3 ms·

I'm a brainlet or maybe it's the 10+ hours of work I did today but I just can't grasp it right now. There are a lot of ways to take integrals though, so I woul

I'm a brainlet or maybe it's the 10+ hours of work I did today but I just can't grasp it right now. There are a lot of ways to take integrals though, so I wouldn't doubt your method could work. Generally though the holes in a lot of these integration methods is their inability to work with functions that are pathological: https://en.wikipedia.org/wiki/Pathological_(mathematics) https://en.wikipedia.org/wiki/Pathological_(mathematics) |group1 = Types of integrals |list1 = * [[Riemann integral]] * [[Lebesgue integration|Lebesgue integral]] * [[Burkill integral]] * [[Bochner integral]] * [[Daniell integral]] * [[Darboux integral]] * [[Henstock–Kurzweil integral]] * [[Haar measure|Haar integral]] * [[Hellinger integral]] * [[Khinchin integral]] * [[Kolmogorov integral]] * [[Lebesgue–Stieltjes integration|Lebesgue–Stieltjes integral]] * [[Pettis integral]] * [[Pfeffer integral]] * [[Riemann–Stieltjes integral]] * [[Regulated integral]] |group2 = Integration techniques |list2 = * [[Integration by substitution|Substitution]] * [[Trigonometric substitution|Trigonometric]] * [[Euler substitution|Euler]] * [[Weierstrass substitution|Weierstrass]] * [[Integration by parts|By parts]] * [[Integration by partial fractions|Partial fractions]] * [[Integration using Euler's formula|Euler's formula]] * [[Integral of inverse functions|Inverse functions]] * [[Order of integration (calculus)|Changing order]] * [[Integration by reduction formulae|Reduction formulas]] * [[Integration using parametric derivatives|Parametric derivatives]] * [[Leibniz integral rule#Evaluating definite integrals|Differentiation under the integral sign]] * [[Laplace transform#Evaluating improper integrals|Laplace transform]] * [[Contour integration]] * [[Laplace's method]] * [[Numerical integration]] * [[Simpson's rule]] * [[Trapezoidal rule]] * [[Risch algorithm]]

yarg 4y ago
(No idea if you'll ever see this, I also have no damned idea what I'm talking about.) Looking at the Weierstrass function, it's a chaotically convergent series with a non-convergent gradient series. (The standard differential forms don't apply, because the assumption of linearisation at the scale of 𝛿x is invalid, due to the scale invariant properties of the fractal.) In that case the integral function would be even more convergent since the 'noise' (pretty much) adds to zero. So can you differentiate via integration, and get a gradient for a related function (with less noise)? I messed around with this, taking the areas of a pair of little triangle approximations and adding them to get a ~rectangle. Then divide that area by 𝛿x to get 𝛿y, and divide that by 𝛿x to get 𝛿y/𝛿x. This was the form that I reached: let y = f(x) let F(x) = ∫f(x)𝛿x 𝛿y/𝛿x = (F(x + 𝛿x) + F(x - 𝛿x) - 2 * F(x))/((𝛿x) ^ 2) At very least, it passed my polynomial sanity check: let F(x) = x^6 𝛿y/𝛿x = 30(x^4) + 30(x^2)(𝛿x^2) + 2(𝛿x^4)
- goldenkey 4y ago
  That's awesome. Glad to see your method worked. If you ever have a chance to encounter a serious mathematician, they can probably give you the proper name of it. Or perhaps we'll have to name it the Yarg integral :-)
  - yarg 4y ago
    I've just been thinking of it as differentiation via integration. It took me until the end to realise that all I'd done was derive a form for the second derivative of F(x).
    - goldenkey 4y ago
      Ah yes, this is why epsilon squared equals 0 in the dual numbers -- to automatically remove all higher order derivatives when they appear. Exponents on derivatives are indicate of their rank/order.
      - yarg 4y ago
        It's the second symmetric derivative. https://en.wikipedia.org/wiki/Symmetric_derivative#The_second_symmetric_derivative https://en.wikipedia.org/wiki/Symmetric_derivative#The_secon... https://en.wikipedia.org/wiki/Second_derivative#Limit https://en.wikipedia.org/wiki/Second_derivative#Limit > The limit is called the second symmetric derivative. Note that the second symmetric derivative may exist even when the (usual) second derivative does not. > This limit can be viewed as a continuous version of the second difference for sequences. So it actually can do what I intended it to do, nothing original - but there was a degree of satisfaction in deriving it myself.
        goldenkey 4y ago
        Thanks yarg, I just learned something new.

yarg 4y ago

(No idea if you'll ever see this, I also have no damned idea what I'm talking about.) Looking at the Weierstrass function, it's a chaotically convergent series with a non-convergent gradient series. (The standard differential forms don't apply, because the assumption of linearisation at the scale of 𝛿x is invalid, due to the scale invariant properties of the fractal.) In that case the integral function would be even more convergent since the 'noise' (pretty much) adds to zero. So can you differentiate via integration, and get a gradient for a related function (with less noise)? I messed around with this, taking the areas of a pair of little triangle approximations and adding them to get a ~rectangle. Then divide that area by 𝛿x to get 𝛿y, and divide that by 𝛿x to get 𝛿y/𝛿x. This was the form that I reached: let y = f(x) let F(x) = ∫f(x)𝛿x 𝛿y/𝛿x = (F(x + 𝛿x) + F(x - 𝛿x) - 2 * F(x))/((𝛿x) ^ 2) At very least, it passed my polynomial sanity check: let F(x) = x^6 𝛿y/𝛿x = 30(x^4) + 30(x^2)(𝛿x^2) + 2(𝛿x^4)

goldenkey 4y ago
That's awesome. Glad to see your method worked. If you ever have a chance to encounter a serious mathematician, they can probably give you the proper name of it. Or perhaps we'll have to name it the Yarg integral :-)
- yarg 4y ago
  I've just been thinking of it as differentiation via integration. It took me until the end to realise that all I'd done was derive a form for the second derivative of F(x).
  - goldenkey 4y ago
    Ah yes, this is why epsilon squared equals 0 in the dual numbers -- to automatically remove all higher order derivatives when they appear. Exponents on derivatives are indicate of their rank/order.
    - yarg 4y ago
      It's the second symmetric derivative. https://en.wikipedia.org/wiki/Symmetric_derivative#The_second_symmetric_derivative https://en.wikipedia.org/wiki/Symmetric_derivative#The_secon... https://en.wikipedia.org/wiki/Second_derivative#Limit https://en.wikipedia.org/wiki/Second_derivative#Limit > The limit is called the second symmetric derivative. Note that the second symmetric derivative may exist even when the (usual) second derivative does not. > This limit can be viewed as a continuous version of the second difference for sequences. So it actually can do what I intended it to do, nothing original - but there was a degree of satisfaction in deriving it myself.
      - goldenkey 4y ago
        Thanks yarg, I just learned something new.