Mathematics - 2012-09-02 (page 1 of 3)

12:20 AM

I don't like it. It dismisses philosophical inquiries as a waste of time, doesn't acknowledge that the process of mathematics itself may be interpreted differently or more expansively than mere (arbitrary) rules and symbolic manipulations, and makes a claim of unfathomability of mathematics' "true reality" without justification.

Gustavo Bandeira

12:30 AM

@anon "to me"

user19161

I have passed the stage of reading all this mathephilosophical bullshit.

Gustavo Bandeira

I like it.

I don't know much tho.

user19161

You like very different in the two pics @gus.

anon

@GustavoBandeira The phrase "to me" only applies to his perception that the question is philosophical, not to the equivalence made between philosophy and wastefulness.

Gustavo Bandeira

@WillHunting The pseudo face and this new one?

@WillHunting Oh, the red photo.

user19161

12:33 AM

@GustavoBandeira The one I saw at the site you linked.

anon

I would have more sympathy for Wildberger's views of the imprecision in many mathematical concepts if he would explicitly exhibit an instance of two distinct and exclusive meanings being consistent with a given idea as understood in standard mathematics, rather than simply claiming ambiguity is in them.

Gustavo Bandeira

@WillHunting What site, i don't remember?

user19161

@GustavoBandeira You showed me some music.

Gustavo Bandeira

@anon I'm not sure if he meant that it's a "general waste of time".

@WillHunting For a more general view. Me and my frightened cat.

user19161

@gus Don't spend too much time thinking about these things now, you just need to work on basic math first.

user19161

12:37 AM

@GustavoBandeira Ah, I guess this must be quite recent...

Gustavo Bandeira

Yes,Will/Jasper.

@WillHunting 2009.

user19161

I always feel that the undergrads in this chat are going a bit too fast.

user19161

They talk about topics not covered until grad school.

Gustavo Bandeira

I want to become Euler overnight.

Haha

Peter Tamaroff

@anon Link?

anon

12:39 AM

Link to what?

user19161

Euler? Not a big deal IMHO. Try Gauss, much better.

Peter Tamaroff

@WillHunting TO THE BONFIRE!!!

Gustavo Bandeira

@PeterTamaroff youtube.com/…

Peter Tamaroff

@anon "Wildberger's views"

user19161

@PeterTamaroff Sure, I would gladly die in your arms...

anon

12:40 AM

He has many youtube videos, I don't have a particular one in mind.

Gustavo Bandeira

Peter, the link is the one I provided

Peter Tamaroff

@anon Does he criticize? What is going on?

Gustavo Bandeira

Will, I was trying to read "What is mathematics" but I guess I should abandon it for now.

Peter Tamaroff

@GustavoBandeira Yaeh, thanks.

@WillHunting You would die tied to a post, swallowed by vicious flames.

Not in my arms.

user19161

@GustavoBandeira Just read a little of this stuff and move on.

Gustavo Bandeira

12:42 AM

@WillHunting He begins with some proofs, but I usually get lost.

user19161

@GustavoBandeira You just need to know how proofs work, simple things like sets and logic.

Gustavo Bandeira

@WillHunting Why Gauss?

user19161

@GustavoBandeira Well, he seems the greatest ancient mathematician to me.

Peter Tamaroff

@WillHunting Why not Zoidberg?

Gustavo Bandeira

Whos the greatest modern mathematician?

user19161

12:47 AM

@PeterTamaroff Unheard of.

user19161

@GustavoBandeira Me, but yet to arise.

Gustavo Bandeira

Zoidberg

Doctor John A. Zoidberg is a fictional character in the television series Futurama. He is a lobster-like alien from the planet Decapod 10, who emigrated to 30th century Earth, where he works as the staff doctor for Planet Express, despite his woeful understanding of human physiology and allusions to his questionable credentials. Zoidberg is voiced by Billy West, who performs the character with a Yiddish-inflected accent inspired by actors George Jessel and Lou Jacobi. Character creation Zoidberg is named after an Apple II game that David X. Cohen created in high school called Zoid, simil...

I don't want my pizza burning

Peter?

I made this question:

1

Q: Under what conditions does a cubic polynomial have two equal roots?

I'm reading Sawyer's Prelude to Mathematics, here: I can't understand what's the meaning and application of "condition" here. Also when he gives the example on the cubic equation, stating that the condition is: $$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$ I can understand that it is $b^2-4ac=0$ (I h...

I guess MJD altered the title of the question to something that wasn't my question.

Peter Tamaroff

1:06 AM

@GustavoBandeira What is the original question?

Gustavo Bandeira

Where are the variables inside the parentheses coming from

Someone comented that it's the discriminant of the equation.

But I don't know how to find it.

I made a little experiment with Mathematica.

Peter Tamaroff

@GustavoBandeira OK. I don't know much about cubics, but let's see if I can help.

Do you know the general way to solve a cubic?

Gustavo Bandeira

Yes.

Peter Tamaroff

@GustavoBandeira I mean, the formula for the roots.

Say $x^3+ax^2+bx+c=0$

Gustavo Bandeira

Oh, no. I know only for quadratics.

Peter Tamaroff

1:09 AM

Do you know how to depress that cubic?

It is not that crazy.

Say I have $x^3+ax^2+bx+c=0$

I want to get rid of the $x^2$ term.

What change of variables should I perform?

Gustavo Bandeira

Divide everything by $x^2$?

Peter Tamaroff

@GustavoBandeira That would fuck things up! =D

@GustavoBandeira Think about what we do when we complete the square

We "get rid" of the term $bx$

in $x^2+bx+c=0$

Right?

Gustavo Bandeira

Yes.

Peter Tamaroff

Well, here we want to do something similar.

And we want to substitute $x=z-M$ for some suitable $M$ to get rid of the $ax^2$. That means a $-ax^2$ should appear somewhere

But that cn only happen in the cube, yes?

Gustavo Bandeira

Isn't it $cx$?

Peter Tamaroff

1:14 AM

Try $x=z-\frac{a}{3}$

Gustavo Bandeira

@PeterTamaroff This is the same technique for the quadratic equation, where we do: $x=y-\frac{b}{2}$, right?

Peter Tamaroff

@GustavoBandeira Yes.

So basically we get a new equation we'll call a depressed cubic: $x^3+Bx+C=0$

And that is "easy" to solve now.

Gustavo Bandeira

Proceed.

Peter Tamaroff

Well, the starting point is that $$(a-b)^3+3ab(a-b)=a^3-b^3$$

So let's think about the equation $x^3+mx=q$, with $m,q>0$

We want then $x=a-b$, $m=3ab$ and $q=a^3-b^3$

Yes?

Gustavo Bandeira

Yes.

Peter Tamaroff

1:20 AM

Now comes the fun part. We have that $$(m/3)^3=a^3b^3$

And then

$$q{b^3} = {\left( {\frac{m}{3}} \right)^3} - {b^6}$$

Now let $b^3=\alpha$

We get

$${\alpha ^2} + q\alpha - {\left( {\frac{m}{3}} \right)^3} = 0$$

Yes?

And we can solve this quadratic:

$$\alpha = - \frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} $$

..

$$b = \root 3 \of { - \frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} } $$

$$a = \root 3 \of {\frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} } $$

and $x=a-b$

For the case $x^3=mx+q$ with $,q>0$ we exploit $${\left( {a + b} \right)^3} = 3ab\left( {a + b} \right) + {a^3} + {b^3}$$

@GustavoBandeira Are we good?

Gustavo Bandeira

Nope, I'm still lost.

Peter Tamaroff

@GustavoBandeira Where did I lose you?

Gustavo Bandeira

@PeterTamaroff Here. Oh

anon

I am working on an explicit computation using the theory of elementary symmetric polynomials and a recursive algorithm. There is a good chance I will get bored in the middle of the tedious calculation and do something else, like eat or watch whose line or read more galois theory, however...

Gustavo Bandeira

I got it, i got it.

BenjaLim

1:27 AM

@anon Hi

anon

hello

BenjaLim

Can you help me with something?

You're familiar with matrix exponentials?

anon

no I do not have an answer to your balls question

Peter Tamaroff

@GustavoBandeira Good.

@anon Hahahaha

BenjaLim

@anon How did you know I was going to ask about the balls question?

@anon I have tried so many examples

all failed

Peter Tamaroff

1:28 AM

@BenjaLim Because you have a clear ball-fixation issue.

Gustavo Bandeira

@PeterTamaroff From here, we substitute the x for $z-\frac{a}{3}$

BenjaLim

@anon The condition that $||A - I|| < 1$ is a very tight one

Peter Tamaroff

@GustavoBandeira Yepo.

Gustavo Bandeira

Right?

Peter Tamaroff

@GustavoBandeira You'll get many messy coefficients but the idea is you get a new equation of the form $x^3+Mx=Q$ or $x^3=Mx+Q$

anon

1:29 AM

@BenjaLim Can you find an example of such an $A$ with $G=S^1$?

BenjaLim

@anon Ahhh

I have not tried that

Gustavo Bandeira

When the substitution is done, will I get this:

$$b \left(z-\frac{a}{3}\right)^2+c \left(z-\frac{a}{3}\right)+a \left(z-\frac{a}{3}\right)^3+d=0$$

BenjaLim

@anon I was trying all the linear groups yesterdaty

ok

Peter Tamaroff

@GustavoBandeira Huh, wait. Use a simper equation: $x^3+ax+bx+c$.

Keep it monic.

The substitution works for the monic eqn.

anon

My intuition for that is $\log(e^{i\theta}-1)$ should have nonzero real part expect when $\theta$ is an odd multiple of pi/2. (Also, $S^1$ is the only Lie group I can directly visualize...)

BenjaLim

1:31 AM

@PeterTamaroff Lie theory

Peter Tamaroff

@BenjaLim ?¿

Gustavo Bandeira

Then:

$$b \left(z-\frac{a}{3}\right)^2+c \left(z-\frac{a}{3}\right)+\left(z-\frac{a}{3}\right)^3+d=0$$

Peter Tamaroff

@GustavoBandeira You have the letters wrong.

$${\left( {z - \frac{a}{3}} \right)^3} + a{\left( {z - \frac{a}{3}} \right)^2} + b\left( {z - \frac{a}{3}} \right) + c = 0$$

Gustavo Bandeira

They're almost the same, I just switched somehow. Isn't it?

Peter Tamaroff

@GustavoBandeira Yes =)

Gustavo Bandeira

1:36 AM

@PeterTamaroff This point: I can't understand how you came to this starting point.

Peter Tamaroff

@GustavoBandeira What do you mean?

Gustavo Bandeira

You said that this:

$$(a-b)^3+3ab(a-b)=a^3-b^3$$

Is the starting point, I can't figure how you got in this point.

Peter Tamaroff

@GustavoBandeira That is just a nicer way to write the trinomial.

Gustavo Bandeira

How do I got from here:

$${\left( {z - \frac{a}{3}} \right)^3} + a{\left( {z - \frac{a}{3}} \right)^2} + b\left( {z - \frac{a}{3}} \right) + c = 0$$

To here:

$$(a-b)^3+3ab(a-b)=a^3-b^3$$

Factoring?

BenjaLim

@anon I think I have an example

@anon I think it should work

Peter Tamaroff

1:40 AM

@GustavoBandeira Nu nu nu, you're mixing things up. Sorry.

Let's reboot.

First, we note this

$$\eqalign{
& {\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3} \cr
& {\left( {a - b} \right)^3} = - 3aba + 3abb + {a^3} - {b^3} \cr
& {\left( {a - b} \right)^3} = 3ab(b - a) + {a^3} - {b^3} \cr
& {\left( {a - b} \right)^3} + 3ab(a - b) = {a^3} - {b^3} \cr} $$

Yes?

BenjaLim

@anon mwah

@anon love ya

Peter Tamaroff

Now put that aside. Call that "OBSERVATION I"

Gustavo Bandeira

It's almost what I thought - I guess - From the factoring of polynomials:

$$(x + p)^2 \,=\, x^2 + 2px + p^2.\,\!$$

Peter Tamaroff

@GustavoBandeira OK, OK.

Now, let's do "OBSERVATION II".

Let $x^3+ax^2+bx+c=0$; $a,b,c$ reals.

Then the change of variables $x=x'-\frac a 3$ eliminates the $ax^2$ term, and gives a new equivalent equation, namely $x'^3+mx'=q$ or $x'^3=mx'+q$, depending on the signs of $a,b,c$.

Right or left?

Gustavo Bandeira

Thinkng

Thinking

Oh

Right.

Peter Tamaroff

1:47 AM

@GustavoBandeira Just take a piece of paper =)

Gustavo Bandeira

That's a joke. I thought it had something to do with RHS or LHS

BenjaLim

@anon The argument to use I think

would be to find a matrix $A \in SO(2)$

such that no $X$ in the lie algebra maps to it.

Peter Tamaroff

@GustavoBandeira Hahahah it from the BFG.

Rhoad Dhal's BFG.

@GustavoBandeira OK, so let's suppose we get $x^3+mx=q$.

Gustavo Bandeira

What's the meaning of the ´

$´$

Peter Tamaroff

@GustavoBandeira Just that it is a new variable.

Gustavo Bandeira

1:52 AM

It's just a new $x$, isn't it?

Peter Tamaroff

@GustavoBandeira Yes.

Gustavo Bandeira

Yep,I still remember.

Peter Tamaroff

@GustavoBandeira OK. So we want to find $x$ in $x^3+mx=q$.

Let's assume $x=a-b$.

For some $a$ and $b$.

Gustavo Bandeira

@PeterTamaroff From where is this assumption coming?

Peter Tamaroff

Then we use that $${\left( {a - b} \right)^3}+3ab(a - b) = {a^3} - {b^3}$$

@GustavoBandeira Great italian mathematicians.

BenjaLim

1:54 AM

@anon Hmmm

Though my understanding of the problem is correct yes?

anon

no idea

BenjaLim

Even though the theorem says that there are neighbourhoods such that....

Peter Tamaroff

Which means $m=3ab$ and $q=a^3-b^3$; else the equality wouldn't hold, yes?

BenjaLim

it is not necessarily true that the exponential must map an open ball to an open ball

Gustavo Bandeira

@PeterTamaroff Yes.

Peter Tamaroff

1:57 AM

@GustavoBandeira Well, now we'll solve for $b$.

@GustavoBandeira We have that $qb^3=a^3b^3-b^3$ and that $(m/3)^3=b^3a^3$ so we get $qb^3=(m3)^3-b^6$. That is a quadratic in $b^3$.

Namely $$(b^3)^2+qb^3-(m/3)^3=0$$

so that $$b^3=-\frac q 2 \pm \sqrt{\left(\frac q 2\right)^2+\left(\frac m 3\right)^3}$$

Gustavo Bandeira

Proceed.

Peter Tamaroff

@GustavoBandeira Well, that means $$b = \root 3 \of { - \frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} } $$

and since $$q = {a^3} - {b^3}$$

we get

$$a = \root 3 \of {\frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} } $$

so, all in all

$$x = \root 3 \of {\frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} } - \root 3 \of { - \frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} + {{\left( {\frac{m}{3}} \right)}^3}} } $$

=D

Gustavo Bandeira

Well. Don't hope me to understand all now.

But I'm doing my efforts.

This is the way a cubic equation is solved, right?

Peter Tamaroff

@GustavoBandeira YOu should try and write the stuff down. It will help-

@GustavoBandeira Yep.

Gustavo Bandeira

In that question, he tells us that the condition for at least two roots being equal is:

$$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$

I don't understand where are the variables coming from.

This has the same form of: $b^2-4ac=0$

These are the only things I noticed.

anon

2:11 AM

You know where the variables are coming from (they come from the polynomial of course!), what you don't know is how they end up arranged that way and why that condition is equivalent to a double root.

Gustavo Bandeira

@anon Yes.

Sorry for bad question.

Jayesh Badwaik

@GustavoBandeira Haven't you seen how the literature references mathematicians? "Gauss and all the lesser mathematicians" they say!

Gustavo Bandeira

@JayeshBadwaik ?

Jayesh Badwaik

About your discussion of who is the greatest mathematican with Willhunting.

Gustavo Bandeira

Oh, haha.

It's becauseyou pinged tha message. I was like: WTF?!

Jayesh Badwaik

2:20 AM

:P

Peter Tamaroff

@GustavoBandeira Dude, try to work things out with the formulas we obtained.

Gustavo Bandeira

@PeterTamaroff Yep, I'm doing it.

Jayesh Badwaik

@PeterTamaroff Isn't that called the ferrari's method or something? The one you just described?

Peter Tamaroff

2:35 AM

@JayeshBadwaik Yes, Ferrari-Cardano or something similar.

@GustavoBandeira For example.

The solutions for $x^3=mx+q$ are given by $$x = \root 3 \of {\frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} - {{\left( {\frac{m}{3}} \right)}^3}} } - \root 3 \of { - \frac{q}{2} \pm \sqrt {{{\left( {\frac{q}{2}} \right)}^2} - {{\left( {\frac{m}{3}} \right)}^3}} } $$ (same method as the other)

You should be looking at $${\sqrt {{{\left( {\frac{q}{2}} \right)}^2} - {{\left( {\frac{m}{3}} \right)}^3}} }$$

Gustavo Bandeira

2:49 AM

@PeterTamaroff Ok. Thanks for the help.

I'll think a little about it now.

anon

3:08 AM

I applied the theory of symmetric polynomials to tackle the discriminant calculation in a tedious algorithmic fashion. The form in which the author writes the condition however suggests there is a much quicker way to deduce the condition using matrix reasoning.

That is the first time I've actually computed the discriminant instead of taking it for granted, incidentally.

Gustavo Bandeira

You answered also another question I had for sometime, how to do this:
\tag{$\circ$}

Jayesh Badwaik

4:37 AM

http://math.stackexchange.com/questions/189886/reducing-fractions#comment437868_189886

LOL

Gustavo Bandeira

6:09 AM

@JayeshBadwaik Yes.

Gnintendo

6:36 AM

hey

anybody active now?

Fortuon Paendrag

@Gnintendo Perhaps. Whatsup?

Gnintendo

I'm doing my math homework for Calculus III, and I've been getting everything right, but it won't accept my answer for one of the problems

Fortuon Paendrag

fire away.

Gnintendo

it's a T B N problem for a vector through a point

I calculated T and submitted it to confirm I was using the correct format for my answers before I attempted the rest: grab.by/fNLe

it accepted my answer for the y and z coordinate, but even when I formatted my answer for x in several different ways, it's still saying it's wrong

I only have one remaining submission before it's permanently wrong

I've never missed a problem before, hell, I've never even used more than one submission

Fortuon Paendrag

Maybe I'm missing something, but what are T B and N?

Gnintendo

6:42 AM

Frenet–Serret formulas

:"Binormal" redirects here. For the category-theoretic meaning of this word, see Normal morphism. In vector calculus, the Frenet–Serret formulas describe the kinematic properties of a particle which moves along a continuous, differentiable curve in three-dimensional Euclidean space R3, or the geometric properties of the curve itself irrespective of any motion. More specifically, the formulas describe the derivatives of the so-called tangent, normal, and binormal unit vectors in terms of each other. The formulas are named after the two French mathematicians who independently discovered...

the Frenet-Serret formulas

I can find N and B easily, I just want to figure out the x coordinate for the T

I am literally at a complete loss for why it isn't accepting any of my answers

Fortuon Paendrag

Ahh!

How did you compute $T$?

Gnintendo

you want to do it and tell me what you get to x and I'll see if it's mathematically equivalent?

derivative of cos divided by the magnitude of the derivative of the function r

then I plugged in 1

since it's through the point (1,0,0)

Fortuon Paendrag