THAT'S NOT ZERO!!! [MIT Integration Bee 2019]

[@gabrieldogilev1549]

(-1)^n*(x)^n converges when x is between 0 and 1 If you have (-1)^n multiplied by a strictly decreasing series and lim of said series to inf is 0, (-1)^n multiplied by the series converges

[@Silver-cu5up]

Thank you for that!!

[@mohannad_139]

Yeah: −1 < x < 1 » 0 x = 1 » 1 x > 1 » ∞ x ≤ −1 » undefined

[@СтојанЦупаћ]

Sequence x^n where - 1 < x < 0 converges to 0 since it is a product of two sequences: (- 1)^n and (-x) ^n (0 < -x < 1) where the first one is bounded and the other converges to 0. For x < - 1 it also diverges since its two subsequences - (-1)^(2k)*x^(2k) and (-1)^(2k+1)*x^(2k+1) converge to positive and negative infinity respectively.

[@dimastus]

I though, the plot-twist would be about: 1) 'n' which is not specified to be naturals 2) and infinity, which is not 'plus infinity'.

[@joshuaiosevich3727]

Before I even watch the video, I thinks 2 cuz the function converges to the indicator of -1,1

[@ripjawsquad]

what's an indicator?

[@joshuaiosevich3727]

@ an indicator function sometimes notated I_{A) is a function where I_{A}(x)=1 if x /in A 0 otherwise

[@MooImABunny]

same

[@holyshit922]

but can we use Gamma function here

[@Silver-cu5up]

@@holyshit922 idk, itll look nasty maybe

[@AsiccAP]

Why can we justify that integrating from -inf to inf is the same as twice the integral from 0 to inf? From a problem solving standpoint of the integration bee, we of course know an answer exists, therefore any way we treat the bounds should lead to the same results and therefore we can treat -inf to inf as -t to t as t goes to infinity. But let's say we aren't in MIT integration bee and we must first determine if the integral exists at all. Why could we justify integrating from -inf to inf is the same as -t to t and not -t to 2t or anything like that?

[@codyriceandothers]

The function is even so you can make the given intgeral equal to twice the integral from 0 to infinity. The Cauchy Principal Value of the integral equals the integral from -infinity to infinity because you can prove the integral converges.

[@AsiccAP]

That is basically my point. I know that for the cauchy principle value, we could use the result with even functions, and the integral converges therefore it should not matter how we approach infinity and we can use the Cauchy principle value. My question is why it converges in the first place, which is skipped in the video because obviously if the integral is presented in the integration bee, it definitely exists. I'm not good at analysis, so I couldn't figure out a way to prove it's convergence. Maybe using a larger function and proving its convergence, but I couldn't figure out that function.

[@koopakidlarry8408]

It converges because the constant function 1 dominates exp(-x^(2n)) for x in [-1,1], and exp(-x^2) dominates exp(-x^(2n)) for x in (-∞,1) and x in (1,∞). It is well known that exp(-x^2) is an integrable function over the whole real line. This can be shown using polar coordinates. Therefore, by the dominated convergence theorem, we can swap the limit and the integral, and the integral exists for all n. Because the integral exists, the limit exists, and therefore the symmetry argument still holds even for the limiting case, as limits are unique.

[@AsiccAP]

@@koopakidlarry8408 thanks, that is the method I was trying to go for. I figured f(x)=1 for x in [-1, 1] is an upper bound, but couldn't figure out the rest of the domain. Turns out you just use the case for n=1 itself as an upper bound. what a clever method.

[@soreto314]

"Be very careful" 0:05 3:14 4:33 4:47

[@Silver-cu5up]

LMAO

[@aleksandervadla9881]

LMCT

[@축복-l1l]

asnwer=1dx isit

[@축복-l1l]

asnwer=2dx what isit