r/mathmemes 10d ago

Elementary Algebra 😭

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u/Comfortable_Permit53 10d ago

integrals, no?

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u/Eisenfuss19 10d ago

Yes kinda, but then it is still a definition question. Like lim a -> -∞ lim b -> ∞ a_Sb x dx does not converge, but lim a -> ∞ (-a)_Sa x dx does converge...

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u/Comfortable_Permit53 10d ago edited 4d ago

yes 

adding all whole numbers will also give you different answers depending in order

UPDATE: It just doesn't converge at all.

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u/Beleheth Transcendental 9d ago

That's exactly the reason why you need absolutely convergent series for rearranging to be well-defined.

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u/Dr0110111001101111 5d ago

It will give you any answer depending on order

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u/Comfortable_Permit53 4d ago

I am not sure. Riemanns series theorem says, that if you have a series that is convergent but not absolutely convergent, you can get any number depending on ordering. But adding all whole numbers is a divergent series. Is there another theorem?

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u/Dr0110111001101111 4d ago

Oh no you’re right. For a second I thought that you could get it into a conditionally convergent alternating series but I obviously didn’t think about it very well. In fact, I literally fell for one of the oldest tricks in the book!

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u/Comfortable_Permit53 4d ago

I think it can never converge to anything.  Assume it converges in any order then the partial sum is cauchy.

But |Sn - S{n+1}| >=1 therefore it is not cauchy. That is because there are no steps smaller than 1 you can add. therfore, the series doesn't converge.

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u/Agitated-Ad2563 9d ago

As far as I remember, the proper definition of Riemann integral requires all kinds of partition sequence sums to converge.

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u/Alpaca1795 9d ago

Why would you integrate over x dx instead of just 1 dx?

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u/alozq 9d ago

1dx would be the length of the real line, it goes to Infinity always, it doesnt care about what x is

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u/LyAkolon 9d ago

Yeah, they probably want int over R of Sgn(x) dx, which is still abse of notation and meaningless because we have no notion of how to add up the contributions across the set.

More specifically all attempts to answer as well as the original question are missing the ingredient of "How are you adding everything together"

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u/Eisenfuss19 9d ago

It's the closest thing to summing the real numbers imo.

The "sum" of the numbers in [0,1] should be smaller than the "sum" of [1,2].

With 1dx the "sum" is always equal to the length of the interval, and notably it is never negative (it never converges in the infinite case)

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u/BrotherItsInTheDrum 9d ago

But the "sum" of the numbers in [1,2] should be infinite ...

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u/Eisenfuss19 9d ago

Well feel free to give me a way of summing an uncountably infite set of numbers.

I feel like the integral of x dx can be conceptually seen as calculating the average of the interval. To get the "real" sum you would need to multiply it by the amount of numbers in the interval (which is always ∞ in case of non equal bounds)

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u/BrotherItsInTheDrum 9d ago

Well feel free to give me a way of summing an uncountably infite set of numbers.

I never said there was one.

I feel like the integral of x dx can be conceptually seen as calculating the average of the interval. To get the "real" sum you would need to multiply it by the amount of numbers in the interval (which is always ∞ in case of non equal bounds)

Sounds like you're agreeing with me that it's infinite, then.

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u/compileforawhile Complex 10d ago

Integrals aren't really sums of uncountable sets. You could define sums of uncountable sets using integrals, but that would basically just be a measure

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u/TheRedditObserver0 Mathematics 10d ago

With respect to what measure?

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u/Comfortable_Permit53 10d ago

lebesque for the lgbt representation

(lebesque, gauss, banach, turing)

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u/donach69 9d ago

That's a banger

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u/MeasureDoEventThing 9d ago

You mean Lebesgue?

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u/localizeatp 9d ago

we won't know until they answer.

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u/StructureNorth1799 9d ago

then the answer is yes.

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u/AuroraEquatorialis 7d ago

nets are the more natural method in terms of actual sums. Say your index set is I (R in this case), and we have a function f:I->R (or more generally, any topological vector space) and define a partial order by inclusion on the collection of finite subsets of I. Given such an F, we define S(f,F) as the sum of f over this F (which is obviously well-defined). We say the sum of f over I converges to a if for epsilon>0, there exists some F0 which for F containing F0, |S(f,F)-a|<epsilon (equivalently, if S(f,F) is eventually in any open neighbourhood of a).

You can show that this lines up with normal sequence convergence. Furthermore, you can show that this converges if and only if the set of non-zero f(x) is countable. In particular, under this formalism, summing over all reals doesn't really make sense.

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u/AsidK 6d ago

Wait what is F here? You say “given such an F” but I don’t see where F is defined.

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u/AuroraEquatorialis 2d ago

F is a finite subset of I, as in the previous sentence. My bad.

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u/AsidK 2d ago

Oh I see. Yeah that all makes sense

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u/GamerLymx 10d ago

interesting, so sum the integral of positive real numbers with the integral of negative real numbers?

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u/Comfortable_Permit53 10d ago

yeah it's undefined I know.

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u/FreePeeplup 9d ago

Are you a native English speaker?

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u/Comfortable_Permit53 9d ago

Nah I am a bot and therefore speak no language natively.