This is like knowing there's two possible answers, but choosing to pick a 100% wrong answer instead

this is like knowing there's two possible answers, but choosing to pick a 100% wrong answer instead.

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No one believes that this series is equal to 1/2. It's non-convergent meaning it can't be represented by a number.

no its not you stupid american.

> it can't be represented by a number
Yeah it can. It just doesn't converge to a number

>someone stupid must be American
Das raciss

The 1/2 is what the series tries to converge to but never does.

There are different ways to derive the 1/2.
>common sense: the series just jumps back and forth between 0 and 1
>set the Sum equal to S and play with S
>playing with integration of sinusoidal and exponential functions

Reasoning is something like this
[math]1-1+1-1+1-... = x[/math]
[math]1-x = 1-(1-1+1-1+1...) = 1-1+1-1+1-1... = x[/math]
[math]\Rightarrow 1 - x = x \Leftrightarrow 1 = 2x \Leftrightarrow \frac 1 2 = x[/math]

>you can do operations with series like you can do with humber

[math] \dfrac { 1 } { 1-z } = 1 + z + z^2 + z^3 + z^4 + \dots [/math]

[math] z = -1 + \epsilon [/math]

with

[math] \epsilon > 0 [/math]

So

[math] \dfrac { 1 } { 2 - \epsilon } = 1 - 1 + \epsilon + (-1 + \epsilon)^2 + (-1 + \epsilon)^3 + (-1 + \epsilon)^4 + \dots [/math]

or

[math] \dfrac { 1 } { 2 - \epsilon } = 1 - 1 + O(\epsilon) + 1+ O(\epsilon) -1 + O(\epsilon) + 1 + O(\epsilon) + \dots [/math]

There is a way to get 1/2 as a number to *relate* to this series, but all three of those methods are bullshit.

It's just analytic continuation
terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

>America leads the world in mathematics
>they must be idiots

Lolno it really doesn't.

Another way to see this sort of behavior is by working in reverse: start with a power series that is already well defined by the standard rules of convergence.

Let [math]f(t) = 1 + t + t^2 + \ldots [/math] where we restrict [math] -1 < t < 1 [/math] in the reals. We know that the power series given here is well defined for these values of [math] t [/math] so we have a well defined function and domain.

Next we recall that this series *agrees* with the function [math] \frac{1}{1-t} [/math] on this domain. The next step is to say that if we want to extend [math] f(t) [/math] to further values of t we can just use this identity to "extend" our original function.

If we view this extension as preserving the original "series structure" of [math] f [/math] then we can plug stuff into both sides and say we have an equality (as long as our extension is well defined there: so not at t = 1.)

So, thinking [math] 1 + t + t^2 + \cdots = \frac{1}{1-t} [/math] for all values of t not equal to 1 we get things like (plugging in -1)

[math] 1 - 1 + 1 - 1 + \cdots = \frac{1}{1-(-1)} = \frac{1}{2} [/math]

So while you can see where such an equality might be suggested, it's not really there in the way you might worry it is (in the sense of partial sums giving a well defined limiting value)

That's still more medals, faggot.

>my shithole country is smaller, therefore it's better!

The integration method is pretty legit and has more rigor than the other two methods. Again, I never said that the series ever convergences but the 1/2 is the value the series tries to converge to.

>medals per capita

nice scam-post

P = NP
P/P = N
N = 1

Make that 2 for Canada.

what is the number then famaladindong

>>set the Sum equal to S and play with S

This is wrong because the moment you set it equal to S you are already implicitly assuming that such an S exists.

The only correct answer here is saying that this sum converges in the Cesaro way to 1/2.

Everything else is wrong popsci bullshit made up by Numberphile to teach retards.

Well the S sum certainly isn't rigorous.

I still like the integration method though.

Given any epsilon>0, the sum of the infinite epsilons diverges to infinity. So your right side diverges while your left converges.

You fail.

>dividing by something that can be 0
Canadians confirmed for brainlets.

>You fail.
Sounds like 2005. Are you trying to be rude?

Also, you're wrong. The geometric series converges for z in (-1,+1), in which z=-1+epsilon falls.

O(epsilon)'s doesn't mean all expression look the same, or even have the same sign.

when retards who don't study math learn math from shitty clickb8 pop math channels

How the fuck did they get to 1/2?

>hurr i dont need no educationz i have vsauce

>lol i lerned so much more from this video than my teachers have ever taught me! teachers are sooo le dumb, amirite??? XD

It's just a sum that tries to converge to 1/2 but never actually does.

>America leads the world in mathematics

HAHAHAHhahahahahahahahahaha

Thanks for the laugh brainlet. When compared accordingly, the US isn't even in the top 20 countries with the best math scores. Congrats though on winning the Google Mathematical Olympiad these last two years, I think China got bored of winning 14 times during America's drought of medals. And even then, it's always fun seeing the diversity in American teams. Pic related, true Americans.

Z-transform would give 1/2

>skinny little faggots who think they're smart

Let's see how many of them grow up to be failures and end up committing sudoku.

>tries to converge to 1/2 but never actually does
What? Do you know what "convergence" means?

With each step, the sum never actually makes any progress to get to 1/2.

Take a sum like 1+1/2+1/4+1/8 +... . Yes, there are an infinite number of steps but each step is one step closer to the convergence value.

Was this before or after students started rioting because teachers had to start caring about cheating

Surely after. When the challenge appeared, they had to dedicate their entire days studying and learning in order to stay on their level. Explains why they look like they're in such a terrible state right now.

The middle guy looks weird and awkward. Small head and big ass legs.

>one canadian is brainlet
>therefore all are

*your country* confirmed for brainlets.

lol dude that's not the same as OP's sum, his didn't have that big backwards 3 thing

Delete this post.

>The middle guy looks weird and awkward. Small head and big ass legs.

Y-yeah, totally. I'd probably kill myself if I looked that goddamn ugly...

You did a good job mating, congratulations

...

why

integrate and average

Year 2036
>tfw u will never be a physicist with a PhD in String Theory
>tfw you will never make a bunch of extraordinary findings using the -1/12 result
>tfw u will never have a mid-life crisis when a mathematician disproves it
>tfw u will never feel so much agony that u shoot yourself in the head while burning your useless research papers

It does not converge. I'm American and even I know that.

The very first thing they teach you about sequences is that if a sequence has 2 subsequences which converge to different values, then the sequence itself does not converge.

How does it feel to be dumber than me?

Thank you. I remember meeting some students in joint honours physics & math who tried to tell me the 'amazing' fact that this sum 'equals' 1/2 (as well as the infuriating 1+2+3... = -1/12 one). I felt like pulling my goddamn hair out when they disregarded this same explanation.

I can't believe how many people just absorb popsci and then regurgitate it.

>tfw some faggot in first year of college unironically tried ot argue that 0.9999... =/= 1
>"dude, it's a bit smaller than 1"

>based France
Frexit when????

b-brazil got one too