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Joined 1 year ago
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Cake day: June 14th, 2023

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  • Oh, I’ll try to describe Euler’s formula in a way that is intuitive, and maybe you could have come up with it too.

    So one way to think about complex numbers, and perhaps an intuitive one, is as a generalization of “positiveness” and “negativeness” from a binary to a continuous thing. Notice that if we multiply -1 with -1 we get 1, so we might think that maybe we don’t have a straight line of positiveness and negativeness, but perhaps it is periodic in some manner.

    We can envision that perhaps the imaginary unit, i, is “halfway between” positive and negative, because if we think about what √(-1) could possibly be, the only thing that makes sense is it’s some form of 1 where you have to use it twice to make something negative instead of just once. Then it stands to reason that √i is “halfway between” i and 1 in this scale of positive and negative.

    If we figure out what number √i we get √2/2 + √2/2 i

    (We can find this by saying (a + bi)^(2) = i, which gives us (a^(2) - b^(2) = 0 and 2ab = 1) we get a = b from the first, and a^(2) = 1/2)

    The keen eyed observer might notice that this value is also equal to sin(45°) and we start to get some ideas about how all of the complex numbers with radius 1 might be somewhat special and carry their own amount of “positiveness” or “negativeness” that is somehow unique to it.

    So let’s represent these values with R ∠ θ where the θ represents the amount of positiveness or negativeness in some way.

    Since we’ve observed that √i is located at the point 45° from the positive real axis, and i is on the imaginary axis, 90° from the positive real axis, and -1 is 180° from the positive real axis, and if we examine each of these we find that if we use cos to represent the real axis and sin to represent the imaginary axis. That’s really neat. It means we can represent any complex number as R ∠ θ = cos θ + i sin θ.

    What happens if we multiply two complex numbers in this form? Well, it turns out if you remember your trigonometry, you exactly get the angle addition formulas for sin and cos. So R ∠ θ * S ∠ φ = RS ∠ θ + φ. But wait a second. That’s turning multiplication into an addition? Where have we seen something like this before? Exponent rules.

    We have a^(n) * a^(m) = a^(n+m) what if, somehow, this angle formula is also an exponent in disguise?

    Then you’re learning calculus and you come across Taylor Series and you learn a funny thing, the Taylor series of e^x looks a lot like the Taylor series of sine and cosine.

    And actually, if we look at the Taylor series for e^(ix) is exactly matches the Taylor series for cos x + i sin x. So our supposition was correct, it was an exponent in disguise. How wild. Finally we get:

    R ∠ θ = Re^(iθ) = cos θ + i sin θ








  • To add, let’s do some math!

    Let s be the total annual salary of every employee using Adobe. Our goal is to find the productivity ratio r such that changing to Gimp and open source more generally is a net positive from the standpoint of productivity and labor.

    s/r will be the total annual salary after changing over, because (for instance) if r = 0.8 then LTT will need to either hire or work his existing hires 1/0.8 times longer, giving (at best, ignoring overtime and so on) s/r as the new labor cost.

    We then subtract the current labor cost to get the switching cost s/r - s, and if this is greater than $10,000 then the switch is not worth it.

    For instance, let’s say LTT employs 1 person at $50k/year. He’s a bit of a skinflint. We solve for r and arrive at a ratio of 5/6 or 83.33%.

    If we have a different world where LTT hires 10 people and pays each of them $100k, we solve for r and get about 99%.

    In other words, the switch is worth it only if the labor cost is small, so the extra labor is not very expensive, or the difference between the two software is negligible.







  • Most closely matches the behavior of actual SNES consoles.

    This requires very careful emulation of the timings of the various buses and co-processors, as well as on-cart chips which may or may not be present. For instance, a Speedy Gonzales game has a button in the final stage which crashes almost every emulator because enters an infinite loop reading from an open bus and waiting for the value to attain a specific pattern. However reading from an open bus is generally specified to be the last value loaded into the bus, which in this case is the load instruction itself, $18. So the value is read to be $1818 by most emulators, which doesn’t match the pattern expected.

    However, this is only if you’re emulating with instruction level accuracy. It is possible for the value of the bus to change in between the instruction being loaded and the value of the bus being loaded due to an HDMA load being triggered, but this requires a cycle accurate emulator.