Mathematics - 2014-05-29 (page 1 of 5)

AndrewG

12:00 AM

Ah

robjohn

@Hippalectryon I'm working on it, but I have not had much time yet.

Hippalectryon

Ok thank you

Pedro Tamaroff

@seaturtles Looks cool. I guess it isn't too surprising once you know a bit of it.

Shahar

12:15 AM

hello

whatup G's

Ted Shifrin

@Andrew: Note that one is invariant under change of orientation (switching sign on $n$) and the other isn't. Gauss's Theorema Egregium says you can in fact compute $K$ knowing just the first fundamental form.

Shahar

$$\frac{d^2x}{dt^2}=G\frac{M}{(R-x)^2}$$ How do you solve that? $x(0)=0$ and $x'(0)=0$ if that helps

Ted Shifrin

@Shahar: Multiply both sides by $dx/dt$.

Shahar

What will the left side be? $\frac{d^3x}{dt^3}$?

Sorry I don't really know differential equations

sea turtles

or derivatives apparently

Mike Miller

12:21 AM

@TedShifrin I wouldn't mind making it through my life without spending much time on finite spaces, but I think the pseudocircle is very cute.

sea turtles

@Shahar $x''=a/(b-x)^2$ multiplied by $x'$ is $x''x'=ax'/(b-x)^2$, both sides are easy to integrate

Shahar

easy?

$$\frac{d^2x}{dt^2}\cdot \frac{dx}{dt}=G\frac{M}{(R-x)^2}\cdot \frac{dx}{dt}$$

Ted Shifrin

@Mike: I'm sort of surprised this is the first time in my 44 years of math to see it!

Shahar

So basically

is this possible:

wait actually

Ted Shifrin

Yes, @Shahar: you should know how to integrate with substitutions.

Shahar

12:25 AM

I don't think I do ^_^

Mike Miller

@TedShifrin It's a very cute counterexample. I wish I knew a way of explicitly showing that its universal cover is contractible. (The answer cites an arXiv article.)

Ted Shifrin

Have you had second semester calculus?

Shahar

no

Karl Kronenfeld

or equivalently recognize the result the result of the chain rule when it's in plain sight @Shahar

Shahar

that's the thing

well I did

I only had super basic differential equations

Mike Miller

12:27 AM

Yes I have, @TedShifrin

Ted Shifrin

Go relearn basic calculus!

AndrewG

@TedShifrin Stupid question, but...That's something else I'm confused about: why is the first fundamental form considered intrinsic? Both the first and second are calculated, in practice, by first parameterizing a surface, then doing the derivative song and dance, yeah? Isn't that just the same as embedding the surface and working in $\mathbb{R}^3$? I think my problem is pretty basic and I just don't understand the whole extrinsic v.s. intrinsic thing.

Shahar

That's not basic calculus

Ted Shifrin

Shaddup @Mike

Mike Miller

@TedShifrin No joke, I'd expected exactly those words with exactly that spelling.

Ted Shifrin

12:28 AM

What is the derivative of $x(t)^2$, @Shahar?

Shahar

$$2x(t)\frac{dx}{dt}$$

Ted Shifrin

Good, @Shahar ... Now think about going backwards. Do it similarly with each side of our equation.

Shahar

Yes that's what I was thinking

but I get nowhere

So i have $$GM(R-x)^{-1}$$

on the right side, working backwards

Ted Shifrin

@Andrew: You're asking great questions, not stupid ones. One can measure distances and angles intrinsically on the surface once one knows the first fundamental form. For example, a cylinder and a plane look the same locally. It doesn't depend on the embedding in $\Bbb R^3$. But the principal curvatures are measuring bending of the surface as it sits in $\Bbb R^3$: That's extrinsic.

OK, @Shahar. Now what would give you $x'x''$ when you differentiate? Also, don't forget your constant of integration.

AndrewG

Hrmmm

Ted Shifrin

12:32 AM

I tell my differential geometry students that "intrinsic" means something an ant living on the surface could understand/compute.

Shahar

$\frac{(x')^2}{2}$?

Ted Shifrin

Yup @Shahar. See, that wasn't so bad!!

Shahar

Dude

what the heck

I had that before

But then I couldn't solve it

(I also had that using energy)

Ted Shifrin

Well, there's a constant of integration you need. Yes, this is the "energy" trick we used to solve the problem.

Shahar

but then when I do an anti-derivative

Ted Shifrin

12:34 AM

You have $(x')^2 = 2GM(R-x)^{-1} + C$.

Shahar

I have like an impossible integral

look so let's find $C$

$x'(0)=0$

Mike Miller

@TedShifrin What if it's a really big ant?

Shahar

So we have $$0=2GM(R)^{-1}+C$$

Ted Shifrin

It's a point ant, like all animals in mathematics, @Mike.

AndrewG

@MikeMiller Hehehe, maybe he can move that rubber tree plant.

Shahar

12:35 AM

And our equation is $$\frac{dx}{dt}=\sqrt{2GM\left(\frac1{R-x}-\frac1{R}\right)}$$

Ted Shifrin

OK. So you can simplify it a bit and then separate variables.

Mike Miller

@AndrewG :D

Shahar

Right so I have

$$\frac{dx}{\sqrt{\frac1{R-x}-\frac1{R}}}=\sqrt{2GM}\; dt$$

Can't integrate it now

Not possible

Ted Shifrin

Simplify the algebra. It is possible.

Shahar

undefined

No I'm saying

x=0

AndrewG

12:37 AM

@TedShifrin In physics we have spherical cows. And then by Gauss's law, I suppose they're point cows.

Shahar

I had this before

and Wolfram didn't give an answer

Ted Shifrin

It will appear to be an improper integral, @Shahar. Tough, Wolfram is wrong.

Stop being a stubborn mule and do as I say. Or I'll quit helping.

Well, a hyperbolic cow would just make a mess @Andrew.

AndrewG

Oh jeez lol

Shahar

actually never mind, it does work. It treated $R$ as a variable.

If Wofram didn't mess me up...

Ted Shifrin

You young people need to learn to do algebra and not be lazy.

The typical physics problem like this will lead to hyperelliptic integrals, but this force won't.

Shahar

12:40 AM

Wolfram was invented

so that we don't have to do algebra

Ted Shifrin

Well, fine. Go screw off elsewhere.

AndrewG

No wai, it was invented so you can check yourself after you algebra.

Ted Shifrin

Before I get banned, I'm leaving.

Shahar

bye?

Hippalectryon

@TedShifrin NUUUU STAY

Shahar

12:44 AM

lol the anger

btw it doesn't work

at x=0

it's undefined

David Kirby

$GM=3\pi V/P^2$ ?

Hippalectryon

1:08 AM

@robjohn Yeah it's $c^{n+1}$, it was an editing mistake

robjohn

@Hippalectryon I figured. Just wanted to make sure. I have to go out for a while... I have the answer in terms of determinants, I just need to convert it to the form you have.

David Kirby

2:32 AM

"... given a collection of identical dominoes, it is clearly possible to stack them at the edge of a table so that they hang over the edge of the table without falling. The counterintuitive result is that one can stack them in such a way as to make the overhang arbitrarily large, provided there are enough dominoes" - from the mu epsilon journal (circa sometime in the 60s) ... .. hmm...

abuse of notation.

David Kirby

2:46 AM

so is $$H_{-\frac{1}{4}}-H_{-\frac{3}{4}}=\pi$$

eventually this series should diverge right?

Studentmath

3:13 AM

I really need to expand my knowledge of notations. Seen this H too many times

David Kirby

this is the more common notation : $$\psi ^{(0)}\left(\frac{3}{4}\right)-\psi ^{(0)}\left(\frac{1}{4}\right)$$

Studentmath

Ah!

David Kirby

... and a functional approach to this series : $$-\frac{1}{4} \cos (n z) \psi ^{(0)}\left(\frac{\pi -2 n z}{4 \pi }\right)+\frac{1}{4} \cos (n z) \psi ^{(0)}\left(\frac{3}{4}-\frac{n z}{2 \pi }\right)-\frac{\pi }{4}$$

Studentmath

3:33 AM

$sin(y^3)$ is rather tough to integrate, correct?

David Kirby

yup, you got it

this is messier -- the gibbs phenomenon is more pronounced

but it compiles to a closed form in base 4, making it really useful for signal processing

Ted Shifrin

@Studentmath — Have you been awake all night?!

Studentmath

Hey Prof. @Ted - Just woke up actually :P

Ted Shifrin

Ah ...

Studentmath

I'm not that insane, yet

Ted Shifrin

3:37 AM

i dunno about that

Studentmath

Went to sleep early after being disappointed with some work in Graph Theory.. haha

Ted Shifrin

Well, my answers haven't been getting much appreciation, but I'm pondering one to write in the morning ...

Studentmath

People are usually ungrateful, nothing new there

And really interesting @David

Ted Shifrin

I blew my stack totally at a student here earlier ...

Studentmath

Yeah I saw when I scrolled up :P

I get your annoyance with it, but I don't think it's worth it - if it doesn't bring you joy to answer that way simply don't continue helping there

Ted Shifrin

3:44 AM

Yup, I know. It's @Pedro's fault for luring me to chat 10 months ago or so ...

Ok, get some work done. Good morning/good night.

Studentmath

Nah, the chat overall is nice, bar the horrible ping sound

David Kirby

i turned off the ping

Studentmath

Good night for you I assume, @Ted!

Ted Shifrin

No ping on my iPad ... But loud bong on the desktop :)

Studentmath

I should too - I guess I like the thrill of being pinged at 3 am just as I think I see someone outside my window

Ted Shifrin

3:46 AM

LOL ... I won't go there ... :D

David Kirby

hah, i'm usually listening to some bizarre sound art loudly in headphones and the ping is pretty intense when you're having a psychedelic experience.

Smoke louder? @David

Hi

Hi/bye @Mike

Bye

3:49 AM

Hello @Mike

greetings

Hi

Hi guys.

I am really weak with integration..

David Kirby

4:05 AM

conceptually or functionally?

Studentmath

Functionally

Not enough practice I gues

Plus long time since Calculus I

David Kirby

ah, .. yeah, i think it has a lot to do with practice.

skullpatrol

Use it or lose it, as they say.

David Kirby

there's definitely a feel to it though --- it's like playing feedback on the guitar -- you've got find that sweet spot (er, leaky integrator)

right now my big struggle is learning the 'lingo' ... i

i've been doing math my whole life, but it's been in a bit twiddling vacuum

Studentmath

Say, $\int^{32}_0 \int_{\sqrt[5]{x}}^2 f(y) dy dx$ is the same as $\int^{32}_0 \int_0^{y^5} f(y) dx dy$, right?

Finally!

David Kirby

4:17 AM

see, my intuitive approach is to try and solve that iteratively -- either in my or head, or by jotting down a few lines of c code and pressing compile.. but i'm not satisfied with that anymore.. i want PROOF. haha.. so much to learn.

Studentmath

There are all these jokes about coder's approach to such problems vs. math approach :P

David Kirby

they are so true...

i used to think 'numerical recipes in c' was the friggin' bible.. heh

with that said, i still think there is a lot of interesting ground explored by software developers, but there's no peer review -- no standards -- no way to sift through muck ... so while some finnish guru might have solved the next big problem, he probably doesn't know it, and his code is probably so damn obfuscated that he wouldn't be able to retrace his steps anyway.

Studentmath

4:42 AM

They still argue over how you should declare and explain your codes, right?

David Kirby

oh, there's no argument. ;) "the one true brace style" is the correct style.

Studentmath

5:00 AM

Hiii I managed the integral! Back on track.

David Kirby

rad

David Kirby

5:20 AM

integrals are so beautiful... i love the transformations -- dirichlet, laplace, z transforms ... still learning -- keeping a little notebook of "keepers" hehe

eXtremiity

What's $c_0$ again in the context of Banach sapces?

AndrewG

Are there any surfaces with fundamental group $Z_2$ besides $RP^2$? Maybe something that can be embedded into $R^3$? I can't seem to come up with any.

eXtremiity

Dw, I got it.

Chris's sis

5:55 AM

Greetings

r9m

@Chris'ssis Greetings :)

Chris's sis

@r9m How are you doing?

r9m

@Chris'ssis under a blanket cover .. I am running a fever cartel :P

Chris's sis

:-)

@r9m What do you think about that double inequality involving integrals? That one is a candy you might like I guess.

r9m

@Chris'ssis I'm on it .. couldn't make any progress though

Studentmath

7:19 AM

Gah. I got to some expression with $\int arctan(0.5x) dx$. Shouldn't be too tough - I thought it is $x*arctan(0.5x)-2ln(sqrt{1+0.25x^2})$ but.. it's not. No idea why not.

Or could it be that wolframalpha is wrong and I am right..

Studentmath

7:56 AM

Seems wolfram had it wrong

skullpatrol

8:16 AM

►

Studentmath

Anyone minds helping me understand how to approach some coordinative double-integral question?

Hippalectryon

8:43 AM

@DavidKirby H is digamma ??

G.T.R

9:02 AM

@R9m do you remember the game ? See this maa.org/sites/default/files/pdf/cmj_ftp/CMJ/November%202010/…

r9m

@G.T.R wow .. thats nice :) .. I read about it a few years back in a Spectrum Magazine (its an Indian magazine) :D

Hippalectryon

@G.T.R T'as pu regarder ma conjecture bizarre ?

G.T.R

@Hippalectryon pas encore, je dois faire la fiche synoptique de mon TIPE

Hippalectryon

@G.T.R Ah oui les TIPE :=) Tu le fais sur quoi ?

G.T.R

sur l'approximation des fonctions continues

Hippalectryon

9:13 AM

Tu parle de quoi dans le TIPE? (le sujet est vaste non ? )

G.T.R

je parle d'interpolation et d'approximation :P

Hippalectryon

C'est vague xD

Si j'étais en MP(SI) j'aurai pu faire un TIPE sur la création d'approximation bien plus rapides à partir d'approximations déjà existantes de la forme f(x,h,c), et utiliser la formule bizarre que j'essaie de démontrer mais que personne n'as prouvée jusq'à maintenant bien que je l'ai posté sur 3 forums différents et ait demandé à 2 profs de maths :/

Au départ c'est en travaillent là dessus que je l'ai trouvée lol

G.T.R

9:40 AM

@Hippalectryon tu fais de la recherche alors lol

Hippalectryon

Non même pas

C'est juste une approche faire par moi et un amis

J'aurai aimé que quelqun l'ai trouvé avant comme ça j'aurai peut-être eu la démo lol

skullpatrol

Estouro de Pilha

(╯°□°)╯︵ ┻━┻

2

G.T.R

Vous faites quoi en math là ? @Hippalectryon

Hippalectryon

Matrices et déterminants

Après on fera je crois les espaces euclidiens

Puis les probas (les probas huehue ^^)

Ah il nous reste aussi les séries avant les espaces euclidiens

G.T.R

lol les séries en sup

Hippalectryon

9:49 AM

Juste le début lol

Bah il ont tellement enlevé de choses ...

G.T.R

il me semble qu'en PCSI vous faites que les déterminants 3*3 ?

Hippalectryon

Non

Déterminants de matrices carrées, toutes tailles

G.T.R

avec la théorie des signatures, et des formes n-linéaires alternées ?

Hippalectryon

(quoique, notre prof va un peu eu delas du programme donc ... pas sûr) que ça soit au programme

Hmm je ne pense pas

On verra bien :D

@G.T.R desmos.com/calculator/8vjsvlv8xr pas mal comme meilleur approximation :)

D'après mes calculs ça converge ~1500x plus rapidement

* afk 20 minutes *

Hippalectryon

10:13 AM

* déjeuner le + rapide du monde *

G.T.R

à la cantine ? :P

Studentmath

The French won the war and I've been asleep?

If anyone could guide me a bit here, will appreciate it.. I am lost now:

I need to find - God. I am stupid.

G.T.R

?

Hippalectryon

Oui

@Studentmath We french guys took over the chatroom muhahahaha

G.T.R

t'es en internat ?

Hippalectryon

10:20 AM

Oui

Studentmath

Nah, still lost. I need to find the space inside the following curve:
${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=\frac{x^2}{a^2}-\frac{y^2}{b^2}$

skullpatrol

what's the biggest difference between French and Spanish?

Hippalectryon

@skullpatrol Spanish is easier ?

Studentmath

So, I set $y=brsin\alpha$, $x=arcos\alpha$ and got to the final equation: $r^2=cos(2\alpha)$ where alpha obviously variates..

skullpatrol

easier for who?

Studentmath

10:22 AM

Now I have no idea what to do. How do I compute the space there?

Hippalectryon

Easier to learn as a 2nd language

skullpatrol

starting with what language as the first?

Hippalectryon

Any. From my point of view, french is harder to learn than spanish as a 2nd language

G.T.R

"Spanish
For language learners, a great feature of Spanish is its shallow orthographic depth – that is, in most cases, words are written as pronounced. This means that reading and writing in Spanish is a straightforward task.

Pronunciation is also fairly easy for native English speakers, with only ten vowel and diphthong sounds (English has 20), and no unfamiliar phonemes except for the fun-to-pronounce letter ñ. Grammatically speaking, Spanish has fewer irregularities that other Romance languages.

also, theeasiestlanguagetolearn.com

skullpatrol

Thanks @G.T.R and @Hippalectryon :-)

Studentmath

10:32 AM

Gah, just don't get what I should do there.

G.T.R

@Studentmath there's a formula for these sorts of things

Studentmath

Yeah, just don't get how to implement it

G.T.R

It's trivial though

Studentmath

Obviously use some short of double integral, but on what ranges, and on $r^2$?

G.T.R

nope

Studentmath

10:38 AM

Hm, why the half and - how do I decide max and min theta?

Hippalectryon

@robjohn You there ?

robjohn

@Hippalectryon I just posted an answer

Hippalectryon

@robjohn yeah i saw but i don't get it all so i have some questions :)

Studentmath

The $cos2\theta$ should come into play here, no?

robjohn

@Hippalectryon okay.

Hippalectryon

10:40 AM

@robjohn what exactly is a shift operator ? it's the first time I see that

Is it a function ? a constant ?

G.T.R

@Studentmath this formula is very famous, I think you won't have trouble finding references about it online. I have a link that explains it (but French)

Studentmath

Not allowed to use it anyhow, only text-book formulas

robjohn

@Hippalectryon It is just as defined in the answer... $S$ applied to a sequence advances the indices... S applied to $1,2,3,4,5,6,\dots$ would give $2,3,4,5,6,7,\dots$

Hippalectryon

@robjohn so it's a function of i ?

robjohn

@Hippalectryon it is a function on sequences.

Hippalectryon

10:43 AM

So when you write $S$ it's like $S(i)$ ?

robjohn

@Hippalectryon No. It operates on a sequence and the result is another sequence

G.T.R

@Studentmath sad then :P

Studentmath

Yeah :P

Thanks anyhow

skullpatrol

@Studentmath which textbook?

Studentmath

Translation of Howard Anton's calculus

Hippalectryon

10:44 AM

@robjohn why do we have $\prod\limits_{k=1}^n\left(c^kx-1\right)=\sum\limits_{p=0}^na_{n,p}x^p\tag{3}$ ?

robjohn

@Hippalectryon Since the recursion is $U_n=\frac{c^nS-I}{c^n-1}U_{n-1}$, we want to look at that polynomial

Studentmath

Think I got how to do it with given formulas though! Thanks again

Hippalectryon

But why is the equality true ?

How are the $a_{k,k'}$ defined ?

robjohn

@Hippalectryon well, it is how the $a_n$ are defined. I spend most of the answer determining what the $a_n$ are.

Hippalectryon

Ooooh ok

How do you deduce $(4)$ from the roots ? (i'm not used to doing it with matrices) @robjohn

robjohn

10:52 AM

Plug $x=c^{-k}$ into $$\prod_{k=1}^n\left(c^kx-1\right)=\sum_{p=0}^na_{n,p}x^p$$ for $1\le k\le n$.

the left side gives $0$, the right side gives one of the rows of the matrix equation

G.T.R

A new mathematical field is born: MathGolf math.stackexchange.com/questions/812949/…

Hippalectryon

Ooooh right then we have $x^p=c^{-pk}$

robjohn

@Hippalectryon indeed

Hippalectryon

@robjohn But since it's $\sum\limits_{p=0}^na_{n,p}x^p\tag{3}$ shouldn't it be AX=0 and not XA=0 ?

robjohn

@Hippalectryon it is $Ax=0$, however, the equation $(3)$ does not dictate how the matrices are oriented.

Hippalectryon

10:56 AM

Then why $\begin{align}
0
&=\begin{bmatrix}
1&c^{-1}&c^{-2}&\cdots&c^{-n}\\
1&c^{-2}&c^{-4}&\cdots&c^{-2n}\\
1&c^{-3}&c^{-6}&\cdots&c^{-3n}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&c^{-n}&c^{-2n}&\cdots&c^{-n^2}
\end{bmatrix}$ ?sss

$
\begin{align}
0
&=\begin{bmatrix}
1&c^{-1}&c^{-2}&\cdots&c^{-n}\\
1&c^{-2}&c^{-4}&\cdots&c^{-2n}\\
1&c^{-3}&c^{-6}&\cdots&c^{-3n}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&c^{-n}&c^{-2n}&\cdots&c^{-n^2}
\end{bmatrix}
\begin{bmatrix}
a_{n,0}\\
a_{n,1}\\
a_{n,2}\\
\vdots\\
a_{n,n}
\end{bmatrix}\\
\end{align}$ ?

That looks like 0=XA to me @robjohn

robjohn

@Hippalectryon I guess that depends on what $X$ is and what $A$ is

Hippalectryon

A = [an,k]

I guess

robjohn

@Hippalectryon what is important is what each row says

Hippalectryon

Well to me the equation is just $a_{n,0}c^{-k}+a_{n,1}c^{-2k}+...=0$

robjohn

@Hippalectryon we have to fix $n$, so $a_{n,p}$ is a vector.

Hippalectryon

11:02 AM

Owait that's exactly the matrix one with $k$ changing xD

Okay i got $(4)$

Studentmath

@G.T.R actually appears to use double integration after all

Hippalectryon

@robjohn 'That is, c−pan,p is proportional to (−1)p times the determinant of the submatrix gotten my removing column p from the matrix in (4)' ew i'm not familiar with determinants yet

robjohn

@Hippalectryon Ah, that is a method of finding a vector perpendicular to $n-1$ others in $\mathbb{R}^n$

Hippalectryon

And where do you see that 'c−pan,p is proportional to (−1)p times the determinant of the submatrix gotten my removing column p from the matrix in (4)' ?

robjohn

@Hippalectryon Unfortunately, it depends on knowing how determinants operate.

Hippalectryon

11:08 AM

Uh but i need to understand the proof xD

Chris's sis

@r9m I have something to show you ...

robjohn

@Hippalectryon well, I could try to describe determinants here, but perhaps it would be easier after you have looked at the section in Wikipedia on determinants. You do know about matrices, right?

Hippalectryon

I do.

r9m

@Chris'ssis I am all eyes :D

Hippalectryon

I know like the det of a 3x3 matrice but that's about all regarding determinants

robjohn

11:11 AM

@Hippalectryon then look at the section in Wikipedia about determinants

@Hippalectryon They are multilinear functions on matrices...

@Hippalectryon if two matrices are the same except for one column, then adding those columns, leaving the other columns alone, will add the determinants...

@Hippalectryon multiplying a column by a scalar multiplies the determinant by that scalar

@Hippalectryon swapping two columns changes the sign of the determinant

that means that a matrix with two identical columns has $0$ determinant

Hippalectryon

Yeah that's logical

robjohn

which means that adding one column to another does not change the determinant

Hippalectryon

How much should I read ? The whole page ?

robjohn

The first sections of the Definition and the Properties will help.

You can read about 2x2 and 3x3 determinants, but it shouldn't be needed for this.

Hippalectryon

Ok i read that already then :)

But i still don't see why 'c−pan,p is proportional to (−1)p times the determinant of the submatrix gotten my removing column p from the matrix in (4)'

Studentmath

11:24 AM

Wouldn't it be, after I get the curve to the form of $r^2=cos(2\theta)$, finding the enclosed area be the same as computing: $\int_0^{2\theta} \int_0^{cos(2\theta )} r^2 dr d\theta$? Or am I having it wrong - after all it's the complete circle around the axis, limited by this function

Chris's sis

@r9m are you now?

r9m

@Chris'ssis O-O (eyes wide open)

Chris's sis

@r9m Take this part ...

r9m

:15788244 got it ... thanks :D !! .. but won't you answer the question on the main ? :)

Chris's sis

@r9m There is just half a way. ;)

Hippalectryon

11:27 AM

@robjohn

robjohn

note that the vector $$\begin{bmatrix}
c^{-0}a_{n,0}\\
c^{-1}a_{n,1}\\
c^{-2}a_{n,2}\\
\vdots\\
c^{-n}a_{n,n}
\end{bmatrix}$$ is perpendicular to all the rows of the matrix.

Hippalectryon

What do you mean 'perpendicular' for matrixes ?

robjohn

@Hippalectryon this is normal vector perpendicular

Hippalectryon

What's the definition of a perpendicular vector again ?

Studentmath

Which gets me to 0 which doesn't make sense..

robjohn

11:32 AM

@Hippalectryon the dot product of the vectors is $0$

Hippalectryon

Uh but if i take the row [1,1,1,...] how is it perpendicular to [c-0an0,c-1an1,...] ? Isn't the vector product $c^{-0}a_{n,0}+c^{-1}a_{n,1}+...$ ?

Ooooh wait

That equals 0

right -_-

@robjohn Okay it's perpendicular. What about it ?

robjohn

@Hippalectryon so that vector is perpendicular to all the rows. Now lets look at what happens when we add an $n+1^\text{st}$ row to the matrix and compute the determinant. This may involve the section on Laplace's formula.

Hippalectryon

Doesn't Laplace's formula involve removing a row and a column ?

robjohn

write the $n+1^\text{st}$ row as $u_0,u_1,u_2,\dots,u_n$

@Hippalectryon yep, and it expands the determinant in terms of the elements of the removed row.

Hippalectryon

Ok so $
\begin{align}
\begin{bmatrix}
1&1&1&\cdots&1\\
1&c^{-1}&c^{-2}&\cdots&c^{-n}\\
1&c^{-2}&c^{-4}&\cdots&c^{-2n}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&c^{1-n}&c^{2-2n}&\cdots&c^{n-n^2}\\
u_0&u_1&u_2&\cdots&u_n
\end{bmatrix}
\end{align}$ ?

How do i use Laplace's formula here ? the number of columns is n, so ...

@robjohn

robjohn

11:45 AM

The determinant is then $$(-1)^n\left(u_0\det(M_0)-u_1\det(M_1)+\dots+(-1)^nu_n\det(M_n)\right)$$ where $M_n$ is the matrix deleting column $n$

Hippalectryon

Uh right ... what about it ?

robjohn

@Hippalectryon No, the number of columns is $n+1$

Hippalectryon

RIght n+1

But what about that det

robjohn

@Hippalectryon the det of the whole matrix is given by the formula I gave above:

1 min ago, by robjohn

The determinant is then $$(-1)^n\left(u_0\det(M_0)-u_1\det(M_1)+\dots+(-1)^nu_n\det(M_n)\right)$$ where $M_n$ is the matrix deleting column $n$

Hippalectryon

But what do I do with it ? How is that proportional to $(-1)^p$ times $c^{-p}a_{n,p}$ ?

Studentmath

11:47 AM

Hrm, perhaps I am supposed to take into consideration the symmetry..

robjohn

@Hippalectryon hang on... you have to learn to walk before you can run.

Hippalectryon

@robjohn Uh what shoud I do

robjohn

@Hippalectryon I am still explaining...

Hippalectryon

Ok

robjohn

@Hippalectryon you should have patience :-)

Hippalectryon

11:50 AM

I just thought you had finished @robjohn so i was like 'uh i'm still lost' :D

robjohn

remember when we said that if two columns were the same, the determinant was $0$? well, the same is true for rows. If the $u_k$ are one of the other rows, the determinant is $0$

Thus, the vector $(\det(M_0),-\det(M_1),\det(M_2),\dots,(-1)^n\det(M_n))$ dotted with any of the rows of $M$ is $0$.

Hippalectryon

I get the fact that if there are similar rows then the det is 0. I don't see why it implies that (det(M0),−det(M1),det(M2),…,(−1)ndet(Mn)) dotted with any of the rows of M is 0

robjohn

@Hippalectryon if we use any of the rows for the $u_k$ the det is $0$ right?

Hippalectryon

Yeah

robjohn

12 mins ago, by robjohn

The determinant is then $$(-1)^n\left(u_0\det(M_0)-u_1\det(M_1)+\dots+(-1)^nu_n\det(M_n)\right)$$ where $M_n$ is the matrix deleting column $n$

That means that any of the rows put in there will give $0$

and that is just the dot product of the row with that vector I mentioned above

5 mins ago, by robjohn

Thus, the vector $(\det(M_0),-\det(M_1),\det(M_2),\dots,(-1)^n\det(M_n))$ dotted with any of the rows of $M$ is $0$.

Hippalectryon