Mathematics - 2017-07-06 (page 10 of 10)

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SteamyRoot

22:00

Did I mess up somewhere? >.<

Sha Vuklia

but what are the directions; so we have our stationary point, and we move now along $x$ right

Ted Shifrin

You can have a saddle point more generally.

Sha Vuklia

but if we keep $y=1/2$ fixed, then the function stays zero

so it seems we always have to shift from $y$

Daminark

How's everything going?

Ted Shifrin

@Sha: The definition of saddle point is very simple — in every neighborhood of the point, the function is both larger and smaller than its value at the point.

SteamyRoot

22:01

Right, French terminology.

By increasing and decreasing I mean $\geq$ and $\leq$

@TedShifrin I did mention that this reasoning was only the case for this special kind of function, though, and not in general.

Sha Vuklia

I still don't get it. So they say that $(x,1/2\text{ with }x\in(0,1))$ are local weak maxima. Why don't they mention the minima?

Typhon

@TedShifrin Also wandered into constructive feedback. I can definitely say that that room is ran quite strangely.

Sha Vuklia

or should they have mentioned those too?

Ted Shifrin

Oh, I missed that, Steamy. @Sha: I think that's wrong.

Typhon

anyways

@TedShifrin you're good with geometry, right?

SteamyRoot

22:03

That answer indeed sounds wrong.

Ted Shifrin

Oh, I see.

SteamyRoot

Oh, wait, no.

the factor with $x$ makes it negative

Ted Shifrin

Typhon, leave me alone, please.

SteamyRoot

So any change to $y$ will increase the $(y-1/2)^4$ factor

Sha Vuklia

yes

ohh hm okay

Typhon

22:04

sorry

SteamyRoot

Or, you could just play with signs and (non-)zeroness, I guess.

$f(x,1/2) = 0$

and for any point close enough to $(x,1/2)$, $(x-1)$ will still be negative and $(y-1/2)^4$ will always be positive.

So the function is $\leq 0$ close to $(x,1/2)$ with equality if and only if the $y$-coordinate is $0$.

Sha Vuklia

okay one last attempt: so say we have $x\in(0,1)$. Then we know that the function is negative. This means that if we move along the $y$ direction, that the function will decrease. However, if we move along the $x$ direction, the function could either increase of decrease

SteamyRoot

Which is why you maximum is weak

Sha Vuklia

oh I have to read what you wrote

ohh I think that it "begint te dagen"

I'm so glad I did this exercise tonight, I've never done one before.

sooo many thanks for the patience.

SteamyRoot

No problem

Astyx

22:09

@Ted too easy :(

Sha Vuklia

alright, I'm off to bed, see ya!

SteamyRoot

Good luck on the test!

Sha Vuklia

thanks!! (I'll need it :P) bye!

Mary Star

22:40

Hello!! I am looking at the following:

A lottery wheel contains five red, blue and yellow balls. For each color the balls are labeled with the numbers $ 1,2,3,4 $ and $ 5 $. A good fairy draws from the lottery wheel two successive balls. A ball which has already been drawn is not moved into the drum.

To calculate the probability that two balls with the label-sum $4$ are picked do we have to calculate the probability $$P((\text{ Ball with label } 1) \cup (\text{ Ball with label } 3))+P((\text{ Ball with label } 2) \cup (\text{ Ball with label } 2))$$ ?

Is someone of you familiar with that?

Typhon

@AkivaWeinberger I give up. please give me a hint. Does induction work?

Typhon

23:19

Q: Proving that if $a^n$ and $(a+1)^n$ are both elements of a polynomial field, then $a$ is also an element of that polynomial field?

In the main MSE chat @AkivaWeinberger proposed the following conjecture. If $a^2$ and $(a+1)^n$ are both rational, then $a$ is a rational number. I have tried several times to prove it and I just wish to know how to prove it. Since I was told this was a previous unlinked question, let me a...

I posted a broader conjecture here

please help

this is driving me nuts not knowing how to solve it. XD

^exaggeration.

Simply Beautiful Art

23:32

Well, its been an hour and 16 view, so I hope my advertising here is not a problem for anyone.

Q: Tighter bounds on the fast growing hierarchy?

Not a dupe of this question, as I'm searching for tighter bounds. We define the fast growing hierarchy for finite values as follows: $$f_k(n)=\begin{cases}n+1,&k=0\\f_{k-1}^n(n),&k>0\end{cases}$$ where $$f_k^a(n)=\underbrace{f_k(\dots f_k(}_an)\dots)$$ For example, $$f_2(2)=f_1(f_1(2))=f_...