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7:00 AM
Any of you know anything about fractional calculus? It's come up in some papers I've had to read and I can't find anything solid about it.
Besides the wiki, which hasn't helped too much with some things.
 
lol
 
@KajHansen struggle life. but it's rather common in this age with them 'darn techno-gizmos
@Fargle Funny. It's just past 2 am here, and I have to wake up some time before 7. I'll probably all-night it though
 
Ugh
 
Deleting an answer doesn't remove the downvotes you received does it?
 
@Brody, a lot of it was that school was taking up a very, very large chunk of my life, and I felt like I needed time to do things that I was interested in. I did not like school very much at all.
 
7:09 AM
@KajHansen I see. so your sleep took the hit, eh
 
Yeah
 
I'v spent the last 3 hours making this answer better but I'm starting to wonder If I should just give up or not.
 
It shouldn't be mandatory
 
@Balarka I slept for five hours but I did get 18 pounds back and so I feel great. Maybe try getting a refund.
 
I could've passed the GRE in 8th grade.
It's a little infuriating actually, thinking about all the squandered time :/
 
7:11 AM
Impressive
 
yeah, they could really work on accessibility to higher-level programs for gifted students
 
@MikeMiller Haha
 
I think I'm an idiot and that's no joke. You people are sometimes way younger than me and know eons more and yet here I am rambling about nonsense nobody cares about. Perhaps I'm just not meant to be on this community I realize that now. Sorry for annoying you all these weeks and for wasting your time. Goodbye.
 
Most people feel that way
 
7:16 AM
Don't leave please @TheGreatDuck
 
How did you lose them in the first place?
 
Most people that know enough to not be Dunning-Krugering themselves at least
 
Hey there
 
hey
 
7:18 AM
Hello
 
@KajHansen yeah but all I do is ask tings that are inevitably vague and unclear anyway. I have nothing to contribute here.
 
@TheGreatDuck Literally that's exactly how I feel.
 
That takes practice to improve on
 
Definitely. A big part of the craft of mathematics is learning how to ask the right questions. It just takes time, and asking a lot of questions.
 
Having experts to interact with in person helps too. I'd be a lot worse off without people like Ted. I really sort of doubt I could've improved anywhere near as quickly alone.
 
7:26 AM
@Kaj I don't consider all the time I spent playing video games to be squandered.
 
Yeah except I occasionally have the uncontrollable urge to just want to rip everything apart. Figuratively that is. I mean, there's questions I ask that get a bit of attention and then I just feel bad for wasting people's time like that.
 
In fact, I might play more this morning.
 
I don't have time for video games. Math is far more important.
 
The fact at least three or four frequents in this chat are middle to high school age and know significantly more than me can be inhibiting or it can be motivating. It really is an attitude thing.
 
Neither do I actually @Mike
 
7:27 AM
I find it to be neither @Brody. It just is.
 
Yeah but I mean this site is only for people who know real analysis. So im not supposed to be here.
 
The most inhibiting thing right now is that I don't have my laptop.
 
Why not?
 
On the bright side you don't have to write
or at least have a good excuse not to
 
Because only people who know real analysis are supposed to use the site.
 
7:29 AM
I dunno analysis
 
i was point blank told to either learn it right away or leave.
 
I left it at airport security. I'll be able to get it again on a tuesday.
 
Duck feathers, ink and parchments, like the greatest mathematicians!
 
LOL, what's that all about @TheGreatDuck ? I'm pretty shit at analysis myself. I know the basic dealios behind differentiation, integration, uniform continuity & convergence....and that's about it
 
I had an idea I want to work out but I can't exactly find references for related things ::
 
7:30 AM
My posts are like 40% algebra
 
@user1952009 told me this site was only for the level of real analysis or higher?
 
@KajHansen Ted is a great teacher. But I am too ashamed to ask my generally stupid questions to him
 
Me too Balarka :P
Heck, I was ashamed to come to office hours
 
Idk. It seems lately like everything I do on this site has a negative impact.
 
Ted's just too good at spotting nonsense said by me
 
7:32 AM
@TheGreatDuck Fortunately, that user isn't the gatekeeper of this site. We don't discriminate.
 
Was perhaps a good thing in its own right though. I was motivated to figure things out myself
 
@KajHansen so you are his student?
 
@Ramanujan ted is retired.
 
@KajHansen Yeah.
 
I took linear algebra/multivariable analysis and differential geometry from him a few years ago
 
7:33 AM
:/
 
Did you ace it @KajHansen if you don't mind me asking
 
Idk. I think I just really need to think about why I'm here. I want to know cr
ertain things and I ask a lot of questions but on the flip side it seems like I just keep undermining myself.
 
You really shouldn't be ashamed of asking questions. The job of a good student is to be stupid long enough to learn not to be stupid.
10
But how are you going to do that if you're never willing to be an unabashed idiot?
 
"Absolutely not" would be the answer for the second semester of that multivariable analysis @thoughtforfood. It got incredibly abstract. Doing vector calc with differential forms instead of the traditional route on very-hard-to-visualize stuff (sometimes >3 dimensions) and got into stuff like proving the Hairy ball theorem & Brouwer fixed point with techniques we picked up.

I did not-horrible in first semester though, and not-horrible in differential geometry. I have a general difficulty with visualizing things though. A lot moreso than most people it seems. I'm very depended on mathe
 
You're right. I think I really am going to ask him about moving frames today.
It's all very confusing for me.
 
7:42 AM
I don't think it's abstract at all! But it does take time to understand it.
 
I personally like frames on a curve
 
@KajHansen interesting...
 
I see where you're coming from @MikeMiller. I guess it was abstract for me, someone fresh out of high school and hadn't seen a math course in 3 years
 
Though I probably have no clue what I'm talking about anyway.
 
@KajHansen I think I am quite good at visualizing stuff because topology is my arena, but I have hard time doing anything with diff. forms, and differential geometry.
 
7:43 AM
I think it would've been a lot easier had I taken a Spivak-oriented single variable course
 
That'll do ya.
 
@KajHansen like a transition buffer?
 
Ted won't believe me, but I don't like abstraction. You need a reason for everything you do.
 
There's something called the Frenet frame @TheGreatDuck. Interesting stuff really
 
I think any good concept has a clean and comprehensible origin and meaning.
(My usual phrasing for this is: I don't believe in magic.)
 
@Mike reminded me of just that
 
Every time I tell someone my slogan I get that song stuck in my head.
 
Yeah @Brody. For example, Riemann integration in $n$ dimensions is a natural extension of Riemann integration in single variable
 
I think if there's ever a rapture, I'm going to find someone else still around and start freaking out: "Everyone's gone! Everyone's gone!"
"Everyone's gone... surfin. Surfin USA."
 
Well @Kaj, you took the courses I wish to have taken. I would've loved the more theory-intense courses at UGA, even if they'd kill me
 
7:54 AM
@KajHansen I know about the Frenet Seret
@MikeMiller that's a horrible joke
 
That's awesome @TheGreatDuck. Have you looked at any surface theory? That's where frenet applications get interesting
 
Nah
I just made a half-broken tube plot in the graphics classs
I have not taken differential geometry
im undergrad
 
I will be the first person murdered (as opposed to accidental death) as a consequence of the rapture.
 
Oh? Why's that?
 
Are you the antichrist or something?
 
8:00 AM
I just meant the above joke.
 
Cause last I checked everyone keeps putting Donald trump on that list.
granted
those people are idiots
 
@MikeMiller Hey, get anywhere on that puzzle?
 
They are the majority.
Like it or not
 
God may perhaps smite you Himself for that, @Mike
 
No I mean the people who literally believe Donald trump is the antichrist
@MikeMiller I'm pretty sure the first murders will be from the serial killers realizing they have nothing to stop them from rampant killing. Honestly I wonder how many people would die solely from the sheer insanity of people flipping out and going on a killing spree.
I'm too cowardice for that. Randomly deleting my favorite posts is enough chaotic fun for me. Though I suppose it's not fun after the fact.
 
8:13 AM
It's interesting how much you can learn about a book from just browsing the index :P
 
@thoughtforfood I've thought that from time to time too; but usually upon closer inspection I haven't learned anything from it.
 
It really depends how much effort the author has put into it, no? @user400188
Also the table of contents, of course.
 
@thoughtforfood I was under the impression that index was just the title and location of somehting in a publication. I guess if you added a description of the subject in the index it would teach you something.
 
8:30 AM
Descriptive titles is what I was talking about @user400188
and descriptive phrases too
 
user228700
8:42 AM
Hello, everyone :-)
 
user228700
I have a quick question about the composition of functions. For two functions $f$ and $g$, which are inverse of each other, how is it that $f\circ g= g \circ f$?
 
@Kaumudi.H By definition of being inverse both of those are the identity
 
user228700
$f \circ g$ gives $y$ where as $g \circ f$ gives $x$, no?
 
Domain and range of $f$ and $g$ are same in your case, yes?
 
Ohh, I assumed that they were function from some set to itself
otherwise clearly they will never be equal
@BalarkaSen You want a hint on that puzzle btw?
 
user228700
8:46 AM
@BalarkaSen Dyou mean to ask "Domain of $f$ = Domain of $g$ etc., yes?" ?
 
I really gave up on it, but sure.
@Kaumudi Yeah. $f, g$ are supposed to be functions $A \to A$.
 
@BalarkaSen Remember that the puzzle also has a name which might give a hint to what it is about
 
user228700
@BalarkaSen Right. OK, it makes sense, now. Thanks :-)
 
@TobiasKildetoft Had to google it. Sounds like a technical terminology in cinematography.
 
8:59 AM
@BalarkaSen Ahh, I had not actually looked up the term, I just went with what it wounded like it should be about
 
I haven't thought about it; rather I've been thinking about more mathematical puzzles
 
Those are probably time better spent
 
user228700
I'm afraid I have a bit of a homework-tsy question.
 
user228700
I'm trying to find whether or not the functions $y=x^2+2x$ is onto.
 
user228700
The domain of this function is restricted such that it is bijective: $x \ge -1$
 
user228700
9:11 AM
Since there is a minima at $-1$, it is safe to conclude that this function is monotonically increasing from this point, because of which we know that this function is one-one.
 
user228700
Now for the onto part. So I've learned to do this by finding a function that gives me $x$ when I input $f(x)$ into it, such that $x \in$ Domain of $f$.
 
user228700
Usually I just do this by switching around the terms and manipulating the original function.
 
user228700
In this case, completing the square and all, I get $y= \pm \sqrt{x+1} -1$ as that function.
 
You just literally need to know when a solution to $x^2 + 2x = a$ exists, given an $a$ in whatever codomain you have.
 
user228700
...but you'll be quick to notice that that, in fact, is not a function because of the $\pm$.
 
user228700
9:14 AM
@BalarkaSen Hang on, let me really quickly finish.
 
user228700
But, OK, hear me out. My argument is that look, I manipulated the original function to get $x= \pm \sqrt{y+1} -1$
 
user228700
But the thing is, the minimum of the original function is $-1$, for which I get $x=-1$ and then for $0$, I get $0$ etc. I'm never going to use that $-$ case because including those will give me $x$s which aren't in the domain of the original function.
 
user228700
...by which I conclude that the negative sign shouldn't be there and that yes, I've found a function so yes, it's onto...aaand I'm starting to suspect that this whole line of reasoning is wrong so...is it?
 
user228700
Oh crap.
 
I don't even understand your line of reasoning. The point is simply that a solution to $x^2 + 2x - a = 0$ exists iff $(2)^2 + 4a \geq 0$, aka, $a \geq -1$. I don't know why you're trying to find an inverse function - onto functions don't always have inverses. This is a special case because it's bijective.
 
user228700
9:22 AM
For $0$ I get either $0$ or $-2$ and $-2$ isn't there in the domain of the function so wait, what am I supposed to do? Just ignore that negative sign?
 
@BalarkaSen you need $\geq$
 
Fixed.
 
@Kaumudi.H plot a graph
 
user228700
@BalarkaSen I'm still trying to understand the first part of the message but as to the second part, I know it's one-one. I'm trying to prove that it's bijective so I can find the inverse function.
 
user228700
9:24 AM
@DHMO I won't always be able to do this in my exam, so I'm trying to figure this out without having to plot a graph.
 
@Kaumudi.H plot it by hand
 
I agree that you should plot. It's very easy to do it by hand.
 
Learn the skill^
 
user228700
Sure, yeah, but even then, it probably won't give me the whole picture. There may be holes, asymptotes, etc.
 
@Kaumudi.H learn to tell if there are holes, asymptotes, etc.
 
9:26 AM
Hello!! I want to show that the function $f:[0,\infty)\rightarrow [0,\infty), f(x)=(x^{\frac{1}{3}}+x)\sqrt{x}$ is bijective.

So, we have to show that it is injective and surjective.

To show that the function is injective we have to show that for $f(x)=f(y)$ it follows that $x=y$.
We have that $(x^{\frac{1}{3}}+x)\sqrt{x}=(y^{\frac{1}{3}}+y)\sqrt{y}$. What can we do now? Do we square that equality?

Then to show that the function is surjective we haveto show that for each $y$ there is a $x$ such that $f(x)=y\Rightarrow (x^{\frac{1}{3}}+x)\sqrt{x}=y$.
 
It's a parabola. There's no holes and asymptotes.
 
@MaryStar show that it is monotonic
Actually, $f(x) = x^{\frac 5 6} + x^{\frac 3 2}$.
 
user228700
@BalarkaSen ::Facepalm::. Dude, I know. Never mind plotting for now. Can u please tell me if the way I did it makes sense?
 
$f'(x) = \dfrac 5 {6 x^{\frac 1 6}} + \dfrac {3 \sqrt x} {2} > 0 + 0$ (assuming $x > 0$)
 
It didn't make sense to me, but I also don't care to try and read that whole thing. Maybe someone else will help.
 
9:28 AM
you can deal with the case $x=0$ separately.
 
user228700
@BalarkaSen I don't understand the point you're trying to make by saying that $a \ge 1$.
 
@MaryStar Actually, since both $x^{\frac 5 6}$ and $x^{\frac 3 2}$ are monotonic, it follows that their pointwise sum is also monotonic.
 
hi, if i want to proof marginal independence and conditional dependence, is the understanding below correct:
 
@Kaumudi.H Simply the thing that for $a \geq 1$, there is a solution to $x^2 + 2x = a$. That's your proof of surjectivity.
 
user228700
@BalarkaSen Yeah, I don't get why that's my proof.
 
9:30 AM
P(A=a, B=b) = P(A=a).P(B=b) for marginal independence?
 
It's not your proof, it's just the proof I am giving.
"your" was not meant to be literal.
 
user228700
:-P Yeah, no, why does that count as a proof is what I'm asking.
 
Because having a $x$ in domain such that $f(x) = a$ for all $a$ in codomain is precisely what onto means?
That's what I showed.
 
and P(A=a,B=b|C=c) does note equal to P(A|C).P(B|C) to proof conditional dependence? Thanks
 
(your domain and codomain both are [-1, infty), right?)
 
9:32 AM
Ah ok. Could we justify that the function is monotonic as follows?
For $x<y$ we have that $x^{\frac{3}{2}}<y^{\frac{3}{2}}$ and $x^{\frac{5}{6}}<y^{\frac{5}{6}}$. Therefore, $x^{\frac{3}{2}}+x^{\frac{5}{6}}<y^{\frac{3}{2}}+y^{\frac{5}{6}}\Rightarrow f(x)<f(y)$. So, $f$ is monotonically increasing.
 
user228700
@BalarkaSen Yep.
 
@MaryStar yes, as long as you can justify the first statement.
 
@DHMO You mean that $x^{\frac{3}{2}}<y^{\frac{3}{2}}$ and $x^{\frac{5}{6}}<y^{\frac{5}{6}}$ ?
 
yes
 
user228700
@BalarkaSen Ah, OK. U said everything I was trying to say before much more eloquently. Thanks very much.
 
9:35 AM
@DHMO How could we justify that these functions are monotonically increasing? Using the graph of these functions? Or otherwise?
 
@MaryStar He already write up the derivative and showed it was positive
 
@MaryStar what course are you taking now?
@TobiasKildetoft a more primitive method would be to use the axioms of ordering...
 
Good morning @DanielFischer :-)
 
morning
 
How's it going?
 
9:39 AM
Cold and slowly.
 
Have some tea :)
 
or coffee?
 
Or coffee. I like tea tho
 
me too
lots of sugar tho
 
@DHMO I am not taking a course right now.
So, having that $f$ is strictly increasing, we get that the function is injective, not bijective, or not? @DHMO
 
9:44 AM
@MaryStar then where did you get that question?
@MaryStar check that $f(0)=0$ and $\lim\limits_{x\to\infty} f(x) = \infty$ and it's bijective.
 
@DHMO I find them online.
@DHMO Why does this hold?
 
well you need continuity also
then it covers every value exactly once
 
10:43 AM
Rather off-topic (or at least strange), but I figured you might be interested. A friend of mine (a Historian) is working on old letters by 17th century Dutch politician and also mathematician Johan de Witt. She came across this doodle looking this on the back of one of the letters i.imgur.com/gh3EKOt.jpg and was wondering if I had an idea of what it could be/what he was thinking of. It doesn't look like all that much to me (maybe a lens?) but perhaps it rings a bell for one of you
 
My first impression is similar triangles.
 
How many more ways can 10 juniors running for the positions of president, secretary, and treasurer be selected when compared to 12 sophomores running for 5 identical positions of class representatives
I did 10 c 5 - 12 c 5
But i got something that was not in the multiple chlice
 
@thoughtforfood Yeah, I don't know. I personally think it might just be some doodle or whatever, the middle part doesn't look like something that has meaning, but it would still be interesting to have an idea of what he was thinking of at the time.
 
He could be thinking about some sort of geometry problem.
Perhaps perspective.
 
I have a question: do two parallel lines define a plane? a unique plane? I think yes, And I prove it by showing that there is only one normal possible. is it correct?
 
10:54 AM
I'll ask about what was written on the front of the page, to get some hints.
 
Hellow is it true>
?
 
Well, aren't "parallel lines" defined as two lines in a plane that do not intersect?
 
Indeed, non vertical lines that have the same slope.
 
@thoughtforfood why would it include non-vertical?
 
11:07 AM
Because a vertical line has no slope.
 
Or slope infinity
Depends on how you define slope and in what space you're working
 
just the elementary rise/run
:-)
 
if this were in two dimensions, then the entire question of the lines forming a (unique?) plane becomes very uninteresting
 
Yup
And in three dimension, "slope" is way harder to define.
 
and in more than two dimensions, you need more than a single number as the "slope"
 
11:09 AM
But, anyway, two parallels do define a unique plane.
 
@SteamyRoot Actually, once one defines the slope "correctly" it generalizes to any number of dimensions directly
 
Well, I imagine you can do that, but who decides what the "correct" generalisation is :P
 
rise/run is not "correct"?
:P
 
@SteamyRoot the one that immediately generalizes and has lines be parallel iff they have the same slope
 
Hmmm... I guess
@thought "rise/run" is something relative. You rise/run compared to what?
 
11:12 AM
The axes.
 
@SteamyRoot So you take the unit vector in the direction of the line and require that the first non-zero coordinate is positive
 
Hmmm, sure. I've just never heard anyone call that "the slope"
 
hi, anyone here familiar with probabilities? how to best understand the meaning of marginal dependence, conditional dependence, marginal independence and conditional independence given a contingency table?
 
Wouldn't you also want the first non-zero coordinate to be normalised (i.e. $1$), though?
Oh, nevermind, you used a unit vector.
 
@SteamyRoot either one works
and sure, calling it the slope might not be standard, but it is the closest one gets
 
11:18 AM
:P
In two dimensions @TobiasKildetoft can you say two different vertical lines have the same slope?
 
I get what you are talking thats what we have direction ratios for!
parallel lines so direction ratios are same
 
Indeed.
 
@thoughtforfood Yes, I would say that two vertical lines have the same slope
since rotating everything should not change this fact
 
11:33 AM
parallel lines are lines in the same plane that do not intersect @samjoe
 
Sure, in Euclidean space
 
If you want to include slope in your argument, you must recall not to divide by zero, ie the vertical lines case
@samjoe
Also when you're using perpendicular lines you can't use the product of their slopes equaling -1 if the lines are horizontal and vertical.
In Euclidean space, as Tobias said :-)
 
12:35 PM
math.stackexchange.com/questions/2104428/… don't we have to reckon with no aces being anymore in the deck? or cancvels this out with the chance that all aces are still there, and therefore more likely to draw?
 
> 12 cards are removed from a fair pack of 52 cards. None are aces.
 
meh
thanks, and bye hehe
 
Hey guys
 
user228700
Hi, again :-)
 
user228700
I have a quick question about proving the surjectivity of functions again.
 
12:47 PM
i can do that, i think
 
user228700
I'm trying to prove that the function given by $f(x) = -{(x+1)}^2 -2$ is surjective.
 
what's the function again. the formatting is all messed up
 
@Kaumudi.H no it isn't
 
@ShankRam There is a link on the right on how to get MathJax to work here
@Kaumudi.H on what codomain?
 
user228700
@DHMO Right? My textbook seems to think that it is.
 
12:49 PM
+1 on co domain
 
user228700
Oh, hang on, how did you, @DHMO: Figure that out without knowing what the codomain is?
 
user228700
This is it: $(-\infty,-2)$
 
@Kaumudi.H well you didn't say the codomain so it defaults to the reals
 
well, without the co-domain, it is not possible to say for sure
 
@Kaumudi.H then the function isn't defined when x=-1
 
12:50 PM
quadratic is not surjective on all reals
 
user228700
Not all quadratics but this one, yes.
 
here, since the co domain is (-infinity, -2), you can say it is surjective
 
all quadratic is not surjective on the reals
 
-(x-1)^2 varies from (-infinity , 0)
 
no it doesn't
 
user228700
12:52 PM
Hang on, let me explain what I did. I followed along the lines of Balarka's proof earlier today.
 
oh, my bad
dhmo, it does
 
no it doesn't
 
(x-1)^2 is a concave upwards parabola with vertex at 1
just flip it. what does it vary from?
 
well, from 0
 
oh, okay. (-infinity,0]
 
12:54 PM
yes
 
user228700
The equation $y= -{(x+1)}^2 -2$ will have real roots of $x$ only if $y\ge2$ (Unless I have made a silly calculation error somewhere).
 
don't go along those lines
 
@Kaumudi.H no, only if $y \le -2$
 
let g(x) be -(x-1)^2
 
user228700
@DHMO Argh, I've made a mistake then.
 
12:55 PM
find the interval of g(x)
subtract 2 from that
 
First, consider the negative of the function instead to stop confusing yourself and all of us with positive versus negative signs
 
i meant on both the end points of the interval
 
and replace the codomain by the set of negatives of its elements
 
user228700
@DHMO I have absolutely no idea what I did wrong >.<
 
now, if your co domain matches that, the function is surjective
 
12:56 PM
@Kaumudi.H and I have absolutely no idea what you did
 
lol
 
user228700
@DHMO Lol, hang on.
 
@Kaumudi.H Are you familiar with the intermediate value theorem?
 
@TobiasKildetoft i personally think that will only make stuff complicated at this level as continuity and differentiabilty starts after functions, according to indian syllabus
i assumed she is indian
 
@ShankRam no need for differentiability here
but sure, if he has not heard of continuity then this is a bad direction
 
user228700
12:58 PM
@TobiasKildetoft I googled it and now I'm wondering why this had to be explicitly stated.
 
yes, that would probably be the case
you need to first prove the function is continuous
 
@Kaumudi.H You mean the intermediate value theorem?
 
user228700
@TobiasKildetoft Yep.
 
user228700
@DHMO And u arrived at this how..?
 
Because it could fail if this was something other than the reals (such as the rational numbers)
 
12:59 PM
you acctually dont need that for proving surjectivity of a function
 
@Kaumudi.H its vertex is (-1,-2) and it opens downwards
 
graphing is one way to solve this problem
probably the easiest
 
user228700
@DHMO Right, well I tried to do it algebraically by trying to find all the values of $y$ for which the aforementioned equation will have real roots; $b^2-4ac \ge 0$
 
@Kaumudi.H show me your steps
 
user228700
OK, let me double-check and take a photo...
 
1:02 PM
@ShankRam graphing is not proving
nor is resorting to the codomain of x^2
 
user228700
Yep, silly calculation error it is.
 
user228700
Alright, thanks, guys :-)
 
@Kaumudi.H that is also a way to do that. since x belongs to reals, you can say the discriminant is >=0 and you will get the range of the function. if that is the same as codomain, the it is onto
*then
 
user228700
Um, yes.
 
are you using that calculus book by scp?
*DCP
 
user228700
1:04 PM
Nope.
 
well, that is a real good one if you are looking for reference books (mainly for exam preps)
 
user228700
I see. Thanks.
 
Perhaps not the most academic of questions, but I have some data (i.imgur.com/oidCweM.png and i.imgur.com/Sr1N8p7.png are two zooms) and I'm trying to guess a functional dependence. It looks more or less like it is initially linear, and then smoothly transitions into another linear part with a smaller slope. It doesn't look like something familiar on LogLog to me, but I did find that $\frac{a}{1+b/x}$ is pretty close; just not quite.
 
Hmmm
sqrt?
 
Sqrt rises too quickly initially it seems, it's a bit too concave
 
1:12 PM
Hmmm, okay
 
I suppose context might be relevant tho. It's a solution to something with a finite bandwidth; for infinite bandwidth it has the initial slope all throughout, while when the bandwidth is comparable to the amplitude we get this second regime i.imgur.com/S6ZLt4D.png
 
You can always take the life sciences approach. A straight line near the origin, a straight line away from the origin, and the point where your data "bends" becomes a "transition value" :P
 
Yeah, that is definitely true
 
1:27 PM
I will personally be quite interested in that round bend near 0.1, whether there is some kind of "phase transition" going on involving your system
 
+1
 
ok nvm, seemed steamyroot have spelt out a possible mechanism
 
@Secret hi
 
hi
 
Does every endomorphism in linear algebra have a matrix representation
 
1:33 PM
Yes
Between finite dimensional vector spaces
 
What about infinite
 
@Secret have you proved that the derivative of x^n is nx^(n-1) before?
 
only via first principles back in year 1 and high school via limits?
I am not sure whether they have gone through the real case, as we have gone through integers and fractions
 
@user379685 Still yes, but you need to expand the notion of what a matrix is to something much less useful
 
@Secret Indeed. I can intuitively explain the deviation from the initial trend, but it is a quantum physics problem, so I'm not sure if you'd be interested. It has to do with timescales becoming comparable, so the fact that there is a change is not surprising. The form is a bit odd though, because it just goes from a1*x to a2*x where a2<a1, which indeed is more in the direction of a phase transition
 
1:39 PM
@Secret turns out the real case is quite simple if you know the definition
 
DHMO: Well first thing that came to mind will be basically repeating the proof on a cauchy sequence (since these define the reals), but I suspect you are talking about something even simpler than that
 
hello
what are you talking about?
 
@Secret yes I am
how do you define power?
 
in the abstract algebra level, it is defined by a certain number of times of multiplying the same element n times
 
well, how do you define 2^pi?
 
1:46 PM
for that we can use exp
...wait, so you are suggesting $x^n=e^{n\ln x}$ and then d/dx this by exploting the fact that e^x is an eigenfunction of d/dx?
 
yes
 
hey
I would appreciate some ideas here
im thinking about two complex varieties such that their product has class group distinct to the product of their class groups
 
@user129412 I like quantum problems. Unfortunately my background on that is not very strong thus depending on the specifics of the problem, I may or may not be able to give any advices. Your data, as interpreted by stemyroot, seemed to suggest you obtain these from some laser tuning experiment. Are you working on some quantum computing or solid state matter type research?
DHMO: Ah I see, yes, that's indeed a way to go
 
Currently my schedule is free but plain: In the coming week (more or less), I am going to be preacticing for that driver knowledge test for a license and it is so bornign especially some laws simply cannot be derived by doing maths
 
Apology. The study of ap.
 
1:58 PM
someone should write an article about appology
 

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