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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]

#61

Post by hasbroslasher »

unholiness wrote: post #52

This explanation strikes me as a little too aggressive in throwing out the notation with the bathwater, only reaching its result by redefining the terms we already have intuition for into space-stretching operations that don't work like arithmetic does in my head. In that sense I think I have the same problem with this proof that I do with the standard one, where you add the Maclaurin series of cos(θ) and i * sin(θ)…

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 math isn't in knowing, it's in proving.

Anyway: as for proof methods, I like the Maclaurin series trick with this one. Too bad the internet hasn't come up with a universal/nice way of writing math notation in a readable format yet, I'd like to sketch that proof here...

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

#62

Post by jeffwass »

cousin_it wrote: post #3

1) e^x is a function whose derivative is equal to its value. 2) e^ix is a function whose derivative is equal to its value rotated by 90 degrees (ie^ix). 3) As x goes from 0 to pi, the trajectory of e^ix always has a velocity vector perpendicular to its current position. For example, when x = 0, the current position is 1 and the velocity vector is i. 4) So the trajectory a circle arc of length pi, which ends at -1.

When I was a physics TA, I used similar arguments with double derivatives show the connection between imaginary exponentials and sines/cosines.

For f(x)=e^(kx), double derivative d^2f/dx^2 = k^2 f(x)

Meanwhile for g(x)=sin(kx), d^2g/dx^2 = -k^2 g(x), and similarly for cosine.

So if k is imaginary, from a differential equations point of view, the exponential behaves exactly like a sine or cosine.

That shows the general idea, and further consideration of boundary conditions gives e^ix = cos(x) + isin(x).

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

#63

Post by mtourne »

gizmo686 wrote: post #60
moresilenter wrote: post #47

Earlier quoted context omitted.

as a masters in mathematics, this is exactly what i came here to write. thanks! To have a notion that multiplication by imaginaries causes rotation, you'd need Euler's formula. I honestly think the best way to get a visual sense for why multiplication by r exp(i theta) is to look at the first few terms of the taylor series added together and see that the adders combine into a spiral that converges on r cos(theta) + i…

>To have a notion that multiplication by imaginaries causes rotation, you'd need Euler's formula. No you don't. You might notice this by simply working with imaginary numbers. You might invent imaginary numbers for rotation [0]. Alternatively, consider the multiplication (x + yi)(a + bi), as the value (a + bi) performing a transformation on (x + yi). We want (x + yi)(a + bi) = xy - by + ayi + bxi. If we consider (x +…

Yay skew symmetric matrixes.

It's also fun to introduce e as a matrix operator for 3d rotations.

It's useful for kinematics and having compact representations for axis angle notations. It's a little far in my head but at some point it felt like a ha-ha moment with the Euler identity, in the "2d version".

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

#64

Post by dnautics »

godmodus wrote: post #48
kmill wrote: post #5

Earlier quoted context omitted.

In a linear algebra course I helped teach last semester, I had the students go through the exercise of writing down the matrix for "multiplication by i," where the complex numbers are thought of as a two-dimensional vector space with {1,i} as a basis. Then I asked them to recognize the matrix (it's a 90-degree rotation of the plane), and then I asked them for an interpretation of that matrix squared (a 180-degree rot…

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

the standard IEEE floating point uses a 'sign and magnitude' system. Negative numbers are basically made by flipping a single bit. I've been playing around with a number system that expresses real numbers as two's complement (like how integers work) - xor all your bits and then add one. In this system, 0 is -0 and the number that is weird for signed integers: 0b100000...0000 is infinity.

Now, there's an exponent and a fraction, too. I've been playing around with how to do these numbers with logic gates (and verilog!) and you can either two's complement the whole thing and work with absolute values, or you can keep the fraction part as a two's complement...

So I just redid multiplication using two's complemented fractions! And addition/subtraction too, which for floating points in general is significantly harder than multiplication. The nice thing about two's complemented floating points is you don't need separate algorithms for addition and subtraction; you can just do everything with one algorithm.

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

#66

Post by chombier »

gizmo686 wrote: post #60
moresilenter wrote: post #47

Earlier quoted context omitted.

as a masters in mathematics, this is exactly what i came here to write. thanks! To have a notion that multiplication by imaginaries causes rotation, you'd need Euler's formula. I honestly think the best way to get a visual sense for why multiplication by r exp(i theta) is to look at the first few terms of the taylor series added together and see that the adders combine into a spiral that converges on r cos(theta) + i…

>To have a notion that multiplication by imaginaries causes rotation, you'd need Euler's formula. No you don't. You might notice this by simply working with imaginary numbers. You might invent imaginary numbers for rotation [0]. Alternatively, consider the multiplication (x + yi)(a + bi), as the value (a + bi) performing a transformation on (x + yi). We want (x + yi)(a + bi) = xy - by + ayi + bxi. If we consider (x +…

Alternatively if you don't like coordinates, you may notice that multiplication by a unit complex number is linear and preserves the complex norm (i.e. modulus), so it defines an orthogonal transformation of the plane.

Topologically, unit complex numbers are the unit circle which is connected, and 1 is part of this circle so all of these transformations must have determinant 1: these are rotations.

No Euler formula.

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

#67

Post by espeed »

dnautics wrote: post #64
godmodus wrote: post #48

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.

the standard IEEE floating point uses a 'sign and magnitude' system. Negative numbers are basically made by flipping a single bit. I've been playing around with a number system that expresses real numbers as two's complement (like how integers work) - xor all your bits and then add one. In this system, 0 is -0 and the number that is weird for signed integers: 0b100000...0000 is infinity. Now, there's an exponent and…

Have you seen "Stanford Seminar: Beyond Floating Point: Next Generation Computer Arithmetic"?

https://www.youtube.com/watch?v=aP0Y1uAA-2Y

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

#68

Post by chombier »

recursive wrote: post #44
snogglethorpe wrote: post #39

Earlier quoted context omitted.

Is this just the contents of the video in text form? In any case, thanks! I don't need to watch the video now, as you've very clearly explained it in only four lines of text!

It's not really any clearer to me. I cannot generalize my understanding of exponents to anything that deals with imaginary numbers.

For me what made it clicked is realizing that complex numbers are 2D matrices: z = x + i y = [[x -y][y x]].

So really we should be writing z = x * [[1 0][0 1]] + y * [[0 -1][1 0]], but since it's tedious we just call 1 == [[1 0] [0 1]] the 2x2 identity matrix and i == [[0 -1][1 0]], and check that i^2 = -1.

Then no more magical i number, the complex product can be derived from the matrix product, the exponential becomes the 2x2 matrix exponential, and so on.

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