should I use Taylor expansion to show it's not continuous? I get this: $$ \frac{x^2y^2-x^2O(y^6)}{x^2+y^4}, $$ so I'm guessing I should show that $\begin{aligned}\frac{x^2y^2}{x^2+y^4} \end{aligned}$ is not continuous at $\vec 0$. I can't use homogeneity. Can I use the fact that: $$ \left\vert\frac{x^2y^2}{x^2+y^4} \right\vert\geq \left\vert\frac{x^2y^2}{x^4+y^4} \right\vert, $$ for $\vert x\vert,\vert y\vert<1$. Because we know that the limit of $\begin{aligned}\left\vert\frac{x^2y^2}{x^2+y^4} \right\vert
Oh, I see what you were thinking. I'm thinking $g(x,u) = f(x,\sqrt{|u|})$.
Anonymous
Does $$y=(x-1)(x-2)^2(x-3)^4(x-4)^3$$ have an inflection point at $x=3$ or not? I think it should have since it has even number of roots $x=3$ (i.e. 4) [Can we say it without calculating the second derivative explicitly?] Ideas anyone?
@TedShifrin I think Leeb teaches some undergrad stuff most semesters. This semester the topology of surfaces that I'm grading for, but sometimes also easier stuff.
I forget how to do that. Can I overkill and use that it must preserve geodesics? If every straightline is sent to straightlines I think you should be able to see it's linear.
And @Alessandro so a Vitali covering of a (let's say bounded) set $A\subset \mathbb{R}^n$ is a cover with balls such that for any $x\in A$ and any $\epsilon$, there's a ball of radius less than $\epsilon$ containing $x$
@TedShifrin now I know we were discussing what the avatar of yours is in terms of geometry, but what software did you use to make it? I've seen quite a few models with that same sort of look on here and I'm curious what makes them. :-)
The argument (in brief) is that you can take those sets of radius larger than, say, 1, and throw them out, since every point is covered by an arbitrarily small ball. Now, you take a ball with radius greater than 0.9 (or any constant less than 1) times the sup of the radius of the balls, then take another at least 0.9 times the sup of everything disjoint from the first, and keep going
So you get a sequence of balls $B_1,B_2,\ldots$, with radii tending to $0$ (since $A$ is bounded). Given some $x\in A\setminus \bigcup_i B_i$, you choose some ball $B$ in the covering containing it. We know that there is some $B_k$ that intersects $B$, since otherwise we would have chosen it instead of the way smaller ones down the chain. Take the first one that does it, $B_k$, and note that you can scale $B_k$ to contain $B$, say by a constant $c$.
Then $cB_1,cB_2,\ldots$ will necessarily cover all of $A$, so that $\sum_i \mu(B_i) \ge \frac{1}{c^n}\mu(A)$ (where $n$ is the dimension), so that there exists an $N$ such that $\mu(B_1) + \ldots + \mu(B_N) \ge \frac{1}{2c^n}\mu(A)$.
Actually wait I forgot one step in the algorithm which was to throw out the balls that were disjoint from $A$. But yeah, so we then have that finitely many of the balls will cover some fixed portion of $A$ (since $c$ only depends on the constant we choose, in this case 0.9), so then you repeat the process for $A\setminus (B_1 \cup \ldots \cup B_N)$ and continue, covering almost all of it with countably many sets
@Ted Oh, I see. $f(x) \cdot f(\mathbf{e}_i) = x \cdot e_i$. Since $f(\mathbf{e}_i) = A \mathbf{e}_i$, $x \cdot e_i = f(x) \cdot A \mathbf{e}_i = A^T f(x) \cdot \mathbf{e_i}$. $A^T = A^{-1}$ (follows from orthonormality), so you have $A^{-1} \circ f$ is identity.
@Ted I think it should be done by the geodesic preservation though. That not only says straightlines goes to straightlines, but that your map is affine on the straightlines.
@TedShifrin looking around at other concepts that might be relevant to a problem I'm trying to solve. Has anyone in the past used other sorts of differential algebrae to solve differential equations through some connection between their solution sets?
How can I show that $f$ is differentiable at $\vec 0$, where $$ f(x,y)=x^2\log(x^4+y^2), $$ and $f(0,0)=0$. I’ve already shown that $f$ is continuous at $\vec 0$. I started off by calculation the first partial derivative: $$ D_1f(\vec 0)=\lim_{t\to 0}t\log t^4, $$ however, I don’t know how to calculate this limit even. I looked at the plot, and it seems that $D_f(\vec 0)=D_2f(\vec 0)=0$, so apparently $t$ goes faster to zero then $\log t^4$ goes to minus infinity. How can I show this? Can I use Taylor? Should I evaluate then at $x=1$? This would yield:
@ShaVukila You want to show $\lim_{t \to 0^+} t\log(t) = 0$, right? Note that $\lim_{t \to 0} t^t = 1$, and $t \log(t) = \log(t^t)$. You can push the limit inside the log (why?)
A bit of a self-imposed challenge: express the number 13256278887989457651018865901401704640 in 40 characters or less using standard mathematical terminology (+, *, ^, etc.)
Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.
@Semiclassical oscillatory discontinuity is a bit of a grey area for some people and this function shouldn't have any oscillation so we can rule that out. That generally occurs with sine or modulo or something periodic. I assumed we were dealing with discontinuity of the second kind.
@BalarkaSen It's not as clear what counts as that type of discontinuity, and the discussion of it is irrelevant here anyways. We're trying to deal with the other kinds. That doesn't even have any relevance here.
@BalarkaSen besides, I was here first helping the person and you kind of jumped in.
@BalarkaSen really? I was telling the person that once they proves continuity of the original function, they could ignore that piecewise defined issue they were saying stopped them from just taking the derivatives...
Either you have to include an additional argument as to why this function is well-behaved at zero (i.e. rigorously exclude the possibility of oscillations) or you just proceed directly from the limit definition.
if I have one point as a separate case, but then prove the outer function is continuous and has a limit approaching said value... is there any need to have it be piecewise defined anymore?
@Semiclassical except that we know that the multiplication of continuous functions is a continuous function and yada yada yada. Couldn't one just use that as an argument?
@Semiclassical and I was which is why I was saying the continuity thing as I was presuming that the asker would understand i was referring to jump discontinuity which was their primary concern (since that is the potential discontinuity in the original function). It already went without saying that there was no oscillation. A simple graph can show that one is definitely not occurring.
the asker is asking whether it is continuous or not. We don't have to be 100% rigorous in all of it. We're just conveying to them why it works. And intuition of that it most likely doesn't occur leads to a simple proof: just look at the fact that the partial derivatives are compositions of continuous functions in that neighborhood.
you're not bothering us. I was just pointing out that maybe we're all overthinking it? :p
most problems in calculus classes tend to be simple and procedural if only to hammer the concepts into ones head through repetition. Of course, that can vary depending on the instructor.
Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.
