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Why does e to pi i equal -1? (2015) [video]

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Re: Why does e to pi i equal -1? (2015) [video]

#111
post #34

Earlier quoted context omitted.

Honestly, after spending months studying the subject, I don't think it's really possible to "get" complex numbers. I just view them as affine transformations written in an unusual notation. I don't think they make sense as anything but a recontextualization of R^2.

The problem with that is that complex numbers initially emerge as roots of polynomials with real coefficients. Getting to affine transformations from there seems a much bigger leap than asserting there is a square root of -1.

The video linked in this post will only make sense if you accept that x e^(theta i) and x k are respectively the rotation part and the scaling part of a linear transformation of x. I'm not aware of any other way to intuitively grasp an expression like e^(pi i).

Re: Why does e to pi i equal -1? (2015) [video]

#112
post #104

Earlier quoted context omitted.

So what don't you get? Are you implying there is something more to complex numbers?

What I don't get is why someone would bother with complex numbers and their silly notation when linear algebra would work perfectly fine. I know that in certain contexts they are useful. E.g. to avoid losing information when solving polynomials. But that's a minuscule fraction of their range of application. Nearly every practical use I've seen of complex numbers just use them as a vector representation.

Well the two uses I'm most familiar with are AC circuit analysis and quantum mechanics. They can both be reformulated without complex numbers of course, since nothing is special about i.

Yet the complex versions are a lot easier to work with, because even in manifestly real formulations, the complex structure is still there, but in disguise:

- http://www.scottaaronson.com/democritus/lec9.html

- http://physics.stackexchange.com/questions/32422/qm-without-...

Re: Why does e to pi i equal -1? (2015) [video]

#113

I had never understood how imaginary numbers had any bearing on physics (probably because I'd never been taught). But lately I've thinking about how all mathematics must have some natural physical equivalent or relation. E.g. how does multiplication happen in nature.. what does it mean really to multiply something?...how does this happen in nature? For the most part, we take these things as facts and learn the mechan…

It's an interesting and unresolved question whether imaginary numbers are fundamental to the nature of physics or just a handy tool to calculate stuff. They are certainly handy for calculating - they are an easy way to represent oscillations.

Re: Why does e to pi i equal -1? (2015) [video]

#114
post #97

Earlier quoted context omitted.

Speaking of number sets. I never understood why complex numbers are considered one. I mean, yes they "are" a set, but besides that they are completely different. All number sets I learned about did fill some gaps in one dimension, but complex numbers somehow added a new dimension. Like real numbers stood in an entirely different context to rational numbers than complex numbers stood to real numbers.

C (complex numbers) is a field, just like R (real numbers). It's an algebraically closed field, unlike R, so in some sense it's actually the most natural set to call "numbers". https://en.wikipedia.org/wiki/Algebraically_closed_field

I'm not trying to argue for R being more a set than C.

In my head a one dimensional thing like R is fundamentally different from a multidimensional thing like C.

Re: Why does e to pi i equal -1? (2015) [video]

#115
post #109

In Lie theory, given a Lie group, there's a general notion of an "exponential function", which maps elements in the tangent space to the identity to "full" elements of the group. In this case, our group is the unit circle in the complex plane. This is not circular logic, by the way. If a, b are complex numbers on this circle, then |a| = |b| = 1 and so |ab| = |a||b| = 1. The identity of the group is the complex number…

I was about to write a similar story, but then I realized you need a Riemannian metric (and exponential) for everything arclength-related.

IIRC since the unit circle is a compact Lie group, there is a bi-invariant Riemannian metric whose exponential is the Lie group exponential and we land back on our feet, but the Lie group structure alone is not sufficient.

Re: Why does e to pi i equal -1? (2015) [video]

#116

I had never understood how imaginary numbers had any bearing on physics (probably because I'd never been taught). But lately I've thinking about how all mathematics must have some natural physical equivalent or relation. E.g. how does multiplication happen in nature.. what does it mean really to multiply something?...how does this happen in nature? For the most part, we take these things as facts and learn the mechan…

You should read about Maxwell's original equations for electromagnetism in component form (1865), then in quaternion form (1873), and Heaviside's vector-based rewrite (1888), then look at how they can be rewritten in geometric algebra.

If you invent enough crazy new mathematical constructs, eventually one of them will mirror a natural physical phenomenon. And then a pile of equations collapses into something incredibly simple.

Re: Why does e to pi i equal -1? (2015) [video]

#117

Earlier quoted context omitted.

I agree, the suggestion that anyone could/would be able to just throw out everything they know about arithmetic is sketchy at best, and even it's sketchier to make it seem as though a result like Euler's formula could purely be derived from geometric intuition. Sure, it "makes sense" in the video, but I don't think it's possible to truly grok and appreciate the result without having done any calculus. The real joy of…

The Maclaurin trick is what I'd call the "standard proof", and I think it's the source of a lot of the lack of intuition surrounding e^i... It's exactly what you call it, a "trick", where you prove two rabbits are identical by transforming them both into an infinite number of hats. On the one hand, it's all sound logic and algebra, but on the other hand, I certainly don't blame Randall Munroe for seeing that proof in…

Love that xkcd, but I found the Maclaurin "trick" to be extremely common-sensical... If you're already finding the area of infinitely many small boxes under a curve then Taylor series don't seem that weird. Once you are convinced that Taylor series are O.K. then this proof is really cool. I also think it's weird to show high-schoolers this, as about 1% of high school seniors have the mathematical knowledge needed to grok this level of math.

Anyways - people have differing levels of magic tolerance, and some people don't have a lot of background in calc, so I don't blame them for looking for other explanations.

Re: Why does e to pi i equal -1? (2015) [video]

#118
post #97

Earlier quoted context omitted.

I love this answer. For some reason, I've been dreaming about negative numbers lately. I think they deserve their own number set notation.

Speaking of number sets. I never understood why complex numbers are considered one. I mean, yes they "are" a set, but besides that they are completely different. All number sets I learned about did fill some gaps in one dimension, but complex numbers somehow added a new dimension. Like real numbers stood in an entirely different context to rational numbers than complex numbers stood to real numbers.

You are mixing up a couple of things. This "dimension" you're talking about is probably the dimension of a vector space. The real numbers, as a vector space over the field of real numbers (recall that vector spaces are defined over fields of numbers), have dimension 1. However, you can consider the vector space of real numbers over the field of rational numbers. What is the dimension of this vector space? Well, the dimension of a (finite-dimensional) vector space is given by the number of elements in a basis for it. So how do we go about finding a basis for R over Q? A basis for R over R consists solely of the number "1", since given a real number "x" there exists an element from the field (in this case R), namely x itself, such that multiplying it by 1 will give you the number x. Yes, this sounds obvious, but that's how you prove that R over R has dimension 1. So, going back to R over Q, we see that "1" cannot be a basis, since if we pick, say, pi, there is no rational that we multiply 1 by to give pi. Moreover, there cannot be any finite number of real numbers that would make up a basis for R over Q, since then R would be countable (look this up if you don't know what it is). So the dimensionality of a vector space depends on which field you're considering. By the way, the vector space of complex numbers over the reals has dimension 2 (a basis is {1, i}), but over the complex numbers it has dimension 1. So, there's an "extra" dimension only if you consider it over the reals.

Re: Why does e to pi i equal -1? (2015) [video]

#119
post #114

Earlier quoted context omitted.

C (complex numbers) is a field, just like R (real numbers). It's an algebraically closed field, unlike R, so in some sense it's actually the most natural set to call "numbers". https://en.wikipedia.org/wiki/Algebraically_closed_field

I'm not trying to argue for R being more a set than C. In my head a one dimensional thing like R is fundamentally different from a multidimensional thing like C.

Well I was answering the question about why is C considered numbers - it's a field, hence elements of C behave exactly like numbers. Moreover, it's an 'algebraic closure' of R, hence a more natural choice for "all numbers", and it's a maximal one at that (i.e. quaternions and such are no longer fields and don't really behave like numbers, while C still does).

Edit: FWIW, I consider the terminology of 'real' vs 'imaginary' completely stupid and misleading. This terminology didn't really make it to other languages, e.g. in Russian it's 'material' vs 'complex' numbers, but they don't use the term 'imaginary'.

Re: Why does e to pi i equal -1? (2015) [video]

#120
post #114

Earlier quoted context omitted.

I'm not trying to argue for R being more a set than C. In my head a one dimensional thing like R is fundamentally different from a multidimensional thing like C.

Well I was answering the question about why is C considered numbers - it's a field, hence elements of C behave exactly like numbers. Moreover, it's an 'algebraic closure' of R, hence a more natural choice for "all numbers", and it's a maximal one at that (i.e. quaternions and such are no longer fields and don't really behave like numbers, while C still does). Edit: FWIW, I consider the terminology of 'real' vs 'imagi…

We do: вещественная часть и мнимая часть.
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