Mathematics

12:34 AM

Alright, here's a puzzle: given an nxn matrix A with coefficients randomly selected from {-m,...,m}, what is the probability that det(A) is in {-m,...,m}?

Rithaniel

12:55 AM

(Though, "puzzle" might not be the best word. I don't know if the answer is known by anyone.)

Rumplestillskin

1:07 AM

Can somebody help a tired old brain out. I have an integral of the form $\int^{a}_{0}\int^{x}_{0} f(y)dy dx$ and apparently I can change it to $\int^{a}_{0} \int^{a}_{y} f(y) dy dx$?? This is probably very trivial but I just can't seem to get my head in to it!!

user76284

@Rumplestillskin y is both the integration variable and one of the endpoints? Doesn't seem right.

Rumplestillskin

O my dizzy days!! That is what I thought as well and the notes I am looking at just have it written down wrong! I think the person has just reversed the order of integration except forgot to actually change the order. Like they have changed the limits but not changed the order of integration. Does that sound righT? sorry I am having a slow day

user76284

Perhaps what they meant is y goes from 0 to a and then x goes from 0 to y? The region of integration forms a triangle between (0,0), (0,a), and (a,a).

Sorry, x goes from y to a.

Yeah, just switch dx and dy.

Rumplestillskin

Yep that's precisely what it is! @user76284

user76284

@Rithaniel Not sure what the answer is, but the empirical distribution looks like this

Secret

1:34 AM

looks surprisingly tidy

Leaky Nun

7:15 AM

let $f : \Bbb R \to \Bbb R$ continuous

is $C[a,b] \to C[a,b] : g \mapsto f \circ g$ continuous with the sup norm?

Silent

7:55 AM

T/F: The polynomial $x^6-27x^4-13x^3+11x^2+24x$ has exactly one real root.

The sol i have says: polynomial is degree 6, so it may have 6 roots at most. since complex roots come in pairs, it can't have exactly one real root.

But, how do we know that multiple roots do not occur?

Tobias Kildetoft

@Silent They might, but that would not count as a single real root

Silent

oh :( so this question was kind of open to interpretation

Tobias Kildetoft

@Silent Well, does it have a multiple root?

Silent

yes, i saw in desmos

Tobias Kildetoft

so no interpretation needed

You could also have checked this by hand if you needed to

Silent

8:04 AM

how?

Tobias Kildetoft

by finding gcd with its derivative

Silent

i am sorry. but gcd of derivative? i have heard about that first time

Tobias Kildetoft

gcd with its derivative, i.e. $gcd(f,f')$

Silent

ok, i googled it, found some articles, reading. thank u

Larry Eppes

m.wolframalpha.com/input/…

i just test wolfram alpha

Tobias Kildetoft

8:09 AM

You don't need to. It has been thoroughly tested already :)

Larry Eppes

the ans is two complex root of your equation

Silent

@TobiasKildetoft, can you please suggest reading for that gcd (f,f')? The ones i googled already assume some familiarity with material.

Tobias Kildetoft

@Silent It does require some familiarity with the abstract theory of polynomials

but the proof is fairly straightforward, and the result simply says that multiple roots are the same as shared roots between a polynomial and its derivative (and to find whether they have shared roots, you can find their gcd)

Silent

ok

nbro

10:20 AM

> Fathers of the Deep Learning Revolution Receive ACM A.M. Turing Award: Bengio, Hinton and LeCun Ushered in Major Breakthroughs in Artificial Intelligence

Akiva Weinberger

@Silent Prove that if $(x-a)^2$ divides $f(x)$ then $(x-a)$ divides $f'(x)$

By the way, note that we don't need calculus to define derivatives in $k[x]$. We just need to define the derivative of $\sum a_nx^n$ to be $\sum a_nnx^{n-1}$.

In fact, if we're working in a finite field like $\Bbb Z/p\Bbb Z$, it doesn't even make sense to define it with calculus and limits and things like that.

You should check that this definition still follows the product rule no matter what $k$ is (i.e. $(fg)'=f'g+fg'$).

(Also: if $k=\Bbb Z/p\Bbb Z$, then $(x^p)'=px^{p-1}=0$. So there can be nonconstant things with zero derivative.)

Akiva Weinberger

12:07 PM

"Abel prize is certainly more difficult to get than Fields Medal." - Cedric Villani

Rumplestillskin

Does any one know where and why the absolute value shows up in the accepted answer here

4

Q: Simplifying a double integral to a single integral

In a recent cross-validated post a comment was left by Dilip Sarwate that stated that the following double integral: $$\int_0^L \int_0^L \rho(t-u)dudt$$ Could be simplified to the following single integral: $$\int_{-L}^{L} \rho(s)(L-|s|)\,\mathrm ds$$ Where $\rho(t-u)$ is the correlation func...

Akiva Weinberger

12:33 PM

It's a bit like rotating a square 45 degrees and cutting it down the diagonal @Rumplestillskin

geometrically

Rumplestillskin

12:47 PM

@AkivaWeinberger I’m not sure I follow?

Akiva Weinberger

Think of $\int_0^L\int_0^L$ as integrating over a square

Draw a plane where one axis is the $u$ axis and the other axis is the $t$ axis. You're integrating over the region where $t$ and $u$ both range from $0$ to $L$

Since you're integrating $\rho(t-u)$, it doesn't matter quite where on the square you are as much as how far away you are from the diagonal $u=t$ line

For any point in the square, we can draw the line of slope $1$ (parallel to the $u=t$ line) and see how much of that line lies in the square

and that's gonna depend on the absolute value of your distance away from the diagonal $u=t$ line

@Rumplestillskin You know what would be a more algebraic (but less geometric) way of doing this

Define $r=t+u$ and $s=t-u$

and do the change of variables

noting that $dudt=\frac12drds$ (using the Jacobian)

Silent

2:13 PM

@AkivaWeinberger Thank you!

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$Mathematics$ Mathematics