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12:00 AM
Ah, point.
 
I've seen stuff like that, but it's been a while.
 
> In this talk we will introduce the ordinary power series generating function and consider its applications to binomial identities, recursion relations and combinatorial enumeration. Along the way we will encounter famous integer sequences such as the Fibonacci and Catalan Numbers. If time permits, we will discuss other kinds of generating functions.
That's what's on today in uni
 
@Secret Nice. Have you read generatingfunctionology?
 
Not yet, I am mostly busy on trying to get my chemistry DFT calculations to work, and after advice from my collegues, realised I forgot to substitute 32 electrons for one effective charge, thus end up using too much memory and failing the calculation
The only bit of generating function I have read so far is when DHMO and 2017 discussed about the method of snake oil
 
12:07 AM
Flajolet's book on Analytic Combinatorics has some good stuff, but it's got a lot of formalism.
 
[Random thought] Formal power series with continuous parameter n:
$$\sum_{n \in \mathbb{R}}c_nx^n=\int_a^b c(n)x^n dn$$
 
Compare that with $\int_0^\infty f(x)e^{-s x}\,dx$.
 
Laplace?
 
The analogy is exact, in fact, if you take $(a,b)=[0,\infty)$ and $x=e^{-s}$.
 
hmm, looks like a generalisation of Laplace transforms
 
12:20 AM
Quite.
Pretty much, yeah.
So generating functions are to discrete sequences / difference equations what Laplace transforms to continuous functions / differential equations.
 
Magic?
 
Eh, it's just taking what @Secret suggested and putting it in a more recognizable form.
Once you have that motivation, the analogy between discrete and continuous systems is obvious.
 
I see
 
How would you begin proving this?
 
in The Periodic Table, 11 mins ago, by Hexacoordinate-C
I remember some months ago someone ask about Absolute Zero. Now third law of thermodynamics gets its proof https://www.sciencealert.com/after-a-century-of-debate-cooling-to-absolute-zero-‌​has-been-declared-mathematically-impossible so it looks tricky now the reach it :)
This is an example where if $\aleph_0$ physically exists, it will be interesting
In short: No you cannot mathematically reach absolute zero without an infinite reservior and infinite time
 
12:33 AM
There's the additional complication with the third law that the ground state of a system may itself be degenerate at absolute zero, in which case the entropy needn't be zero.
 
O yeah, amorphous solids love to do that
 
Yeah.
 
An airplan is flying on a bearing of 335 degree at 530 mph. Find the component form of the velocity of the airplane
to find the component form i did <530cos335,530sin335>
 
335 degrees relative to what?
Typically they'd specify that.
 
got points <480.34,-223.987>
maybe the north side
doesn't say it in the question
 
12:36 AM
Well, that obviously makes a big difference.
If you fly 335 degrees counterclockwise relative to north, you're basically going north-northwest.
 
but isn't 335 degrees down at the 4th quadrant
wont that be south?
 
That's if you take it to be counterclockwise relative to east.
That may be reasonable, but they should say it.
 
hi
i'm feeling down. someone give me a challenging math thing
 
(also an addenum: Negative heat capacity systems (like a clusters of population inverted sodium ions), while physical, are actually hotter because their entropy energy relationship (temperature) is of the opposite sign)
 
Yeah.
Hence why magnetic cooling is a thing.
 
12:38 AM
so if they don't do I assume its going to be north west
so 90+335 or 90-335?
 
Probably best to say why it's ambiguous. (Also, I should've said "335 degrees clockwise from north.")
 
oh I'll try by adding cuz subtracting will give me an negative degree
and I don't think its right
 
One thing to recognize is that, while this does matter in determining which component is which here, it doesn't matter in the magnitudes of those components.
 
@Dragneel That doesn't make sense
 
If you do 30 degrees north of east, for instance, and got $(v_x,v_y)$
then 30 degrees east of north with the same speed would just be $(v_y,v_x)$
 
12:42 AM
$16 = 15+1 = 3 \cdot 5 + 1 = 2^4$
 
The signs can be different, and which goes with x/y can be different.
The first prime isn't 3, it's 2. @Lozansky
So that example should be 2*3*5+1=31.
 
Oh they're in order
 
magnitude is the velocity in this question right
 
Magnitude of the velocity is the speed they give.
By magnitude of the components, I just mean "ignoring +/- signs"
 
oh
so the component form of the velocity would be 530cos425 , 530 sin425
 
12:46 AM
Right.
If you're doing counterclockwise from north.
If you're doing clockwise from north, it'd be 90-335 instead.
 
oh bit in one case the component alue could be negaive right so do i ignore the signs?
 
cause the magnitude I get will be the same?
 
The magnitude is already given by the problem to be 530.
If you ignored the signs, you'd end up having it having the right lengths but not the right direction.
compare the vectors (3,4) and (3,-4) to see what I mean.
 
but like the values would be different right, the x and y would be flipped as u showed
ones going up the other down
 
12:49 AM
Sure. But the signs would come out from the sine/cosine directly.
You shouldn't need to insert them by hand if you've done the angle properly.
 
ya but they would be different if its in different quadrants
 
It'll depend on where you start from, sure. But that should be reflected in your angles.
 
oh ok
 
1:05 AM
@Meow What types of problems are you looking for?
 
topology, linear algebra, anything
 
Here's a kind of puzzle
Let's say you have some polynomial that starts with the terms $x^{17} + 5x^{16} + 13x^{15} + \ldots$
Prove that not all of its roots are real
 
looks vieta-y
 
Ojemba is sitting on a sled on the side of a hull inclined at 60 degrees. The combined weight of ojemba and the sled is 160 pounds. What is the magnitude of the force required for Mandisa to keep the sled from sliding down the hill
I don't know if this is how Im supposed to do it
 
Hi @Semiclassical
 
1:12 AM
Try it and see @Meow
 
the sum of the roots must be -5
 
@MATHASKER The correct answer is "too much force" because Ojemba is malnourished
 
cyclic sum of roots is 13
wtf @MATH the name literally changed halfway through the problem
 
Ya lol I know the book has the problem I'm just copying and pasting what it says lol
 
1:30 AM
Don't question this beautiful book that uses racially diverse names of males and females instead of blindly asserting the white male patriarchal and racial dominance over society
(jk im sane, and you never said where the other person is applying their force)
 
anyone knows the Runge Kutta methods
 
1:46 AM
@Simple works great for physics simulators
I don't know how to prove the order of their errors, but they are pretty simple to implement lol
 
yes
I am trying to show math.stackexchange.com/a/240478/127116 why we just require the sum of $b$ equals to 1
 
I think implicit methods are where it's at
 
I mean in general
 
I don't know a good answer, though it seems intuitively obvious that it's not a weighted average anymore if it doesn't total up to 1
 
The text book I am using, it has a additional condition, $\sum a_j=c_j$ where $a_j$ is the row entries of the matrix $A$
Then, I find out the additional condition is not necessary
Do you have a good example to prove the additional condition is not necessary
 
2:00 AM
Is it really not necessary though?
 
2:23 AM
hi! Can someone please help me with my question? I would appreciate it so so much math.stackexchange.com/questions/2188735/…
 
@MickLH yes
Runge–Kutta methods are methods for the numerical solution of the ordinary differential equation d y d t = f ( t , y ) {\displaystyle {\frac {dy}{dt}}=f(t,y)\,} which take the form y n + 1 = y n ...
 
2:51 AM
Hey, hey! Jessy's come to bother you.
0
Q: How do we know all Sylow $p$-subgroups of a given order are distinct?

ALannisterRight now, I am in the process of showing that a group $G$ of order $35$ must be cyclic, for the purpose of then showing that there is a unique group of order $35$, namely $\mathbb{Z}_{35}$. To that effect, I was interested in applying an argument seen in the answer to the linked question given...

@BalarkaSen the star spammer is back?
 
3:10 AM
@arctictern here ?
 
yes
 
I have 1 question. So I am trying to prove the following thing The sequence $M_1 \rightarrow M_2 \rightarrow M_3$ is exact if for all modules $N$ $0 \rightarrow Hom(M_3,N) \rightarrow Hom(M_2,N) \rightarrow Hom(M_1,N)$.
so I am almost there
I proved everything but now I want to show $\ker(Hom(M_2,N) \rightarrow Hom(M_1,N))$ is equal to the $im(Hom(M_3,N) \rightarrow Hom(M_2,N))$
I showed one inclusion that is im is included in kernel I am almost there for the other side. But here is my issue
Let $\phi \in ker(Hom(M_2,N) \rightarrow Hom(M_1,N))$
call that map by $\bar{u}$ that is the map induced $M_1 \rightarrow M$
 
Hey @Adeek and @arctic
 
@Adeek aren't you hypothesizing that ker equals that im for all N?
 
hey @Daminark
 
3:18 AM
and trying to prove $M_1\to M_2\to M_3$ is exact
hello
 
no it is the other way around
we have that $M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0$ is exact
and we want to prove that passing to hom is left exact
we have $\phi \circ u = 0$ so that means that $im(u) = ker(M_2 \rightarrow M_3) \subset ker \phi$
can we conclude we have a map from $\eta : M_3 \rightarrow N$ such that $\phi = \eta \circ v$ where $v : M_2 \rightarrow M_3$ ?
I think that is universal property of kernels or something ?
@arctictern ?
 
You said you want to prove $M_1\to M_2\to M_3$ was exact if $0\to\hom(M_3,N)\to\hom(M_2,N)\to\hom(M_1,N)$ for all $N$
did you mean it the other way around?
 
yeah sorry I meant it the other way around
 
anywho not in the mood for modules
 
I proved it for the above that wasn't bad
oh okay
 
3:42 AM
It is true yeah @arctictern More generally Given three modules $M_1,M_2, and M_3$ and two morphism $\phi_1 : M_1 \rightarrow M_3$ and $\phi_2 : M_1 \rightarrow M_2$ such that $ker(\phi_2) \subset \ker(\phi_1)$ then we have $\phi_3 : M_2 \rightarrow M_3$ making diagram commute in general it is defined as $\phi_3(m_2) = \phi_1(\phi_2^{-1}(\{m_2\})$ and this makes sense because $\phi_1$ fixes the fibers and we can prove this thing is A-module homomorphism
 
4:32 AM
When solving differential equations $dy/dx = x$ are we allowed to rearrange the operators to $dy = x dx$ , or is this just a quirk of physicists?
 
What you're really doing is just the FTC: $y(x)=\int\frac{dy}{dx}\,dx+C.$
(There are things that can be said of differential 1-forms, but for the purposes of ODEs it's enough to just take it as a formality.)
The main reason to do it is because it makes things easier to write stuff out. For instance, suppose $dy/dx=x/y$. One can rewrite this as $y(dy/dx)=x$ and then integrate both sides to get $\int y(x)\frac{dy}{dx}\,dx=\int y\,dy = \int x\,dx$.
 
@Semiclassical that formal way is what I would normaly do actualy.
 
Yeah.
 
I have always thought of it as really doing that first line; but I have been unsure why the other way doesn't work. I don't know much of anything about differential 1-forms; so the best I can do is to say that $dy$ on its own is not infinatesimal with respect to anything.
 
It has the virtue of being a bit more efficient; it 'automates', for instance, the use of the substitution rule in integrating $y$.
 
4:42 AM
not sure what you mean by automates here.
 
Well, if you're doing it by integration you have to stop for a moment and say "okay, I can use the substitution rule to trade one of the integrals for one over $y$."
With the differential notation, you take that for granted.
But, I'm someone who is used to the (probably abusive) differential form of it. So I don't really think about it much in practice.
 
an integral to me is an $\int$ symbol followed by what to integrate accross. What to integrate accross on its own seems a bit meaningless.
 
That's not an uncommon perspective, I think, especially from the math POV. There's probably some discussion of it on the main site.
 
Hmmm. I'll have to look into it later.
Anyway; thanks for the help @Semiclassical I've got to go now.
 
mmkay
 
 
1 hour later…
6:14 AM
Everybody is welcome to join the NCAA bracket challenge Password is quora.
For added banter:

 The Clubhouse

General discussion for sports.stackexchange.com
 
6:30 AM
Brackets lock in 8 hours.
 
7:10 AM
@everyone what year in undergrad are you supposed to take linear algebra and diffeqs?
 
Here linear algebra is in the first year as well as a bit of of ODEs in the analysis course, the actual diffeqs courses are a PDE one in the second year and a non mandatory ODE one in the third
 
ok cool
 
Both must be completed by the end of second year.
Most people take them in second year.
For math and physics/engineering majors.
 
7:32 AM
Hi all!
what's clubhouse meant for sports right , is it for wrestling ??
wrestling is also a sport
na
:)
 
@Julian Here it's linear algebra first/second year (it's weird), ODE & PDE are not mandatory, but has a prereq of analysis (so third year). Sometimes basic results on existence/uniqueness feature in analysis
 
Oh I see
I just want to know because I am trying to apply for an internship and wanted to know what year you usually take those classes.
 
I think the "standard" is that you take linalg and diffeq second year ish?
(Assuming you start from scratch with calculus)
 
ok ty
 
If you had calc in high school you'll prob do linalg, then multi
In first year
 
7:45 AM
oh i took multi then now taking linear
and diffeqs sec sem
 
@Semiclassical how could I show that Lorenz attractor is not energy conserving!any reference !
 
@BAYMAX yeah it's a sport even though they took it out of the Olympics because of lack of interest :-(
 
aahhh...WWE .. IWGP..NJPW..TNA..ROH..UFC..??
@skillpatrol
 
8:08 AM
yeah! nice @skillpatrol
You watched the movie pele
 
Is there something like a "maximal normal neighborhood" of a point in a Riemannian manifold? I.e. a maximal open neighborhood $U$ of $p$ such that the exponential map $\exp_p$ is a diffeomorphism from $\exp_p^{-1}(U)$ onto $U$?
 
yeah! nice @skillpatrol
 
@BAYMAX yup
 
@abenthy Sure, but it would be highly non-unique. (Take the poset of open subsets of $T_pM$ for which the exponential map is a diffeomorphism onto its image and run Zorn's Lemma.)
 
Right..., too bad :| I would like to be able to say something like "the geodesic symmetry of a riemannian manifold is $s_p = \exp_p \circ (-\text{id}) \circ \exp_p^{-1}$, but it seems like we always have to put in the choice of a normal neighborhood $U$ of $p$. Is there a nice mathematical concept to put all these objects $(s_p,U)$ into "one object" so that I can rigorously speak of the geodesic symmetry at $p$?
 
8:20 AM
that was a pure motivational movie @skillpatrol
also now a days I am in searchof similar motivational movies!
 
For sure pal.
 
@abenthy Perhaps doing something germinally (instead of maximally).
 
8:56 AM
Anyone can help me on this question on fixed point iteration?
I dont understand why $g'(\xi_{n}) \geq 2\cdot ln(2)$
 
@Mike Yeah, I think so too, one can probably formulate it elegantly in this language with something similar to sheafs/stalks. It's probably not worth the effort for me sadly, just curious :)
 
@LittleRookie You forgot absolute value there.
 
yes with absolute
 
You have $|g'(x)|=2^x\ln 2$ and for $x\in[1,2]$ you get $|g'(x)|\ge 2^2\ln 2$.
Simply from $2^x\ge 2^2$.
oops I am saying non-sense
 
This is only true for certain value of x
 
9:09 AM
Correction: For $x\in[1,2]$ you get $|g'(x)|\ge 2^1\ln 2=2\ln 2$.
 
Which is a consequence of $2^x \ge 2^1$. (Since the function $x\mapsto 2^x$ is increasing.)
 
which means that $\xi_{n} \geq 1$ for $n=1,2,3,..$
 
Well, that's exactly what you wanted - with $x=\xi_n$.
 
and thats the part i dont understand why
can u show me how $\xi_{n} \geq 1$ ?
 
9:11 AM
I am not sure what is $p$ in what you wrote.
But if you get $\xi_n$ from mean value theorem, then you know that it is between $p$ and $p_0$.
 
$p$ is the fixed point in [1,2]
then i can tell $p_{1} \in [1,2]$
But for $p_{2}$ onwards, i cant tell
 
I see.
At least now I understand what is the problem.
If fact, how you know that $p_1\ge1$?
If we can choose arbitrary $p_0\in[1,2]$, then we could get for $p_0=2$ the value $p_1=g(p_0)=g(2)=0$.
 
i mean $\xi_{1}$ not $p_{1}$ sorry
i cant tell for $\xi_{2}$ onwards
 
Didn't you have some kind of general theorem which says that if $|g'(p)|>1$ then the fixed point iterations do not converge?
You could use that instead of what you wrote in the image.
I'll have to leave, sorry. Maybe somebody else will pick this up. I have also mentioned this in calculus room - perhaps somebody will notice the problem there.
 
Thanks
I only have theorem for <1
 
 
2 hours later…
11:11 AM
Well, chat is pretty dead today :O
 
yup
perhaps, too much pi
we've crossed the feynman point
 
 
2 hours later…
12:55 PM
Well, I did eat too much pie on Tuesday...
But that only seemed to expand the diameter of my tummy, not remove my desire to chat about mathematics :P
 
1:07 PM
@SteamyRoot did you know about the Feynman point?
 
1:17 PM
Uhhh
If I remember correctly, it's a point where some digit repeats a few times
And it happens sooner than you would statistically expect it to happen (assuming $\pi$ is normal)
 
yup, 9s repeat six times
starting at the 762nd decimal place
 
Could you please share your answer to this question? Thanks...
131
Q: Counterintuitive examples in probability.

Leila HatamiI want to teach a short course in probability and I'm looking for some counterintuitive examples for it. Results that seems to be obviously false but they true or vice versa... I already found some things. For example these two videos: Penney's game How to Win a Guessing Game In addition I'v...

 
1:54 PM
@Leila St. Petersburg paradox
 
@Lozansky Thanks very much. But It has mentioned in answers...
 
 
1 hour later…
3:07 PM
 
Interesting
 
I have to find a way to decode that message
I will not adapt of any change while doing that however
 
Haha
Can someone help me out with this Discrete Math problem. I have to prove the following.
 
3:46 PM
@Dragneel Hint: If $k=l$, then how can you express $k$ in terms of $n$
If $l>k$, then how does $n=k\cdot l$ compare to $k^2$?
 
Do you like my t-shirt? @Danu
 
No
(I'm in a grumpy mood; no offfence intended)
 
np pal
:-)
 
If every subgroup of a finite group G is normal, then G is solvable.
Any link to solve this!
 
@Dragneel Incidentally, that's why when you check a number for its prime factors, you only need to check up to the greatest prime lesser than the square root of the number.
 
4:01 PM
Thanks for the tips guys. I'm still trying to solve it as we speak.
Should I be using Direct Proof?
Or should I approach this using another method? (such as Contradiction, or Contrapositive, or Induction)
 
The contrapositive and induction are a lot more effort than its worth. A contradiction is pretty straightforward but could also be easily rephrased as a direct proof
try manipulating the inequality $k \le l$
 
I'm not sure how to express $k$ in terms of $n$
 
4:20 PM
You do, in case $l=k$
So first solve that case, then study the case $l>k$ and see why there's nothing left to do
 
@Dragneel Hint- the inequality $k \le sqrt(n)$ is equivalent to $k \times k \le n$. Now remember that $n = kl$ and $1 \le k \le l <n$. How do you go from the given data to the equivalent conclusion?
 
So, in the case where $l=k$, then I square both sides of $k ≤ sqrt(n)$ which becomes $k^2 ≤ n$. Which is $k^2 ≤ kl$. Dividing both sides by $k$, the inequality becomes $k ≤ l$.
Would that be sufficient as a proof?
 
4:43 PM
You're starting with $k\leq \sqrt n$. But that's not right. You should start with $k\leq l$ (or $k=l$ or $k<l$) and then show $k\leq \sqrt n$.
But the steps are basically the same ones you just did, but run in reverse.
 
Ahh, got it.
 
5:02 PM
Is it possible to follow an user profile?
*at this site, of course :P
 
@Anderson someone had developed an extension to do that, but he deleted his profile and his work
 
That's a pity... So there is not a way of doing that... It would be interesting!
 
@AndersonFelipeViveiros he came back recently as Normal Human and it seems he no longer supports the extension I mentioned (see normalhuman.github.io)
 
I see... Thank you anyway, @LeGrandDODOM !
 
5:22 PM
Let $f$ be a polynomial with nonnegative integer coefficients. If $f(1)=7$ and $f(7)=7597$, what is $f(10)$?
^ A nice puzzle
 
@MikeMiller Fixed typo
And no
 
But I see your thinking
 
haha cute problem
haven't done the calculation but the trick is cute
did the calculation but don't want to spoil
 
5:30 PM
Nice
@MikeMiller It also answers the question, "Suppose I have a polynomial with nonnegative integer coefficients. How many values of the polynomial do you have to ask for in order to figure out what the polynomial is?"
 
Yup.
 
(Assuming you can't ask for $f(\pi)$ — only values at integer points)
 
note the f(1)=1 edge case
 
Hi chat
What trick ?
 
@MikeMiller Right, yeah. You still only need two values though
 
5:45 PM
yup
 
What is the trick you are talking about ?
 
@Astyx See my puzzle above
 
I saw the puzzle, thus my question :)
 
Does anyone have any experience with affine transformations
I have some questions regarding transformations between two triangles
Least squares fit
and Umeyama
and ICP
 
[Method of snake oil thought]
Let
$$L_n(f(n,m))=\sum_{n\in I}f(n,m)=a_m$$
Then consider generating function:
$$F(x)=L_m(a_mx^m)=\sum_{m \in J}a_mx^m=\sum_{m \in J}\sum_{n\in I}f(n,m)x^m$$
Now by linearity of $L$
$$F(x)=L_m(a_mx^m)=L_m(L_n(f(n,m))x^m)=L_m(L_n(f(n,m)x^m))$$
$$F(x)=L_n(L_m(f(n,m)x^m))$$
$$F(x)=L_m(a_mx^m)=L_n(L_m(f(n,m)x^m))$$

If $\displaystyle{L_m(f(n,m)x^m)=G(x)=L_n(b_nx^n)}$, then

$$F(x)=L_m(a_mx^m)=L_n(b_nx^n)$$
$$a_k=b_k$$
So method of snake oil works as long the inner sum that is resulted after interchange becomes a generating function, thus allow the coefficients to match up
If $G(x)$ does not exist, then method of snake oil fails
 
5:54 PM
To prove this, I can prove it in two parts right? First I prove $p|n$, then I separately prove $p \le sqrt(n)$ right?
 
@Akiva So what's a nice way to show the result (I got it, but not sure wether my way is the best) ?
 
[Method of snake oil thought] Now what about:
Let
$$L_n(f(n,m))=\int_B f(n,m) dn= a(m)$$
Then consider some "general power series transform":
$$F(x)=L_m(a(m)x^m)=\int_C a(m)x^m dm=\int_C\left(\int_B f(n,m) dn\right) x^m dm=\int_C\int_B f(n,m) x^m dndm$$
Suppose Fubini's theorem holds, then
$$F(x)=\int_B \int_C f(n,m) x^m dmdn$$
Now if $\int_C f(n,m) x^m dm=G(x)=\int_B b(n)x^n dn$, then
$$F(x)=\int_C a(m)x^m dm=\int_B b(n)x^n dn$$
Finally (condition to be found)
$$a(k)=b(k)$$
Note unlike the discrete case, whether $C=B$ matters (not sure how)
Also as mentioned by semiclassical. Let the subtitution $x=e^{su}$ and you recover laplace transform version of the above
 
 
1 hour later…
7:38 PM
Is this person's answer correct? math.stackexchange.com/a/1208758/249166
I got confused when he wrote $(a+2)-a=2$. Why would he subtract?
 
7:50 PM
hi
 
Evening
 
8:33 PM
@Dragneel It's known that, for any $d$, $x$, and $y$, if $d$ divides both $x$ and $y$, then $d$ divides $x+y$ and $d$ divides $x-y$.
That is, if you divide two things, you also divide their sum and difference.
So, that's why he could subtract them. As for why he subtracted them, well, it's 'cause it helped solve the problem
 
how are you all :(
 
8/10
Tired though
 
ok
 
@AkivaWeinberger Thanks for the explanation. I came to realize that shortly after asking though. Thanks anyway :D
I'm trying to figure out how to recognize perfect $n$th powers from their prime decompositions.
 
@Danu I fixed the typos
Guys, my music assignment is one measure short
What do I do
 
8:40 PM
one measure of rest
 
Time to improvise!
 
Desperate times call for desperate measures.
 
what do you play? @Akiva
 
Piano, but this is being written on my computer
 
make the time signature 128/4
so you can play a whole song
in one measure
 
8:41 PM
The assignment is to do a fugue with three voices. Most people are doing a spoken fugue, but we were allowed to do a musical fugue as an option
and that's what I'm doing
 
Do you have any idea about this contradiction?
 
Hm, I can duplicate one of the measures
I mean, I already had this measure play twice in a row, but hey, now it's three times
 
I have an integral from - inf to + inf of 1/(x + i0). If I regularise it with an exponent like e^{- i 0 x} and use residue theorem, I get - 2 pi i. But if I just use Sokhotski formula, I get - pi i
 
Hi chat
 
8:45 PM
Lol I literally don't understand this definition
 
The first bullet point is the definition
It's a number that's an $n$th power of another number
 
How is $5$ a multiple of $32$
 
@Dragneel In that example, $n=5$
So 5 is a multiple of 5
For example, $1728=2^63^3$
So, because $6$ and $3$ are both multiples of $3$, we have that $1728$ is a perfect third power
Er, a perfect cube
(In fact, $1728=(2^23)^3=12^3$)
@Dragneel
 
So, in your example, $n$ would be $3$?
 
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