Yo yo yo my name is joe I integrate backwards and derive down low

@MatheinBoulomenos was denkst du uber Bühnensprache?

idk

ein stimmhafter alveolarer Vibrant klingt heutzutage ziemlich seltsam

unlike in Spain: youtube.com/watch?v=04kGcQ_gIVk

@MatheinBoulomenos brilliantly scripted

is there a bijection $f:\Bbb N \to \Bbb Q$ such that $\displaystyle \sum_{n=0}^\infty f(n) 2^{-n} = 0$

Only by computing $dx^2$, I got sooo many terms already. Or is there any easy way?

is there an inverse for the factorial function?

Do all trigonometric identites also work for trigonometric identities with complex numbers?

Do all trigonometric identites also work for complex numbers is what I meant to say xd

$\cos(x)=\mathrm{Re}(e^{ix})$ holds for $x\in\mathbb{R}$, but fails in general for complex $x$.

yeah I see

@Thorgott but all of the identites for multiple angle , shifts and angle sums hold for complex numbers ?

Yes, the addition theorems still hold

thanks

@JamesGroon it's even better than that

$e^{i(x+y)} = e^{ix} e^{iy}$ naturally

and from that you can derive the trigonometric identities

oh and the connection is $e^{ix} = \cos x + i \sin x$ which works for complex $x$

and instead of $\cos(x) = \Re(e^{ix})$, you have $\cos(x) = (e^{ix} + e^{-ix})/2$

in a sense $\exp$ is more beautiful and should come before $\cos$ and $\sin$, but that's just my opinion

wouldn't that imply learning real analysis without introducing sine and cosine ?

seems like a tough job for me

well they can stay

but like I don't know maybe don't be so afraid of complex numbers is what I'm trying to say

i see but I don't want to use something in which I'm not 100% positive and also proving all those identities ..

i rather ask here

to be somewhat sure

how can we write x^x in some form of the gamma function?

In the first analysis lecture I heard, we defined the exponential as complex power series, sine and cosine as it's odd and even part and then deduced their properties from there. It's not much more work to extend the basic theory of sequences, series and continuous functions to complex domains and that's already all you need for that. I do think that approach is both more elegant and insightful.

@LeakyNun what UK maths institutions you would recommend for Bsc degree ? Does not have to be top-tier though

Cambridge

You know of anything where entry standards are not so high ?

I'm mature student, got some (not best) qualifications, but want to continue studies in UK

no idea, sorry

ok

thanks

Is there a big difference between Russell group uni and the one which is not in R. Group ?

How to show that $1+\langle3+i\rangle$ has additive order 10 in $\Bbb Z[i]/\langle3+i\rangle$? Does the observation $i=-3=7$ mod $\langle3+i\rangle$ help in that?

$\Bbb Z[i]/(3+i) = \Bbb Z[X]/(X^2+1,X+3) = \Bbb Z[X]/(X+3,10) = (\Bbb Z/10\Bbb Z)[X]/(X+3) = \Bbb Z/10\Bbb Z$

Man! I have to learn so much.

doesn't everyone

name a function whose integral is always a perfect square

name a function whose integral isn't always a perfect square

$\int_{0}^{\infty} xxxxxxxxxxxxxxxxxxxxxx dx $

name no more than three fruits from India

easy to state but hard to prove

In $\Bbb Z \oplus \Bbb Z$, let $I = \{(a, 0) | a \in\Bbb Z\}$. Show that $I$ is a prime ideal but not a maximal ideal.

I see from $(\Bbb Z \oplus \Bbb Z)/I\cong\Bbb Z$ that indeed this is the case. But, while proving directly that $I$ prime, I go like 'let $(a,b)(c,d)=(ac,bd)\in I$ that is $bd=0$, so $b=0$ or $d=0$ hence $(a,b)\in I$ or $(c,d)\in I$'

where am i wrong?

[Random]

@Silent if $R/I$ is an integral domain iff $I$ is prime

can some one help me in math.stackexchange.com/questions/3194837/…

@MatheinBoulomenos Yes, I see that $I$ prime from $(\Bbb Z \oplus \Bbb Z)/I\cong\Bbb Z$. But, can't see directly that it is prime.

you already wrote a proof above, there's nothing wrong with it

ok. thank you very much.

Hello

Are k-cycles the only cycles with $\sigma ^k=i_n$ ?

No, @FuzzyPixelz. What if $k$ is even?

Well alright, what if add $\sigma^{k-1} \neq i_n?$ making k its order ?

the condition $\sigma^{k-1} \neq i_n$ doesn't guarantee that $k$ will be the order of $\sigma$. But it is true that the order of a cycle is equal to its length

Should it be for all $l \lt k$ ?

right

hi chat

@FuzzyPixelz Like Ted said suppose $k$ is even. What happens to a 2-cycle?

Yes it doesn't hold for transpositions you should get the identity for k=4 for example, ruining the case for 4-cycles

I don't know how to prove that the order of a cycle is its length

Well if we consider any cycle

If we composing by $\sigma$ on the first element, we wouldn't get it back until k iterations..

If it had more or less than k elements that wouldn't hold

Am I missing something ?

What exactly are you trying to prove? That the order of a cycle is its length?

Yes?

The order of $\sigma$ is the least positive integer $k$ such that $\sigma^k = i$, right?

Yes

Since $\sigma$ is a $k$-cycle then $\sigma^k = i$.

Furthermore, $\sigma^l \neq i$ for any $0 < l < k$.

It can't be an &l&-cycle for all those values

Suppose $\sigma = (n_1, n_2, \dots, n_k)$. Suppose $\sigma^l = i$ for some $0 < l < k$. Then what would $\sigma^l(n_1)$ be?

$n_1$?

And also n_l

Right.

But $n_1$ cannot be $n_l$.

By definition, otherwise the cycle would have repeated elements.

Actually, $n_{l+1}$.

Off-by-one error.

Yes, but can it be an $l$-cycle for $l \gt k$ ?

What do you mean "can it be an $l$-cycle"?

Hi. Given $n$ points in a unit hypercube, I'm looking for an algorithm to partition it into $k$ hyperrectangles, for a specified $k$, so that each hyperrectangle contains approx. $n/k$ points. Does this problem have a name?

We proved that it can't be an $l$-cycle for $0 \lt l \lt k$, what about the rest?

I thought you wanted to prove that the order of a $k$-cycle is $k$. Our proof is complete.

We only had to prove that, if $\sigma$ is a $k$-cycle, then $\sigma^k = i$ and $\sigma_l \neq i$ for any $0<l<k$.

We don't have to worry about $l > k$ at all.

My bad, I'm still getting the definition wrong

howdy @Mathein

The last message was posted $1$ hour ago.

3 years later...

Does every single variable polynomial over a field have a splitting field?

what is the relation between a limit point of a subset $S$ in a metric space $(S,d)$ and the limit of a convergent sequence that is entirely contained in $S$?

If $x$ is a limit point of $S$, then there is a sequence in $S$ that converges to $x$. The reverse implication holds under the additional assumption that the sequence is not eventually constant.

@Thorgott so, if we assume that there is a convergent sequence that is entirely included in $S$, then one can show that the limit of the sequence is a limit point of the set $S$, right?

It is not necessarily a limit point of $S$. It definitely is contained in the closure of $S$ though.

The crux is that if $x$ is a limit point of $S$, then every neighbourhood of $x$ contains a point of $S$ that is not $x$ itself. Consider $S:=[0,1]\cup\{2\}\subset\mathbb{R}$. Then the sequence $(2)_n$ is contained in $S$ and converges to $2$, but $2$ is not a limit point of $S$.

i see your point! so it will not be good enough then if one uses the fact that if $S$ is closed then the limit of the sequence $x_n\in S$ MUST be in $S$?

i thought i could argue that if the limit of the sequence is a limit point then it must be in $S$

That is still true. The limit of a convergent sequence in $S$ is a limit point of $S$ or a point of $S$ (as the point $2$ in the above example). To see why this is true, think about the following: a convergent sequence in $S$ must either be eventually constant (in which case the limit is a point of $S$) or it's limit must actually be a limit point of $S$.

could you provide a hint on how to show the last part? namely, the limit of a convergence sequence in $S$ must actually be a limit point of $S$

Well, the additional assumption that the sequence is not eventually constant is crucial here (if that is not satisfied, the above counter-example holds). Under that assumption, use the definition of a limit to see that eventually all terms of the sequence lie in an arbitrary neighborhood of the limit and that since the sequence is not eventually constant, the arbitrary neighborhood must contain a point of the sequence distinct from the limit. Then the definition of limit point is satisfied.

@Thorgott Thanks a lot! :)

np

Anyone familiar enough with mathematica to know how to export a table into latex as a tabu with horizontal and vertical separating lines at strategic locations? I'm familiar with teXform, but not enough to know if such precise manipulation of the resultant table is even possible.