Mathematics - 2013-05-03 (page 4 of 5)

@Lord_Farin There's the forlorn hope that we wake up and figure what they, and more importantly we, are doing wrong.

@Vrouvrou Yes, that seems to be correct (presuming $H$ is continuous).

yes $H$ is continuous of cors

we don't have probleme with the continuti of $r_t$

guys: i (hypothetically) have a programming assignment for my students. Can I write a program that looks at their code and tells me if they've done it correctly (and also tells me if they've done it incorrectly)?

7:07 PM

@Lord_Farin

@QuinnCulver I think the usual procedure is to run their programs on judicuously-chosen inputs.

@QuinnCulver hypothetically?

@QuinnCulver I'd consider this infeasible. The only thing you could do is for the program to require standardised input and output.

@Charlie I mean, i don't really have such an assignment, this is just a question for fun.

7:08 PM

hmmmmmmmmmm

Correctness (computer science)

In theoretical computer science, correctness of an algorithm is asserted when it is said that the algorithm is correct with respect to a specification. Functional correctness refers to the input-output behaviour of the algorithm (i.e., for each input it produces the correct output). A distinction is made between total correctness, which additionally requires that the algorithm terminates, and partial correctness, which simply requires that if an answer is returned it will be correct. Since there is no general solution to the halting problem, a total correctness assertion may lie much deepe...

Geez! why don't I find problems to answer like this

@Vrouvrou For a subspace $B \subseteq A$ we say a $f: A \to B$ is a retraction iff $f(b) = b$ for all $b \in B$, and $f$ is a continuous surjection.

@J.M. mathematicians should be seen and not herd ;-)

@robjohn I do believe rec math is a most inappropriate tag...

7:11 PM

@Lord_Farin yes but we need defformation retract

@robjohn :D

@Vrouvrou We say a retraction $f: A \to B$ is a deformation retraction iff there is a homotopy $H: A \times [0,1] \to A$ (relative to $B$, i.e. $H(b,t) = b$ for all $t \in [0,1]$) such that $H(-,0) = \operatorname{id}_A$ and $H(-,1) = \iota \circ f$, where $\iota: B \to A$ is the inclusion.

@J.M. I saw that one coming...

yes

Additionally we say that $B$ is a deformation retract of $A$.

7:12 PM

@J.M. I don't see anything in that article that answers the question

yes

@robjohn :) Well, certainly not arithmetic, but I'm hard pressed to think of any other more appropriate tag.

@Vrouvrou So what exactly is your question?

but befor that i want to prove that $r_t$ is a defformation retract

(In the present case, we have $H(q,t) = r_t(q)$.)

7:13 PM

@J.M. well, it's certainly cutting edge mathematics

@robjohn Something one can really get one’s teeth into?

@Vrouvrou Hence we have to show that $H$ is continuous in order to deduce $r_1$ is a deformation retraction from $M^b$ onto $M^a$.

http://mathworld.wolfram.com/UniformSumDistribution.html if P_n^(1) = integral from 1 to n of PX1...Xn(u)du - integral from 1 to n-1 of PX1...Pn-1(u)du tells the probability of getting a sum over 1 after n turns, how can i turn this to figure out the expected value of just the last turn?
as opposed to the whole sum

@BrianM.Scott Yeah and get a handle on as long as one doesn't get board

user19161

@robjohn bored

7:16 PM

@QuinnCulver Hmm, should've linked to the formal verification article. There's software for that already, if memory serves (under the assumption of precise input specifications), but I seem to be having a hard time searching.

@robjohn If it’s a cross-cut saw, maybe we can use it to make a cross-cap.

@JasperLoy read the question before you start correcting me.

@JasperLoy No, board.

@robjohn Ah, that explains all the buzz it's been getting...

user19161

@robjohn Oh noes!

user19161

7:17 PM

I will just pretend that I was testing you, LOL

@JasperLoy ;)

@BrianM.Scott ...at the huge risk of the saw cutting itself, tho.

Oh!

@J.M. Hmmm, I wonder if i'm being understood properly. Let's be more specific: suppose I want my students to write a program that takes (any) input and always outputs 1. Can I write a program that looks at and analyzes their code to determine if they've done it correctly?

(and also determines if they've NOT done it correctly)

@J.M. Ein *Klein*es Risiko!

7:19 PM

@QuinnCulver The "any" seems to be a sticking point, I think. Without restrictions, I don't believe so.

@BrianM.Scott If I have a size 11 set, the number of 3-uples I can make of the form $a,a,b$ with $a\neq b$ is $11\times 10$; yes?

@QuinnCulver There are many ways to write correct code, so probably not unless you merely want to automate inputting students' code and then checking to see if it compiles / outputs the correct values

@PeterTamaroff Yes.

And the number of 3-uples $a,b,c$ with $a,b,c$ distinct is $11\times 10\times 9$.

OK

(Of course, one could be doing any number of useless garbage in between before proceeding to produce the correct output.)

7:19 PM

Then my final answer to the problem is:

@PeterTamaroff Yes. But don’t forget that there are $3$ ways to decide which vertex gets the odd color.

@J.M. the "any" really isn't the restiction. what I'm asking is equivalent no matter what assigment i give them.

@BrianM.Scott Yes, yes. Two are the same, but one is different.

@QuinnCulver Well, take the hypothetical case I mentioned in my previous line. Would that be a correct program by your requirements?

@Hayaku but can such a thing be automated?

7:21 PM

yeah

@J.M. pardon, but what was your 'hypothetical case'?

$${10^5}11 + 2\frac{{11!}}{{9!}}{10^3}{9^2} + \frac{{11!}}{{9!}}{10^2}{9^3} + \frac{{11!}}{{8!}}8\;{9^2}{10^2}$$

3 mins ago, by J. M.

(Of course, one could be doing any number of useless garbage in between before proceeding to produce the correct output.)

@PeterTamaroff You’re missing a factor of $3$ in the third term.

@BrianM.Scott Oh, FUUUUUUUUUU

7:24 PM

how does one find the expected value of something on condition like the above problem? i can't find any online literature that explains it either

@J.M. Ahhh, I didn't see that. Yes, many correct programs could have 'useless garbage' (aka 'padding'). And this is the heart of why it's not possible for me to write such a program.

@BrianM.Scott See, that's the kind of stuff I hate.

@J.M. how did you link the Wikipedia article so nicely before?

$$\huge \rangle 8($$

@QuinnCulver It's the magic of this chat room, just pasting in a Wikipedia link works. The room does the formatting for you.

7:26 PM

Rice's theorem

In computability theory, Rice's theorem states that, for any non-trivial property of partial functions, there is no general and effective method to decide whether an algorithm computes a partial function with that property. Here, a property of partial functions is called trivial if it holds for all partial computable functions or for none, and an effective decision method is called general if it decides correctly for every algorithm. The theorem is named after Henry Gordon Rice, and is also known as the Rice-Myhill-Shapiro theorem after Rice, John Myhill, and Norman Shapiro. Introduction ...

@PeterTamaroff hahahaha

@QuinnCulver Nailed it.

@BrianM.Scott @anon Talked about chromatic polynomials.

@Lord_Farin i asked the question

see

Defformation retract

@Vrouvrou One "f" suffices.

7:27 PM

@Hayaku Did you see the result of Rice's theorem that I linked?

user19161

@PeterTamaroff I see you have become GN!

thank you

@JasperLoy I have always been one!

hhhmmmm

@PeterTamaroff Yay. That makes two of us. :)

user19161

7:28 PM

@PeterTamaroff We are one, LOL.

i dont understand

what is this for

But @Peter, why don't you take it out on all those questions on main? The amount of GN work there grows faster than you can counter it :).

@soma Hies!

Hi

:)

@somaye how are you?

7:30 PM

how are you?

...and, I now have to bow out of this room. Been fun, as always. See y'all later.

hehe fine thanks you?

@somaye I'm fine

@J.M. byes

@Lord_Farin I edit every now and then.

thanks for your email

7:30 PM

@J.M. Bye.

@somaye :)

user19161

@PeterTamaroff Lord Farin seems not to know who THE SE GN is, lol.

@JasperLoy Hehe, no.

what are you doing?

@charlie

@somaye I'm studying (trying?) some algebra

7:35 PM

which algebra?

which book?

@somaye it's more like number theory... it's a brazilian book

Ok

you have exam?

@BrianM.Scott @robjohn Do you believe in the Feynman philosophy of "know how to solve every problem that has been solved"?

@somaye no

i do not disturb your study

:)

7:39 PM

@somaye no, no, it is ok

hey charlie thanks for your email

@somaye hehe no problem Somaye

@BrianM.Scott

Hello @skullpatrol

Brian's fans will post a video on youtube:LEAVE BRIAN M. SCOTT ALONE!

7:42 PM

@somaye Hello, how are you?

@skullpatrol Feynman wasn't around for Wiles' proof :-)

fine thanks

you?

@somaye Fine thanks.

nice

@robjohn So that means no?

7:44 PM

@amWhy you deleted and added the same comment. Is the acceptance that important?

@skullpatrol I don't think that anyone could possibly do it, but it would be a nice goal.

@skullpatrol I do not, Skull

bye for now:) every one

@somaye Bye, Sommy!

Later

:)

7:46 PM

:)

stats = impossible

starting to understand why everyone (in general) is bad at it

I'm leaving for the night. Bye.

bye

have a good night

@Lord_Farin

@Lord_Farin Bye, F

@Vrouvrou Thanks; the same to you.

@Charlie Bye.

7:52 PM

@Lord_Farin if you can tomorrow answer my question

thank you

0

Q: Deformation retract

how to prove that $r_t$ is a deformation retract $M^a=\lbrace q\in M ; f(q)\leq a\rbrace$ $r_t$ is a difformation retract if $r_t$ is a continius ,onto application and $r_t(q)=q , \text{for }q\in M^a$ , and there existe an continuous application $H: M^b\times [0,1]\rightarrow M^b$ such that $H...

Lots of writing for a binomial identity. I wonder if someone will come up with a shorter proof.

@robjohn Brian will interpret it combinatorially in some crazy manner, I guess.

@PeterTamaroff I would really like that :-)

@BrianM.Scott: get to it :-)

@robjohn Today I counted the ways to colour a graph using $11$ colours.

Was interesting.

@BrianM.Scott didn't tell me his way to solve it.

@PeterTamaroff Don't crayons usually come in boxes of 12? :-D

7:59 PM

@robjohn =)

@PeterTamaroff looks almost as mean as I do :-)

@PeterTamaroff Did you see my answer to Chris's Sister's sum?

can anyone explain briefly the general logic behind "expected final of a final toss/roll/etc" in terms of setup and conditioning?

@Hayaku I am not sure what you mean by expected final and what you want in the way of setup and conditioning.

do you mean expected total?

see mathworld.wolfram.com/UniformSumDistribution.html where they set up the probability of rolling the sum over 1 after n terms. This helps you eventually find the EV of n, but I want to know the EV of just the last turn

equation (7)

or say the probability that the last turn will be between some two values a and b

@Hayaku do you know about convolution?

8:10 PM

yes

@Hayaku Convolution will answer a lot of questions about summing random variables

somewhat

yes but i am after the last random variable in that series

The convolution of two uniform densities is the sawtooth (one tooth, or triangle) density

yes

what do you mean the last random variable in the series? Do you want to know a certain duration?

8:13 PM

the final turn that finally puts it over 1

the page explains how to get EV of the sum of X1 + X2 + blah + XN, I want to look at XN

are these uniform independent random variables?

in [0,1]

yes

i can simulate it obviously and have done so so i already know the answer but i want to solve it analytically

The probability that $1$ variable is less than $x$ is $x$

right

Then the probability that the sum of two variables will be less than $x$ is $\int_0^x(x-t)\,\mathrm{d}t=\frac12x^2$

8:21 PM

may be more than 2 though

could be inifinitely many, n just becomes less likely over time

@Hayaku perhaps I am going somewhere with this...

?

sorry i do not follow what you mean

Okay, I will simply skip to the end. The probability of the sum of $n$ dice summing to less than $1$ is $\int_{\sum x_i\le1}\,\mathrm{d}x$

that is a multiple integral

and if you carry out the integration you get $\frac1{n!}$

yes

so the probability that the sum will be less than $1$ is $\frac1{n!}$

8:28 PM

on mathworld they have (1-1/n!) to represent the probability of exceeding 1 after n turns

(subtracting out the probability of not exceeding 1 after n-1 turns)

now you want to get when the sum exceeds 1

@robjohn Yeah. Kicking-ass 24/7!

how long does it generally take for these elections to get off the ground?

@AlexanderGruber hi Grubs!

@AlexanderGruber Ah?

8:30 PM

@Hayaku If you carry out the same argument, for $x\le1$, the probability that the sum does not exceed $x$ after $n$ trials is $\frac{x^n}{n!}$

@Charlie charlie, hi. :)

@PeterTamaroff i want to see our list of new potential emperors.

@AlexanderGruber wassup?

@AlexanderGruber I wanted to be mod... once.

"I used to be a mod like you. Then I took a Bill Dubuque to the knee."

9

@PeterTamaroff HAHAHAHAHAHAHHAHAHAHHAHAA

@Charlie i'm giving you a chinese name.

8:37 PM

@robjohn I am trying to see if I can get an EV out of this, sorry for my late response

THAT WAS THE BEST, @peter

@AlexanderGruber oh!!!

@PeterTamaroff i'd vote for you.

seems like i have to maybe do a summation to infinity for all the different probabilities minus the probabilities of what came before <= 1?

not giving the right value though

@Hayaku So to compute the probability that the sum exceeds $1$ on trial $n$ is $\frac1{(n-1)!}-\frac1{n!}$

right

8:39 PM

@Charlie here it is: 查雳

@AlexanderGruber hahah thank you!

it's pronounced Cha Li, and it means the thunderous researcher!

3

@PeterTamaroff is bill dubuque famous?

@Hayaku So the expected value is $$\sum_{n=1}^\infty n\left(\frac1{(n-1)!}-\frac1{n!}\right)$$

@AlexanderGruber oh! really?

8:40 PM

yes

which is e

@Charlie yup. :p

@AlexanderGruber how cool!

@Hayaku indeed

which is all good but i can't get the final value of XN or its probability/EV

@Hayaku what I gave is the expected value of how many trials it takes to exceed 1

8:44 PM

yes

what do you mean by XN then?

from mathworld link too

the final variable involved in the sum

not the entire sum

well entire sum is just .5e i mean

expected number of turns e

i want xn, the final variable in x1+x2+x3+etc+xn

xn the final variable that pushes the sum over 1 for the first time

(it should be around .64 etc) but i don't know how to get it mathematically

The probability that trial $n$ sends you over is $\frac1{(n-1)!}-\frac1{n!}$

yes

but i am after xn and the probability of its value range

@alexandergruber do you speak chinese?

8:55 PM

@Hayaku are you giving me $n$?

n could be anything

because there are many ways to get over 1

infinitely many

it could take 2 rolls to get over 1 or 1 trillion, but the former is much more likely

then why are you calling it XN if you don't know N that is confusing.

what else is it called?

"the last value rolled to get the sum of the variables over 1"

I don't know ... you're giving the terminology and it was not clear what you meant

remember that I said before that the probabilty that the sum of $n$ was under $x<1$ was $\frac{x^n}{n!}$

hi

8:59 PM

sure, but i don't know what you mean by that exactly

how does that help in this situation?

The probability that $n$ rolls sum to $\le x$ is $\frac{x^n}{n!}$

sorry i am just getting frustrated, you keep saying that but i don't see how that applies to my question

please give me a chance to explain

if we supposed the integral $F(x)=\int_x^{2x} \dfrac{1}{\ln(1+t^2)}dt$

I was simply asking if you remembered me explaining that

hang on

9:04 PM

Hi

@Nimza Hi

@pourjour hi

@Charlie hi Marilie4ka

@Nimza how are you Leshe4ka?

@Charlie My advisor doesn't understand proof of one theorem in my thesis. Today I will meet him and explain :(

@Charlie and how are you?

Hi all

9:08 PM

Hi Old John

@Nimza Hi there

@Nimza I'm fine

@Nimza Oh no!

@OldJohn Hi, John!

Hi @Charlie

Hm, $L_1([1,+\infty), \frac 1 t) \subseteq L_2([1,+\infty),\frac 1 t)$?

@OldJohn how are you? and what about your garden?

9:13 PM

@Charlie yeah i speak some mandarin

@AlexanderGruber fascinating

i lived in china for 6 months or so

@Charlie I'm fine, thanks - and the garden is doing great

@OldJohn awesome!

@Charlie will post some pics on FB when I get some done

9:14 PM

@OldJohn Ah, perfect!

homework analysis functional-analysis manifolds

need help for this please

0

Q: Deformation retract

how to prove that $r_t$ is a deformation retract $M^a=\lbrace q\in M ; f(q)\leq a\rbrace$ $r_t$ is a difformation retract if $r_t$ is a continius ,onto application and $r_t(q)=q , \text{for }q\in M^a$ , and there existe an continuous application $H: M^b\times [0,1]\rightarrow M^b$ such that $H...

hah, "there existe" - it's frenglish, right @Vrouvrou? :D

oh , yes

thank you

@Charlie hi

ok so if we supposed the integral $F(x)=\int_x^{2x} \dfrac{1}{\ln(1+t^2)}dt$
how can I prove that it's odd

@pourjour I think you can just write out what $F(-x)$ is as an integral, and then use standard properties of integrals to show that it it equal to $-F(x)$?

9:23 PM

@OldJohn but what I'm going to change in the integral is it the "t" or "x"

Hi guys, I am confused with an statement from my probability book that has to do with Markov chains. I hope someone could clarify that, if possible....Consider a Markov chain for which $P_{11} = 1$ and $P_{ij}=\frac{1}{i-1}$. Let $T_{i}$ be # of transitions needed to go from state i to state 1. A recursive formula for $E[T_{i}]$ can be obtained by conditioning on initial transition: $ E[T_{i}]=1+\frac{1}{i-1}\sum_{j=1}^{i-1}E[T_{j}]$. Could anyone tell me how this formula is derived? I am lost.

@pourjour $t$ does not change - just the limits become $-x$ and $-2x$

By the way I am not sure of the rules here in this chat, as far as questions or number of lines allowed...

@robjohn just seeing if you are still here

@pourjour $F(-x)=\int_{-x}^{-2x} \dfrac{1}{\ln(1+t^2)}dt = \int_x^{2x} \dfrac{1}{\ln(1+\tau^2)}(-d\tau)$ (when you make the substitution $\tau = -t$).

9:34 PM

@OldJohn ok thanks

@Hayaku I am afk for a short while

I think I am having a problem in visualizing how E[X}= E[X|A]P[A]+E[X|B]P[B]+... is being used (if it is the case).

@HeberSarmiento "visualizing"?

@PeterTamaroff what do you think about about proving that the integral is odd

@pourjour Which one?

9:48 PM

@PeterTamaroff. Conditioning on initial transition made me think on that formula.

@PeterTamaroff $F(x)=\int_x^{2x} \dfrac{1}{\ln(1+t^2)}dt$

@pourjour Old John provided you with an answer. Think about it.

Make an effort to understand what he wrote.

@HeberSarmiento I think the "1" is the initial transition. You spend 1 move "transitioning" plus the expected value of all the other possible transitions

@PeterTamaroff I did a mistake I told him to prove it's even but in fact it's odd as the question said

@pourjour He is showing it is odd.

9:50 PM

@PeterTamaroff I don't think so ($-\tau = t$)

@pourjour Yes. Trust me.

so he proved $F(-x)=F(x)$