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Feb
9
comment Statistics: Where did this function for normal distribution come from?
They are axioms for the purposes of this answer (i.e. they're assumptions that I am making without any justification). There isn't a name for them that I'm aware of.
Feb
6
comment How to find remainder modulo $n$, when $n$ is a large number
Dear editor - not only did you screw up the math in this post by editing it, you also screwed up the formatting. I have reverted your edit.
Oct
4
comment Choice Problem: choose 5 days in a month, consecutive days are forbidden
Thanks, edited :)
Sep
26
comment Consider the problem minimize $f(x)= x^4 −1. $
The proof of quadratic convergence for Newton's method uses the fact that the first derivative of the function whose zero you are finding (i.e. $f''(x)$ in your case) does not vanish on some interval surrounding the root. But you have $f''(x) = 12x^2$ which is zero at the root $x=0$, so the proof of quadratic convergence fails.
Sep
26
comment Solve linear system with matlab
Also, your text specifies $m\geq n$ but your example has $m < n$ (although I don't think this is relevant - probably just a typo).
Sep
26
comment Solve linear system with matlab
You have more equations than unknowns; why do you expect to be able to find an exact solution?
Sep
11
comment Pseudo Proofs that are intuitively reasonable
@MJD Yes, good to mention the Andreas Blass paper too. I'm not sure whether I knew about it or not when I wrote this answer. If I did, then it was surely an oversight.
Aug
7
comment Coordinate descent with constraints
Thanks Michael!
Aug
6
comment Coordinate descent with constraints
Thanks for this answer Michael. Is it easy to see that when dealing with coordinate-wise bounds, thresholding to $l_i\leq x_i\leq u_i$ at each step results in the global minimizer for the problem I posed (convex function with separable non-smooth constraints and coordinate-wise bounds)?
Aug
1
comment What are some conceptualizations that work in mathematics but are not strictly true?
@BCLC a physicist would say $f(r,\theta)=r^2$ whereas a mathematician would say $f(r,\theta)=r^2+\theta^2$.
Jul
28
comment Interview puzzle with a deck of cards, some cards upside-down
@Wonder Yes, it is the same answer (+1) - I saw that another answer appeared just as I was finishing mine, but since I'd already written it I hit 'post' anyway.
Jul
18
comment What numerical methods are known to solve $L_1$ regularized quadratic programming problems?
I'll certainly give it a shot. I've already tried several general-purpose solvers (open source and commercial) which have been okay, but not as fast as a couple of hand-rolled special-purpose solvers. I've not tried CVX yet though...
Jul
18
comment What numerical methods are known to solve $L_1$ regularized quadratic programming problems?
Thanks Michael, this is really helpful - I'm exploring several different options you suggested. I never have to solve a particularly tricky problem (mostly quadratic programming with an $L_1$ term and sometimes linear inequality constraints) but I need to solve a lot of them, so speed is a factor for me. Thanks for your help!
May
21
comment A fair coin is flipped 2k times. What is the probability that it comes up tails more often than it comes up heads?
It seems likely that the "2k" in the question refers to a general even number $2k$ rather than "2000".
May
7
comment Help with 2 questions my professor gave us
For (i) assume there are two solutions $a,b$ s.t. $a^2=r$ and $b^2=r$ with $a\neq b$. Then either $a>b$ or $a<b$. Can you derive a contradiction?
May
6
comment How to draw contour lines for a bivariate Gaussian distribution by hand!
Are you talking about a 2d gaussian distribution?
Mar
5
comment Game Theory/Bayesian approach to a bluffing game
@DanielR Pretty sure (although open to being proved wrong). Player 1's expectation, if he bluffs with probability $q$, is $E = 0.2(2p + (1-p)) + 0.8(q(-p + (1-p)))$ which simplified to $E=0.2(1+p) + 0.8q(1-2p)$. Therefore if $p>0.5$ the second term is negative, so $q=0$ maximizes the expectation. If $p<0.5$ the second term is positive, so $q=1$ maximizes the expectation. This assumes that player 2 never changes their strategy, of course.
Mar
4
comment What is equivalence of $(p \vee q) \wedge \neg (p \vee q)$?
Let $a = p \vee q$. Then you have $a \wedge ¬ a$.
Feb
21
comment How do you describe your mathematical research in layman's terms?
@Arthur Feynman was speaking about theoretical physics, which deals with quarks, gluons, hadrons, bosons and other things that most people aren't familiar with. His book QED succesfully imparted the flavour of quantum electrodynamics and the path integral formalism to me as a sixteen year old. I don't think we should shrug off our responsibility as mathematicians so easily.
Feb
18
comment Standard Deviation Annualized
@kookster Yes, still true (note that with only 30 data points the confidence intervals on your measured standard dev will be quite wide).