Mathematics desk
< March 18	<< Feb \| March \| Apr >>	March 20 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 19

Angle bisector proof

In triangle ABC, draw angle bisectors AD and CE, where D is on BC and E is on AB. If angle B is 60 degrees, show that AC=CD+AE.

I've figure out that if the intersection of AD and CE is F, then <CFA and <EFD are 120 degrees and <DFC and <EFA are 60 degrees, but I'm not sure what's next. Thanks, 76.248.244.196 (talk) 01:06, 19 March 2008 (UTC)[reply]

Your four angles around F are all the angles there are around F and you have them adding up to 60+60+30+30 = 180. Surely this should be 360, so some of your figuring has gone astray. Present what working you have done so far and maybe it will become clearer... -- SGBailey (talk) 07:24, 19 March 2008 (UTC)[reply]

I've correct my numbers above, but I'm still not sure where to go next. 76.248.244.196 (talk) 01:36, 20 March 2008 (UTC)[reply]

OK, so you know ALL the angles in the diagram (<ABC=60, <BAD=a, <BCA=c=120-a, <BAD=a/2, <BDA=120-a/2, <BEC=120-c/2=60+a/2 <EXD=360-60-(120-a/2)-(60+a/2)=120). Now you can use the law of cosines. -- SGBailey (talk) 10:25, 20 March 2008 (UTC)[reply]

Algebra question

Given any three natural numbers, show that there are two of them, a and b, that a^3b-b^3a is divisible by thirty.

I have no clue how to prove it, please help. Thanks, 76.248.244.196 (talk) 01:09, 19 March 2008 (UTC)[reply]

Note that "divisible by thirty" is equivalent to "divisible by two, three, and five". I would first factor the polynomial as

a^{3}b-b^{3}a=ab(a+b)(a-b)

. If either a or b is even, then this product is even, and if they are both odd, then (a+b) is even so the product is still even. Therefore

a^{3}b-b^{3}a

is even for every pair of natural numbers. A similar argument can be used to show that it is also a multiple of three (work modulo three). So the polynomials for all three of the pairs are multiples of six, and you can use the pigeonhole principle to show that at the polynomial for at least one of the pairs is also a multiple of five (work modulo guess-what). —Keenan Pepper 01:29, 19 March 2008 (UTC)[reply]

I don't understand the modulo part. Where do I go with this? 76.248.244.196 (talk) 01:52, 20 March 2008 (UTC)[reply]

If you haven't learned modular arithmetic yet, it'll be quite challenging. It's really not that complicated, though. "Working modulo n" just means considering the remainder you get when you divide by n. For n=2, you split numbers up into even and odd. For n=3, you split them up into three categories: multiples of 3, numbers that are one more than a multiple of 3, and numbers that are two more than a multiple of 3. Just consider all the possibilities, and use the fact that if you have n+1 numbers, then at least two of them must have the same remainder when you divide by n (because there are only n possibilities; that's the pigeonhole principle). Good luck! —Keenan Pepper 05:26, 20 March 2008 (UTC)[reply]

Divisibility curiosity

If n is a positive integer that is not divisible by 6, (n^2-1) seems to be always a multiple of 24. Why? Imagine Reason (talk) 02:54, 19 March 2008 (UTC)[reply]

It only works if neither 2 nor 3 divides n. Difference of two squares says a^2-b^2 = (a-b)*(a+b). Can you use that to rewrite n^2-1 and make arguments about where 2 and 3 must divide? PrimeHunter (talk) 03:18, 19 March 2008 (UTC)[reply]

You're right, it's about 2 and 3. So n^2-1 = (n-1)(n+1), which when n is not a multiple of 2 equals a multiple of 4. And since n isn't a multiple of 3, (n-1)(n+1) must be, but that's as far as I got. Imagine Reason (talk) 19:05, 19 March 2008 (UTC)[reply]

You need to think slightly harder to show n²-1 is a multiple of 8, not just of 4. If n is odd, then it is either one more or one less than a multiple of 4. Algebraist 19:34, 19 March 2008 (UTC)[reply]

Oh, two consecutive even numbers must include one that is a multiple of 4. Imagine Reason (talk) 02:10, 20 March 2008 (UTC)[reply]

Gödel's Incompleteness Theorems: Is The Math Reliable?

Many sciences depend on the math to prove something and use it for rigorous study. But Gödel's incompleteness theorems states:

For any consistent formal, computably enumerable theory that proves basic arithmetical truths, an arithmetical statement that is true, but not provable in the theory, can be constructed.1 That is, any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete.

Therefore, I would like to know are all the theories we use (for biology, chemistry, physics, medicine, computer science, etc.) considered to be consistent theories themself? And are all of maths we learn from elementary school to university considered to be reliable and don't contradict each other? - Justin545 (talk) 07:00, 19 March 2008 (UTC)[reply]

What do you mean by "reliable"? I would say the mathematics underlying biology, chemistry, etc is far less likely to be in error than the biology and chemistry themselves. But if you're looking for apodeictic certainty -- the sort of thing that, by its nature, cannot be wrong -- well, sorry, we don't have any of that. In my humble opinion, anyway. We'll settle for being right; we don't have to be completely certain.

Or as the Eagles put it -- "I could be wrong, but I'm not". --Trovatore (talk) 07:18, 19 March 2008 (UTC)[reply]

Math is used as a tool for studying many sciences. If the tool itself is "problematic" or "questionable", the consequences of employing it are very likey to be wrong! "reliable" means "consistent" and "don't contradict". Incompleteness theorems, in other words, states: if every arithmetical statement that is true and is provable in the theory, the theory is inconsistent but it is complete. So what I want to know is: the math we use is either

(1) consistent but not complete, or

(2) complete but not consistent - Justin545 (talk) 07:48, 19 March 2008 (UTC)[reply]

Well, we don't know for certain, but the general view is that we are in the consistent but not complete case, which is really not as bad as it sounds at first. If you know any group theory, consider that there are plenty of facts about groups that cannot be deduced from the axioms for a group alone -- the theory of groups, as given by the most basic group axioms, is not complete. In some sense this is because there are different models, different groups, that all meet those basic axioms, and thus have truths that are not derivable from just those axioms. You can think of arithmetic as being similar, with different models, just with the proviso that, unlike groups, we haven't found models that disagree on any arithmetic facts you or I would generally care about. -- Leland McInnes (talk) 12:11, 19 March 2008 (UTC)[reply]

Could you give an example of what you mean, there? I've done quite a bit of group theory and have never come across something that's true but can't be proven from the axioms of a group (together with ZF). --Tango (talk) 13:39, 19 March 2008 (UTC)[reply]

>> "which is really not as bad as it sounds at first"

It sounds bad to me... since we are not able to justify our math.

>> "there are plenty of facts about groups that cannot be deduced from the axioms for a group alone...the theory of groups is not complete."

I don't know any of group theory, but: Could those set of un-deducible facts themself be considered as axioms? Will group theory be complete if we make those facts axioms? - Justin545 (talk) 02:50, 20 March 2008 (UTC)[reply]

What I mean is that the group axioms don't uniquely define the group, but rather a whole slew of possible objects each of which satisfies the axioms of being a group. Thus there isn't a unique model of "group" specified by the axioms, but rather each and every different group is a different model that satisfies the basic group axioms. There are things that are true of particular groups that you can't deduce from just the group axioms -- you need more information (more axioms in essence) to pin down which group (or class of groups) you are talking about. Thus there are truths that occur in systems that fulfill the group axioms that are not provable from the group axioms alone. Does that make more sense? -- Leland McInnes (talk) 17:26, 19 March 2008 (UTC)[reply]

Right, but arithmetic (and set theory) are quite a different case from group theory. Arithmetic is not the study of models of arithmetic; it's the study of numbers. All models of arithmetic have (copies of) all the true natural numbers, but some of them also have fake natural numbers. The one true Platonic intended model of arithmetic has only the true ones, and none of the fake ones, and is unique up to a canonical isomorphism. There's a limit to what we can find out about the behavior of the true natural numbers from a fixed set of axioms and first-order logic alone. That doesn't mean we have to stop there. --Trovatore (talk) 17:49, 19 March 2008 (UTC)[reply]

Let me insert a response here: essentially, yes. I was going for a loose analogy suggesting that incompleteness isn't really a horrible thing. As to models of arithmetic, there is the question of what the intended model is, and, for the sufficiently messy cases where we can't practically distinguish is from some fake model, whether it even matters. I would liken it (again, an analogy, so don't take it too literally) to science trying to model (in a different sens of the word) some objective reality -- we can't know the objective reality, only our model of it, but as long as we can't tell the difference between our model and the reality (i.e. where our model hasn't been falsified) we may as well consider our model as true. -- Leland McInnes (talk) 20:55, 19 March 2008 (UTC)[reply]

Sure, there are plenty of things that can't be proven using just the axioms of a group, but those things aren't true.

gh=hg\quad \forall g,h\in G

can't be proven just from the group axioms, because it isn't true in general. That's not incompleteness, it's just a false statement. If you want it to be true, you have to add an additional assumption (that the group G be cyclic, say). If the statement can be stated in terms of only the group axioms, and is true, then it can be proven using only the group axioms. If it can't be stated using only those axioms, then it being impossible to prove isn't a case of incompleteness. A framework is incomplete if there are unprovable true statements within that framework. --Tango (talk) 18:06, 19 March 2008 (UTC)[reply]

Tango, I think you have not thought these things through terribly well. At least it isn't clear to me what you mean by a framework, or unprovable but true within a framework. Is a framework a first-order theory, or a model, or what exactly? --Trovatore (talk) 18:19, 19 March 2008 (UTC)[reply]

Let me be a little less Socratic and hopefully more constructive (took me more time to figure out how to say this than it did to ask a question). Let's take a specific example. Peano arithmetic neither proves nor (we suppose) refutes the claim "Peano arithmetic is consistent" (the claim is usually abbreviated Con(PA). Therefore there are models of PA in which Con(PA) is true, and there are models of PA in which Con(PA) is false. So we can make an analogy with your example statement "multiplication is commutative": There are models of group theory (that is, groups) in which "multiplication is commutative" is true, and there are other models of group theory in which it's false.

Here's the big difference: There's no such thing as "the intended group", the group that defines the truth value of "multiplication is commutative in group theory". We're interested in Abelian groups, and we're also interested in non-Abelian groups, and you just have to specify which ones you're talking about.

But Peano arithmetic (we suppose) really is consistent. The models of PA that think otherwise are wrong about that. That's not to say they're not interesting (people devote whole careers to them), but merely by their opinion on this one issue, they prove that they are not the intended model. --Trovatore (talk) 18:35, 19 March 2008 (UTC)[reply]

Ok, I think I understand what you're saying now. I'm not sure I agree, though. Group theory is defined in terms of set theory. Once you've determined a model of set theory, your model of group theory is completely determined (a group is simply a set together with a function - both concepts defined outside of group theory). Is there a (reasonable) model of set theory in which all groups are abelian? --Tango (talk) 18:56, 19 March 2008 (UTC)[reply]

Whoah, we have to be careful here -- the phrase "group theory" is being used in two different ways (my fault, probably). When I say "model of group theory==group", I'm using "group theory" to mean the first-order theory defined by the three axioms (identity existence, existence of two-sided inverses, associativity). That's different of course from "group theory" as in "the study of groups", which is not a formal first-order theory at all. Please re-read my remarks keeping this clarification in mind -- they won't have made any sense at all if you were thinking of "model of group theory" as meaning "model of the study of groups(?)". --Trovatore (talk) 19:02, 19 March 2008 (UTC)[reply]

Ok, but I think my point still stands. Group theory, in that sense, is still built on set theory. Any model of group theory must be a model of set theory, since it has to satisfy ZF plus the 3 axioms of a group. Can you have such model of set theory in which all groups are abelian? For example, set theory provides all kinds of methods of combining sets to produce sets - those method can be used to combine groups and produce other groups. Is there a model in which all such possible combinations are abelian? --Tango (talk) 19:22, 19 March 2008 (UTC)[reply]

No, of course not. It's a theorem of ZF that there exist non-Abelian groups. But you're still mixing things in a confusing way -- whatever a "model of group theory" is, it's certainly not something that "satisfies ZF plus the three axioms of a group"; that doesn't even make sense; the ZF axioms are in a different language from the group axioms. If by "group theory" we mean the three axioms, then "model of group theory" means precisely "group", and does not imply that the model satisfies the ZF axioms. That's the sense in which I was using the phrase "model of group theory". --Trovatore (talk) 19:47, 19 March 2008 (UTC)[reply]

[edit conflict] I think you're still misunderstanding. Sure, it's possible to define groups as a special kind of set in ZF set theory. But that is not what we are talking about here. You are probably confused by the fact that ZF is an immensely more complex system then the meager 3 axioms of groups (to which I will refer as GP). But they are the same thing for this discussion. Each of them is a collection of rules governing a world of objects. A bag of objects can either satisfy these rules, in which case it is called a model of the theory, or not. In the case of ZF, the models are very complicated and hard to point out, but I think Godel's constructible universe is an example of one. For GP, every simple little group is a model, and the elements of the group are the basic objects. In ZF, you can have models that satisfy choice, and models that don't; in GP, you can have models that satisfy commutativity, and models that don't. -- Meni Rosenfeld (talk) 19:55, 19 March 2008 (UTC)[reply]

Ok, I get you. So, if I'm understanding your definition of completeness correctly, a theory being complete is basically equivalent to there only being one model satisfying it? Since, if there are two models of the theory, they must differ in some way and that way gives rise to a statement which is true in one model and not true in the other. --Tango (talk) 20:05, 19 March 2008 (UTC)[reply]

Not quite. It's possible for two models to satisfy all the same first-order statements, but to be nonisomorphic. For example the theory of torsion-free abelian groups is complete, but there are nonisomorphic torsion-free abelian groups. --Trovatore (talk) 20:38, 19 March 2008 (UTC)[reply]

If memory serves, all torsion-free abelian groups are of the form

\mathbb {Z} \times \cdots \times \mathbb {Z}

. I sometimes get a little confused with the orders of logical statements, but is

\exists g\in G,\forall h\in G,\exists n\in \mathbb {N} ,h=g^{n}

not a first order statement satisfied by only one of those groups? --Tango (talk) 21:32, 19 March 2008 (UTC)[reply]

It's not a first-order statement in the language of groups. The language of groups has no function symbol for nth power and no symbol for the set of all natural numbers. --Trovatore (talk) 21:40, 19 March 2008 (UTC)[reply]

Ah, good point. I think we've got there, I have no more questions! Thank you. (Well, I'm sure one will come to me at 3am, but that can wait until tomorrow. ;)) You know... I really do wish my Maths dept. had a proper course on logic, it seems a really major topic to miss out (we did a bit in 1st year, but it was really just half a module on set theory in rather vague terms - the phrase "first order logic" did not appear once). I've done some reading on the subject, I should do some more... --Tango (talk) 21:47, 19 March 2008 (UTC)[reply]

For the record, the rationals Q, the reals R, and the p-adic integers Z_p are all torsion-free abelian groups. Your structure theorem holds for finitely generated abelian groups. Tesseran (talk) 16:03, 21 March 2008 (UTC)[reply]

Excellent point. That doesn't change my (nevertheless incorrect) point, though. --Tango (talk) 16:11, 21 March 2008 (UTC)[reply]

Theories in physics (thought to be the trunk of the science "tree") are not necessarily consistent. Helpfulness of established theories begin and end with orders of magnitude. This is why we have semiclassical physics, and the mesoscopic scale, and why we differentiate "Physics in the Classical Limit," Relativity, and quantum theory. Mac Davis (talk) 08:01, 19 March 2008 (UTC)[reply]

Did I misunderstand Gödel's Incompleteness Theorems or the Incompleteness Theorems is really about distinguishing between classical physics and modern physics? I thought Incompleteness Theorems is just all about the math but not the physics. And Incompleteness Theorems should be able to be applied to all kinds of science, not just physics. I'm not offending, just hope someone can clarify the concept. - Justin545 (talk) 09:23, 19 March 2008 (UTC)[reply]

The incompleteness theorems don't apply particularly well to the kind of math you're probably familiar with. That is, they aren't relevant. They claim that a specific very sensible, very general way of justifying things doesn't work very well in certain contexts. That doesn't mean that what we were trying to justify is wrong, just that we'll have to look somewhere else for confidence in it. It also throws essentially no doubt on actual arithmetic, which deals only with fairly small numbers and can be justified by direct experience and some common sense. Black Carrot (talk) 08:10, 19 March 2008 (UTC)[reply]

Theorems are proved based on axioms. Experience in proving theorems made mathematicians conjecture that every true statement could eventually be proved. This conjecture turned out to be naive. The incompletenes theorem states that the conjecture is not true: the fact that some statement cannot be proved does not imply that the statement is false. The incompleteness theorem does not threaten the reliabilty of mathematics. Bo Jacoby (talk) 11:06, 19 March 2008 (UTC).[reply]

>> "The incompleteness theorem does not threaten the reliabilty of mathematics."

I think you mean the mathematics we use is consistent but not complete since there are still some true statements can not be proven by mathematics, and also you said the mathematics is reliable. But your opinion sounds a bit different with the other. For example, some one said "mathematicians believe that mathematics is consistent". Which means mathematicians "can not prove" mathematics is consistent. - Justin545 (talk) 01:20, 20 March 2008 (UTC)[reply]

Before the incompleteness theorem mathematics was supposed to be consistent and complete. After the incompleteness theorem mathematics is known to be incomplete. The incompleteness theorem does not clarify whether mathematics is consistent or not. So I do not say that mathematics is reliable as a consequence of the incompleteness theorem, nor do I say that mathematics is unreliable as a consequence of the incompleteness theorem. Bo Jacoby (talk) 05:03, 22 March 2008 (UTC).[reply]

In general, mathematicians believe that mathematics (however we may choose to define that term) is consistent. This is mainly because we have not found an inconsistency (a statement P such that both P and not-P can be proved). We can even express this "conjecture" as a (humungously complex) arithmetical statement. Problem is that we also know, thanks to Gödel, that we cannot prove this statement - at least, not without stepping up to some more powerful axiom system, which then leads a "turtles all the way down" type of regression. Bottom line is, most mathematicians say "that's interesting and slightly weird" but they don't lose sleep worrying that mathematics might be inconsistent. On a scale of rational evidence-based confidence, you can put the consistency of mathematics right up at the 99.99% mark. Gandalf61 (talk) 12:21, 19 March 2008 (UTC)[reply]

>> "(a statement P such that both P and not-P can be proved) ... thanks to Gödel, that we cannot prove this statement"

I believe Gödel used "logic" to build his Incompleteness Theorems. But isn't logic a kind of mathematics? If logic is a kind of mathematics, Gödel was using a tool, about which its consistency can not be sure, to prove his Incompleteness Theorems. In other words, Incompleteness Theorems is questionable since the logic itself is questionable. - Justin545 (talk) 01:57, 20 March 2008 (UTC)[reply]

It sounds like we can not use logic to justify the logic itself. It's meaningless! If we are doubt of the logic, we should also doubt of the natural language, such as English, Chinese,... etc., we use, since the logic is just a symbolization of our natural language. We can do inference by the logic and we can also do inference by our language. - Justin545 (talk) 02:15, 20 March 2008 (UTC)[reply]

Learn to live with uncertainty. (Like you have a choice....) --Trovatore (talk) 02:23, 20 March 2008 (UTC)[reply]

Maybe, learn to live with confidence in the logic and the language. - Justin545 (talk) 02:28, 20 March 2008 (UTC)[reply]

Confidence is one thing; fully justified certainty is quite a different thing. You seem to be looking for the latter. You're not going to find it. --Trovatore (talk) 02:33, 20 March 2008 (UTC)[reply]

Knowing the incompleteness theorems somewhat shakes my confidence in the logic and math. I was just trying to find my confidence in them by this discussion. I'm not pursuing the absolute certainty. Imperfection is allowed. - Justin545 (talk) 03:01, 20 March 2008 (UTC)[reply]

Ah, I see. Well, I suppose maybe they should shake your confidence. Just not very much. The take-away message is that mathematics is not really different in kind from the experimental sciences -- you can have confidence in it because it's observed to work, not because it's built up from an unassailable foundation via unassailable steps. The latter idea never really did make sense, even before Gödel -- there was always an infinite regress built into it, as you've noticed. But Gödel does seem to have made people come to terms with this more. --Trovatore (talk) 03:13, 20 March 2008 (UTC)[reply]

Literally, your prior response "I would say the mathematics underlying biology, chemistry, etc is far less likely to be in error than the biology and chemistry themselves." seems to contradict "mathematics is not really different in kind from the experimental sciences". Well, just my picking hobby, I'm not trying to "offend" you again. And excuse my English, I don't know why you use "kind" in italic.

>> "not because it's built up from an unassailable foundation via unassailable steps."

The foundation may not be unassailable, but the stpes is unassailable I think. That's why I like deduction more than induction.

>> "The latter idea never really did make sense"

The latter idea? - Justin545 (talk) 03:40, 20 March 2008 (UTC)[reply]

I said the mathematics was "far less likely" to be in error. That's a difference in degree, not a difference in kind. Your English seems to be pretty good, but I see on your user page that you're not a native speaker -- are you familiar with the phrases "different in degree" and "different in kind"? --Trovatore (talk) 04:15, 20 March 2008 (UTC)[reply]

Sure, I know what are "different in degree" and "different in kind". Maybe I understand your point now. Well, thanks for the compliment. But actually I can not write articles without a dictionary. Besides, my English grammar is questionable. - Justin545 (talk) 05:03, 20 March 2008 (UTC)[reply]

Poincare Video

I tried to access the video at [1], but it doesn't work. Does anyone know where I can find a good copy of it? Black Carrot (talk) 08:14, 19 March 2008 (UTC)[reply]

Have you tried contacting Clay Mathmatics Institute through the contact us link? The page http://www.claymath.org/millennium/Poincare_Conjecture/Lecture_by_Cameron_Gordon_at_UT.ram is still there, but the real audio file at rtsp://zb.msri.org/UT-poincare.rm is no longer available. SayCheeeeeese (talk) 07:55, 23 March 2008 (UTC)[reply]

Diophantine equation

I've got the equation n(2n+1)=a(a-1), both n and a to be positive integers. The solutions (10,15) and (348,493) came by searching, but nothing higher has been found. I'm wondering if the absence of further solutions can be shown from the fact that as a and n increase, their ratio will tend to the irrational root 2?—81.132.237.15 (talk) 13:16, 19 March 2008 (UTC)[reply]

Next two solutions are (11830, 16731) and (401880, 568345). As you conjecture, there is indeed a connection with rational approximations to the square root of 2. Gandalf61 (talk) 15:50, 19 March 2008 (UTC)[reply]

So there will be an infinite number of solutions?—81.132.237.15 (talk) 19:26, 19 March 2008 (UTC)[reply]

Yes, there are an infinite number of solutions. Gandalf61 (talk) 20:03, 19 March 2008 (UTC)[reply]

You can prove this by solving your equation using the quadratic equation (using n or a as the bound variable doesn't matter). Then solve the limit for n/a as a -> infinity:

$n={\frac {{\sqrt {8a^{2}-8a+1}}-1}{4}}$ (solve for n. this solution is used because they must be positive integers)

$\lim _{a\rightarrow \infty }{\frac {{\sqrt {8a^{2}-8a+1}}-1}{4a}}$ (solve the limit for n/a)

$=\lim _{a\rightarrow \infty }{\frac {\sqrt {8a^{2}-8a+1}}{4a}}$ (lim_a->inf (1/4a) is just 0 so we can leave it out)

$={\sqrt {\lim _{a\rightarrow \infty }{\frac {8a^{2}-8a+1}{16a^{2}}}}}$ (put everything inside sqrt and apply power rule in limits)

$={\sqrt {\lim _{a\rightarrow \infty }{\frac {8a^{2}}{16a^{2}}}}}$ (do the same thing, limit of (-8a+1)/(16a^2) is 0 as well so we can leave it out)

$={\sqrt {\frac {1}{2}}}$ (answer)

I think that's right...

--wj32 ^t/c 05:48, 20 March 2008 (UTC)[reply]

Yes, that proves that if there are integer solutions then the ratio n/a approaches 1/sqrt(2). It does not prove that there are infinitely many integer solutions, or indeed that there are any integer solutions at all. You could apply the exactly the same analysis to the equation 2n² = a², which has no integer solutions. The integer solutions to the original equation n(2n+1)=a(a-1) can be derived from the convergents of the continued fraction expansion of sqrt(2). Gandalf61 (talk) 09:41, 20 March 2008 (UTC)[reply]

Complex Analysis

Determine, with motivation, all funtions that is analytic on {z/ |z| < 4}, with f( 0 ) = i and |f( z ) <= 1 on {z/ |z| < 4} —Preceding unsigned comment added by 163.187.240.51 (talk) 13:33, 19 March 2008 (UTC)[reply]

Simple tip: When trying to get other people to do your homework, don't include phrases like with motivation. Also, you'd have better chances if you explained specifically what you've tried so far, and what you're having trouble with. —Keenan Pepper 15:23, 19 March 2008 (UTC)[reply]

Oh, what the heck. I'm feeling nice, so I'll practically give you the answer: Maximum modulus principle. —Keenan Pepper 15:28, 19 March 2008 (UTC)[reply]

>> Thanks very much!!!! It is actually not my homework but anyway thanks! And I will remember all your advise for the future.......

How do you explain the fractional root of a number?

I can't find an answer to this because I'm not sure if I'm using the right terminology. What I mean is, how does one explain what the "1/2"th root of a number is? Or the "2/3"rd root? I understand that the nth root of A is that number x which when multiplied by itself n times yields A, but how do I explain a fractional root in similar terms? Thanks in advance, Narxysus (talk) 17:22, 19 March 2008 (UTC)[reply]

the n th root of a is in fact a^1/n eg sqrt(10) = 10^1/2

Therefor the a/b th root of n is n^b/a. That's a consistent intepretation. So the 1/2 root of x is in fact x squared.83.100.183.180 (talk) 18:24, 19 March 2008 (UTC)[reply]

It just occured to me that you will be able to work this out if you have knowledge of logarithms. Or at least they offer a method of understanding the problem - suggest reading about them.83.100.183.180 (talk) 20:28, 19 March 2008 (UTC)[reply]

It looks like you are trying to do two conceptual steps at once here. First you should be asking what is the fractional power of a number, such as power of 1/2, and we can try to answer that question if you do. Only then should you ask about fractional roots, and the answer is that the 1/2th root of A is that number x such that

x^{1/2}=A

. -- Meni Rosenfeld (talk) 18:29, 19 March 2008 (UTC)[reply]

The (A/B)th root of a number C is what, multiplied by itself A times, gives C^B. Black Carrot (talk) 22:16, 19 March 2008 (UTC)[reply]

Ah, I understand it now - if I want ½th root of A, what I really want is A^(1/(1/2)) which is really A^2. It's a little tricker to explain, but I understand it, anyway. Thanks everyone! Narxysus (talk) 04:49, 20 March 2008 (UTC)[reply]

There's one little catch, it's not always correct to assume that a radical like a 4th root can be expressed as a 1/4th power, this has to do with even roots of negative values, for which the answers are not real numbers, but complex numbers. A math-wiki (talk) 22:51, 20 March 2008 (UTC)[reply]

It's not just even roots of negative values - except for square roots of positive real numbers, some of the roots are always going to be complex.

{\sqrt[{n}]{x}}

is generally taken to mean the positive real root when x is a positive real number. If x isn't a positive real number, the notation doesn't have an obvious meaning, it can mean any of the n roots.

x^{1/n}

is usually taken to be multivalued, with no attempt to define which root you're talking about. --Tango (talk) 02:41, 21 March 2008 (UTC)[reply]

problem setting up the equation

1. In this problem, we will analyze the profit found for sales of decorative tiles. A DEMAND EQUATION shows how much money people would pay for a product depending on how much of that product is available on the open market.

A. Suppose that a market research company finds that s price of p=$20, they would sell x=42 tiles each month. If they lower the price to p=$10, then more people would purchase the tiles, and they can expect to sell x=52 tiles in a month's time. Find the equation of the line for demand equation. Write your answer in the form p=mx+b (hint: write an equation using two points in the form (x,p)) A company's revenue is the amount of money that comes in fromsales, before business costs are subtracted. For a single product, you can find the revenue by multiplying the quantity of the product sold, x, by the demand equation, p.

B Sustitute the result you found from part a into the equation R=xp to find the revenue equation. simplified answer. —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 19:52, 19 March 2008 (UTC)[reply]

Answer effectively here you need to find the line that connects (20,42) and (10,52) ie sales (x) is a function of cost (p) - you can assume the relationship is a straight-line. —Preceding unsigned comment added by 83.100.183.180 (talk) 20:20, 19 March 2008 (UTC)[reply]

To make things a bit more obvious, your looking to start with Point-slope form of a line A math-wiki (talk) 23:42, 21 March 2008 (UTC)[reply]

Graphing calculator peculiarity

Why does the graph of sin(e^x) appear to be a sine wave when graphing on the interval (19.9999, 20}? Zooming in furthur on the function, the graph returns to the chaos I expected to see. I'm wondering why this occurs at all. I'm assuming that my calculator's low resolution (I'm using a TI-84 Plus, which has a screen about 94 pixels wide and 62 pixels high) is partly to blame, But I'm wonding if there is a better, more mathematical reason.

First you should remember that it is (approximately) a sine wave in this interval, only a very high frequency one. If it looks like a low frequency sine wave, the effect is known as aliasing and results from the signal frequency being close to a multiple of the sampling frequency. -- Meni Rosenfeld (talk) 22:51, 19 March 2008 (UTC)[reply]

(ec) It should look like a sine wave on a small enough scale - e^x is a differentiable function, which basically means it looks roughly like a straight line on a small enough scale, so sin(e^x) is going to look roughly like sin(ax+b), which is a sine wave. The chaos you see on a large scale is caused by the low resolution - the calculator just picks a few points to calculate and joins them with straight lines, which doesn't work if the function is changing direction very frequently, which this one does (for large values of x). --Tango (talk) 22:53, 19 March 2008 (UTC)[reply]

It should be emphasized that (19.9999, 20) is not a small enough scale - (19.9999999, 20) is. -- Meni Rosenfeld (talk) 23:02, 19 March 2008 (UTC)[reply]

True, if I've done the sum right (which I didn't think to do before answering), there should be around 8000 periods in that interval, so I doubt 94 pixels is going to be enough to resolve them! --Tango (talk) 23:19, 19 March 2008 (UTC)[reply]