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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. $f(x) \geq f(p)/2$.