 1}}} \right| + \cdots + \left| {{a_m}} \right| < \varepsilon $$

user19161

@PeterTamaroff Si, si! Via triangle inequality. QED.

Peter Tamaroff

1:11 AM

@JasperLoy I'm getting wound up in something simple. Suppose $\{a_n\}$ is summable, and it is non negative and decreasing. Then $\lim\; na_n=0$.

Now, using the Cauchy criterion,

$$\varepsilon > {a_{m + 1}} + \cdots + {a_{2m}} > {a_{2m}} + \cdots {a_{2m}} = m{a_{2m}}$$

But I wanna get $ma_m$.

Charlie

If I could use one other tool , I would solve this

Peter Tamaroff

@Charlie Those problems are boring.

@MarianoSuárez-Alvarez Mariano, sabés de algun libro que trate productos infinitos? Spivak y Landau tienen algunas cosas, pero son bastante poco detalladas.

Charlie

@PeterTamaroff If I could use it...just one second

Peter Tamaroff

En realidad, me interesa la definicion formal.

Landau tiene los teoremas que hablamos el otro día.

Por un lado, Spivak especifica que $b_n>0$ para definir el producto,mientras que Landau solo pide la existencia del limite de los "productos parciales."

somekindarukus

How can I develop a generating function for the number of partitions of $n \epsilon \mathbb{N}$ into summands that cannot occur more than 5 times?

Peter Tamaroff

1:20 AM

@somekindarukus You can use \in for $\in$

@somekindarukus What do you mean "summands that cannot occur more than $5$ times"?

Say $n=10$.

somekindarukus

So we have partitions of 10 like:
10 = 5 + 5,
10 = 5 + 4 + 1,
10 = 5 + 3 + 1 + 1, etc

Peter Tamaroff

@somekindarukus But not $1+1+1+1+1+1+1+1+1+1$ right?

somekindarukus

summands cannot occur more than five times, so
10 = 5 + 1 + 1 + 1 +1 + 1 would be valid, but not
10 = 4 + 1 + 1 + 1 + 1 + 1 + 1

correct

Peter Tamaroff

@somekindarukus Yes.

@somekindarukus Well, there is a general procedure to get taht I think. For example, for $10$.

(Give me time to write)

somekindarukus

Most of the examples include functions like $f(x) = (1 + x+x^3+x^5+..)(1+x^3+x^9+..)(1+x^5+x^{15}+..)..$ ( which is the generating function for partitions using only odd summands

Peter Tamaroff

1:26 AM

@somekindarukus Yeah, that is nice.

somekindarukus

Now are each of the polynomials enclosed in brackets representing one summand?

Charlie

@Jasper Hi Jasper!

Peter Tamaroff

@somekindarukus That is a product of series.

user19161

@Charlie Yo yo.

Peter Tamaroff

@somekindarukus Does that make sense? It is pretty "algorithmical"

I missed a few, wait.

Charlie

1:29 AM

@JasperLoy Wassup?

somekindarukus

Ah wait, I've found something

"If $n \in \mathbb{Z}^+$, the number of 1's we can use is 0 or 1 or 2 or.. The power series $1 + x + x^2 + x^3 +.. $ is used to keep track of this number. In the like manner, $1 + x^2 + x^4 + x^6 +..$ is used to keep track of the number of 2's we can use... " etc.

So i guess the answer to my question would then be $(1 + x + x^2+x^3+x^4+x^5)(1+x^2+x^4+x^6+x^8+x^{10})(1+x^3+x^6+x^9+x^{12}+x^{15}$)..

Would $(1 + x + x^2+x^3+x^4+x^5) = \dfrac{1-x^5}{1-x}$ ?

Argon

@somekindarukus Try $$\frac{1-x^6}{1-x}$$

Peter Tamaroff

@somekindarukus $$1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$$

Argon

1:44 AM

Note that there are $n+1$ terms on the LHS.

somekindarukus

@Argon ** RHS?

@Argon oh wait. nevermind, $x, x^2, .. , x^n and 1$?

Argon

@somekindarukus Yes

somekindarukus

So the following is correct:
$$(1+x+x^2+..+x^5)(1+x^2+x^4+..+x^{10})(1+x^3+x^6+..+x^{15})..$$
$$=\dfrac{1-x^6}{1-x}*\dfrac{1-x^{11}}{1-x^2}*\dfrac{1-x^{16}}{1-x^3}..$$
$$=\prod^\infty_{i=1}\dfrac{1-x^{5i+1}}{1-x^i}$$

peoplepower

I don't think so.

somekindarukus

can you point out my error?

peoplepower

1:56 AM

In the second expression, for instance, set $y=x^2$ calculate it for $y$, replace $x$ back in.

somekindarukus

.. i don't really understand

peoplepower

You have $1+y+y^2+\dots+y^5$

somekindarukus

yes

peoplepower

You know its value: $\frac{1-y^{5+1}}{1-y}$

somekindarukus

Yes.

peoplepower

1:58 AM

Now, putting $x^2$ back in for $y$: $$\frac{1-(x^2)^6}{1-x^2}$$

Gives $1-x^{12}$ instead of $1-x^{11}$ in the numerator.

somekindarukus

But why would we be subbing in $x^2$? Are you just making a point that it should work for any $y$ and it clearly doesn't?

peoplepower

That's the value of $1+x^2+x^4+\dots+x^{10}$.

somekindarukus

Ah I see. Sorry, I'm so confused by these generating functions

No wait, I won't be subbing in $x^2$, will I?

$1 + x^2 + x^4 + .. + x^{10} = \dfrac{1-x^{11}}{1-x^2}$ .. I don't have to sub in $x^2$ anywhere

Peter Tamaroff

@somekindarukus Careful.

@somekindarukus We wrote that $$1 +x+ {x^2} + {x^3} + \cdots + {x^n} = \frac{{1 - {x^{n + 1}}}}{{1 - x}}$$

But you're dealing with $2,4,6,8,10$

You're forgetting about some terms.

peoplepower

This might be a better way to look at it (no random $y$ entering the scene). $$1+x^k+x^{2k}+\dots+x^{5k}=x^k(1+x+x^2+\dots+x^5)$$

Peter Tamaroff

2:05 AM

Now, if $y^2=x$,$$1 + {y^2} + {y^4} + \cdots + {y^{2n}} = \frac{{1 - {y^{2n + 2}}}}{{1 - {y^2}}}$$

somekindarukus

That's what @peoplepower was just saying and I didn't really understand

Taking out $x^k$ makes sense..

peoplepower

It's also an error.

somekindarukus

As does the $\dfrac{1-y^{2n+2}}{1-y^2}$

Peter Tamaroff

@peoplepower What are you referring to?

peoplepower

My most recent display equation.

Peter Tamaroff

2:07 AM

@peoplepower Yes, that is fatal...

=P

peoplepower

@PeterTamaroff In that I died laughing when I asked myself "Hey, how do I factor out an $x^k$ from 1?"

somekindarukus

So would it be

$$\dfrac{1-x^6}{1-x}*\dfrac{1-x^{12}}{1-x^2}*\dfrac{1-x^{17}}{1-x^3}..$$
$$=\prod^\infty_{i=1} \dfrac{1-x^{5i+i}}{1-x^i}$$

or for the purposes of the question, to figure out the partition sizes of $n \in \mathbb{Z^+}$:

$$=\prod^n_{i=1} \dfrac{1-x^{5i+i}}{1-x^i}$$

peoplepower

The $17$ should be $18$, but your expression using $\prod$ is correct.

somekindarukus

@peoplepower I love that I said your equation up there made sense as well... I didn't even notice the mistake sigh

peoplepower

@somekindarukus Here's an algebraic way to do it: $1+x+x^2+\dots+x^{11}=(1+x^2+x^4+\dots+x^{10})+x(1+x^2+x^4+\dots+x^{10})=(1+x)(1‌+x^2+x^4+\dots+x^{10})$.

Since the right side is $$\frac{1-x^{12}}{1-x}$$ Our desired sum is just that over $1+x$ as we saw earlier.

somekindarukus

2:17 AM

Do you think in the following question they are asking for no more than 12 summands total, or summands no greater than 12 in value: What is the generating function for the number of partitions of $n \in \mathbb{N}$ into summands that cannot exceed 12 and cannot occur more than 5 times.

I don't quite understand how you got from
$$(1 + x)(1+x^2+x^4+...+x^{10}) = \dfrac{1-x^{12}}{1-x}$$

peoplepower

I meant left.

I'm left-handed, so my left hand is usually right for the task.

I meant to say "Since the left side is [that fraction]".

Mariano Suárez-Alvarez

3:24 AM

@PeterTamaroff There is a little book by Knopp (Infinite sequences and series) which has the basics and a bi tmore. Books usually do not say much about infinite products because there is not much to be said (that has not been said about series, that is)

leo

4:31 AM

@PeterTamaroff @skullpatrol it is never enough spanish! :-)

Mariano Suárez-Alvarez

4:57 AM

how do you call a word which only contains one letter, possibly repeated, like aaaaaa

if letters were colors, it would be monochromatic...

leo

@MarianoSuárez-Alvarez have you seen spanish.SE?

Mariano Suárez-Alvarez

yes

but I need an english word :-)

leo

@MarianoSuárez-Alvarez en libros que hablan sobre palabras, lenguajes y esas cosas, (digamos donde una palabra es una funcion de $\{0,\ldots,n\}$ en un conjunto $A$), no tienen nombre esas?

Mariano Suárez-Alvarez

5:16 AM

no recuerdo ninguna :-/

user19161

5:39 AM

@MarianoSuárez-Alvarez I don't know of any word better than "one-letter"!

user19161

It is only three syllables and easy to read too. So one can just say this is one-letter, two-letter and so on.

user19161

In general, one can even use n-letter to refer to a word with exactly n letters or at least n letters. Just define it at the start for a technical paper which I suppose is the reason you are asking.

user19161

@MarianoSuárez-Alvarez Actually, the book is quite big!

leo

bad comment here:

Finally $h_n\to h$. To prove this remember that since $f$ is continuous:$$\begin{array}{l}\text{1.it's uniformly continuous and }\\ \text{2.attains its extremums}\end{array}$$ over kompact intervals. — leo Sep 22 at 0:59

Mariano Suárez-Alvarez

6:06 AM

@JasperLoy, but «two-letter word» already has a meaning!

aa is a two-letter word :-)

hence the conundrum :D

user19161

6:17 AM

@MarianoSuárez-Alvarez Actually, if so then one-letter already has a meaning, for it refers to "a" and "I"!

user19161

Hmm, perhaps something else would be better.

user19161

Hello @shifty you look huge there!

Chris's sister

Hi.

user19161

@Chris'ssister Hey! Hope you had a great weekend!

Chris's sister

Not really :-(

How about what W|A says here "$\int_{0}^{\infty} x dx- \int_{0}^{\infty} x dx = (\text{integral does not converge})$"?

Jayesh Badwaik

6:22 AM

@JasperLoy I If you have only one type of letter in a word, wouldn't it be a monosyllable?

user19161

@Chris'ssister That's because the integral itself already does not exist! So one cannot say the right side is zero.

user19161

@JayeshBadwaik A syllable is not a letter. Apple has two syllables but four letters.

Chris's sister

I'm not convinced that it is correct.

Jayesh Badwaik

@JasperLoy Dude. Five! Yup. But all words of the form $a^*$ are mono-syllable. All monosyllables are not necessarily $a^*$ though.

user19161

Hmm, in the first place, what is the definition of the left side? It involves taking the limit as the upper limit of the integral goes to infinity right?

Chris's sister

6:25 AM

Right.

user19161

Hey, anyway, what I say in chat is not necessarily right. Actually the same goes for elsewhere too!

Jayesh Badwaik

@JasperLoy Of course dude, we are all here to share our lack of understanding.

Chris's sister

But $\int_{0}^{\infty} x dx- \int_{0}^{\infty} x dx = $\int_{0}^{\infty} (x - x) = $\int_{0}^{\infty}0$

user19161

@Chris'ssister So what would be 1/0-1/0? I would say it is undefined and not 0.

Chris's sister

am I wrong?

user19161

6:26 AM

@Chris'ssister The first equality is not true strictly speaking.

Chris's sister

Could you prove that?

user19161

Notice that when people state theorems like the integral of the difference is the difference of the integrals, it is assumed that the two integrals exist!

user19161

There is nothing to prove here because this is a matter of definitions!

Jayesh Badwaik

${}{}$

Chris's sister

@JasperLoy: I had this question in my mind ...

I understand your point.

Jayesh Badwaik

6:28 AM

@Chris'ssister The individual integrals do not exists, but the integrals of the subtraction does. Its like the limits
\begin{equation}
\lim\limits_{x \to \infty} (x - x) = 0
\end{equation}
however,

\begin{equation}
\lim\limits_{x \to \infty} x -\lim\limits_{x \to \infty} x
\end{equation}

does not exist.
since neither of the two limits exist.

user19161

lim n - lim n does not exist, while lim (n-n)=lim 0=0.

user19161

So lim n - lim n is not equal to lim (n-n).

user19161

End of story. Everyone grab a beer!

Chris's sister

:-)

The point is that I wanted to go beyond the definitions.

user19161

@Chris'ssister Sure, one can go beyond the definitions, but definitions cannot change. In the same way, one wants to solve difficult problems, but the problems themselves do not change.

Chris's sister

6:37 AM

@JayeshBadwaik: thanks!

I like these simple questions :D

Jayesh Badwaik

@Chris'ssister welcome.

Chris's sister

@JasperLoy: thanks for explanations! Are you teacher?

Jayesh Badwaik

The space jump has been made, those who have not seen it, should see it.

Chris's sister

@JayeshBadwaik: Oh, nice jump! :-)

Jayesh Badwaik

@Chris'ssister Yup.

Chris's sister

6:43 AM

10 minutes in the air!

Jayesh Badwaik

@Chris'ssister Yup!! I am yet to find a complete full video of the jump. Searching for it. Saw it live yesterday/today.

Chris's sister

@JayeshBadwaik: I saw it at news.

user19161

@Chris'ssister No, I am currently a jobless nutcase. =)

Chris's sister

@JasperLoy: you're not the only one. :-)

user19161

@Chris'ssister Erm, I really am a nutcase, but I won't go into the details...

Jayesh Badwaik

6:48 AM

You're not the only one, You're not the only one. (November Rain)

Chris's sister

lol :-)

user19161

How do we know if you are Chris or his sister @Chris'ssister?

Chris's sister

Sister is here.

@JasperLoy: that is a funny question. :-) It's hard to answer this.

user19161

@Chris'ssister I am a very funny guy. You will know if you hang out here more.

Jayesh Badwaik

When I first heard of Joe Kittinger, I had a dream of doing a space jump some day. Sadly, I am not a fighter test pilot, and I do not have that much money. :-(

Chris's sister

6:52 AM

@JasperLoy: I like to spend my time on SE, and I like people here. I'm really glad that such a place exists.

user19161

Hey why did someone star her message but not mine? I am jealous.

user19161

@Chris'ssister Have you started undergraduate studies?

Chris's sister

Yes.

Jayesh Badwaik

@JasperLoy "Sister is here" is a delightfully ambiguous message.

user19161

@JayeshBadwaik Ah I see, amazing!

user19161

6:54 AM

@Chris'ssister I get the feeling that you are Chinese by race. Am I right?

Chris's sister

I live in Romania.

(I'm Romanian)

user19161

Haha, OK. Because your English is very Chinese.

Chris's sister

lol

How is that?

"very Chinese"?

user19161

Well, I don't know. I have chatted with various Chinese people on SE, and somehow you all sound similar to me.

Chris's sister

:-)

Are you from US?

user19161

6:57 AM

No, I am not.

user19161

Sometimes I feel I am not human. Perhaps I was born on the wrong planet.

Chris's sister

I feel this too. :)

user19161

I also feel that my entire life is one big mistake, a mistake that I am still trying to correct, but without much success yet.

user19161

@Chris'ssister Do you know about chatjax?

Chris's sister

No.

brb

user19161

7:02 AM

@Chris'ssister math.ucla.edu/~robjohn/math/mathjax.html

user19161

Install the first bookmark on your browser.

user19161

Every time you come to this chat, click on it to render LaTeX.

user19161

You only need to click once after each refresh of chat.

user19161

Thanks to our very own, the great robjohn aka mean square.

4

user19161

7:15 AM

@jay So next time you say "Jay is here" that has two meanings?

Jayesh Badwaik

@JasperLoy Yup. It can mean, I am here. Or it can mean "Jay is here (in this house, in this room, in this country who knows), but I may not be Jay." :P

user19161

@JayeshBadwaik She must have been kidnapped...

Chris's sister

@JasperLoy: I think that life is more than that. We usually say " this is good / this is bad" or "this correct / this is wrong". Well, I'm trying to go beyond these words. Actually, the way we feel comes from our mind. Therefore, we should be the landlord of our mind and carefully analyze all our thoughts. In the end, it's your decision the way you feel.

There is no mistake, it's just an interpretation.

I need to leave.

Thanks for the nice chat!

Henning Klevjer

8:29 AM

Quick question that might not fit here:

Say I have a mathematical function with a computational complexity of 2^63 (or take 2^63 "time"). Will f(f()) have complexity 2^63^2?

or is it 2^(63*2) ?

user19161

12:10 PM

Haha @jay I just answered a rep whoring question on TeX!

Jayesh Badwaik

12:39 PM

@JasperLoy :-) Link?

skullpatrol

Hi @WannabeMiniMathematician

Wannabe Mini Mathematician

@ThePersonIThinkIsJohnJunior: Hi

user19161

6

Q: What does 'texmf' stand for?

Just a (simple) short question: Why is the LaTeX root directory called texmf? Is there a meaning of the mf ending?

texmf

Wannabe Mini Mathematician

MF... if you know what I mean @JasperLoy

user19161

@WannabeMiniMathematician Yes, I know.

Jayesh Badwaik

12:42 PM

@JasperLoy And you reach 1902. 100 more to go.

Wannabe Mini Mathematician

My inner devil is laughing at the thought of the pervert guy who made LaTeX.

Jayesh Badwaik

@WannabeMiniMathematician Wait till you learn quantum mechanics.

user19161

@JayeshBadwaik Yes, then I can leave this world...

Jayesh Badwaik

@JasperLoy You are going nowhere.

user19161

@JayeshBadwaik Anyway, the interesting thing about the TeX site is that just when you think nobody else will answer you get a few more answers.

skullpatrol

12:45 PM

What do you find interesting about this site?

Wannabe Mini Mathematician

I find everything interesting on this site, except that guy named John Junior. His presence annoys me. @skullpatrol

skullpatrol

John Junior is gone.

Wannabe Mini Mathematician

Thank God.

skullpatrol

What annoyed you about him?

user19161

@skullpatrol How some questions are so unclear that the OP can even think of them.

skullpatrol

12:49 PM

@JasperLoy You find unclear questions interesting?

user19161

@skullpatrol Yes. In a non-serious way of course.

skullpatrol

@JasperLoy How about in a serious way?

user19161

@skullpatrol Hmm, how fast most questions get answered then.

skullpatrol

Well? @WannabeMiniMathematician

Wannabe Mini Mathematician

His presence annoyed me.

skullpatrol

12:53 PM

What is it about his presence?

Wannabe Mini Mathematician

@skullpatrol: You ask a lot.

skullpatrol

@WannabeMiniMathematician He must have said something to annoy you, no?

user19161

Is this some kind of joke you two are playing?

Wannabe Mini Mathematician

Why, @JasperLoy?

user19161

@WannabeMiniMathematician Never mind.

Wannabe Mini Mathematician

12:58 PM

Hey guys... do you know about this guy? I forgot his name.

I think his first name was William.

user19161

@JayeshBadwaik Only 70 more to go, haha!

Jayesh Badwaik

@JasperLoy Nice!! Check your mail. It is moot now, but still.

@WannabeMiniMathematician I don't remember any William here.

Wannabe Mini Mathematician

OK, but yeah, I think it was something like that.

user19161

@JayeshBadwaik What was that mail for?

Wannabe Mini Mathematician

He had a blue display picture (similar to Jasper's).

skullpatrol

1:01 PM

I'm asking the "wanna be" why he did not like "Junior"?

and all he can come up with is his presence.

Jayesh Badwaik

@skullpatrol No we were talking about this. Check out the revisions.

user19161

@WannabeMiniMathematician So what do you think of that William?

Wannabe Mini Mathematician

@JasperLoy: He was smart.

user19161

@WannabeMiniMathematician Hmm, I think he is mysterious. He is always saying mysterious things...

Wannabe Mini Mathematician

@JasperLoy: Wait... what? Just a minute back you said that there is no William.

Jayesh Badwaik

1:13 PM

@skullpatrol I missed what you said. was AFK.

skullpatrol

@JayeshBadwaik Never mind it wasn't important :)

Jayesh Badwaik

@skullpatrol :-)

skullpatrol

No more important than a 12 year old kid who you tried to help in math telling you that he is annoyed by your presence.

user19161

@WannabeMiniMathematician Hmm, that was Jayesh not me. I recall him now that you mention it.

Wannabe Mini Mathematician

@skullpatrol: I'm 13, and you're not John Junior.

Heh... I was just playing. :)

skullpatrol

1:22 PM

Hi @somekindarukus @spernerslemma

@WannabeMiniMathematician

Wannabe Mini Mathematician

Aw kid don't cry.

skullpatrol

Wannabe Mini Mathematician

Aw kid don't poop.

Aw kid don't breathe.

somekindarukus

@skullpatrol Just failed an assignment, for sure..

skullpatrol

1:38 PM

@somekindarukus Too bad, did you ask any of the questions on this site?

somekindarukus

@skullpatrol Yeah, quite a few. I hopefully didn't fail, I'm guess ~60% .. Thanks to all these stackers.

@skullpatrol I just can't get my head around combinatorics.. I don't know, I always think I'm counting things the correct way and then it turns out I'm just looking at it completely wrong.. When people show me the correct way I understand it.. It's just that I can't seek it out myself. I guess that just comes with practice, and I clearly haven't been giving my course enough effort.. So I'm gonna step things up

@skullpatrol Anyways, gonna go hand this assignment in!

Thanks for all the help everyone !

N3buchadnezzar

2:36 PM

Hi

@JasperLoy, I need to solve $e^z = -2$ where $z$ is complex. Rewriting the equation gives me the conditions $ e^x \cos y = -2$ and $e^x \sin y = 0$, the second equation gives me that $y$ has to be on the form $\pi n$

Using this in the first equation I get $e^x \cos( n \pi ) = -2$

Now this equation is only solvable iff $\cos(n\pi)=-1$, which happens when $n$ is odd.

So the solution to the equation is
$$ z = \log 2 + i \pi (2n+1) $$

But my book and maple, says that the answer should be $z= \log 2 + 2 i n \pi$, where is my mistake? =)

Jayesh Badwaik

@N3buchadnezzar there is no mistake. Log of complex number is a many-one function. (You have given a general solution, while the book as given the principle solution.)

user19161

@JayeshBadwaik You answered before I finished reading...

Jayesh Badwaik

@JasperLoy :-)

N3buchadnezzar

@JayeshBadwaik :-)

Definitively time for water/vodka

Jayesh Badwaik

@N3buchadnezzar This is similar to the $\mathrm{asin} \frac{1}{2}$. You can have multiple answers.

N3buchadnezzar

2:51 PM

Yeah

hic

Jayesh Badwaik

@Jonas See this. Especially see the semantic file browser there. I guess the application will come soon to KDE mainline. For now, it seems, I won't have to implement my tagged filesystem. (Thought I am not sure how good the tag management will be in itself though.)

anon

the answer given by the book and maple would only be true if the branch cut was taken at the real axis, with the real axis exponentiating to negative real numbers...

that would be highly nonstandard

user19161

Hmm, softwarers are not mathematicians.

Jayesh Badwaik

@JasperLoy Hmmpff.

anon

the softwarers that make computer algebra systems are by and large mathematicians

user19161

2:56 PM

Hmm, now I need to scratch my head.

user19161

As usual, the great anon has left us dumbfounded.