xavierm02
Reputation
5,814
Top tag
Next privilege 10,000 Rep.
Access moderator tools
1 8 22
Impact
~108k people reached

# 1,050 Actions

 1d awarded Yearling Mar 16 comment Proof of uniqueness about distribution in Rudin's I'd guess that $u \mapsto (\phi \mapsto u * \phi)$ is linear so he just says that the kernel of that map is $0$, hence proving the injectivity of that map (and so the uniqueness of $u$). Feb 26 comment Can I divide entire equations by each other when trying to solve a system of non-linear equations? $0=x$ and $0=y$ imply $x=y$ that clearly allows new solutions. You're in that case with F instead of 0. Feb 26 comment Can I divide entire equations by each other when trying to solve a system of non-linear equations? In most cases, you won't be able to reduce the number of equation: when you combine two equations to create a third one, you have to keep at least one old one. A method is to find solutions to one of the equations and the search for solutions of the second equation in by plugging in the shape of solutions if the first one. Feb 26 comment Can I divide entire equations by each other when trying to solve a system of non-linear equations? No. For example, if you have $x=6$ and $y=3$, you can deduce $x/y=2$ but $(x,y)=(2,1)$ is a solution of $x/y=2$ that is neither a solution of $x=6$ nor a solution of $y=3$. Feb 26 comment Can I divide entire equations by each other when trying to solve a system of non-linear equations? Another thing is that $(1) a = b$ and $(2) c = d$ imply (assuming $c\not=0$) $(3) a/c=b/d$ but to have an equivalent system, you need to keep either ($(1)$ and $(3)$) or ($(2)$ and $(3)$). All solutions to $(1)$ and $(2)$ will satisfy $(3)$ but some solutions of $(3)$ may satisfy neither $(1)$ nor $(2)$. Feb 26 comment Can I divide entire equations by each other when trying to solve a system of non-linear equations? If you have $a=b$ and $c=d$ and you know that $c\not=0$, then you have $a/c=b/d$. Here, you just do that where $a$ and $c$ are $F$. So your logic is ok, unless $F=0$. Feb 25 comment Evaluate the limit $\lim_{x\to \infty}\frac{(x+1)^1+(x+2)^2+(x+3)^3+…(x+100)^{100}}{x^{10}+10^{10}}$ In your case, you have exactly $100$ terms in the sum. He had $x$, which causes the problem. Feb 25 comment Evaluate the limit $\lim_{x\to \infty}\frac{(x+1)^1+(x+2)^2+(x+3)^3+…(x+100)^{100}}{x^{10}+10^{10}}$ The property you have about continuity is that if $f$ and $g$ are continuous, then $h:x\mapsto f(x)+g(x)$ is continuous (). By using that a finite number of times, you can know that if $f_1,\dots,f_n$ are continuous, then $h:x\mapsto f_1(x)+\dots+f_n(x)$ is continuous. What happens in the question you link is that you define $g:x\mapsto f_1(x)+\dots+f_x(x)$. So now you have to use () $x$ times to prove the continuity of $x$. But you see it's really weird to have the number of functions depend on the argument and that won't work. Feb 25 comment Evaluate the limit $\lim_{x\to \infty}\frac{(x+1)^1+(x+2)^2+(x+3)^3+…(x+100)^{100}}{x^{10}+10^{10}}$ There is no problem here because the number of terms in the sum does not depend on $x$. $\cfrac{(x+1)+\dots +(x+n)}{x}\underset{x\to\infty}{\longrightarrow}n$ but $\cfrac{(x+1)+\dots +(x+x)}{x}\underset{x\to\infty}{\longrightarrow}+\infty$. In fact, it's much easier to see by simplifying a bit first: $\cfrac{(x+1)+\dots +(x+n)}{x}\underset{x\to\infty}=\cfrac{nx+\frac{n(n+1)}{2}}{x}{\longrightarrow}n‌​$ but $\cfrac{(x+1)+\dots +(x+x)}{x}=\cfrac{x+\frac{x(x+1)}{2}}{x}\underset{x\to\infty}{\longrightarrow}+\‌​infty$ (where I used that $1+\dots + k = \cfrac{k(k+1)}{2}$). Feb 16 comment Sum of two subspaces is equal to the span of their union Unless I'm mistaken, the span is the set of (finite) linear combinaisons, not just of finite sums. The argument remains the same though, you just have to add coefficients in front of the $u_i$s and $w_i$s. And having as many elements in $V_1$ and $V_2$ feels a bit weird, even though it doesn't really matter since we can take $k$ large enough and just add zeros. Feb 16 answered Sum of two subspaces is equal to the span of their union Feb 12 comment Closed subsets of $\mathbb{C}^*$ proper for multiplication Yeah. I edited and added an overline. But I'm starting to think that you can never have $0\in\overline{S_1}$: $S_1$ is closed so if it has a sequence converging to $0$, it contains $0$ but it can't because it's a subset of $\mathbb C^*$. And so $\overline{S_1}$ can't contain $0$... Feb 12 comment Closed subsets of $\mathbb{C}^*$ proper for multiplication Unless I'm mistaken, $(0,\infty)\in \overline{S_1}\times \overline{S_2}$ means exactly $0\in S_1$ and $S_2$ is not bounded. And $S_1\times S_2$ being proper means that for every $r\ge 0$, $S_1\cap \left\{\frac{r}{z}\mid z\in S_2\right\}$ is compact. So your theorem says that if $0\in \overline{S_1\cap \left\{\frac{1}{z}\mid z\in \overline{S_2}\right\}}$, then it's not compact... Whereas it probably should be $\infty\in \overline{S_1\cap \left\{\frac{1}{z}\mid z\in \overline{S_2}\right\}}$. Are you sure it's not $(\infty,0)$ instead of $(0,\infty)$? Nov 22 comment Is $f:\Bbb Q\rightarrow \Bbb R$ continuous? @ManolisLyviakis Yes. The only problems are with neighbourhoods of $\pm\sqrt{2}$ but since your sequence can't get arbitrarily close to $\pm\sqrt{2}$, by taking a small enough neighbourhood of $\pm\sqrt{2}$, you can make it so that all your sequence is outside of it, and so you avoid the "problem". Nov 22 comment Is $f:\Bbb Q\rightarrow \Bbb R$ continuous? The only "problems" are at $\pm\sqrt{2}$ (because everywhere else, you can find a neighbourhood where your function is constant so your function is continuous). Now, let's take a sequence $(u_n)_n$ that converges in $\mathbb{Q}$. To prove that $f$ is continuous, you just need to show that $f(u_n)$ converges (for $u$ arbitrary). But since $u_n$ converges in $\mathbb{Q}$, it can't converge to $\pm\sqrt{2}$ so you're "far enough" from the "problems". Nov 18 comment why is automaton with a queue is more powerful than an automaton with a stack? The problem is the one that prevents ${a^nb^nc^n}$ from being context-free: You can only use stored information (the number of $a$) once (to count the $b$s and so then you can't count $c$s). Whereas with a queue, you can use it and immediately put at back in the queue to not forgetting it, while still being able to read the rest of the information. Nov 12 comment Let X be an infinite set with a topology T, such that every infinite subset of X is closed. Prove that T is the discrete topology. @zermelovac $T$ is the set of open sets. Since $\{x\}$ is finite and $X$ is infinite, $X\setminus \{x\}$ is infinite and is therefore closed. And saying that $X\setminus \{x\}$ is closed is the same thing as saying that $\{x\}$ is open. So $\{x\}\in T$. Oct 4 comment What is the property of the eigenvalues of this matrix? $A^*A=(I-A)A=A(I-A)=AA^*$ so $A$ is normal and you can diagonalize it. Sep 25 comment Order of operations in polynomial with exponent You can also factor a $n^2$ out to further simplify.