eⁱˣ = cos(x) + i sin(x)pic.twitter.com/4fCuF4LOY3
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A natural question is: Why is this animation smooth? The answer is that you can define the so-called fractional derivative, which generalizes nth derivative to a real number of derivatives The fractional Taylor series smoothly interpolates between integer derivativespic.twitter.com/757q5yvu7k
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I learn so much from almost every post you make. - thank you for that :)
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Well, I don't think you need fractional calculus at all, it just makes things unnecessarily complicated. I made the exact same animation simply by adding the next term in the Taylor series weighted by a number going from 0 to 1.
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yes, you can interpolate any way you wish you can also make it by just replacing the n! with Γ(1+n) and summing over reals.. using the fractional derivative is more general it just happens that the fractional derivative of e^x is simple
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Oh I assumed it was linearly interpolating, that's way cooler
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Bruh Speak English
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what's the dot in the curve that lands at -1?
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I believe it's the point in the approximation where theta=pi. It slowly converges to -1, since e^(i*pi)=-1.
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Wait, so does that mean the nth order Taylor approximation for non-integer n is the floor(n) Taylor approximation plus f^(n) (0) x^n/gamma(n+1)? It seems like Taylor's theorem would break down when used like that, since the powers of the terms aren't evenly spaced.
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Like wouldn't there be a jump discontinuity at integer values of n
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