Neal
Reputation
17,274
Next privilege 20,000 Rep.
Access 'trusted user' tools
 2h revised In $n>5$, topology = algebra made cw because it's really not *my* answer 9h answered In $n>5$, topology = algebra Apr 17 comment Solving exp integral in closed form? Typo: $r^2 = x^2 + y^2$. For 2) try substituting $u = x/m\sqrt{2}$ and $v = y/n\sqrt{2}$. Apr 13 comment Length of a differentiable curve with respect to a Riemannian metric. @JohnHughes To your first comment, consider a differentiable homeorphism $f:[a,b]\to [c,d]$ such that $f'$ limits to zero at $a$ and $b$. Apr 5 comment Solving the wave equation $u_{tt}=\alpha^2 u_{xx}$ given initial condition $u_t(x,0)=\sin{\frac{\pi x}{L}}+0.5 \sin{\frac{3 \pi x}{L}}$ Do you know how to take a Fourier transform? Apr 5 answered Surface integral (Please help me) Apr 4 comment Finding $H_3$ and $H_2$ for $S^3 \setminus K$ where $K$ is a knot $H_3$ will be zero because the complement of $K$ will deformation-retract onto some $2$-complex, which can be constructed by taking loops corresponding to generators of $\pi_1$ of the knot complement and gluing disks according to the relations. Apr 2 comment If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. A concrete counterexample, considering $D^2$ as the set if complex $z$ with $|z|\leq 1$: $f(z) = e^{i4\pi x}$. The range is in fact the unit circle, and is certainly arcwise connected. Apr 2 comment If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. (Yes, it is because $D^2$ is contractible, but treating it as a nullhomotopy is particularly cute. :) ) Apr 2 comment If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. @takecare I edited it into my answer. Is that a bit clearer? Apr 2 revised If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. added 372 characters in body Apr 2 comment If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. The map $\tilde{f}$ is actually a nullhomotopy! Parametrize $D^2$ with polar coordinates to see explicitly. Apr 2 comment A Soft Question on Differential Geometry @QiaochuYuan Exotic spheres may be a good example of a topological submanifold $M$ of $\Bbb{R}^n$ given a smooth structure whose inclusion is not smooth. ... is it known whether an exotic $S^7$, say, admits any smooth embedding into $\Bbb{R}^8$. Apr 2 answered If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. Apr 2 comment If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. If it misses a point, then it deformation retracts to a map $S^1\to S^1$ which must be nullhomotopic. Apr 1 comment $\pi \not\in \mathbb{Q}$? @MarianoSuárez-Alvarez Well of course not, that's why I said "in some sense" :P Apr 1 answered $\pi \not\in \mathbb{Q}$? Mar 30 comment Estimate Δf using the Linear Approximation and use a calculator to compute the error. @pewpew Writing homework problems like this. The idea is usually to find the value of $f$ near an "easy" value. Here, we know what $\sqrt{9}$ is and we can find the derivative at $9$, so we can use linear approximation to estimate square roots of numbers close to $9$. In this case, I guessed, $\sqrt{9.5}$. Mar 30 comment Estimate Δf using the Linear Approximation and use a calculator to compute the error. Maybe you're supposed to find $\sqrt{9.5} using linear approximation and then determine the error. Mar 30 comment Show for all continuous$f: (\mathbb{N}, \tau_{c}) \to (\mathbb{R}, \tau_{ST})$,$f\$ is a constant function @oliverjones That's not quite what I said. An interval with nonempty preimage contains all but finitely many elements of the image. Its preimage must be open, hence the complement of the preimage must be finite.