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23:00
all finite extensions of finite fields are normal... if F is finite and F(a)/F is finite and a has multiplicative order m, then F(a) is the splitting field for x^m-1.
Can anyone help me out as to how to see all the expression in understandable form?
An explanation for why can't $\sin(a+b)=\sin(a)+\sin(b)$ ?
@AnimeshAshish the number of expressions that are understandable is probably in the millions or tens of millions...
$\sin(a+b)$ can be $\sin(a)+\sin(b)$, for example if $a=0$
It looks obvious from the diagram, but I don't get why do we multply by $\cos$
23:03
@Mahmoud Trivial. If $a=b=0$, the statement is false.
if a=b=0 then the statement sin(a+b)=sin(a)+sin(b) is true not false
I know right, but I want to have a geometric intuition behind the fuzzy formula.
well you have a diagram
with something of size $\sin(a+b)$ (the dotted arrows)
And a weird text,
The full height of the blue triangle (sin(a)sin⁡(a)) can’t be used, since the red triangle doesn’t extend as far. (Why? When we add angle bb, we’re moving at a steeper angle with the same hypotenuse. We gained vertical distance and lost horizontal distance.) We’re effectively “sliding back” sin(a)sin⁡(a), reducing it by a factor of cos(b)cos⁡(b).
The full height of the red triangle (sin(b)sin⁡(b)) can’t be used either, since it’s at an angle. We’re “turning” sin(b)sin⁡(b), reducing it by a factor of cos(a)cos⁡(a).
something of size $\sin(a)$ and something of size $\sin(b)$
and they are not positioned in a way that they match up
23:06
@Mahmoud let's start with the bottom. the blue triangle. the right side has height sin(a). do you understand why the dotted line has height sin(a)cos(b)?
@arctictern No, I don't
do you understand why it's shorter than the gray arrow
@Mahmoud look at the hypotenuse of the blue triangle. the whole hypotenuse has length 1 by assumption. how long is it from the origin to the top of the dotted line segment?
@arctictern : can you tell me why does the "dollar sign" or rather "$" at the start and end of the expression is used? AND THE FORWARD SLAshes/backslashes in the mathematical expression?
cos(b) ?
23:08
besides, if $\sin(a+b)$ were equal to $\sin(a)+\sin(b)$ you would get $0 = \sin \pi = 2 \sin (\pi/2) = 2$
@AnimeshAshish it's called LaTeX, it's how we typeset mathematical equations. see "LaTeX in chat" in the room description, upper right corner.
@Mahmoud correct. so the smaller blue triangle within the whole blue triangle is scaled down by a factor of cos(b). in particular that means the height sin(a) is scaled down to cos(b)sin(a). make sense?
$\LaTeX$ Is kinda cool, and powerful @AnimeshAshish
@Mahmoud : DO I have to learn it for the chat? Or is there a software to convert al the LaTeX written programs into understandable format?
@AnimeshAshish See the link I told you to click. You don't need to understand LaTeX to read it (your browser will turn it into math automatically once you follow the directions), but you'd need to know LaTeX in order to type some yourself.
@arctictern Yes, $\sin(a)=\frac{\sin_2(a)}{\cos(a)}$
23:12
$\sin_2(a)$??
The smaller height we're interested in.
I guess you mean the height of the dotted blue line segment.
okay. now work on figuring out the sin(b) to sin(b)cos(a) part.
@AnimeshAshish It won't take a lot of time, en.wikibooks.org/wiki/LaTeX/Mathematics
@Mahmoud : Gonna start asap!
@arctictern Why are all finite extensions of finite fields normal?
23:17
@arctictern How ?
@Mahmoud the right side of the red triangle and the red dotted line form a smaller triangle. the angle at the bottom they make is a (do some basic geometry to figure this out). since the hypotenuse of this small triangle is sin(b) and its angle is a, its base (the dotted red line) is cos(a)sin(b).
@AnimeshAshish You learn something new everyday
unless you don't
um\
if I may interject
if $\sin(x)>0$ and $\cos(x)>0$, then
@arctictern Thanks $\displaystyle{\huge{:)}}$
23:22
$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)<\sin(x)+\sin(y)$
@Mahmoud $\displaystyle\huge\ddot\smile$
Oh good one.
@Mahmoud you mean you didn't already know the algebraic proof?
@SimpleArt $\displaystyle\huge ت$
@arctictern I did, but It doesn't help memorize it
23:26
@KajHansen thanks a lot for that link. I didn't know about his videos. They are really something.
I will be going through them.
@Adeek What videos ?
Hey there
Hey.
$\displaystyle\huge\heartsuit$
23:33
"And here we have a classic example of the Butt-Curve, can someone calculate its B-value?"
Bye.
Can anyone tell me how I render latex on my smartphone? I am using Chrome Browser.
Hi @Mahmoud and tern ....
BTW, @Mahmoud, if you're interested, I have a different proof of the addition rule for $\sin$ based on areas. If you email me, I can send you that.
hi @Steamy, @Null, et al.
is a non-stationary turning point, defined by f'(x) != 0 and f''(x) = 0 a point of inflection that is not an extrema?
as opposed to a stationary turning point defined by f'(x) = 0 and f''(x) = 0
23:43
But even the "stationary turning point" you just gave needn't be a turning point.
Try $f(x)=x^3$ or $f(x)=x^5$.
A point of inflection is only guaranteed by $f''(x)=0$ IF the second derivative actually changes sign at the point.
$$\huge\color{red}{\heartsuit}$$
3
for some reason dollar signs are showing up in your answers
Yes, that's how we type math in here.
See "LaTeX in chat" over there >>>> ^^^^
@frostedcake Please check the LATEX in chat on the top right
@SimpleArt: Is that a general declaration?
23:45
what is?
the red heart
Oh, yeah
Can latex be rendered in smartphone browsers?
Yes, but getting that to bookmark is trickier (at least it was for me) ...
but according to my textbook a non-stationary point of inflection can be defined as above
Can anybody teach me how?
There is no 'latex in chat' thing coming from my smartphone.
@AnimeshAshish Sorry, I actually never figured out how to do it on phone
Can you edit bookmarks?
@AnimeshAshish: I did figure it out and I think robjohn gave instructions on the webpage. This is the link: math.ucla.edu/~robjohn/math/mathjax.html
Like, open up options in whatever browser and find the bookmark editting option?
23:48
Yes, I do
Then copy and paste the block of text at the bottom of TedShifrin's link into the URL of a bookmark
@TedShifrin Sorry I meant why wouldn't my rule necessarily be a point of inflection
then, whenever you come visit us, you can click that bookmark to get the math looking fancy
and you'll see my pretty heart.
@frostedcake: I was warning you about needing to change from concave up to concave down to have an inflection point. What about $f(x)=x+x^4$? I don't think 0 is an inflection point. Look at it.
23:51
Never mind. I was right. Look at $f''(x)$, @frostedcake. It's 0 at 0, but positive everywhere else. So no change of concavity.
Its the end of the world!!!!! $\color{white}{\text{dies dramatically in the background}}$
sorry for which graph?
Btw, hit that "select all text" to see my invisible messages./
For $f(x)=x+x^4$.
@SimpleArt : I am sorry, are you telling me?
23:53
@AnimeshAshish Please see comments above.
Heya, @MikeM. I escaped icemageddon (it cost me $175 to change my Delta ticket — those shysters were letting people traveling today and tomorrow change for free, but not with my Monday ticket date).
@TedShifrin why concave up instead of convex?
@robjohn: I prefer convex and concave, but 99% of today's calculus books say concave up and down.
Hi, btw. :)
@TedShifrin hey there
Congrats.
23:56
Yeah, I'm too old for all the stress, @MikeM. How did your lecture for all the dignitaries go? :)
Let $p$ be a prime, $n\in \mathbb{N}$ and $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ irreducible. We have that $a\in \overline{\mathbb{F}_p}$ is a root of $f$.
We have that $\mathbb{F}_p(a)$ is a finite extension of $\mathbb{F}_p$.
How can we compute $|\mathbb{F}_p(a)|$ ?
It was ok. 10 people came. I was nervous and not satisfied with what I did.
Well, another time or two and you won't get nervous.
@MaryStar If a is a root of a polynomial f(x) irreducible over F, then F(a)/F has degree deg(f)
@TedShifrin that's really weird
23:58
Teachers/books make mistakes, @frostedcake ... or sometimes students misinterpret.
@robjohn Haha, a funny pain of my life. When I tried to first learn calculus...
@arctictern Yes, we have that $[\mathbb{F}_p(a):\mathbb{F}_p]=\deg f$, but can we calculate also $|\mathbb{F}_p(a)|$ ?
@TedShifrin is it because maybe they work for a certain degree of functions? as in up to x^3 ?
@MaryStar If K/F has degree n then K is isomorphic to F^n as a vector space over F.
23:59
That's not a good excuse, @frostedcake. There are zillions of functions out there (including trig functions and exponentials, etc.).
Hi again, tern :)
hello ted

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