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The Riemann integral was a mistake

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Thread replies: 14
Thread images: 4

File: perverted pos.png (31KB, 600x600px) Image search: [Google]
perverted pos.png
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The Riemann integral was a mistake
>>
y tho?
>>
File: PkAiP[1].png (45KB, 330x320px) Image search: [Google]
PkAiP[1].png
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>>8549119
Don't you mean the Tai method?
>>
>>8549138
i'm sure i do
anti differentiation is cancer
>>
>>8549172
Why
>>
>>8549173
because they're called integrals, fool!
>>
File: IMG_1622.png (227KB, 1280x960px) Image search: [Google]
IMG_1622.png
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>>8549138
>>8549119

> not using this circles
> Being this bad at integration
>>
>>8549200
looks interesting. the circles are local approximations to the function you want to integrate?
>>
File: circle_integral.png (12KB, 1533x748px) Image search: [Google]
circle_integral.png
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>>8549200
>>8549329
I implemented it in matlab now.
Seems like a pretty shitty quadrature rule though. It's nonlinear so not really useful in any numerical analysis context. It doesn't integrate any polynomials exactly and it doesn't work in the case of 3 colinear evaluation points (for example in linear functions)
>>
>>8549119
Lebesgue integration is the only real thing.

Vive la France.
>>
Numerical analysis is not real math
>>
>>8549545

The key is the correct partioning of circles. Start at a point, find the tangent line, move a specific distance along a the polynomial, and find the tangent line again. Keep averaging the new tangent lines until you get one that deviates significantly from the continuous average. This marks the end of a circle and the start of a new one. Now you just have to find the radii to calculate their areas, then subtract them from trapezoids.
>>
>>8549119
Measure Theory was a mistake.
>>
>>8549622
An integral that doesn't work with arbitrary partitioning is not really a good idea though.
Also that still doesn't solve the problem that you can't integrate linear functions.
All of that doesn't even consider that it's pretty computationally expensive to evaluate (finding midpoints and radii takes a lot of operations and exact evaluation of the area under the circle needs inverse trigonometric functions etc)
Thread posts: 14
Thread images: 4


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