11–20 of 62 posts
(i^i)^i is pretty easy to do with Euler's formula: it's just i^(i*i)=i^(-1)=1/i=-i.
If the x in e^x = (1+x/N)^N is understood as some transformation, then e^x is essentially repeating an infinitesimal transformation lots of times. So it's like a for-loop where the body of the loop is some infinitesimal transformation.
I tried to define the integration operator in terms of e^x. The 1 + x/N needed to be one "infinitesimal" iteration of integration, that adds an extra infinitesimal rectangle to the area. But it didn't seem to work out.
Ultimately that helps to explain Euler's formula. For large N, 1+ix/N is an "infinitesimal" rotation by angle x/N. Repeating it N times produces a rotation of angle x. That's essentially what TFA says. And it's a special case of the Lie theoretic view of e^x.
When was this posted? I think i read exactly that explanation like 7-8 years ago and it really made it click for me. It was a wonderful insight!
See https://hn.algolia.com/?query=Intuitive%20Understanding%20of...
e^x is a function whose value is its rate of change. (De^x=e^x). Now imagine the unit circle by taking a point an unit away from O, and set "rate of change" perpendicular to that vector. You will end up with Df(x) = i f(x), which really only works when f(x) = e^ix, supposing i means perpendicularity.
I once had an interesting thought about the function e^x. I think this is a key idea in the theory of Lie groups. If the x in e^x = (1+x/N)^N is understood as some transformation, then e^x is essentially repeating an infinitesimal transformation lots of times. So it's like a for-loop where the body of the loop is some infinitesimal transformation. I tried to define the integration operator in terms of e^x. The 1 + x/…
The exponential of a matrix X is an infinite sum just like that of the normal exponential function except with operations being matrix multiplication, addition, and scalar multiplication (I is the identity matrix):
e^X = I + X + X^2/2 +X^3/6 +...
X = [0 -1; 1 0].
Now you can compute that the matrix exponential
e^(tX)
[cos(t) -sin(t); sin(t) cos(t)].
[cos(t) -sin(t); sin(t) cos(t)]*[x;y] = [x*cos(t) - y* sin(t); x*sin(t) + y*cos(t)]
To say briefly how this is a special case of the exponential map in Lie theory: the 1-d vector space spanned by X is the Lie algebra of the unit circle (which is a group) and the exponential map sends an element tX to e^(tX).
The article is quite right that multiplying by i gives a rotation. But it doesn't quite explain the reason for this: it's because that's the whole point of defining imaginary numbers in the first place! Remember you start off wanting to find a solution for the equation: i^2 = -1 This is actually easier to think about if you multiply it by a general real number r: r i^2 = -r In other words you want i such that if you…
People used imaginary numbers for a long time before Cartesian coordinates even existed.
i^i is actually multi valued - it depends on which branch of log(z) you choose. The rotation analogy is still correct.
(i^i)^i is pretty easy to do with Euler's formula: it's just i^(i*i)=i^(-1)=1/i=-i.
I think you have to be careful here when using rules that hold for real number for complex numbers. In fact, Wolfram|Alpha says, that -i is just one of multiple results: http://www.wolframalpha.com/input/?i=(i%5Ei)%5Ei
The article is quite right that multiplying by i gives a rotation. But it doesn't quite explain the reason for this: it's because that's the whole point of defining imaginary numbers in the first place! Remember you start off wanting to find a solution for the equation: i^2 = -1 This is actually easier to think about if you multiply it by a general real number r: r i^2 = -r In other words you want i such that if you…
> it's because that's the whole point of defining imaginary numbers in the first place! People used imaginary numbers for a long time before Cartesian coordinates even existed.