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Apr
25
comment Prove by mathematical induction: $n! < n^n$ for $n\geq2$
Hint: $(n+1)^{(n+1)} = (n+1)(n+1)^n \geq (n+1)n^n$.
Apr
24
comment Bounds on $L^2$ and $L^{\infty}$ norms in terms of $H^1$-seminorms for functions attaining a zero in a domain.
There are similar Poincare inequalities in higher dimensions that apply to functions that have trace $0$: en.wikipedia.org/wiki/Poincar%C3%A9_inequality.
Apr
12
comment Poincaré-like inequality
Should $\Omega$ be connected as well? What if $\Omega$ is disjoint two balls and $\Gamma^1$ and $\Gamma^2$ are the boundaries of the separate balls. With effectively no conditions on $\Gamma^2$, I don't see what stops $\xi = 0$ on the first ball and $\xi = C$ on the second ball.
Apr
11
comment Counting linear maps over finite fields
Any linear transformation is uniquely determined by its action on a basis. So the answer is $\# W^{\dim V}$.
Apr
11
comment Difference between $\sum_i\frac{a_i}{b_i}$ and $\frac{\sum_i a_i}{\sum_i b_i}$
What happens to the second formula if you replace $a_i/b_i$ by $2a_i/2b_i$ for some fixed $i$? You would like there to be no change if you are only interested in the value of $a_i/b_i$. The second expression is not a well-defined function of the ratios $a_i/b_i$.
Apr
11
comment Prove that $|k(x)|\le C|x|^{-n}$ under suitable hypothesis on $k\in\mathcal{C}^1(\Bbb R^n\setminus\{0\})$
It occurs to me that the mechanism here is a Poincare inequality: mean value $0$ allows the derivative to bound the $\sup$ norm.
Apr
11
comment Is the Cauchy principal value “invariant” under change of variables?
@uranix It might be enough for $dG(v)$ to be a multiple of the identity w.r.t. some orthogonal basis, since $dG(v)$ should control the sets $\{ \lvert G(v) - G(w) \rvert < \varepsilon \}$ by Taylor's theorem. I haven't worked out whether first-order control is enough though.
Apr
11
comment Is the Cauchy principal value “invariant” under change of variables?
Your $G$ is not a diffeomorphism. The derivative is not bijective at $0$. It seems to me that there should be no counter-example in one dimension, since the shape of the sets $\{ G(v) - G(w) > \varepsilon \}$ is determined locally by the differential of $G$, which is symmetric in $1$ dimension.
Apr
11
comment Is the Cauchy principal value “invariant” under change of variables?
@uranix If $\{ \lvert G(v) - G(w)\rvert < \varepsilon \}$ is a ball for each fixed $w$, $\varepsilon$ small enough, it should work.
Apr
9
comment Strong convergence on the unit sphere of $l_2$
@Tryss Yes. WLOG they can be taken to be $e_j = (0,0,\ldots,0,1,0,0,\ldots)$ with the $1$ in the $j$th place.
Apr
7
comment Let $X,Y$ be rvs with $E[X] = a, E[Y] = b$. How can I tell if they are independent?
If you were to prove that all such random variables are independent, e.g., then you would not be able to find specific $X$ and $Y$ with $E[X] = a$ and $E[Y] = b$ that are not independent.
Apr
7
comment Let $X,Y$ be rvs with $E[X] = a, E[Y] = b$. How can I tell if they are independent?
To prove you can't tell given information: come up with two independent events and two non-independent events that both satisfy the hypotheses.
Mar
8
comment From a pack of 52 playing cards, three are drawn at random. Find the probability of drawing a king, a queen and jack.
The book has the wrong answer (?). Yours seems fine to me.
Mar
4
comment Is there a dictionary for math notation?
Perhaps useful (pdf), Handbook for Spoken Mathematics
Mar
2
comment Curls and Gradients
A vector field with nonzero curl is not conservative, while the gradient of a scalar field is always conservative. You can prove curl(grad f) = 0 as an identity from the definitions. I am looking for a good explanation online. This link seems good (pdf).
Mar
2
comment Showing two topologies are equal
Possible strategy: two topologies are equal if given an basis element $U$ of one topology, you can find a basis element $V$ of the other topology contained in $U$.
Mar
2
comment Curls and Gradients
The curl of $F$ is $0$ if $F$ is the gradient of a twice differentiable function.
Mar
1
comment Understanding bounded linear operators
It might also be instructive to see what a not-bounded operator looks like. For example, the derivative operator $C^1(\mathbb{R}) \to C(\mathbb{R})$, where $C^1(\mathbb{R})$ is equipped with the $\sup$ norm inherited from $C(\mathbb{R})$. You can make the operator bounded by equipping $C^1(\mathbb{R})$ with the norm $u \mapsto \| u \|_\infty + \| u' \|_\infty$.
Mar
1
comment How can there be an inner product space when inner product yields a scalar?
The inner product on $\mathbb{R}$, for example, is just multiplication $(a,b) \mapsto a \cdot b$. The elements of $\mathbb{R}$ are "vectors" in that they are elements of a vector space.
Feb
29
comment The convex combination of two unequal numbers is strictly between them
When you multiply by $x-y$, you are multiplying by a negative element, so you have to reverse inequality signs. The general relation $s + r < s$ when $r < 0$ holds, for any $s,r \in F$, so see if you can apply this to the expression $y + a(x-y)$ to see that it is less than $y$.