Mathematics - 2015-01-12 (page 4 of 4)

evinda

22:02

@DanielFischer I haven't undrstood how we constructed the function $h$. Could you explain it further to me? :/

Daniel Fischer

We map $0$ to the smallest element of $B$ (the minimum), then we map $1$ to the next larger element, the second smallest element of $B$. And so on, while we have not yet exhausted $B$, we map $k$ to the $k+1^{\text{st}}$ smallest element of $B$.

evinda

@DanielFischer I thought that we were looking for a map from a m to f(B).. Am I wrong?
If I am right, then why do we take the minimum element of B, and then the second minimum and so on?

Daniel Fischer

35 mins ago, by Daniel Fischer

@evinda That doesn't make sense. Let's forget about $f$ and suppose that $A = n$, hence $B \subset n$ to simplify notation. Then we map $\min B \mapsto 0,\, \min (B\setminus \{\min B\}) \mapsto 1$ etc.

"Let's forget about $f$ ..."

Or take $f = \operatorname{id}$.

evinda

So B corresponds to f(B) right? @DanielFischer

@DanielFischer So if we take $f = \operatorname{id}$ can we then generalize in order that it holds for each function?

Daniel Fischer

@evinda Yes. If you want to keep $f$, then replace $B$ with $f(B)$ in the construction, so $h(0) = \min f(B)$ - assuming $B\neq\varnothing$ - and $h(k+1) = \min \{ u \in f(B) : u > h(k)\}$.

@evinda "taking $f = \operatorname{id}$" is just notation. We identify $A$ and its subsets with $n$ and its subsets via $f$.

evinda

22:21

@DanielFischer And then why do we have to show that $h(k) \geq k$ ?

Arkamis

There needs to be a reason to close, "user refuses to put any damn thought into the question what-so-f***ing-ever."

Daniel Fischer

@evinda So that we know that the domain of $h$ is contained in $n$. And hence we can conclude that $B$ is finite.

@Arkamis That would be useful indeed.

Ted Shifrin

I'm looking at fewer and fewer and fewer and fewer and ... questions these days.

Arkamis

To wit: math.stackexchange.com/q/1101883/31475

evinda

@DanielFischer We are looking at the function $h:m \to f(B) \subsetneq n$.
So don't we have to show that $h(k)<n, \forall k \in m$ ? Or am I wrong?

Ted Shifrin

22:27

Yup, totally standard easy calculus problem ... Sigh.

Mike Miller

hello

Ted Shifrin

G'bye.

Kevin Driscoll

You say goodbye, but I say hello

Ted Shifrin

You're the reincarnation of the Beatles, @Kevin?

Mike Miller

Lots of people do that, @Ted, they just made it popular.

Ted Shifrin

22:30

I prefer just to leave without ceremony.

Kevin Driscoll

@TedShifrin Imagine how confusing it is having all 4 in my head at once......

Ted Shifrin

You're a physicist, @Kevin. You can handle it.

Mike Miller

That's weird.

You're weird.

That's to Ted, of course. In my experience voices in your head are completely normal.

Daniel Fischer

@evinda Yes, we know $h(k) < n$. Together, $k \leqslant h(k) < n$ tells us the domain of $h$ is contained in $n$.

evinda

@DanielFischer From $k \leqslant h(k) < n$ don't we get the range of h? Or am I wrong?

Daniel Fischer

22:35

@evinda We know the range, it's $f(B)$.

evinda

@DanielFischer I see... But why do we get from this inequality $k \leqslant h(k) < n$ the domain of h?

Daniel Fischer

@evinda We don't get the domain from the inequality, only a restriction on the domain. We know the domain is some $m\in \omega$, or $\omega$ itself. The inequality tells us the domain is not $\omega$, and more, it is an $m\in \omega$ with $m\leqslant n$.

evinda

@DanielFischer Do we see from $k \leqslant h(k) < n$ that $m\leqslant n$ because of the fact that we are looking for a bijective function?

Daniel Fischer

@evinda The inequality - in fact the inequality $k < n$ for all $k\in \operatorname{dom} h$ that follows from it - shows that $\operatorname{dom} h \subset \{0,1,\dotsc,n-1\} = n$. We get $k \leqslant h(k)$ from the construction, which was motivated by the goal of getting a bijection.

Alizter

@TedShifrin

user159870

22:49

Alizter

or @MikeMiller

Mike Miller

I've only got a few minutes

Alizter

Oh OK then I won't bother you

I shall bother @Ted instead

Mike Miller

Well, if your question can be resolved within a few minutes... :)

Alizter

I was going to ask about nested polynomials

Say with have a ring of polynomials over a ring $R[X]$

Can we talk about sets that result from nesting/iterating in a meaningful way?

I mean take for example the polynomial $X^2+1$ can I talk about $R[X^2+1]$ as a thing?

Mike Miller

22:52

like, $p(R[x])$, where $p$ is a given polynomial?

Pedro Tamaroff

@Alizter Yes.

Mike Miller

ok pedros got you

Alizter

Yup @PedroTamaroff

evinda

@DanielFischer How from $k \leqslant h(k)$ do we get $k < n$ ?

Pedro Tamaroff

$R[X]$ is a ring. $R[X^2+1]$ is the subring generated by $R$ and $X^2+1$. So consists of "polynomials in $X^2+1$".

evinda

22:53

@user159870 Is it you in the picture that you sent us?

Daniel Fischer

@evinda We get that from $k \leqslant h(k) < n$.

Alizter

@PedroTamaroff Can we compute this ring exactly? Perhaps as a quotient ring?

Pedro Tamaroff

@Alizter How is "polynomials in $X^2+1$" not computing the ring exactly?

It is isomorphic to $R[X]$.

Alizter

Or perhaps rephrased as: Can I write $R[X^2+1]$ as $R[X]/x$ with $x\in R[X]$

Pedro Tamaroff

@Alizter No.

Alizter

22:56

@PedroTamaroff Really?

evinda

@DanielFischer And how do we show that $h(k) \geq k$ ?

Pedro Tamaroff

Yes.

Alizter

@PedroTamaroff Is this independent of the polynomial chosen?

Daniel Fischer

@evinda Quasi by induction. Just we can take only finitely many steps.

Alizter

So $R[g]\cong R[X]\quad \forall g \in R[X]$

Pedro Tamaroff

22:58

@Alizter As long as it is monic non-constant. Constant polynomials give you $R$.

evinda

@DanielFischer We don't know what elements f(B) contains... How can we know if its values are greater than k or not?

Pedro Tamaroff

@Alizter Let $Y=X^n+\cdots$ a monic polynomial.

And suppose that $\sum_{i=0}^k a_i Y^i =0$.

Let's show each $a_i=0$.

We can work by induction on $k$.

If $k=0$; there is nothing to prove.

Now, using the binomial theorem, we can expand this in terms of $X$; and since the polynomial is monic, we get it has degree (in $X$) equal to $nk$, and it's leading coefficient is $a_k$.

This means that $a_k=0$.

By induction on $k$, every other $a_i=0$.

Alizter

Yes

Pedro Tamaroff

This means that the map that sends $X\to Y$ from $R[X]\to R[Y]$ is injective, and it is obviously surjective.

Hence the claim.

If $R$ is a domain, you can do the same without assuming your polynomial is monic.

Daniel Fischer

@evinda We know $h(0) \geqslant 0$ - why? And then the construction of $h$ gives us $h(k+1) > h(k) \geqslant k$. But $h(k+1) > h(k)$ means $h(k+1) \geqslant h(k) + 1$, and hence $h(k+1) \geqslant h(k) + 1 \geqslant k+1$.

Pedro Tamaroff

23:03

But for non-domains things can be different. For example, if we have a nilpotent element $a\in R$, the $R[aX]$ is not isomorphic to $R[X]$.

Here I'm assuming $R$ is commutative.

Alizter

Ah yes.

@PedroTamaroff So do similar properties hold for formal power series? i.e. $R[[s]]\cong R[[X]]\quad \forall s \in R[[X]]$ and s obeying some restrictions of sort.

Pedro Tamaroff

@Alizter Well, first there are some problems when you want to make sense of powerseries in $1+X$ for example, because there are infinitely many "carries" and we cannot sum infinitely many terms in an arbitrary ring.

So for starters you want a series with zero constant term.

In that case, think about it.

Alizter

@PedroTamaroff So it's not really well defined

Pedro Tamaroff

@Alizter Aha. Like $\exp\exp x$.

Yet $\exp(\exp x-1)$. This is a known powerseries.

Alizter

$\exp \exp x$ then is a power series

Pedro Tamaroff

23:08

Formally, it is not a powerseries.

evinda

@DanielFischer f(0)=0 because f(B) is a set and so its least element will be the empty set $\varnothing=0$, right?
I have understood that it will be $h(k+1) > h(k)$, because h(k) will be the k-th smallest element of f(B) and h(k+1) will be the k+1-th smallest element of f(B). Right?
But how do we conclude that $h(k) \geq k$ ? Or is it our induction hypothesis?

Alizter

@PedroTamaroff I was trying to think of functional equations in terms of iterates. Such as $af^2(x)+bf(x)+c=0$ where the exponent is iteration. I was trying to imagine what the rings would look like but what you said is pretty substantial. If they are isomorphic to each other it makes them harder to study. Weird.

Chris's sis

@robjohn Did you see this one? :-) $$\int_0^{\infty } \log \left(\frac{1}{1-e^{-x}}\right) \cos \left(\log \left(\text{Li}_2\left(e^{-x}\right)\right)\right) \, dx=\frac{\pi ^2 }{12} \left(\cos \left(\log \left(\frac{6}{\pi ^2}\right)\right)-\sin \left(\log \left(\frac{6}{\pi ^2}\right)\right)\right)$$

robjohn

@Chris'ssis I don't believe I have. It doesn't seem like one that has been tossed around recently.

Chris's sis

@robjohn I created it today.

robjohn

23:17

@Chris'ssis you seem to like writing $\cos(\log(\zeta(2)))+\sin(\log(\zeta(2)))$ cryptically.

Ted Shifrin

@Alizter: To plug a power series into a power series and get something that makes any sense, you need the constant term of the plugee to be $0$. :)

Chris's sis

@robjohn I used the way Mathematica made the computations.

robjohn

@Chris'ssis yes, it will simplify $\zeta(2)$ to $\pi^2/6$

Chris's sis

@robjohn Yeap.

robjohn

The question I have is why did it invert to $6/\pi^2$

Alizter

23:20

@TedShifrin Have you seen equations like I described above?

robjohn

I got two downvotes today... one on an old answer, and one on a recent answer. They were pretty close in time, so they may be related.

I modified the newer answer a bit, making it more of a suggestion than an answer. Perhaps that is what the downvoter objected to.

Heh... if they take back their downvote, I'll break 140K :-)

Basant Sharma

Can someone please answer this question.Kindly read the comments if the question is not clear.If it's still unclear I would be happy to explain again

Well the homomorphism that you are explictly given must be on the list of all possible homomorphisms. That is all that is being said. — Derek Holt Jan 9 at 15:13

Chris's sis

@robjohn Will you? I think you already have 140K :-)

I'm out.

evinda

23:37

We have the function $f: \{ a\} \overset{\text{1-1 and surjective}}{\rightarrow} 1$, $f(a)=0$. Why can we write the function $f$ like that: $f: \{ \langle a,0 \rangle\}$ ?

Alex

What speed covers 78 miles in 52 mins?

s=d/t
= 78 / (52/60)

is that correct ? it must be my working that is wrong

evinda

Hey @111
Could you take a look at this? http://math.stackexchange.com/questions/1101953/the-sum-of-three-colinear-rational-points-is-equal-to-o

robjohn

23:55

@Chris'ssis Good night! Thanks.

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