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 Apr 29 comment Let $f: [0, 1] \to \mathbb{R}$ s.t $f(0)=f(1)=0$ then measure of $A = \{h \in [0, 1] \mid \exists x \text{ such that }f(x+h) =f(x)\} \geq 1/2$. Why is it wrong? Clearly $0\in A$ and $1\in A$, and clearly $A\subset [0,1]$. If you can prove $A$ is connected, then you are done. And I can't think of a function $f$ that makes $A$ disconnected. Apr 25 comment Find critical point by graph observation The derivative is defined as the limit as $h$ goes to zero of $\frac{f(x+h)-f(x)}{h}$, provided that the limit exists and is finite. Now, you know that you can identify the derivative (if it exists) with the slope $m$ of the tangent line $y=mx+q$ (strictly speaking, "slope" makes sense only when $m$ is finite). From the graph, you can say that the tangent line is vertical, so there is no slope ($m$, in the formula $y=mx+q$). Hence, there is no derivative. Apr 25 comment Find critical point by graph observation Inflection points are not, per se, critical points... Apr 25 answered Find critical point by graph observation Apr 19 comment I need to figure out how to fit an $x^3$ curve to fixed endpoints, but a variable middle Or you can fix one of the coefficients. For instance, you may take $a=1$, if you want your cubic function to grow roughly as fast as $x^3$... Apr 18 answered I need to figure out how to fit an $x^3$ curve to fixed endpoints, but a variable middle Apr 17 comment If $\sum_{1}^{\infty}(a_n)^3$ diverges, does $\sum_{1}^{\infty}(a_n)$? Uh, right. Thanks. Apr 15 comment If $\sum a_{n}$ is convergent then what about $\sum a_{n}^{2k+1}.$ My bad. I implicitly assumed $a_n$ to be, if not positive, at least alternating, in which case $\sum a_n^{2k+1}$ converges. But those are only particular cases. Apr 15 comment If $\sum_{1}^{\infty}(a_n)^3$ diverges, does $\sum_{1}^{\infty}(a_n)$? @ThomasAndrews I just stumbled here, more than 4 years later, so maybe you won't see this. Your counterexample does not seem to work. Sure, the sum of three consecutive terms is zero, but that does not mean that the partial sums are all zero. Indeed, the partial sums $s_n$ for the series $\sum a_n$ do not converge either, since it is an oscillating sequence. Apr 15 comment If $\sum a_{n}$ is convergent then what about $\sum a_{n}^{2k+1}.$ Hint: comparison test. Apr 14 answered $\limsup_{n\to \infty}(n^n/n!)^{1/n} = e$ Apr 14 comment $\limsup_{n\to \infty}(n^n/n!)^{1/n} = e$ Are you allowed to use Stirling's approximation? en.wikipedia.org/wiki/Stirling%27s_approximation Apr 14 answered Find the equilibria Apr 14 comment Find the equilibria That's right. Those are the equilibria. If your initial condition is, say, $s=1$, then the solution will not change for all times, which is why it's called "equilibrium". Apr 14 comment Find the equilibria Well, an equilibrium is a stationary solution, meaning that time derivatives are all zero. Perhaps step 1 can help you finding them... Apr 14 comment Find the equilibria Well, do you know what an equilibrium is? If yes, have you tried something? Apr 14 comment Mean of binary random variable with probability $.75$ of getting $1$ You're right. The answer sheet is wrong. Apr 11 comment Why complex variables used for Laplace equation? Uhm, I'm not looking forward to share my email here. I don't even know if it's allowed. Perhaps on chat.stackexchange.com Apr 6 comment Finding the radius and the interval of convergence. That's my point. With ratio test you don't need to use L'Hopital. You just need to evaluate a limit of a rational function, which is really easy and fast. Apr 6 answered Suppose $f:E \rightarrow \mathbb{R}$ is continuous at $p$. Prove that if $f(p) > 0$, then there is $\delta>0$ s.t. $f(x) \geq f(p)/2$.