EEVblog Electronics Community Forum
General => General Technical Chat => Topic started by: Things on March 14, 2013, 01:12:04 pm
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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:
(https://dl.dropbox.com/u/203420/Equation2.png)
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
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it'd require you to cube root -27
the cube root of -27 is -3
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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)=-27
Hence (-27)^1/3 = -3
Try again.
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must've beaten you by milliseconds
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Hah, of course! That's what happens when you study at midnight. Cheers :D
(https://dl.dropbox.com/u/203420/Equation3.png)
Looks good to me :D
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how do you make these nice formula's on a PC
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how do you make these nice formula's on a PC
I would recommend installing wxMaxima and checking some literature on this application. Great piece of mathematical software.
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Qno you should search "latex" or "daumn" equation editor
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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 job :)
Just 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.
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But from what I can see, it'd require you to cube root -27, which isn't possible being a negative number.
There's actually no need to take the cube root of a negative number if you apply the powers in a suitable order:
(https://www.eevblog.com/forum/chat/simple-maths-question/?action=dlattach;attach=41300;image)
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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 :)
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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 = 1
Now 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)-2/3 is not single valued, but can be any one of a multiple (can be infinite in other cases) number of values just as 11/2 is double valued (+1 and -1).
For the simplest case of n=0,
(-1)-2/3 = -0.5 + i*sqrt(3)/2
For the case of n=1,
(-1)-2/3 = 1
...
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None of the answers are complete. The key is to understand what is (-1)-2/3.
If you consider the implicit expectation that only real answers are required then the answers above are quite complete.
Consider that (-1)-2/3 may be expressed as (-1)(-1)x(2)/(3). We have:
(-1)-1 = 1/(-1) = -1
(-1)2 = 1
(-1)1/3 = -1
There 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 9th or 10th grade high school question.
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how do you make these nice formula's on a PC
LaTeX. 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.
At what educational level are complex numbers usually introduced?
The 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.
-1 = (-1)^1 = (-1)^2/2 = [(-1)^2]^1/2 = 1^1/2 = 1
This 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.
1^1/2 is double valued (+1 and -1).
It 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.
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if someone of you need to do advanced math, just use wolfram|alpha
i think it's the most userful sites in the world after google and wikipedia. (p**n sites excluded :-DD)
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Qno you should search "latex" or "daumn" equation editor
+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
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yeah i use "daum equation editor" for mac, but it write code in latex
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If you consider the implicit expectation that only real answers are required then the answers above are quite complete.
No, 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:
-1 = (-1)1 = (-1)2/2 = [(-1)2]1/2 = 11/2 = 1
And that can't be right, RIGHT?
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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:
-1 = (-1)1 = (-1)2/2 = [(-1)2]1/2 = 11/2 = 1
You really can't.
In the case above, someone is trying to use the fact that the equation
x2 = 1
has two real solutions, x = 1 and x = -1.
However, this does not occur in the original problem. In that case the negative number has a 1/3 power, not a 1/2 power. The analogous equation would be
x3 = 1
for 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...
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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.
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math is the last remaining field where argumentum ad verecundiam is really unwanted, unneeded and unwelcomed.
What then can we say about Latin?
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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.
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.
I think we speak of the same thing.
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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.
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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? :)
Here's what WolframAlpha gives, but it may not be as much help as you hoped for ;)
(https://www.eevblog.com/forum/chat/simple-maths-question/?action=dlattach;attach=41369;image)
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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.
The 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.
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The restriction to non-negative numbers is as arbitrary as many other things in math are;
Nothing is arbitrary in mathematics. I wonder how you got that impression.
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.
This 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.
you just can't do anything with the expression from the first post, as -27 is obviously negative.
Contrary 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.
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The restriction to non-negative numbers is as arbitrary as many other things in math are;
Nothing is arbitrary in mathematics. I wonder how you got that impression.
I 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.
This 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.
It 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.
Contrary 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.
In 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).
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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.
Of course we can.
If we write,
x = (-27)2/6
then,
x6 = (-27)2
and we find that x = -3 satisfies that relation, consistent with the way that x satisfies the original relation
x3 = -27
We know that when we replace 1/3 with 2/6 we 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.
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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.
Of course we can.
If we write,
x = (-27)2/6
then,
x6 = (-27)2
and we find that x = -3 satisfies that relation, consistent with the way that x satisfies the original relation
x3 = -27
I 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)2
And you find x=-3 satisfies x6 = (-27)2 but 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)2 and it doesn't).
We know that when we replace 1/3 with 2/6 we 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.
And THAT is the whole point, the moment you say we need to be careful when replacing 1/3 with 2/6 because you get something with 1/3 and some other thing or set of things with 2/6. 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.
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I don't understand why you write as an equation
Because 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 + 2.598i being picked.)
And you find x=-3 satisfies x6 = (-27)2 but this says NOTHING about whether it satisfies x = (-27)2/6, it might or might not be the case.
Actually that is not correct. Bearing in mind that 2/6 is an improper fraction, and we should more properly write,
x = ((-27)2)1/6
then any x which is a solution to
x6 = (-27)2
is also a solution to
x = ((-27)2)1/6
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Because that's how we do mathematics.
It 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.
any x which is a solution to
x6 = (-27)2
is also a solution to
x = ((-27)2)1/6
You are totally wrong and 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.
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any x which is a solution to
x6 = (-27)2
is also a solution to
x = ((-27)2)1/6
-3 is a solution to the equation above but NOT one for the equation below.
I think you will find that it is.
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Looooool! You guys still at it? If you just rewrite the original question in euler notation, then you can reduce the amount of miscommunication. :P
Unless of course the main objective is a math pedantics pissing match, in which case carry on. XD
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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 :D