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02:00 - 16:0016:00 - 00:00

02:00
@Teddy No, but I hope to learn it if I go to grad school.
@TedShifrin How was dinner? I just woke up here.
hello!
@Omen Nice eye.
why thank you!
how are you?
okay then..
02:21
Hi @Jasper. Welcome back to your old persona :)
 
2 hours later…
03:56
@JasperLoy Ssup?
@JayeshBadwaik Hi. I am trying to solve my mental problems by the end of next year. I hope I don't delay anymore.
@JasperLoy Okay. I hope you've started on it though? You cannot get 100% cured at some point. You have to start small.
@JayeshBadwaik Also, I have promised not to delete my email or SE accounts anymore, lol.
@JayeshBadwaik Yes, I started on it 15 years ago.
@JasperLoy That's a start! Good!
@JayeshBadwaik I have finalised my reading list. It has 12 books. Let me share with you.
03:59
Sure
Marsden and Weinstein: Calculus I, Calculus II, Calculus III; Cohn: Classic Algebra, Basic Algebra, Further Algebra; Rudin: Mathematical Analysis, Real and Complex Analysis, Functional Analysis; Lee: Topological Manifolds, Smooth Manifolds, Riemannian Manifolds
There will not be any more changes to this reading list either, I promise.
Hmm, I'm almost complete with the Rudin part of the series, and for Algebra, I'm studying Lang, so I guess, that covers the Cohn part.
I have gotten all the books, except the last one, whose second edition I am waiting for.
Manifolds I'll be doing next semester.
Do you have a gf now? =)
@TedShifrin Thank you, I promise never to change my username again.
04:03
no comments. :-P
OK. I am still single, lol.
I just answered a lhf, lol.
Nice, nice.
@JayeshBadwaik Are you doing a Masters or PhD?
Amazing that a wrong answer is upvoted and a right answer is not, lol. Just answered another lhf.
Actually, I don't take this site seriously anymore.
4
04:24
That's a good idea.
That's a red idea.
r u srs?
Yes and not at the same time.
Why isn't there a word for yes and no at the same time?
that would be contradictory
What's the problem?
A paraconsistent logic is a logical system that attempts to deal with contradictions in a discriminating way. Alternatively, paraconsistent logic is the subfield of logic that is concerned with studying and developing paraconsistent (or "inconsistency-tolerant") systems of logic. Inconsistency-tolerant logics have been discussed since at least 1910 (and arguably much earlier, for example in the writings of Aristotle); however, the term paraconsistent ("beside the consistent") was not coined until 1976, by the Peruvian philosopher Francisco Miró Quesada. == Definition == In classical logic (as well...
04:36
18 mins ago, by Jasper Loy
Actually, I don't take this site seriously anymore.
is it true that for any random variable X, Prob(X<mean(x))=1/2?
no
that would be the median (by definition), not the mean
for a counterexample, consider e.g. X uniformly distributed in [0,k] U [n,n+m] where k>m but n is sufficiently big to make the mean in between k and n
thanks, you are right!
@anon I've seen your answer on sets of notes and lectures.
@anon In my country, there is a very good book about binary relations.
btw, @anon, if you are good with probability i would be very grateful if you could take a look at this problem: math.stackexchange.com/questions/959988/…
04:44
yeah
It's list of subjects is the following: pastie.org/9623982
Do you know something similar to it?
It's a book in portuguese, and the information I provided in the link is also in portuguese. But you can easily figure out the cognates.
the only difference between it and a generic algebra text is that it focuses on binary operations specifically it looks like
Yes.
This book is very rare in here. It would be good to have something similar.
@EricGregor I think you mean a in the subscript of the inf
@anon yes, you are right. fixing it now
@anon i am trying the case where the constants are equal, a' is the median. i think that is easy to show
04:52
Heya @Vibhav
@Vÿska Whoa. That's just like a survey on groupoids with extra structures give or take
@BalarkaSen Yes, I guess.
Hey @BalarkaSen
Hello @Vibhav
@Vÿska There is more to algebra than that.
@BalarkaSen I don't get it.
if $\theta_n \in \mathbb{R}$, is it possible to choose 13 real numbers such that $\sum_{k = 1}^{13}e^{i\theta_k} = 0$?
05:00
What are those 13 real numbers to be chosen?
@BalarkaSen any real numbers
the only $e^{i\theta_k}$ I can think of are the k'th roots of unity
Yes, roots of unity would be it.
so their sum would be zero
anything else than that?
Just use Euler
euler?
05:02
And evaluate the real and imaginary parts (both zero)
@VibhavPant $$e^{i\theta} = \cos(\theta) + i \sin(\theta)$$
yeah, I know that
@VibhavPant You knew that. Saying that you know after he just said it to you is quite obvious.
@VibhavPant Use it.
05:04
@anonp did you take a look?
or rather, did you take a crack at it
@Vÿska Im not sure how would I use it though
I guess setting $\sum\cos(\theta) = \sum i\sin(\theta) = 0$
mmhmm
you can cancel out the $i$s
perhaps the addition formulas?
that would be quite a disaster, no?
I found this book quite insteresting:
haha, @Vibhav, yes
05:07
:D
also, if you look at it geometrically, the complex numbers need to be opposite to each other on the argand plane so that they cancel each other
yes, well the obvious choice are the 13th root of unity.
yeah, guess thats the only answer
@BalarkaSen Why?
That's because it's philosophy. It is, in it's whole, a big BS
05:17
@BalarkaSen Oh, you are stupid. I see.
@BalarkaSen Gauss was a great mathematician. He wasn't a great at cooking for example. Also, notice that a random quote at the internet does not prove anything.
Well, I am a Gaussian.
=P
maybe he was just joking
I believe his philosophies about philosophy.
@Vÿska It's a well-known quote.
@VibhavPant I am trying to get myself kicked out, aren't I? =P
05:21
@BalarkaSen It still does not prove anything.
I am not trying to prove.
I am merely trying to show that I am a Gaussian.
@anon you there?
Hey @Nick
Quickly, what the heck is $\mathrm dy = (\mathrm dx)^2$ ?
05:24
physics? AKA blatant abuse of notation
What's the context, @Nick?
@VibhavPant: No, it's math... some strange differential thing that I wish I never saw
@Nick non-standard analysis maybe
@BalarkaSen: Assume for a second that there is no context. That you found this written on a wall. How would you comprehend it?
I don't know. It might be some stuff about differential forms. Never studied.
05:26
@VibhavPant: It's analysis... it's definitely non-standard
1
Q: What is the meaning of $dy=dx^2$?

lanse2ptyWhen I read the mathematical analysis ,I think if the differential is $dy=Adx^2$ $A$ is a function about x, what will happen? Maybe, it is not proper defined ,but I think the "function" meet $dy=A(dx)^2$ will has some fractal structure.

@Nick You found it written in a wall and want us to guess what it is?
@Vÿska: Yes... Is that such a bad thing?
@Nick No. But it indicates that people write weird things nearby your location.
Well, it isn't as insane as the things I may have already asked the people in this room.
@Vÿska Well if you don't want to guess... don't.
That's what I am doing.
=P
05:29
Well, it was worth the ask :D
@Nick Have you given a thought on Map(X)?
? Map(X) is a set
Set of bijective maps from X to itself
Yeah, I better go think about it, then.
toodles
:D
@Nick You familiar with subgroups?
@BalarkaSen: a group whose members are all members of another group, both being subject to the same operations.
No, I just met him.
05:54
$\text{SubGroup} \subset \text{Group}$
 
1 hour later…
07:23
@N3buchadnezzar hi pal
That guy is bound to be a crank.
Hello @N3buchadnezzar.
07:44
Hello my friends @DanielFischer and @AlexanderGruber
Good time-of-day-where-you-are, @IceBoy.
@AlexanderGruber!
@DanielFischer!
Afternoon, @Balarka (or is it already evening?)
Nah it's 1:23 PM
I have forgotten what afternoon or evening is...
07:55
there is no clear dividing line between the two
4:30
that's the dividing line
and he vanishes as quickly as he came
so 4:31 is evening?
Out of nowhere the wind blows on the chimes to form the following words
Depends on how bright the outside is but when indoors, it's a good convention to start saying Good Evening
08:11
Evening in its primary meaning is the period of the day between afternoon and night. Though the term is subjective, evening is typically understood to begin when the temperature has noticeably fallen and other accompanying weather changes have occured, such as increased wind speed and change in cloud types and sky color, and lasts until an hour or so after sunset, when maximum darkness has been reached. The evening is also characterized by the activities which usually occur during this time, such as preparing and eating dinner or having more-formal social gatherings and entertainment with friends...
anyone here know some decision theory? i'm still stuck on this problem (math.stackexchange.com/questions/959988/…)
The main reason we have greetings that include the time of day or denoting period such as "Good Day/Morning/Afternoon/Evening/Night" and even "Happy Birthday/ Special Event" is because it is beneficial to reminding people, without watches or calendars, what time it is. It also acts as a statement that says "Hi, I respect you enough to recognize your existence at this moment of time which I am reminding you of". Evolution gave society all these automatic bonus features.
hi. off topic: How can a complex valued function depend on $z$ and not on the conjugate of $z$? I don't see how a holomorphic function can depend on $z$ but not on the conjugate of $z$ since $z$ and its conjugate depend on one another.
PS: Notice that my total points on MSE is a year which lived through!
@TheSubstitute No, $f(z) = \bar{z}$ is not holomorphic. So roughly $z$ doesn't "holomorphically" depend on $\bar{z}$
Well, no holomorphic in $\Bbb C$. It is, for example, in $\Bbb R$ where this is just the identity function/
08:23
I'm not sure what you mean. For example, if $f(z)=z$, then $f$ is holomorphic, so that it does not depend on the conjugate of $z$. However, $f$ depends on $z$ and $z$ depends on its conjugate, so $f$ depends on the conjugate of $z$.
In that case, holomorphicity is lost
my definition of homorphic is that the partial derivative with respect to the conjugate of z is 0
Partial derivative is 0? Huh?
Holomorphicity over C is defined by the Cauchy-Riemann equations, if I haven't forgotten all the complex analysis I have ever done.
Claim : $z \mapsto \bar{z}$ is not holomorphic.
08:27
I agree
Thus $z$ can't technically "depend" on $\bar{z}$ preserving holomorphicity
oops i gotta go. sorry.
ah, thank you
09:02
@BalarkaSen," Let $X_1,X_2,\dots$ be a countable collection of closed subsets of $\mathbb R^k$, and let $a_1,a_2,\dots$ be a sequence of nonnegative numbers such that $\sum_na_n<\infty$. For each $x\in X$, define $$U(x)=\sum_{\{n=1,2,\dots | x\in X_n\}}a_n.$$"
Then $U$ is an upper semi-continuous function.
I can't understand what does U mean.
@TheSubstitute I suppose it's bad notation. Take $f(z)=|z|^2$. You can represent $f(z)$ as the specialization of a function $g(x,y) = xy$ with $x=z, y=\bar{z}$. Note that $x,y$ are allowed to be complex and that $g(x,y)$ is holomorphic in $x$ for fixed $y$ and holomorphic in $y$ for fixed $x$. Mixing this together beyond recognition leads to expressions like $\partial f/\partial\bar{z}$. The intended meaning is $\partial g/\partial y$ applied to $x=z, y=\bar{z}$.
@JasperLoy Why don't you take this site seriously anymore?
@Committingtoaname I think he said because wrong answers are upvoted and right answers are downvoted
09:17
yes
( ͡° ͜ʖ ͡°)
My beautiful face
You have no nose
Why do I need one
09:33
To breath
I am a mouth-breather
ok
I didn't know that
I thought you knew me
Maybe we're not meant for each other
Perhaps
10:13
... What an intense moment.
$\nabla$ & Ice, why have you changed your pictures?
I haven't changed mine in a while
So you inspired Ice?
How often can you change your avatar?
10:21
unlimited
I will never change my avatar. I am a human :D
Also, is the following notation wrong?
$$(x \in X) = (y \in Y) $$
Is it better to just say: $$x = y\quad ,\text{ where } x \in X, y \in Y $$
@nablablah: What does $\nabla$ usually mean in math? (Other than that curly thing)
10:59
@Nick Other than the gradient function?
@Committingtoaname: I have no idea what the curly thing or gradient function is
@Nick $\nabla = \left(\cfrac{\partial}{\partial x}, \cfrac{\partial}{\partial y}, \cfrac{\partial}{\partial z}\right)$ Or the components are actually really dependent on the function. So $f(x,y,z)$ would give that for example.
@JasperLoy: Greetings wise guy who is no longer Will Hunting
@Nick Rather, I am mad guy who may never recover, but I will keep trying.
@Committingtoaname: That makes no sense to me. Why are there partial derivative operators stuck in a coordinate pointer?
@JasperLoy: You should know that you will never ever be totally sane. No one is.
11:03
Sorry was still editing.
The partials are with respect to all of the functions variables, and given $\nabla f(x,y,z)$ we would have $f$ on the top of all of them\
@Nick Maybe finding a gf would help, lol.
... $\nabla f$ called a gradient function?
@JasperLoy: Yes, but $g$old$f$ish poop a lot.
I haven't heard it be called a gradient function. But it is the gradient of the function
ah
You can also have the gradient of a vector field $\mathcal{V}$
E.g. $\nabla \cdot \mathcal{V}(x,y,z) = \left(\cfrac{\partial \mathcal{V}}{\partial x}, \cfrac{\partial \mathcal{V}}{\partial y}, \cfrac{\partial \mathcal{V}}{\partial z}\right)$
11:09
So, we were talking about the curly thing!
In vector calculus, the curl is a vector operator that describes the infinitesimal rotation of a 3-dimensional vector field. At every point in the field, the curl of that field is represented by a vector. The attributes of this vector (length and direction) characterize the rotation at that point. The direction of the curl is the axis of rotation, as determined by the right-hand rule, and the magnitude of the curl is the magnitude of rotation. If the vector field represents the flow velocity of a moving fluid, then the curl is the circulation density of the fluid. A vector field whose curl is zero...
Curl is a little different.
@JasperLoy: Seriously, catfish are a better investment. They clean up after themselves.
That is the cross product of the gradient and a vector field
So above we have the gradient of a vector $\nabla f$
But this is the cross product of the gradient with a vector field, say $\mathcal{F}$
@Nick
$\nabla \times \mathcal{F} = \begin{vmatrix} i &j&k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \mathcal{F}_x & \mathcal{F}_y & \mathcal{F}_z \end{vmatrix}$
Greetings
ooooh...
11:12
@Committingtoaname Because the votes mean nothing, and many questions are poorly written.
@robjohn have you seen this one?
$$\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(\psi)}{ \cos(a \cos(\psi) \cos(\varphi))} d\psi d\varphi =\frac{\pi}{ 2a}\log\left( \frac{\displaystyle 1+ \tan \left(\frac{a}{2}\right)}{ \displaystyle 1-\tan\left(\frac{a}{2}\right)}\right)$$
@Chris'ssis: and a Happy Greeting to you too :D
@Nick :D
Where $\mathcal{F} = \left(\mathcal{F}_x,\mathcal{F}_y,\mathcal{F}_z\right)$
@Committingtoaname: I have to look into this one day. Thanks for making it less alien for me :D
11:15
It is actually not very hard to learn :). It is often taught like an algorithm
11:28
@Committingtoaname: Isn't $\frac{\partial}{\partial x}$ an operator?
@Nick I prefer learning systematically from books rather than randomly from the internet.
Hi @eric you look very pink.
@JasperLoy: I respect your preferences.
have you guys had a chance to consider the problem i've been stuck on?
@JasperLoy @Nick?
@EricGregor: Do you mean:
3
Q: Probability/Decision problem

Eric GregorLet $X$ be a random variable over $\mathbb{R}$ with finite first moment (mean). Let $H$ be a piecewise function defined such that $H_a=c_1(x-a)$ for $x>a$, and $H_a=c_2(a-x)$ for $x<a$, with $c_1,c_2>0$. Let $a'$ be a number such that $P(X<a')=c_1/(c_1+c2)$, and $\mathbb{P}(X>a')=c_2/(c_1+c_2)$....

yes, @Nick
11:42
No, I have not considered it. It is most likely that Jasper hasn't considered it either.
i don't know if it's up your alley, @Nick, but for some reason this problem is really stumping me and i would really appreciate it if someone could give me a clue what i'm missing
it shouldn't be this hard!
It's on a different planet for me. Sorry. You should probably wait until you can open a bounty. Put it up for 50 points. Then, atleast someone will try harder to answer it.
What do you mean an operator @Nick?
It is a partial derivative in need of a function or vector space :)
@EricGregor: Your question has only been viewed 42 times. When atleast 500 people have seen your it, then you have a better chance of getting a good answer.
@Committingtoaname: Yeah, I referred a book. My terminology is outdated. Thanks for the help, btw :D
Toodles
12:04
i hope so, @Nick
Greetings!
12:21
Greetings :D
(Glad to know we all mimick Chris's sis)
@rehband: So, how's your day going?
@Nick Quite good, drinking green tea, eating grapes and evaluating some sums. Doesn't get much better than that. How about u?
@rehband: Well, I haven't seen @u in a very long time. So, I don't know.
@Nick :D
12:29
Going back to university to take a course is a good way to get your self back to earth, realizing that your own invented mathematics is not all there is.
@rehband: So, what grape sums are you drinking? (As you can see, I'm a very good listener)
@Nick Hahaha
Is the area of a simple polygon with integer coordinates also an integer? I was reading shoelace formula and this occurred to me.
@MatsGranvik: How do you know I have homework?
@rehband: Seriously, though, what sums?
@saadtaame Take a triangle with vertices $(0,0),(1,0),(0,1)$.
12:34
@Nick I can try to give you feedback. But I have a matrix problem I can't solve. I just guessed you have home work based on your discussion with Balarka Sen
@MatsGranvik: Oh yeah, he told me to think about Map(X) ... I forgot why. Better go look at the transcripts.
@Nick Trying to prove that $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)...(k+n)} = \frac{1}{n \cdot n!}$$ where $n$ is a natural number.
How do I show that if the matrix equation $AX=B$ has more than one solution, then it has infinitely many solutions?
@rehband: I have never been able to prove things like that. Any tips on how to do it?
@rehband You see $(k+n)!$ and $(n+1)!$?
12:41
@Nick It always depends on the problem...most of the time I don't know how to start either and then I just do simple cases like $n=1,2,3$ to see if I can see what's going on :P
@Committingtoaname Sorry?
They are denominators for each
Opps sorry the right one was a typo, but the left is good
$$\sum_{k=1}^{\infty} \frac{1}{\prod_{j = 0}^n (k+j)} = \frac{1}{(n+1)!}$$ is just as tough. lol
I meant $\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)...(k+n)} = \sum_{k=1}^{\infty} \frac{1}{(k+n)!}=\frac{1}{n\cdot n!}$ Which looks nicer though
12:48
$$(k+n)! = (k+n)(k+n-1)(k+n-2)\dots (k + 2)(k+1)k!$$
right?
To be more clear:
$$(k+n)! = (k+n)(k+n-1)(k+n-2)\dots(k+n-n)(k+n - (n+1))(k+n - (n+2))\dots 3\cdot 2 \cdot 1$$
@Nick Yes
@rehband: My only point is $k(k+1)(k+2)...(k+n) \neq (k+n)!$
Solving your equation is too hard for me :(
@Nick Indeed
@rehband: You can try to just prove it by induction, right?
lol
@Nick I think the best way would be to do something partial-fraction esque. Haven't proved it yet either though :P
12:59
@rehband: That sounds messy
Is learning really the best way to learn math?
Hi @DanielFischer If you have a chance please view post. I am only interested if the question allows the use of Abels Theorem, I am aware of how to solve it using other methods. Thanks for any assistance.
@Nick What do you think?
@JohnJack By Abel's theorem, you mean that if $f(z) = \sum a_n z^n$ has radius of convergence $1$, and $\sum a_n$ converges (conditionally or absolutely), then the sum is the non-tangential limit of $f(z)$ as $z\to 1$ in the unit disk? Or which?
$$\frac{1}{k(k+1)...(k+n)} = \frac{1}{n \cdot k(k+1)...(k+n-1)} - \frac{1}{n \cdot (k+1)...(k+n)}$$
13:11
I meant $\frac{(k+n)!}{(k-1)!}$ Sorry, I am really tired, and am going to bed now.
@MatsGranvik: My thought: Osmosis of information through the semipermeable membrane that is your skull from book of high information density to your brain of low information density is the most reliable way of data transfer. In English: Try using your book as a pillow.
@Committingtoaname Don't let the bed bugs bite
$\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)...(k+n)} = \sum_{k=1}^{\infty} \frac{(k-1)!}{(k+n)!}=\frac{1}{n\cdot n!}$
If you ever get bed bugs, pay the professionals first thing. You CANNOT deal with them by yourself. Spend the $400, and save yourself a year of your time, and $1000 in do it yourself methods.
Night :)
13:33
@DanielFischer I'm using this definition from wiki entry.
@JohnJack That's the one I mentioned, okay. I don't see how one could use that. One would need a relation between $\sum x_n z^n$ and $\sum \frac{x_n}{1+x_n}z^n$, and I'm not aware of such a relation in general.
@DanielFischer Do you know of another result by Abel used to show convergence of innfinite series?
Hello everony :)
everyone*
@JohnJack There's Abel's criterion that if $\sum a_n$ is convergent, $(b_n)$ is monotonic and bounded, then $\sum a_n b_n$ converges.
Not sure where to ask this question but a separate thread is too much so I thought I'd ask here
I'm starting to do some self-study about mathematical logic and my current university textbook doesn't seem enough, did some searching around and found this answer: math.stackexchange.com/a/11894/6899
I'm however not sure which one to start with, if someone would please recommend a book to buy it'll be greatly appreciated :)
13:48
@DanielFischer Okay thanks. Do you know if $x_{n}$ is positive, $\sum x_{n}$ is divergent $x_{n} \rigtharrow 0$ then does this imply $x_{n}$ is decreasing?
@JohnJack It doesn't. $x_n > 0$ and $x_n\to 0$ implies that the jumps $x_{n+1}-x_n$ when $x_{n+1} > x_n$ must tend to $0$, but the sequence need not be monotonic.
@DanielFischer And the additional assumption that $\sum x_{n}$ doesn't add anything?
@JohnJack No. It allows larger jumps up than if the series were demanded convergent, even.
14:15
Okay thanks. @DanielFischer Can I also confirm that the integral test can't be used in my post.
@KhaledNassar The best logic book is the one by Mendelson. I am not sure if I spelled correctly.
@JohnJack Yes, you need monotonicity for the integral test. If you add such a premise, it would become applicable.
@JasperLoy Alright thanks, will give it a shot then
14:30
@DanielFischer Would you recommend me any English-German dictionary?
@DanielFischer jilbounet mentioned that the integral test can not be used since we do not know $x_{n}$ explicitly and the integral test requires $f(n)$ defined and continuous on an interval of $\mathbb{R}$. So then monotonicity would still not suffice?
@JasperLoy I'm not acquainted enough with any to recommend one. By and large, I'd say if you take one from one of the traditional big houses, it should be not too bad.
@JohnJack Of course, to actually apply a convergence test, you need something concrete. But monotonicity is the condition that makes the integral test applicable in principle. The continuity of $f$ isn't necessary, the monotonicity suffices. But of course it must be defined on $[1,\infty)$ with $f(n) = a_n$. However, for every monotonic sequence $a_n \searrow 0$, there are many - even smooth - functions with the required properties.
14:45
any folks here know decision theory, or probability?
Hi. I have quick question, not sure if worth posting on the site--I haven't done prob and stats in 7 years, so, while I think I have the right idea, I might be completely off.

Basically, I want to calculate the expected roll increase from having the option to reroll a 1d6. In other words, given the option, what's the expected increase, and when should I reroll?

(I can post my calculation/thought process, if anyone would like to see that first)
15:06
@Khashir do share :)
(Your idea as to how to approach it)
Also, I assume reroll once only, so on?
(yes, only reroll once)

So, if X is a random variable that represents the increase from rerolling, the average is the weighted average of each possible increase. In other words, if you roll a 5, then the average increase will be sigma(p * [roll-5]). So:

1/6*(1-5) + 1/6*(2-5) + 1/6*(3-5) ... + 1/6*(6-5).
Which comes out to -1.5
My confusion comes from a fellow SE-er, who then calculates that the expected increase would be 1/6 * 2.5 (expected when rolling a 1), 1/6 * 1.5 (when r 2) ... 1/6*-1.5 when rolling a 5, and he then concludes that on average, you're better off rerolling anything but a 6
which seems counterintuitive
I don't get his calculations. Do you have a link?
It seems as if he did the expectation of increase without knowing what the first roll is.
the only way i can interpret that is: rather than re-rolling on everything, it's better not to re-roll on a six. which is true, but it doesn't follow that that's the best strategy (only that it's better than always rerolling)
That's what I thought too
Then you weight 1/6 each possibility of first roll. But then again, his conclusion couldn't have came from that, and his conclusion seems wrong to me.
15:15
http://rpg.stackexchange.com/questions/48988/how-to-calculate-the-expected-damage-increase-from-empowered-spell/48996?noredirect=1#comment98195_48996

it's the post that starts "So we have to ask ourselves"
A question good to be upvoted
0
Q: Evaluating $\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(\psi)}{ \cos(a \cos(\psi) \cos(\varphi))} d\psi d\varphi $

Chris's sisCan we avoid the use of the geometric interpretation combined with polar coordinates change of variable for proving that $$\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(\psi)}{ \cos(a \cos(\psi) \cos(\varphi))} d\psi d\varphi =\frac{\pi}{ 2a}\log\left( \frac{\displaystyle 1+ \tan \left(\frac{a}{2}\...

man, I feel dumb, comparing that question to mine =P
and reminded of how awe-inspiring Mathematics is.
<3
Ahh I think I get what he is doing.
Nope I don't get what he is doing.
It seems false one way or another. I think your way is the right way.
But let me make sure I get this right - when you choose to reroll, whatever comes out becomes the new number you get
Not some sort of average with the older roll, or anything alike?
Anyhow, the expectation of a new roll is 3.5, one way or another. If you roll 3, your expected increase by reroll is 0.5. If you roll 4, your expected increase is -0.5. So on, so on.
Yeah, whatever number you get is the result
Basically, he's weighing the increase of the roll by the probability
and adding them up
So, 1/6*(3.5-1) + 1/6(3.5-2) ... + 1/6(3.5-5)
(the sum doesn't include 1/6 (3.5 - 6) because he won't be rerolling the 6)
15:31
it seems silly to re-roll on 4 and 5, though.
Agreed... and my claim is that the math he's done doesn't reflect this very clearly
for 3 and less, you're more likely to go up or break even. for 4 and up you're more likely to go down or break even
That's what I thought too
Awesome guys, thanks :)
Much appreciated
15:57
any probability theory people here? math.stackexchange.com/questions/959988/…
02:00 - 16:0016:00 - 00:00

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