"The law that entropy always increases - the Second Law of Thermodynamics - holds, I think, the supreme position among the laws of physics. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations - then so much the worse for Maxwell's equations. If it is found to be contradicted by observation - well, these experimentalists do bungle things from time to time. But if your theory is found to be against the Second Law of Thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation." — Sir Arthur Eddington
Lifted from Wikipedia, and really too long for my quotes page, so here it is.
To see cancellation issues really go off the deep end, review the 1999 paper by Cuyt, Verdonk, Becuwe, and Kuterna.
However, don't review that paper if you aren't willing to go off the deep end yourself. Gahhhh.
I thought I blogged this long ago, but then again I thought I learned this long ago, and rediscovered this morning that I needed to recreate it: how to solve A x^2 + B x + C = 0.
We use a trick called "completing the square" to solve the equation above for X
A x^2 + B x + C = 0
x^2 + B/A x + C/A = 0
x^2 + 2B/2A x = - C/A
x^2 + 2 (B/2A) x + (B/2A)^2 = (B/2A)^2 - C/A
(x + B/2A)^2 = ( B^2 - 4AC ) / 4A^2
x + B/2A = ± ( B^2 - 4AC ) ^ .5 / 2A
x = - B ± ( B^2 - 4AC ) ^.5 /2A
Multiplying both sides by:
1 = [ -B ∓ (B^2 - 4AC) ^.5 ] / [ -B ∓ (B^2 - 4AC) ^.5]
x = 2C / [ -B ∓ (B^2 - 4AC) ^.5 ]
So we actually have two ways of writing x. Why would we need more than one? In a world of exact numbers, we can get by without. But when there are limits to the precision of our numbers, choosing the wrong operation can be catastrophic.
Consider a problem of the form:
x^2 - b x + 1 = 0 (note the minus sign).
The exact solution is
x = [ b ± Q ] / 2.
Q = sqrt( b^2 - 4 )
What happens when we use subtraction here? Let's try an example, using 16 digits of precision:
b = 10000.00000000000
b^2 = 100000000.0000000
b^2 - 4 = 99999996.00000000
Q = 9999.999799999998b+Q = 19999.99979999999
b-Q = 00000.00020000001
Contrast the last two operations. When we added, we lost a little bit of information off the far right side, but we still have 16 useful digits. When we subtracted, the first 8 digits cancelled; and that precision is lost.
What effect does this loss of precision have on our solution?
2/(b+Q) = 0.000100000001
(b-Q)/2 = 0.000100000005
And when we turn around and plug these values back into the original equation, to get a sense for how big the error is
e( 2/(b+Q) ) = +0.000000000000000200000001
e( (b-Q)/2 ) = -0.000000039999998999999975
The precision being lost is essentially that to which b and Q agree. We can use a Taylor expansion to approximate Q
Q = b - 2/b + O(b^2)
So when b is of order 10^n, b-Q is of order 10^-n; losing 2n digits of precision. The wrong approach gives you x = 10^-7, which has an error 10^-14. The right approach gives you x = 10^-7 + 10^-21, and the error is of order 10^-28.
Oh baby, I wanna read wit-chya,
cause your Bible's got pick-chyas....
I don't think I'd mind going to hell nearly so much if the lines to get in weren't so damned long.
From Wildernesse
My definition of well-educated for laypersons is whether you can explain a frequency-dependent selection model, a phylogenetic tree, and why humans are taxonomically classified as an ape. If you can’t do those things, what makes you think you know enough? I am arbitrary and I love it.
Wikipedia on frequency-dependent selection. Huang, Singh, Kojima, 1971 [pdf].
Wikipedia: phylogenetic trees, including examples.
"By your leave, Mother, the trilogy is not yet done."
Adicts will already have a copy, and those who prefer stories with an ending should hold off for at least one more book. But if you are thinking about holding off because the last few books were disappointing, there is no need for such concerns - this one is pretty good.
Events of note:
The interminable "Faile is lost" arc is finished [finally!]
The interminable "Elayne has a bath" arc is finished [finally!]
The courtship of Mat and Tuon is complete, and well done (just a little bit loose at the end).
The battle of "Egwene vs. Elaida" takes off in a startling direction; and Jordan writes it very well.
Robert Jordon wrote at his blog
For those who have read the book and believe you have identified the "gasp" moment, congratulations. For those who have read the book and still don't know what the "gasp" moment is, my sympathies. I mean that in all truth. You failed to see something that really should have made you gasp. I think I am fairly hardened, but occasionally something happens that makes me mutter, "Where are you, God? Are you sleeping? Are you blind?" This is fiction, but even so, I had to pause a couple of times in writing about it. Of course, I get deeply immersed in my work so that it becomes real to me while I am writing, but I hope to pull the reader into that level of realness, too. Either I failed completely in this instance, or some of you have become way too hardened.
I think I recognized his gasp moment. If so, he failed completely. It's possible that another thing was his gasp moment. That one didn't work either. There's a very weak third possibility for a gasp moment, if that's the one, he's deluding himself.
UPDATE: reportedly from the Boston book signing
Q: [Why write the gasp moment?]A: Because Hell on Earth is about to break loose! Haven't you people
got that in your heads yet? The Shadow is winning, it's the fifteenth
round, our boy is on the ropes and the Shadow feels great and has
tremendous punching power! Three minutes to go and there's only one
chance. The Gasp Moment is their reaction to this.
So it's the first one. And he missed, big time.
Aram turns out to be a complete waste of ink.
Verin is even more Verin than ever. Whatever that means - we get no closer to finding out.
Jordan has been telling us for years that the books won't return to the Seanchan home land. Now we know why.
Mat makes a brilliant realization about the 'finn. There are still some lose bits there, but the core idea was excellent.
It's a damn shame that Min foresaw the healthy birth of Elayne's children, because she desperately needs a drastic attitude adjustment, and losing the kids might barely suffice.
I'm more impressed with the Dark One than ever. That he even got close to destroying the world with the Chosen as his instruments is most impressive. This lot couldn't organize a car wash if you spotted them varsity cheerleading, two Hooters franchises, and the statutory version of Traci Lords.
Which speaks volumes about the competance of Lews Therin and the knights who said Ni!
Question: why haven't we heard anything about Moiraine since she battled Lanfear? Answer: because the author said so.
The Archimedes Death Ray. Pirates should avoid MIT on sunny days.
OK, if we really understand what is going on at this point, we ought to be able to get a rough cut at a system with three down spins.
So following the approach outlined by Feynman in the lecture, we scribble out the Hamiltonian. This is going to have six terms generated by the spin exchange; one for each coordinate l,m,n.
Now we assume, as before, that we can write the solution as the product of a time component, expressing the energy of the system, and a position component. That is, we expect to find solutions of the form
exp[iEt]a(l,m,n)
Following the earlier insight, we consider the three spins to be far enough away that they could be treated independently of each other:
a(l,m,n) = exp[ib lx]exp[ib my]exp[ib nz]
where x,y,z are wave numbers, as before. Operate on this with the Hamiltonian, and we get
E = 6A - 2A cos (bx) - 2A cos(by) - 2A cos(bz)
Now, there's no particular reason that x and l be related; the following expression also works, and gives the same expression for the energy:
a'(l,m,n) = exp[ib ly]exp[ib mx]exp[ib nz]
In fact, there are six expressions that look like this, which are each satisfactory, providing the same energy. So what we really want is some linear combination of these
a(l,m,n)
= exp[ ib(lx+my+nz) ] + B exp[ ib(lx+mz+ny) ]
+ C exp[ ib(ly+mz+nx) ] + D exp[ ib(ly+mx+nz) ]
+ F exp[ ib(lz+mx+ny) ] + G exp[ ib(lz+my+nx) ]
Now you can work these through the Hamiltonian, for each of the cases where l = m+1 etc. Or alternatively you can recognize that what we really need is
a(l,m,n) = 0 whenever l=m, m=n, or l=n.
In other words, we need the solution to be anti-symmetric when we exchange two coordinates. This is pretty easy:
a(l,m,n)
= exp[ ib(lx+my+nz) ] - exp[ ib(lx+mz+ny) ]
+ exp[ ib(ly+mz+nx) ] - exp[ ib(ly+mx+nz) ]
+ exp[ ib(lz+mx+ny) ] - exp[ ib(lz+my+nx) ]
I may spoil something, so off behind the break....
Short summary: it's not Firefly, so much as "based on the characters created by...."
Did anybody else think "midichloridians" when the reavers were explained?
I think Mal Reynolds got Hollywooden, and basically screwed up as a character. The central elements of Mal in the television series were the loyalty he inspired in his subbordinates, and the precautions featured in most of his planning (exhibits: the pilot, the Train Job, Trash). Instead, the movie version has inherited a bunch of brawler traits that really belonged to Jayne Cobb, with the charisma of a schoolyard bully.
Similarly, I think River as insane unstable witch worked much better than River as insane unstable assassin. Boy, that safeword would have come in handy several times during the series, though.
And the resolution of this story really felt like it ripped the rug out from under the foundation of the mystery of the series.
I'm hoping the movie does well enough commercially that it begets sequels (by preference, rather than prequels) so I can discover where Joss is going next. But if he does, I expect him to use these characters, rather than the ones from the television program that I had learned to like.