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18:03
yes, these matrices are isomorphic under swapping of rows or columns @Agawa001, the same happen for the matrix $\left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$ or $\left( \begin{array}{ccc} 0 & 0 \\ 0 & 1 \end{array} \right)$
indeed exist some papers about avoiding matrices of these kind... it maybe related with what I want but not completely
18:16
@SohamChowdhury well the roots of 0 is only zero, so would that be how i explain there being only one solution ?
You would state that the solution set is {0} @Dave
well i mention the discriminant = 0 so only one solution but i don't think saying the solution is {0} would really be explaining why its only one solution
What do you mean by "explaining 'why' it is only one solution"? @Dave
i have to explain why a discriminant = 0 means theres only one solution
not in numbers but just like a sentence as if explaining to a student
@Dave You mean why the discriminant of a quadratic being 0 implies there is just one root?
18:29
well it says on solution rather than one root here
The quadratic formula gives you values for which the given quadratic equation is equal to zero, right?
@Dave discriminants are associated to polynomials, not equations (but people tend to mix them up quite a bit)
@Dave But anyway, discriminant 0 means there is at least one root with multiplicity at least 2, and since a quadratic can have at most two roots, that means there is only the one
basically b^2 - 4ac = 0. so one solution. i there for stated there is only one y-intercept
but i dont think 1 y -intercept is really explaining it in my mind
@Dave y-intercepts are not the solutions, those are the x-intercepts
sorry x-intecept
18:31
a function can never have more than one y-intercept
We have that $x=r \cos \phi$.

Why does the following stand?

$dx=d(r \cos \phi)=\cos \phi dr-\sin \phi d\phi$

Why isn't it as followed?

$dx=d(r \cos \phi)=\cos \phi dr-r\sin \phi d\phi$
that was a typo
or rather a mix up
What does the question ask for? @Dave
Hello @pjs36 !! Do you have an idea about my question?
The only real explanation you can give is that the value of x = 0 will, when substituted into the given equation will reduced to the true statement 0 = 0. Try it and prove it for yourself :-) @Dave
18:40
@skillpatrol the question says find the discriminant (which i got already as 0), what does this tell you about the number of solutions of the equation, explain briefly your answer.

Then it says, what does this tell you about the graph of y = 9x^2 -12x + 4
Hello @DanielFischer !! Do you have an idea about my question above?
so, the graph goes through (0, 0) @Dave
Wait, what about the -b/(4a) part @dave?
18:56
whats that ?
So not (0,0) but (0,4).
i thoutht the discriminant was via b^2 - 4ac which gets me 0
oh your on about the vertex
It does
But there is more to the formula, right?
no thats the equation given
9x^2 -12x + 4 = 0
Yes, the vertex must be on the y-axis.
18:58
yes
is that what the question is asking me to mention regarding the graph ?
the vertex position
i can't really tell what else they want me to say
ah okay thank you :)
Thanks for asking :)
@Dave What do you mean by the vertex? The lowest point?
well the point where it crosses the axis of symmetry
which could be the highest point for an n-shaped parabola
19:01
@Dave that is not on the y-axis for this specific example
it is on the x-axis
its on 0 for the x-axis and +4 for y i think
which ill mention in my answer when i substitute x for 0
@Dave You don't find the vertex by setting x=0
no i find y
wait ive got myself confused
@Dave You mean the y-intercept?
if the discriminant is 0 then the y axis of the vertex is 0 right ?
so i actually need to find x
19:05
@Dave You mean the y-coordinate
yes
ok i get it now
i shall isolate x to verify the vertex co ordinates
Nico Temme, in a talk on numerical computation, "...so by spending more money - I mean, by increasing the precision..."
@MaryStar Are you familiar with the method of characteristics?
19:11
What is your question? @evinda
1
Q: General solution of Transport equation

evindaGeneral solution of Transport equation (homogeneous): Method of Characteristics $$u_t+cu_x=0 (\star)$$ We know that if $f: \mathbb{R} \to \mathbb{R}$ is differentiable then $u(x,t)=f(x-ct)$ is a solution of $(\star)$. We will show that each solution of $(\star)$ is of the form $u(x,t)=f(x-ct)...

user2016963 - if you want to discuss your question any further, I'm kinda loitering in here for a moment to do so.
@felipa How did you compute the probability of getting an inner product $0$?
I computed it, but it is not as simple as the previous question
@evinda Is there an initial condition of the problem?
@MaryStar No...
19:58
@robjohn hi... I think of it as a symmetric random walk
and you are asking for the probability of being at the origin
@robjohn there is a comment by Byron Schmuland from which I learned how to get the asymptotics easily
20:14
can anyone tell me why this doesn't yield an answer, but asking for rotation along y axis gives and answer?
an*
@pilot I am impressed by wolfram alpha!!
you can ask on mathematica.se
20:29
yea it's useful and I keep getting impressed by it as time passes
 
1 hour later…
21:58
hey guys
I am currently doing some research on algebra
and I am reading representations and characters of groups
sigh I"m disappointed
I was thinking to take a look at hungerford book
the graduate version
it is readable ?
easy to read I mean and organised ?
Why do so many people make silly comments
"A question: Why is this better than simply solving f=ma ?" - really, do you really want me to show that using acceleration (and the inherent problem of not having infinite small time steps in simulations) is not working to simulate orbits that should be repeatable?
22:19
@alkabary I don't know anything about Hungerford, but I learned character theory from Isaacs. It's very terse, if you're into that sort of thing. He has a book on character theory, and an Algebra for grad students book (called Algebra) that I believe does some representation theory.
Not to say it's better than Hungerford or anything, just an option you can look into
Could someone help me at the following?
1
Q: The functional is not continuous in respect to the strong norm

evindaLet $V=C^1([a,b])$. If $J$ is a continuous functional for the norm $\|\cdot\|_\infty$ then it is continuous for the norm $\|\cdot\|_1:= ||y||_{\infty}+||y'||_{\infty}, y \in V$. But the converse is false. In my book there is the hint that we can use the functional of arc length, in order to show...

Quick TeX question, I'm getting a syntax error which I've isolated to a line where the only non-standard text is my attempt to use the $\LaTeX$ symbol. It is returning a "you can't use \spacefactor in math mode." Is there some specific package I need to call to use the symbol?
(This is in an actual .tex document, not a MathJax page)
22:34
The LaTeX and friends room should know :-)
just a suggestion
The error stemmed from habits formed using MathJax too much. bleh
Removing the dollar signs fixed the issue.
0
Q: The functional is continuous

evindaShow that the functional $J(y)=\int_a^b (\sin^3 x+y^2) dx$ is continuous in respect to the $||\cdot||_{\infty}$ norm, at any $y_0 \in C([a,b])$. Let $y_0 \in C([a,b])$. Then for $y \in C([a,b])$ we have: $$|J(y)-J(y_0)|=\left| \int_a^b (y^2-y_0^2) dx\right| \leq \int_a^b ||y-y_0||_{\infty} |y+y...

I just finished grading and typesetting responses to homework for a section of 'distance learning' students in a course I'm the grader for. In an effort to encourage them to type their work and not send in literal photographs of their work (you can even see the guy's thumb in every single shot), I'm posting responses in TeX and providing a link to miktex.org
Could someone explain to me why it stands that $\partial{W}=\partial{W_0} \cup \partial{W_1}$ at the picture of the question physics.stackexchange.com/questions/187579/… ?
22:46
Have you tried the physics chat room? They call it the h-bar.
Yes, but most of the times they don't answer me...
That's cute. Heh. Reminds me of a bar near the college my sister went to named "the library"
"... Where am I? Why... I'm at the library of course. No... I'm not slacking off and not doing my homework, why are you asking?"
You just have to ask the right people in there @MaryStar
😃
Happy hour at the library
Yes... Do you maybe know someone that is often there and who is good in fluid mechanics? @skillpatrol
22:53
Dan and a curious mind are great.
They help everybody
"A curious mind" is the best.
Ok... I will ask A curious mind, he is online now... Thanks for the information!! :-) @skillpatrol
Np pal :-)
Bib
Bib
does anyone have examples of full idempotents in the matrix ring $M_n(R)$?
I read somewhere that $E_{11}$ the matrix with 1 in the upper-left corner is full idempotent, but I have my doubts
@Bib a matrix, $A$, is defined to be idempotent iff $A=A^2$, yes? Is there some difference with "full idempotent"? The definition I'm reading is that an element, $e$, in a ring, $R$, is full idempotent iff $ReR=R$. That is the definition you are working with?
Bib
Bib
23:08
@JMoravitz That's correct.
hmm, there are several examples of idempotent elements, the question of if they are full idempotent or not. Trivially, the identity satisfies the property.,
Another easy to prove one should be projection matrices, though I don't believe they would satisfy the "full" requirement.
Bib
Bib
do you think $E_{11}$ as I noted above satisfies "full"?
$\begin{bmatrix}1&0&\cdots\\0&0&\\\vdots&&\ddots\end{bmatrix}$?
Bib
Bib
yes, that's the one
T.Y. Lam mentions it here books.google.ca/… but I feel like it doesn't make sense
Yes, I believe it is.
It certainly is in the $2\times 2$ and $3\times 3$ case (just worked it out on my whiteboard)
Should be easy enough to formulate a proof for the general case. The sketch goes as follows:
$\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots \\ a_{2,1} & a_{2,2} & \cdots \\ \vdots & & \end{bmatrix} \cdot \begin{bmatrix}1&0&\cdots\\0&0&\\\vdots&&\ddots\end{bmatrix} \cdot \begin{bmatrix} b_{1,1} & b_{1,2} & \cdots \\ b_{2,1} & b_{2,2} & \cdots \\ \vdots & & \end{bmatrix} = \begin{bmatrix} a_{1,1} & 0 & \cdots\\ a_{2,1} & 0 & \cdots\\ \vdots & &\end{bmatrix}\cdot \begin{bmatrix} b_{1,1} & b_{1,2} & \cdots \\ b_{2,1} & b_{2,2} & \cdots \\ \vdots & & \end{bmatrix}$
Continue through the remaining multiplication to get what appears to be a full matrix. I'm starting to secondguess myself though, if the location of one zero forces too many conditions on the other entries of the matrix
Bib
Bib
23:23
The issue I have is that it seems hard to guarantee that every matrix can be made like this
because every element in the second row of the resultant matrix necessarily has $a_{2,1}$ as a left factor, for example
Bib
Bib
and can't we of course find a matrix in $M_n(R)$ whose elements on the second row don't all share a common left factor except 1?
yea... looking at the 3x3 case, the result is (sorry for mixing notation) $\begin{bmatrix} aa'&ab'&ac'\\ba'&bb'&bc'\\ ca'&cb'&cc'\end{bmatrix}$, but if we want a zero in the middle, that implies that either $b$ or $b'$ are zero, forcing either the entire second row or the entire second column to be zero
thus, we can't form $\begin{bmatrix}1&1&1\\1&0&1\\1&1&1\end{bmatrix}$
Bib
Bib
so I guess $E_{11}$ is not full idempotent?
what the heck Lam
So... what does that leave us with? Here's a thought. If $e$ is invertible and is idempotent, then we have $e=e^2$ and that $1=e^{-1}e=e^{-1}e^2=e$
So, the only idempotent invertible matrix is the identity
suppose now that $e$ is not invertible, that implies that $det(e) = 0$
however, note that $\det(aeb) = \det(a)\cdot \det(e)\cdot \det(b) = 0$ since $\det(e)=0$
so, there is no way to have $ReR = R$ as an element on the righthandside if it were to be equal to $aeb$ must have determinant zero
thus, the only full idempotent matrix is the identity.
If someone wants to check my argument, I don't think I missed anything (I could rephrase it a bit better)
since $ReR\subsetneq R$ whenever $e$ is singular
Perhaps we should double check the definition that Lam uses for $E_{1,1}$ and what ring in particular he is referring to. If perhaps he intends that $E_{1,1}$ is infact the diagonal matrix of ones, or if perhaps the operation is not multiplication, then it could still be salvaged.
Bib
Bib
23:45
how odd
Hi, does someone has an idea about how I could solve the following recurrence ?
u(1)=1, u(2)=1, u(3)=1, u(4)=2, u(5)=6, u(6)=14, u(7)=28, u(8)=56
u(n)=2*u(n-1)-u(n-2)+2*u(n-3)+u(n-4)+u(n-5)-u(n-7)-u(n-8)
One of the easier (in my opinion) methods is to think of it via matrices and linear transformations @iluso You have $T\cdot\begin{bmatrix}u(n-1)\\u(n-2)\\u(n-3)\\\vdots\\u(n-8)\end{bmatrix} = \begin{bmatrix} u(n)\\u(n-1)\\u(n-2)\\\vdots\\u(n-7)\end{bmatrix}=\begin{bmatrix} 2u(n-1)-u(n-2)+2u(n-3)+\dots\\u(n-1)\\\vdots\end{bmatrix}$
You can then find a matrix representation for $T$. By diagonalizing the matrix, you can come up with a representation for $T^k$, and applying that to your initial vector $\begin{bmatrix} 56\\28\\14\\6\\2\\1\\1\\1\end{bmatrix}$ will give you a solution
In effect, the matrix argument is the same as the other approaches, the numbers which get raised to powers happen to be the eigenvalues of the matrix representation for $T$.
Unfortunatley, with a matrix that large, I wouldn't want to diagonalize it by hand, but computers can do so rather easily.

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