Recent content by suyver

yes, I thought about that one too. But I got confused when I did a bunch of numerical simulations. Sometimes, the sum got very large (>1000). So I worry about the case where the sum may actually diverge: theoretically this is clearly possible (e.g.: sum of 1/n). Do we need to prove that this is...

I choose a random number p_1 \in [0,1) and a subsequent series of (increasingly smaller) random numbers p_i \in [0, p_{i-1}). Then I can calculate the sum \sum_{i=1}^\infty p_i. Naturally, this sum is dependent on the random numbers chosen, so its particular result is not very insightful...

OK, that makes sense. I had originally expected that both roots would converge to -c/b, but now that I think about it, it seems clear that the fact that b can be negative (so that I cannot just say \sqrt{b^2}=b) will spoil this. Oh well... too bad (for what I wanted).

But for the other root, I get: \frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-b-\sqrt{b^2-4ac}}{2a}\cdot \frac{-b+\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}= \frac{b^2-(b^2-4ac)}{2a(-b+\sqrt{b^2-4ac})}=\frac{2c}{-b+\sqrt{b^2-4ac}} which seems ok... but I don't think it is! After all, I multiply by...

I was wondering something that is so simple that it baffled me... When I have the equation a x^2+b x+c=0 this obviously has the solutions x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} And when I have the equation b x+c=0 this has the solution x=\frac{-c}{b} My problem now is the limiting case...

Good point! That must be why you are not allowed to choose a zero in the link that I gave.

OK, I think I completed the answer to my own question (great forum, this is! ;-) This trick is that the website divides the number that I imput by 9 to see what remainder is left. If it's 0, the missing digit had to be 9. If it's 1, the missing digit had to be 8. If it's 2, the missing...

OK, I got slightly further: My number Z=X-Y (X>Y) must be devidable by 9, right? Is it then just the case that because the website knows that the number I imput plus the unknown digit must be dividable by 9, the value of the unknown digit can be determined? I.E.: Because it knows that the...

Another "play with numbers" website http://digicc.com/fido/ It's probably related to the fact that we have a base-10 number system, but I fail to see how it works... This website does the following: 1) You write down a 3 or 4 digit number, say X=x1 x2 x3 x4 (where the x's denote the...

But I like LaTeX'ing! :)

No, this is not correct. You are saying that \pi + (1-2 i)\pi^2=-1 which would imply that \pi=\left(\mp\frac{1}{10}\pm\frac{i}{5}\right)\left(\mp 1+\sqrt{8i-3}\right) which is clearly not the case...

Even though I am born and raised Dutch, I find this insulting. Shame on you.

No, also 1/2. Negative spins are not possible: only the projection of the spin on a (z) axis can be <0.

They are a particle: (almost) no mass, no charge, spin 1/2. No electromagnetic or strong interaction, only weak interaction. Typical strength of the interaction: you'd need about 1000 lightyears of lead to have a 50% probability for an interaction between a neutrino and the lead! On Earth...

Tachyons do not exist.