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>The whole point of inventing the complex numbers was to have a set of numbers for which any polynomial has a root Think about what this implies. You have an
by ActorNightly 8mo ago
>The whole point of inventing the complex numbers was to have a set of numbers for which any polynomial has a root
Think about what this implies.
You have an operation, like exponentiation, that has limits. Something squared can never be negative if you are talking about any real number.
In terms of Sets, you essentially have an operation that produces results only in a finite subset of the overall set. And so the inverse of that operation, when applied to the complement of that finite subset, is undefined.
However you can introduce another (ordered) set in complement to your original set and combine them to form a new set, with operations that define how you move around the values of those sets. So in the case of imaginary numbers, you basically redefine all your reals as "real number + 0 i". And now you have a way to apply that inverse operation to the complement of the finite subset, which means you can get answers to the roots of the polynomial.
And in defining the operation of multiplication, you essentially define a way to move around the 2 dimensional set now. And moving around 2 dimensions is exactly the same thing as rotation+scaling.
And note that when you say sqrt(-1) = i, you basically assume that the complex plane is 2d. There is nothing that is stopping you from making a complex plane 3d or 4d or nd. So sqrt(-1) can also be j, or it can be k. To know what it is, you have to specify the axis of the plane when you specify the sqrt operation, which again, brings it back to the concept of rotations.
And thats my whole point, there is nothing special about i, its simply just a construct that bakes in rotations through any way you wanna define it.
>our whole point about Taylor series is also wrong, as Taylor series are not approximations, they are actually equal to the original function if you take their infinite limit
Looking back at what I wrote, I worded it very poorly.
I don't have a problem with any math involved, not trying to say that Eulers identity is not valid.
What Im trying to say is that all the definitions sort of assume that if you have some operation that you can do on real numbers, and have a result, if you just plug in a complex value, it all works out, so people think that i behaves like a number. I personally don't think that this is the case, specifically about i behaving like a number just because those results work out.
For example, even without Taylor series, you can prove Eulers identity using the limit formula for e(x). The idea is that you have (1+xi/n)^n as n goes to infinity, but because you baked in the rotation as a multiplication in the definition, all you are doing is starting at 1+0i and doing smaller and smaller rotations to get to some value, and the limit of that value is essentially the unit vector rotated by a certain angle. So naturally the cos and sin equivalence arises.
My issue is that the limit equation for e, in the case of the reals, take e x times in multiplication and then compute the limit equation, and you get equivalence. But in the case of the complex, you don't really have any idea what it takes something to ith power, but you can compute the limit equation, and so you end up with a definition of what it means to take something to the ith power.
My argument is that its not really applicable - not that its wrong, but the fact that its not defining exponentiation to the ith power in the sense that i has "number like" qualities like real numbers do. You would have to prove that an equivalence
What is really happening is that you never really escape the real numbers, and your complex numbers are just simplified operations that rotate/scale a number, like rotation matricies do through multiplication, and that in the nature of the definition of those rotations, you get stuff like Eulers identity, which is somewhat pointless because like I mentioned - the value of e (i.e 2.7) is never really used to compute anything in regards to complex numbers in polar form of re^ix, all you care about is r and the x which is the angle.
And for this reason, I don't consider i a number, so the analytic/smooth interpretations to me are meaningless.
- tsimionescu 8mo ago> And in defining the operation of multiplication, you essentially define a way to move around the 2 dimensional set now. And moving around 2 dimensions is exactly the same thing as rotation+scaling. Again, this is not how complex numbers were defined. The only original goal was to come up with a number that can solve the equation x^2 + 1 = 0, so that (R + {this number}, + , *), becomes an algebraically closed field. Once you've set this goal, there is really a single simple choice for how the operations will work, because everything else is already constrained. If x^2 + 1 = 0, we already know that: (a + bx) (c + dx) = [polynomial multiplication] = ac + (bc + ad)x + (bd)x^2 = = ac + (bc + ad)x + bd(x^2 + 1 - 1) = = ac + (bc + ad)x + bd(x^2 + 1) - bd = [x^2 + 1 = 0 by definition] = ac + (bc + ad)x + bd * 0 - bd = = (ac - bd) + x(bc + ad). So the formula for complex number multiplication comes out of the arithmetic of real numbers, extended with this extra entity defined simply by being a root of x^2 + 1. The fact that this operation happens to represent a rotation in the RxR plane is "an accident" (I'm sure there are deep ties that make this necessary, probably related to the structure of polynomials themselves). And while you can define other algebraically closed fields that include the reals as a subfield, the complex numbers are the simplest such set. R^n for n>2 is clearly more complex, for example. So there is a clear reason to prefer sqrt(-1) = i, and thus ending up with a 2d vector space. > What Im trying to say is that all the definitions sort of assume that if you have some operation that you can do on real numbers, and have a result, if you just plug in a complex value, it all works out, so people think that i behaves like a number. I personally don't think that this is the case, specifically about i behaving like a number just because those results work out. Again, complex multiplication is not some intentional construction, it is baked into how we defined the complex numbers the first time we did. Again, our goal was to find (well, define) the solution(s) of the equation x^2 + 1 = 0. The fact that we can plug in this x to any formula that involves other numbers just falls out of this goal, it's not an additional assumption. In the case of e(x), this is simpler to see with the power series formula: e(nx) = sum(1 + nx + (nx)^2/2! + (nx)^3/3! + (nx)^4/4! + ...) but, by definition, x^2 + 1 = 0, so x^2 = -1, so the formula becomes: e(nx) = sum(1 + nx - n^2/2! - xn^3/3! + n^4/4! + ...) = = sum(1 - n^2/2! + n^4/4! ...) + sum(xn - xn^3/3! + xn^5/5! ...) = = sum(1 - n^2/2! + n^4/4! ...) + x sum(n - n^3/3! + n^5/5! ...) = = cos n + x sin n Note that this falls out of the properties of e(x), cos(x), and sin(x) for real numbers, and the single property of i that it is a solution of x^2 + 1 = 0. I also think that the definition that e(x) is "take e x times in multiplication and then compute the limit" is any more intuitive. I certainly don't think that `e^2 = e(2) = lim (1 + 2/n) ^ n, with n-> infinity` is any more intuitive than the definition of `e(2i) = lim (1 + 2i/n)^n, with n -> infinity`. > which is somewhat pointless because like I mentioned - the value of e (i.e 2.7) is never really used to compute anything in regards to complex numbers in polar form of re^ix, all you care about is r and the x which is the angle This is also not really true, because the value of e is deeply tied to the values of cos x and sin x. This also becomes visible if you want to compute 2^(ix). 2^(ix) = e ^ ix (log_e 2) = cos (x log_e 2) + i sin (x log_e 2), using log_e to denote the natural logarithm to make it clearer that it is related to e. So the value of e itself is still there in the formula, even if we discount the relationship between e and cos and sin.