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01:00 - 22:0022:00 - 00:00

01:47
I was reading about space-filling curves, and I was wondering how they fit in with the Jordan curve theorem (that a closed simple curve has an inside and an outside). Do space filling curves not satisfy some condition for the Jordan curve theorem to apply to them? Or does it?
Looking for someone to verify a proof.
@detly Space-filling curves are not injective.
02:31
@MikeMiller That reminds me. I need to write up my stuff on $n$-dimensional space filling curves.
So I'm working with functions of the form p1(z)exp(a1*z) + p2(z)exp(a2*z) where a1 and a2 are distinct complex numbers and p1 and p2 are polynomials. I'm trying to characterize when these functions have no zeros. Can someone give me a hint? Maybe Rouche's theorem would be helpful
The existence of space-filling curves for $n>2$ always surprises me because random walks in those higher dimensions don't always return to the origin
@MikeMiller because they self-intersect?
Matlab question: I have a vector of points like [0 2 6 9 12] is there a function to find in which area an input numbers falls into?
02:50
@detly It depends what you mean by 'self-intersect'. The answer is either 'yes, by the definition of injective' or 'no, self-intersection doesn't make sense for a curve so pathological'.
hah
The better answer is probably: no, it's not for any 'geometric' reason, but rather for topological ones. Suppose a space-filling curve means a surjection $[0,1] \to [0,1]^2$. It's a theorem that any injective map from a compact metric space to another metric space is a homeomorphism - it preserves certain topological properties. But (another theorem) $[0,1]$ is not homeomorphic to $[0,1]^2$.
because there's no breaking point
Right.
 
1 hour later…
04:22
@MikeMiller Hello Mike, not sleeping yet?
04:36
can someone go here on the transcript and answer my question above?? chat.stackexchange.com/transcript/36?m=19533852#19533852
12 hours ago, by Combinatorics
i was trying to measure if for the above question, and i dropped it out the window(high) and he has clipped wings
Huy
Huy
05:17
Wow, did you know there is going to be a third season of Twin Peaks in 2016, @MikeMiller?
I uploaded the Navy's NUPOC study guide to my github. Not all the sections are completed and I am looking for contributors that would like to work on providing solutions in their free time. Everyone who helps, will be credited with their work. Also, a proof read of the sections currently complete would be great too! github.com/dwsmith1983/NUPOC-Solutions
I didn't, @Huy. I hope it's good.
05:51
@iwriteonbananas The interesting point being the point that has no semilocally simply connected nbhd?
 
2 hours later…
07:27
Hello,everyone,someone can interesting this problem: if $a+b\sin{c^{circ}}=1+4\sin{10^{\circ}}$,where $a$ postive integer,and $b,c$ integer,show that a=1,b=4,c=10
@robjohn
1
Q: If such $1+4\sin{10^\circ}=a+b\sin{c^\circ}$ How find this integer $a,b$

math110Interesting problem: Assmue that: $$1+4\sin{10^\circ}=a+b\sin{c^\circ}$$ where $a>0$,and $a,b,c$ are integers and $0<c<90^\circ$, show that $$a=1,b= 4,c=10$$ is unique solution Different Old Question: see Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ I ...

08:01
[Is this OK?][i.imgur.com/j44qoVC.png)
(damn you markdown)
user61230
How do you all search for MathJax on Math.SE? I'm trying to search a specific integrand, but I have no clue as to how I should actually find it.
08:37
@NajibIdrissi I've no problem with adding tags that one's created. But... what kind of a tag description is this?
09:08
@NajibIdrissi Nothing wrong with bumping multiple posts.
09:25
Ugh, this problem is such a pain.
@Emrakul There isn't really a good way to search for mathematical expressions. You can try to type up a question and see what pops up as a possibly related question. One user has begun to compile an index, but this is still a work in progress.
09:45
Have you seen this SE site @KajHansen?
I'm a member on there @skullpatrol
Greetings
@Emrakul You can find some useful tips in various posts on this site, e.g., here, here here and here.
Greetings
09:47
@skullpatrol How is it going?
@Emrakul In particular, in this answer I have posted examples showing attempts to search for a specific integral. You can judge for yourself to which extent it was successful/efficient.
Fine thanks, how are you my friend? @Chris'ssis
10:05
@skullpatrol Not that bad, thanks. You?
I'm ok.
10:35
@math110 haven't seen a way yet.
Oh,@robjohn,hello,you can see this post,math.stackexchange.com/questions/1102198/…
@math110 That is what I was commenting on, if you follow the link.
I think it's interesting,maybe $c=10$ is clear,But can't prove it,$c$ must equal to 10
and you can consider this problem,the key to prove $c=10$
I think it's interesting,How about you?
3 mins ago, by robjohn
@math110 haven't seen a way yet.
3
Q: If such $1+4\sin{10^\circ}=a+b\sin{c^\circ}$ How find this integer $a,b$

math110Interesting problem: Assmue that: $$1+4\sin{10^\circ}=a+b\sin{c^\circ}$$ where $a>0$,and $a,b,c$ are integers and $0<c<90^\circ$, show that $$a=1,b= 4,c=10$$ is unique solution Different Old Question: see Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ I ...

This post,Now you can see it?
10:51
@robjohn,do you know which paper have reaserch there problem?
11:46
@math110 no.
12:03
Hi pal @robjohn
12:42
@BalarkaSen yes !
13:11
@robjohn I saw an amazign question these days ...
If $f(x)=1/\tan(x)-1/x$, then $$f^{(2015)}(x)<0, \space x \in(0,\pi)$$
where $f^{(n)}(x)$ is the $n$th derivative of $f(x)$
hi @all
@Danielfischer One quick question, again with regard to [the post](http://math.stackexchange.com/questions/774552/convergence-of-characteristic-functions-on-hypercube). In the answer to this the post it states that if we can prove that $$\langle g,f_{m} \rangle \rightarrow \langle g,f \rangle$$ for all $g$ is some set $A \subset L^{1}$ with norm-dense linear span. then given any $h \in L^{1}$ it follows that $|\langle h,f_{m} \rangle-\langle h,f \rangle| \rightarrow 0$. I understand the reason given for this but I tried a different proof. Please see if my proof is also fine, it is similar.
@JohnDoe It will be fine if you justify the exchange of limits. The easiest way to justify that is, I believe, the argument given in the answer.
@DanielFischer Can the exchange of limits not simply be seen since the integral is a continuous operator? Does that make sense at all?
13:28
@JohnDoe Take $$a_{n,k} = \begin{cases} 1 &, n \geqslant k \\ 0 &, n < k.\end{cases}$$ Then you have $$\lim_{n\to\infty} \bigl(\lim_{k\to\infty} a_{n,k}\bigr) = \lim_{n\to\infty} 0 \neq 1 = \lim_{k\to\infty} \bigl(\lim_{n\to\infty} a_{n,k}\bigr).$$ You need to show that something like that (or worse) cannot happen in the given situation.
@DanielFischer What is your profession?
humanitarian
Jun 23 '14 at 21:01, by Daniel Fischer
Contract killer.
Still waiting for my first contract to kill.
7
@DanielFischer Unrelated question, do you know if the function $g(y) := \int_{a}^{y}f(x)$ is continuous if and only if $f$ is continuous?
@DanielFischer Too bad, Neo already killed Agent Smith.
13:36
@JohnDoe $f$ need not be continuous. $g$ is continuous if and only if $f$ is locally integrable, that is, if and only if $g$ is well-defined.
@Oracle Hey, I'm killing contracts, not agents.
@DanielFischer Understood, only continuous, not uniformly continuous?
@JohnDoe For uniform continuity, you need a bit more, "uniform local integrability" I'd call it. Integrability suffices for that. Or boundedness.
@DanielFischer Kewl thanks a lot.
De rien.
@DanielFischer What do you mean integrability suffices for that you mean if $f$ is integrable and not just locally integrable then $g$ is uniformly continuous?
13:44
@JohnDoe Yes. And if $f$ is bounded and measurable, then $g$ is also uniformly continuous (even Lipschitz continuous).
@DanielFischer Kewl :).
@DanielFischer Most people use term 'integrable' to mean $\int f dx < \infty$, I was taught that this is called summable. If the integral exists then $f$ is integrable.
@JohnDoe Most people I know use "$f$ is integrable" to mean $f \in \mathscr{L}^1(\mu)$ resp. $[f] \in L^1(\mu)$, where $\mu$ is the measure under consideration, and $[f]$ is the equivalence class of $f$ modulo "$\mu$-a.e. equal".
Please, somebody knows what is the name of the theorem that allows you to complete a set of linearly independent vectors into a basis? I only know its name in portuguese. Need a wikipedia page for it.
@DanielFischer So we don't know the same people....mostly at least.
@lucas Gram-schimdt
@JohnDoe but Gram-schimdt is for creating new basis. I need the theorem that allows you to complete a set to a basis
13:56
@LucasZanella Finite-dimensional spaces? That would be [Steinitz'] replacement lemma if I remember the English name correctly.
For infinite-dimensional spaces, usually one uses Zorn's lemma or Hausdorff's maximality principle.
@DanielFischer thanks! Steinitz exchange lemma says: "If {v1, ..., vm} is a set of m linearly independent vectors in a vector space V, and {w1, ..., wn} span V then m ≤ n and, possibly after reordering the wi, the set {v1, ..., vm, wm + 1, ..., wn} spans V."
@DanielFischer that's what I needed. Thank you.
@JasperLoy I disagree. The front page was filled with these questions. Bumping 2-3 questions OK, eight in a row is too much...
 
1 hour later…
15:20
Hello!!!
 
1 hour later…
16:23
0
Q: The sum of three colinear rational points is equal to $O$

evindaShow that in an elliptic curve $E/\mathbb{Q}$ the sum of three colinear rational points of it is equal to $O$ exactly when the neutral element of the group $E(\mathbb{Q})$, $O$ is an inflection point of the curve. Ifound the following in my notes. Let $C$ be a cubic curve that is defined over...

Could you take a look at this?
Hey @DanielFischer
I am looking at the algorithm MERGE that is the following:

MERGE(A,p,q,r)
  n1=q-p+1
  n2=r-q
  for i=1 to n1
       L[i]=A[p+i-1]
  for j=1 to n2
       R[j]=A[q+j]
  L[n1+1]=oo
  R[n2+1]=oo
  i=1
  j=1
  for k=p to r
       if L[i]<=R[j]
          then A[k]=L[i]
                  i=i+1
          else A[k]=R[j]
                  j=j+1
@DanielFischer There is a comment that says that we create the subarrays L[1... n1+1] and R[1... n2+1].
Why? Doesn't it hold that the arrays have sizes n1 and n2 respectively?
17:28
Not a big fan of 9-gag, but @MikeM @TedS you gonna like this: 9gag.com/gag/ae3BvNB
17:56
@DanielFischer @robjohn @ArthurFischer @KajHansen
This is the algorithm of heapify: http://pastebin.com/C5ZJLWt7

According to my book, the time complexity of the algorithm at a subtree of length n corresponds to the time Θ(1) that is required in order the elements A[i], A[LEFT(i)], A[RIGHT(i)] to get the right relations plus the time complexity of HEAPIFY at a subtree. Each of these subtrees has a length that is $\leq \frac{2n}{3}$- the worst case happens when the last line of a tree is exactly semicompleted.
That's a bit outside of my range of knowledge, unfortunately
A ok..
@Studentmath, I'm sure @TedShifrin has had similar thoughts grading my work. :)
18:21
@evinda After the array elements, this implementation adds a $\infty$ entry as a sentinel, so one needs an extra slot in the temporary arrays. If you don't do that, you need to check $i$ and $j$ to prevent out-of-bounds access to the arrays.
@DanielFischer I see.. Thanks a lot!!! :) Have you also taken a look at my other question?
@Kaj I know for certain at least one prof. who said that about my test
18:45
@DanielFischer In order to prove the correctness of an algorithm, do we have to state a statement and then prove it by induction?
@evinda There are many ways to prove correctness. For a lot of algorithms, proving that some invariant holds at all steps by induction is a good method to prove correctness.
@DanielFischer So can we only use induction proving that some invariant holds?
How else could we prove correctness of an algorithm?
In any way that works. Really, one can't make a list, there are way too many possibilities.
@DanielFischer Ok, but suppose that we have as input an array of length n.
Could we maybe then prove the correctness of the algorithm, using induction on n, i.e. say at the base case what happens for n=1, suppose at the induction hypothesis that it is true for length=n-1 and then show that it holds for lenth=n ?
@evinda Yes, for that kind of situation, induction is usually a good strategy.
19:00
morning
Evening
printf("Good %s.\n", local_time_of_day);
@DanielFischer Could you explain me why in order to show the correctness of BUILDHEAP that is the following:

BUILDHEAP(A){
for (i=floor(size(A)/2); i>=0; i--)
HEAPIFY(A,i);


it suffices to show that at the beginning of each for loop, each node i+1, i+2,...,n is the root of a max-heap?
Evening. Are (x,y,z)/(x,y) and k[x,y,z]/(x,y) isomorphic as k[x,y,z]-modules?
Dunno, but I don't think they should be
19:10
@evinda Because HEAPIFY(A,i) (it's very ugly to use all-caps for function names by the way, and violates convention: all-caps is for preprocessor macros) makes A[i] the root of a max-heap without destroying the heap property of its children (and grand-children, ...) and doesn't touch the part of the array belonging to trees not below A[i]. Of course, these properties of of HEAPIFY must also be proved.
19:59
Mister @MikeMiller Do you want to hear something mathematical? ;)
@user159870 Yes, please.
I only function in my domain. @Anonymous-agroup @MikeMiller
Do you like it? @Anonymous-agroup @MikeMiller
@Chris'ssis Hello
@Hippalectryon Bienvenue
@Chris'ssis What's a good looking result for $\displaystyle\int\dfrac{t^n}{1+t+t^2}dt$ ?
20:08
@Hippalectryon Who knows?
I thought you would q_q
@Hippalectryon a good looking result?
As in, not that V
@Chris'ssis I wonder if this can be handled using $\pi\cot(\pi x)=\sum_{k\in\mathbb{Z}}\frac1{k+x}$
@Chris'ssis Actually, I think that is just the tool.
@Hippalectryon Yeah, I think it might have a good looking result.
@robjohn Yes, that's the way.
20:13
@Chris'ssis Do you also have one for the generalization, $\displaystyle\int\dfrac{t^n}{1+t+t^2+\dots+t^q}dt$ ?
@Hippalectryon ooo, wait, I didn't work on these ones, but at least the one you mentioned above seems pretty interesting for some reasons.
@Chris'ssis Yeah... $\frac1{\tan(x)}-\frac1x=\sum\limits_{k\ne0}\frac1{k\pi+x}$
@Chris'ssis Ok, let me know if you find something :)
@robjohn Yeap.
@Hippalectryon Use recurrence relations and you're immediately done.
@Chris'ssis How so ?
20:15
@Hippalectryon Well, think about it. I didn't put anything on paper yet.
@Hippalectryon I would break that $n$ into $3$ sets: $n=3k, n=3k-1, n=3k-2$ (well, that's just another starting way)
@DanielFischer At the initial check, we say the following: Before the first iteration, we have $i=\lfloor \frac{n}{2} \rfloor$. All the nodes \lfloor \frac{n}{2} \rfloor+1, \lfloor \frac{n}{2} \rfloor$+2,..., n$ are leaves and thus each of them is the root of a trivial max-heap.
Then at the check of maintenance: in order to show that the invariant holds, we notice that the children of the node i have greater numbers of position than i. Therefore, based on the invariant, both of these nodes are roots of max-heaps.
@DanielFischer How do we conclude the last proposition? :/
@Hippalectryon I guess that was an irony (maybe Mathematica's irony). :-)
@Chris'ssis ?
20:32
@Hippalectryon Let's suppose we have $$\displaystyle\int\dfrac{t^{3n}}{1+t+t^2}dt$$
@Hippalectryon How do we proceed further? How can we rewrite $1+t+t^2$?
$$\int \frac{(1-t)t^{3n}}{1-t^3} dt$$
@BalarkaSen Good!
Then start partial factorizing like crazy.
20:35
@Hippalectryon Then? Recurrence relation, remember!
Hello!! We have an equation which has the roots $r1, r2$. We are looking for the appropriate iterative method $x_{n+1}=\phi(x_n), n=0,1,2$ such that the sequence $(x_n)$ converges ro the root $r2$, $\forall x \in [a,b]$. When we have four possible $\phi(x)$, how can we know which we have to choose?
@Chris'ssis Recurrence.... wrt which variable ? If it's wrt $n$, i'm not sure what to do here for a recurrence...
$$\int \frac{(t-1)t^{3n}}{t^3-1} dt=\int \frac{((t-1)t^{3n}-(t-1)t^{3n-3})+(t-1)t^{3n-3}}{t^3-1} dt$$
Ooh that
@Hippalectryon :-)))
20:38
nice trick
@BalarkaSen Thanks! :-)
@Hippalectryon I'll create some problems using that. Thank you for asking! :-)
No problem :)
Using that result I just created a crazy series.
Show me :D
@Hippalectryon Wait a bit ... :-)
$$\frac{1}{1/2+1}-\frac{1}{2(3/2+1)}+ \frac{1}{4(7/2+1)}- \frac{1}{5(9/2+1)}+ \frac{1}{7(13/2+1)}-\frac{1}{8(15/2+1)}+\cdots $$
Maybe it's not that crazy ...
21:05
Do I see a @BalarkaSen and an integral together? rubs eyes
I have a question. I hope someone can help me!

A dreamy cat doesn't notice a flower pot, which at first flies upwards and then downwards by an open window. The flower pot has been seen for 0.50 seconds, the height of the windowsill till the top edge of the window is 2,00 m. 2,00 m.

Which height over the Top edge of the window does the flower pot get?
@user159870 What have you tried ?
@Hippalectryon I don't know how to solve it and I don't know how to find $v_0$. Do you have any idea?
@user159870 I'm not asking you to find the solution, i'm asking what you have tried so far.
@user159870 To me your English is probably harder than the problem itself. :-)
21:15
@Chris'ssis Aren't you fluent in English ?
3
@Hippalectryon I starred your message.
:D
:O
Maths everywhere
@Hippalectryon :-)
I just noticed that "Hippa" reversed sounds like "Happy" :D
Evening, people.
21:22
Evening, my Lord :P
Lord of the Farin HAHA
(Farin in French sounds like flour)
@Hippalectryon Ah, yes, farine.
All hail the Lord of the Flour
I'm surprised to remember that.
It's been eight years since I last touched French.
$$\frac{1}{(1+1)^2}- \frac{1}{2(2+1)^2}+ \frac{1}{4(4+1)^2} -\frac{1}{7(7+1)^2}+ \frac{1}{8(8+1)^2}-\frac{1}{10(10+1)^2}+\cdots=?$$
@Chris'ssis Omoshiroi
21:26
@Chris'ssis I fail to see the pattern for the denominators.
$$\frac{-6 \left(\sqrt{3} (\log (3)-4) \log (3)+\Re\left(4 \left(\sqrt{3}-i\right) \text{Li}_2\left(\frac{1}{6} \left(3+i \sqrt{3}\right)\right)+4 \left(\sqrt{3}+i\right) \text{Li}_2\left(\frac{1}{6} \left(3-i \sqrt{3}\right)\right)\right)\right)+6 \sqrt{3} \pi ^2-4 \pi (10+\log (3))}{48 \sqrt{3}}$$
@Chris'ssis You meant $\dfrac{-1}{6(6+1)^2}$ didn't you ?
@Lord_Farin Maybe I should add some more terms.
Well .... something is wrong there ...
$$\frac{1}{1(1+1)^2}- \frac{1}{2(2+1)^2}+ \frac{1}{4(4+1)^2}- \frac{1}{5(5+1)^2} +\frac{1}{7(7+1)^2}- \frac{1}{8(8+1)^2}+\frac{1}{10(10+1)^2}-\cdots=?$$
@Chris'ssis I think some algebraic manipulation and writing $1 = \ldots$ in smart ways should work. Unfortunately, no time to fully explore this at the moment; I have to prepare an important presentation for tomorrow :(.
@Lord_Farin No worry. I only posted it for fun.
21:38
I know. I don't like to not have time for fun :).
:(
@Lord_Farin :-(
@Chris'ssis Well, today was a lousy day anyway.
@Lord_Farin What happened?
@Chris'ssis Several things. I heard that an apartment I applied for hiring had gone to someone else. I had trouble obtaining some stupid authorisations at my work. Also, my patience with slow-paced people has worn out today.
And as said, I have to prepare a presentation in the evening hours.
@Lord_Farin Just as a personal opinion, nothing can be worse than attending an interview where you're asked personal questions you don't wanna discuss about. :-)))
21:47
@Chris'ssis It's more like a monologue of vitriol :).
@Lord_Farin lol, kind of. :-)
Now, let me take my responsibility and remove myself from this room, so that I can actually work.
See you around.
@skullpatrol Thanks skull, you always make me smile :).
21:53
@Lord_Farin Here, to compensate
STAAP!!!
Sad kittens win
1 min ago, by skullpatrol
STAAP!!!
recursion
2 mins ago, by skullpatrol
STAAP!!!
21:57
7 secs ago, by Hippalectryon
recursion
5 secs ago, by Hippalectryon
7 secs ago, by Hippalectryon
recursion
Olol
11 secs ago, by Hippalectryon
5 secs ago, by Hippalectryon
7 secs ago, by Hippalectryon
recursion
How far does it go ?
Infinity
Have you seen this one? Let $x\in (0,\pi)$ and $\displaystyle f(x)=\frac{1}{\tan(x)}-\frac{1}{x}$. Prove that for $n
= 0
,
1
, . . .$
$$f^{(n)}(x)<0$$
@Chris'ssis Nope
It's too nice!
01:00 - 22:0022:00 - 00:00

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