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A new day, a new Artist Training Grounds on Equestria Daily. Today’s prompt is Draw a pony warming up / Draw a pony stretching its limits. I’m sure I’m not the only one to make a “Limits” joke of this calibur today, but seeing as I am a fan of mathematical concepts I felt like I had to. This also gave me a good excuse to start practicing drawing Spike, which I clearly need. Also, screw Twilight’s castle and making it more detailed; I couldn’t be bothered after all the time spent transcribing limit problems off of Google Images.
Hope you enjoy; I know I’ve been having a lot of fun so far! Looking forward to the next prompt!
Twilight Sparkle, Spike, and My Little Pony belong to Hasbro and DHX Media.
Hope you enjoy; I know I’ve been having a lot of fun so far! Looking forward to the next prompt!
Twilight Sparkle, Spike, and My Little Pony belong to Hasbro and DHX Media.
L’Hospital would be my first try too. But if we look at this problem differently and if we try to minimise our steps, maybe the shown way is faster. So I think to differentiate could cost some FLOPs. I’m not an expert at numerical analysis yet, so I could be wrong.
I’m not sure if you can get a situation where I(x) = f(x) + AI(x) with integration by parts because you’d always get f(x) = 0 and A = 1. Otherwise, you’d get something like I(x) = f_n(x) + A_nI_n(x), and then you’d continuously iterate to get an infinite series expansion if possible.
I guess to get I(x) = f(x)/(1 - A) from integration by parts, you’d have to find an integral that would take the form I(x) = f(x) + AI(x) after being put through the formula. Would be a nice exercise to prove if there is a non-trivial function g(x) = u(x)v(x) who’s integral I(x) = int[g(x)] dx produces the form I(x) = f(x)/(1 - A) when put through the integration by parts formula int[u(x)v’(x)] dx = u(x)v(x) - int[v(x)u’(x)] dx.
That actually wouldn’t be that bad, providing that the new undeterminate form is somewhat “proportional” to the previous one. Because then you have a recursion relation in hand, and you can solve for it ;) Kinda similar to when you’re solving an integral by parts and you get something + the same integral you began with (let’s call it “I”). Then you can replace its occurrence with “I” and get the recursive formula
I = somethng + constant·I, so you subtract it from both sides to getI – constant·I = something, or(1–constant)·I = something, and finallyI = something / (1–constant)and you have solved for the integral :) Keep this trick in mind anytime you deal with recursive formulas like these ;)Edited because: minor
I also wonder if you’d just get indeterminate forms no matter how many times you used the rule.