[QUOTE=Dead Madman;43497816]I seriously hope none of you guys ever ask why you're virgins[/QUOTE]
What on earth does that have to do with anything?
[QUOTE=Dvd;43497423]Hi. I'll just leave this here.
[url]http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation[/url]
[url]http://en.wikipedia.org/wiki/Ramanujan_summation[/url]
This video came up at my work. I consider it a form of intellectual dishonesty -- they're not making it clear that the method of summation is entirely different from what any normal person would consider 'summation'.
Their answer to 1-1+1-1+1... is likewise suspect outside of Cesaro's method. That is a nonconvergent series, meaning that it does not have a finite sum.[/QUOTE]
They have done a video on Grandi's series before.
[QUOTE=Dead Madman;43497816]I seriously hope none of you guys ever ask why you're virgins[/QUOTE]
Please, I get all the ladies doing math.
"Would you like to see the exponential growth of my natural log?"
[QUOTE=Goodthief;43497377]no it works fine. you just forgot that you have to do 0-1 at the end, making the total 0. e.g.
(1+1+1+1)-(1+1+1+1) is the same as
1+1+1+1+0 -
0+1+1+1+1
notice how you have 1-0 at the start and 0-1 at the end, this totals up to 0.[/QUOTE]
[QUOTE=Zyx;43497402]Except in "Theoretical end" of the non-shiftet S there will be a +0, making the result 0 again. Will there not?[/QUOTE]
What end?
[editline]11th January 2014[/editline]
Also, how about this? What do you do at "the end"?
[IMG]http://i.imgur.com/u2WIV3a.png[/IMG]
[QUOTE=Quiet;43497985]What end?[/QUOTE]
Like I said, the theoretical end. Shifting numbers like that doesn't change the result.
Just like [code](1+1+1)-(1+1+1)[/code] is the same as [code](1 +1 + 1) -
(1+ 1 + 1)[/code]and as well as this
[code]
(1+1+1 -
(1+1+1)[/code]
doesn't equal 3?
this is retarded. it does not fucking equal -1/12
wow everything I know is a lie this is the worst day of my life
[QUOTE=Zyx;43498125]Like I said, the theoretical end. Shifting numbers like that doesn't change the result.
Just like [code](1+1+1)-(1+1+1)[/code] is the same as [code](1 +1 + 1) -
(1+ 1 + 1)[/code]and as well as this
[code]
(1+1+1 -
(1+1+1)[/code]
doesn't equal 3?[/QUOTE]
The point I'm trying to make here is that talking about infinite sums has no meaning unless the concept is given a definition. Sure you can display an operation whichever way you wish without bearing on the result, but talking about the end of an infinite sum is absurd.
Care to comment on my other example?
There must be a mistake somewhere in there. It can't equal -1/12
[QUOTE=Quiet;43498176]Care to comment on my other example?[/QUOTE]
Nope, no idea.
[QUOTE=Laserbeams;43498177]There must be a mistake somewhere in there. It can't equal -1/12[/QUOTE]
yeah this guy is an idiot you can't just average out some infinitely repeating equation (which is already theoretical and not practically useful in any way) and then get -1/12
if you can just add anything to anything and do whatever you want I think we should really work out what 1/0 equals I have been dying to know
[QUOTE=Penguiin;43498268]yeah this guy is an idiot you can't just average out some infinitely repeating equation (which is already theoretical and not practically useful in any way) and then get -1/12
if you can just add anything to anything and do whatever you want I think we should really work out what 1/0 equals I have been dying to know[/QUOTE]
uh yes you can
this is how divergent series work and it is actually practical and useful in physics
[QUOTE=Penguiin;43498268]yeah this guy is an idiot you can't just average out some infinitely repeating equation (which is already theoretical and not practically useful in any way) and then get -1/12
if you can just add anything to anything and do whatever you want I think we should really work out what 1/0 equals I have been dying to know[/QUOTE]
No, that's not quite what I mean. I can't call him an idiot, he's probably a really good mathematician, and there's some string theory stuff that seems to support this, but it's still pretty unbelievable. Something has to be wrong, it just doesn't make any sense
[QUOTE=Laserbeams;43498288]No, that's not quite what I mean. I can't call him an idiot, he's probably a really good mathematician, and there's some string theory stuff that seems to support this, but it's still pretty unbelievable. Something has to be wrong, it just doesn't make any sense[/QUOTE]
because you're not talking about how all of the numbers equal this one specific value, thats not how these series work
The mistake is claiming that a non-convergent infinite sum has a value. It does not, the result is undefined in the same way that dividing by zero is undefined.
[QUOTE=Dead Madman;43497816]I seriously hope none of you guys ever ask why you're virgins[/QUOTE]
get a load of this tool hahahahahha
[QUOTE=Ziks;43498347]The mistake is claiming that a non-convergent infinite sum has a value. It does not, the result is undefined in the same way that dividing by zero is undefined.[/QUOTE]
but it does
"In mathematical analysis, Cesàro summation is an alternative means of assigning a sum to an infinite series. If the series converges in the usual sense to a sum A, then the series is also Cesàro summable and has Cesàro sum A. The significance of Cesàro summation is that a series which does not converge may still have a well-defined Cesàro sum."
they are just applying this to the infinite series and calculating the results based on this base result
Think about the series S(n) = 1-1+1-1+1-... with n many components. The range for the result is the set {0,1} for any n >= 0. If you claim that S(infinity) = 0.5, this contradicts the fact that the range is {0,1} as 0.5 is not a member of that set. So S(infinity) != 0.5, it is undefined.
the fundamentally weird part of all of this is the [URL="http://en.wikipedia.org/wiki/Cesàro_summation"]sum of the Grandi series[/URL]. it's not a "proper" summation, but if you do it on other convergent series, you get the answer you would with "proper" summation, so it follows that it has some kind of mathematical usefulness. if you can believe that, then the rest follows without too much trouble.
[QUOTE=Ziks;43498398]Think about the series S(n) = 1-1+1-1+1-... with n many components. The range for the result is the set {0,1} for any n >= 0. If you claim that S(infinity) = 0.5, this contradicts the fact that the range is {0,1} as 0.5 is not a member of that set. So S(infinity) != 0.5, it is undefined.[/QUOTE]
this is a divergent series, meaning you take the average of both results
this is not normal arithmetic; this is mathematical analysis, you're arguing against a type of math
[QUOTE=Loriborn;43498392]but it does
"In mathematical analysis, Cesàro summation is an alternative means of assigning a sum to an infinite series. If the series converges in the usual sense to a sum A, then the series is also Cesàro summable and has Cesàro sum A. The significance of Cesàro summation is that a series which does not converge may still have a well-defined Cesàro sum."
they are just applying this to the infinite series and calculating the results based on this base result[/QUOTE]
The Cesàro sum may be useful in some situations, but it is not sound. It's just a tool used for dealing with non-convergence.
[editline]11th January 2014[/editline]
Let's say I introduced a new tool called the Ziks Principle, which defined 1/0 to be 0. You could use it to produce whatever result you liked, and it may have some uses for avoiding undefined results, but it would lead to contradictions in regular arithmetic just as the Cesàro sum does.
[editline]11th January 2014[/editline]
So basically you agree that 1+2+3+4+... does not equal -1/12 when using standard arithmetic, but only if you apply an analysis tool that is only useful in certain situations but leads to contradictions in others and so is not universally valid?
[QUOTE=Ziks;43498416]The Cesàro sum may be useful in some situations, but it is not sound. It's just a tool used for dealing with non-convergence.
[editline]11th January 2014[/editline]
Let's say I introduced a new tool called the Ziks Principle, which defined 1/0 to be 0. You could use it to produce whatever result you liked, and it may have some uses for avoiding undefined results, but it would lead to contradictions in regular arithmetic just as the Cesàro sum does.
[editline]11th January 2014[/editline]
So basically you agree that 1+2+3+4+... does not equal -1/12 when using standard arithmetic, but only if you apply an analysis tool that is only useful in certain situations but leads to contradictions in others and so is not universally valid?[/QUOTE]
You realise that you are arguing against a professor of physics on this? And against the scientific literature that have documented this?
Apparantly this result has been used in physics and returned results that confirm that it's valid, how would you argue against that?
Seriously, a friend of mine studies math at uni, the crazy shit he gets lectures on is insane, math is not a simple world and the more you play with infinity the less intuitive it gets.
[QUOTE=Laserbeams;43498288]No, that's not quite what I mean. I can't call him an idiot, he's probably a really good mathematician, and there's some string theory stuff that seems to support this, but it's still pretty unbelievable. Something has to be wrong, it just doesn't make any sense[/QUOTE]
Maths doesn't care what makes sense and what doesn't.
[QUOTE=Quiet;43498176]The point I'm trying to make here is that talking about infinite sums has no meaning unless the concept is given a definition. Sure you can display an operation whichever way you wish without bearing on the result, but talking about the end of an infinite sum is absurd.
Care to comment on my other example?[/QUOTE]
[IMG]http://i.imgur.com/u2WIV3a.png[/IMG]
This example? If so, then you have to imagine an end, this end is not the usual end we are familiar with but it works exactly the same. For the top line, you will get 1+2... and since this goes forever, it goes up to infinity (actually -1/12 as evident by this video, but using infinity will demonstrate easier), so your end number is (with the shift method) infinity + 0. The bottom line is same, only 0 + infinity. Right now imagine no sum, just what we have. So it is
[∞+0]
-[0+∞]
=
∞-∞
(You may also notice the 0's don't matter if they are there or not, but this goes well with the shifting so are there for easier understanding)
we now have a cancel out which always
= 0
Even though this is technically false, we will get the same answer, using
[(-1/12)+0]
- [0+(-1/12)]
subtracting a negative is positive (and we are taking it away from 0) so
=-1/12+(1/12)
cancel out so it
= 0
You can also visualize this without "imagining a start and an end of an infinite sum" like I assume you are thinking. Imagine an infinite line, now imagine you cut it in half, you now have 2 infinite lines, but both have ends. Now subtract 1 infinite line from the other, and you have 0, like always shown with that equation. Shifting does not cause any issues.
[QUOTE=judgeofdeath;43498629]You realise that you are arguing against a professor of physics on this? And against the scientific literature that have documented this?
Apparantly this result has been used in physics and returned results that confirm that it's valid, how would you argue against that?
Seriously, a friend of mine studies math at uni, the crazy shit he gets lectures on is insane, math is not a simple world and the more you play with infinity the less intuitive it gets.[/QUOTE]
If 1+2+3+4+5... = -1/12 is a universal truth I would like to know exactly what about my perception of mathematics is incorrect, which is why I'm voicing my understanding.
Surely something is only true if it does not implicate a contradiction? 1+2+3+4+... = -1/12 introduces contradictions, so it is not true. I'm fairly sure you could prove that 0 = 1 using that result.
Also side note incase I made that too jumbled, you shifted the bottom line and therefore added a zero (which like I said doesn't matter except for visualization purposes), so if S=S we can add 0 to the top line as well. Addition doesn't matter in which order they are added as we will always get the same "sum", so we can add the zero to the theoretical "end" and we will get 0. This is the same as my above equation.
[QUOTE=supersoldier58;43498787][IMG]http://i.imgur.com/u2WIV3a.png[/IMG]
This example? If so, then you have to imagine an end, this end is not the usual end we are familiar with but it works exactly the same. For the top line, you will get 1+2... and since this goes forever, it goes up to infinity (actually -1/12 as evident by this video, but using infinity will demonstrate easier), so your end number is (with the shift method) infinity + 0. The bottom line is same, only 0 + infinity. Right now imagine no sum, just what we have. So it is
[∞+0]
-[0+∞]
=
∞-∞
(You may also notice the 0's don't matter if they are there or not, but this goes well with the shifting so are there for easier understanding)
we now have a cancel out which always
= 0
Even though this is technically false, we will get the same answer, using
[(-1/12)+0]
- [0+(-1/12)]
subtracting a negative is positive (and we are taking it away from 0) so
=-1/12+(1/12)
cancel out so it
= 0
You can also visualize this without "imagining a start and an end of an infinite sum" like I assume you are thinking. Imagine an infinite line, now imagine you cut it in half, you now have 2 infinite lines, but both have ends. Now subtract 1 infinite line from the other, and you have 0, like always shown with that equation. Shifting does not cause any issues.[/QUOTE]
I think applying finite arithmetic rules [URL="http://en.wikipedia.org/wiki/Riemann_series_theorem"]to sums that can be rearranged to diverge to infinity (or any real number for that matter)[/URL] is quite pointless.
Also, you can't manipulate infinity like a finite number so that it "cancels", consider the following function for x tending to infinity:
x^2 - x,
naively you'd think that this becomes ∞ - ∞ =?= 0,
factoring out x^2 gets you:
x^2 * (1 - 1/x)
taking the limit now gives you +∞.
The counter-intuitive results with infinite sums come from the fact that you think you can handle them like they're real numbers, which you can't.
For a,b and c > 0 and finite you can always say
a + b = c
but taking a=∞ you get
∞ + b = ∞, switching back to variable letters you get
a+ b = a, which is nonsense unless you take b equal to zero, which contradicts our prior conditions.
That, and as Swebonny says, they didn't use a technique that also works for absolutely convergent series for the initial alternating series. Any results derived from that will therefore not follow the usual finite arithmetic rules.
I think what many people should understand is that it's not a traditional summation of the natural numbers, however the way they presented may have made it seem like one and can be confusing.
The sum of all natural numbers from 1 to infinity is still going to be infinite. And this is what the majority of us will use and think of.
But:
The [B]Ramanujan summation[/B], as someone mentioned, of all natural numbers from 1 to infinity is -1/12 or this copy paste from Wiki:
[IMG]http://i.imgur.com/LfKHq1I.png[/IMG]
[QUOTE=supersoldier58;43498787][IMG]http://i.imgur.com/u2WIV3a.png[/IMG]
This example? If so, then you have to imagine an end, this end is not the usual end we are familiar with but it works exactly the same. For the top line, you will get 1+2... and since this goes forever, it goes up to infinity (actually -1/12 as evident by this video, but using infinity will demonstrate easier), so your end number is (with the shift method) infinity + 0. The bottom line is same, only 0 + infinity. Right now imagine no sum, just what we have. So it is
[∞+0]
-[0+∞]
=
∞-∞
(You may also notice the 0's don't matter if they are there or not, but this goes well with the shifting so are there for easier understanding)
we now have a cancel out which always
= 0
Even though this is technically false, we will get the same answer, using
[(-1/12)+0]
- [0+(-1/12)]
subtracting a negative is positive (and we are taking it away from 0) so
=-1/12+(1/12)
cancel out so it
= 0
You can also visualize this without "imagining a start and an end of an infinite sum" like I assume you are thinking. Imagine an infinite line, now imagine you cut it in half, you now have 2 infinite lines, but both have ends. Now subtract 1 infinite line from the other, and you have 0, like always shown with that equation. Shifting does not cause any issues.[/QUOTE]
∞-∞ does not equal 0. These paradoxes are really fucking with my mind
[QUOTE=Penguiin;43498147]this is retarded. it does not fucking equal -1/12[/QUOTE]
I bet 0.99999... doesn't equal 1 neither right?
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