Simple maths question? - Page 1
Forum section: electronics / general

author: Things
author_rank: Regular Contributor
timestamp: March 14, 2013, 01:12:04 pm
content: Hey guys, let me start off by saying I suck at maths pretty bad, and this is probably just some really simple index law I'm overseeing, but I have this:And it apparently = 3.But from what I can see, it'd require you to cube root -27, which isn't possible being a negative number.I've tried putting this into wolframalpha and it doesn't really interpret correctly, so hoping someone here might be able to help me out?Dan

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: March 14, 2013, 01:22:19 pm
content: Quotethe cube root of -27 is -3

author: ee851
author_rank: Regular Contributor
timestamp: March 14, 2013, 01:22:33 pm
content: Not the cube. The cube root. You know a negative number times a negative number yields a positive number. Hence a negative number times a negative number times a negative number yields a negative number.3*3*3 = 27(-3)*(-3)*(-3)=-27Hence (-27)^1/3 = -3Try again.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: March 14, 2013, 01:24:10 pm
content: must've beaten you by milliseconds

author: Things
author_rank: Regular Contributor
timestamp: March 14, 2013, 01:26:02 pm
content: Hah, of course! That's what happens when you study at midnight. CheersLooks good to me

author: qno
author_rank: Frequent Contributor
timestamp: March 14, 2013, 04:18:41 pm
content: how do you make these nice formula's on a PC

author: McMonster
author_rank: Frequent Contributor
timestamp: March 14, 2013, 06:07:06 pm
content: Quote from: qno on March 14, 2013, 04:18:41 pmI would recommend installing wxMaxima and checking some literature on this application. Great piece of mathematical software.

author: ddavidebor
author_rank: Super Contributor
timestamp: March 14, 2013, 07:07:01 pm
content: Qno you should search "latex" or "daumn" equation editor

author: Things
author_rank: Regular Contributor
timestamp: March 15, 2013, 01:14:12 am
content: Or you can just make them in MS Word's "Equation editor". It's a bit clunky but it gets the job done if writing formulas isn't your day jobJust watch out though, as sometimes it can try correct your layout. For example in my image there, the first line it looks like the number is actually 9 and -3/2, but it's actually 9^-3/2, as seen on the 2nd line. Gets a bit irritating at times.

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2013, 03:11:23 am
content: Quote from: Things on March 14, 2013, 01:12:04 pmThere's actually no need to take the cube root of a negative number if you apply the powers in a suitable order:

author: Things
author_rank: Regular Contributor
timestamp: March 15, 2013, 03:38:03 am
content: Hmm, so basically applying the power laws in reverse. Didn't really think about it that way, although I think generally the answer is reached in around the same number of steps anyway

author: onlooker
author_rank: Frequent Contributor
timestamp: March 15, 2013, 05:11:06 pm
content: None of the answers are complete. The key is to understand what is (-1)-2/3.If one is not careful, one can easily find oneself asserting -1=1 as in the derivation,-1 = (-1)1= (-1)2/2= [(-1)2]1/2= 11/2= 1Now what is (-1)-2/3? Some complex analysis is in order.(-1)-2/3= [ei*(2*n+1)*PI](-2/3)= e-i*PI*(4*n+2)/3= cos(PI*(4*n+2)/3) - i*sin(PI*(4*n+2)/3)where n can be any integer.This means (-1)-2/3is not single valued, but can be any one of a multiple (can be infinite in other cases) number of values just as 11/2is double valued (+1 and -1).For the simplest case of n=0,(-1)-2/3= -0.5 + i*sqrt(3)/2For the case of n=1,(-1)-2/3= 1...

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2013, 06:32:50 pm
content: Quote from: onlooker on March 15, 2013, 05:11:06 pmIf you consider the implicit expectation that only real answers are required then the answers above are quite complete.Consider that (-1)-2/3may be expressed as (-1)(-1)x(2)/(3). We have:(-1)-1= 1/(-1) = -1(-1)2= 1(-1)1/3= -1There is a single real answer in every case.At what educational level are complex numbers usually introduced? I imagine it is at college level for most, and then only in a math or engineering program. The opening question in this thread is clearly more like a 9thor 10thgrade high school question.

author: Anquietas
author_rank: Regular Contributor
timestamp: March 15, 2013, 07:06:01 pm
content: Quote from: qno on March 14, 2013, 04:18:41 pmLaTeX. All professional documents in STEM should be typeset in LaTeX. I can't stand the looks of documents done in Word or similar WYSIWYG editors.QuoteThe concept of complex numbers is usually introduced in high-school, at least if you go to a high-school that focuses on STEM subjects. I didn't, so I learned about complex numbers in college. Complex analysis (of one variable) is done at university, usually during the second year for all undergraduate students. In mainland EU at least.QuoteThis is incorrect. You are not using the domain of definition of the complex root properly. There is no need to pay particular attention to this, if you don't know about the basics of complex numbers, and if you do you know what to pay attention to, so in my opinion this is a non-issue and given the simplicity of the exercise, it is clearly not what is expected of the student.In a sense you are right that there is a certain multi-valuedness at play here, but this only leads to a relation, not a function, and in order to define a function you introduce more machinery that eventually makes your equations invalid - as it should.QuoteIt is not. Why would you think that the square root is not a well-defined map on non-negative real numbers? In any case, if you want to be overly formal, you should start by concretizing branch cuts and/or Riemann surfaces.

author: ddavidebor
author_rank: Super Contributor
timestamp: March 15, 2013, 08:23:10 pm
content: if someone of you need to do advanced math, just use wolfram|alphai think it's the most userful sites in the world after google and wikipedia. (p**n sites excluded)

author: mrflibble
author_rank: Super Contributor
timestamp: March 15, 2013, 08:37:57 pm
content: Quote from: ddavidebor on March 14, 2013, 07:07:01 pm+1 for latex. Latex rules for this sort of thing, as well as for more complicated formulas. There's even a mediawiki plugin for all your math nerd documentation needs. XD

author: ddavidebor
author_rank: Super Contributor
timestamp: March 15, 2013, 08:43:54 pm
content: yeah i use "daum equation editor" for mac, but it write code in latex

author: rr100
author_rank: Frequent Contributor
timestamp: March 15, 2013, 09:49:34 pm
content: Quote from: IanB on March 15, 2013, 06:32:50 pmNo, if you stick to reals the answers are just wrong, the correct answer being that for negative numbers you can only have integer powers and the rest are undefined (or whatever the formal word is for "you just can't do that"). Otherwise as it was already correctly pointed out, if you use precisely the same technique used by Things and IanB in the nice pictures posted before you can as well prove that -1 = 1:Quote from: onlooker on March 15, 2013, 05:11:06 pmAnd that can't be right, RIGHT?

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2013, 10:08:02 pm
content: Quote from: rr100 on March 15, 2013, 09:49:34 pmYou really can't.In the case above, someone is trying to use the fact that the equationx2= 1has two real solutions, x = 1 and x = -1.However, this does not occur in the original problem. In that case the negative number has a1/3power, not a1/2power. The analogous equation would bex3= 1for which only one real solution exists, x = 1.On top of that, Anquietas knows something about mathematics, so I don't think it would be wise to argue...

author: rr100
author_rank: Frequent Contributor
timestamp: March 15, 2013, 10:23:59 pm
content: There is no equation there, just some simple calculations following precisely the same rules you used. Plus the fact that 1 = 2/2 which I hope is clear and of course we used your (false) assumption that you can just raise negative numbers to fractional powers without all hell breaking lose.If you think there is some other trick involved, except the one of allowing negatives to be raised to fractional powers just say so, preferably pointing out precisely which one of the 5 "equals" is wrong. One MUST be wrong, otherwise -1 =1, right?As for Anquietas, no offence but math is the last remaining field where argumentum ad verecundiam is really unwanted, unneeded and unwelcomed.

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2013, 10:33:43 pm
content: Quote from: rr100 on March 15, 2013, 10:23:59 pmWhat then can we say about Latin?

author: Anquietas
author_rank: Regular Contributor
timestamp: March 15, 2013, 10:41:30 pm
content: For the sake of argument, let's assume that our point of view is that of "multi-valued functions" (this terminology really should be avoided, because single-valuedness is one of the major properties that singles out functions among general relations in set theory) and we know about the rudiments of complex numbers. How do you then, by onlooker's comment, interpret (-1)^(-2/3)?Note that we want a well-defined object. So (-1)^(-2/3) stands for the set { [exp(i*(2*n+1)*PI)]^(-2/3) | n \in N }. And suddenly onlooker's equations are all correct, because = denotes equality on the set level and the symbols stand for their respective images under the multi-valued map. Therefore 1 = -1 is perfectly valid on the set level. So there is nothing to discuss. Obviously nobody really does it that way in vivo because of ambiguity. On the other hand, if you require the symbols to denote the actual possible values under the multi-map, you can't put an = sign there any more and the equations become invalid.In fact, in order to be able to speak of maps at all, you at least have to define the underlying relation, which requires you to specify domain and codomain. None of this has been done by onlooker, rendering his initial motivation to complete the other responses rather moot. If you want to speak of well-defined functions in this particular context of complex analysis, you are going to have to invent branch cuts at least, or ideally Riemann surfaces.QuoteI think we speak of the same thing.

author: Anquietas
author_rank: Regular Contributor
timestamp: March 15, 2013, 10:49:01 pm
content: My point is that the domain of definition is handwavingly assumed to be the non-negative real numbers. Everything else would require three years of undergraduate calculus to define, discuss and practice in order to understand. This would simply go ridiculously beyond a course that requires you to algebraically simplify an expression like the one posted by the OP.

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2013, 11:04:46 pm
content: Quote from: Things on March 14, 2013, 01:12:04 pmHere's what WolframAlpha gives, but it may not be as much help as you hoped for

author: rr100
author_rank: Frequent Contributor
timestamp: March 16, 2013, 07:34:23 am
content: Quote from: Anquietas on March 15, 2013, 10:49:01 pmThe restriction to non-negative numbers is as arbitrary as many other things in math are; but the important point is that it's a necessary restriction and can be nicely explained without going beyond the undergraduate level. The explanation being based on the "-1=1" (or any similar) manipulations with an argument like: if we allow negatives see what happens so we can't do that.And of course with this arbitrary (but absolutely needed) restriction to non-negatives you just can't do anything with the expression from the first post, as -27 is obviously negative.

author: Anquietas
author_rank: Regular Contributor
timestamp: March 16, 2013, 10:38:27 am
content: Quote from: rr100 on March 16, 2013, 07:34:23 amNothing is arbitrary in mathematics. I wonder how you got that impression.QuoteThis is incorrect, because onlooker's "derivation" of -1 = 1 uses invalid manipulations. -1 and 1 are on different branches of the ^(1/2) function, you can't put an = sign there.QuoteContrary to the ^(1/2) that onlooker used to "show" -1 = 1, -27 is raised to the power ^(1/3), which denotes the unique real solution to x^3 = -27, as remarked by IanB.

author: rr100
author_rank: Frequent Contributor
timestamp: March 16, 2013, 11:52:13 am
content: Quote from: Anquietas on March 16, 2013, 10:38:27 amI don't know how you got the impression that I've got the impression that something is arbitrary - or not - in mathematics. I mentioned just that the restriction (also mentioned by you but somehow for whatever reason not followed) to non-negatives is nothing out of the ordinary, as arbitrary or "handwavingy" as many other things.Quote from: Anquietas on March 16, 2013, 10:38:27 amIt is just as valid or invalid as the manipulation (-27)2/3= [(-27)2]1/3(which was used earlier). Just the numbers are different, -1 instead of -27, 2 is still 2 and 2 instead of 3.Quote from: Anquietas on March 16, 2013, 10:38:27 amIn this context there is no such thing as-27 raised to any fraction, pure and simple, unless you define this notation as a totally different beast, for example you say ^(1/3) is just another way of saying "cube root". Apart from the fact that you are hijacking standard notation and using it from something else it would work, for whatever the cube root would work. BUT if you adopt that notation then the consequence is that 1/3 used in this context is not the fraction/rational/real number we're used to, is just a part of the notation. If it would be then we should be able to replace it with 2/6 (=1/3) and we can't do that in any meaningful way. If we say (-27)^(2/6) is just the 6-th root of 27^2 then that's 3 and it should be -3 to be consistent (equal with cube root of -27). Or if you say (-27)^(2/6) is the square of the 6-th root of -27 then it's even worse as there is no such thing in this context (even root from negative number).

author: IanB
author_rank: Super Contributor
timestamp: March 16, 2013, 02:51:28 pm
content: Quote from: rr100 on March 16, 2013, 11:52:13 amOf course we can.If we write,x = (-27)2/6then,x6= (-27)2and we find that x = -3 satisfies that relation, consistent with the way that x satisfies the original relationx3= -27We know that when we replace1/3with2/6we have introduced more possible solutions and we have to be careful with that, but when we account for the consequences of what we have done we still have meaning.

author: rr100
author_rank: Frequent Contributor
timestamp: March 16, 2013, 03:27:19 pm
content: Quote from: IanB on March 16, 2013, 02:51:28 pmI don't understand why you write as an equation but whatever, let's go with it. Keep in mind that those equations aren't equivalent, when you raise to the power 6 both sides of the equation the implication is only one-sided. As in:x = (-27)2/6=> x6= (-27)2And you find x=-3 satisfies x6= (-27)2but this says NOTHING about whether it satisfies x = (-27)2/6, it might or might not be the case.Now if you found some number that DOES NOT satisfy x6= (-27)2, for example 37, yes, it would be correct to say you proved 37 does not satisfy x = (-27)2/6(because if it would then it should satisfy as well x6= (-27)2and it doesn't).Quote from: IanB on March 16, 2013, 02:51:28 pmAnd THAT is the whole point, the moment you say we need to be careful when replacing1/3with2/6because you get something with1/3and some other thing or set of things with2/6. But it's not the moment to "be careful", it is the moment to realize everything is broken and go back to the drawing board and rethink the assumptions.

author: IanB
author_rank: Super Contributor
timestamp: March 16, 2013, 04:40:55 pm
content: Quote from: rr100 on March 16, 2013, 03:27:19 pmBecause that's how we do mathematics.If we are considering, for example, how to evaluate (-27)1/3(cube root of -27) then we are in fact looking for a value of x that satisfies the relation:x3= -27(We are looking for a root of that equation, which is why it is called the cube root--it is the root of the cube.)Since there are actually three roots to that equation we should introduce a rule governing which one to pick when we ask for "the" cube root (the principal root), and once we have such a rule ambiguity is removed and consistency is preserved. (The usual rule in complex analysis would result in 1.5 + 2.598ibeing picked.)QuoteActually that is not correct. Bearing in mind that 2/6 is an improper fraction, and we should more properly write,x = ((-27)2)1/6then any x which is a solution tox6= (-27)2is also a solution tox = ((-27)2)1/6

author: rr100
author_rank: Frequent Contributor
timestamp: March 16, 2013, 06:23:51 pm
content: Quote from: IanB on March 16, 2013, 04:40:55 pmIt doesn't matter that many operations we have (including division, subtraction, n-roots, logarithms and many more) are actually coming from solving some equations. Once you have the operations defined you don't need to bring any equation into play. Unless you want to play tricks like "ok, this isn't something defined but I'll call it x anyway". Then even if it's a beast outside any rule I'll just go ahead and apply the rules I know for well-formed objects and not take into account any contradiction I reach.Quote from: IanB on March 16, 2013, 04:40:55 pmYou aretotally wrongand this is what I meant by the implication that goes only in one direction, precisely NOT the one you need. If x=3 (this is basically the expression below) of course x6= 36. But the converse doesn't hold, -3 is a solution to the equation above but NOT one for the equation below.

author: IanB
author_rank: Super Contributor
timestamp: March 16, 2013, 06:34:26 pm
content: Quote from: IanB on March 16, 2013, 04:40:55 pmQuote from: rr100 on March 16, 2013, 06:23:51 pmI think you will find that it is.

author: mrflibble
author_rank: Super Contributor
timestamp: March 16, 2013, 07:11:17 pm
content: Looooool! You guys still at it? If you just rewrite the original question in euler notation, then you can reduce the amount of miscommunication.Unless of course the main objective is a math pedantics pissing match, in which case carry on. XD

author: Things
author_rank: Regular Contributor
timestamp: March 18, 2013, 01:23:13 am
content: Bloody hell, I'll be back in a couple years time to re-read this when I have a clue what you're all on about