Mathematics - 2023-05-21

user 858770

00:02

Hi

leslie townes

my students always wanted homework to be worth as much as possible in determining the grade. past a point, however (the point being roughly "if it isn't worth enough, people won't do it"), i don't recall the percentage contribution having any appreciable effect on student incentives, or on outcomes.

Ted Shifrin

In most of my courses it was around 35% …

user 858770

Homework grades were taken into account for those on the border between two grades, no?

Ted Shifrin

Different teachers grade very differently.

user 858770

True true.

Ted Shifrin

00:08

Different levels, different types of courses.

I think in most of Europe homework is not graded and exams are once every year or two. Most American students would die in that system .

user 858770

Not vice versa? They would die in the American system.

Ted Shifrin

Much more mature and disciplined. I prefer our system, but I disagree. They might be bored in most US schools.

user 858770

Ok, die of boredom :-)

With no all or nothing exam at stake at the end.

D.C. the III

00:39

Don't get irritated Ted, but I'm trying to put together the understanding of what you said with regards to why the coefficient of $y$ is nonzero. So I related things to the Vandermonde matrix I posted above with the $c_i$ terms. In that one since the $c_i$ are all distinct, reducing it to echelon form all my terms will be of some form of $c_j - c_i$ with a and since all the $c_i$ are distinct, none of these will be zero so I can reduce the matrix to a reduced echelon form albeit the reduction will be messy but it can be done. Which means the nullspace will only be the zero vector.

@TedShifrin

Ted Shifrin

You expanded the determinant along the first column. That’s where all these interesting $3\times 3$ determinants came from.

D.C. the III

Ok..Yes....So doing that, my determinant for the $y$ coefficient will only involve $x_i$ values and as just noted again since these $x_i$ are all different I can't get a determinant of $0$ in this specific $3 \times 3$ matrix.

I really fudged up thinking this through....

That's why my $y$ coefficient won't be $0$

I could use this same sort of reasoning to understand why the $x^2$ term will be nonzero as well

Ted Shifrin

00:54

That’s the Vandermonde determinant. No, the other determinant is quite different.

Show that matrix is singular iff the three points are collinear.

D.C. the III

Ok. I'm going to do this now while I go and eat. I do see one flaw in what I did was not isolating the individual determinants to explain them.

D.C. the III

01:39

Proving that: "if the matrix is singular, then the three points are collinear":

First we reduce the matrix:

$$\begin{pmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3
\end{pmatrix}
\Rightarrow
\begin{pmatrix}
1 & 0 & 0 \\
x_1 & x_2-x_1 & x_3-x_1 \\
y_1 & y_2 - y_1 & y_3 - y_1
\end{pmatrix}
$$

If our matrix is singular, then this means that $\det(M) = 0$. This means that one column is a scalar multiple of another column. By the transformed matrix this means that the last two columns are parallel iff $P_1P_2 || P_1P_3$. But this can only be the case iff the three points are collinear.

@TedShifrin

Ted Shifrin

You keep asserting the same wrong crap about determinants. Forget manipulations. What does singularity mean from the viewpoint of linear equations?

one potato two potato

@TedShifrin I thought any composite number can be a counterexample. Something's wrong...

D.C. the III

from the view of linear equations in terms of matrices it means the rank of the matrix is less than the number of columns $n$

for a given $n \times n$ matrix

Ted Shifrin

Come on. That’s not about linear equations.

D.C. the III

01:55

The only other notion of singularity coming to mind is the notion form analysis. So the point where the object is not defined or doesn't exist.

one potato two potato

I'll ask my professor about the problem.

D.C. the III

In terms of linear equations after reviewing it means that there is a non-trivial solution to the system of equations from $Ax = 0$.

one potato two potato

@TedShifrin You think my counterexample is wrong or what? Is there somewhere I'm misunderstanding the situation?

The problem statement itself is: If $R$ is UFD and $a\in R\setminus\{0\}$ then there exists a unique maximal ideal $\mathfrak{m}$ of $R$ such that $R/(a)\simeq R/\mathfrak{m}$ as an $R$-module.

Ted Shifrin

03:51

@D.C.theIII Right. Use this.

D.C. the III

Yea I'm using it, just got caught up watching this boxing fight

I'll report back tomorrow probably....also with tears because there is no reason this should've taken this long.

Ted Shifrin

@onepotatotwopotato Clearly false.

one potato two potato

Yeah, I e-mailed my professor 2h ago. Maybe it's a typo.

D.C. the III

also after rewatching the video you specifically said to make sure to that singualrity and invertibility are two different things

Ted Shifrin

04:12

Nonsingularity and invertibility are equivalent by theorem, but I wanted the students to know the definitions.

copper.hat

04:46

@TedShifrin Fonda's on Solano:-)

one potato two potato

05:38

If $M$ is an $R$-algebra ($R$ is a commutative unital ring) and $S$ is a commutative unital $R$-algebra then $S\otimes_RM$ has an $S$-algebra structure?

I know it is $S$-module and commutative unital ring but don't need to care about some kind of compatibility?

It has $S$-algebra structure.

Koro

06:26

0

Q: Every collection of non trivial intervals is a countable subcollection of the collection.

Let $\scr A$ be a collection of non trivial intervals (i.e., an interval with at least two points) in $R$. Then, it is to be proven that there exists a countable subset $A\subset \scr A$ such that $\bigcup_{I\in \scr A} I= \bigcup_{I\in A} I$, where $I\in \scr A$. I tried to prove it the followin...

Curio

08:41

if (X1, Y1) and (X2, Y2) are indepedent random vectors, is it true that X1 and X2 are independent and Y1 and Y2 are independent?

leslie townes

5

Q: Intuition on Independence of Random Vectors

Intuitively, two random variables $X$ and $Y$ are independent if knowing the value of one of the random variable provides zero information about the other. The same holds true for two random vectors $\mathbf{X}=(X_1,X_2,\cdots, X_m), \mathbf{Y}=(Y_1,Y_2,\cdots, Y_n)$. But does it also mean that $...

probability self-study correlation random-variable independence

one potato two potato

09:55

Professor says to think of $\Bbb C[x]$... I don't know what he's trying to say. He thinks the statement is correct. Hmm

northerner

10:56

What is (1)^(infinity) equal to? wolframalpha.com/input?i=%281%29%5E%28infinity%29

robjohn

11:07

here is my take.

schn

11:29

I'm struggling understanding the proof of Theorem 3.54 in Rudin's book. I still do not understand why $\beta_1>0$ is a necessary condition. What does it ensure? If anyone would be so kind to explain, I'd be very grateful.

冥王 Hades

schn

@schn The OP claims we can even allow $\beta_1=0$, but I disagree.

schn

11:54

If $\beta_1=0$, then I guess the LHS of $P_1 + \cdots + P_{m_1} > \beta_1$ would simply be $P_1$, since $P_i>0$ by definition. Then $m_1=1$ and $|x_1-\beta_1|=P_1\le P_{m_1}$ would hold, so $\beta_1=0$ is probably possible after all.

noballpointpen

17:19

I am curious, if it is the case that $\vdash \mathscr C_1 \lor \mathscr C_2$ but not $\vdash \mathscr C_1$ and not $\vdash \mathscr C_2$, does it always mean that one formula is a negation of the another?

copper.hat

17:54

@noballpointpen Let the first formula be $x \lor y$ and the second $x \lor \lnot y$.

noballpointpen

Makes sense. Thanks.

copper.hat

i drew up a truth table that satisfies the formula assertions (you need at least 3 entries), added a 4th and then figured out what the formula must be for each.

geocalc33

18:22

What is the enclosed volume of my gravatar?

copper.hat

18:47

depends on how it is defined

noballpointpen

19:12

@copper.hat may we then further argue that these formulas being disjuncted introduce a subformula of the form $p \lor \neg p$? For the ease of the question, we assume the formulas do not have conditional connectives. I am solving a question where this property seems to be a key to the answer.

A "counterexample" I found would be $(x \land \neg x) \lor \neg (y \land \neg y)$, but the second disjunct is equivalent to $y \lor \neg y$.

Oh, nvm, it's not a "counterexample". We can derive the second :) Does not meet our conditions.

Seems like your formula would work as that proposed "counterexample" if we negate each disjunct.

Or not... hm...

No it won't, of course not.

Koro

19:34

Let $(X, S, \mu)$ be a measure space. Let $A_1,A_2,...,A_k$ be an S- partition P of X: this means that $Ai$'s are elements of the sigma algebra S and that $\bigcup{1\le i\le k} Ai=X$. Let $f:X\to [0,\infty]$ be S-measurable. Lower Lebesgue sum is defined as $L(f, P)=\sum_{j=1}^k \inf_{x\in A_j} f(x)\mu (A_j)$. Then, $\int f , d\mu := \sup \{L(f,P): P$ is an $S-$ partition of $X$}

With this definition of integral, how can I show that the counting measure on $Z^+$ is the summation?

(i.e., if $(b_n)$ is a sequence of non negative integers and $b: Z^+\to [0,\infty]: n\mapsto b_n$, then $\int b \,dc=\sum_n b_n$, where c is the counting measure.)

It is not known yet that $\int (f+g) d\mu= \int f +\int g$.

Koro

20:31

nvm, I think I figured it out.

:-)

user4539917

:-)

noballpointpen

@Koro, are you pursuing a Ph.D.?

Ted Shifrin

@geocalc33 $\pi$, of course.

noballpointpen

20:52

Seems like my claim can be shown to be false with $[p \land q \land \neg r] \lor [(p \land \neg q) \lor \neg p \lor r]$. The formulas in square parentheses are not derivable, the whole formula is derivable, but no subformula of the form $p \lor \neg p$ here.

mick

21:36

0

Q: When is a solution $P(f'(x)) = Q(f(x))$ periodic or double periodic?

Consider the differential equation $$P(f '(x)) = Q(f(x))$$ Where $P(x),Q(x)$ are polynomials. Examples are $f'(x) = 1 + f(x)^2$ where we get a tan solution and $f'(x)^2 = 4 f(x)^3 - g_2 f(x) - g_3$ where we get a Weierstrass elliptic function solution. One of them is periodic , the other double p...

ordinary-differential-equations polynomials fourier-analysis periodic-functions elliptic-functions

got this on my mind today

D.C. the III

Big wall of text incoming......................

I went back and did exercises in Sec 4.1 that you didn't assign. Specifically #23 which was to establish the claim of three points $P_1, P_2, P_3$ are collinear iff the equation $Ax = 0$ has a nontrivial solution (i.e the matrix $A$ is singular).

First suppose that matrix $A$ is singular where matrix $A$ represents the coefficients from the equation of the line $c + ax + by = 0$:

$$\begin{pmatrix}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{pmatrix}
\begin{pmatrix}
c \\
a \\
b
\end{pmatrix}$$

Eliminating the first column of $1$'s:

$$
\begin{pmatrix}
1 & x_1 & y_1 \\
0 & x_2 - x_1 & y_2 - y_1 \\
0 & x_3 - x_1 & y_3 - y_1
\end{pmatrix}
$$

Observe as this is the equation of the line and we assumed the system was singular, $a$ and $b$ are not both $0$. Without a loss of generality suppose $a \neq 0$. Also by singularity row 3 has to be a multiple of row 2. Transposing this reduced matrix we can see that the coordinate vectors of column 2 and column 3 would be parallel, which implies collinearity.

$$\begin{pmatrix}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{pmatrix}

$$

Row reducing again as before:

$$
\begin{pmatrix}
1 & x_1 & y_1 \\
0 & x_2 - x_1 & y_2 - y_1 \\
0 & x_3 - x_1 & y_3 - y_1
\end{pmatrix}
$$

But this time we assumed the points are collinear so row 2 in terms of coordinates would be a scalar multiple of row 3. This means we would have a row of zeros in our matrix:

$$
\begin{pmatrix}
1 & x_1 & y_1 \\
0 & x_2 - x_1 & y_2 - y_1 \\
0 & 0 & 0
\end{pmatrix}
$$

This implies that we will have a non-trivial solution to the linear system $Ax = 0$. Which means $A$ is singular.

Back to the original issue which was validating why the coefficient for the $x^2$ term is not zero. The three points used in the parabola are distinct and not collinear. This means the matrix we have set up for $C_{3,1}$ will be nonsingular by what we just proved. Since this matrix is nonsingular, then $\det(C_{3,1}) \neq 0$. And this is why the $x^2$ coefficient is not zero.

@TedShifrin I believe this is how you want it reasoned out. No rush whenever you are free

If you are willing to read through it as well

Ted Shifrin

You have got to get past this tendency to be mechanical rather than conceptual. If $Ax=0$ has a nontrivial solution $(a,b,c)$, this means the points all lie on the line $a+bx+cy=0$. Done.

Oh, maybe I need $A^\top$ there.

Wait. You started with this. You’re done. Why all the computing?

D.C. the III

I do "kinda" get what you mean by me having a tendency to be more mechanical. But can you expound on it a bit because. I'm aware I tend to do this but getting out of that habit is harder than I expect

I was trying to be through

perhaps it is practicing applying the ideas.

Ted Shifrin

No. We know nonsingularity is equivalent to only the trivial solution of $Ax=0$.

Row reduction is not needed.

D.C. the III

Ok. Got what you mean

Time to write a reflection on this question for the trouble it caused, then to the putnam question.

Ted Shifrin

21:54

Reflection across what plane? 🤷‍♂️

D.C. the III

The plane of enlightenment

also a caccio e pepe break with mucho caccio

Ted Shifrin

Mixing Spanish and Italian?

D.C. the III

I could've not been lazy and actually looked for the right word in Italian, but I let my knowledge of the romance languages guide me even though I knew it was incorrect.... 🤷🏿‍♂️

noballpointpen

22:23

I guess if $\vdash \mathscr C_1 \lor \mathscr C_2$, not $\vdash \mathscr C_1$ and not $\vdash \mathscr C_2$, then we must have $\mathscr C_1 \lor \mathscr C_2$ to be an instance of a tautology... hm... Holds with the completeness theorem, but I do not assume it here.

Ah, it's the soundness theorem. And I have it... then... hm...

Nvm, just thinking.

Wolgwang

23:04

For p1, why (100×100×20)/(100×100×100) doesn't work?

Ted Shifrin

@D.C.theIII Molto

D.C. the III

Makes sense. And I should've recognized that with all the Italian Soccer I watch.

Ted Shifrin

@Wolgwang Because you are counting ordered selections rather than subsets

D.C. the III

Since you're here I did have two questions: One went back to computing the integral over the region bounded by $y + x^2 = 0, x-y = 2, x^2 -2x + 4y = 0, x\geq 0$..... What was the motivation for/ how did you know to choose $x = u + v$ and $y = v - u^2$?

Ted Shifrin

Oy, that infernal problem.

D.C. the III

23:13

The second question was with the problem I just did. Where does the idea of expressing the equation of the parabola as a determinant come up. Why did we do this?

Ted Shifrin

I made this up back in 1974 or so. I think I started with the mapping and a region and saw what I ended up with in the $xy$-plane.

Start with the lines and planes and then say to yourself, “Self, generalize.”

It shows up with any number of algebraic hypersurfaces. Take a basis for the monomials you’re working with.

D.C. the III

I'll ponder and generalize.

Wolgwang

23:51

@TedShifrin Oh!

So will this work:

$3×\dfrac{(100×100×20)}{(100×100×100)}+3×\dfrac{(100×20×20)}{(100×100×100)}+\dfrac{(20×20×20)}{(100×100×100)}$

Ted Shifrin

You are overcounting everywhere!

Wolgwang

:(

The solution first finds complementary probability (80×80×80)/(100×100×100). I can't see how to find without it.

Ted Shifrin

That is the easy way to do it.

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$Mathematics$ Mathematics