Mathematics - 2017-04-16 (page 2 of 6)

Balarka Sen

12:00

There's a typo here, of course. I meant $\sum b_i y_i^2$ (here $y_i$ are variables which are linear combinations of $x_i$).

Topologicalife

simplify* sorry.

Balarka Sen

@Alessandro Agreed.

Anonymous

@BalarkaSen I noticed. Nvm, I'm watching the MIT OCW video on diagonalization.

Anonymous

Thanks a lot!

Balarka Sen

Great. No problem :)

Secret

12:01

If $x$ is real, then $x^2=(x^{\frac{2}{n-2}})^{n-2}$ and $x^yz^y=(xz)^y$

Topologicalife

Oh, right, thanks.

s.harp

i wouldnt call that a simplification

Secret

so given such conditions, one can bubble the $x^2$ term into the power

Topologicalife

Oh, but, I'm trying to put together the term $x+1$ and $x$ in one term

$(x+1)^{n-2} \cdot x^2 = x^{\text{something}}$ or $(x+1)^{\text{something}}$

I need to do it in order to get a recurrence relation in my problem.

Anonymous

@Topologicalife That doesn't seem possible....

DHMO

12:09

@Topologicalife you can't

@Topologicalife can you show us your whole problem?

Topologicalife

Yeah.

Anonymous

Anonymous

How did they apply LMVT with $f'(\theta h)$ ?

DHMO

@blue I'm not trusting your source anymore

Anonymous

I have been solving these papers for the last 4 hours. I'm really fed up with the questions too =P.

Anonymous

12:11

hehe

DHMO

@blue $\theta h$ is just a way to parameterize a number between $0$ and $h$

$0 < \theta < 1$

$0 < \theta h < h$

Topologicalife

Given the ODE $(1-x^2)y'' -2xy' + \lambda y = 0$ find a solution given a power series in the variable "(x+1)" centered at $x = -1$.

DHMO

so let $u=x+1$

we want a power series in the variable $u$ centered at $u = 0$.

Topologicalife

i.e: the solution is in the form $P(x)=\sum_{k=0}^{\infty} a_k (x+1)^k$

Anonymous

@DHMO Hmm, I understand that...but in LMVT how can they write $$f'(\theta h)=\frac{f(h)-f(0)}{h-0}$$

DHMO

12:13

@blue that is LMVT itself

And the ODE is $(2u-u^2)y'' - 2(u-1)y' + \lambda y = 0$

Topologicalife

@DHMO I have to do it specificaly centered at $x = -1$.

Balarka Sen

$\theta h$ lies in $(0, h)$

Anonymous

@DHMO Oh wait, $c=\theta h$

Anonymous

Darn

Balarka Sen

Ya

DHMO

12:13

@Topologicalife yes, which means $u=0$

Anonymous

my head isn't working

Anonymous

gotta go and sleep

Anonymous

thanks people!

Topologicalife

@DHMO yeah, I know! I mean I want to do it (compute it) centered at $x=-1$ without to that subtitution.

DHMO

@Topologicalife why wouldn't you want that substitution?

it's so convenient

you can substitute it back after it

Topologicalife

12:15

Because then I get the step where I'm stuck and I want to solve it, not avoid it.

DHMO

substitution ain't avoiding nothin'

why would you want to complicate things

Topologicalife

I.e: $P'' (x) = \sum_{k=0}^{\infty} k(k-1)a_k (x+1)^{k-2}$, then how I do simplify $x^2\cdot P'' (x) $

DHMO

none of your variables match

you use $n$ and then $k$ and then $x$

when they should be one single variable

Topologicalife

sorry

DHMO

that's better

alright

$x^2 = (x+1)^2 - 2(x+1) + 1$

(that's what you get for not using substitution)

Topologicalife

12:18

That's really clever.

DHMO

lesson: USE SUBSTITUTION

Topologicalife

:) Yeah, but I need to check if I get the same solution (it is given without to do that subtitution). Thanks, so much appreciated.

DHMO

not using substitution is ridiculous

Topologicalife

Well, I do it because one reason.

DHMO

what reason?

Topologicalife

12:21

They gave me this recurrence relation: $\dfrac{a_{k+1}}{a_k}=\dfrac{k^2+k-\lambda}{2k^2+4k+2}$, and then they apply the Gauss criterion for convergence of series...

DHMO

alright

I'm sure using $u=x+1$ will give you the same relation

Topologicalife

But they can't apply it since the first coefficient of the greater power (i.e: 2) is not 1.

Actually it doesn't.

(I already did it)

DHMO

alright

Topologicalife

so now I will check if they gave a wrong recurrence relation.

@DHMO uhmm actually it should give me the same recurrence relation. It can't depends of a change of variable, right?

DHMO

Right

Topologicalife

12:25

I got this recurrence relation: $a_{k+2} =\dfrac{(k+1)k - \lambda}{(k+2)(k+1)} a_k$

Which doesn't seems to be the same since my odd terms are all 0.

DHMO

Maybe you could show us your steps

Topologicalife

well, if their "$k$" is even...

DHMO

for both approaches

Topologicalife

I checked my recurrence relation with the solution a book gives to Legendre differential equation and it is the same.

Now I have to do this approach we just talk about.

chat.stackexchange.com/transcript/message/36741598#36741598 There is $\sum_{k=2}^{\infty}$ instead of $\sum_{k=\color{red}{0}}^{\infty}$

DHMO

Is $\Bbb Q[x]/\langle{x^{200}+1}\rangle$ even an extension?

@BalarkaSen

Kane

12:40

$f(x,y) = \begin{cases}
& \frac{1}{3}\text{ if } (x,y)= (1,1), (1,1), (2,2)\\
& 0\text{ if } x= otherwise
\end{cases}$

$cov(X,Y)=E(XY) - E(X)E(Y)$
$=(0\times 0\times \frac{1}{3})+(1\times 1\times \frac{1}{3})+(2\times 2\times \frac{1}{3})$
$=\frac{5}{3}$

Anbody good with statistics?

Does that look ok for covariance?

DHMO

What is $f(1,3)$?

Kane

oops i forgot to -E(x)E(y)

@DHMO double oops, the first (1,1) should be (0,0)

DHMO

there's a lot of typos in the first line (definition of $f$)

Balarka Sen

@DHMO Sure, that's a field extension of Q. Why not.

DHMO

@BalarkaSen but it's reducible

Balarka Sen

12:43

x^200 + 1 is not reducible in Q...?

DHMO

@BalarkaSen x^200 + 1 = (x^40)^5 + 1

Balarka Sen

Ah. Well, shit.

Then it's not a field, yeah.

DHMO

Is it true that if $F=\Bbb Q[x]/\langle{P(x)}\rangle$ is a field-extension, then $[F:\Bbb Q]=\deg P$?

Kane

$f(x,y)=\begin{cases}
& \frac{1}{3}\text{ if } (x,y)=(0,0), (1,1), (2,2) \\
& 0\text{ if } (x,y) = otherwise
\end{cases}$

DHMO

that's better

Kane

12:47

haha

DHMO

but can be improved

Kane

oh yeah?

Balarka Sen

@DHMO Sounds right to my ears.

I have forgotten algebra tho

DHMO

$f(x,y) = \begin{cases}
\dfrac{1}{3} & \text{ if } (x,y)=(0,0), (1,1), (2,2) \\
0 & \text{ if } (x,y) =\text{ otherwise}
\end{cases}$

Kane

@DHMO much better

DHMO

12:48

@Kane I agree with you for once

Kane

@DHMO I'm trying to find the covariance of X and Y

DHMO

I know

Kane

I think it's 2/3

DHMO

but you need to show your steps

Kane

@DHMO

cov(X,Y) = E[XY] - E[X]E]Y]
= (0*0*1/3) + (1*1*1/3) + (2*2*1/3) - 1 * 1
= 0 + 1/3 + 4/3 - 3/3
= 2/3

DHMO

12:52

alright

Kane

I'm actually trying to find correlation, but this is the first step

var(X) = E[X^2] - (E[X])^2
= (0^2 * 1/3) + (1^2 * 1/3) + (2^2 * 1/3) - 1^2
= 0 + 1/3 + 4/3 - 1
= 2/3

is that correct

$corr(X,Y)=\frac{cov(X,Y)}{\sqrt{var(X)\times var(Y)}}$

$corr(X,Y)=\frac{\frac{2}{3}}{\sqrt{\frac{2}{3}\times \frac{2}{3}}}=1$

Secret

Joint probability question: How to plot the following probability distribution where every letter you saw here is either 0 or 1?

$$P(a,b\vert A,B)=\left\{\begin{matrix}\frac{1}{2},a+b \mod 2 = AB \\ 0, \textrm{otherwise}\end{matrix}\right.$$

DHMO

I'm seeing more letters than I should

Kane

how can my correlation be one? that can't be right

Secret

sorry, I mean a,b,A,B

DHMO

13:02

@Kane how did you calculate $\operatorname{var}(X)$?

Kane

var(X) = E[X^2] - (E[X])^2

Astyx

König-Huygens formula

Kane

@DHMO is it possible to have a correlation of 1 if the probability function is always 1/3?

Astyx

Or theorem even, some might say

DHMO

@Kane I have no idea

Alessandro Codenotti

13:16

@DHMO yes, it's homeomorphic to $\Bbb Q[\alpha]$ where $\alpha$ is a root of $P$ and simple (field) extensions have the degree of the minimal polynomial of the element you're extending with iirc

DHMO

@AlessandroCodenotti thanks

but what is $P$ has no root?

Alessandro Codenotti

$P$ has roots somewhere (in the field extension actually)

DHMO

are you considering complex roots?

Alessandro Codenotti

Not necessarily

Ah, wait, you started with $\Bbb Q$ as base field, then yes, the roots will be complex at worse

Topologicalife

if $\displaystyle \sum_{j=0}^\infty \frac{f^{(j)}(c)}{j!} (x - c)^j = 0$ shall I prove that $f^{(j)}(c) = 0$? or is it trivial?

Alessandro Codenotti

13:26

I mean $\Bbb Q[x]/(x^2+1)$ is isomorphic (thanks to the first isomorphism theorem) to $\Bbb Q[i]$

LegionMammal978

I've been thinking of something

DHMO

@AlessandroCodenotti I see, thanks

@LegionMammal978 I would be quite shocked if you haven't been thinking of anything

@Topologicalife it is trivial.

Astyx

It's not that trivial

If you've done the work before, it becomes trivial

DHMO

I thought "trivial" is just a shortcut for tedious proofs

Topologicalife

Okay, thanks. Another related question. If I have $a + \displaystyle \sum_{j=1}^\infty a_j x^j = 0$, then why $a = 0$ and $\displaystyle \sum_{j=1}^\infty a_j x^j = 0$?

Astyx

13:35

That's not true

Topologicalife

I understand that $\displaystyle \sum_{j=0}^\infty a_j x^j = 0$ implies $ a_j = 0 \forall j$.

Astyx

Take $a=-1$, and $a_i = \delta_{i,1}$

Topologicalife

Sorry :D

DHMO

@Astyx I thought $\delta$ is not a function

Topologicalife

Now?

DHMO

13:36

@Topologicalife differentiate both sides

Astyx

Evaluate at $x=0$

LegionMammal978

Take a continuous, total function $f:[0,1]\rightarrow\Bbb R$. Now, how could one determine the length $L_f$ of a string obtained by plotting the function over $[0,1]$?

For instance, if $f(x)=x$, then $L_f=\sqrt2$. As another example, if $f(x)=c$ for any constant $c\in\Bbb R$, then $L_f=1$.

Astyx

The sum is zero, so that implies $a=0$

The rest follows

Topologicalife

@Astyx that is because is a functional equation, right?

Astyx

@DHMO Why wouldn't $\delta$ be a function ?

@Topologicalife I guess so

DHMO

13:38

> From a purely mathematical viewpoint, the Dirac delta is not strictly a function, because any extended-real function that is equal to zero everywhere but a single point must have total integral zero.

Astyx

You'd normally have quantifiers before, to determine what $x$ is

DHMO

From Wikipedia

Topologicalife

I mean, it must be true for all x.

$\delta$ Is a distribution.

DHMO

@LegionMammal978 what is a total function?

Astyx

Oh that's not the Dirac function, which does not exist

DHMO

13:39

what is it then

Astyx

It's the Kronecker delta

Topologicalife

Kronecker delta I think.

DHMO

$\begin{array}{rcl}
a + \displaystyle \sum_{j=1}^\infty a_j x^j &\equiv& 0 \\
\displaystyle \sum_{j=1}^\infty j a_j x^{j-1} &\equiv& 0 \\
\displaystyle \sum_{j=0}^\infty (j+1) a_{j+1} x^j &\equiv& 0 \\
\end{array}$
Then use the previous theorem

Topologicalife

Uhm and the coefficientes are unique because the identity principle.

Astyx

@Topologicalife Then again you need quantifiers around, else it becomes ambiguous what you mean

LegionMammal978

13:40

@DHMO Here, let's just say that $f(x)$ is defined for all $x\in[0,1]$

So no singularities

Astyx

@DHMO you don't really need to differentiate, you only need to evaluate at $0$

DHMO

@LegionMammal978 that is just arc length

$L_f = \displaystyle \int_0^1 \sqrt{1+\left(\dfrac{\mathrm df}{\mathrm dx}\right)^2}\ \mathrm dx$

Alessandro Codenotti

@LegionMammal978 take the sup of the lengths of the polygonal paths inscribed in the graph of your function as a subset of $\Bbb R^2$, this will give you an integral formula that you can surely find if you google arc length

DHMO

And find another person if your function is not differentiable

@Astyx alright

LegionMammal978

Okay, originally thought arc length was only for circular arcs :)

DHMO

13:42

@AlessandroCodenotti what if the function is not differentiable

@Astyx what is this

Astyx

What is what ?

Topologicalife

Okay, thanks guys :)

DHMO

@Astyx never mind

Alessandro Codenotti

As long as it's at least piecewise linear you're fine

Otherwise the arc length won't really work

Mats Granvik

Riemann Siegel theta:
$$\vartheta (t)=\pi \left(\left\lfloor \frac{t \log \left(\frac{t}{2 \pi \exp (1)}\right)}{2 \pi }+\frac{7}{8}\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)\right)-\frac{\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)}{\pi }\right)$$

DHMO

13:45

@AlessandroCodenotti piecewise linear?

Mats Granvik

Apart from differing singularities.

DHMO

@AlessandroCodenotti I can't have $x^2$?

Astyx

What's wrong with my first sentence $DHMO ?

Alessandro Codenotti

@DHMO piecewise differentiable*

DHMO

@Astyx $\sqrt \cdot$ is well-defined even for negative numbers?

Astyx

13:56

Meh :p

Dodsy

Happy holidays to those who celebrate

Astyx

Yeah, happy easter

Dodsy

I was in bed last night thinking and came up with something pretty trivial

Astyx

It's still better than nothing at all

Dodsy

Let's say you bisect an equilateral triangle horizontally, is the top triangle made also equilatera?

DHMO

14:03

@Dodsy yes

provided that the base of the original triangle is horizontal

Dodsy

I was thinking about finding the length of the bisecting line

Which I thought about for an hour I'll have to admit before I realized how easy it was

DHMO

that happens

Dodsy

it was still nice to think about I suppose

So then the length of the bisection would be equal to one minus the height of the bisection divided by the height of the equillateral triangle multiplied by the base, right?

DHMO

you skipped too many details and assumptions

Dodsy

[1-(y/h)]b

DHMO

14:10

alright

Dodsy

You're right first we have to note that all sides of the triangle are the same and that the base of the triangle is horizontal

To find the height of the equilateral we take $b^2 - 1/2b^2$

DHMO

you don't need all sides of triangle being the same

Dodsy

No this could also work with a right handed triangle

So any triangle with a horizontal base

Good point

Astyx

Challenge : prove any closed set of $\Bbb R$ is the set of the zeros of a $C^{\infty}$ function

DHMO

What is $C^\infty$?

Dodsy

14:14

Okay my thought process started as thinking that the smaller triangle made will have a ratio to the larger triangle it was 'cut from'

DHMO

@Astyx Consider the function $x \mapsto d(x,S)$ where $S$ is the closed set

Topologicalife

Uh, @DHMO, I got the same recurence relation they gave me.

DHMO

@Topologicalife nice

Dodsy

Which means the sides will also have some ratio to the sides of the original triangle

Astyx

Infintely differentiable @DHMO

DHMO

14:15

Oh, never mind

Even the cantor set?

Astyx

It is closed, isn't it ? :)

I'll let you ponder on that, I have to go now !

DHMO

salut

Topologicalife

Now I have $\dfrac{a_{k+1}}{a_k}=\dfrac{k^2+k-\lambda}{2k^2+4k+2}$, but I think the Gauss criterion for convergence requires the first coefficient of the greatest power must be 1 (both in denominator and numerator). So Can't I apply it? This is the Gauss Criterion: gyazo.com/f4b3a9ce24374809b862219ad15e28e7

Dodsy

Anyways it turned out to be trivial

If you google it it doesn't even show up because it's probably something a 5th grader who has taken a geometry course could solve

I was telling zach that up until 8th grade I only ever did multiplication sheets in school :P I had a lousy education in public school

Koolman

14:34

-1

Q: Question on quadratic equations .

A quadratic equation with roots $m$ and $n$ where $mn=4$ satisfies the equation $\frac {m}{m-1} + \frac {n}{n-1}=\frac {a^2 -7}{a^2 -4}$ . We have to find the number of integral values of $a$ for $m$, $n \in (1,4)$. By comparing the given I found the quadratic equation as $x^2 -(a^2 +1)x +4 =0...

In this how they got inequalty 1 and 2

Astyx

Hi @Daminark

How's it going ?

Daminark

Hey @Astyx! I'm alright, how about you?

Astyx

Fine, eating too much chocolate probably :p

Daminark

There's no such a thing as too much chocolate!

Lolol

Astyx

i guess you're right - gets some more

Topologicalife

14:49

Is the same the recurence relation $a_{k+2} = \dfrac{(k+1)k - \lambda}{(k+2)(k+1)} a_k$ for $k=0,1,2,\dots$ and $\dfrac{a_{k+1}}{a_k}=\dfrac{k^2+k-\lambda}{2k^2+4k+2}, 2a_1 + \lambda a_0 = 0, 8a_2 + a_1 (\lambda -2) = 0$?

I don't think so.

it doesn't give the same $a_2$. In the first case is given by the recurrence relation and in the second one is given by the pair of two equations given at the end.

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