I took all of my notes throughout university (including engineering courses) using OpenOffice.org. The equation editor in OpenOffice is easy-to-learn, fast (as in, no mouse use required and the keystrokes are all sane), and the completed equations look great. (By default, there isn't a keyboard shortcut for inserting a new equation, so you'll need to manually assign one—I used Ctrl-Shift-F, if I remember correctly.

Your example would almost work as is; it would be entered as:

f_x (x) = int from -infinity to infinity f (x, y) dy

Or, if you prefer your parentheses to stretch (in case you have fractions inside, or what have you):

f_x left ( x right ) = int from -infinity to infinity f left ( x, y right ) dy

Either way, it comes out looking very nice. The one thing that takes some getting used to is that you need to make liberal use of whitespace (e.g. between f and the opening parenthesis of the function), otherwise things will occasionally come out looking a little strange. The best part is, when you don't know what you need to type for a particular symbol, you can select it from the menu and OO will insert the plaintext code, which makes it very easy to learn the code for new items.

From the linked article: "The books published by The Espresso Machine will have a recommended sales price of $8 per copy, although the final decision will be left to each retailer."

Possibly, but this at least eliminates the need for shipping the printed books (which weigh a LOT) all over the world. Plus, for me at least, I can only spend maybe five minutes at a time reading from a computer screenâ"and I'm even a Gen Y computer programmer. There's something about reading from a computer screen that is just a lot harder than reading from a paper.

Okay, you're right in one sense, because I wasn't completely accurate in my previous comment (I wrote it hurriedly during the last few minutes of my lunch break at work)--the probability does in fact increase somewhat with the sample size. However, your description doesn't describe the situation accurately at all; the correct analogy would be that, each of 100 stars has a 1 in 100 chance of having a planet spiraling into it; in this case, viewing all 100 stars does not yield a probability of 1 of seeing a planet spiraling into it. Let me try another simplified description. First, consider a single, six-sided die. One of the six sides has a one on it; if you examine a single side of the die (roll it), you have a 1 in 6 chance of seeing the one; if you examine three of the six sides, you have a 1 in 2 chance of seeing the one; if you examine all six sides, you have a 1 in 1 chance (probability of 1) of seeing the one. This is analogous to your "1 out of 100 stars" situation above, but it is not analogous to real life. Now consider six normal, six-sided dice. If you roll all six of them, what is the probability that at least one of them will come up with a one? You will probably immediately realize the probability is not 1, but calculating it is a bit of a math problem--it's been a while since my college statistics class, but if I remember correctly, the correct way of finding it is to calculate the probability that none of the dice will be a one, and then subtract that from 1, thus:

• For each die, the probability of it not being a one is 5/6;
• Thus, the probability of none of the six dice being a one is (5/6)^6, or about 0.335;
• Thus, the probability of at least one of the six dice being a one is 1 - 0.335, or about 0.665, which is significantly less than 1.

Going back to your 1 in 100 probability, if there are 100 stars and each has a 1 in 100 chance of having a planet spiraling into it, then the probability of any of the 100 stars having a planet spiraling into it is 0.634. Examining only 50 of the 100 leaves a probability of 0.5 * 0.634 or only 0.317. Now, we're making some huge assumptions about the probabilities of this event occurring; but just for the sake of discussion, let's just say that, for any given star, there is a 1 in 10^15 (one in a trillion) chance that, at the present time, it has a planet spiraling into it. (Given the relatively small number of stars we know of that have any planets at all, I suspect that number is a significant overestimate, but I'll use it.) Using your estimate of 100 billion stars in the universe, that makes the probability that any star exists, anywhere in the universe with a planet spiraling into it about 0.0000999, or 1 in 10,000, which is pretty small. Now I'll assume that we have examined 1 billion of those stars closely enough that we would be able to detect this occurrence (which I would guess is a gross overestimate); that makes the probability of any star that we have examined being in the midst of the occurrence about 0.000000999, or 1 in a million. So, being what I would say is quite generous with all of the numbers, we have a 1 in a million chance of seeing this.

No: This is a common misunderstanding of probability. The probability of a particular event occurring does not increase with the number of non-occurrences (well, if the event is random, which in this case it is). Think of it on a smaller scale: Flipping a coin. Let's say I flip a coin and get heads five times in a row (not a likely occurrence, but very definitely possible). On the sixth flip of the coin, what is the chance of getting heads again? It's 1 in 2, or 50%. Regardless of the sample size, the probability of a given flipped coin being heads is 1 in 2.

So extend that concept to larger numbers: If the probability of a given occurrence is 1 in 1 billion, then the probability remains 1 in 1 billion regardless of whether there are 100 or 100 billion samples.

This is why statisticians have the concept of a mathematical impossibility: If the probability of an event is lower than a certain number (I think it's 1 in 10^50, if I remember right), it is considered mathematically impossible, regardless of the sample size! Now, I don't know what the probability of this is consideredâ"I doubt it's low enough to be considered a mathematical impossibility. But the point is, increasing the sample size does not make the event more likely to occur.

That's my vote. :-)

Perhaps they should reconsider their evaluation of its age...

It seems to me there are several other implications of this approach:

1. Overhead costs should go down - anytime a company can standardize or reduce variability in their products, overhead goes down; if Lexmark can translate that into lower prices for the consumer, that will be very good for them.
2. Quality should go up - by the same token as number 1, reducing variables usually leads to higher build quality; if Lexmark can get both of these going in their favor, they will have a huge advantage.
3. They might be giving up some market share - last time I checked, there were several printers on the market for significantly less than $99; by choosing not to offer a truly budget-model printer, Lexmark may be cutting itself out of a pretty significant segment of the market.

It will be interesting to see if they can make the pros outweigh the cons.