Reputation
Next privilege 250 Rep.
View close votes
Badges
8
Newest
 Scholar
Impact
~584 people reached

  • 0 posts edited
  • 0 helpful flags
  • 6 votes cast
Jul
19
comment Let $A,B$, and $C$ be sets. If $A\subseteq B$, $B\subseteq C$, and $C\subseteq A$, then $A=C$.
I'm trying to hint that you may already know a little proposition about arbitrary sets $X$ and $Y$ that reduces the proof of what you want to just invoking that proposition. What propositions do you already know about the subset relation?
Jul
19
comment Compute Surface Integral
As @Ivo Terek indicates in his (accepted) answer, when you integrate a scalar field over a surface, you do not take its dot product with the normal—as indeed you note would make no sense.
Jul
19
comment Tangent plane and tangent lines to curves through a point
Can you offer of such an example where there are no such smooth curves??
Jul
18
comment Show the function is continuous in $\Bbb R^2$
To compute the partial derivatives at $(0,0)$, you'll need to go back to their definition in terms of limits—and this would seem to be no simpler than dealing showing $\lim_{(x,y) \to (0,0) f(x, y) = 1$.
Jul
18
comment Compute Surface Integral
If the question is to integrate $x^2 + y^2$ then you are not integrating a vector field but rather the scalar field $F(x, y, z) = x^2 + y^2$.
Jul
18
comment Let $A,B$, and $C$ be sets. If $A\subseteq B$, $B\subseteq C$, and $C\subseteq A$, then $A=C$.
Do you already have at your disposal the little result that if $X \subset Y$ and $Y \subset X$, then $X = Y$?
Jul
18
comment If we have an embedding $f:X \rightarrow A$, where $A \subset Y$, do we have to show $f^{-1}$ is continuous?
"Embedding" may be used in more restrictive ways in a particular context, e.g., an isometric embedding. But if all you have are topological spaces, or all you're considering is topology, then "embedding" is synonymous with "topological embedding".
Jul
18
comment Does this intuition for “calculus-ish” continuity generalize to topological continuity?
As to your students' question as to why the projections should be continuous: Is that not a natural generalization of the situation the projections from the cartesian plane onto each of the coordinate axes?
Jul
18
comment Does this intuition for “calculus-ish” continuity generalize to topological continuity?
The issue for a function $f: X \to Y$ between topological spaces is: what is the meaning of $\lim_{x \to c} f(x)$? One may use the notion of "net" to formulate the definition and then, with that definition, the condition you state is equivalent to continuity at $c$, provided you state it a bit more precisely: for every net $(x_i)_{ \in I}$ that converges to $c$ in $X$, the net $\bigl((f(x_i)\bigr)_{i \in I}$ converges to $f(c)$ in $Y$.
Jul
6
comment How can $f(x)=x^4$ have a global minimum at $x=0$ but $f''(0)=0$?
@Whyka: right, $a > 0$ implies $a^b > 0$ for all real $b$. In your example, $f(x) > 0$ for all $x \neq 0$ and $f{0} = 0$, which is why I gave the condition $b > 0$.
Jul
6
comment How can $f(x)=x^4$ have a global minimum at $x=0$ but $f''(0)=0$?
@Whyka: Certainly if $a > 0$ and $b > 0$, then $a^b > 0$.
Jul
5
comment Solve 4 A (L^(3/4)) - wL (((24 - L) w)^(-3/4)) = -(((24 - L) w)^(1/4)) for L? Using Mathematica
The code has also a second error: wL should be w L.
Jul
5
comment How can $f(x)=x^4$ have a global minimum at $x=0$ but $f''(0)=0$?
"maximum" should be "minimum" in the answer at hand. How show this function has a minimum at $x=0$? simple: use the definition of "minimum"!
May
1
comment Traditional axes in 3d Mathematica plots?
"Traditional axes" to me sounds like the way mathematicians would typically sketch a 3D plot, no just with axes emanating from the origin, but with arrows at their positive ends and axis labels, too.
Jan
3
comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval?
I just obtained full-text access to reference #2, where I could read the cited Theorem 3. And yes, the condition I asked about in my preceding comment is indeed the one shown there to suffice. Thanks!
Jan
3
comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval?
I cannot access reference #2. For reference #1, I cannot access the relevant section beyond page 372. So would you mind quoting the theorem(s) to which you refer? And note that the Riemann-Darboux criterion in Theorem 2 of #2 involves controlling the difference between the Riemann sums using the maximum value and minimum value, respectively, of $f$ in each subinterval; but that's not the condition I asked about.
Jan
3
comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval?
I'm not sure what the final sentence means. Does it mean the following? $f$ is integrable if: there exists some sequence of partitions $P_n$, with mesh tending to 0, for which, no matter what sample points in the subintervals of $P_n$ are used, the corresponding sequence of Riemann sums converges, and that limit is independent of the sample points chosen. (I knew the problem wasn't banal; I just didn't say what I meant the first time 'round.)
Jan
2
comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval?
Taking the min $m$ of the non-regular partition's lengths is what I immediately thought of before I asked the question. The trouble is that you won't necessarily get a regular partition from it: $\Delta x = (b-a)/m$ need not be an integer. So the final subinterval need not have the same length as the others.
Jan
2
comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval?
Sorry, I meant to allow arbitrary choice of sample points in the subintervals, not just existence of sample points there. I've edited to indicate that. Yes, of course the characteristic function of the rationals on the unit interval is a counterexample to what I originally wrote; in fact, as I had in mind in my 2nd note, it's the well-known counterexample to allowing only endpoints as sample points.
Dec
27
comment Easy example why complex numbers are cool
@Mario Carneiro: My $(x - 2)$ was a typo, caught too late to edit. My point was not that determining the radius of convergence of some series expansions could not be facilitated by using complex numbers. Rather, my point was simply that the originally proffered example of $\frac{1}{1+x^2}$ was treatable easily by purely real methods. In any case, it's doubtful that power series expansion provides examples "explainable to someone only knowing high school mathematics" (which is what the OP requested).