a question about (-1)^x

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Dmitry
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a question about (-1)^x

Postby Dmitry » Sat Jun 02, 2012 4:35 pm UTC

I've been thinking a lot recently about (-1)^x, can somebody clear my mind of these... questions?

1) a^b = c, I would assume that a = b root of c...
but a^(even integer) = c means that a = +- b root of c...
can someone explain more rigorously why? because while it logically makes sense, I wish to understand it better.

2) I cannot fathom the concept of (-1)^x, this would mean e^(xln(-1)) and ln(-1) is not real... O_o

3) Why is it that for any noninteger value n, my calculator refuses to calculate (-1)^n... Each time I try to do it manually, i end up dealing with imaginary numbers, which makes my head hurt.

4)
I also had many difficulties dealing with integrals and derivatives of it...

Many thanks to those who help me get this problem out of my head
Cheers!

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Re: a question about (-1)^x

Postby gmalivuk » Sat Jun 02, 2012 5:04 pm UTC

You need to use complex numbers to do this, since for instance x = 1/2 doesn't give any results in the reals. And then once you're including complex numbers, you have to accept that exponentiation isn't single-valued.
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Re: a question about (-1)^x

Postby eSOANEM » Sat Jun 02, 2012 5:40 pm UTC

You really do need a good understanding of complex numbers in order to answer these questions.

Dmitry wrote:1) a^b = c, I would assume that a = b root of c...
but a^(even integer) = c means that a = +- b root of c...
can someone explain more rigorously why? because while it logically makes sense, I wish to understand it better.


There are (for integer b) always b solutions for a to the equation a^b=c. The reason you only see 1 or 2 of them is that you're only interested in the real solutions but the rest of the solutions are spread evenly around a circle in the complex plane.

If you have an odd number of points spaced evenly on a circle, only one can lie on any given diameter whereas with an even number of points, two can lie on that diameter (but you cannot have just one). If you're taking the root of real numbers, the diameter of the circle you're using is the real number line. These facts about the geometry of circles then manifest as the fact that you only see 1 root for odd roots and two for even roots. The reason the real even roots sum to 0 is that the circle is centred at 0+0*i so, two points opposite each other, must sum to 0.

Interestingly, the sum of all the bth (b=/=1) roots will also be zero. This is because the solutions will be evenly spaced around a circle and so their displacements from the origin will all cancel out.

Dmitry wrote:2) I cannot fathom the concept of (-1)^x, this would mean e^(xln(-1)) and ln(-1) is not real... O_o


Correct.

Euler's identity easily rearranges to e^(i*pi)=-1 suggesting that ln(-1) is, in some sense i*pi (there are actually infinitely many other values, but i*pi is the principle one).

Dmitry wrote:3) Why is it that for any noninteger value n, my calculator refuses to calculate (-1)^n... Each time I try to do it manually, i end up dealing with imaginary numbers, which makes my head hurt.


Using this, you get that (-1)^x=e^(i*pi*x) which, when you throw in Euler's formula, goes to (-1)^x=cos(pi*x)+i*sin(pi*x) (which explains why you can't get real values for non-integer values of x).

Dmitry wrote:4) I also had many difficulties dealing with integrals and derivatives of it...


You should be able to differentiate/integrate it as e^(xln(-1)) by treating it as e^(kx).

Alternatively, as I showed above, (-1)^x=cos(pi*x)+i*sin(pi*x) which may be easier for you to differentiate/integrate.

gmalivuk wrote:you have to accept that exponentiation isn't single-valued.


Huh? The logarithm and root certainly aren't single-valued, but I'm fairly certain exponentiation is.

Generally z=e^(r+i*theta) so z^a=e^(r*a+i*theta*a) as multiplication is single-valued, this leads to z^a=e^(r'+i*theta') which will only have one value (being related by Euler's formula to the trig functions which are definitely single-valued).
Last edited by eSOANEM on Sat Jun 02, 2012 10:46 pm UTC, edited 1 time in total.
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Re: a question about (-1)^x

Postby gmalivuk » Sat Jun 02, 2012 5:43 pm UTC

eSOANEM wrote:
gmalivuk wrote:you have to accept that exponentiation isn't single-valued.
Huh? The logarithm and root certainly aren't single-valued, but I'm fairly certain exponentiation is.

Generally z=e^(r+i*theta) so z^a=e^(r*a+i*theta*a) as multiplication is single-valued, this leads to z^a=e^(r'+i*theta') which will only have one value
Sure, when you start out picking one value for theta, you end up with one value after raising z to a power. But the point is that there's no reason to pick that value for theta rather than any of the infinite other values that differ by 2k*pi for integer k.
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Re: a question about (-1)^x

Postby Radical_Initiator » Sat Jun 02, 2012 6:06 pm UTC

gmalivuk wrote:
eSOANEM wrote:
gmalivuk wrote:you have to accept that exponentiation isn't single-valued.
Huh? The logarithm and root certainly aren't single-valued, but I'm fairly certain exponentiation is.

Generally z=e^(r+i*theta) so z^a=e^(r*a+i*theta*a) as multiplication is single-valued, this leads to z^a=e^(r'+i*theta') which will only have one value
Sure, when you start out picking one value for theta, you end up with one value after raising z to a power. But the point is that there's no reason to pick that value for theta rather than any of the infinite other values that differ by 2k*pi for integer k.


But don't all of the other values for theta (that differ from any given theta by 2*k*pi) lead to the same value of z^a? It seems to lead back to the idea that roots aren't single-valued, but I don't see how that means exponentiation isn't.

My math isn't as advanced, though; I may be missing something.
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Re: a question about (-1)^x

Postby Ben-oni » Sat Jun 02, 2012 6:52 pm UTC

Radical_Initiator wrote:
gmalivuk wrote:
eSOANEM wrote:
gmalivuk wrote:you have to accept that exponentiation isn't single-valued.
Huh? The logarithm and root certainly aren't single-valued, but I'm fairly certain exponentiation is.

Generally z=e^(r+i*theta) so z^a=e^(r*a+i*theta*a) as multiplication is single-valued, this leads to z^a=e^(r'+i*theta') which will only have one value
Sure, when you start out picking one value for theta, you end up with one value after raising z to a power. But the point is that there's no reason to pick that value for theta rather than any of the infinite other values that differ by 2k*pi for integer k.


But don't all of the other values for theta (that differ from any given theta by 2*k*pi) lead to the same value of z^a? It seems to lead back to the idea that roots aren't single-valued, but I don't see how that means exponentiation isn't.

My math isn't as advanced, though; I may be missing something.

No, you're fine. Exponentiation is single valued. Let θ' = θ + 2, for some integral n. Then er'+iθ' = er'+i(θ + 2) = er'+iθe2inπ = er'+iθ.

However, over the complex numbers, exponentiation stops being one-to-one, which is important to internalize.

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Re: a question about (-1)^x

Postby Twelfthroot » Sat Jun 02, 2012 7:30 pm UTC

Ben-oni wrote:No, you're fine. Exponentiation is single valued. Let θ' = θ + 2nπ, for some integral n. Then er'+iθ' = er'+i(θ + 2nπ) = er'+iθe2inπ = er'+iθ.

Unless I'm mistaken, the subject at hand is raising arbitrary complex numbers to powers, not specifically the exponential function. For complex w in general, w^z := exp(z log w), which is multi-valued because the complex logarithm is multi-valued.

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Re: a question about (-1)^x

Postby gmalivuk » Sat Jun 02, 2012 7:33 pm UTC

Yeah, I'm not talking about the exponential as in exp(z), I'm talking about the operation of exponentiation, i.e. raising a number to a power. 1^(1/n) has n values, ergo exponentiation is not single-valued.
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Re: a question about (-1)^x

Postby eSOANEM » Sat Jun 02, 2012 10:46 pm UTC

Yeah, I didn't think that through.

I was thinking "but if I raise z^asflkjsdaf I get a single z" and didn't really think about how that works for |asflkjsdaf|<1 because the fact that theta is modulus 2*pi and therefore doesn't not generally have single-valued multiples.

I then handwaved the the z^1/a case in my head along the lines it not actually being the ath root of z similarly to how their is just one primitive of f(x) but a family of integrals to f(x).

So yeah, I didn't quite think that through, Sorry.
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Re: a question about (-1)^x

Postby TwistedBraid » Sun Jun 03, 2012 3:49 am UTC

I think it's good to start as simple as we possibly can, and only get more complicated as we need to. Let's really think about the equation q = b^x. The only thing we really know from the start is how to calculate q if b is a real number and x is a positive integer: multiply b times itself, x times. So far so good. This immediately implies that b^(x+y) = b^x*b^y, (b^x)^y = b^(xy), and (bc)^x = b^x*c^x. These laws are really pretty and they basically uniquely determine what ^ means. So if I'm going to extend my definition of ^ to other numbers, what I care about most is making sure these laws still work. Phrased differently, I'm going to assume these equations work when x and y aren't positive integers, and deduce how things should be defined from there.

Well, b = b^1 = b^(1+0) = b^1*b^0 = b*b^0, so I know b^0 = 1 for any (non-zero) b. Then 1 = b^0 = b^(x-x) = b^x*b^(-x), so we know b^(-x) = 1/(b^x). Now we know what b^x equals when b is any non-zero real and x is any integer. Nothing interesting yet really, but now let's try x = 1/2. b = b^1 = b^(1/2*2) = (b^(1/2))^2. So whatever q = b^(1/2) is, it ought to satisfy q^2 = b. That's the only thing we know about it. This tells us that when b is negative, q can't possibly be a real number. Let's ignore that for a second, and focus on the case when b is positive. In that case the equation q^2 = b has two solutions, one positive and one negative. (The fact that it has a solution relies on some inherent properties of the real numbers: notice this is false if b=3 and we wanted q to be rational. I don't wanna go into axioms/constructions of the real numbers but roughly the idea is this: any equation which has arbitrarily accurate approximate solutions must actually have a real solution.)

There's no way to get around the fact that this equation q^2 = b has multiple solutions, so let's drop math and talk philosophy. One possible way to view things is that b^(1/2) is not actually a number, it is in fact a pair of numbers. Nothing is wrong with this in principle, but it makes me really uncomfortable as a mathematician. Now all of our equations aren't going to be written in numbers any more, they're going to be sets of numbers. All of a sudden everything gets way harder to even think about. So instead, we'll take a cheap way out. b^(1/2), by definition, is the positive solution to q^2 = b. It kinda feels like we're cheating but there's no way around the problem of multiple solutions so we go with what works.

Now, whenever b is a positive real number, we can define b^(1/n) to be the unique positive solution of q^n = b. We can then define b^(m/n) = (b^(1/m))^n. We're doing pretty good, now we know what b^x is for any positive real b and any rational x. Then we can extend things further to every real x: if we want to calculate 2^pi we can calculate 2^3.1, 2^3.14, 2^3.141, 2^3.1415, ..., eventually this gets closer and closer to some real number and we define 2^pi to be whatever number this approaches. (Again I don't want to go too deep into why this works, but basically "continuity" is the answer.) Our three fundamental laws still hold for any real numbers x and y, so we're pretty happy.

But when b is negative we haven't made much progress...we only know what b^x is when x is an integer. Since math has failed us again it's back to philosophy. I'm going to just make up a number called "i" and its defining property is that i^2 = -1. Obviously i can't be a real number but I don't have a problem with that. Now we have a whole new set of numbers x + y*i where x and y are real numbers, let's call them the complex numbers. We can define 1/(x + yi) = (x - yi)/(x^2+y^2), and it's a good definition because (x + yi) * 1/(x + yi) = 1. So we know how to add, subtract, multiply, and divide complex numbers, which is a really nice thing to know.

So know what is (-1)^(1/2)? Even with the complex numbers, we have the same problem we did before: i^2 = -1 but (-i)^2 = -1 also. And our old cheater ways don't work so well: saying i is a "positive imaginary number" is a bad thing to say, because i*i*i = -i. And if (positive)*(positive)*(positive) = negative then I have no idea what "positive" and "negative" even means. We've made a bit of progress, we at least have candidates for what (-1)^(1/2) could mean, but this whole non-uniqueness business is really biting us in the ass. (-1)^(1/3) has three possible definitions, (-1)^(1/10) has ten possible definitions, and (-1)^x when x is non-rational? Things are starting to get really murky, so let's look somewhere else and maybe it'll help us come back to this problem.

The case when b was positive seemed really nice before, so let's go back to it. What happens if we try something like 2^i? And now we come to Euler's magical formula: e^(xi) = cos(x) + i*sin(x). We can take this as a definition, but just like everything above we have to prove our definition is a good one. There's a number of proofs but my personal favorite is to calculate that the derivative of (cos(x) + i*sin(x))/e^(ix) is zero. Thus this function is a constant and by letting x = 0 we see that this constant is 1.

Back to b = -1. By the magic formula e^(pi*i) = -1 so we can say (-1)^x = (e^(pi*i))^x = e^(x*pi*i) = cos(pi*x) + i*sin(pi*x). That's a fairly nice definition, it gives a unique, simple to calculate complex number for any real x. Note that if x is an integer it gives the correct answer. From here, everything is easy to calculate, derivatives and integrals are easy.

But this is an incomplete picture, and the story gets so much better. With the magical formula, all of our non-uniqueness can be described by the non-uniqueness of solutions to trig functions. e^(pi*i) = -1, but e^(3pi*i) = -1 also. Really what we want to say is that for any integer n, e^((1+2n)pi*i) = -1. Now we've captured all the solutions: (-1)^x = cos((1+2n)pi*x) + i*sin((1+2n)pi*x). For x = 1/2, this gives both solutions. For x = 1/10 this gives all 10 solutions, no more, no less. And when x is a non-rational number? This gives us infinitely many solutions, forming a dense set of points in the unit circle.

You can do the exact same trick to get a complete picture for any (non-zero) complex b, but b = -1 is basically as interesting as the general case. It's all really pretty I think, enough to write a huge wall of text about anyways.


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