Mathematics - 2016-11-12 (page 6 of 6)

TheGreatDuck

21:00

@Hiro really hiro. using programming is basically just going to be a way of automating the process of rolling the die and doing all the calculations yourself

Astyx

If the sum had remainder 1, then the first throw was 1 or 25 (because $k\ge1$)and the last throw was 1, 9 or 25 because that means you've got to an even number.

TheGreatDuck

unless you can somehow collect the data many times over and know how to do analysis of that large data pool, it's basically useless for the problem.

Astyx

And all throws in between are 36s

TheGreatDuck

and really, I'd imagine it would be more useful for seeing how many levels it goes down to

not necessarily what the final value would be.

tl;dr the software will be useless for this problem. Use algebra and calculus to solve it.

Astyx

Thus the probability of getting k throws in this case is ${2\over 6}{3\over6}{1\over36^{k-2}}$

Hiro

21:03

so that's the probability of getting k throws, so how can you manipulate that to get a sum? (# of throw * values obtained)?

Astyx

Now if the sum had remainder 2, then you did the following scheme :
first 1 or 25
then a certain number of times you did 36
then you did either 4 or 16
then you did 36 a certain number of time again
then you did 1,4,9,16 or 25 and this leads you to a multipls of 3 or an even number

And the probability of making only one throw is ${4\over 6}$

TheGreatDuck

Is there a way to make my question clearer?

-2

Q: How can I define alternate versions of the indefinite integral operator rigorously using set theory?

I am trying to define a generic version of an operator that acts exactly like an anti-derivative but takes two extra parameters defining certain policies it should follow. The first parameter has the slight minor change of someone reaching into its innards and telling it that some particular set ...

Hiro

and does that work using the formula above?

TheGreatDuck

I actually have an example of one of these operators in practice somewhere else. Someone write a paper with my permission describing how it works for that particular V and E and how to evaluate it. Would that be something I could include in my question to make it clearer what I want?

Astyx

Therefore the probablity of getting k throws is ${6\over 36^{k-1}} + {(k-2)}{10\over36^{k-2}}$ if $k\ge 2$ and $2\over3$ otherwise

TheGreatDuck

21:07

I managed to answer my own question with a bit of inspiration; however, I would still like my answer to sound clear to people reading it.

Astyx

Wait a bit, my solution does not seem to work

Hiro

yeah that's what i was wondering :P

rather than getting k=1, i am getting k=1.6 lol

Ughhh... this question is annoying -.-

Astyx

Oh my stupid me

it's $1\over 6$ not $1\over 36$

Adeek

hey @BalarkaSen still here ?

Hiro

so ${6\over6^{k-1}}+ {(k-2)}{10\over6^{k-2}}$?

Astyx

21:15

Okay so it works

No

${1\over 6^{k-1}} + {(k-2)}{20\over6^{k}}$

if $k\ge 2$ and $2\over3$ otherwise

Hiro

and this represents what?

Astyx

And this works

The probability of getting $k$ throws, ie the probability that you stop at step $k$

Hiro

ok so we calculated the probability of getting k throws, now what about the probability that the sum is odd?

Astyx

That's nearly the same proof

Hiro

Could you explain what you did step-by step?

Cause I didn't quite understand what you did lol :P

Astyx

21:20

It's $(k-2){2\over6} {2^2\over6^2} {1\over6^{k-3}}$ summed for $k\ge2$

ie $(k-2){8\over6^k}$

Okay so here it goes

@Hiro The remainder of the sum you had just before you threw the last dice was odd, and had either remainder 1 or 2 mod 3 right ? (for $k\ge2$, where $k$ is the step at which you stop)

Hiro

why is it odd before the last roll?

Astyx

Otherwise you would have stopped before

Since you stop when the sum is either odd or has remainder 0 mod 3

Hiro

you stop when the sum is nonzero, either even or a multiple of 3 and is odd

Astyx

Yes

Hiro

or rather you stop when the sum is nonzero, and either even or a multiple of 3. Then you need to compute the probability that when you stop given that condition, the sum is odd

meow-mix

21:27

what books should i use to study field theory / galois theory in preparation of commutative algebra?

Astyx

If the sum was even before you throw the dice, you wouldn't throw it again since you would have stopped then

meow-mix

ive already read the general topology section of Munkres

and taken abstract algebra up to some basic ring theory

but nothing past ideals, and the polynomial rings

Hiro

But if it was odd, it could also possibly be a multiple of 3

meaning that you can also stop even if it's odd

so if the sum was 9 before the last roll, you'd stop because it's nonzero and a multiple of 3, not necessarily even

Astyx

Let's move to another room no ?

Hiro

oh we can do that?

how to move?

user3670473

21:30

Hey all, I have a probability problem and I would like to know the method of solving it.

meow-mix

post away

user3670473

10 people are presented with a binary choice with Option A and Option B. Person 1 has a 25% chance of picking Option B, Person 2 has a 30% chance, and so on, incriminating 5% each person (for simplicity sake). What is the total probability, in percentage, that at least 3 of these people will pick Option B?

Or simpler, *exactly three

I asked this in the forums but am also asking it here

Mary Star

22:04

We have that a mapping is bijective iff it has an inverse.
We have that the inverse of $\phi$ exists and it is $\phi^{-1}: g\mapsto x^{-1}g$, since $\phi \circ \phi^{-1} = \phi^{-1} \circ \phi = Id_G$.
Why exactly does it follow that when $g$ runs through $G$, so does $\phi(g)$ ? Is it because since $\phi$ is bijective we have that $G=\phi(G)$ and that $xg_1\neq xg_2$ for $g_1\neq g_2$? @robjohn @TobiasKildetoft @Astyx

Astyx

@MaryStar Because for every element $g\in G$ there is another one $g'\in G$ such that $g=g'x$

Because $\phi$ is surjective

Adeek

do we have partition of unity for complex manifold ?

Mary Star

Ah ok... I see!! Thank you very much!! :-) @Astyx

Astyx

It's my pleasure ! @MaryStar

meow-mix

hey guys, what text should i use to study field theory and galois theory, in preparation for commutative algebra

ive already studied point-set topology w/ munkres

and group theory and basic ring theory in another book

particularly, i'll be reading A-M

for commutatve algebra

Astyx

22:23

@meow-mix what language ?

meow-mix

22:34

english

Astyx

I don't have references for lectures concerning that topic in english, sorry

meow-mix

alright

robjohn

@MaryStar what could be missed? $\phi\!\left(x^{-1}g\right)=g$

Alessandro

I want to prove that $|\bar{F}:F|=\infty$ implies that the degree of the irreducible polynomials in $F[x]$ is unbounded, I know that's true but I have no idea where to begin with a proof, any hint?

Balarka Sen

23:24

@Adeek You should ping me with your question than asking everytime if I am available.

Adeek

so I solved that question for the splitting it is easy.

Balarka Sen

Because if I am not, I'd just answer to it later. Otherwise you'll just waste unnecesary chatspace.

Adeek

just choose inner product on V. Then consider $W$ as subbundle of V then $V = W (+) orth(W)$

$orth(W) = V / W$

I have now a question about definition of a vector field.

what is that a(p),b(p) ?

is that an interval ?

Balarka Sen

@Adeek Well you need to do it fiberwise, that's why you choose a metric.

Adeek

yeah I understand the details etc but lazy to write it haha

Balarka Sen

23:28

@Adeek Yes, and interval in $\Bbb R$ which depends on $X$. You're defining the flow on a small neighborhood of $0 \times X$ in $\Bbb R \times X$

Adeek

so what does it mean if a vector field is complete intuitively ?

I understand vector field in regular euclidean spaces.

let us look at $\mathbb{R}$ what does it mean that every vector field in $\mathbb{R}$ is complete ?

Balarka Sen

@Adeek Flowing along a vector field $X$ on a smooth manifold $M$ means you can move each point $p \in M$ along the integral curve of $X$ starting at $p$ and get a diffeomorphism in this process. There's no reason it can be done "for all time", which is to say, it's possible the integral curves will be sufficiently meddled up "far away" that you can no longer flow along it.

If you can, the vector field is complete.

Adeek

oh ok that is very visually nice.

Danu

Also, on compact manifolds every vector field is complete.

Balarka Sen

More generally any compactly supported vector field is complete.

Danu

23:36

Hi @MikeMiller!

Balarka Sen

@Adeek Eg, in $\Bbb R$ flowing along the constant vector field "$X(p)$ = vector at $p$ of length 1 pointing to the right" is just translation, as it should. But not every vector field in $\Bbb R$ is complete (can you give an example?).

Adeek

the flow of $x^2 = dx/dt$ is just $\psi_t(x) = x/(1 - tx)$ right ?

Balarka Sen

Flow of the vector field $X(x) = x^2$ is that, yes.

But can you visually see why it's not complete, without doing that computation?

Danu

Bob Dylan - Hurricane (Desire Outtake 1975)

►

Adeek

yeah because as you said we could get meddled up as we move p along an integral curve in $\mathbb{R}$

there are many places to move.

Balarka Sen

23:43

That doesn't seem like a very precise picture to me.

Mike Miller

Hi @Danu.

Adeek

I don't see it visually yet @BalarkaSen I think I need to understand this more.

meow-mix

does anyone have any recommendations for an introduction to field theory / galois theory? particularly in preparation for M-A's commutative algebra

Adeek

@meow-mix Allufi chapter 0, Hungerford, Dummit foote, artin. There are many books.

I really like allufi chapter 0 if you like category theory.

Balarka Sen

23:58

@Adeek I have to go to sleep now but what does the field $X(x) = x^2$ look like?

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