Proof of specific integral

[silver hollow]

there're at least two different methods of proving this

1. subinterval endpoints in arithmetic progression (i..e. equal subinterval width)
2. subinterval endpoints in geometric progression (i..e. consecutive subinterval widths has a common ratio)

for the second approach, use the technique in this question: #1183754217404514364 message

[uncut abyss]

@silver hollow what about induction for integers n

[silver hollow]

nice trick 🙂

[silver hollow]

we give hints/guidance, and it's you to solve the problem

[uncut abyss]

what is this

[silver hollow]

there're rules of this server above yours. if you don't like that, close it.

[uncut abyss]

@silver hollow

bro

this isn't a help post

it's a "discussion" post

[uncut abyss]

both of me and Vin are 13

it was an obvious joke

both of us are above 24

Anyways about the integral

first off, do you know what an upper/lower reiman sum is

and also the fact that continuous functions on (a, b) are reiman integrable on [a, b]

@plucky yew what are your preliminary ideas?

on this

do you know what "length" of an interval is

and the norm of a partition (biggest length of subinterval)

I think I get what you mean, never called it "delta..."

anyways

we define the length of an interval $I$ with $\sup I = b, \inf I = a$ as $\abs{I} = b-a$

[lucid lintel]

a domesticated coffey

[uncut abyss]

@plucky yew

it's not the same thing as distance

now let's get to the actual integral

$\int_{[0,a]}x^n$

[lucid lintel]

a domesticated coffey

[uncut abyss]

you wish to compute this without anti derivatives right?

without the fundamental theorem of calculus

okay

first off could you prove these for me

$U(f, P) \geq U(f, P')$ Where $P'$ is a refinement of $P$

[lucid lintel]

a domesticated coffey

[uncut abyss]

and also prove that, $\abs{\abs{P}} \geq \abs{\abs{P'}}$

where the ||.|| is the partition norm

[lucid lintel]

a domesticated coffey

[uncut abyss]

i think so too yes

after this prove the sufficient and necessary condition for P being the partition such that $\inf_{\text{K parts I}} U(f, K) = U(f, P)$

[lucid lintel]

a domesticated coffey

[uncut abyss]

it's $\abs{\abs{P}} \to 0$

[lucid lintel]

a domesticated coffey

[uncut abyss]

which one exactly

like how to use the latex bot?

just use latex commands within "$"

it'll work anywhere :3

[uncut abyss]

Try copy pasting my code

[lucid lintel]

Jackson_Gunshot

[uncut abyss]

anyways did you manage to prove it?

I'm assuming you know analysis that's taught prior to reiman sums

do you know how U(f, P) is defined?

the upper reiman sum, we just find the supremum of our function in each of the sub intervals and multiply it to the length of the subinterval and add all of these up

$\sum_{J \in P} \sup_{x \in J} (f(x)) \abs{J}$

[lucid lintel]

a domesticated coffey

[uncut abyss]

this is the upper reiman sum

the lower one is just the Infimum

the sum can seem weird so think of it geometrically

@plucky yew

you don't need the max quality, it does the job

so do you get what the lines are?

the blue are the upper sums

the red the lower

[uncut abyss]

you should actually need anything except this

okay

do you agree that if there exists a partition $P$ such that $\abs{U(f, P), L(f, P)} < \varepsilon$ for all such positive epsilon, the function is reiman integrable

[lucid lintel]

a domesticated coffey

[uncut abyss]

i believe so

nope

that's the worst thing you could do while learning stuff you'll actively use ij the proof

@plucky yew ill ask again, have you done the analysis before they teach reiman sums? then ill get straight to the proof

alright

you need a monotone increase sequence $(a_n)$ such that $a_0 = a, a_n = b$ this sequence will partition our interval $I = [a, b]$

[lucid lintel]

a domesticated coffey

[uncut abyss]

$I = [a_0, a_1)\cup [a_1,a_2)\cup \ldots \cup [a_{n-1},a_n]$

[lucid lintel]

a domesticated coffey

[tribal gust]

can't we just say

[uncut abyss]

now you will clearly see that the P-norm here is $\sup \qty{\abs{a_{i+1}-a_i} : 0 \leq i \leq n-1}$

[lucid lintel]

a domesticated coffey

[tribal gust]

can't we just say $\int x^n \sim \sum x^n \sim \frac{x^{n+1}}{n+1}$

[lucid lintel]

cute rizzly bear (nom nom nom)

[uncut abyss]

yes?

although I wouldn't say that

[tribal gust]

actually, this is probably how you would derive the asmptotic growth of the sum

which is less obvious than the integral

so it would go the other direction

[uncut abyss]

we can do this for integers

with the recurring power sum equation

[tribal gust]

like can you derive that $\sum_{n=1}^{N} n^e \approx \frac{N^{e+1}}{e+1}$

[lucid lintel]

cute rizzly bear (nom nom nom)

[tribal gust]

withuot using integral

[uncut abyss]

but I doubt he knows what asymptomatic growth is

yes I do

I don't know it myself 🤣

if you did know what. zfn said is the instant proof

anyways

continuing

[uncut abyss]

so we just wish the make the partion very very fine

we can do this with an arithmetic sequence or an geometric sequence both

i mean you should know why before knowing how

anyways

take a arithmetic sequence

from a to b

$a_k = a + (b-a) \cdot \frac{k}{n}$ Where $n \geq k \geq 0$

[lucid lintel]

a domesticated coffey

[uncut abyss]

also you can take any value of the function in the specific intervals

don't ask me why

while taking the limit

and our sum here would be

$\int_{[0,b]}x^{\alpha} =\lim_{n \to \infty} \sum_{k=0}^n \left( \frac{b-a}{n} \right) \cdot \left( \Delta _k\right)^\alpha$ Where $\Delta k \in (a{k-1},a_k)$

now, what value of delta_k should we choose in our interval

what about it specifically?

this is the limit you wanted

[lucid lintel]

a domesticated coffey

[uncut abyss]

i mean sure

$(b-a) \lim_{n\to \infty} \sum_{k=0}^n \frac{\left( a + \frac{(b-a) k}{n} \right)^{\alpha} }{n}$

[lucid lintel]

a domesticated coffey

[uncut abyss]

I should leave the computation to the reader 🫡

adieú

@plucky yew set a = 0

and you'll get the integral from 0 to b

$b \lim_{n\to \infty} \sum_{k=0}^n \frac{\left(\frac{b k}{n} \right)^{\alpha} }{n}$

[lucid lintel]

a domesticated coffey

[uncut abyss]

$b^{\alpha +1} \lim_{n\to \infty} \frac{1}{n^{\alpha +1}}\sum_{k=0}^n k^{\alpha}$

[lucid lintel]

a domesticated coffey

[uncut abyss]

recheck it just it case I made some error

[uncut abyss]

@plucky yew we just use this here

[uncut abyss]

what??

use asymptomatic behavior

i mean

you can derive a formula for the power series right?