Not really, no. I loved the series when I was younger, but there isn't much that I loved about it still alive today. I may give it a try later, but the franchise as it stands today is a lot like seeing your childhood hero laying drunk and unwashed in the gutter of some red light district.
But seriously, this is a completely different mode of play; I've been waiting for a game like this since Super Mario RPG (though I would also be stoked if a proper sequel for this game came out and featured Sonic in some way).
At least Opera is free nowadays. Some years ago it was not free, which was the greatest obstacle for many.
I remember when I first heard about it, back then it had become free but the free version was still ad-supported so I didn't think much of it.
Then one day after I had gone a bit too hog-wild with adding extensions to Firefox (I think this was shortly after FF2 came out), I heard that Opera had become both free and ad-free, and then I tried it out for a while and the only reasons I ever considered going back were that it is not as customizable (but it turns out to have a great deal of features available natively and with minimal performance impact) and that it is not open-source.
On the other hand, a few are amazing, like Tiny Menu, Stop-or-Reload Button, and Download Statusbar, and I also enjoy Mr. Tech's suite of addons (along with the standard-issue security extensions like Adblock Plus and NoScript).
Then again, it's been a long time since I've used Firefox or any of its derivatives on a regular basis; I briefly switched back when FF3 came out just to try it out but it didn't wow me away from Opera.
This reminds me, I'm surprised that Flock still hasn't migrated to being based on FF3 and saddened that Netscape stopped developing its excellent browser (which in its later years was based on and an improvement over Firefox) just months before FF3 was released; Flock contains the latest fixes in the FF2 line (2.0.0.16), but Netscape 9 is still based on Firefox 2.0.0.12!
It's best if you can get wireless router capability right in your modem, but those solutions are also more expensive than cheap router+cheap wireless modem.
^^That reminds me of when I frequently become unable to use my DHCP-allocated IP address and have to input one manually, and then I can't Remote Desktop into my desktop from my laptop.
I also have a crappy old D-Link router.
PLEASE post WIPs! ...Particularly of the don't-have-to-download-emulator-and-correct-rom variety.
Oh, and, uh...cue "Hallelujah Chorus".
What you're thinking about is called an "encode"; this is thanklessly made by one of the regulars using the "emulator-and-correct-rom" along with the WIP emulator input file and something like x264.
This isn't SDA here: There they look for videos, here we look for input files, and encoding into video is a mere courtesy.
I have a hardware firewall, and do not bother with anti-virus at all on my laptop. My desktop still uses AVG, but mainly because it's already there and keeps up with updates, and because my desktop is DMZ.
Unfortunately, the only software firewall that impressed me (Sygate) was discontinued. It was the only software firewall I ever used that would halt the worm running around on Bellsouth a few years ago. Damnedest thing; on a fresh installation of XP, if I plugged the network cable in before installing Sygate, within ten minutes my CPU usage would max out and never go down. Ever. Nothing in the task manager using resources, just a rapidly overheating processor running under full load for no apparent reason. Only way to stop it would be to format, reinstall the OS, and install Sygate before getting on the network. Never did find out what it was, other than it only occurred on Bellsouth, and if I wasn't firewalled. Didn't even have to open a browser window, just being connected spelled certain doom. Oh well, no such troubles now.
Since that "few years ago" XP Service Pack 2 was released, and now you won't get swarmed with malware just by connecting to the Internet.
Exact form requires solving the transcendental equation 2^x=2+1/x, which I don't feel like doing, and suspect isn't possible anyways.
You forgot the word "algebraically" before "possible".
Well, yes. The fact that I solved it numerically immediately before saying that meant that that should have been implied.
Anyways, nice jobs, even remembered the branch cuts. Not sure about analyticity, but it's something to look into. In the meantime, here's another question.
Does |z^z| go to infinity as |z| goes to infinity in all directions (like x^a), or does it go to a finite or zero value for some direction (like a^x)? Shouldn't be too hard now that we have closed form expressions, but still an interesting question.
From my polar-coordinate expression for z^z, |z^z|=|z|^ℜ(z)/e^(ℑ(z)arg(z)).
Let z<0 and z→-∞, then z^z=(-z)^z*e^iπz and |z^z|=|z|^z=e^(z*ln|z|); in fact, z^z for z<0 spirals around the origin at a constant angular rate, and |z|^z at first glance looks like a very smooth function on ℝ.
Now, as in the proof that lim(z→0)z^z=1, use L'Hôpital's Rule to determine that lim(z→-∞)ln(-z)/(1/z)=lim(z→-∞)(-1/z)/(-1/z²)=lim(z→-∞)z=-∞, so lim(z→-∞)z^z=0.
Of course, it's obvious that z^z→+∞ as 0<z→∞ (an ever-larger base raised to an ever-larger exponent, or more analytically, the derivative is strictly positive for z>1/e), but now let's see what happens if ℜ(z)=0 and y=ℑ(z)≠0; then |z^z|=e^-(π|y|/2) on the principal branches, but in general |z^z|=e^-((2n+1/2)π|y|) where n∈ℤ. I should mention that this view of z^z is sufficient to show that it is not analytic at 0 and that if it has a limit at 0 its value must be 1; also it is obvious here that if z is pure imaginary and |z|→∞ then z^z→∞ for those branches with n<0 and z^z→0 otherwise. In either case z^z spirals ever faster as it becomes infinite or approaches 0, because arg(z^z)=y*ln|y|.
(This is an abuse of notation, because the branches with y<0 and y>0 for a given n are the same only when n=0; generally the branch with y>0 for a given n is also the branch with y<0 for -n.)
What if, more generally, Arg(z)=k, a constant? Then if r=|z|, |z^z|=r^(r*cos(k))*e^-((k+2nπ)r*sin(k)) and arg(z^z)=r*ln(r)sin(k)+(k+2nπ)r*cos(k) where n∈ℤ. The behavior of z^z as r→∞ then depends on ln|z^z|=r*ln(r)cos(k)-(k+2nπ)r*sin(k); the derivative is ln(r)cos(k)+cos(k)-(k+2nπ)sin(k), and in the previous special cases, k=π and this derivative is -1-ln(r)<0 for r>1/e, k=0 and this derivative is 1+ln(r)>0 for r>1/e, and k=±π/2 and the derivative is 2nπ+π/2<0 for n<0 and positive otherwise. In general, that derivative is positive for ln(r)>(k+2nπ)tan(k)-1 if |k|<π/2 and negative for such r if |k|>π/2, and as long as k≠±π/2 there will be a sufficiently large r for each n beyond which this derivative is always positive or always negative, respectively, so ln|z^z| approaches +∞ in the former case and -∞ in the latter.
Therefore, if Arg(z) is constant and |z|→∞, then z^z→0 if |Arg(z)|>π/2 (negative imaginary part), z^z→∞ if |Arg(z)|<π/2 (positive imaginary part), and either case is possible if Arg(z)=±π/2 (pure imaginary), depending on the branch chosen.
Of course, unless z>0, z^z spirals out to infinity or inward to 0; the derivative of arg(z^z) is ln(r)sin(k)+sin(k)+(k+2nπ)cos(k) where n∈ℤ. Once again, in the special cases I mentioned at the beginning, k=π and z^z spirals toward 0 at a constant angular rate dependent only on the branch, k=0 and z^z stays on the positive real axis, and k=±π/2 and z^z spirals faster and faster toward infinity or 0 depending on the branch, but the angular rate does not depend on the branch chosen.
Therefore, except for the special cases in which k=0 or k=π, the angular rate is either always increasing or always decreasing for sufficiently large r on any branch, so for imaginary z, z^z spirals faster and faster toward the origin or infinity as z→+∞ with constant argument.
Somehow I suspect that there exists a spiraling path toward infinity in the complex plane along which z^z approaches a finite nonzero value, and furthermore that such a path exists for any c∈ℂ; even more exciting would be to extend that to the extended complex plane, finding a path toward infinity along which z^z approaches any directed infinity, not just +∞.
As long as we're having fun with x^x, lets shed our dependence on positive real numbers. For complex z=x+i*y, find a closed form expression for z^z that has no complex exponents except of the form e^(i*a), where a is real.
z^z=e^((x+i*y)ln(x+i*y)).
ln(x+i*y)=ln(x²+y²)/2+i*tan⁻¹(y/x)+2nπi where n∈ℤ.
(x+i*y)ln(x+i*y)=x*ln(x²+y²)/2-y*tan⁻¹(y/x)-2nπy+i*(y*ln(x^2+y^2)/2+x*tan⁻¹(y/x)+2nπx).
Therefore, z^z=|z|^ℜ(z)*e^-(ℑ(z)arg(z))*e^(i*(ℑ(z)ln|z|+ℜ(z)arg(z))).
By the way, |z|=√(x²+y²), ℜ(z)=x, ℑ(z)=y, and arg(z)=tan⁻¹(y/x)+2nπ where n∈ℤ, with the principal value (n=0) as Arg(z)=tan⁻¹(y/x); it would be interesting to see where z^z is analytic, if anywhere.
On another note, Unicode is the greatest invention since sliced bread.
Use an encoder like x264 or mencoder to better encode information within YouTube's 300kbps limit (which is a serious hamper on video quality IMO).
Oh yeah, by "limit" I mean "If your video has a higher bitrate than that then YouTube will automatically downgrade it to 300kbps, though it may also give an option to view in high quality."
Okay, attempt number 2!
lim(x->0)x^x = lim(x->0)(1+(x-1))^x = (use the taylorserie for (1+y)^a with y=x-1 and a=x) =
1 + a1*y + ... + ai*y^ + ...
where ai = x*(x-1)*(x-2)*...*(x-i)/i which closes in on 0 when x closes in on 0 for all i. So all ai are close to 0.
y^i are all numbers that are close to -1 for odd i and close to 1 for even i (because y=x-1 closes in on -1 when x->0). Either way, they are finite and pose no problem when multiplied with a number very close to 0.
So! The sum looks something like this now:
1+ 0*-1 + 0*1 + ... + 0*(+-1) + ... = 1
so the sum, and therefor the expression lim(x->0)x^x, approaches 1.
EDIT: If you don't like it that a is not fix, which it usually is in the taylor serie, just divide the problem into lim(a->0)lim(y->-1)(1+y)^a, making a fix while you compute (1+y)^a and then making a approach 0 afterwards.
That expansion that you used only works if a is constant; however, you can try something like this near r>0 if f(x)=x^x:
Sum(d(f(x),x,r,n)(x-r)^n/n!,n,0,infinity)=
=r^r+r^r*(1+ln(r))(x-r)+r^r*((1+ln(r))^2+1/r)(x-r)^2/2+...
This is a bit simpler if r=1:
Sum(d(f(x),x,1,n)(x-1)^n/n!,n,0,infinity)=
1+(x-1)+(x-1)^2+(x-1)^3/2+(x-1)^4/3+(x-1)^5/12+(x-1)^6*3/40+...
Then just determine the pattern of a_n, the sum of the coefficients of terms in the nth derivative of f with no logarithmic factor, and use the ratio test to determine the radius of convergence: |a_n/a_(n-1)*(x-1)/n|<1 iff |x-1|<n>infinity)n*a_(n-1)/a_n.
Somehow I suspect that the radius of convergence for this series is 1 (because the first derivative, x^x*(1+ln(x)), approaches -infinity as x->0), and for the more general Taylor series it is r, but it is a challenge to determine this.
lim(x->0+)x^x=e^(lim(x->0+)x*ln(x))=e^(lim(x->0+)ln(x)/(1/x))=e^lim(x->0+)(1/x)/(-1/x^2))=e^(lim(x->0+)-x^2)=e^0=1 by L'Hôpital's Rule; in fact, it can be proven that if f and g are meromorphic functions of a complex variable z and f(c)=g(c)=0≠f(b) for some b and c in their common domain, then lim(z->c)f(z)^g(z)=1.
I used to have the gamepad in the OP (well I still do), but when I wanted to have enough buttons for N64 gaming I went with this one from Saitek: http://www.saitek.com/uk/prod/p990.htm
More observations about NYC in general:
-- Women have big tits. This may sound weird coming from me, but really, compared to Europeans, women in NYC have bigger tits. Though in all fairness, they have larger asses too. Nevertheless, even though I'm not a person to stare at women's breasts (in fact, I'm less interested of that particular section in woman anatomy than an average male is), I can't help but be unsettled when I notice a young (black) woman whose chest around the breasts is literally twice as big as it's 15 cm down from that section. I wonder how she manages her life. She must be stared at a lot.
This thread is worthless without pics.
Seriously though, I think it's the high rate of obesity; did you notice that the women with big tits also tended to be fat?
arf...lechery?