16,687 reputation
12133
bio website math.ntnu.no/~hanche
location Trondheim, Norway
age
visits member for 2 years, 3 months
seen 6 hours ago

6h
comment Re-interpreting double integral as a Type II Region $\mathrm{d}y\,\mathrm{d}x$ vs $\mathrm{d}x\,\mathrm{d}y$
Have you tried drawing a picture?
1d
comment Proving that $\mathrm{rank}(P_1+P_2) = \mathrm{rank}(P_1)+\mathrm{rank}(P_2)$
But if $P_1+P_2$ is also a projection, that's another story.
1d
comment Proving that $\mathrm{rank}(P_1+P_2) = \mathrm{rank}(P_1)+\mathrm{rank}(P_2)$
You can't, because it is not true. For a counterexample, take $P_1=P_2$.
2d
comment Is a probability density function necessarily a $L^2$ function?
For the mean squared error to be defined, don't you rather need to have $\int_{\mathbb{R}}x^2 f(x)\,dx<\infty$? In any case, the answer to the question as asked is no. For a counterexample, assume $f(x)$ is proportional to $x^{-1/2}$ for small positive $x$.
Apr
15
comment left Haar measure and right Haar measure on ax+b group
Hint: Compute the Jacobian at the identity element of left and right multiplication by a fixed element given by $(a,b)$.
Apr
12
comment Finding the sixth roots of $-8i$
@user136088 Yes.
Apr
12
comment Finding the sixth roots of $-8i$
@NasuSama Why is that necessary?
Apr
12
comment Finding the sixth roots of $-8i$
@Léo Because that is cumbersome when $\theta$ is a complicated expression. Though you could use $\exp(i\theta)$ instead.
Apr
12
comment Finding the sixth roots of $-8i$
Actually, $\operatorname{cis}\theta=e^{i\theta}=\cos\theta+i\sin\theta$.
Apr
12
revised Finding the sixth roots of $-8i$
Better LaTeX
Apr
11
comment For any prime $p>3$ show that 3 divides $2p^2+1$
I think this is silly. The two (eerrr, three) answers given so far, which do use modular arithmetic, are trivially transformed into proofs that do not. I.e., put $p=3k+1$ or $p=3k+2$ and see where this gets you.
Apr
9
comment Show that $\sum_n \frac{1}{a_n}\lt90$
@Lucian I should have guessed. Voted to close.
Apr
9
comment Show that $\sum_n \frac{1}{a_n}\lt90$
Oh, and one more thing: You are not really removing a small fraction of the natural numbers. You are keeping $(9/10)^n$ of all $n$-digit numbers, which is a vanishingly small fraction when $n$ is large.
Apr
9
comment Show that $\sum_n \frac{1}{a_n}\lt90$
Here is an observation that I strongly suspect will lead to a solution (but I haven't got the time to actually try it): There are nine times as many $n+1$ digit numbers with no zeroes as there are $n$ digt numbers with no zeroes. You get the former by taking each of the latter, call it $k$, and forming $10k+1$, … ,$10k+9$. Now use the fact that each of the latter is $>10k$. I smell the sum $\sum(9/10)^n$ in there …
Mar
30
comment Understanding the proof of completeness of $L^1$.
The answer to your last question is: The integral of $g$ is finite by the monotone convergence theorem, and in particular, the pointwise limit is finite a.e. But it may well be infinite at some points. I added some detail to my answer.
Mar
30
revised Understanding the proof of completeness of $L^1$.
More detail.
Mar
29
answered Understanding the proof of completeness of $L^1$.
Mar
24
comment Which probability distribution is this?
$F$ would be the cumulative distribution function of $y$, and $f$ the corresponding density (i.e., $f=F'$).
Mar
6
answered Convergence of Integrands and Integrals
Feb
18
comment Sum of four squares not a prime
The only thing I can think of is to add $0=2ab-2cd$, so that you have the sum of two squares: $(a+b)^2+(c-d)^2$. Not sure where to go from there, however. But perhaps it gives you an idea.