10:49 AM
@XanderHenderson o/

2 hours later…
12:21 PM
5

If I have a differentiable function $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(x+1)-f(x)=f'(x)$ and $\lim_{x\to\infty}f'(x)=A$. Can I show $f(x)=ax+b$?

@amWhy why is it a duplicate?

@LeakyNun I think the case is quite clear. And I think your question sounds, to me anyway, a bit disingenuous. It's just hard for me to take your question seriously. Perhaps you can start by telling me why it isn't?

If the case was quite clear, it wouldn't have 3 reopen votes now. And I don't think it's a duplicate because the answer cannot be found in the question linked, since this question requires the extra assumption that the limit of the derivative exists.

The duplicated question acknowledges that $f(x) = ax+b$ provides a solution, but not the only solution. Other solutions are offered in answers. So, the OP's question is answered. $f(x) = ax+b$ is a solution among many others. So it can be a solution, but it is not the case that it is the solution.

The other solutions to the delay differential equation have not been shown to have a derivative that have a finite limit, so the OP's question has not been answered.

You are free to disagree, @LeakyNun. And you are free to, and already have, vote(d) to reopen. So we simply have to agree to disagree. Cheers!

12:32 PM
@amWhy There is the additional assumption that the limit of $f'$ exists, which is not present in the older one. That does make a difference, I think.

@amWhy I presented an argument and all you can offer is "we simply have to agree to disagree" instead of destructing my argument?

You must have missed my comment before that, @LeakyNun. Please keep this conversation friendly/respectful. Thank you.

You said that "The duplicated question acknowledges that $f(x)=ax+b$ provides a solution, but not the only solution. Other solutions are offered in answers. So, the OP's question is answered", which I responded by saying that OP requires $\lim f'(x)$ to be finite, which the other solutions have not been shown to satisfy, so the OP's question has not been answered.
It is not a matter of opinion that the other solutions have not been shown to satisfy the requirement that $\lim f'(x)$ be finite.
Also, if you find any part of this conversation unfriendly or disrespectful, you are free to flag the relevant messages and let the mods decide whether you are just reading disrespect into my comments.

1 hour later…
1:53 PM
Even though the question is reopened, I think that question still lacks context in the sense that it doesn't include OP's attempts as pointed out by @Did here.
2

3 hours later…
4:23 PM
1, 2, 3, 4
5, 6, 7, 8
9, 10, 11, 12
13, 14, 15, 16

5:31 PM
@Xander I agree

5:48 PM
@JohnMa Please include poor questions that you answered two, three, four, five years ago, in your campaign to delete poor questions others have answered. No reason you and the poor questions you've answered should be spared, hey?
You can start by voting to close questions you've answered at the same level of quality at which you vote to close questions. Just asking for some consistency wrt which questions you target.

6:24 PM
Is this a good question?
0

While reading about graph theory I wondered if there is a process of rotating graphs in graph theory. After searching around I could not find anything. So I will try to explain what I mean: There is a graph $G =(E,V)$ fixed in place. Then the graph is copied and rotated by an infinitesimal amo...

@amWhy
@XanderHenderson

6:40 PM
@geocalc33 No, I don't think so. I have not idea what you are trying to do. You seem to be embedding your graph into $\mathbb{R}^2$, then asking what rotations do to the graph.
As per the first comment, rotations do nothing to the graph (since the embedding of the graph is irrelevant to the structure of the graph
)
In reply to that, you suggest that you are copying the graph? Then rotating it? So you have two isomorphic copies of the graph, which basically means that you have two copies of the graph, peried
Again, with respect to the structure of a graph, I don't see that you are doing anything except making a copy.

Someone thinks it's a good question
algebraic anonymous
consider unvoting to close

@geocalc33 The fact that one person thinks that the question is good has no effect on my opinion that it is not good.

well it is a good question

Again, I disagree. I've explained why I think it is a poor question. You asked for my opinion, and I gave it. Do with it as you like (either nothing, or edit the question to make it clear what you are asking).

@geocalc33 Please read advice on what good questions actually look like: math.meta.stackexchange.com/questions/9959/…

6:51 PM
It's not clear
apparently...

And read carefully. And in comparison to the advice given on many points in the link I just provided, I'd have to agree with @XanderHenderson, that once again, you've asked a very poor question. I'd encourage you to take some tips from the linked post, to improve your question.

I'm going to make it so clear
that no one can vote to close it for that reason
my example is crystal clear

Have you read the post I linked above? I encourage you to do so, because being clear is merely one expectation of questions.

do i need a picture
?
Honestly I spent a lot of time on that question and I'm struggling to understand what changes you want!

@geocalc33 When you ask a question and someone tells you that they don't understand it, there are two possibilities: either the failure is on the part of the reader who doesn't understand, or on the writer. When multiple readers claim confusion, the probability is that the problem is with the question.

7:00 PM
I provided the link to help you learn about what users on this site suggest wrt posting a question. If you choose not to read it, to get a wider view than your immediate need to get an answer, and NOW, for this question, that's your choice. But I'll still hold you responsible when you fail to incorporate expectations and expressed suggestions. You're no longer a "new user".

Right now, I see at least four readers who are confused (two who posted comments, and two in here)
Simply asserting that your question is clear does not make it so.
To reiterate my confusion:
(1) You seem to be embedding the graph into $\mathbb{R}^2$ so that you can rotate it. You do understand that this embedding does nothing to the structure of the graph, right?
(2) When you "rotate" the graph, you appear to be making an exact copy of the graph, then rotating the embedded version. You aren't adding any other new nodes or edges. From the point of view of graph theory, all you are doing is making a copy of the graph. The rotation is irrelevant (as was the embedding). It isn't clear what you are trying to do here.
(3) You are rotating by a parameter $\delta$, yet your question is about "continuous" rotation. You can't have it both ways. Either you rotate by some discrete quantity, or you rotate continuously. Since it is not at all clear what the rotation is meant to be doing in the first place, I don't know what you actually want to do here, but you should try to clear that up.
(4) A couple of commenters (@amWhy and B. Mehta) have suggested that you include an image. That might help to clear up the confusion.
Let me also note, since text is a poor medium for conveying nuance, that my conclusion that the question is poor is not reflective of my feelings toward the asker. Taking an idea that is clear in your mind and writing it down in language that is clear to everyone is hard---that is a huge part of what mathematics is. What determines you quality as a mathematician is your ability to take the constructive feedback that is given and make adjustments.
I am honestly trying to help you improve your question. Take my advice as you will, but do please note that if you don't clarify your question, I am not going to retract my close vote, and am going to be much less inclined to offer advice in the future.

Thank you

7:16 PM
@LeakyNun lmao. Just goes to show how some people don't even bother to read questions in detail before they vote to close, and won't admit they were wrong afterwards

7:27 PM
@amWhy@XanderHenderson added a picture, was it helpful in improving the question?
Okay I also dropped the continuous rotation part
The thing is, I am adding new nodes, lots of them
so I'm confused why people say there's no nodes being added

@geocalc33 The only new nodes that you seem to be adding are nodes that correspond to a copy of the graph. That is, you are doing exactly what I outlined in a comment. You now have one graph with two connected components.

I still don
understand
You're saying all I have is one graph?

7:43 PM
Suppose that $G$ is a graph with two vertices and one edge connecting them: |
I make a copy of $G$, and put is next to the original. This gives me a new graph ||

okay

this graph has four vertices and two edges.
I could also put the new copy on top of the original (after a rotation) to get +

how about you rotate it and make a graph with 5 vertices
and 4 edges

But, from a graph theory point of view, the two graphs || and + are the same graph.
They are isomorphic
each of the two graphs has exactly four vertices and two edges.
The fact that one of them looks like a parallel set of line segments and the other like an orthogonal set of line segments is irrelevant to the structure of the graph.
These are just embeddings of the graph into Euclidean space.

7:48 PM
As Chandler Watson noted, the "position" and "orientation" of a graph in $\mathbb{R}^2$ is not a notion in graph theory. Graphs don't live in Euclidean space (they can be represented or embedded in Euclidean space, but the structure of a graph has nothing to do with Euclidean space).

so this graph is a triangle and then a copy is made and rotated
isn't this a completely new graph than the orignal triangle?
where each intersection is a node

GAH! This is the very first time that you have suggested that every intersection of segments is meant to be understood as a new node!
That completely changes the question...
But then your example of the square is wrong...

oh sorry
why?

You said that you start with a square, then copy it to get 8 vertices and 8 edges
in fact, you get a lot more than that

oh yeah
i forgot the other nodes

7:53 PM
I'm going to add a fifth point to those above: (5) Why? What is your motivation for this procedure? If you tell us why you are doing this, it might help us to understand exactly what you are attempting to describe, and provide better answers.
Anywho, I need to go get some work done. I'm out.

2 hours later…
10:04 PM
One more close vote needed here. Note the asker's comments below their question.

10:21 PM
Wee also this PSQ
I meant "See also" immediately above. Yikes! :-)

@amWhy Please let me know how I can see my questions of a particular tag.
like I want to see my calculus questions, how do I search them?

10:39 PM
@Abcd I just tried to check if I can do the same with my answers. There's no direct feature to do so, but I believe there's a way to search for one's post, along with the name of the tag, but I am not at all an expert on how to do so. I'll see if I can find @MartinSleziak, to ask him.

okay

@Abcd I'm thinking it will need to be done through the "search bar" on the top, left bar on most MSE site pages, with specific formatting to designate both user (you), and the tag of interest. I've asked @MartinSleziak over in the mod's office; and linked him to your question here. Hopefully, he'll get back, and surely with far more information than I can provide. But let's try to be patient, since it's Sunday (weekend), and give him a chance to check in.

@amWhy hmm. user:me [calculus] works (gives answers and questions both together though)

@Abcd Great! That's a great start!

@Abcd you can add is:answer or is:question to restrict the type.
4

Is there a way to sort questions I have asked by categories? For example: "number theory", "Linear Algebra", "Calculus". And each question will correspond to its topic.

10:47 PM
Oh thanks both!

Thanks, @quid!!

You are (both) welcome!