@KhallilBenyattou hmmm, I didn't try that yet.

@Chris'ssis Smeagol loves riddles ! :D

@r9m :D

@Chris'ssis $$ \int_{0}^{\infty} \dfrac{\log^{2015} (x+1)}{2x^2 + 3x + 1} \text{ d}x = \int_{0}^{\infty} \dfrac{\log^{2015} (x+1)}{(2x+1)(x+1)} \text{ d}x \overset{u=\log(1+x)}= \int_{0}^{\infty} \dfrac{u^{2015}}{2e^{u} - 1} \text{ d}u $$

The last of which seems even more obscure than the first!

I'm confident you can finish it off, @Chris'ssis! ^_^

@KhallilBenyattou Yes, you're right. How can you rewrite $2$ in the denominator?

My self-nomination as moderator of M.SE:

$1+1$, @Chris'ssis?

If you answer the title, rather than reading the long body of work the OP has written, I will not only delete your answer, I will come to your home and eat your cat.

@KhallilBenyattou It's not useful. You may consider $e^{-\log(1/2)}$ and then make use of the Bose-Einstein Distribution integral. Q.E.D.

My vote goes to @ThomasAndrews just to see if he actually comes through with his promise.

A person who loves cats could flag that remark @ThomasAndrews

@Studentmath Something like this: Five vertices are labeled 1,2,3,4,5. In how many ways can edges be drawn between some pairs of these vertices so that the result is a connected graph?

Personally, I love cats. With a side of asparagus.

Well, you already know the minimum number of edges to that the graph is connected, and the maximum number of edges that can be, right?

can someone upvote this answer ! (not mine) I can't bear to watch him get a Naruto Hat =P

Minimum number: 7 @Studentmath

I've reached my voting limit :-(

I've up-voted him, @r9m!

@r9m did you used to be wut lol?

Right. And how many edges can there be in total?

@r9m Did you see why it's so easy? (apart from the way stated above using Bose-Einstein Distribution integral)

Or in other words, how many pairs of vertices are there?

@Studentmath 34?

@KhallilBenyattou thanks pal ! ;)

I'm sleepy ... :-(

@Studentmath oops, typo. 4*

@skullpatrol ? what does that mean ?! :o

Nah, you have 5 vertices. It's $\binom{5}{2}=10$ pairs of vertices, right @Mathy?

@Chris'ssis I have got no idea ... lemme put pen to paper ! :)

Although I've never heard of it before, here goes nothing @Chris'ssis: $$ \begin{aligned} \int_{0}^{\infty} \dfrac{\log^{2015} (x+1)}{2x^2 + 3x + 1} \text{ d}x & \overset{u=\log(1+x)}= \int_{0}^{\infty} \dfrac{u^{2015}}{2e^{u} - 1} \text{ d}u \\ & = \int_{0}^{\infty} \dfrac{u^{2015}}{e^{u-\log(1/2)} - 1} \text{ d}u \\ & = \Gamma(2016) \ \text{Li}_{2016} (e^{\log(1/2)}) \\ & = 2015! \ \text{Li}_{2016} \left( \frac{1}{2} \right) \end{aligned}$$

@r9m did you used to have a different username?

why C(5,2)? @Studentmath

@skullpatrol no .. I was and always will be r9m ;)

You have five vertices, and between every pair there may be an edge. The Complete graph on 5 vertices has 10 edges

What does $\text{Li}_{s} (\mu)$ mean, @Chris'ssis?

oh ok. derp. @Studentmath

@KhallilBenyattou mathworld.wolfram.com/Polylogarithm.html

Okay, so we know the minimum number of edges is 7, and there are 10 'slots' in total for edges. How many ways can you order 7 edges in 10 'slots'?

Ok @r9m so it wasn't you I was talking about yesterday and the division question, sorry pal :-)

I couldn't sleep, so I am back.

Welcome.

@skullpatrol ah ! okay :-)

I'm terribly sleepy ...

@Studentmath C(9,6)?

C(10,7)!

Why 9,6?

@Studentmath sorry. still stuck in "distribution of items" mode. i was working on that problem prior.

Oh, alright.

Well, each one is a different graph on 7 edges. But you may also have 8, 9 or 10 edges and it will still be connected and different.

So the sum of these different conf. will get you the right answer, right?

So C(10,7) is the case for the min. of 7 edges?

@Studentmath ^

Correct.

C(10,9) the case for 9 edges, which we know to be connected, C(10,8) for 8, etc.

@Studentmath And we had...10 edges max?

Correct

10 pairs of vertices, 10 possible edges. Any more and the graph will no longer be simple.

@MikeMiller: I must have made a mistake in my notes, I can't reproduce the calculations used in the proof I want to show. I'll check again tomorrow evening, I'm too tired to finish it today. :(

@Studentmath So it would be... C(10,7) + C(10,8) + C(10,9) + C(10,10)?

Precisely :)

If I'm not mistaken, that's it. Don't think I am.

@Studentmath hmm...i entered in 176 and apparently it was incorrect

What was the question again?

@Studentmath "Five vertices are labeled 1,2,3,4,5. In how many ways can edges be drawn between some pairs of these vertices so that the result is a connected graph?"

Well yes, we still miss the possible variations that form some connected graph, but not always necesserily do. That gets to a bit uglier counting.

Since 6 edges can also form few connected graphs

@Studentmath How would we calculate that then? C(10,6)?

No, as it would include options not necesserily forming a connected graph.

I see you updated Review+, @Behaviour. Thanks.

Okay, that's a nice hint. Can you work it out for the cases of 6 and 5 edges?

Can someone help me at this exercise: cs.stackexchange.com/questions/35456/… ??

(4 can only form one connected graph - the path $P_4$, and any less can't form a connected graph as we need at least $n-1$ edges to form a connected graph on $n$ vertices)

@Studentmath I have a problem for you.

@Pedro shoot

Let $G$ be a graph with $n$ vertices and suppose that the minimum degree of $G$ is at least $n/2$. Then $G$ contains a $1$ factor.

@Studentmath So therefore, I would additionally need to find the cases of 4, 5, and 6 edges?

@Mathy yes, which are not $C(10,6)$ etc. as there are some cases of 6 edges where the graph isn't connected. For 4 it's simple - only one case, the path.

@Pedro let me think

@Studentmath yep, i see. Would individually drawing it out be effective?

@Studentmath I have no guarantee this is easy.

Not really @Mathy, it's possible since it's 5 vertices but it's not effective, nope

@Studentmath ah. ok. i have to go in a few min, but when i come up with an answer, i'll ping you :)

@Mathy okay :)

@Pedro it's not immediate, that's for sure

@Studentmath For the case of 4 edges, what do you mean by "only one case, the path"?

Actually, the path can be formed in different ways, so scratch that. But what I meant is that the only connected graph with 5 vertices and 4 edges is $P_4$, the path with 4 edges. But the order of the vertices in the path may vary.

@Studentmath oh ok

@Pedro we know it has a hammilton cycle

Considering $n\ge 3$

@Pedro so if $n$ is even we are done, just gotta consider odd $n$

Which doesn't seem to make it too complicated either, I think.

@Pedro yep, it's doable with the hammilton cycle. Odd $n$ requires a bit more explanation but it seems fine as well

@PedroTamaroff I changed my mind about my votes, but I will still be voting for you, lol.

@Studentmath Wow you have a small hat :3

Shameless advertising

@Hippa I take pride at it

I wonder if I should change my name to Sanic-sensei :3

I've a question. If $f:X\to Y$, $g:Y\to Z$. If $gf$ is open, I would think all I need for $f$ to be open is that $g$ would be continuous. But it seems I also need it to be injective. Why?

Maybe @Behaviour posted that line to get the 10 stars or some secret hat.

@Studentmath You should ask on main, I'm surprised it hasn't been asked before

Hm, actually it's a good question for the main, yeah.

@Studentmath You want "Hamilton"

=D

So the generalization is that if the minimum degree is at least $(1-k^{-1})n$; then $G$ has a $k$-factor.

Oh, nice.

The proof, I have no idea.

I just learned that yesterday,.

Maybe some tweaking with the Hamilton cycle? Gets messy for high k's though. Maybe induction.

My bet is on induction

Was it given as exercise or as a theorem @Pedro?

Ugh. Bombed my analysis final

@user130018 Bombed means?

Failed

Didn't know how to prove $f: \mathbb{N} \rightarrow \mathbb{R}$ is continuous using the definition of sequential continuity

Really? Americans use interesting language.

@Studentmath As a theorem, in a talk. =)

@user130018 Are there any convergent sequences in $\Bbb N$?

Any mapping with domain a discrete space is continuous.

@PedroTamaroff Constant ones I guess

Because in a discrete space, every set is open.

@Pedro interesting. I did a quick google search, there are actually quite a few papers relating minimum degree to k-factor. But your result seems the strongest.

@Studentmath Yes, it is tight.

@PedroTamaroff I think you are just confusing him with the topology, lol.

@user130018 It's alright, you'll do better next time.

Interestingly the original problem doesn't appear on google (except as a byproduct of the stronger proofs, not mentioned)

Once you know the proof let me know! I tried to play a bit with the Cycle or induction, but can't seem to think of an elegant way

I am taking topology next semester

I think I should send out a bunch of emails on Christmas eve.

G'night!