« first day (3928 days earlier)      last day (1086 days later) » 
00:00 - 18:0018:00 - 00:00

6:00 PM
@PM2Ring: Let me take a look at that as well.
@leslietownes: So since there’s no meaning in comparing that (infinity). My question doesn’t make sense. Hmm.
*every non identity element.
 
yeah, and same with R.
 
6:18 PM
I got it. Many thanks @leslietownes :) I also saw the link shared by @PM2Ring: it says the same thing that you were saying earlier: sequence of rationals converging to any real number.
 
@Koro and the negative of identity element
 
@robjohn: Hi. Identity is 0 for both additive groups R and Q.
 
6:38 PM
@BalarkaSen Of course! And congrats, although you know I had complete confidence in you!
 
Hi all. I have a doubt in the video : youtube.com/…
from 1-18 minutes , the professor is discussing Gaussian elimination and suddenly around at 25 minutes he is trying to obtain the same results through matrix multiplications . Why ?
 
@Koro Oh, I thought you were talking about multiplicative order
 
im probably not going to watch the video but i will say that you can encode gaussian elimination in matrices. namely, elementary matrices
 
@Bhavay Quin has it. The operations of Gaussian Elimination are the operations of elementary matrices.
 
Lets try to push mathjax here! If we want to add 2 times the 2nd row to the first of an arbitrary 2x2 matrix:
i may need to delete this if i fall on my face!
$$\left[\begin{array}{cc}1&2\\0&1 \end{array}\right]\cdot \left[\begin{array}{cc}a&b\\c&d \end{array}\right] = \left[\begin{array}{cc}a+2c&b+2d\\c&d \end{array}\right]$$
 
6:51 PM
@robjohn But why is he doing that way ? any particular reason ?
 
@Bhavay because matrix operations are so much more versatile and abstractible
you can compose and invert matrices
 
@Quin How did you know that the particular left matrix will produce the desired result which you wanted.
 
you basically do the thing you want to the identity matrix
 
experience is valuable here
 
so i added 2x(second row) to the first row in the identity. Then multiplied on the left (left is row operations, right is column)
sorry, had left/right backwards
 
6:55 PM
@robjohn I thought there was some trick involved ..
Is there way , I can know that Gaussian elimination will fail before attempting the question ?
 
@Bhavay no tricks, just work for a while with matrices (eg doing multiplication problems) and it will eventually become clearer
 
from the spot in the video linked onwards, he is explaining just this
 
@Bhavay if the determinant of the matrix is 0
 
Thank you @robjohn @Quin.
 
$$ \begin{align} 2x+3y+4z&=1\\ \phantom{2}x+\phantom{3}y+\phantom{4}z&=0\\ \phantom{2}x+2y+3z&=1 \end{align}\implies \begin{bmatrix} 2&3&4\\ 1&1&1\\ 1&2&3 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} 1\\0\\1 \end{bmatrix} $$
@Bhavay the determinant of that matrix is $0$, so there may be no solution or infinitely many.
 
7:06 PM
he gets paid per chatjaxed line
 
@leslietownes there... one line :-p
 
the vibe of row reduction using elementary matrices is definitely one that is only acquired with experience. i found it helpful to use a computer algebra system where you can rapidly input matrices and see what the products work out to. nothing more fun than stacking up a product of three or four elementary operations on the basis of your mental view of how you'd row reduce, hitting the matrix with it, and seeing it work.
it's like a video game.
 
7:26 PM
@Bhavay If you want an alternative to Strang, you can try some of my YouTube lectures. I can figure out which specific one(s) if you are interested.
 
hi everyone!
 
7:44 PM
@TedShifrin Yeah , I did talk this with you before starting the series. You said that your lecture series is more proof oriented , but then I get the advantage of asking doubt from the instructor who taught it. how cool is that ? Also , i find strang lectures to be a bit fast paced for me...
 
Hello! Does the condition $\varphi(s):L_{\nu,q}(R^+)\to\mathbb{C}$, where $\nu,q\in\mathbb{R}$ and $q\ge2$ suffice, to ensure that the Inverse Mellin-Transform of $\varphi$ exists?
 
Right. I remember. I personally think my lectures are clearer than Strang's, although I admire him deeply. My lectures may be fast, too. I'll look up the numbers for this topic.
Try Lectures 33, 34 here. Go back to 30-32 if you want.
 
Hi, is it possible to extract eigen vectors of a three-dimensional matrix?
 
@enthu Of course.
Do you know the definition of an eigenvector?
Have you found the eigenvalues?
 
@TedShifrin sure, it is some equation like this: [B]-g*{I}=[0]
 
7:52 PM
Um, no.
What are all these crazy []{}?
 
:)) I wanted to show matrix and vector!
 
Where is a vector?
What is $g$?
 
[]n*n matrix and {}n*1 vector
g: eigen values
 
I stands for identity matrix.
This equation makes absolutely no sense.
 
all right
 
7:55 PM
How do you find the eigenvalues? Let's start there.
I assume $B$ is the $n\times n$ matrix.
 
yes
 
So how do you find the eigenvalues of $B$?
 
@TedShifrin Thanks I will look into it.
 
My feelings won't be hurt if you prefer Strang, @Bhavay :)
 
I do not remember exactly, it was 10 years ago in my bachelors when i read the topic. But I remember that the solution of those equations lead to eigen values and vectors...
 
7:57 PM
You had better do some more reading now.
 
all right...
 
patient man
 
@TedShifrin You only have 6-7 lectures on the topic , right ?
 
@copper I basically sent him away, so ...
 
you were very polite in doing so. i was close to apoplexia and i was not even involved.
to be fair, i do not understand people.
 
8:00 PM
@Bhavay There are more in different places, but as you've noted there are more lectures on calculus/analysis.
 
8:12 PM
@leslietownes Here's an actual Gauss-Jordan elimination game: people.math.harvard.edu/~knill///pedagogy/webeducation/…
 
1999 called and they want their web design back
is it time for me to code up a 2021 reboot of the elimination game
 
Maybe...
 
 
3 hours later…
11:10 PM
Hi I have a question about solvability of a problem. Given a triangles angles x, y and the side c shared by both angles and knowing that sin(y)*a^2 = sin(x)*b^2 where a and b are the sides opposite to y and b respectively can we solve for all angles? imgur.com/a/Yd87ahD image of the situation. It comes from a part of a real life physics problem I'm trying to solve and I think in theory it should be solvable based on physical intuition, but I can't seem to find a solution.
 
the law of sines comes to mind. this might be an overdetermined problem. decidedly not a geometer, so don't take my word for it
 
so you only are interested in finding the last angle? is this triangle just in the plane? if so, you might just need that the angles in a triangle sum to 180 degrees
 
i was thrown off by 'c' at the bottom in text. is something special about c or is it just c?
 
it's known
 
11:17 PM
and the relationship between a y x b is also known
I tried using both the law of sines and cosines to no avail
 
that is the worst kind of avail.
 
so you only know the relationship and not the values of x,y,a, or b?
 
Yes
Here's my attempt imgur.com/a/LRM7Hjt
E_A and E_B are known
 
if so, i think you can conclude from law of sines that $a^3=b^3$ which would make $x=y$ and the triangle isosceles. then i you may be able to get that there is a whole family of "solution triangles" with the family given by the height of the triangle with $c$ fixed
 
It shouldn't be isosceles
To give context, C is a point of light and A and B measure the intensity, and we know those readings and the distance between the sensors and I'm looking for the angle X at which the light source is tilted
But solving for the base angles is enough.
 
11:31 PM
from law of sines, $\frac{\sin(x)}{b}=\frac{\sin(y)}{a}$ and your formula (solving for $\sin(x)$) gives $\frac{a^2}{b^2}\sin(y)=\sin(x)$. Subbing into law of sines, we get $\frac{a^2\sin(y)}{b^3}=\frac{\sin(y)}{a}$ and doing some algebra, we get $\frac{a^3}{b^3}=1$.
i may have made a mistake, in which case im very sorry! but thats what it seems to give
if this is correct (check me please!) then we get $a=b$ and the triangle is isosceles and the height seems to be able to be anything
 
Hmmm
You're right.
 
i dont really know enough about optics to tell if there may be a problem in the setup but thats where id check
 
If you could take a look at the formula here hyperphysics.phy-astr.gsu.edu/hbase/vision/Areance2.html
Oh yeah
I simplified it too much for the sake of presentation
a/b = constant here, sorry xd
But if you look at the image on the link it's the same sitiuation r1, r2, |x_2 - x_1| are the sides only the triangle is tilted.
 
from the hyperphysics link, you have "E" readings at points $A$ and $B$ on a line from a light source at point $C$ and the goal is to find the angle line $AB$ makes with line $CD$ where $D$ is the midpoint of $A$ and $B$? Is this correct? And the value of $I$ from the light source is known?
 
Yes the statement of the problem is correct, and no the I from the light source is unknown.
But if you think about having three sensors on a flat table and you move a lightbulb above them, the readings should give you enough information about the bulbs azimuth and altitude angles or is my intuition wrong?
We don't need the distance from the bulb to the table
So only two out of three coordinates in the spherical system
 
11:50 PM
i have no intuition unfortunately.. i may need a sec to think. unless someone else is quicker than me :)
 
let me set up some of my sensors on my desk and see
 
00:00 - 18:0018:00 - 00:00

« first day (3928 days earlier)      last day (1086 days later) »