Wikipedia:Reference desk/Mathematics

Wikipedia:Reference desk/headercfg

September 5

Differentiation

I encountered a differential equation problem recently, and, being unable to solve it, turned to the answers. The first step involved substituting ${\frac {d}{dy}}\left({\frac {dy}{dx}}\right)$ for ${\frac {d^{2}y}{dx^{2}}}\times {\frac {dx}{dy}}$. I can see that these are equal, but the explanation of the answer says that the first expression is differentiated to give the second using the chain rule. None of my math teachers have been able to figure out how this is done. Could someone give me some help? --superioridad ^(discusión) 10:18, 5 September 2007 (UTC)

The chain rule would say that we can do this (I think!);

${\frac {df(x)}{dy}}={\frac {df(x)}{dx}}\times {\frac {dx}{dy}}$

if you then substitute dy/dx in for f(x), then you'll have;

${\frac {d({\frac {dy}{dx}})}{dy}}={\frac {d({\frac {dy}{dx}})}{dx}}\times {\frac {dx}{dy}}$

${\frac {d({\frac {dy}{dx}})}{dy}}={\frac {d^{2}y}{dx^{2}}}\times {\frac {dx}{dy}}$. Richard B 11:43, 5 September 2007 (UTC)

Quine's method?

Prove that the following proposition is a tautology using Quine's method and standard logical equivalences.

$(P\to Q)\land (Q\to R)\to (P\to R)$

So,

$(P\to Q)\land (Q\to R):(P\to R)$

$(P\to R)\to (P\to R):??$

a) What is Quine's method? Can't find much on it

b) Did I do anything right there —Preceding unsigned comment added by 91.84.143.82 (talk) 12:43, 5 September 2007 (UTC)

Possibly Quine–McCluskey algorithm is what you need. Try to express implications with or, and and not operators, then use the algorithm to minimize the whole expression. If it minimizes to true then it is a tautology. --CiaPan 13:40, 5 September 2007 (UTC)

Question about zero

I am trying to find out if the number 0 (zero) is an even number or is it considered to be an odd number. My friends and I are in a major debate about this question. Thank you very much for your time and assistance in this matter. Please forward your answer directly to my email address: (email removed) thanks again. —Preceding unsigned comment added by 4.239.234.26 (talk) 18:13, 5 September 2007 (UTC)

Apologies, we answer questions right here, as noted at the top of the page. This gives everybody a chance to see the answer (and correct it if necessary). Also if you had looked up Zero here on Wikipedia, you'd have been shown along to 0 (number), in which it is stated that zero is considered an even number. --LarryMac | Talk 18:36, 5 September 2007 (UTC)

I confess I'm curious about how there could be much debate. Most people would agree that 3, 5, and 7 are odd, and that 2, 4, and 6 are even. Continuing the pattern of separation by two implies that 1 is odd and that 0 is even. A number is even if it is a multiple of 2, and odd if it is not; since 2×0 = 0, zero must be even. Notice it does not matter what integer is doubled, whether even, odd, positive, negative, or zero; thus −6 is also even (2×(−3) = −6), and −7 is odd. (Often negative numbers are not discussed as odd or even, but there are times when they are important; an example is modular arithmetic.) --KSmrq^T 19:25, 5 September 2007 (UTC)

I believe the logic is that the even numbers are multiples of 2, and the odd numbers are the others. In that case, if you figured a number could only be a multiple of 2 that was at least 2, it would be hard to categorize 0. It's not at all natural to consider 0 even, when working with positive numbers. It's only when working with all integers that it seems obvious. Black Carrot 18:14, 7 September 2007 (UTC)

Actually, one natural option would be (working in the nonnegative integers) to call positive integer multiples of 2 even, all other positive numbers odd, and 0 neither. It's the same way we deal with primes - primes 2 and up are primes, the rest are composites, 1 is a unit, and 0 is none of those. It's fairly arbitrary that we consider 0 automatically even. Black Carrot 18:16, 7 September 2007 (UTC)

Nonsense. It is anything but arbitrary that zero is even; it is essential! That's one reason I mentioned modular arithmetic. Throughout mathematics we depend on the rules for multiplying, adding, and subtracting evens and odds. If zero were not even, the consequences would be painful indeed. --KSmrq^T 20:03, 7 September 2007 (UTC)

Absolutely - in the more general context. How many people have heard of modular equivalence, though? For that matter, how many people are really all that familiar with the integers, other than those aspects related to basic arithmetic? Not so many, especially at the level I assume the poster is at. What I said was, considering only the counting numbers (1,2,3... with 0 tacked on as an afterthought) it makes as much sense to consider 0 a separate case, which is where the confusion seems to come from. Black Carrot 23:50, 7 September 2007 (UTC)

I think that's a stretch. The reason 0 is a special case in the unit/prime/composite thing is that it doesn't have a unique factorization, which is getting reasonably deep into the multiplicative structure of the naturals. For even/odd you don't need to do that; you just notice that the numbers are arranged, as it were, boy-girl-boy-girl. It's about the simplest pattern you can imagine that would ordinarily be called a pattern, and it unambiguously assigns 0 to the evens.

My personal theory as to why so many people are confused about this (even a biology PhD I won't name!) is that they're remembering being taught that 0 is neither positive nor negative, and confusing the two dichotomies. --Trovatore 21:26, 8 September 2007 (UTC)

Another good example. I would take it as supporting my case, though, not refuting it. My point was that, from the perspective of someone who doesn't yet have a bird's-eye-view of the whole thing, it all seems a bit arbitrary, so why not one more to toss on the pile? Another example that a lot of people have trouble with, until they accept it and move on, is that negative times negative is positive, not negative. Another, from a bit later in schooling, is that .999~ = 1, or that the square root of a negative number exists. That final one is really only justified by the fact that it works, which only experience can show, and that it forms an algebraic closure, which is an advanced concept. I myself didn't know whether zero was even or not until several years ago, because nobody really said whether it was at all clearly, and it wasn't obvious by itself. It wasn't until I saw it rigidly defined as "absolutely any integer multiple of 2, including 0" that I could be sure. Kids have seen enough patterns randomly broken in math by that time that it's hard to be certain unless someone says it clearly. Black Carrot 03:35, 10 September 2007 (UTC)

I can allow that that may be correct, if you're offering it as an explanation of why many people get this wrong. But it doesn't remotely support your previous claim that "It's fairly arbitrary that we consider 0 automatically even". All it supports is the notion that the confusion is not, after all, too surprising a mistake, given the large number of peculiar-sounding things that students are asked to accept. --Trovatore 17:56, 10 September 2007 (UTC)

It has sometimes happened that I've been with mathematical nonexperts and have had to address the question of whether 0 is even or odd. They always seem more surprised by my assertion that there is absolutely no controversy about the answer than they are by the answer itself. So I conclude that for nonmathematicians the answer is really not obvious! -- Dominus 16:45, 14 September 2007 (UTC)

Math Riddle

The following is a simplification of a math riddle I have came across:

"Bob met Alana on the street and forgot how old each of her kids are. She has 3 of them. Alana told Bob that the product of their ages is equal to 36. The sum of their ages is 1 higher than the house number of the white building across the street. Bob went home and couldn't figure it out and phoned Alana. Alana told him that she forgot to give him one more clue: the oldest one is a girl. Upon receiving that clue, Bob immediately figured out their ages."

Using some logic with that last hint, it seems to infer that two of the kids are twins, while the older one is not because the last clue seemed to make a huge difference in Bob solving the riddle. If that is the case, then the possible combinations are 1-1-36, 2-2-9 and 3-3-4.

I can not figure out whether the clue about the white building is relevant.

Could someone point me in the right way?

Thanks. Acceptable 21:59, 5 September 2007 (UTC)

The idea behind this riddle is that it is much easier for Bob than it is for us -- Bob knows what the number is, and we don't -- but we can work out the number, and their ages, based on Bob's reactions (whether or not he could figure it out at some point.) Try to work out what Bob is thinking, and cross off possibilities as you go. That list you've got will not help you. Gscshoyru 22:05, 5 September 2007 (UTC)

Is the clue about the house number relevant? Acceptable 22:10, 5 September 2007 (UTC)

Yes. He couldn't solve it after those two hints -- what does that mean? Gscshoyru 22:14, 5 September 2007 (UTC)

Is there one definite answer or are there multiple answers? Acceptable 22:20, 5 September 2007 (UTC)

Just one. Gscshoyru 22:24, 5 September 2007 (UTC)

Oh, then in that case it means that before the "oldest one is a daughter clue", there must have been multiple answers? Acceptable 22:28, 5 September 2007 (UTC)

So that means if I find the sums of all of the possible combinations of the 3 ages, then the answer is the one with the repeating sum? Acceptable 22:32, 5 September 2007 (UTC)

Yes. Gscshoyru 22:44, 5 September 2007 (UTC)

Is the answer 9-2-2? Acceptable 23:13, 5 September 2007 (UTC)

Sorry, Didn't realize it was a question. Zain Ebrahim 07:44, 6 September 2007 (UTC)

Okay if anyone wants the answer it's on [1] at the bottom somewhere. Just Ctrf+F "insurance salesman". Zain Ebrahim 10:56, 6 September 2007 (UTC)

That link gives me a "page not found" message. DuncanHill 10:59, 6 September 2007 (UTC)

Try it now. I'm not quite sure how to link to other websites. Zain Ebrahim 11:02, 6 September 2007 (UTC)

Thanks, that works now. DuncanHill 11:03, 6 September 2007 (UTC)

Thanks guys. Acceptable 23:04, 6 September 2007 (UTC)

If I understand this right, the solution hinges on the idea that two people can't be born in the same year and still have one that's older. Even if there's twins, there's an older one, and there doesn't have to be twins. — Daniel 02:28, 7 September 2007 (UTC)

Wow. I've known this riddle for a while and have never considered this point. This can be remedied by rephrasing it as "the one with the greatest age", though this sounds artificial. Of course, clarifying that ages are taken to be integers is necessary for the riddle to make sense, and is all the more important with this phrasing. -- Meni Rosenfeld (talk) 11:26, 7 September 2007 (UTC)

"...the oldest one is a girl" tells us that 1-6-6 is wrong and 2-2-9 is right (without that clue it could be either). I don't quite see how the problem proposed by Daniel would be remedied by using "...the one with the greatest age is a girl". Unless I'm completely missing something here.
Zain Ebrahim 11:36, 7 September 2007 (UTC)

Is Daniel's problem unclear, or is it my solution? 1-6-6 isn't wrong because it could be that one was born exactly 6 years ago, and one was born 6 years and 10 months ago. This one would be the eldest, though both their ages are 6. With my phrasing, there will be no single person of greatest age and thus 1-6-6 is indeed wrong. -- Meni Rosenfeld (talk) 11:41, 7 September 2007 (UTC)

If you really want to continue down this road, you'd want to say "the one with the greatest age, when expressed as a truncated integer," or some such. But come on, guys, it's a riddle, not an exercise in how finely you can slice a problem (with or without Occam's Razor). --LarryMac | Talk 11:56, 7 September 2007 (UTC)

Actually, I have already mentioned that ages must be integers. But I think I disagree with your point. As a mathematical exercise, this problem is trivial. Its only virtue is the challenge of noticing the mathematical details hidden in the real-world explanations. One part of this is extracting "there is a unique greatest number" from "there is an oldest child" (which itself is extracted from "the oldest is a girl"). But in fact this deduction is not correct, so we are cheating our listeners if we use the original phrasing and expect them to make it. -- Meni Rosenfeld (talk) 12:03, 7 September 2007 (UTC)

Oh okay, I get it now. Sorry, Meni, I'm really having a slow day here. Btw: listeners?
Zain Ebrahim 12:22, 7 September 2007 (UTC)

The best word I could think of for "the people to whom we are posing the riddle". I'd be glad to hear better suggestions. -- Meni Rosenfeld (talk) 12:36, 7 September 2007 (UTC)

I don't know if it's better but . . . Audience? :) Zain Ebrahim 12:39, 7 September 2007 (UTC)

Call me crazy, but "readers" has always been standard in newspaper and magazine articles. Black Carrot 18:11, 7 September 2007 (UTC)

Oh, I get it. I didn't mean the readers of the reference desk, but rather people (friends, etc.) to whom we (as in, anyone) might pose the riddle in person. -- Meni Rosenfeld (talk) 18:33, 7 September 2007 (UTC)

Aahhh... Indeed. <rubs chin thoughtfully> Black Carrot 23:46, 7 September 2007 (UTC)

Mad mathematician

Hi all, hope you can help. There was a BBC programme recently about a mathematician who spent his declining years (I think in the '30s of the last century) trying to solve some famous mathematical problem. He would write to his publisher announcing that he was close to a solution, only to write again a few weeks later to say he had failed, this behaviour repeated several times, and he eventually ended in an insane asylum (possibly in Switzerland). I must admit that I wasn't really paying attention at the time, but if anyone could identify the chap I would be grateful - it's been nagging away at the back of my mind ever since! DuncanHill 22:00, 5 September 2007 (UTC)

Sounds like Georg Cantor to me. Gscshoyru 22:06, 5 September 2007 (UTC)

I'm HUGELY impressed! Thank you :) DuncanHill 22:18, 5 September 2007 (UTC)

Could it be this? PrimeHunter 22:23, 5 September 2007 (UTC)

Yes that's it, here it is on the BBC website [2]. Thanks again. DuncanHill 22:25, 5 September 2007 (UTC)

If it was Cantor, then someone has the story wrong, or at least is giving it the wrong nuances. Cantor was never insane, only depressed. I don't mean to minimize depression; I know it's a cruel and debilitating illness, but in my amateur estimation it's unlikely to lead to crankish behavior. Cantor was greatly disappointed by his inability to prove the continuum hypothesis, so it's conceivable the story derives from that.

Kurt Gödel, on the other hand, did go a bit further round the bend, but again I don't know of any problem that he kept claiming to be on the verge of solving.

Then there's Alexander Abian. Who knows, maybe we just didn't understand him. (Certainly we at least didn't understand him.) --Trovatore 22:36, 5 September 2007 (UTC)

As I said, I didn't really pay much attention to the programme, so it is entirely possible that they said he died in a san, and I misrembered this as being an asylum. DuncanHill 22:45, 5 September 2007 (UTC)

According to the article, Cantor did in fact end up in a sanatorium. And he wasn't being a crank -- he truly thought he had the answer a few times... though we know that's impossible. Gscshoyru 22:44, 5 September 2007 (UTC)

Well, no we don't really "know that's impossible". We know that it's impossible to prove or refute from ZFC. That's not what Cantor was trying to do (in fact ZFC had not even been formulated). --Trovatore 22:48, 5 September 2007 (UTC)

September 6

Looking for a fallacy in statistics

I'm looking for the proper name of a fallacy in statistics. It goes something like this:

You work in a factory that manufactures six-sided dice. Your factory manufactures one thousand dices a day. After each die is created, it is rolled 5 times and the results recorded.

As you are walking through the factory one day, you noticed a die with the result of five 6s in a row. You suspect that the die is "unfair" and then calculated the odds to be 1/(6^5) = 1/7776 and thus so you threw that die into the defective bin.

The fallacy is "observing the data" to form a belief and then using the exact same data to confirm that belief. What is the proper name of this fallacy?

The reason, I'm asking this is related to a thought experiment where a police notices a muslim behaving suspiciously. The muslim performed 5 suspicious activities, the police then calculated that the probability of 5 suspicious activities happening by coincidence to be 1/7776 thus the police concludes that muslim must be a terrorist.

202.168.50.40 01:17, 6 September 2007 (UTC)

Given that all dice were tested, this is a case of the Prosecutor's fallacy, specifically the multiple testing variety. Testing all as such will quite likely give a small number of unusual rolls so singling them out is tricky.

An additional problem: the nominally operative probability should be that of seeing a set of 5 identical rolls (we assume that if, for example, 5 "3"s were observed, the reaction would be the same). This is six times 1/7776 or 1/1296. This issue is roughly testing hypotheses suggested by the data, although that article could use some work. Baccyak4H (Yak!) 02:25, 6 September 2007 (UTC)

BTW. Suppose you see another die, rolled five times with a result of 1-2-3-5-6. The probability of this result is also (1/6)^5 = 1/7776. So this die should be considered "unfair", too, and land in a defective bin, together with the previous one. Am I right?
CiaPan 09:32, 6 September 2007 (UTC)

That is the probability of those rolls in that order. But I suspect any set of 5 rolls without a repitition would be so flagged as unfair. Here the operative probability is (6!/1!)/(6^6), or about 1.5%. Baccyak4H (Yak!) 13:27, 6 September 2007 (UTC)

Two further issues: 1) a better way to assess QC in this manner is to exploit the empirical observation that when there is a manufacturing problem, it tends to linger until noticed and the process fixed. Testing only a small number from the entire day's batch greatly reduces the multiple testing issue, and loses very little effectiveness (assuming a good sampling scheme). 2) The application to the thought experiment is problematic in that people generally are (and one could argue police should be) Bayesian, not frequentist (which my above replies assume). But that philosophy is subjective: one person's valid but imprudent application could end up locking up a lot of silly suspects (e.g., elderly people who become disoriented), while another's would lock up nearly every single olive-skinned person with a pulse. So there still is some ambiguity as to how such operations should work. Baccyak4H (Yak!) 13:37, 6 September 2007 (UTC)

Something a bit different, but along similar lines, would be confirmation bias. It doesn't fit the dice example very well, but it fits the other example quite a bit better, in a way I think you haven't explored yet. Black Carrot 23:45, 6 September 2007 (UTC)

One other detail on the thought experiment - there are such things, at least in America, as "reasonable doubt" and "just cause". No argument the officer gave for detaining someone on those grounds would fly. Black Carrot 18:07, 7 September 2007 (UTC)

Solving Inequalities

I'm working with deltas and epsilons and limits and I'm trying to solve an inequality. I have -0.75<1/x^2<1.25. Do I need to change the direction of the inequality when I take the reciprocal of all the numbers? And how do I do a square root when one of the numbers is negative?

Thanks! —Preceding unsigned comment added by 130.49.11.65 (talk) 02:49, 6 September 2007 (UTC)

Manipulating inequalities is much trickier than manipulating equations. With an equation, you can apply any function to both sides, and the resulting equation is still true. With inequalities, this does not work unless the function is monotonic. For example, the function f(x) = 1/x is not monotonic, so you cannot apply it to both sides of the equation, even if you change the direction:

-1 < 1 < 2

but

1/(-1) < 1/2 < 1/1

The new values are not in the same order, and they're not in the opposite order either. Their ordering has been shuffled around by the action of the non-monotonic function f(x) = 1/x.

Sometimes the best approach to solving an inequality is to find the conditions in which two of the expressions are equal, and then determine the signs of the inequalities in each of the resulting regions. And use common sense. As you know, real numbers have non-negative squares, so if x is a (nonzero) real number, then 1/x^2 >= 0 and the left part of your inequality is always satisfied. Isn't that a nice simplification? —Keenan Pepper 03:43, 6 September 2007 (UTC)

Another thing you can always do is mulptiply all sides by some positive number. You know that x must be nonzero (otherwise you can't divide by it) so $x^{2}$ is positive, so you get $-0.75x^{2}<1<1.25x^{2}$. As mentioned, only the $1<1.25x^{2}$ part actually conveys information.

You do not need to take a square root of a negative number here, since the mere usage of inequalities suggests we are dealing with real numbers, so this is impossible. If you are interested in this anyway, take a look at imaginary unit and complex number. -- Meni Rosenfeld (talk) 11:34, 6 September 2007 (UTC)

September 7

Chatin, turing machines and origin in coordinate systems.

In this article [3] Greg Chaitin makes a comparison between the choice of a Universal Turing Machine and the choice of origin in a coordinate system. Here is the relevant quote:

Algorithmic information theory focuses on individual objects rather than on the ensembles and probability distributions considered in Claude Shannon and Norbert Wiener's information theory. How many bits does it take to describe how to compute an individual object? In other words, what is the size in bits of the smallest program for calculating it? It is easy to see that since general-purpose computers (universal Turing machines) can simulate each other, the choice of computer as yardstick is not very important and really only corresponds to the choice of origin in a coordinate system.

Can anyone shed any more light on this interesting aside? Perhaps computation can be understood as traversal of some sort of convolution space in information? (no university education in mathematics, so please don't parse terms in technical senses).
Thanks, 86.132.15.29 02:07, 7 September 2007 (UTC)

Let M be some Turing machine, and let U be a universal Turing machine. Then there is a program P for U that will make it simulate M. In other words, giving input J to M is the same as giving input P+J to U, in which P+J consists of concatenating P and J on the input tape. If J is the shortest encoding of some object, then the Kolmogorov complexity of that object, using M as the machine, is |J|; call this number n. Using U as the machine, we find an upper bound |P+J| = |P| + n on the complexity. So using U as the measuring stick, we add at most a fixed number, |P|, to the complexity. Now if M is also a universal machine, it simulates U with some program |Q|, and so, by the same argument, using M as the measure of all things instead of U adds at most |Q| to their complexity. Putting m := max(|P|, |Q|), we see that the two measures using M and U never differ by more than the fixed value m.  --Lambiam 05:35, 7 September 2007 (UTC)

I don't think he's trying to make a formal analogy between computability and geometry. I think he's simply saying that just as the choice of coordinate system is unimportant in geometry, so the choice of Turing machine is unimportant in algorithmic information theory. (edit: I do understand the reference. What I said still stands.) -- BenRG 10:49, 7 September 2007 (UTC)

This illustrates something I worry about when I write: An analogy is only helpful if readers understand the reference!

Take a blank sheet of paper as a representative of a Euclidean plane. If we want to lay out a Cartesian grid for coordinates, we must first choose some point as the origin. Maybe we choose the bottom left corner; maybe we choose the center of the page. Whichever point we decide to use, we can later change our mind and adjust the coordinates accordingly. Almost nothing we do is sensitive to the choice of origin.

Take a collection of universal Turing machines. Each one can simulate any Turing machine (since that's what "universal" means). In particular, each can simulate every one of the other universal machines. We can choose any machine we like as our "base" machine for measuring complexity. If we later change our mind, we can adjust. Almost nothing we do is sensitive to the choice of base. --KSmrq^T 14:51, 7 September 2007 (UTC)

You could even take it a step beyond that, and consider the thing being measured as completely independent of the coordinate system, as a rectangle is the same rectangle no matter where you put the origin and axes - you're just measuring it with different rulers, not measuring something different. In the same way, the thing being considered in terms of complexity is independent of what machine you implement it on - it's the same algorithm, being perfomed equivalently on different hardware, like a program running the same on any PC. It's not even relevant which one it's on, as long as it's being run correctly. On the subject of well-chosen analogies: this was apparently published in the International Journal of Theoretical Physics. It's likely that its intended audience was comfortable with symmetries of coordinate systems. In fact, I think that would count in this case as translating the concept into lay terminology. Black Carrot 18:02, 7 September 2007 (UTC)

Curiously this was the classical approach to geometry. The whole notion of a coordinate system is quite a modern notion. I know the Cartesian coordinate system was only introduced in 1637 by René Descartes. Does anyone know more of the history of the idea? --Salix alba (talk) 22:16, 8 September 2007 (UTC)

Well, here's my thoughts on that and perhaps some explaination of why i think the analogy might of consequence. The choice of origin makes no difference to the results of measure in some space but it does make a difference to the process of measure. In particular, if one is to measure the length of an object from a particular Point of Origin, then the difficulty will vary as a function three points: the origin, and two endpoints of the length in question. For example, the optimal origin to measure the distance between two points is their centre. Starting further away requires greater precision, thus more calculation. Incidentally, this also makes incompleteness intuitive as a result any origin being unable to measure the length of two points whose extension passes through it. (Remember that such distance or complexity is only defined for self-delimiting programs, i.e. those which do not halt. You can't distinguish finite and infinite distances along a line you can't measure.) 62.56.116.66 02:34, 9 September 2007 (UTC) (question poser)

What we're doing in "measuring" algorithmic complexity is counting the number of bits in a string. Since the strings involved, when using different "origins" (universal TM's), are about equally long (see above), and we use mathematical reasoning that ignores constant differences anyway, rather than actual counting, the choice can hardly be of influence on the "difficulty". The whole point of Chaitin's remark is in fact that the choice of universal TM is a non-issue.  --Lambiam 03:39, 9 September 2007 (UTC)

Graphing Calculators

Which of the following graphing calculators is best in terms of its funcionality, price, reliability, etc...:

• TI-83
• TI-83+
• TI-84

Thanks. Acceptable 20:59, 7 September 2007 (UTC)

The TI-83+ is basically a faster version of the TI-83 and has more memory; unless you have a lot of applications and games, it's not worth the extra money. The TI-84 also improves in the same way; it's faster and has more memory. Personally, I'd just go with the TI-83 and save yourself the money. Dlong 21:04, 7 September 2007 (UTC)

I created a page [4] to address this issue when I was teaching math. For most of my students I recommended a basic scientific calculator and not wasting their money on an 83/84/89. Donald Hosek 00:49, 9 September 2007 (UTC)

Thanks. As well, what additional functions does the TI-89 have that bans it from some nationalized math tests? Acceptable 21:06, 7 September 2007 (UTC)

The TI-89 basically extends the functionality of the TI-83 and -84 to include variables. For instance, the 83 and 84 for can find the value of a derivative at a specific point. For instance, if you ask for the derivative of $x^{1/2}$ at x=3, the 83 and 84 gives you the numerical approximation, .28867514; The TI-89 gives you the exact value, ${\frac {\sqrt {3}}{6}}$; additionally, it will give you the derivative in terms of x, ${\frac {1}{2{\sqrt {x}}}}$. The TI-89 improves over the TI-83 and 84 in this way with regard to solving equations, using trigonometric functions, finding limits, and finding integrals, and many other things. Additionally, the TI-89 provides many other features, such as calculating Taylor series, finding the arclength of an equation, finding improper integrals, solving differential equations, and solving algebraic equations for complex solutions as well as real. Dlong 21:39, 7 September 2007 (UTC)

Last I checked, on tests (like the SAT) the TI-89 isn't banned... but the Voyage 200 and the TI-92 are, because of their qwerty keyboards. As for calculators, I don't think they make TI-83's anymore, as the 84 line has superseded it, and the best TI-84 is the TI-84 plus silver, but if you're looking for cheaper and won't be making/adding lots of programs it the TI-84 is probably what you want.

Speaking from experience, the TI-89 titanium/Voyage 200 (same OS, different shape) is one of the best calculators out there... but if you don't need all of its functionality, go with the TI-84 series. Gscshoyru 22:08, 7 September 2007 (UTC)

The reason that the TI-89 is often banned, as far as I know, is that it can easily [more easily?] store large amounts of text. Students can save a list of useful formulas, for example, then access it during the test, giving them an unfair advantage. This was a big problem in my chemistry class in high school (though this was before TI-89s) -- some of the kids had found the entire periodic table somewhere online and downloaded it onto their calculators as a program. (It would fail it you tried to execute, but all you had to do was look through the "code".) Tesseran

Good times. Why would that be a problem for the TI-83, though? I thought it could hold that. There's only, what, 200 elements? And each packet of information would be just a few characters long. Black Carrot 23:44, 7 September 2007 (UTC)

Even a regular TI-83 could hold way more text than you need -- I think it's just the ones with qwerty keyboards that are banned from stuff like the SAT, so you can't type the questions into it fast enough to store them all. (This is fallacious, as it would be simple to train yourself to do so, but, whatever.) Gscshoyru 00:00, 8 September 2007 (UTC)

You don't even need to do that. There's a TI83+ asm program that lets you turn your calculator on the side and type as if the calculator keys are a qwerty keyboard. Bit cramped but you can get used to it. Not that it's very useful for anything. Always amuses me when I read "No QWERTY keyboards!" when obviously any array of buttons large enough can be turned into one. BungaDunga 06:55, 11 September 2007 (UTC)

Can the regular TI-84 solve algebraic equations? For example, if I entered "3x^2+2x+3=0" (don't know if this quadratic eqn actually works) would it spit out the answer "x=...." ?Acceptable 02:42, 8 September 2007 (UTC)

Well... no calculator will know what you want if you do that, you have to explicitly tell them to solve. Under normal circumstances, I don't think TI-84's can solve quadratic equations and return all the solutions... but they may come with an add-on that does. Plus they have a solver, that will give you one of the answers. But if you want all the answers, TI-89 titanium is the way to go. Gscshoyru 02:50, 8 September 2007 (UTC)

Is the TI-84 PLus Silver Edition worth the extra $40? Acceptable 03:09, 8 September 2007 (UTC)

The 84 Silver is basically identical to the original except with even more memory (and possibly speed?). Unless you want to run an epic roleplaying game, or read Paradise Lost on a 96x64 screen or something, you don't need it. You can do just fine in school with an 83+ non-silver off EBay. 83+s only fit four or so applications (distinct from "programs" like games). I'm running a 3D/derivative Grapher, Omnicalc (a general functionality booster), Symbolic (extra math functions- arclength and derivatives mainly) and PrettyPrint, which displays expressions a lot like latex (equations you see on Wikipedia and how you'd write them). I could probably squeeze in a Tetris clone and maybe Asteroids. Note that an 83+ doesn't include a computer transfer cable (or a usb interface, so transfers are slooooww). Obviously if price isn't a huge issue an 84+ is better; from the ones I've tried the graphing speed (which is where you notice clock speed most, really) is far and away faster. Plus they have a USB port, which is theoretically hackable to interface with a USB memory stick and pull files off.BungaDunga 06:55, 11 September 2007 (UTC)

How many googols in a googolplex?

for the magic: the gathering players: i use a dryad arbor enchanted with utopia sprawl (creating blue mana) and freed from the real to create infinite green mana. i use this mana to play wurmcalling with buyback, putting exactly one googol green mana into x. this creates a 10¹⁰⁰/10¹⁰⁰ wurm token. considering i still have infinite mana at my disposal, i do this so that the sum of the tokens' respective powers is equal to a googolplex of damage. how many wurm tokens do i have to put into play to get this?

for the non-magic players: 10¹⁰⁰x = 10^{10^100}. see headline. DJRaveN4x 22:40, 7 September 2007 (UTC)

I don't think that in the old days it was possible to obtain unbounded mana. All those new cards with their fancy effects really messed up the game.

Anyway, applying the rule ${\frac {a^{b}}{a^{c}}}=a^{b-c}$ gives:

$x={\frac {10^{10^{100}}}{10^{100}}}=10^{10^{100}-100}=10^{9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999900}$.

But the shortest way to describe this number is simply "googolplex over googol". -- Meni Rosenfeld (talk) 23:06, 7 September 2007 (UTC)

i was really hoping the number was something a bit more tangible than that...but thanks a lot for the help! DJRaveN4x 23:18, 7 September 2007 (UTC)

Meni... I'm pretty sure there have been infinite mana combos in existence for a long, long time. Not sure when the first one came about. Or, if you're talking about the fact that you can't have "an infinite" amount of mana in your mana pool, you can't -- if you can do an action repeatedly, choose a number and do it that many times. And yes, I am magic rules freak. Oh well. Gscshoyru 23:58, 7 September 2007 (UTC)

The fact that you can only have unbounded, not infinite, mana, is something I have considered (and reflected in my phrasing) but not what I complained about. The time I was playing a lot was over 10 years ago - are you sure such tricks existed then? -- Meni Rosenfeld (talk) 07:11, 8 September 2007 (UTC)

It's all a consequence of the fact that 10ⁿ-n is still pretty darn close to 10ⁿ. So

10-1=9

10²-2=98

10³-3=997

10⁴-4=9996 etc.

So dividing a googolplex by a googol hardly makes a dent in it ! Of course, if you are looking for seriously large numbers, a googolplex is a tiddler compared to, say, Graham's number. Gandalf61 13:30, 8 September 2007 (UTC)

It takes away 99.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999% Isn't that some dent?  --Lambiam 03:44, 9 September 2007 (UTC)

Well, yes, to be precise I should have said dividing a googolplex by a googol appears to hardly make a dent in it. This is because the ratio of the logs of a googolplex and googolplex/googol is very close to 1. 1000000000000000000 is only 1% of 100000000000000000000, but they appear to have a similar magnitude. Gandalf61 09:07, 9 September 2007 (UTC)

Completely irrelevant to the mathematics, but I know I worked out an infinite mana combo that was valid around 4th edition. Given how broken some of the earlier cards were before they got errata'd or restricted, I wouldn't be surprised if it was possible in Alpha or Beta. Confusing Manifestation 01:19, 12 September 2007 (UTC)

Syntax of multiple derivatives

When you take the 2nd derivative of something, say x(t), it is written d²x/dt². Why is it not written dx²/dt² or d²x/d²t? Why are the superscipts not in the same place in the numerator and denominator? jwillbur 22:58, 7 September 2007 (UTC)

The short answer is that the squares don't have the same meaning in the numerator and denominator. Informally speaking, in the numerator your square the differential operator (in other words, you take x and apply the operator d to it twice) and in the denominator you square the differential itself (you apply d to t, and multiply the result by itself).

Another informal way to motivate this notation is that

$x''(t)={\frac {d\left({\frac {dx}{dt}}\right)}{dt}}={\frac {d(dx)}{dt\cdot dt}}={\frac {(dd)x}{dt^{2}}}={\frac {d^{2}x}{dt^{2}}}$.

-- Meni Rosenfeld (talk) 23:13, 7 September 2007 (UTC)

Also, for any well-behaved x, ${\frac {dx}{dt}}{\frac {dt}{dx}}=1$, but in general ${\frac {d^{2}x}{dt^{2}}}{\frac {d^{2}t}{dx^{2}}}\neq 1$. The notation is consistent with this. (For single derivatives they "cancel" but not for higher orders.) —Keenan Pepper 05:53, 8 September 2007 (UTC)

Another way of looking at it is viewing d/dt (differentiation with respect to t) as an operator, and the meaning of the expression d²x/dt² is then

${\frac {d}{dt}}\left({\frac {d}{dt}}\,x\right).$

Using the convention that f²(x) is short for f(f(x)), we can rewrite this to:

$\left({\frac {d}{dt}}\right)^{2}x\,\!,$

from which it is a small step to

${\frac {d^{2}}{dt^{2}}}\,x={\frac {d^{2}x}{dt^{2}}}.$

If I were to encounter dx²/dt² in mathematical writing, my first assumption would be that this means the same as (dx/dt)², which is something entirely different.  --Lambiam 04:04, 9 September 2007 (UTC)

September 8

No questions today

This is truly our darkest hour... — Kieff | Talk 03:51, 9 September 2007 (UTC)

September 9

Maths is about numbers

Is it true that Maths is about numbers just like Agatha Christie murder mysteries is about the English alphabets? I heard someone use that phrase but I thought that they were just jesting. Is there any truth to that phrase? 211.28.131.182 09:07, 9 September 2007 (UTC)

I wouldn't say this phrase is very accurate, but it has a point and I do not believe it was said in jest. The point is that math is about much more than just numbers. Granted, numbers of all sorts are very important in almost any aspect of mathematics, but not everything revolves around them. -- Meni Rosenfeld (talk) 11:03, 9 September 2007 (UTC)

To be absolutely clear, Agatha Christie's murder mysteries are not about the English alphabet, even though that alphabet does play a major role in how they are written and read. See also our Mathematics article, which tries to get across what maths is about.  --Lambiam 12:14, 9 September 2007 (UTC)

The A.B.C. Murders? —Blotwell 14:52, 10 September 2007 (UTC)

I suspect that if the OP's doubts had been about the second part, he would have asked the question at Wikipedia:Reference desk/Humanities. -- Meni Rosenfeld (talk) 12:26, 9 September 2007 (UTC)

But perhaps (what do I know?) the questioner assumes that the truth of the second part is raised above all doubt, and wonders if the first part is, likewise, true.  --Lambiam 14:07, 9 September 2007 (UTC)

I would say that it is certainly true that modern mathematics is not just about numbers, and that numbers play a very small role in much of modern math. The quote perhaps exagerates the case, but I think the sentiment behind it is valid. Saying what mathematics actually is about is rather hard, but I would suggest that words like "abstraction", and "structure", and "structural relationships" are far loser than "number". See, for instance, category theory where the fundamental things under consideration are the (directed) relationships between objects, and the structures that result, and then note that category theory provides a language and perspective that is applicable to almost all of modern mathematics. Numbers don't need to enter into it. -- Leland McInnes 15:28, 9 September 2007 (UTC)

Number bases

How do you solve this equation? xyyz(base 5) + xyyz (base 7)=xyyz (base 8) —Preceding unsigned comment added by 124.82.104.162 (talk) 15:42, 9 September 2007 (UTC)

I take it that each of those numbers is a digit? Well... a number in base n is represented by the digit times a power of n, so the first number would be z * 5^0 + y * 5^1 + y*5^2 + x*5^3 in base 10... does that help? Gscshoyru 15:52, 9 September 2007 (UTC)

Sounds like constrain logic programming.

x*5^3 + y*5^2 + y*5^1 + z + x*7^3 + y*7^2 + y*7^1 + z = x*8^3 + y*8^2 + y*8^1 + z

x in {0,1,2,3,4}

y in {0,1,2,3,4}

z in {0,1,2,3,4}

Concentrate on z where z + z = a * 8 + z where a in {0,1}

We have z in {0} so z must be zero.

alternatively you can use the brute force method which is not that hard as there is only 125 different combinations. 202.168.50.40 00:48, 10 September 2007 (UTC)

"z + z = a * 8 + z" is wrong. PrimeHunter 03:28, 10 September 2007 (UTC)

Simplfying the above we get

14y + z = 44x

Clearly z must be even; and so must be 0, 2, or 4. Also, increasing (or decreasing) x and y by 7 and 22, respectively (since gcd(14,44) = 2), or any multiple thereof will not change the truth of the equation.

If z = 0, then x = y = 0 is a solution (but we don't want x to be 0). The next higher solution is (7, 22, 0) which is outside our bounds.

If z = 2, then 44x - 14y = 2. By the extended Euclidean algorithm we get x = 1, y = 3. (1, 3, 2) is a solution. Increasing or decreasing x by 7 will put us outside our bounds.

If z = 4, then 44x - 14y = 4. Doubling the above, we see that x = 2, y = 6 works. But y = 6 is outside the bounds. And increasing or decreasing x by 7 will also go outside the bounds. So this doesn't work.

So we conclude the only solution is x = 1, y = 3, z = 2. --152.75.18.108 22:38, 10 September 2007 (UTC)

Chess books

For a number of years I have been playing chess casually and have taken a liking to the game, and I wish to take my game to the next level. However, I have no theoretical background. I was hoping that you guys could recommend some good chess books to improve all aspects of the game (opening repertoire especially). Thanks a lot! —Preceding unsigned comment added by 70.52.45.191 (talk) 19:25, 9 September 2007 (UTC)

You could try the references at chess openings (disclaimer: I know nothing of chess) Algebraist 13:30, 10 September 2007 (UTC)

Re openings, generalising, there are three types of books about chess openings. There are the encyclopedic (of which Batsford's Modern Chess Openings is probably the best, I think it's now in its 14th edition), those that deal with a "family" of openings and those that pick over the bones of a single line. I'd recommend you start with Batsford, but try and find a 2nd hand copy in good condition - theory of the main lines doesn't move on so fast that an out of date edition makes a huge difference. Use Batsford to develop at least a basic understanding of the most common openings - then use your new knowledge to find the right specialist books to take you on. See also Category:Chess books and Chess endgame literature --Dweller 13:55, 10 September 2007 (UTC)

TI-84+ equation solver

When I'm using the TI-84 Equation Solver (Math>Solver) after I enter in an equation and enter a guess then hit Alpha+Solve, it gives me an answer but one line down, it says "left-rt=0". What does "left-rt=0" mean? Thanks. Acceptable 23:03, 9 September 2007 (UTC)

er? Lefthand side minus righthand size equals zero? 202.168.50.40 00:42, 10 September 2007 (UTC)

Lol, thanks, all those words on the screen confused me. Acceptable 00:59, 10 September 2007 (UTC)

September 10

Vector product

why is the product of two vectors , perpendicular to those two? if it's an axiom why isn't it self evident? —Preceding unsigned comment added by 203.145.156.9 (talk) 02:00, 10 September 2007 (UTC)

It is a definition, rather than an axiom. It happens to be a very convenient definition. Phils 02:51, 10 September 2007 (UTC)

It's definitely not an axiom, just part of the definition of cross product. See Cross product#Cross product as an exterior product. It's not fundamentally a vector (oriented line element), but an oriented area element (bivector). But a plane can be defined by a vector perpendicular to it, so there you go. —Keenan Pepper 04:18, 10 September 2007 (UTC)

Quintics and Up

I've been reading about Galois and the invention of group theory, and I keep running across the same thing that never gets developed - once they knew most large polynomials couldn't be solved in the radicals, what did they do next? It's implied that they quickly found other ways of getting answers, but nothing ever explores what those new ideas were. What did they come up with? Black Carrot 03:40, 10 September 2007 (UTC)

I think your history is a little mixed up. Numerical approximation methods such as fixed point iteration and Newton's method were known long before the Abel-Ruffini theorem was published. Solving high-order polynomials in radicals never had much practical application. Even for a cubic equation it's often much easier to form a sequence that converges to the solution than to go through the whole process to find an exact answer in radicals. Knowing that an exact answer is possible only for degree less than five is mostly of theoretical/philosophical interest.

If instead you're asking about ways to represent exact roots of polynomials that cannot be expressed in ordinary radicals, see Bring radical. —Keenan Pepper 04:44, 10 September 2007 (UTC)

Probabilities

No, this is not a homework question; I'm not in school.

Say you have a pizza shop that has 11 toppings: Pepperoni, mushrooms, onions, green peppers, pineapple, sausage, tuna, olives, ham, artichokes and anchovies. A pizza can have zero, 1, 2, 3, 4... up to all 11 toppings. For example, it can have just mushrooms; olives and onions; pepperoni, green peppers and tuna, etc.

How many different combinations of toppings are there? Please explain how you come up with the number. Thanks -- Mwalcoff 07:31, 10 September 2007 (UTC)

The order in which the toppings are applied does not matter (a pepperoni & mushroom pizza is the same as a mushroom & pepperoni pizza). So you want the count the number of subsets of the set of 11 toppings - not forgetting to include the possibility that there are no toppings at all, which is the empty subset. The set of all subsets of a set is called its power set. Read our articles on subset and power set and you will be able to work out the answer for yourself. Gandalf61 08:07, 10 September 2007 (UTC)

Easy! First assume that the pizza can only have two toppings, mushroom and/or olives.

The solution is 4 types. (no topping), (mushroom only), (olives only) and (mushroom and olives).

So the answer is mushroom(yes/no) * olives(yes/no) = 2*2 = 4

So for 11 toppings the answer is

2*2*2*2*2*2*2*2*2*2*2 = 2^11 = 2*2^10 = 2*1024 = 2048

Solved. 211.28.131.182 13:28, 10 September 2007 (UTC)

Do you assume that no-one wants a pizza with olives and, erm, more olives? Not a safe assumption with pizzas. --Dweller 13:47, 10 September 2007 (UTC)

So, let's allow double olives ... triple olives ... triple olives and hundred-fold mushrooms ... looks like we've discovered multisets. These pizzas sure are educational ! Gandalf61 14:31, 10 September 2007 (UTC)

Thanks. -- Mwalcoff 22:47, 10 September 2007 (UTC)

wording Exponential_function#Formal_definition

e is defined in the article e , but at the above link it states that e^x is defined as the power series given; shouldn't this be a proof that the two are equivalent (for reals), and not a definition.

Additionally is there notation that distinguishes between e^0.5 = |sqrt(e)| and +/- |sqrt (e)| ?87.102.77.35 13:04, 10 September 2007 (UTC)

The usual (because easiest) approach to exponentials is the following: define e^x by its power series. Define e to be e¹. Define log to be the inverse of e^x. Define a^x to be e^{x log a}. Check that this agrees with the previous definition for a=e, and that exponentiation has all the required properties, in particular a^{x + y}=a^x a^y. Thus the power series for e^x is central to the definition of exponentiation in general.

There is an alternative approach, which you might be happier with: define aⁿ as a multiplied by itself n times, then define a^n/m as the mth root of aⁿ, then extend to all real x by continuity. Then you can define e however you want, and define e^x in accordance with your previous definition of exponentials. However, I've never seen a book that actually does this, and I believe it to be substantially more effort to prove all the relevant facts. Algebraist 13:26, 10 September 2007 (UTC)

But by the definition given e^x is singly valued - which is(seems) wrong since e^x can be multiply valued - it shouldn't be quoted as a definition, but as being equivalent for the positive real values of e^x (when x is real) (when x is imaginary only a single value results so that doesn't matter here)87.102.77.35 13:49, 10 September 2007 (UTC)

e^x is single-valued. It has no branch points. Now x^0.5 might be multi-valued, but that is irrelevant. --Spoon! 15:17, 10 September 2007 (UTC)

What I said above concerned only real numbers. For complex numbers, one defines the function sometimes called exp by the power series for e^x, then defines the multi-valued inverse function Log, then the multi-valued functions a^x. One then has to distinguish between the functions exp (which is a single-valued function given by a power series) and e^x, which is multi-valued and is by definition exp^{x Log e}=exp^x(1+2niπ). Algebraist 17:44, 10 September 2007 (UTC)

I'm really confused here. If e^x is single value then what is the single value of e^0.5? Doesn't it have two values? ie. 1.64 and -1.64 202.168.50.40 00:45, 11 September 2007 (UTC)

There are two things commonly called e^x. One is also called exp, and is a well-defined function, given by a well-known power series. The other is a multi-valued function on the complex plane. exp(0.5) is 1.65 or thereabouts, while e^0.5 is 1.65 or -1.65, depending on which branch you choose of this multi-valued 'function'. Algebraist 01:13, 11 September 2007 (UTC)

I wonder, is $e^{x}$ really a "multi-valued function" in the common sense in complex analysis? Sure, you can take different n and end up with different single-valued functions, but for any particular choice there are no branch points, branch cuts or anything similar. Not like $x^{0.5}$, for which you can clearly define a Riemann surface and obtain different branches by taking different sheets of it, or by choosing a branch cut. Perhaps this term is more general than I am accustomed to?

As for the definition of the exponential function, defining it as a power series seems too artificial. My favorite is defining Log as an integral and then Exp as its inverse. My other, similar, choice, would be to define Exp by a differential equation. -- Meni Rosenfeld (talk) 08:14, 11 September 2007 (UTC)

How do I distinguish between e^x as a single valued function and 2.7...^x which can also be written e^x which can have more than one value83.100.251.220 08:42, 11 September 2007 (UTC)

coordinates of a point

Hello. I'm not sure that this question has the complete information but anyway. Let X and Y be two points in R^n and Z be a point on the line segment joining X and Y such that |X-Z| = a. What will be the value of t if Z=tX + (1-t)Y and 0<t<1? Thanks.--Shahab 16:10, 10 September 2007 (UTC)

Assuming X and Y are at least distance a apart, there is indeed a unique solution for t. To find it, we have |(t-1)X + (1-t)Y|=a, i.e. (1-t)|X-Y|=a, so t=1-a/|X-Y| Algebraist 17:48, 10 September 2007 (UTC)

Modular arithmetic

Whilst attempting a question involving mod 5 arithmetic, I noticed that for a select few numbers

if

$a+b=5\!$

then

$a^{2}\equiv b^{2}{\text{ (mod }}5)$

Is this always true and, if so, can it be generalised to other mods and powers of 'a' and 'b'? asyndeton 19:14, 10 September 2007 (UTC)

As difference of two squares says: $a^{2}-b^{2}=(a+b)\cdot (a-b)$. PrimeHunter 20:02, 10 September 2007 (UTC)

Well think about this, we have $a+b\equiv 0{\pmod {n}}$ so $a\equiv -b{\pmod {n}}$ Can you get the rest of the way to proving the general case for your hypothesis? Donald Hosek 21:30, 10 September 2007 (UTC)

Quadratic formula

• How do i show that this: $x={\frac {-3\pm {\sqrt {(5^{2})-4x-3x2}}}{2x3}}$ = two thirds or minus five thirds
• How do i use the quadratic formula,$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ and these numbers a=1, b= -3, c= -10 to make the answers x=5 x=-2
• How do i use the quadratic formula to solve the equation X²+4x+2=0, whilst leaving the answer in surd form,

I dont understand my homework questions, was wondering if any of you could help me, thank you

For the second question, just plug in your values for a, b, and c. --LarryMac | Talk 20:02, 10 September 2007 (UTC)

• The plus and minus thing confuses me, can somebody help? —Preceding unsigned comment added by 82.36.182.217 (talk) 20:11, 10 September 2007 (UTC)

You want two answers, right? You know that. The answer is either one thing or the other... so what do you think the plus or minus means? Gscshoyru 20:16, 10 September 2007 (UTC)

• Does it mean i have to both add it and subtract it? —Preceding unsigned comment added by 82.36.182.217 (talk) 20:28, 10 September 2007 (UTC)

Does that sound helpful to you? Not sure what you mean. If by both adding and subtracting you end up with two different numbers, then yes. If you mean adding and subtracting from the same number, than no -- adding and then subtracting something from the same number will get you the same thing back, of course. There's a reason it says or, and not and. Basically, in one answer you'll add, the other you'll subtract. Gscshoyru 20:38, 10 September 2007 (UTC)

The symbol "plus or minus" (±) means that you perform both operations (the plus and the minus) (separately and independently).

So, for example, consider this mathematical equation: x = 100 ± 25

In English words, you would say: "x equals one hundred plus or minus twenty-five"

And, that is just a combination of the following two English sentences:

• "x equals one hundred plus twenty-five" OR "x equals one hundred minus twenty-five"

So, back to mathematical symbols: x = 100 ± 25

That statement is simply a short-hand way to combine the following two separate, independent mathematical statements:

• x = 100 + 25 (in other words, x = 125 as ONE answer/result)

OR

• x = 100 - 25 (in other words, x = 75 as ANOTHER answer/result)

So, in other words: if x = 100 ± 25 ... then that means that x is equal to BOTH the value of 125 as well as the value of 75.

Thus, in correct mathematical wording, we would say: "x = 100 ± 25" means that x = 125 OR that x = 75.

Does that make sense to you? (Joseph A. Spadaro 20:41, 10 September 2007 (UTC))

• I think but when i finished my GCSE's i took a gap year, so i forgot lotsa stuff :-( —Preceding unsigned comment added by 82.36.182.217 (talk) 22:45, 10 September 2007 (UTC)

If it's easier to understand, you can eliminate the ± by considering two equations: $x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$ and $x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$. This is all the ± means anyway. Tesseran 02:15, 11 September 2007 (UTC)

Generally YES, BUT... If there are two or more $\pm$ or $\mp$ symbols in the same equation, they do not work independently, so you do not get 2^{no. of symbols} cases. Instead take two solutions, one for all upper signs and the other for all lower signs. For example $100\pm 20\pm 5$ means $100+20+5$ OR $100-20-5$ but NOT $100+20-5$ or $100-20+5$. Another example is a trigonometric identity:
$\tan(\theta \pm \phi )={\frac {\tan \theta \pm \tan \phi }{1\mp \tan \theta \tan \phi }}$
which means either
$\tan(\theta +\phi )={\frac {\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }}$
or
$\tan(\theta -\phi )={\frac {\tan \theta -\tan \phi }{1+\tan \theta \tan \phi }}$
CiaPan 06:58, 11 September 2007 (UTC)

Multiplication of Negative Integers

Let's say that I want to offer an "intuitive" example to illustrate integer multiplication. Example 1: For (+3) x (+5) = +15 ... I can intuit adding 3 dollars (a gain of 3 dollars, thus +3) into my bank account every day for 5 days. Thus, I will have gained 3 dollars 5 times and thus I will have gained 15 dollars (that is, +15). Example 2: For (-2) x (+4) = -8 ... I can intuit subtracting 2 dollars (a loss of 2 dollars, thus -2) from my bank account every day for 4 days. Thus, I will have lost 2 dollars 4 times and thus I will have lost 8 dollars (that is, -8). What is a good, clear, real-life, concrete, intuitive example to illustrate why a negative number multiplied by a negative number yields a positive product? How can I visualize or intuit why, say, (-7) x (-6) = +42 ...? Yes, mathematically, it makes sense and everything falls into place (with regard to negation and opposites and integer rules, etc.). But, can anyone offer an intuitive example (perhaps analogous to my two bank account examples) that clarify and concretely illustrate why a negative times a negative should yield a positive? Thanks. (Joseph A. Spadaro 22:33, 10 September 2007 (UTC))

Perhaps you could say that you know how much money you have in your bank account today, and you know that for the past six days you've been subtracting 7 dollars per day. Then from (–7)×(–6) you know you had 42 dollars more in your bank account then than you do now. Strad 23:03, 10 September 2007 (UTC)

I didn't follow. I got confused. In your example, what exactly does the -6 represent? ... what exactly does the -7 represent? ... and what exactly does the +42 represent? Thanks. (Joseph A. Spadaro 23:15, 10 September 2007 (UTC))

6 days ago. As in, if we plug in -6 for time. Gscshoyru 23:18, 10 September 2007 (UTC)

So, to translate the integer operation of (-6) x (-7) = +42 ... you are saying (in plain English): "Withdrawing seven dollars (-7) for the past/previous six days (-6) yields a balance that initially was 42 dollars greater (+42) than today's balance." Is that it? OK. It would still take an elementary student a bit of doing to wrap their mind around that. (I think.) But, it's OK. Does any one have anything else perhaps more "obvious" / easier to "see"? Thanks. (Joseph A. Spadaro 00:40, 11 September 2007 (UTC))

For each tree that you bring into the Biodome research station, the daily production of oxygen in the sealed facility is increased by 3 litres. Bring in 5 trees and the resultant increase in daily O₂ creation is (+5 trees) × (+3 litres/tree) = +15 litres. Each mule in the Biodome consumes 7 litres of O₂ per day. So if you bring in 4 mules, the resultant increase in daily production of oxygen is (+4 mules) × (-7 litres/mule) = -28 litres, or a net loss of 28 litres/day. But if you remove 6 mules, then the effect on the rate of O₂ creation is (-6 mules) × (-7 litres/mule) = +42 litres, or a net gain of 42 litres every day. PaulTanenbaum 02:37, 11 September 2007 (UTC)

And I might add that removing 3 mules has the exact same effect as introducing 7 trees. Saying that algebraically, (-3 mules) × (- 7 litres/mule) = (+7 trees) × (+3 litres/tree) = +21 litres. Either one of these changes to what organisms are in the Biodome will increase the oxygen level within at the rate of 21 litres/day. PaulTanenbaum 02:51, 11 September 2007 (UTC)

Paul Tanenbaum -- Thanks. That is an example that I think will work well to illustrate the concept that negative times negative yields positive. Can you offer any type of similar analogy to the adding/subtracting money from the bank account examples? Thanks. (Joseph A. Spadaro 06:17, 11 September 2007 (UTC))

Rotations

So, I have a vector in 3d space, the initial point at the origin. The terminal point can be rotated around the world's Z axis and around another vector, which would lie along the X-axis except it is rotated around the Z axis the same amount as the vector; a sort of internal x-axis, wingtip-to-wingtip on an airplane as it rolls.

In truth, this is an object in a game-engine and these rotations are easily achieved using "SetAngle, Z" and "SetAngle, X". For some reason SetAngle X does operate on the object's own X axis and SetAngle Z on the world axis. I have no idea why, but it does make it easier to have one object point toward another.

I have another point sitting somewhere else; I want to transform it along with the original object so it stays in the same angle relative to the object as it rotates. Somewhat like someone sitting on an airplane that can pitch and roll (rolling around an axis perpendicular to the ground).

Trig functions are relatively costly (or at least annoying, I have to approximate them myself) so an answer that minimizes them or answers in terms of sin(or cos)(θ or Φ), (θ and Φ being how much to rotate around the Z and internal-X axes) would help because I can hard-code those. -=BungaDunga 01:50, 11 September 2007 (UTC)

Well, the reason it rotates about the objects "own" X axis is that if you are doing two separate rotations, one has to be done first. In this case, the X rotation is done first (at the time, it is in world space), and then the Z rotation is done afterward, which also rotates that original X axis.

Now, if you want an object B to be attached to object A, you can apply object A's rotation to its position... so keep track of B's position relative to A (position and two angles), and use that combined with A's position/rotation to find the position in world space:

B's world position = A's world position + ( B's relative position rotated by A's rotation )

B's world rotation = B's local rotation rotated by A's world rotation.

The second step is trickier than it looks if you're just learning about rotations; if both X and Z axes are used by A, the rotation isn't going to be easy to specify in terms of X and Z (this is hard to explain in words, it's easier with something physical to show you, but you take a look at Rotation, Euler angles, and Rotation matrix perhaps). If you don't know the math involved yet, you might want to stick to just Z for now. If you are using just the rotation about Z, it is much simpler:

B's world Z angle = B's local Z angle + A's world Z angle.

A note about optimization though: in a game situation, unless you are needing to use "sin" or some other trig functions a thousand or so times in a frame, it's not generally worthwhile trying to speed it up--it might be better to keep the accuracy. Usually it's best to get it working correctly first, and then afterwards try to figure out what can be sped up, rather that trying to do them both at once (it's very often not obvious where the speed bottleneck is going to be). How many objects do you actually have to rotate? - Rainwarrior 06:22, 11 September 2007 (UTC)

Also, unless the rotation is by some special angle (e.g. 90 degrees), you need to do at least two trig calculations (one sine, one cosine, or something equivalent) per rotation. What language are you writing in that you need to write your own trig functions anyway? - Rainwarrior 07:08, 11 September 2007 (UTC)

September 11

Square bracket or small bracket

This is really embarassing but I needed to ask you something for my friend. Can you tell me which bracket to use for 0<=x<=5 is it a. (0, 5) or is it b. [0, 5] ?

I appreciate your swift and honest answer. (this is really an emergency) Thank you.

Regards, Kushal --Click me! write to me 01:24, 11 September 2007 (UTC)

Our article on interval notation has the answer, but I'll just outline it here. When writing an interval, [ means the element is included in the interval, whereas ( means that it is not included. I'll be kind and say the latter expression you wrote is correct. Splintercellguy 01:32, 11 September 2007 (UTC)

Thanks a lot. --Click me! write to me 02:01, 11 September 2007 (UTC)