Antiderivative

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Anonymous

Antiderivative 2015-01-18 08:43:20 Post No. 7017342

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Antiderivative 2015-01-18 08:43:20 Post No. 7017342

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In which cases and why can't we find any primitive?

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if our IQ is too low

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All rational functions have an elementary anti derivative.

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>>7017342

find the values where x3+x2+1=0, go complex, shift the zeroes by i \epsilon , find a contour which vanishes and use the residue theorem to inegrate.

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All functions have an antiderivative, it is simply not always computable in terms of elementary functions

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>>7017354

>All functions

no

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>>7017359

why?

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>>7017365

Please take a math class and learn what an integral is.

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>>7017365

Weierstrass functions

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>>7017354

If the function is absolutely continuous (on a compact interval), then there is an antiderivative. In any other cases, you can't know it.

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>>7017371

You are retarded. The Weierstrass function is continuous, hence has an antiderivative.

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>>7017365

Function has to be continuous.

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>>7017378

Are you fucking retarded? Absolute continuity is a much stronger condition than continuity.

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>>7017342

Try to break it down in simple elements.

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>>7017382

>Are you fucking retarded?

Yes, I got confused with another property. Sorry.

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File: A discontinuous Riemann integrablefunction with an antiderivative..jpg (43 KB, 471x374)
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43 KB, 471x374

>>7017381

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>>7017397

how about f(x)=1 of x=0, f(x)=0 for every other x

this can't be the derivative of any g(x), because then f should assume every value between 0 and 1

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>>7017354

>>7017359

>>7017365

>>7017367

>>7017371

>>7017378

>>7017380

>>7017381

>>7017382

>>7017397

If g is a Lebesgue-integrable function on some interval [a,b] and we define f as:

f(x) = \int_a^x g(t)dt

then:

1. f is absolutely continuous

2. f is differentiable almost everywhere

6. it's derivative coincides almost everywhere with g(x)

So, if g is Lebesgue-integrable then, almost everywhere, f' = g (i.e it has an antiderivative).

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>>7017401

Your function f is a weak derivative of every constant function.

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>>7017406

so?

why does that imply if f is a weak derivative of some g than there is a h such that f is the strong derivative of h?

we were looking for functions that dont have antiderivative

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>>7017405

>almost everywhere

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