Can someone explain to me how the sum of an infinite number of

it is zero.
it's the sequence:{0,0,0,0,0...} which converges to zero

>>8527339
Why wouldn't it be zero? What is happening in this board? Is /B/ raiding us?

>>8527348
>>8527371
Doesn't integration require the sum of an infinite number of zeros not to equal zero, though. If you imagine integration as a Riemann Sum as the area of the box approaches zero, then you're just summing a bunch of zeros together and getting a non-zero number.

Or, since I'm more familiar with statistics, we know that, given a continuous distribution, the probability of getting any one number in particular is 0%, but the probability of getting a range of scores is non-zero. Thereby, summing an infinite number of zero probability events, gives you a non-zero probability.

>>8527380
that's not how integration works

>>8527380
Integration isn't the same as summing a series.

Also, the thing that you are talking about is basiclly one of Zeno's paaradox:
https://en.wikipedia.org/wiki/Zeno%27s_paradoxes#Paradox_of_the_Grain_of_Millet
Anyway, if you want to know the sollution to this paradox check out Lebesgue measure.

File: Screenshot_20161209-113142_1.jpg (44KB, 1072x341px) Image search: [Google]

44KB, 1072x341px

>>8527383
>>8527391

As authoritative a source as Khan Academy is, this is how they define an integral. I also have two calculus textbooks that both define integrals that way.

So, at the very least, it seems to be a very popular misconception that integration is done by taking the sum of an infinite number of zeros. Though if you can show that it's not, I'm happy to be convinced.

>>8527399
the thing is, for every Riemann sum you have a finite amount of terms with finite value each. not the sum of infinitely many zeros.

>>8527339
Only the sum of an uncountable amount of zero's can be non-zero. Which is (informally) the case in the Riemann integral.

>>8527399
Its a bit more complicated, notice that this definition depends on the way that you choose the partintion of the interval that you integrate over and also on the way that pick the points to evaluate f(x_i), so by this definition an integral of a function f over [a,b] could be two different numbers.

>>8527418
This is why the Riemann integral is just the walmart edition of the Lebesgue integral.

>>8527418
it works for continuous functions and that what most people find anyway.
I like the definition on Baby rudin tho, for bounded functions and upper and lower sums

>>8527423
this isn't even the Riemann integral. it's the naive integral

>>8527427
but the poimt is, this definition here>>8527399
is not even the real definition of the Riemann integral. Plus the Lebesgue integral is not strictly better then Riemann, there are functions that are Lebesgue integrable but not Riemann integrable.

>>8527339
This is how

0 = 0 + 0 + 0 + ... = (1 - 1) + (1 - 1) + ... = 1 + (-1 + 1) + (-1 + 1) + ... = 1 + 0 + 0 + ... = 1 + 0 = 1

Boom.

>>8527436
That WOULD make the Lebesgue integral better than the Riemann integral.
One integral isn't necessarily better than another.
The Lebesgue doesn't totally supersede the Riemann integral either.

>>8527470
which function is Riemann integrable but not Lebesgue integrable?
no improper integrals pls

>>8527470
eeh i ment the other way around, and yes you re right, it works in both ways

>>8527471
On closed bounded intervals, they're the same.
Infinite intervals is where things mess up.

>>8527473
Yeah, this.

>>8527471
well if no improper integrals, then Riemann integrable implies Lebesgue.

>>8527476
>>8527477
but Riemann integral isn't defined on non compact sets innit?

>>8527487
true, but you can easy extend the definition to arbitral interval, even infinite.

>>8527487
It can be defined for infinite intervals. Improper Riemann integrals exist and are perfectly valid.
But it's with infinite intervals that there are issues with them matching up.

>>8527525
but on finite interval they match "only in one direction", in the sense, that you have functions that are Lebesgue integrable but not Riemann.

>>8527454
So then the sum is 1/2?

WAIT

0 = 0 + 0 + 0 + ... = (2 - 2) + (2 - 2) + ... = 2 + (-2 + 2) + (-2 + 2) + ... = 2 + 0 + 0 + ... = 2

But since 2 is arbitrary, we could replace 2 with infinity

But then we would have an infinite number of sum within an infinite sum, so like infinity^infinity^infinity^infinity... to infinity

>>8527399
They sums approach an epsilon, and epsilon is always greater than 0.