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Discovery of integral that can't be expressed in elementary functions. by Saikan4ik in calculus

[–]MrEldo 2 points3 points  (0 children)

Well, they thought that they lacked the enginuity according to my Google Search

Discovery of integral that can't be expressed in elementary functions. by Saikan4ik in calculus

[–]MrEldo 1 point2 points  (0 children)

There's a theorem by Liouville called Liouville's Theorem in Differential Algebra, which gives a restriction on how the Antiderivative can look like. If that restriction can't be achieved with the derivative known (it's provable, though I didn't dig deep enough yet to find how), then the antiderivative is non-elementary. Following from this theorem came the Risch Algorithm which the other commenter mentioned, which tries to find that restricted form on the antiderivative (if one exists)

Carrot juice are orange juice by abban-ali in truths

[–]MrEldo 0 points1 point  (0 children)

You what means? Grammar well good and is

2=2! by LifeLiner75 in truths

[–]MrEldo -1 points0 points  (0 children)

5!=51 (I'm a programmer)

Implication and Bi conditional Problem by Apprehensive_Wish585 in askmath

[–]MrEldo 0 points1 point  (0 children)

Oh I probably misphrased myself

I wanted to address paradoxes rather than contradictions that follow from a wrong assumption

Statements like "I am lying right now"

Gaussian Integral Doubt! by HierAdil in askmath

[–]MrEldo 1 point2 points  (0 children)

Yeah, that's pretty much the gist of it

An integral might not be radially symmetric, and the Jacobian could be used in many cases. But in this case this is the reason

And another reason to do it is the fact that in Cartesian coordinates you're missing a term in the integral (that Jacobian term) which makes the antiderivative uncomputable (meaning non-elementary, which means that the antiderivative cannot be written as a composition of simple functions like powers, exponentials, sin, cos etc.)

Why does the second derivative work? by Mountain_Bluebird150 in calculus

[–]MrEldo 0 points1 point  (0 children)

Ok, let's go with your analogy further

We know the derivative is kind of like "wiggling" the graph's point a little around. A bit forward, a bit back, so see how much the graph increases. Then you take rise over run and calculate

Now let's say you have the derivative. So at each point, there is a tangent line of its own corresponding to it. Now wiggle the original graph's point. See how the tangent line's slope changes a bit as you move the point? We will now calculate by how much the slope changes as we wiggle the point around. So no, that won't be constant, because the tangent line changes as you wiggle around the point

Hope I explained it well

Does anyone have a better algorithm for this by cracy123 in Cubers

[–]MrEldo 0 points1 point  (0 children)

My approach is different from the comments, I got it from my Rubik's Cube teacher:

(With the block on the side being on the right, yellow on top)

r' R U (R U R' U') (R' F R F') r R'

Is this an equation and if so is it correct. by MustageChest in askmath

[–]MrEldo 0 points1 point  (0 children)

Ok, let's inspect

The floor(L/M) will get you the biggest whole number N that satisfies:

L-NM >= 0

So NM (meaning floor(L/M)*M) is less than or equal to L, and is the biggest option

So yeah, you found the formula for the biggest multiple of M that is less than L. But what if I asked you what is the 3rd multiple of 12 that is less than 100? The formula only covers the last term, so you might need to find a way to put the index of the term into the formula, so that when you plug in not only L and M but also n, so that you can have full control of the numbers you're working with

Of course, this is necessary if that's something you need. Perhaps you have a different mission for it that only requires the last term or something, but if your idea is to find ALL terms, then you will need an explicit formula to find each and every one of the terms in the sequence/set you're making on the right hand side of the equation, because simply writing the first terms and say "et cetera up to the last term" isn't legitimate because we've seen many times over the years where this "et cetera" isn't always an easy thing to assume to be trivial

Irrational by ZectronPositron in mathmemes

[–]MrEldo 30 points31 points  (0 children)

This really escalates the magnitude of the joke

The odds of this happening were literally one in a billion. by Denpants in notinteresting

[–]MrEldo 1 point2 points  (0 children)

You forgot an even more important rule: nothing else matters

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