Math peeps explain this to me

[Bird222]

https://www.youtube.com/watch?v=w-I6XTVZXww

I can buy that S=1-1+1-1..... equals 1/2 but is there something in math that says you can shift one of the equations over one digit when adding to equations together? To me that seems where the flaw in this is. Maybe it is perfectly 'legal'. Can someone explain how if it is?

 

[lozina]

yeah I remember seeing this for the first time. cool stuff, video explains it very well it's just that the conclusion is so incredible

 

[Mark R]

It's "legal" if you change the definition of "sum". Look up "Ramanujan Summation".

Some of the algebra using the series is also dubious - it works in this case - but adding and subtracting infinite series doesn't work reliably and you can end up proving that 1=2 or some other nonsense.

 

[Squisher]

I don't think I'm going to sleep tonight.

 

[highland145]

That shift looks convenient to give the desired results.

 

[cbrunny]

Yeah you can do the shift because there is no multiplication or division.

e.g.

x = 1 + 2 + 3 + 4 = 10
y = 1 - 2 - 3 - 4 = -8

x + y will always equal 2, no matter what order you add the numbers in.

 

[TwiceOver]

"We don't know so we take the average"

I'm sure they are right, far smarter than I am, but that just feels like "Eh, math, close enough". LOL

 

[Bird222]

That shift looks convenient to give the desired results.

See that's what I thought.

 

[highland145]

See that's what I thought.

"Eh, math, close enough".

this.

 

[blinblue]

"We don't know so we take the average"

I'm sure they are right, far smarter than I am, but that just feels like "Eh, math, close enough". LOL

Basically funky things happen with infinity alternating sums. That bit where he says S_1 = 1/2 is true enough, but you can also make it "equal" to any number you like.

Now I have no idea what what the implications in physics are, there is likely some reason why 1/2 is a useful number to use. But mathematically you could make it "equal" whatever you want.

Here's a much shorter video using the same sort of thing
https://www.youtube.com/watch?v=kIq5CZlg8Rg

 

[Ferzerp]

Ramanujan sums are not sums in the traditional sense. Your video doesn't address this, hence this confusing result.

It's easier to think of it like this. 1-1+1-1... doesn't *really* have a sum. But if we were to accept it is 1/2, what are the ramifications? It's sort of a completely different kind of sum where 1-1+1-1... is defined as 1/2.


edit: I may have the nuances wrong, so any mathematicians here should feel free to clarify. It's been ages since I looked at this stuff. The gist of it though is what I have posted.

 

[tmc]

yep, this exact 1+2+3+...=-1/12 equation was in Ramanujan's first letter to Hardy.

 

[steppinthrax]

https://www.youtube.com/watch?v=w-I6XTVZXww

I can buy that S=1-1+1-1..... equals 1/2 but is there something in math that says you can shift one of the equations over one digit when adding to equations together? To me that seems where the flaw in this is. Maybe it is perfectly 'legal'. Can someone explain how if it is?

1+2+3+4+5+6..... Can be represented like this ----> Summation of 1 to infinity of n.

This is a divergent series, meaning it is infinite and can't be calculated.

However, if you graph the series you get this...

http://www.wolframalpha.com/input/?...y&a=*FVarOpt.1-_**-.***Sum.sumvariable---.*--

The sum diverges. However, that summation can be represented by a partial sum (also on the wolfram page).

If you graph the partial sum for Y you will get a parabola.

http://www.wolframalpha.com/input/?i=graph+of+y+=+1/2x(x+1)

If you use calculus you can find the area of the portion underneath the Y-axis on the left side of the graph. This area is equal to -1/12.

THE ROOTS ARE -1 AND 0.

-1/12 is represented by the integral of the partial sum.

http://www.wolframalpha.com/input/?i=integral+of+1/2+x(x+1)+from+-1+to+0

The integration of the partial sum yields -1/12.

Since both sides of the parabola are divergent, you solve for the area underneath the y-axis and set the boundaries of the integral to the roots of that area.

 

[disappoint]

I'll tell you if shifting is "legal" or "legit":

If you don't shift it, or if you shift it by one more, or 2 more, do you get the same answer? If you do get the same answer, then why did you have to shift it in the first place? If you don't get the same answer, then how is it legit to do so?

 

[DrPizza]

https://www.youtube.com/watch?v=w-I6XTVZXww

I can buy that S=1-1+1-1..... equals 1/2 but is there something in math that says you can shift one of the equations over one digit when adding to equations together? To me that seems where the flaw in this is. Maybe it is perfectly 'legal'. Can someone explain how if it is?

The entire proof is based on you buying that 1-1+1-1... = .5
It's not a convergent series.

 

[DrPizza]

I'll tell you if shifting is "legal" or "legit":

If you don't shift it, or if you shift it by one more, or 2 more, do you get the same answer? If you do get the same answer, then why did you have to shift it in the first place? If you don't get the same answer, then how is it legit to do so?

Most of the time, the shifting is merely an application of the commutative property of addition and the associative property of addition.

 

[Drako]

That first guy needs a chin.

 

[Bird222]

Most of the time, the shifting is merely an application of the commutative property of addition and the associative property of addition.

So is it legitimate them shifting that one equation over?

 

[dighn]

The entire proof is based on you buying that 1-1+1-1... = .5
It's not a convergent series.

if anything it should be lim (sum(1 to n)/n) as n -> infinity = 0.5

 

[Dari]

1+2+3+4+5+6..... Can be represented like this ----> Summation of 1 to infinity of n.

This is a divergent series, meaning it is infinite and can't be calculated.

However, if you graph the series you get this...

http://www.wolframalpha.com/input/?...y&a=*FVarOpt.1-_**-.***Sum.sumvariable---.*--

The sum diverges. However, that summation can be represented by a partial sum (also on the wolfram page).

If you graph the partial sum for Y you will get a parabola.

http://www.wolframalpha.com/input/?i=graph+of+y+=+1/2x(x+1)

If you use calculus you can find the area of the portion underneath the Y-axis on the left side of the graph. This area is equal to -1/12.

THE ROOTS ARE -1 AND 0.

-1/12 is represented by the integral of the partial sum.

http://www.wolframalpha.com/input/?i=integral+of+1/2+x(x+1)+from+-1+to+0

The integration of the partial sum yields -1/12.

Since both sides of the parabola are divergent, you solve for the area underneath the y-axis and set the boundaries of the integral to the roots of that area.

this is nonsense.

 

[Schmide]

You can't just take the avg of an infinite sum. It either converges or diverges.

 

[dullard]

A) S1 = 1 - 1 + 1 - 1 + 1 ... = [0] or [1] depending on how many terms you use.
Thus, S1 for n terms is 0 if n is even or 1 if n is odd.

The average result is 1/2, but you never ever get the average result, you get either 0 or 1.

B) S2 = 1 - 2 + 3 - 4 + 5 - 6 = [n / -2] if n is even or [(n+1) / 2] if n is odd.

The average result is even stranger. If n is even, the average result is 0. If n is odd, the average result after many terms approaches 1/2. So the average of the average is 1/4. But again, you never actually get 1/4. You get [n / -2] or [(n+1) / 2].

In the video, they added S2 with an even number of terms to S2 with an odd number of terms. So, they aren't actually playing by the rules. They are using a sum of terms which is neither odd nor even. Either n is even or n is odd, but they use both.

C) From [S - S2 = 4S] they get [-S2 = 3S]. But, you can only do that math if S is not infinite. In other words think about this: let Y = X + 1 and Z = X + 2. If X is infinite, then you get Y is infinite and Z is infinite. Thus Y = Z. Thus, X + 1 = X + 2. Subtract X from both sides, and you get 1 = 2. The problem is that you can't add or subtract infinity from both sides (just like you can't divide by zero on both sides and get anything meaningful). So they are fudging again if S happens to be infinite.

So lets assume that S is not infinite. In that case, S = -S2 / 3. Lets look at -S2 / 3. That value would be -S2 / 3 = [n / 6] if n is even or [(n+1) / -6] if n is odd. With an infinite number of terms, n, you get [-S2 / 3 = infinity] or [-S2 / 3 = negative infinity]. Thus S is infinite, which violates our assumption. So, you can't go from [S - S2 = 4S] to [S = -S2 / 3].

In conclusion, if you average a number S1 that can't be averaged, add in the average of an average S2 that also can't be averaged, then forget whether you are using an even number of terms or an odd number of terms, and finally subtract infinity from both sides, you will end up with 1+2+3+4+5+... = -1/12.

That or I've had too much wine.

 

[boozzer]

"We don't know so we take the average"

I'm sure they are right, far smarter than I am, but that just feels like "Eh, math, close enough". LOL

nope, the guy clearly explained. depends on where you stop, even or add number, you get 0 or 1. that is why you avg the 2 results and get 1/2.

 

[dullard]

Oh and one last thing. Most of the proof relies on adding averages to averages (or subtracting them). That is usually a bad thing.

Take this set of numbers for example: 0, 0, 0, 0, 5, 6, 10. The average of that set is 3.

But what if you break it up into two pieces? Say piece A is 0, 0, 0, 0, 5. And piece B is 6, 10. The average of A is 1. The average of B is 8. But adding or subtracting or averaging the average pieces just leads to useless numbers. A + B = 9, which is useless. A - B = -7, also useless. The average of A and B is (A + B) / 2 = 4.5, also useless. No matter what you do with the piece averages, you won't get the true entire set average (3) back.

 

[steppinthrax]

this is nonsense.

How? Explain