Mathematics - 2021-01-23 (page 3 of 3)

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11:00 PM

Which part are you doubting?

All of it, basically

Let me type it out from the beginning..

sure

sometimes that's a necessary thing to do

If $f(0) = 0$ or $f(1) = 1$, then we are done. Define $g(x) = f(x) - x$. Let $c \in \mathbb{R}$ such that $0 < c < 1$. Then notice that $g$ is continuous on $[0,1]$, so notice that $g(0) = f(0)$ and $g(1) = f(1) - 1$. Since $0 \le f(0) \le 1$ and $0 \le f(1) \le 1 \implies -1 \le f(1) - 1 \le 0$.

Before I continue

I think something needs to be done about the cases so we have a strict inequality chain

You may assume $f(0)\neq0$ and $f(1)\neq1$, since in those we are done (as you already noted). Then what can you say about the signs of $g(0)$ and $g(1)$ based on what you just said?

But didn't you say I can't do that?

Since if I let it be $c$ for some $c \in [0,1]$

Then it could still be $0$ or $1$

Let me just continue I suppose.

11:11 PM

you don't need $c$

the $c$ in what you just wrote serves no purpose, you might as well delete that sentence

@TedShifrin subscribe to a french vnp and you can watch netflix as if you were there :-).

Hmm, ça coute combien?

@copper.hat Are you sponsored by NordVPN as well?

If f(0)=0 or f(1)=1, then we are done. Define g(x)=f(x)−x. Let c∈ℝ such that 0<c<1. Then notice that g is continuous on [0,1], so notice that g(0)=f(0) and g(1)=f(1)−1. Assume that $f(0), f(1)$ are not fixed points. Then $0< f(0) < 1$ and $0 < f(1) < 1 \implies -1 < f(1) - 1 < 0$, and therefore, $0 < g(0) < 1$ and $-1 < g(1) < 0$. We see that $g(0) > 0$ and $g(1) < 0$, so they have opposite signs.

Then, by Bolzano's Theorem, there exists a $k$ in $[0,1]$ such that $g(k) = 0.$ So $g(k) = f(k) - k = 0$ and so $f(k) = k.$

@Astyx: Do his various sponsors lead to a conflict of interest?

11:13 PM

No, it's just that everywhere I go I see people sponsored by NordVPN

@Astyx no :-). are they good? i have my own vpn at home which is use to watch netflix when i am traveling.

i have a similar setup in the uk for various reasons. i am tired of stupid regional restrictions.

NordVPN is not good, because they advertise everywhere. You don't want a privacy firm spending all of their budget on advertising. Or at least I wouldn't.

They also setup scam deals that say that "they end in 10 minutes" each time you visit the site.

there is little pure in the world...

But I digress

Back to the grindstone, @polite.

11:15 PM

No idea if it's good, I've never had use of a VPN. However they have been having large publicity campaigns (is that the right terminology) lately

you can't even trust whatspp & microshaft these days

@TedShifrin In your video, you touched upon $X^\top X$ being invertible if... Suppose I have the system of equations $y_i=a+bx_i+ce^{x_i}+\epsilon_i$, where $E(\epsilon_i)=0$. I have estimated $a,b,c$ using the method of least squares. I'd like to know under which condition $X^\top X$ is invertible. Is there a way to find that out by looking at $a+bx_i+ce^{x_i}$ and realizing it can have at most two roots...

or faecesbook

@TedShifrin Curious, how long do you think it takes a person to learn mathematics, say like, how long do you think it takes someone to go through baby rudin or equivalent by themselves?

@politeproofs yup, that's correct

11:16 PM

@Thorgott Thank the lucky stars...

note that you never had to use $c$

I said this in the lecture. It's invertible if and only if $X$ has linearly independent columns. So look at your column vectors.

@TedShifrin Right.

The first column is all $1$'s, the second column is $x_i$, the third is $e^{x_i}$.

Correct.

11:19 PM

The $x_i$ are distinct, right?

Indeed.

So the first two columns are linearly independent. If the third were a linear combination of the first two, you'd be saying that the exponential function is linear.

You can have a linear relation at one point, but not at more than one :)

Yes, so the columns are very independent.

Now...

Do you have proof that $X^\top X$ is independent iff columns in $X$ are? :)

Or some link to a good proof.

I meant invertible not independent.

That is, $X^\top X$ is invertible iff columns in $X$ are independent?

I already gave you all the steps for the proof earlier.

Will check it out.

11:26 PM

I probably did the proof in that very lecture, but maybe I skipped it because they'd done it in homework.

$X y = 0$ iff $y^TX^T Xy = 0$ iff $X^TXy = 0$

Gosh, this book gives such little motivation!

i.imgur.com/8mpjokr.png How did he find what to set as the minimum?

First line, on the second page

That's the proof I suggested earlier, @copper. There are others, of course.

presumably by looking at the alluded figure 3.12

@Thorgott Well, then how does one come up with figure 3.12 then?

11:37 PM

I don't know what figure 3.12 is

user image

Here it is

try proving it by yourself first.

off for some exciting grocery shopping...

Well, we wish to prove that $\forall \epsilon > 0, \exists \delta > 0, |f(x)| < \delta \implies |x - x_0| < \epsilon$

Indeed, it doesn't seem clear how to get that minimum without some magical picture appearing before my eyes

@copper.hat will try, thanks though

(Also doesn't help that $f(x_0 - \epsilon) \not \mapsto y_0 - \epsilon$

11:47 PM

@schn: Did you go back and read all the steps I told you? Write it down and work on it.

I will definitely quit topology after this lecture. To see that an $\Bbb RP^2$ without the möbius strip is an open disk I had to cut out the pieces of paper, draw arrows on them and see what comes out when I put it together such that the arrows align

this doesn't generalize well to higher dimensions

@TedShifrin Currently watching khanacademy.org/math/linear-algebra/matrix-transformations/… :)

If you think of the hemisphere with opposite points identified on the edge, @user2103480, when you cut away a neighborhood of the boundary (that is a Möbius strip), what's obviously left is a disk.

@schn Why not work on it yourself and learn?

gives up

So you want to find an interval around $y_0$ that $g$ maps into a given interval about $g(y_0)=x_0$. The monotonicity of $g$ means that $g$ maps the interval $(y_0-\delta,y_0+\delta)$ into the interval $(g(y_0-\delta),g(y_0+\delta))$. So we just need to find a $\delta$ such that $g(y_0-\delta)>g(y_0)-\epsilon=x_0-\epsilon$ and $g(y_0+\delta)<g(y_0)+\epsilon=x_0+\epsilon$.
Now since $f$ is monotonous, these are equivalent to $y_0-\delta=f(g(y_0-\delta))>f(x_0-\epsilon)$ and analogously $y_0+\delta<f(x_0+\epsilon)$.

@Thorgott math.stackexchange.com/questions/3997285/… You may want to post it as a response.

11:50 PM

@TedShifrin that is true, although my main problem was the part where I get into my head that the upper part of the hemisphere is a möbius strip. That's how I got to the cutting part

@TedShifrin It was some time ago I studied this. Need a general refresher.

but I kinda get that as well

No, no. The Möbius strip comes from the band around the equator. The identification on the boundary is what gives it to you.

we glue one half of the circle to the other half in opposite direction but meh

Not the opposite direction, but up into the sphere reverses when you get to the second part.

11:52 PM

@TedShifrin that's what I meant. the upper part of the hemisphere is everything above some plane parallel to the xy plane

i.e. the band

But it's the lower part that gives you Möbius, not the upper.

Anyhow ...

is RP^2 the upper hemisphere or the lower hemisphere for you?

In higher dimensions one thinks in terms of attaching cells and attaching maps, which is really what's going on here, in fact.

Upper hemisphere. You're right; I didn't specify.

I thought of the lower one haha

@TedShifrin ah, the cell-decomposition made more sense to me than all the glueings

But I think the picture is clearer. Inverting the function is reflecting the graph about the diagonal. You're searching for an interval about $y_0$ on the $y$-axis (in the right picture) coming from the interval on the $x$-axis under $f$, so why not just take its image.

11:54 PM

it was more "formally obvious"

I don't like thinking of RP^2 as either hemisphere, think about it as the entire sphere. If you cut a strip about the equator out of the sphere, you get two disks. Now identify antipodal at all steps; the sphere becomes RP^2, the strip about the equator becomes a Möbius strip and the two disks get identified.

pure TOP topology

blows up Thor's origin

@Thorgott sorry what did you say

@Thorgott y0−δ=f(g(y0−δ))>f(x0−ϵ), how did you justify this line?

couldn't read that it's too loud here

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