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What makes e natural? (2004)

komal.hu

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Re: What makes e natural? (2004)

#91

Post by kazinator »

Tainnor wrote: post #65
raincole wrote: post #55

Earlier quoted context omitted.

TIL a two dimensional plane isn't natural.

C is a plane with multiplication . I would say, the fact that there is even a way of making this work is surprising (it doesn't work for any R^n, n > 2, at least not if you want a field). The particular way how we multiply complex numbers is also IMHO not self-evident (neither in the algebraic, nor in the geometric interpretation). It just turns out that doing it this way gives us really nice properties.

> how we multiply complex numbers is also IMHO not self-evident

Multiplication of complex number in the x + iy form treats them exactly like any other binomial factors, using the "FOIL" rule:

                       F    O     I     L

   (a + bi)(c + di) =  ac + adi + bci + bdi^2

                    =  ac + (ad + bc)i - bd

                    =  ac - bd + (ad + bc)i
I don't know about self-evident; you pretty much have to do it that way. If you hand (a + bi)(c + di) to someone who knows basic algebra, but has no idea what i is, they will come up with the first FOIL expression above; after that, we have to know that i^2 is -1.

> nor in the geometric interpretation.

That isn't self-evident, but when you gemoetrically work out what the multiplication is doing, it's just "add the arguments (angles), multiply the moduli (distances from origin)".

Very simple!

Re: What makes e natural? (2004)

#92

Post by kazinator »

lisper wrote: post #27
mihaic wrote: post #23

Earlier quoted context omitted.

Exactly the same for me and I'm pretty sure this is how Jakob Bernoulli came to define the number as well, trying to see what the upper bound for infinitesimal compounding was.

That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…

Yes, it does have to do with exponentiating e, though.

e^(iπ) is (e^i)^π.

There is a concrete complex number e^i:

  [1]> (exp 1)
  2.7182817
  [2]> (expt (exp 1) #c(0 1))
  #C(0.54030234 0.84147096)
See, it's around 0.54 + 0.84i. It's on the unit circle.

When you raise this number to pi, you get -1.

  [3]> (expt * pi)
  #C(-1.0 1.2776314E-7)
This means it's the pi-th root of -1; let's try it:

  [4]> (expt -1 (/ pi))
  #C(0.5403023058681397174L0 0.84147098480789650666L0)
and e must be the i-th root of this:

  [5]> (expt * (/ #c(0 1)))
  #C(2.7182818284590452354L0 -4.6847612413106414363L-20)
Yes; it is all literally exponentiation which we can approximate with concrete floating-point numbers that know nothing about the formula we are exploring.

Re: What makes e natural? (2004)

#93

Post by Tainnor »

kazinator wrote: post #91
Tainnor wrote: post #65

Earlier quoted context omitted.

C is a plane with multiplication . I would say, the fact that there is even a way of making this work is surprising (it doesn't work for any R^n, n > 2, at least not if you want a field). The particular way how we multiply complex numbers is also IMHO not self-evident (neither in the algebraic, nor in the geometric interpretation). It just turns out that doing it this way gives us really nice properties.

> how we multiply complex numbers is also IMHO not self-evident Multiplication of complex number in the x + iy form treats them exactly like any other binomial factors, using the "FOIL" rule: F O I L (a + bi)(c + di) = ac + adi + bci + bdi^2 = ac + (ad + bc)i - bd = ac - bd + (ad + bc)i I don't know about self-evident; you pretty much have to do it that way. If you hand (a + bi)(c + di) to someone who knows basic alg…

> you pretty much have to do it that way

You have to do it that way because it's the only way to get a field out of R^2, but I don't think it's at all self-evident. The "we have to know that i^2 is -1" is doing some heavy lifting in your explanation. In the context of "complex numbers are just a 2d plane" there's no a priory reason why you'd think that (0,1)*(0,1) has any particular meaning, let alone that it's equal to (-1,0).

If we interpret vectors in R^n as dislocation (or as forces, or...), then vector addition behaves exactly as one would expect. I think with enough motivation a bright school kid could come up with the formula for vector addition. And it works for every dimension.

Vector multiplication only works for 2d (ignoring the less well behaved cross product that also only works in certain dimensions) and I don't think you'd come up with it naturally.

Also, historically, the 2d plane isn't how complex numbers were discovered. They first appeared as weird quantities that made solving (real) cubic equations simpler but were regarded as just temporary artifacts. The realisation that they can be identified with R^2 came much later.

Re: What makes e natural? (2004)

#94

Post by lisper »

kazinator wrote: post #92
lisper wrote: post #27

Earlier quoted context omitted.

That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…

Yes, it does have to do with exponentiating e, though. e^(iπ) is (e^i)^π. There is a concrete complex number e^i: [1]> (exp 1) 2.7182817 [2]> (expt (exp 1) #c(0 1)) #C(0.54030234 0.84147096) See, it's around 0.54 + 0.84i. It's on the unit circle. When you raise this number to pi, you get -1. [3]> (expt * pi) #C(-1.0 1.2776314E-7) This means it's the pi-th root of -1; let's try it: [4]> (expt -1 (/ pi)) #C(0.540302305…

> There is a concrete complex number e^i:

Yes, that is true, of course. But consulting a Lisp REPL merely demonstrates that it is true. It does not explain why it is true.

You'd get further, pedagogically speaking, by pointing out that (e^i)^i = e^(i*i) = e^(-1) = 1/e, and so e^i is a number that is in some sense "half way" between e and 1/e when you are exponentiating. As an analogy, consider:

e^(1/2) * e^(1/2) = e^(1/2 + 1/2) = e^1 = e

as a demonstration that e^(1/2) is a number that is in some sense "half-way" between 1 and e when you are multiplying, a.k.a. the square root of e. But that still leaves unanswered the question of what "half-way" means when exponentiating rather than multiplying.

Re: What makes e natural? (2004)

#95

Post by kazinator »

lisper wrote: post #94
kazinator wrote: post #92

Earlier quoted context omitted.

Yes, it does have to do with exponentiating e, though. e^(iπ) is (e^i)^π. There is a concrete complex number e^i: [1]> (exp 1) 2.7182817 [2]> (expt (exp 1) #c(0 1)) #C(0.54030234 0.84147096) See, it's around 0.54 + 0.84i. It's on the unit circle. When you raise this number to pi, you get -1. [3]> (expt * pi) #C(-1.0 1.2776314E-7) This means it's the pi-th root of -1; let's try it: [4]> (expt -1 (/ pi)) #C(0.540302305…

> There is a concrete complex number e^i: Yes, that is true, of course. But consulting a Lisp REPL merely demonstrates that it is true. It does not explain why it is true. You'd get further, pedagogically speaking, by pointing out that (e^i)^i = e^(i*i) = e^(-1) = 1/e, and so e^i is a number that is in some sense "half way" between e and 1/e when you are exponentiating. As an analogy, consider: e^(1/2) * e^(1/2) = e^…

Obtaining an intuition for what it means to use an imaginary exponent takes work.

Exponents give us expoential decay/growth in the real number line, but periodicity in the imaginary domain, which is strange.

In physics, this lets us analyze decaying or amplifying oscillations in a unified way. (Laplace transform and all that.)

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