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20:00
@TedShifrin According to u is there any better method than substitution, apparently substitution involves lot of thinking.
The best you can do to start, @Bhavay, is think about how $(2x+1)/2x$ is slighly more than $1$ when $x\ge 1$. So you have $\log 4 + y\log 3$ where $y$ is a bit more than $1$.
@ArjunRana oh common man , maths jax is easy to type.
Now compare to $\log 27$ ...
Hold on sir. May i type the exact question first?
Wait. This isn't the correct equation?
20:03
@Bhavay , it's not that bro, faltu bolta h ye, maine kuch kaha tha kaya is gadhe ko?
boodha ho gaya akal nahi aai!
@Arjun: You will need to be respectful. Enough insults or you will be banned for a while.
The number of value of x for which : $log(4)+ \frac{2x+1} {2x}log(3)=log(\sqrt[x]3+27)$ holds true.
I see. So they don't want you to try to solve it. They want you to understand how many solutions it will have. Are you allowed to use graphing calculators?
@Arjun: You are lazy and rude beyond belief. Go away.
Nothing to do with my ego, sir.
Nope . :-( otherwise why would i ask about period of sin(x+3sinx)
@TedShifrin , what are you gonna do huh, ban me,go ahead!
20:08
@ArjunRana Cut it on man
Hey everyone, a quick question (sorry for the bad terminology...) - for abelian groups, when I see $\bigoplus_{i \in I} A_i$, should this mean the colimit of the discrete diagram ( discrete = no non-identity morphisms) (where this discrete diagram has $I$ many objects, even if $A_i = A_j$ etc.)
@Bhavay , you are in which grade?
@ArjunRana As u wish.
@ArjunRana . Gave 12th paper this year.
@Bhavay , mera matlab tha baat ko samaj
@Bhavay: Do they say any domain for $x$? Positive numbers? $x\ge 1$?
20:11
@Bhavay , bada bhai h tu to apna
@TedShifrin No, i just typed the exact question.
@Bhavay , gave JEE?
So I'm not used to seeing $\root x\of 3$ unless $x$ is an integer. Do you guys write that for general real numbers?
@ArjunRana yes
@Bhavay , percentile?
20:13
@TedShifrin I can type another similar type of problem ,probably simpler one.
So get rid of the logs, @Bhavay. What do you have?
@ArjunRana very -very bad
@Bhavay , i don't know you, speak up
@Bhavay: Here's a suggestion. Set $u=3^{1/x}$. What equation do you have if you remove logs?
The number of integral solutions to this equation- $4log_{\frac{x}{2}}(\sqrt x)+ 2log_{4x}(x^2)=3log_{2x}(x^3)$
@ArjunRana 85
20:18
@Bhavay: Let's finish the first one you gave me. I asked you to remove logs and make a substitution.
Sorry
@Bhavay , not that bad, which coaching?
@ArjunRana self study.
@Bhavay , which place?
@TedShifrin i know $3^{.5x}$ =t
@ArjunRana Delhi
20:20
@Bhavay , should have joined fitjee, my opinion
@TedShifrin But it requires some thinking and manipulation and then finding values of t , is there a procedural way to approach these kinds of problem?
@ArjunRana financial problems.
@Bhavay , sorry to hear that
Yes, you want to get rid of logs and get an algebraic equation in terms of a single variable. I said to let $u=3^{1/x}$.
Figure out why I said that.
@TedShifrin So that variable gets removed from the exponent ?
If you first remove the logs, you should get $$4\cdot 3^{(2x+1)/(2x)} = 27 + 3^{1/x},$$ right? Now try to simplify as much as you can.
You should end up with $3^{1/2x}$ and $3^{1/x}$. So maybe we should use $u=3^{1/2x}$. I was wrong. Why?
20:35
@TedShifrin i think u weren't wrong, it would just make calculations more messy.
Well, best to use the lowest power. Better to have $u$ and $u^2$ than $u$ and $\sqrt u$. :)
I am getting u= 3,9.
olas @TedShifrin
Good. Can both of those be of the form $3^{1/x}$?
@Thorgott that book you referred me to has changed my thesis for the better. I have been reminded how much easier life gets with the right literature.
20:44
Hola, @Sha.
@Bhavay: In particular, if one of your roots were negative, would your answer change?
I dont think so, as graphs of $\sqrt[Z^-]y$ exists .
The point is that $3^{1/x}>0$ for all $x\ne 0$.
@TedShifrin isn't $3^{1/x}>0$ always greater than 0 for the defined domain of x?
Well, they never defined a domain, so I assume the domain is all nonzero real numbers.
But you said the same thing I said :)
The solution states answer is $\phi$ , but we are ending up with two values?
20:56
Ugh.
Well, let's check. You got $u=3,9$, so $x=1$ and $x=1/2$. Go back to the original equation and check those.
Yes ,number of values of x=0
@TedShifrin x=1/4 and 1/2 sir.
Oh, oops :)
Yes, both those values check.
Guess the solutions are wrong.
@TedShifrin In a hint it's also states $\sqrt[x]3$ is only defined for x greater than 2, does it makes sense to u?
@TedShifrin Their or ours?
that's great to hear, @Sha, I'm glad the recommendation was helpful
Oh, see. I asked you what $\root x\of 3$ was supposed to mean. I think they want to allow only integers $\ge 2$ for that.
I personally would write $3^{1/x}$ for general real numbers.
So you have to understand what their notation means for starters.
See, @Thorgott: You were useful :P
21:03
@TedShifrin I don't understand. Isn't $3^{1/x}$ defined for all x?
All nonzero $x$.
Yes. But the point is that the symbol $\root x\of 3$ presumably is defined only for positive integers $\ge 2$.
@TedShifrin Yup.
@TedShifrin Why?
That's what your hint is saying?
How can domain be curtailed without mentioning it in original question?
In the US no one would write that expression, as far as I know.
21:06
@TedShifrin Yes.
It's a matter of knowing what that particular notation means.
Anyhow, you know how to approach the problem correctly now, so try the next one yourself.
So according to u , the correct answer is 2?
If I wrote the problem with $3^{1/x}$ instead of $\root x\of 3$, for sure.
hi, a @Balarka. You missed the fun time of my being sworn at in the room.
I actually kicked someone for the first time ever.
@TedShifrin Sir ,if u don't mind can u explain the difference btw the two?
21:10
@Balarka: Maybe you can translate for me :D
@Bhavay: I have said all I can. I would never write $\root x\of 3$ unless $x$ was an integer $\ge 2$.
I should have wondered why they used that notation.
@TedShifrin Please don't task me with this
LOL
OK, you're untasked.
How's your day?
Okay thank u sir.
@Bhavay: I recommend you learn general problem-solving techniques. One is to know the notation they (you) use.
21:13
@TedShifrin and then the flags on the post rolled in
:P
Try that other problem you typed.
@Thomas: And I didn't even flag it.
@ThomasWard I flagged.
yep
point not withstanding
lurks
That was over an hour ago.
Offensive posts shouldn't exist in the chatlog. That's the point of the flagging system.
21:14
I've encountered crazy people in here numerous times, but never that rudeness.
oh trust me you've not seen it :P
I always thought I was a nice person, but sometimes I do lose patience with people in here (after 6 years or whatever).
@Balarka check section 2.4 here: statslab.cam.ac.uk/~james/Lectures/pm.pdf
well it's handled now, so. shrugs and lurks more
The room apologizes to you, @ThomasWard.
21:17
Nah mods get worked up over non-offensive posts way more than actually offensive posts
This is good work
I don't think my blood pressure needs me to be a mod.
I don't think I've ever said a mean thing to you, Balarka. Maybe I should :P
@TedShifrin not necessary :P
returns to lurk mode
@TedShifrin I am trying to think what to instigate you with
LOL, like insulting Chern?
Maybe if I say your calculus book is worse than Apostol's engineering calculus text?
21:18
LOL @Apostol's engineering text.
God I hate that book
I am instigating myself at this point
Thanks @Thorgott!
Apostol's two-volume calculus book? How could you hate that?
I prefer Spivak, but it's a nice book.
self-inflammatory behavior is fun though isn't it @BalarkaSen xD
(I'm usually never active here, but since I'm here xD)
@TedShifrin I really couldn't digest Apostol vol. 2, from which the instructor taught us multivariable calculus.
It's not at all a bad book. Of course I prefer my own. :)
21:22
one day I shall learn about smooth toposes only to annoy you all by constantly talking about how and why they are definitely the only natural setting for doing differential geometry
Thorgott: You don't have to try that hard to annoy me.
He proves Green's theorem for type 1 and type 2 regions, after introducing type 1 regions, type 2 regions, type 1 and type 2 regions
It's nuts
shrug ... All semi-rigorous calculus books do this. I no longer possess the book, so I can't go back and look.
Hmm I see
I proved it just for rectangles, and then commented, in effect, about singular 2-chains.
21:23
Yeah, that's the best way.
But you're not biased :P
I don't have to, but I sure can (hypothetically speaking)
I think I did make a few explicit comments about non-obvious regions, too (like annuli).
Yup you did
21:25
@TedShifrin drops fifty generic chat-trolls into the chat just to annoy you
Gee thanks, @Thomas.
We already have too many
I never did want to be a room owner, anyhow.
Assume consciousness is the experiential manifestation of thought and that thought and physicality are dual mathematical objects in the spirit of Gromov. $F:P \to T$
21:26
Speaking of which ...
What did I say
I'm not trolling
although it's not in the spirit of gromov at all
@Balarka: You might like this question.
Hm, interesting. We know rational cohomology of Lie groups completely, $H^*(G; \Bbb Q)$ is just an exterior algebra on odd-dimensional generators. So the EMSS should tell something about cohomology of $G/H$, yeah.
21:35
âž•1
I bet the EMSS collapses at the $E^2$ page
Oh I see, I need to use EMSS on $G/H \to BH \to BG$. $H^*(BG; \Bbb Q)$ is a polynomial algebra on even dimensional generators.
So my guess will be $H^*(G/H; \Bbb Q) \cong \text{Tor}_{H^*(BG; \Bbb Q)}(H^*(BH; \Bbb Q), \Bbb Q)$
🧠🧠🧠
That should be computable
Thought you'd like the question :P
And we're back to every invariant form on a symmetric space being closed. :P
Yeah, $H^*(G/H; \Bbb R)$ is the "relative Lie algebra cohomology" $H^*(\mathfrak{g}, \mathfrak{h}; \Bbb R)$ by the same argument.
21:44
Well, something special has to happen in the symmetric case.
(Where you have the $\text{ad}(H)$-invariant complement of $\mathfrak h$, or something.)
Ah yeah $G$ is compact Lie in my last message
But yeah ad-invariant complement of h should be enough
I wasn't worrying about that. Lots of homogeneous spaces with compact $G$ that are not (locally) symmetric.
Of course, my favorite way to prove somebody isn't symmetric is to find a non-closed invariant form :P
Hm yeah
what is something that is symmetric
Spheres, Grassmannians.
21:49
Hey guys
I just discovered the chat
What flavor ice cream cone did you bring us?
I got all sorts of ice cream
which flavor do you want?
Chocolate hazelnut to start.
Wait so is that the kind of soft part of MSE?
hmm, huh?
21:53
It's fascinating that for a compact Lie group $G$ (or a symmetric space in general I suppose but I don't understand these well), its invariant forms complex determines the homotopy type of $G$.
I guess that is what the rational homotopists noted
Yeah, I think that's right, @Balarka.
When asking a question out there on MSE everything is sort of formal and like you can't joke about icecream like we just did
so that's why im a bit surpried
surprised*
yeah this is an informal offshoot chatroom
21:55
more glad actually, that I found this
Oh, I thought you meant it had to be soft ice cream, icecream :D
Humor and friendliness are encouraged. Rudeness, not so much :)
Oh, and math should be typed in MathJax. :)
Will do.
i'm going to stop writing latex just to annoy ted
one chocolate hazelnut ice cream for geocalc33 please
Hey, that's mine!
21:57
I LOVE THIS COMMUNITY
@TedShifrin Hello Professor. I solved the next question. I have another question, whose answer i know but i don't know why is the answer is answer to that question.
So, who are you, icecream?
@Bhavay: So you tried to do the new question and get a different answer?
Yes, different question different answer.
No, I mean you have a solution but it does not agree with theirs?
However the method remains same.
@TedShifrin Nope, it's a different question which i wanted to ask now.
22:01
LOL, I understand that. I'm asking whether you have solved it.
@Thorgott Im gonna try and learn BM
@TedShifrin Solved it.
OK, so what's the question and what's the main part of your solution?
It's more of a doubt.
wait whats BM?
22:04
@TedShifrin I'm a computer science sophomore at the American University of Beirut (rank: 220)
I have a question
LOL: You don't need rank (or serial number) here.
Suppose we have an quadratic equation in Tan(x) ex $tan^2(x) +tan(x) -3=0 $ (it's a hypothetical example).
$=0$?
\tan will help too
OK, sounds like what @Bhavay's been doing.
@Thorgott Black Metal
I mean Brownian Motion, of course
22:06
I don't understand?
Anyway, I used to neglect math in high school. I was one of those that are just - "math is just not for me".
But now, recently, I've started to take interest in it. I'm still a first year computer science student, and I want to give maths a second shot
Right now, I have summer, and I'm not taking any courses $specifically$ for that reason (that I want to start over again.)
Well, you need a certain amount of math for computer science (especially discrete math proof skills).
I have almost 120 days
Yep
I'm willing to spend a lot of effort on this (I'm serious about it).
How can I make the most effective use of my time?
You need to take a year of calculus?
Nah I've taken that course
22:08
Oh.
So your high school math skills (algebra, trig) must be decent.
ah, I see
tough luck man
I've never really gotten into stochastic processes
Yeah I have good skills with algebra, but trigo: i have skills, but i used to rote learn (I know how to use stuff but I dont know what would be going on)
i found some noob probability notes on BM which is more geometry focused (end goal is Chern-Gauss-Bonnet) so hopefully it'll be ok
I even passed the calculus course with not really knowing whats happening
And then I read this:
what does BM have to do with chern-gauß-bonnet?
22:10
theres a probabilistic proof apparently
There are beautiful ideas in calculus and linear algebra, @icecream. If you want to get into computer graphics/games programming, you want some linear algebra understanding for sure.
@TedShifrin can u ping me when u r fnished with chocolate hazelnut?
"It is foolish to answer a question that you do not understand. It is sad
to work for an end that you do not desire." ~ Polya
wow, that sounds crazy
@Bhavay: I was waiting for you to post your question. I am used to multi-tasking.
22:11
This is from the book How to Solve it (G. Polya)
I think we'll cover Chern-Gauß-Bonnet at the end of my manifold lecture this semester
Damn, cool
I never learnt it
I feel guilty. I passed Calc II without understanding it.
A shame
Very cool, @Thorgott. One of my favorite proofs for graduate geometry courses. I've presented a few different proofs.
You're probably not alone, @icecream.
@Balarka: You have a few years left.
22:13
Now the above typed equation will have roots say $\tan(\alpha)$ and $\tan(\beta)$
yeah, we're currently doing de Rham cohomology and the remaining topics are Chern-Gauß-Bonnet and some elementary aspects of Hodge theory, it seems
really exciting
I'm going to try to come up with a proof of CGB. It must come from understanding the top Chern class (the Euler class) by reading it off of the characteristic polynomial for the connection.
How hard can it be, I should be able to prove it
Grr
LOL, @Balarka. Well, Chern was only a bit of a genius.
maybe I'll have you tell me the probabilistic story about BM some day
Of course, you know stuff he didn't know at the time.
22:15
Yeah, I know lots of my friends who got 95+ but without actually understanding the course. They just got that high because they memorized the questions and how to solve them. BUT im not like them. Well, maybe just recently I've been trying to change.
@Bhavay: You should approach that problem the same way we were discussing earlier. What should you immediately do?
What i don't understand why the roots will just be $(\alpha)$ and $(\beta)$ if the equation was something like $ x^2+ tan^2(x) +tan(x) -3=0$
@icecream: It is an indictment of mathematics teaching that students can do well in courses merely by memorizing and regurgitating.
@Thorgott I've set up some reading project with a few friends. So by the end of it hopefully I'll understand
@Bhavay: That equation one cannot solve. You can't mix in polynomials along with trig/exponential and solve without computers/calculators.
22:16
nice
@Ted
oops.
@TedShifrin ooh!
understanding is imperative in mathematics and the good thing is that understanding in almost all cases subsumes memorization
You should figure out my "embedded" proof as well. For bonus, @Balarka, figure out the boundary term for manifolds with boundary. I've actually not done this in courses. Chern's original proof did this.
@TedShifrin yeah, I know
22:18
@TedShifrin Yeah you're right, I should do the extrinsic proof. Let me see. I'll doodle around for a few days and tell you what I have
Anyhow, if you're going to take a discrete math class for your computer science degree, you might like a book like "How to Think Like a Mathematician," by Houston, @icecream.
I have around $123$ days. How many problems should I aim for? (per day)
Or I think maybe I'm asking the wrong question. I dont know
I need your help guys
This is more about language and proofs, less computational.
More about the thinking skills.
I think aiming for a specific number of problems or whatnot isn't particularly helpful
you can solve a 100 problems without learning anything, but you can also solve a single problem and gain tremendous insight
22:21
Nice
Yeah, thoughtful mathematics is different from repetitive high school mathematics.
its both a matter of what problems you're doing and how you engage with them
I see
The reason I said problems is
My math grades in high school were not that good
And so
I think I lack some confidence/self-esteem
Which I may need to regain by doing problems
or is there a better way to rebuild confidence?
If you really want to go back through high school algebra/trig stuff you can, but I'm suggesting a new start that will be more helpful going forward. It doesn't hurt to do a few algebra problems every day "for fun."
Oh okay. Sorry if I sound desperate :p
22:25
Nah.
Just hard to give universal advice.
It's alright.
It's hard to speak broadly, but I think confidence should ultimately come from understanding. What comes from doing certain standard computations over and over is familiarity. Both are important.
I support that message âž•1 ^
NICE, so I need confidence and familiarity
confidence from understanding, and familiarity by practicing, yeah?
Hey, actually, I notice myself getting confident sometimes when I understand a problem. I feel like I did something. I guess understanding is the key.
Yup. Sometimes having someone to explain things to also is helpful for building understanding and confidence. Maybe you can find a (remote) study partner :)
22:34
Where?
I was thinking among your classmates/friends, @icecream.
Hmm
Consistency is important as well, eh?
consistency is the collective action of a sequence of events unified globally by practice
that wasn't a very good definition actually
question #1: Can a machine analyze a sequence of points on a unit disk and tell a mathematician a level of symmetry of the point distribution
like maybe you have 3 points
and the computer says 0.3
in other words, a scalar value provides a measure of the point distribution
and maybe you say boom
124 points
and the computer says okay 0.6
and then you say take 45 points computer. and the computer says okay 0.8
question #2: How would one implement this on a computer?
Here the input is an image
and you upload it and then the computer takes the image as input and then takes time to process it and then it gives an answer
I think Mathematica is the best option
And here's my ultimate question...
Ultimate Question: Can someone here with Mathematica upload an image provided by me to Mathematica and perform just one single operation on it?
22:56
You can't perform operations on images. You need the actual source for the image.
ah okay
actually I might be able to get Mathematica from my school
Is
$log_xx \ always \ 1$ ? (obviously in the defined domain of this expression)
What do you think?
Yes.
23:20
When I have something like $log_k(x)$ I usually say that $k$ to the power what gives $x$. So yeah, $x$ to the power what gives $x$?
That's good, @icecream.
Thanks @TedShifrin.
Logarithm is a tricky concept for many people. A biologist relative asked me what the point of it is; I had to explain it to her using infection growth.
Yeah, biological growth and pH are things you would think she would know.
Yeah chemical equilibrium is a good example
23:26
Well, that's one step harder than pH.
True.
For computer scientists the point is data is written in binary, and there are $2^n$ binary strings of length $n$.
Thus the famous joke that $\log$ for mathematicians is log base e, $\log$ for biologists is log base 10 and $\log$ for computer scientists is log base 2
I don't know why they did pH log base 10.
Yeah it's more natural to do e, because that's what solves $y' = y$
I guess they don't care about constants
My dad says that when mathematicians say $log$ they usually mean $log_10$
and 10 is a better number than e
23:29
$log_{10}$
Never, @icecream.
Hm wonder what economists do, probably e
Nobody really understands e anyway
Weirdo number
It surprises me that u would say that lol
I'm just being honest!
Its a good number to be clear. Just a little mysterious
23:32
yep
I motivate it with infinitely compounded interest.
So then it's natural.
Pun intended.
Yeah that's good
It appears in many nice limiting scenarios
(Proportion of permutations which are derangements, say)
Yeah, that one is a mystery to me.
@TedShifrin It's fun to derive Euler's $e^{ix}=\cos(x)+i\sin(x)$ using $\lim\limits_{n\to\infty}\left(1+\frac{ix}n\right)^n$ in a geometrical setting.
23:35
$e=\prod_{n=1}^{\infty}\left(1-\frac{1}{\tau^n}\right)^{\frac{\mu(n)-\varphi(n)}{n}}$, where $\mu$ is the Möbius function, $\varphi$ is the Euler function and $\tau$ is the golden ratio
2
no questions left
Hmm ... I'm not sure I know what you're referring to, @robjohn.
@Thorgott Definition 100
Makes everything obvious
Top 100 definitions of $e$. Number 72 will surprise you!
@TedShifrin Here is one of my answers.
Ah, I see. Very cute. I've never thought to do that.
23:39
$\sum_{n \geq 1} 1/n!$ is a good practical definition. It says that $e$ can be approximated VERY WELL by rational numbers
Which, ironically, implies it's irrational
Yup, I even stuck that in my algebra book :)
It's funny how $\pi$ is harder to prove irrationality of but it's a more natural number
Math is junk
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