Mathematics - 2018-05-10 (page 3 of 3)

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Jake Rose

22:00

But theres no way to reduce what I have to matrix form? Once you sub each one of the x y and z in to the second equation its just a mess

Ted Shifrin

You don't want matrix form. It is a mess, but you solve for $t$ ...

I didn't make up the stoopid problem.

Perturbative

That partition closely matches how $f$ behaves I think

Ted Shifrin

You need a finite partition that depends on $\epsilon$, @Perturbative.

For example, if I give you $\epsilon = 1/5$, how do you get a partition $P$ so that $U(f,P)<1/5$?

Jake Rose

Okay solving for $\alpha = [s- (pq+qp+rc)/(pl+qm+rn)]l +a $

Ted Shifrin

what's $\alpha$?

Jake Rose

22:02

your t

It was defined at the beginning of the problem instead of t

Ted Shifrin

Oh. So then you need $x,y,z$ in terms of $\alpha$. But your formula doesn't look right at all. How do you get products of $pq$, for example.

Jake Rose

px+qy+rz=s

Ted Shifrin

Yeah, and $x=Lt+a$.

So where do you get $pq$?

Jake Rose

I subbed in for y and z too

Ted Shifrin

I understand.

Jake Rose

22:05

Then a bit of rearranging

Oh the pq should be a pa

Ted Shifrin

$p(lt+a) + q(mt+b) + r(nt+c) = d$.

Jake Rose

I hit the wrong key

Ted Shifrin

So $t=[d-(pa+qb+rc)]/[pl+qm+rn]$.

It's a stooopid problem, as I said.

Jake Rose

yes

Ted Shifrin

Then you put that back in to get $x,y,z$.

Jake Rose

22:09

But then how do I get it in the Ax=d form?

Ted Shifrin

That's the equation of the plane. You're finding the point ...

Jake Rose

No the matrix equaiton

Thats what the problem asks for

imgur.com/a/alnnvcp

Ted Shifrin

It asked where the line meets the plane.

Jake Rose

In the form of a set of eqations in a matrix format

Ted Shifrin

I hate this question. It's truly stoopid.

Jake Rose

22:11

Welcome to cambridge

Daminark

Here's a slick problem. Let $F$ be a non-Galois extension of $\mathbb{Q}$ of degree 4. Show that the Galois closure of $F$ must be $S_4$, $A_4$, or $D_4$, and it's $D_4$ iff $F$ contains a quadratic extension of $\mathbb{Q}$

Ted Shifrin

So take the equations of the line without parameter in it.

Two of those equations give you the line. The third equation is going to be the equation of the plane. You want to solve that system of three linear equations simultaneously to get the $x,y,z$ that satisfies all of 'em. So forget about the parametric way (which is the obvious way any fool would do this).

Evinda

Hello!!! I have the following question

Q: Compute probability using density functions

Suppose that $X$ has the uniform distribution on the interval $[0,2]$ and $Y$ has the uniform distribution on the interval $[2,4]$. If $X,Y$ are independent, I want to find the probability that the difference $Y-X$ is $\leq 1$. I have thought the following. The density function of $X$ is $$p_1...

probability

Jake Rose

What do you mean the third equations gives the plane?

Ted Shifrin

Two linear equations will give that line (use two of the equalities they gave you). The equation of the plane is your third equation. Solve all three simultaneously (in principle) to get the intersection.

Jake Rose

22:16

Would I need to include $\alpha$ in that then?

Ted Shifrin

Nope.

Jake Rose

What do I do about the z in the 3rd equation?

Ted Shifrin

What do you mean?

Jake Rose

So I have $(x-a)/L=(y-b)/m$ and $px+qy+rz=s$

Right?

Ted Shifrin

No, you need another equation to get the line.

Jake Rose

22:19

Oh so use all 3?

Ted Shifrin

Use $(y-b)/m=(z-c)/n$. Of course, rewrite everything with variables on the left.

Jake Rose

Whats wrong with using the x?

Ted Shifrin

To get a unique point, you will need three equations in three variables.

You need to use both equations for the line.

another both ...

Jake Rose

Im confused

Ted Shifrin

The line is given by two intersecting planes. That takes two equations.

You need THREE equations total.

Perturbative

22:21

@TedShifrin $f(0) = 0$ in that question right?

Ted Shifrin

No, $0 = 0/1$, @Perturbative.

Jake Rose

But then why havent you used the equation including x?

Ted Shifrin

Jake, you're not reading what I write.

I said another ... then both

Then I said you need three equations.

Jake Rose

Im not understanding what you write

Ted Shifrin

Well, I give up. I've got other things I need to do.

Jake Rose

22:23

ah okay cool

Perturbative

@TedShifrin Okay so I basically have to make the first point $x$ after $0$ in the partition to be really close to $0$ as the sup of $f(x)$ on $[0, x]$ will be $1$, and similary I have to make the last point $x'$ before $1$ in the partition to be really close to the $1$

MatheinBoulomenos

@Daminark you're using how the subgroups of $S_4$ look like, right?

Daminark

Yup

Perturbative

22:38

Actually no, if $0 < \epsilon < 1$ for a partition $P = \{x_i\}_{i =0}^n$ of $[0, 1]$ since $$M_i = \sup_{x \in [x_{i-1}, x_i]}f(x) \leq 1$$ for all $i = 0, 1, ..., n$ I just need to make the difference $(x_{i-1}, x_i)$ to be less than $ \frac{\epsilon}{n}$

Ted Shifrin

23:34

@Perturbative: Note that that is impossible, because you need to cover the whole interval $[0,1]$, which has length way more than $\epsilon$. :P

There are some real ideas to have here. Hint: Where is $f(x)\ge \epsilon/2$?