9:00 PM
but i shouldn't interrupt your line of argument in any case

@SemiC I would have choices of all elements of the set
i need 5 elements

okay. so you'd have 10 choices for the first element of the permutation.
(shhh overcounting)

they can't be the same set but can have 4 of 5 of the distinct elements right?
so 9?

right. what about the third, fourth, and fifth choices?

if order doesn't matter: 8,7,6
if order does matter the choice will be 10 5 more times?

9:02 PM
not sure what you mean by order mattering

No, you are taking care of the order here. {1, 2} and {2, 1} are distinct.

you've already picked the first two, so order has already been enforced

like if $\{1,2\} \ne \{2,1\}$ or rather they are distinct $m-$cycles

well, keep in mind that you're not picking 12 or 21 by themselves
rather you're doing something like 12345 versus 21345

if $\{1,2,3,4,5\}$ and $\{5,4,3,2,1\}$ are different $m-$cycles then shouldn't there exist 10 choices for 5 of the $m-$cycles, 9 choices for 5 of the $m-$cycles, 8 choices for another 5, etc?

9:04 PM
but you're right that i'm skipping over something. that's the overcounting i'm hushing over momentarily

ok

imagine it more as how many ways I could generate m-cycles, not necessarily distinct. we'll get there.

@Obliv Make it simple. Do the algorithm for choosing a ordered 2-cardinality set out of {1, 2, 3}.

@Balarka Another thing you could do is your own work :)

if i follow that approach, i've got 10 choices for the first element, 10-1 = 9 for the second, 10-2=8 for third...
so 10*9*8*7*6 choices up to the fifth.

9:05 PM
@MikeMiller Ah, well, you're right.
I suppose @SemiC took up the conversation.

thank you for your help and guidance @balarka :)

i think we're nearly there anyways

i'm getting the feel for it now. it's n(n-1)(n-2)...(n-m) for the last one
in the way you describe @semiC

right. and then you're done making choices, because the rest of the elements aren't permuted

Right. Note now that (1 2 3 ... m) and (2 3 ... m 1) are the same as m-cycles. Think about how many of them are same, and what's the correcting factor.

9:07 PM
actually, it's n(n-1)(n-2)...(n-m+1). otherwise there'd be m+1 factors rather than m

yeah that's what i meant

as a check, if it's m=3 you get n(n-1)(n-2)
anyways, if you compare that with what you should be getting as an answer, what are you off by?

there are $m$ different ways to order an $m-$cycle right?

Bingo.

sure. $m$ ways to pick the element which shows up first
those would all look different in the above generation, but they're all really the same m-cycle. so you take the product you had above and divide by m, which is the correct answer

9:10 PM
okay thank you very much to both of you.

note the strategy here. I didn't try to count the set directly. rather, i found a set i could count more easily and then figured out how to obtain the desired count from it.

9:25 PM
hi guys!
I have a silly question. Why the limit of sum equal to integral?
What is the geometric interpretation of this? I know I use Riemann sum to approximate the area under a curve. But why is that area equal to the integral?

it's just a different notation to mean the same thing @AbhishekBhatia except with an indefinite integral, there is a different way to evaluate it

@Obliv by the fundamental theorem of calculus 1?

no part 2
well i meant for a definite integral

yeah, definite integral! The course I did mixes 2 for 1. You are talking about F(b)-F(a) one?

yeah. i mean both basically explain what you can do with an integral (take the antiderivative, evaluate the antiderivative at the lower and upper bound to find the area between them)

9:33 PM
So what meaning does the indefinite integral have? As in it is just the anti-derivative. Is it there any interpretation of it? It is easy to understand something by geometric interpretation.

is a journal more likely to accept a short paper than a long paper?

Eh, I need to get some sleep. Not thinking straight.

an indefinite integral is the non-descript form of a definite integral @Abhishek where no bounds are specified.
oh are you confused by how an antiderivative can model the area of a function between bounds? @abhishek

@Obliv yeah

@AbhishekBhatia Given a function $f$, $\int_a^b f(x) dx$ is nothing but the area of under the curve given by $y = f(x)$, from $x = a$ to $x = b$. Indefinite integral, on the other hand, is just formally the "opposite" of the derivative. It has no apparent geometric meaning.
But fundamental theorem of calculus tells you there are not very different, which is surprising and not "obvious". :)

9:38 PM
@abhishek it's not as intuitive as you would hope I think. math.stackexchange.com/questions/15294/…

Okay, I've nearly got a draft of this MO problem. Could use a second opinion about it, though, especially in stating the precise question.

The basic gist of that is, the man before Newton basically recognized that antiderivatives can be evaluated at bounds to give the area of their derivatives between those bounds.

This is worth watching, I think: youtube.com/watch?v=0Urn4Nrh1BU

So, any volunteers to reading a problem on groupoids? :p
(i'm looking at you, @akiva/@eric/@balarka)

I can read, but I can't guarantee I'll be able to contribute.

9:40 PM
@Abhishek He looked at very simple functions like $f(x) = x^2$ so if you want you can analyze these functions between bounds it might help you see the connection more.

mmkay

Sure, I'm distracted for a while but I'll be slow, but let's see it

i'll put it here:

### Semiclassical's draft

For input on a draft MathOverflow question

@obliv That is scary an answer, I will try understand it. 1 question from what you said: Why do you say the both fundamental theorems are essentially the same?

9:43 PM
@AbhishekBhatia One way to geometrically interpret things is as follows. $f(x)$ be a function, $F(x) = \int f(x)$ be some specific antiderivative (you chose the constant - more precisely you look at $\int_0^x f(t)$ but don't worry about precision). We thus have by definition $F'(x) = f(x)$, so $f(x)$ is like the slope of the graph $y = F(x)$ at a point. So integration is like given a function $f$, coming up with a graph whose slope at point $x$ is $f(x)$.
So you reconstruct the function from it's "infinitesimal informations".
Some careful thinking with this may even lead you to concluding why, "intuitively", $F(b) - F(a)$ is the area under the graph of $y = f(x)$ from $x = a$ to $x = b$.

well if I remember correctly the first part establishes if a function is differentiable and continuous on some interval, call it $f$, a function must exist,call it $F$, such that its derivative is equal to it,so $F' = f$. The second part establishes that the area of $f$ between the bounds that it's continuous and differentiable can be evaluated by the antiderivative $F$ at those bounds. @AbhishekBhatia
sorry if that was a ramble I have to go now. I didn't really understand that answer as much as I wanted to either when I was curious. Try to get what you can out of it. @abhishek

@BalarkaSen I lost you after So you reconstruct the function from it's "infinitesimal informations". Can you explain more on careful thinking?

$f(x) = \int f'(x) dx$ by definition, right?

Okay, it's there now

@BalarkaSen yes

9:51 PM
$f'(x)$ is just the slope of the graph of $f$ at the point $x$. So integrating is the same as reconstructing $f$ from just the information about slope of the tangent at each point on it's graph.

This is more or less the geometric interpretation you'd want to work with. If you try thinking harder, you'll start understanding what antiderivative/indefinite integral has to do with definite integrals.
I won't elaborate on that though.

I'm on the floor laughing (watching a comedy, not because of the stuff I read here).

I don't, but I would love to. tangent is $\delta y$ divided $\delta x$. Now I want to construct y=f(x) from it. How does that relate to the area?

Keep thinking :)
Remember that while doing definite integral a la Riemann sum you piece your region up into "infinitisimal bits" and then sum them up. Something like that is happening here too, but it is a bit less transparent.
It is all about reconstructing something from the infinitisimal informations you have about it at each point.
Alright, gotta go.

10:06 PM
I understand how Riemann sum is equal to the area. What I don't get is FTC essentially?
The proof I saw used FTC2 to prove FTC1 and FTC1 to prove FTC2. That proof makes sense, I don't get the geometric idea of it to better understand it.

Once upon a time there was a theorem called the theorem of Lagrange, and according to it, if we apply it to a function $F$ on a specific interval $[x_{i-1},x_i]$, then there is $\xi_i \in (x_{i-1}, x_i)$ such that $$F(x_i)-F(x_{i-1})=F'(\xi_i)(x_i-x_{i-1})=f(\xi_i)(x_i-x_{i-1})$$
Eventually, the mysterious $F(b) - F(a)$ is the result of a telescoping sum (as one can intuitively guess from what I wrote above).

@AkivaWeinberger We may not have $1=2!$ but at least we have $1=\Gamma(2)$

And, on similar lines, $1=0!$ :P

@user1618033 Can you explain more on the theorem of Lagrange, I can't understand it.

In mathematics, the mean value theorem states, roughly, that given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivatives at points of the interval. More precisely, if a function is continuous on the closed interval , where , and differentiable on the open interval , then there exists a point in such that: A special case of this theorem was first described by Parameshvara ...

10:21 PM
Ah yes! That's the name of it, thank you

Prof. used the same thing in the step I didn't understand. He says, "If two function's have equal derivatives, there integrals differ by a constant. By MVT."
I can't understand how by MVT.

If two functions have equal derivatives on an interval, then they differ by a constant.

How does MVT prove it?

10:36 PM
We both have 0 != 1 and 0! = 1

Given that we talk about $f$ and $g$, why $f-g$ is a constant? (consider $h=f-g$)

(!= means $\ne$ in most programming languages)

OK, time to get some sleep.
First let me see if anyone answered my question on main.
-3

Beginners in calculus may enjoy the following problem: Compute the following integral: $$\int_0^1 \int_0^1 \frac{\displaystyle(1+y) \log\left(\frac{1+x+y-xy}{1-x+2y+y^2-xy^2}\right)}{(1+y)^3+2x(1+y)(1-y-y^2)+x^2(1-y(3-y-y^2))} \ dx \ dy$$ (Since it is intended for beginners, it is of c...

gulp
0

It is well-known that any rational number can be represented using a sum of Egyptian fractions (that is, rational fractions of the form $1/n$ with $n\in\mathbb{N}$). This may be proven by establishing a greedy algorithm that constructs a sequence of such decompositions. For instance, one has \be...

oh, frack, forgot something obvious

@Semiclassical what do you think about my integral on main? Does it seem so evil?

10:50 PM
well, a few things

It's very beautiful, just to see it with the proper eyes.

1) I think what you'd consider a problem for beginners is fairly advanced for them. in particular, i don't think the average reader is going to recognize that.

That's all.

2) There's the usual problem of challenge problems; the main site just doesn't seem to be fond of them, probably because there's so many integral one could write that would simply have no answer at all.

Agree.

10:55 PM
plus people conflating 'no work' with 'must be homework'
sometimes that's true, but certainly not the case here.

@Semiclassical You never know if it's about homework. :D

true :/
I will say I agree with Eric that JMP's edit really was not helpful: The point was not to get a 'simple approach'---you presumably have your way of doing it, and are happy with it---but to present a challenge to the site.
Oh, forgot to say: My question is back up. So anyone who is curious to see me probably making an idiot of myself on a subject that I barely know anything about, well---here's your chance!
@user1618033 I don't know a good way to pose such problems on main

@Semiclassical Well, you can improve it after receiving some feedback.

Is there a shorter answer to this question math.stackexchange.com/q/15294/200649?

I just received an email from a student that tells me he has no idea how to solve that problem with the sequence.

11:03 PM
by that do you mean 1) they say they have no idea, or 2) you can tell they have no idea without them needing to say so?

@Semiclassical Well, not sure they can do it without some help.

@abhishek I suggest you read the part "Second, the area does not turn into "the" antiderivative. What happens is that it turns out (perhaps somewhat magically) that the area can be computed using an antiderivative. I'll go into some more details below."
and below that. That's the part that really concerns your curiosity.

@Semiclassical perhaps these days I was more dedicated to help some stud. but I'll fully return to my stuff.
There is no magic in using the mean value theorem $$F(x_i)-F(x_{i-1})=F'(\xi_i)(x_i-x_{i-1})=f(\xi_i)(x_i-x_{i-1})$$ as I explained above in short.
This tells you very clearly (I doubt a clearer explanation could exist) why you end up with $F(b)-F(a)$ by a simple telescoping sum.

Hey guys! I have a question - why is it that if a matrix A is nxn, and $A^4 = 0$, that it must be the zero matrix?
I know that if the matrix is diagonalizable then it must have eigenvalues of 0, and thus 0^4 would be the only possible matrix that would yield the 0 matrix.
However, the grey area is if it isn't diagonalizable(singular) - how do I prove that $A^4 = 0$ in this respect?

It's not true, Rayny. Even when $n=2$ there are counterexamples
Those variables should find you a counterexample.

11:16 PM
^Right... I supposed the professor made a mistake on the quiz

@user1618033 so the F(..) in between cancel out, leaving only the end-points.

It's not that I'm against telling stories about fact in mathematics, but sometimes you have to tell people things in such a way all become clear.

Oh and another quick question - this is more on the discrete math side... I checked online for solutions for a non-homogenous recurrence relation generally in the form of $a_{n} = c1*a_{n-1} + c2*a_{n-2} + f(n)$
And all the posts were super convoluted - I had no idea what they were talking about. The professor only asked us for $f(n) = C$ where C is a constant, and I don't see any examples of such online.

I think all can be explained very beautifully and very clearly, and everything can be improved (at least in terms of solutions).

@Abhishek I also recommend you do what Barrow did and construct the graphs of $y = x^2$ and $y' = 2x$. Note how the area between $x = 0$ and $x = 3$ of $y' = 2x$ can be given by evaluating $x^2$ at those two points. $y(3) - y(0)$. Notice how the area below $y = 2x$ changes as you widen the bounds. The change in the area is given by the antiderivative. This just happens to be a simple case. Note how the area of the function $\cos(x)$ is represented exactly by $\sin(x)$.

11:22 PM
@user1618033 You mean me?

@OneRaynyDay Sorry, I'm so sleepy I didn't even notice you. :-)
I'm out for some sleep - tomorrow (ah, I mean later today) I have a tremendous amount of work to do.

ah okay no problem!

@Obliv yeah, just reading it. Thanks for the answer reference! This was exactly my question.
@user1618033 Shouldn't there be an answer using MVT for this question: math.stackexchange.com/q/15294/200649.