5:34 AM
can you use the fact that $\exp$ is increasing?
6:17 AM
> Definition 8.20 Let $X$ be a random variable with values in $\mathbb Z_+$. The generating function of $X$ is the function $g_X:[0,1]\to[0,1]$ defined by $$g_X(r):=\mathbb E[r^X]=\sum_{n=0}^\infty \mathbb P(X=n)r^n.$$
Why is the domain of $g_X(r)$ only $[0,1]$? If it has radius of convergence at least $1$, shouldn't the function be defined for at least $[-1,1]$?

2 hours later…
8:27 AM
This is hilarious

2 hours later…
10:41 AM
@psie to be non-negative
No one says you can't plug in values other than that, you can
ok, thanks, I guess it's what you say, for convenience to be nonnegative
I actually don't know why authors restrict themselves like this. One thing that comes to mind is that it might be because $g_X\circ \exp = M_X$ where $M_X$ is the moment generating function of $X$
So to further make clear the duality between the two someone could have done that, since $\exp$ is a positive-valued function.
Of course this equality is valid on $(-\infty, 0]$
11:00 AM
ok 👍
11:21 AM
Another issue might be that they only know Abel's theorem for non-negative coefficients of the power series
So to prove $g_X$ is continuous they make such assumption
But, oddly enough answered by me on this site, there is an alternative approach using Weierstrass $M$-test which shows its continuous in the ball $\{z\in \mathbb{C}: |z|\leq 1\}$
12:22 PM
-1

It is to my understanding that mathematics is in some way the domain of all logical systems. However unconventional, as long as certain criterias are met, they could be considered as part of mathematics. For example, subjects like ring theory or knot theory are quite far fetched from the "normal"...

1 hour later…
1:28 PM
$\int \int_D \frac{xe^{2y}}{y+2}dxdy$ D is the flat domain included between the unit circle with the origin as its center and the ellipse of semi-axes a = 2 and b = 1 and contained in the first quadrant.
How can i find the domain ?
2:13 PM
@BinkyMcSquigglebottom did you draw the region and the equations of the circle and ellipse?

2 hours later…
3:44 PM
@SineoftheTime the First Is $x^2 + y^2 = 1$ , the second is $\frac{x^2}{4}+\frac{y^2}{1}=1$
However, I only have to consider the first quadrant
It should be the area between these 3 points
can you write the domain using the equations?
4:03 PM
$D=\{ x^2 + y^2 ≥1,\frac{x^2}{4}+\frac{y^2}{1}≤1,x≥0, y≥0\}$
@SineoftheTime
In the first quadrant you can write the circle and the ellipse as functions
@SineoftheTime Though, arguably, it might make sense to work in polar coordinates. Otherwise, you have to consider $[0,1]$ and $[1,2]$ separately (which might be fine, too).
4:20 PM
@XanderHenderson Looking at the integrand, I don't know if using polar coordinates it's the simples way. Instead we can write the region as $\{ 1\le x \le 2, g(x)\le y \le h(x)\}$
but we have to consider also $x\in [0,1]$ as you've said
@SineoftheTime Oh, I don't even know what integral we are worrying about. I'm too lazy to read that far back up the thread.
maybe it's best if we fix $y$
@SineoftheTime This doesn't seem to be correct, if we are only interested in quadrant 1.
You need to consider $y$ between the two functions on $[0,1]$, but $y$ between the $x$-axis and the ellipse on $[1,2]$.
$\{0\le y\le 1, a(y)\le x\le b(y) \}$
@SineoftheTime Oh, derp. That's the thing to do.
4:23 PM
By the way, does this type of domain have a name?
Like normal domains?
@SineoftheTime What do you mean by "type"?
And what do you mean by "normal domain" in this setting?
For example $\{a\le x \le b, f(x)\le y \le g(x)\}$
In italian are called "normal domains"
But I've never heard something similar in English
"Normal domain" is a bad translation, as "normal domain" has a more standard meaning in algebra. Google translate suggests a "simple domain".
In any event, I've never heard a terms for such domains. It seems like it might be pedagogically useful, but there is nothing particularly interesting about them mathematically which would necessitate a term (in my opinion).
@BinkyMcSquigglebottom for example, we call this region "normal domain with respect to the $y$-axis" because $y$ is bounded by two numbers and $x$ is bounded by two functions of $y$
Yes
Its clear
4:38 PM
@SineoftheTime I would suggest not calling it such in English. You are likely to confuse people.
4:49 PM
I don't use this term, I was just curious if there's a similar terminology in English
22 mins ago, by Xander Henderson
In any event, I've never heard a terms for such domains. It seems like it might be pedagogically useful, but there is nothing particularly interesting about them mathematically which would necessitate a term (in my opinion).
@SineoftheTime sometimes we call it like that too
but not everyone uses this type of nomenclature
i suspect the same thing exists in English, just Xander doesn't know about it
@Jakobian I mean, that is basically the content of the comment I just quoted...
@XanderHenderson where in your comment does it contain my opinion that it probably exists in English?
Consider the Laplace transform of (the law of) $X$, i.e. $$L_X(\lambda):=\mathbb E[e^{-\lambda X}]=\int_{\mathbb R_+} e^{-\lambda x}\mathbb P_X(\mathrm{d}x).$$ It is claimed $L_X$ is convex. Is it clear to someone why that is?
4:58 PM
That being said, I've taught a lot of calculus, multivariable calculus, and analysis classes out of a variety of books and at several institutions. If such a term were widely used in the US, I imagine that I would be aware of it.
@Jakobian I am not going to engage with you any longer. You are clearly trying to pick yet another fight, and I don't have the energy for it.
@psie Should follow from Jensen's inequality
ah good idea, I'll look it up
@psie actually no, not Jensen's inequality
yeah, Jensen's has as assumption convexity
I think the statement that $L_X$ is convex is a deeper result in convex analysis
@psie what book are u using to study probability / stats (if I can assume thats what youre studying)
5:03 PM
Currently, and this is where my question comes from, I'm using this book.
@psie this should just follow from $\lambda\mapsto e^{-\lambda x}$ being convex for any fixed $x$
I think you're right, but is it because integration preserves convexity? After all, we are integrating $x\mapsto e^{-\lambda x}$.
its linear and preserves inequalities, so yeah
we are integrating $\omega\mapsto e^{-\lambda X(\omega)}$
ok 👍
@XanderHenderson I also don't have energy for it, as I am sick and coughing.
5:14 PM
ok, I guess I should have said, we are integrating $x\mapsto e^{-\lambda x}f(x)$, where $f$ is the density of $X$. Then it would be correct :)
$X$ is has a density, yes
5:49 PM
i was wondering about non zero characteristic fields cuz of this problem in Sheldon Axler LADR: the union of 3 vector subspaces is a subspace if and only if one of them contain the other two. testing it for a vector space containing $a_1,a_2,a_1+a_2,0$ over the field ${0,1}$ the theorem didnt hold, which is interesting
6:04 PM
@nickbros123 yeah, iirc this doesn't hold for finite fields
the proof that union of two proper vector subspaces not containing each other is not a vector subspace requires that your field is infinite
Hi! I was reading a paper. They show $(-\Delta-|x|^{\alpha})u\leq0$ and $(-\Delta-|x|^{\alpha})v>0$ in $\mathbb{R}\setminus B_r(0)$. Then they are using maximum principle on $\mathbb{R}\setminus B_r(0)$ to get $u(x)\leq \frac{u(r)}{v(r)}v(x).$ Does anyone know which maximum principle they are are using?
I think they have rescaled to get 0 on boundary.
6:36 PM
@Jakobian wish you get well soon!
@SineoftheTime so $\{0\le y\le 1, \sqrt{1-y^2}\le x\le 2\sqrt{1-y^2} \}$ ?
What does it mean for "every statement becomes a theorem if a contradiction is true"?
@BinkyMcSquigglebottom yes
@Obliv in this context i think it means if a statement X and its negation not(X) are true, then every conceivable statement becomes true
@SoumikMukherjee thank you, I will eventually
6:45 PM
@nickbros123 yeah i don't understand why
I don't get why in hungerford, he says "A paradox occurs in an axiom system when both a statement and its negation are deducible from the axioms. This in turn implies (by an exercise in elementary logic) that every statement in the system is true"
How come $A\land \neg A$ being true means we can imply $A \land \neg A \implies X$ for all statements $X$ in the formal system?
Don't theorems have to be stated, or can be derived, from axioms? (Or I guess they can be independent statements)
So if a contradiction exists within the system, why does this now "explode" the entire system? Wouldn't it only be limited to the statements connected to the contradiction?
Im stuck here
Pls help , i dont know what to do
Dear all, I need to prove $f(x)=e^x$ is a convex function. I could use the derivative test, but I would like to use the actual definition for practice. The definition says let $\lambda\in[0,1], x_1,x_2\in\mathbb{R}$, a function is convex if
$$f(\lambda x_1 + (1-\lambda) x_2 \leq \lambda f(x_1) + (1-\lambda)f(x_2).$$
In essence, I need to show
$$e^{\lambda x_1 + (1-\lambda)x_2} \leq \lambda f(x_1) + (1-\lambda)f(x_2).$$
To prove the above inequality, let $a\in[1,\infty)$, I start with this
$$idk but I see (1-y^2) and I think of y = \sin\theta maybe @Obliv "with P {\displaystyle P} standing for "all lemons are yellow" and Q {\displaystyle Q} standing for "Unicorns exist". We start out by assuming that (1) all lemons are yellow and that (2) not all lemons are yellow. From the proposition that all lemons are yellow, we infer that (3) either all lemons are yellow or unicorns exist. But then from this and the fact that not all lemons are yellow, we infer that (4) unicorns exist by disjunctive syllogism. " 7:06 PM @nickbros123 Do contradictions come from axioms? Also how can we infer that statement P\lor Q though? In math, let's say we have axioms A_1 through A_9. If we have A_1 true and \neg A_1 is also true, then I can just say A\lor B is true no matter what B is? every statement comes from some string of implications that follow from axioms. But not necessarily directly. In fact, usually this string of implications is large @Obliv contradictions come from what is regarded as "wrong" assumptions, or wrong axioms, so yes, they come from axioms @Jakobian what is meant by not directly? As in, there are many ways to reach the same theorem/statement or something? When we obtain a contradiction then it only means that we found a statement A such that both A and \lnot A are true. This A doesn't have to be an axiom @Jakobian generally when we start a proof by contradiction by "assuming" a particular thing, arent we setting it as an axiom for the time being? 7:11 PM Ah okay. So there is like a path from the axioms to reach A and \neg A. In the shortest path, A and \neg A are axioms. yeah, like a "path" or formal deduction @nickbros123 no Are there multiple paths to the same statement? Or do they just look like different paths/proofs but they're actually the same of course, there can be multiple an example is set theory there is such thing that Kunen (iirc) calls Basic Set Theory or BST Sorry for interrupting, is it difficult to show$$
\lambda e^{(1-\lambda)(x_1-x_2)} + (1-\lambda) e^{\lambda(x_2-x_1)} \geq 1.

The reason is that there's multiple things that can be proved either from power set axiom, or from axiom schema of replacement
for example, you can prove the existence of Cartesian product $A\times B$ just from BST
that is, by assuming either power set or replacement, we obtain two different proofs that it exists
so there can be different "paths" to prove a theorem
7:20 PM
Hmm okay I will look into Kunen, thanks.
@BinkyMcSquigglebottom what is the source of the original integral?
7:36 PM
I haven't seen TedShriffin's messages for so many days.
Is he okay?
@SineoftheTime $\int \int_D \frac{xe^{2y}}{y+2}dxdy$ D is the flat domain included between the unit circle with the origin as its center and the ellipse of semi-axes a = 2 and b = 1 and contained in the first quadrant.
source of the problem?
from an exam trace
I think this integral cannot be done
it seems strange, since the result is related to the exponential integral
I put it on wolfram, I get this notation
How should it be calculated? I didn't understand this notation
7:53 PM
Ei is the exponential integral
$\text{Ei}(x) = -\int_{-x}^{\infty} \frac{e^{-t}}{t} \, dt$ for $x > 0$
I'm not sure if this replay for my question. if so, please take a look at the question in here
https://math.stackexchange.com/questions/4946393/prove-ex-is-a-convex-function-using-the-convexity-definition
@SineoftheTime Indeed
I tried to switch the order of integration but I get the same result, so it's highly improbable we've made a mistake
8:29 PM
so what should I do ? :⁠-⁠(
9:11 PM
@BinkyMcSquigglebottom to clarify, what does "flat domain" mean here?
also, where is said ellipse centered?
my guess would be that the domain is this:
this also recovers the same result found by Pizza (using mathematica)
@Semiclassical this should be the region
yeah
i guess flat might just mean 2D here
but that's redundant here
yes, but it's not necessary
i guess. strange question
9:40 PM
┻⁠┻⁠︵⁠ヽ⁠(⁠`⁠Д⁠´⁠)⁠ﾉ⁠︵⁠┻⁠┻
So it can't be done?
$\ddot{\smile}$
@BinkyMcSquigglebottom It can be done, but the result involves special functions
@SineoftheTime ─=≡Σʕっ•ᴥ•ʔっ
@SineoftheTime But how do I understand where to use the special function?
૮ ⚆ﻌ⚆ა
9:59 PM
@BinkyMcSquigglebottom you usally don't need this knowledge
when solving double and triple integrals