Comment by joppy | Hacker News Reader

Comment by joppy | Hacker News Reader

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> The complex plane is a crutch that gives you pretty pictures.

That’s a very poor outlook on an incredibly useful mathematical tool. Maths is done essentially by finding new ways of viewing an old thing, and restricting the set of tools available seems like a poor idea in general. As for teaching mathematics, finding more ways of looking at something, however imprecise, is a huge aid to understanding. Never be afraid to show someone a picture that’s only 80% true but could potentially aid their understanding in how something works!

I also disagree that the “circular paths” visualisation is misleading - what is misleading about it? This is a completely accurate representation of how a Fourier series is added up to get a real part which varies over time. It even makes it intuitive that multiplying the series by a fixed complex number (corresponding to rotating and scaling the “orbiting” points on the left) will have the effect of translating the resulting signal to the right or the left.

Also for what it’s worth, the product of complex numbers is easy to visualise? Add their angles, multiply their magnitudes.

B/c it actually adding a layer of complexity that's not necessary for the sake of getting a pretty picture. I'll try to illustrate..

The roots of unity are just trivially observed mathematically

Through Euler's identity: e^i2π = cos(2π) + isin(2π) = 1

then obviously (e^i2π/b)^b = 1

So then e^i2π/b is a b'th root of 1. Done!

You don't need a complex plane.. You don't need to explain why you're multiplying 2D points.. and then explain that now you actually have this new operation defined and it's not really a 2D point oh and btw this multiplication is giving you a rotation.. which rotates your point back to 1+i0 (which is maybe the crummiest part b/c 2D rotations are not intuitive and visualizations should leverage our "gut feeling" and intuition). You just skip all of that! The equations are easy. Everyone knows (A^b)^c = A^bc

Then you show that (e^i2π/b)^(b+1) = (e^i2π/b) - so it's "cyclical", the equations starts returning previous values at higher integer inputs.

And then you show orthogonality between basis vector by just taking two basis vectors and doing a dot product. If you write out an example it's really easy to see how the values pair off and cancel out. There is no easy visual equivalent. You can't see how a real/imaginary sinusoid pair is orthogonal to another.. If you try to force a visualization it's only going to be more confusing.

And actually, b/c it's hard to visualize this last step is usually sorta glossed over in intro material. But I'd argue it's the most important step b/c it's the whole reason why are we choosing these extra-challenging e to the i things to build a basis... it's for this orthogonality. If we just wanted sinusoids that are cyclical then we'd be doing everything with real numbers and keeping it simple. You'd cross correlate a sinusoid at different frequencies for instance or something to that effect... but the fourier transform is giving you something much better than that. And as a bonus you see why the chosen frequencies are discrete and not arbitrary.

After that you can look at phase shifts etc...

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