Mathematics - 2014-08-30 (page 3 of 5)

Khallil

3:00 PM

If so, I think that's right.

Gustavo Montano

And it makes sense intuitively, right?

Nick

@Khallil: There are conditions to that rule. You can't evaluate the limits at different times as I did in @toppers question. It's actually wrong to do that in most situations. The tendancy of $x$ has to progress the same way even when you butcher the limit apart.

Gustavo Montano

If they're not continuous they can blow up :p

I like these discussions. Talking about the things they never teach you.

It's great.

Khallil

I really don't know anything about limits.

Nick

@topper: what gustavo implies is that you can do it in your question. The limit has an all road access

Khallil

3:01 PM

I really don't know anything about anything.

Gustavo Montano

"All I know is that I know nothing."

3

Nick

@Khallil: I am with you buddy.

@GustavoMontano: .. then you know something. Meaning it is false to say you know nothing. It's a fallacy I say!

Balarka Sen

@Nick Have you ever heard of prime number theorem?

Khallil

Hey, @BalarkaSen.

topper

@GustavoMontano I assumed that this was stuff that people were taught! For the record, I'm doing a prep module as a requirement for a business degree. So as it's outside school, there's no support, I've "learned" everything online. There was an optional course, but it was very expensive. I'm just going to rock up at the exam...

Gustavo Montano

3:03 PM

en.wikipedia.org/wiki/I_know_that_I_know_nothing ^_^.

Balarka Sen

@Khallil Ahoy.

Khallil

Gustavo Montano

There's so much out there @topper.

My best advice is learn the little rules, then just do as many questions as possible. With pace!

Nick

@BalarkaSen: Is that what PNT stands for?

Balarka Sen

@Nick yes.

Nick

3:06 PM

@BalarkaSen: $$\pi (x) \approx \frac{x}{\ln (x)} $$

Balarka Sen

what about it?

it's unusual for PNT to be described in terms of the difference of $\pi(x)$ and $x/\log(x)$

Nick

I can't find the twiddle!

Balarka Sen

@Nick more or less. depends on how you'd define $\approx$

would you call, say, $50847534 \approx 48254942$?

Nick

@BalarkaSen: I wouldn't. I'd ask my friend Balarka to do it since he's so much smarter than I am.

@BalarkaSen: In physics, yes. In math, there was that $\equiv$ approx. I can't find that either.

Balarka Sen

@Nick but then $\pi(10^9) = 50847534$ and $10^9/\log(10^9) \approx 48254942$

What's going on?

Gustavo Montano

3:11 PM

The formula must hold for only a set of numbers.

An ordered set of numbers.

Balarka Sen

@GustavoMontano what formula?

Gustavo Montano

$\pi(x)$

Balarka Sen

that's just a function. where do you get the formula from?

Gustavo Montano

Oh ok, the function then.

Balarka Sen

The statement of PNT is simply $$\lim_{x \to \infty} \frac{\pi(x)}{x/\log(x)} = 1$$

Gustavo Montano

3:13 PM

Hmmm....

Balarka Sen

It's division that is involved, i.e., order or magnitude, not difference, i.e., error of approximation

Khallil

Be back in a while.

^_^

Gustavo Montano

So what have you done with PNT?

Nick

@Khallil: Toodles

Balarka Sen

@GustavoMontano Just babbling.

AndrewG

3:13 PM

@Balarka the difference can (and does) get arbitrarily large, yes? It's the relative error that goes to nil?

Balarka Sen

@AndrewG Yes.

AndrewG

(just checking if I remember right)

coolsauce

Balarka Sen

In particular the difference is $O(x/\log(x))$

Nick

@BalarkaSen: $O(x)$ ?

Balarka Sen

@Nick Big O notation. google it.

fancy notation stuffs.

Nick

3:16 PM

@BalarkaSen: No, I'll yahoo it! You're not the boss of me.

Balarka Sen

haha

Nick

@topper: Please read and try out tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx

Balarka Sen

there is a cool probabilistic interpretation of PNT

Nick

Yahoo is showing me a rapper named Big O, ... I should probably just google it.

Either way, I'm just gonna read the Wikipedia article... I should have just gone there in the first place.

AndrewG

@Nick it's basically a rough comparison of growth between functions. $e^x$ grows faster than $x^n$ which grows faster than $log^n(x)$ etc. Saying $f(x) = O(x)$ means (I think, Balarka can surely correct me) that f(x) is more-or-less linear.

Mats Granvik

3:20 PM

Big O means that there is a constant "c" times "x/log(x)" such that your expression is bounded by "c*x/log(x)". Am I right?

Nick

... and finally, I realize that it's not you're average run in the mill function.

Balarka Sen

sorry i got disconnected

@MatsGranvik well, almost. the absolute values are taken usually.

Mats Granvik

@BalarkaSen ok

Nick

@BalarkaSen: If this 1860, I'd ask yee if yuz was a robut.

Balarka Sen

i.e., pi(x) - x/log(x) grows almost like something which is bounded, upto absolute values, by a constant multiple of x/log(x)

@Nick wat

@AndrewG right

Nick

3:24 PM

@BalarkaSen: It's about the disconnected statement. Ignore it. Is there anything more to Big O other than growth?

Balarka Sen

saying f(x) is O(g(x)) means that f(x) have atmost the same order of magnitude as g(x).

oops i was thinking of little o

Nick

@Balarka: Wow, so basically little o is what I need to equate the sun and the moon. Cool!

topper

@Nick Thanks, I will, I have a lot of material to get through though, before I start proving :(

Nick

@topper: We're happy here to help :D

topper

@Nick I know, I've been really pleasantly surprised by the vibe in here. Perhaps I'm just used to old-school IRC flame-athons :)

Nick

3:31 PM

@topper: Yeah, it's much more calm, more slow, more natural and everything flows. It doesn't overflow like in IRC's like DeviantArt's dAmn network. Whoo, that's a flood of thought.

@BalarkaSen: You obviously didn't catch that reference about the cosmos but I still have a lot before getting to the O's

@Sawarnik: Arre yaar, what's up?

Sawarnik

@Khallil @GustavoMontano @BalarkaSen @Nick @Topper Hi :)

Khallil

yo

Balarka Sen

@Sawarnik I was reading His Last Bow

Nick

@Khallil: when did you get back?

Sawarnik

@Nick Ow, yaar is used in Malayalam as well?

Khallil

3:35 PM

just now

:)

Nick

@Sawarnik: Nahi... , I don't like Malayalam.

Sawarnik

@BalarkaSen Its good :)

Balarka Sen

@Sawarnik The Dying Detective was cool

Sawarnik

@BalarkaSen The one in which Sherlock fakes a disease? :O

Balarka Sen

yeah

Sawarnik

3:36 PM

It was :)

Only a few stories from the last book were bad, the rest are just as awesome :)

Nick

@BalarkaSen: To this day, you have never realized my Dr.Holmes reference.

Sawarnik

@Nick Where? :O

Balarka Sen

@Nick what reference?

Sawarnik

@Nick Whatdoyamean?

Balarka Sen

@Sawarnik I also have the Case Book Of Sherlock Holmes cobbled up with my mega book

Nick

3:38 PM

@BalarkaSen: Remember when I said Dr.Holmes and "Elementary, my dear Wilson!"

Balarka Sen

he never said that anywhere, btw

it's a fluke.

oh, Wilson?

Nick

I said it.

Sawarnik

actually true.

Balarka Sen

no i never got it @Nick

who the heck is Wilson?

Nick

$$\text{Holmes} \approx \text{Homes} \equiv n\times\text{House} , n \in \mathbb N$$

Sawarnik

3:41 PM

The empty house.
Lestrade.

Balarka Sen

Lame joke

Nick

... You kids never watch TV :( ... mmh, maybe that's a good thing :D

Sawarnik

@Nick Who said? :P

carpet jar

guys, I have a problem: Let $V = \mathbb{R}_3[\cdot]$ be a vector space with dot product $(f|g) = \int_0^1f(x)g(x)dx$. Find orthogonal projection of vector $u(x) = 12x^2\in V$ on subspace $W = \mathbb{R}_1$ and component of vector $u$ orthogonal to $W$.

Following by instructions here: en.wikipedia.org/wiki/Projection_(linear_algebra)#Formulas counted that orthogonal projection should be equal to $((0), (1), (0), (0))\cdot (0, 1, 0, 0) \cdot ((0), (0), (1), (0))$, but that gives me $0$ as a result. Can someone tell me, what am I doing wrong?

Sawarnik

@Nick Who is a kid anyway?

Nick

3:44 PM

Stuart little's dad said you don't watch TV:

Sawarnik

:O

@Nick The match is an interesting situation now :D :D :D :D :D

Nick

@Sawarnik: Basically, Doyle modeled Holmes from a great doctor who investigated incredible medical cases. Dr. House is a doctor on a TV show modeled from Holmes.

Sawarnik

Ok :)

Nick

(Notice how I managed to fit in Doctor Who in my last sentance)

Sawarnik

@nick You watch BBC Sherlock?

Nick

3:50 PM

@Sawarnik: I didn't watch after he faked his death. By the way, ever hear of Wholock (Youtube it, it's awesome) . But I think he did a bad job portraying Khan in Star Trek. And both this Sherlock and Wilson are in the Hobbit! Sherlock is the necromancer.

Sawarnik

:\

Nick

... I need to study more and watch less TV.

Balarka Sen

@Nick Wow

Sawarnik

Coming in 15 mins...

Balarka Sen

Sherlock is the necromancer? @Nick that's cool.

Sawarnik

3:50 PM

@Nick Great :P

@BalarkaSen :D Awesome idea :D

Nick

@BalarkaSen: The actor from the BBC show (who also played the new Khan in Star Trek 2) gives the voice for the dragon thing.

Both those guys were in the movie. (Wilson was the hobbit)

Balarka Sen

You mean the Smaug?

@Nick You mean Bilbo Baggins?

Nick

@BalarkaSen: I never watched the movie nor read that book but I thought Smaug was the necromancer? LOL , I'll go watch it now.

Balarka Sen

That's silly

Nick

Yes Bilbo Baggins is Watson!

topper

3:53 PM

okay, i'm looking to find $$\lim_{x \to 0^+} x \cdot \ln x$$. my first idea was to use the previously mentioned formula and multiply the two limits, but the solution in front of me has the second step as $$\lim_{x \to 0^+} \frac{\ln x} {1/x}$$. is it incorrect to use the formula i.e. $$\lim x \cdot \lim \ln x$$?

Balarka Sen

Necromancer is Sauron. Worms of the West are not human.

@Nick Don't watch the movie.

Lamest thing ever. Try the books.

Nick

I knew that but I thought Smaug was a form of Sauron. That would have been cooler.

Balarka Sen

no no no

However dangerous, Worms never use dark magic.

AndrewG

Jackson turned Hobbit into his own personal fanfiction :( ruined it.

Balarka Sen

yes, he ruined it.

the LOTR movies were lame too

Nick

3:56 PM

@topper: ... now, I have to read the book.

@BalarkaSen: Not to mention LONG! very LONG!

AndrewG

I actually liked those. But all this stuff with Gandalf running off to face the necromancer in the new trilogy...it's just...wut.

Nick

... the book as well, but that was different.

Balarka Sen

You'll faint if you get to see the collected works of Tolkien on LOTR, @Nick

topper

@Nick To give some more info, after that step, they solve it with L'Hopitals. (Of course it needs to be a rational fraction in order to do that.)

@Nick The book of The Hobbit, or some book on solving limits? :P

Nick

@topper: Sorry, but the L'Hospitals didn't work? It should have?

topper

3:59 PM

@Nick It did. I'm just questioning if that's the "best" way to find the limit. Because turning a multiplication by x into a division by 1/x certainly doesn't feel like an intuitive first step

Nick

@BalarkaSen: I've read a treatise on Middle Earth... it was like a Harry Potter Encyclopedia except about the LOTR universe.

Balarka Sen

Among all the popular fantasies I have read, the only thing that survived after "filmification" is Harry Potter

carpet jar

guys, can you please help me with my algebra? I have no idea what I'm doing there.

Nick

@topper: All is fair in love and math. You can't use the product thing because the tendancy has to go in the same way for both guys. What you did is the easiest thing to do.

topper

@Nick "tendancy has to go in the same way for both"?

Nick

4:04 PM

@topper: Imagine you split the limits of of two multiplied functions, if one of them becomes 0, the entire thing becomes 0. We do things in quick solving based on rough intuition. Proper Math is very different.

topper

Maybe I could have used the product thing. Because $$\lim_{x \to 0^+} x = 0$$, and that times whatever $$\lim_{x \to 0^+} \ln x$$ would have been, would have reached the same answer (zero)

Nick

that's what I mean to say, sometimes this logic fails.

The other limit could not exist.

then you'd be doing:

$$0 \times \text{undefined}$$

topper

@Nick "Other limit" being the top or bottom limit I posted?

Nick

Either. I'm not saying it's wrong here in your situation now. But in future, in another question.

topper

Okay, just so I'm very clear on this (because there's no grey area in maths, right?), it would have been valid to split the limits of two functions. i.e. Both approaches work?

Nick

4:09 PM

Here, sadly, it does.

topper

This calls for microwave popcorn and coffee!

Nick

(I'm just saying the splitting a bad approach in many situations)

@topper: I'll give you some good questions if you're up for it.

@topper: Up for it?

Khallil

Up for it, @Nick.

Nick

@Khallil: Seriously? You?

Khallil

What are the questions about?

Nick

4:16 PM

@Khallil: Limits

Khallil

Then seriously, yes!

Nick

@Khallil: I won't give you anything easy, Find the following $$\lim_{x \to a}\frac{x^n - a^n}{x-a} \quad, \space n \in \mathbb Q$$

topper

@Nick Up for it, need to finish some material. Don't post just for me, thanks!

Khallil

na^(n-1)

?

Is that it, @Nick?

Nick

@Khallil: You didn't use a pure limit proof did you...

Khallil

4:19 PM

I did. L'hop.

Nick

@topper: I'll give you relatively easy ones.

Khallil

I hop with L'hop.

=)

Nick

@Khallil: ... I have a book where the derivation is one page long! ... Me in the past didn't know L'Hop

Khallil

Oh. Well all that matters is the current you.

^^

Nick

@Khallil: If you know, L'Hop, here's another easy one $$\lim_{x \to 0} \frac{3^{2x} - 2^{3x}}{x}$$

Khallil

4:22 PM

ahhh

derivative of a^x is ln(a) a^x right?

Nick

@topper: (Notice that in the question above if you take out the numerator using the product split, the 1/x will be infinity. That's why I said that it's a bad approach if done illogically )

@Khallil: Prove that using first principle before you start

Khallil

Ahhh

Boooooring

Nick

But yes

Khallil

urm

Nick

go on, give me an answer

Khallil

4:25 PM

oh

k

$y=a^{bx} \implies \log y = bx \log a \implies \dfrac{1}{y} \ \dfrac{\text{d}y}{\text{d}x} = b\log a \implies \dfrac{\text{d}y}{\text{d}x} = b\log (a) a^{bx}$

2log(3) - 3log(2)

Is that it, @Nick?

Nick

Yup :D

Khallil

That wasn't easy.

That was some tough stuff.

Nick

Actually it was

It's difficult with L'Hosp

Khallil

Diff of two squares?

Nick

You have to use $$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln(a)$$

Khallil

4:30 PM

Why?

-_-

Nick

If you're not using LHosp

Now, I give you the same question as the previous except with $\sin 3x$ as denominator

Khallil

The same as what?

It'll be a third of blah

Nick

huh, okay, I'll type it

$$\lim_{x \to 0} \frac{3^{2x} - 2^{3x}}{\sin 3x} $$

Khallil

1/3 of the answer I got before

Nick

Right :D

Khallil

4:34 PM

that was hard too

Nick

Ok, an easier one then,

Khallil

ok

Nick

$$\lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x\cdot \tan x}$$

Khallil

0

prod rule on denom

1

Nick

... sure?

Khallil

4:36 PM

top derivative log(10/10)

yea

urm

oops

Nick

mhh, I'm getting another answer by another method.

Khallil

double l'hop might do it

Nick

No, I'm right

i rechecked

Khallil

log(10)^2 - log(5)^2 - log(2)^2

?

Nick

No teacher will accept that as a final answer

Simplify

Khallil

4:39 PM

ahh

2log(5)log(2)

Nick

.... Sorry, i tried simplifying that, nope, you're wrong

Khallil

?

It's not 2log(5)log(2)?

Nick

No, you're very close

A slight mistake somewhere

Khallil

really?

Nick

A very small mistake

Khallil

4:43 PM

ahhh

do you just want me to simplify it?

log(5)log(4)

Nick

... lol

Khallil

no?

Nick

You could just recheck your calculation

Balarka Sen

@Nick $\log(5) \cdot \log(2)$

Khallil

hmm?

Balarka Sen

4:45 PM

Easy-peasy

Nick

@BalarkaSen: Yes!

Khallil

i'm just lazy

can't see it ...

Nick

@Khallil: The numerator in the question, simplifies to $(5^x - 1)(2^x -1)$ ... could have saved you a lot of trouble.

Khallil

oh

no

I see my mistake

i think

Nick

@Khallil: My previous statement solves the question in 3 steps.

Balarka Sen

4:50 PM

@Nick That's what I did

Nick

@BalarkaSen: i actually have a limit question that I couldn't do. Care to help me out?

It's pretty easy-peasy but there's something very funda that I missed.

$$\lim_{n \to \infty} \frac{a^n}{n!} \quad, a \in \mathbb R^+$$

@Khallil: Congrats on the last one, see if you can do what I failed.

This came for an MCQ, I had less than 1 minute 30 seconds to do it. Although I got the right answer in the test, I still have no clean proof for it.

Ted Shifrin

@Nick: Try writing that out with $a=5$, say ...

carpet jar

@Nick tried Stirling approximation? The only problem is that I don't remember whether it gives values greater than $n!$ or smaller...

Ted Shifrin

Stirling is killing a fly with a cannon.

Nick

My logic was that $n!$ becomes $\infty$ faster than $a^n$, so the denominator would reach infinity even while $a^n$ is finite. So, since the denominator is infinity, our limit becomes zero.

i know, my logic is absurd

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$Mathematics$ Mathematics