Masacroso

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$Mathematics$ Discussion on question by J.D.: How i…

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Masacroso

Mar 31, 2023 12:39

its a mathematical model, and it "predict" things in a probabilistic way. All the technology is based in mathematical/physics models of reality, however they are just that: models. The accuracy or utility of a mathematical model is limited to some degree of precision, as reality is not as simple as our models

$Mathematics$ Discussion on question by BCLC: Are m…

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Masacroso

Apr 26, 2021 18:07

yes... and I said twice (actually three times if you count this one). Take a look at the definitions and see for yourself, by example here

Masacroso

Apr 26, 2021 18:07

immersed (or regular) submanifolds are a kind of manifolds, they are manifolds. Submanifolds are manifolds from the beginning, they don't need to "become" manifolds... they already have a manifold structure.

Masacroso

Apr 26, 2021 18:07

there is not a concept of "immersed manifold", I meant immersed submanifold, sorry if this confused you. An immersed submanifold is the image of an immersion, yes. The immersion itself defines it manifold structure

Masacroso

Apr 26, 2021 18:07

I dont follow what you are asking: a submanifold (immersed or regular) is a manifold so it have a manifold structure. I add: an immersed manifold is locally an embedding so, locally, the manifold structure of the parent manifold and the immersed submanifold coincide, maybe this clarify your doubts.

Masacroso

Apr 26, 2021 18:07

your question is not so clear. Do you know the difference between an immersed manifold and a regular one?

$Mathematics$ Discussion on question by Steven Hatt…

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Masacroso

Jan 9, 2019 01:12

I think the same, historically it makes more sense to start from one, not from zero. However this is a matter of taste and, nowadays, it is common start with zero by many reasons, one of them is that it is important in informatics start the count on a list from zero instead of one. Other reason could be because it simplifies things a bit in the Peano arithmetic

$Mathematics$ Discussion on question by Detmondyou:…

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Masacroso

May 6, 2018 06:56

@Detmondyou it is not an equation because there are not variables involved where to find solutions. It is an identity, what is a different thing.

Masacroso

May 6, 2018 06:56

your equation doesnt hold because there is not an equation!! You have an identity, and you are replacing values in an identity in hope that the identity remains true. But this is the same to say that the identity $x^{yz}=y^2$ remains true for any $x,y,z\in\Bbb C$, what is easy to see that it cannot be true.

$Mathematics$ Discussion between buddhababe and Mas…

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Masacroso

Apr 21, 2017 05:03

it is possible, yes. I dont know what is the context (the theory that your professor was teaching) to know how to answer your question.

Masacroso

Apr 21, 2017 04:57

the integral is the area under the curve

Masacroso

Apr 21, 2017 04:51

the rectangles are defining a specific partition of the interval $[0,8]$, you said that the width of each subinterval is $2$ so $\Delta x_k=2$.

Masacroso

Apr 21, 2017 04:47

But this is a different thing that find some Darboux sum. This is the reason why your original question was not so clear. The area under the curve is found when the limit that you showed in the heading of your question, where $\Delta x_k\to 0$, is well defined for any tagged partition.

Masacroso

Apr 21, 2017 04:43

I dont understand what are you trying to do. I dont know exactly the context of your original question. My recommendation is that you look again what your professor was teaching the last days.

Masacroso

Apr 21, 2017 04:40

No, I dont know why you want to to do that, but $$\lim_{n\to\infty}n^2-4n+6=\infty$$

Masacroso

Apr 21, 2017 04:36

$M_i$ is the supremum of the function in the subinterval $i$. If the function is continuous then the supremum is the maximum of the function in this subinterval. But this is used to evaluate the Darboux upper sum, you want to evaluate the Darboux lower sum, right?

Masacroso

Apr 21, 2017 04:31

Then use the definition of the Darboux lower sum, see here. I mean, what is your problem in this exercise?

Masacroso

Apr 21, 2017 04:31

the Riemann sums are based in tagged partitions, so it value can be very different if you choose different tags. I assume that your professor (probably) wanted the Darboux upper sum or the Darboux lower sum of the defined partition.

Masacroso

Apr 21, 2017 04:31

the question is not clear, you want to evaluate the area under $y$ in $[0,8]$ or the area of a Riemann sum of four rectangles?

$Mathematics$ Discussion between Masacroso and sama…

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Masacroso

Mar 10, 2017 16:01

are you here Samar?

Masacroso

Mar 10, 2017 15:59

Samar, the distance $\mathcal L$ is not a point, so I dont know exactly how a sequence can converge to it... It depends of the definitions of your book. But I must says that define an imfimum by a sequence is not common.

Masacroso

Mar 10, 2017 15:56

@samankumar it depends on the topology where you are working on. For the standard topology on $\Bbb R$ we can say that if $x\notin[a,b]$ then $d(x,[a,b])>0$... but I cant use a sequence to show that because we dont have defined point $x$ or defined set $[a,b]$, I prove it using the fact that $x>b$ or $x<a$.

Masacroso

Mar 10, 2017 15:56

@samar you have an already stated $\mathcal L>0$ for some pair $x,D$, so it is assumed that this sequence exists. Or you want to prove that the distance $\mathcal L$ is greater than zero for some pair $x,D$?

Masacroso

Mar 10, 2017 15:56

@samankumar you are welcome, you dont need to sorry. Without error there is no learning.

Masacroso

Mar 10, 2017 15:56

@samankumar I dont understand your question. The distance between any point of the boundary of $D$ and $D$ is zero, that is $$d(x,D):=\inf\{d(x,y):y\in D\}=0, \quad\forall x\in\partial D$$

$Mathematics$ Discussion on question by SilenceOnTh…

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Masacroso

Feb 11, 2017 11:02

@SilenceOnTheWire no, it shouldnot. The classic example: $$\lim_{x\to\infty}\frac{\sin x}{x}\neq\frac{\lim_{x\to\infty}\sin x}{\lim_{x\to\infty}x}$$ because $\lim_{x\to\infty}\sin x$ doesnt exists.

Masacroso

Feb 11, 2017 11:02

Your first statement "I know that $\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow c} f(x)}{\lim_{x\rightarrow c} g(x)}$" is not true in general. It is true only in the case that the limits in the RHS exists. The answer is $0$, yes, because the others options goes to a different limit than $2$.

$Mathematics$ Mathematics

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Masacroso

Nov 13, 2016 21:39

if you dont want to says where you read it, it is ok too.

Masacroso

Nov 13, 2016 21:38

I dont undersatnd sorry... moreover, I dont accept that etiquette in the chat ;)

Masacroso

Nov 13, 2016 21:35

where you read it @saturatedexpo?

Masacroso

Nov 13, 2016 21:26

oh, ok I think I understand. Thank you all, you put some light here.

Masacroso

Nov 13, 2016 21:21

but the field of a normed vector space must contain some subset of $\Bbb R$, right?

Masacroso

Nov 13, 2016 21:20

mmm... I think I understand. Having $\Bbb R$ as codomain just mean that some subset of $\Bbb R$ is involved, not necessarily a field, right?

Masacroso

Nov 13, 2016 21:13

the question seems very straighforward but I need some confirmation

Masacroso

Nov 13, 2016 21:12

then Im trying to see if $\Bbb Q$ is a subfield of any normed vector space

Masacroso

Nov 13, 2016 21:12

@Astyx every vector space is over a field

Masacroso

Nov 13, 2016 21:11

Ok, I will open a question on mathexchange after all xD

Masacroso

Nov 13, 2016 21:07

@Fargle oh, right... so, can we say then that $\Bbb Q$ is a subfield of any normed vector space?

Masacroso

Nov 13, 2016 21:05

hi all, I have a little question and I dont want to open a question for this on mathexchange so I come here. The question is this: the set $\Bbb R$ is a subfield of any normed vector space, right? Because the norm is a function $\|{\cdot}\|:V\to\Bbb R$

$Mathematics$ Discussion on question by Lin Ma: con…

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Masacroso

Sep 17, 2016 20:10

I deleted my answer because it doesnt represent the standard point of view about the formalization of random variables. Anyway it is perfectly possible to write a meaning to $Pr[A|B]$ as a function with two variables if we dont take the formula as standard as I did. A probability is a function over a sigma algebra, a measure. So if we define $A|B$ over a sigma algebra then it is a well-defined function.

Masacroso

Sep 17, 2016 20:10

@Did the integral is not under any complete sample space so it is not $1$.

Masacroso

Sep 17, 2016 20:10

@Did I dont know a reason why you cannot define $\Pr[A|B]=\int_{y\in\gamma}\int_{x\in\omega}f_{A|B}(x,y)\mathrm d y\mathrm d x$

Masacroso

Sep 17, 2016 20:10

@LinMa yes, this is right. The problem with the approach to a formal probability theory is that a random variable is defined under measure theory that involve a kind of collection of subsets of the sample space named a sigma algebra, so a random variable $X$ use elements of the sigma algebra not the sample space directly. So generally the formal definition of random variable is omitted for not high courses of probability.

Masacroso

Sep 17, 2016 20:10

By example an event is $X=a$, when the random variable $X$ take the value $a$. In general an event is a subset of the sample space. Check this. By the other way a random variable is a function.

Masacroso

Sep 17, 2016 20:10

What is $A$ and what is $B$? They are events or random variables? If $A$ and $B$ are random variables then $\Pr[A|B]$ is a function of two variables. But generally you see the case $\Pr[A=x|B]$ that is a function of $B$.

$Mathematics$ Discussion between girish kumar chand…

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Masacroso

Jun 7, 2016 08:38

yes

Masacroso

Jun 7, 2016 08:37

In other words: doesnt exist any point $x$ close to $c$ such that $|f(x)-f(c)|<1$

Masacroso

Jun 7, 2016 08:35

yes. Take the function $f(n)=n$. Then try to see what happen with the definition of limit remembering that distance zero is prohibited. For an $\varepsilon\in(0,1)$ we have that doesnt exist a $\delta$ that hold the definition of limit because distance zero is prohibited.

Masacroso

Jun 7, 2016 08:32

this is not true that if a function is continuous to some point then a limit exist.

Masacroso

Jun 7, 2016 08:31

The point is that in a domain as $\Bbb N$ there is no convergent sequence to any point, so you cant define a limit. You need convergent sequences in the domain to define a limit so some point.