Activity for celtschk
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #283593 |
@#53922 Yes, you understand continuous wrong. Continuity does *not* mean that the graph is path connected. Continuity means that you get arbitrary small changes of $f(x)$ for sufficiently small changes of $x$ *inside the domain.* The following would be a discontinuous function:
$$f(x) = \begin{cases... (more) |
— | over 3 years ago |
| Edit | Post #283594 | Initial revision | — | over 3 years ago |
| Answer | — |
A: What is "continuous" in Math? The terms “discrete” and “continuous” are not related to cardinality, but to topology. In particular, you can have a discrete topology on a space of arbitrary cardinality. Note also that “discrete” is a property of a set in a topological space, while “continuous” is a property of functions between... (more) |
— | over 3 years ago |
| Edit | Post #283520 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Is $x=\int \int \ddot{x}\mathrm dx \mathrm dx$? While you can integrate twice for the same variable, your equation is not right. $\ddot x$ means deriving $x$ twice with respect to time, that is, $$\ddot x = \frac{\mathrm d^2x}{\mathrm dt^2}$$ which is very different from deriving twice with respect to $x$ (indeed, $\mathrm d^2x/dx^2=0$). To coun... (more) |
— | over 3 years ago |
| Edit | Post #283449 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Prove that $\int \ddot{x}(t)\mathrm dt=v_0 + \frac{F_0}{m}t$ Your calculation is not correct. What you calculated is $$\\int a\\,\mathrm d\textcolor{red}{a}$$ which is something very different from $$\\int a(t)\\,\mathrm d\textcolor{red}{t}$$ Indeed, in the specific case here, $a = F0/m$ is time independent, that is, it is a constant in respect to time. No... (more) |
— | over 3 years ago |
| Comment | Post #283390 |
I personally learned GR, including the involved mathematics, from the book *Gravitation* by Misner, Thorne and Wheeler. However that's not a math book, therefore if you want mathematical rigour, that book likely isn't for you (e.g. IIRC you won't find any proofs there). Other than that I don't have s... (more) |
— | over 3 years ago |
| Comment | Post #283390 |
I haven't looked into the book (therefore only a comment), but while the topics listed in the title are relevant for quantum theory, they are generally irrelevant or at most tangentially relevant for General Relativity. Therefore I suspect that it is not the best book if your goal is to understand G... (more) |
— | over 3 years ago |
| Comment | Post #283332 |
In your perimeter example, some of the squares the circle crosses are not black; in other words, the pixels do not *cover* the circle. Otherwise you'd get a better approximation. Indeed, the Hausdorff measure is exactly defined by such coverings.
In particular, in your 1st grid, you've got 8 cover... (more) |
— | over 3 years ago |
| Comment | Post #283111 |
This isn't a question about mathematics, but about the specific language used in medicine (I'm not aware of those two terms being used outside that context). It possibly would be on-topic on a medicine site (which Codidact currently doesn't have AFAICT), but not here. It *might* me on-topic on Langua... (more) |
— | over 3 years ago |
| Edit | Post #283261 | Initial revision | — | over 3 years ago |
| Answer | — |
A: After $n - 2$ unchosen doors are opened, how does the probability of the $n - 2$ unchosen doors "shift" or "transfer" to the lone unchosen door? Maybe it's a good idea to start from a situation that's intuitive, and then gradually chance it to the Monty Hall version. So let's start with the following situation: Monty lets you choose a door. Then, after you've chosen it, he offers you to either select the chosen door, or open all the oth... (more) |
— | over 3 years ago |
| Comment | Post #283154 |
Please see the edit, I hope it clears things up. (more) |
— | over 3 years ago |
| Edit | Post #283154 |
Post edited: Fixed a typo |
— | over 3 years ago |
| Edit | Post #283154 |
Post edited: Added explannation why the area of the square is, indeed, 50 |
— | over 3 years ago |
| Comment | Post #283197 |
Yes, now I get the edit window, so that issue seems to be fixed. Thank you. (more) |
— | over 3 years ago |
| Comment | Post #283154 |
Yes, it's a square, with side length (of each of its four sides) $\sqrt{50}$. I tried to add an explanation to the answer, but at the moment for some reason I can't edit it (I've posted a bug report on meta on this). (more) |
— | over 3 years ago |
| Edit | Post #283197 | Initial revision | — | over 3 years ago |
| Question | — |
Editing my answer fails I just tried to edit this answer of mine, but for some reason clicking the edit button only reloads the page. Trying to open in a different tab didn't help either. In case it matters, I'm using Waterfox Classic 2021.02 (64-Bit) KDE Plasma Edition. (more) |
— | over 3 years ago |
| Comment | Post #283154 |
But the side length of the sketched square is not $5$, but $\sqrt{50}$. (more) |
— | over 3 years ago |
| Edit | Post #283154 | Initial revision | — | over 3 years ago |
| Answer | — |
A: How to determine area of square using Calculus in Cartesian coodinate? The correct equation for the square boundary you sketched is $$\left|x\right| + \left|y\right| = 5$$ where $\left|x\right|$ means the absolute value of $x$, that is, $$\left|x\right| = \begin{cases} x & \text{for $x\ge 0$}\\\\ -x & \text{for $x < 0$} \end{cases}$$ Now to calculate the area wi... (more) |
— | over 3 years ago |
| Edit | Post #283087 |
Post edited: Added explanation of dimensional analysis |
— | over 3 years ago |
| Comment | Post #283087 |
Note that the fact that your result passes that test is not a proof that your result is correct, just that you cannot prove it incorrect that way (and indeed, no different power of $x$ would be correct).
On the $x$ in the numerator and denominator: You mentioned as one difference between your resu... (more) |
— | over 3 years ago |
| Comment | Post #283087 |
Dimensional analysis is a quick check normally done on physics equations: If the equation is correct, all terms must have the correct units. Now while your equation for $I$ (probably) didn't come from a physics problem, one can pretend that it does, with $x$ having an arbitrary unit (that would depen... (more) |
— | over 3 years ago |
| Edit | Post #283087 | Initial revision | — | over 3 years ago |
| Answer | — |
A: differentiate under integral sign says something went wrong Assuming you copied the formula from the book correctly, your mistake is simply to assume that your book is right. As a physicist, my first instinct was to make a dimensional analysis. The formula for $I$ gives a dimensionally valid expression as long as $a$ and $b$ have the inverse unit of $x$ (s... (more) |
— | over 3 years ago |
| Edit | Post #282975 | Initial revision | — | over 3 years ago |
| Answer | — |
A: What's wrong with evaluating $n(n-1) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$ at $k = 1$? The informal expression stands for the formal expression $$\prod{j=1}^k (n-j+1)$$ where, since it is a formal expression, you indeed can insert $k=1$, to get $$\prod{j=1}^1 (n-j+1) = (n-1+1) = n$$ as the text states. Now the author of the text obviously wanted to avoid the product notation, ei... (more) |
— | over 3 years ago |
| Edit | Post #282652 |
Post edited: Another interference |
— | over 3 years ago |
| Edit | Post #282652 |
Post edited: Fixed markdown interferring with MathJax |
— | over 3 years ago |
| Edit | Post #282652 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Why rational to be indifferent between two urns, when urn A has 50-50 red and white balls, but you don't know urn B's ratio? Let's assume you know that urn B has 5 balls in it. I deliberately take an odd number, because that way we know for sure that there are not exactly the same number of red and white balls in that urn. Note that since you don't know the content of the urn, you have to assign probabilities to the urn... (more) |
— | over 3 years ago |
| Comment | Post #282343 |
Unless I've missed something, all MathJax I could see on the page (including that in comments) renders fine for me (Waterfox Classic/Linux). (more) |
— | over 3 years ago |
| Comment | Post #281203 |
Thank you for your answer (and sorry for my late reaction). Indeed I was after the space spanned by finite linear combinations of the $\omega_i$; I thought (incorrectly, as I now see) that I had found a base-independent characterisation, which would have meant that it is base independent. So IIUC, th... (more) |
— | over 3 years ago |
| Edit | Post #281016 | Initial revision | — | almost 4 years ago |
| Answer | — |
A: Intuitively, why's X% of Y = Y% of X? Just consider that Z is Z percent of 100. Then you get: > X percent of Y is X percent of Y percent of 100, which is Y percent of X percent of 100, which is Y percent of X. I think that X percent of Y percent of something is the same as Y percent of X percent of something is intuitive. And that ... (more) |
— | almost 4 years ago |
| Edit | Post #281014 | Initial revision | — | almost 4 years ago |
| Answer | — |
A: In "if and only if" proofs, why's 1 direction easier to prove than the other? One thing to consider is that the equivalences are generally based on the validity of certain axioms. And then, one direction may only depend on a smaller set of axioms. Now it is possible that the additional axioms help also in the direction that doesn't need it, but it is also possible that the ea... (more) |
— | almost 4 years ago |
| Edit | Post #281011 | Initial revision | — | almost 4 years ago |
| Answer | — |
A: Why aren't $z_1=f(xy)$ and $z_2=f(x/y)$ functions of 2 variables? You are misrepresenting what Hagen von Eitzen wrote. He did not write that $z1$ and $z2$ depend on only one variable. He wrote that $f$ is a function of only one variable. That's a massive difference. The question this refers to was whether $z1=f'(xy)$ or $z2=f'(x/y)$ are ambiguous, and the correc... (more) |
— | almost 4 years ago |
| Edit | Post #280800 | Initial revision | — | almost 4 years ago |
| Answer | — |
A: Why should a non-commutative operation even be called "multiplication"? In most (but not all) cases where you name an operation multiplication is when you have two binary operations, and one of them distributes over the other, but not the other way round. Then the operation that distributes over the other is called multiplication and the other one addition, in analogy to... (more) |
— | almost 4 years ago |
| Edit | Post #280799 | Initial revision | — | almost 4 years ago |
| Question | — |
Is this topology basis dependent? Consider a topological field $K$ and an algebraic(!) vector space $V$ over $K$, that is, $V$ has not (yet) a topology defined on it. I'm particularly interested in the case where $V$ has infinite dimension. Now be $V^\ast$ the algebraic dual of $V$. Define a topology on $V^\ast$ through pointwise ... (more) |
— | almost 4 years ago |
| Comment | Post #280654 |
Thank you. I did know the term embedding for Riemannian manifolds, but wasn't aware of the purely topological usage. Stated that way, the relation looks natural enough that someone ought to have thought of it before. (more) |
— | almost 4 years ago |
| Edit | Post #280572 |
Post edited: Fixed a typo |
— | almost 4 years ago |
| Edit | Post #280572 | Initial revision | — | almost 4 years ago |
| Question | — |
Does this generalization of path-connectedness also cover general connectedness? I've got the following idea to generalise path-connectedness: Given a topological space $P$ and a subspace $S$, a space $X$ is $(P,S)$-connected if every continuous function $f:S\to X$ can be extended to a continuous function $g:P\to X$. The usual path-connectedness is obtained using $P=[0,1]$ ... (more) |
— | almost 4 years ago |
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