100 Years to Solve an Integral (2020)
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100 Years to Solve an Integral (2020)
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Re: 100 Years to Solve an Integral (2020)
#2[deleted]
Re: 100 Years to Solve an Integral (2020)
#3I know the article is about sec(x) but I want to share this tidbit about its cousin, the hyperbolic secant: sech(x) is its own Fourier transform (modulo rescalings). That’s right, exp(-x^2) is not the only one.
Re: 100 Years to Solve an Integral (2020)
#4If we're playing the map-projection-advocacy game, I'd say the Mollweide projection is underrated among equal-area maps [0]. (For local maps, use whatever you want, appropriately centered.) Sure, it distorts shapes away from the central meridian, but locally it only adds a simple horizontal skew. I'm not a big fan of how many equal-area 'compromise' projections lie about how long the lines of latitude are.
Re: 100 Years to Solve an Integral (2020)
#5The most elegant proof IMHO is the one that avoids the original problem entirely.
Int[csc(x) dx] = 2 Int[csc(2u) du]
= 2 Int[du / (2 cos(u) sin(u))]
= Int[sec^2(u) du / tan(u)]
= log(tan(u)) + C
= log(tan(x/2)) + C
Then Int[sec(x)] = Int[csc(u)] = log(tan(u/2)) + C = log(tan(pi/4 - x/2)) + C.
Of course, this was no use to Mercator, because the logarithm hadn't been invented yet. But you aren't just pulling a magic factor out of nowhere. There is definitely a bit of cleverness in rearranging the fraction — you have to be used to trying to find instances of the power rule when dealing with integrals of fractions.
Re: 100 Years to Solve an Integral (2020)
#6I know the article is about sec(x) but I want to share this tidbit about its cousin, the hyperbolic secant: sech(x) is its own Fourier transform (modulo rescalings). That’s right, exp(-x^2) is not the only one.
The impulse train is another well-known one, though I suppose someone will chime in here to rebut that it's arguably not a function.
Re: 100 Years to Solve an Integral (2020)
#7[flagged]
Re: 100 Years to Solve an Integral (2020)
#8The most elegant proof IMHO is the one that avoids the original problem entirely. Int[csc(x) dx] = 2 Int[csc(2u) du] = 2 Int[du / (2 cos(u) sin(u))] = Int[sec^2(u) du / tan(u)] = log(tan(u)) + C = log(tan(x/2)) + C Then Int[sec(x)] = Int[csc(u)] = log(tan(u/2)) + C = log(tan(pi/4 - x/2)) + C. Of course, this was no use to Mercator, because the logarithm hadn't been invented yet. But you aren't just pulling a magic fa…
This was the one I was taught in my high school. It has some cleverness (e.g., some trig. transformations) but looks less like coming out of nowhere than the original.
Re: 100 Years to Solve an Integral (2020)
#9About how long it'd take me to solve the integral in my calculus finals.
Re: 100 Years to Solve an Integral (2020)
#10It feels like LLMs could be good contenders for solving symbolically integrals. After spending some time, it really feels like translating between two languages.