View Full Version : There Goes Infinity

Hlafordlaes

2014-Jan-17, 09:00 PM

Phil Plait posted this bizarre article and video (http://www.slate.com/blogs/bad_astronomy/2014/01/17/infinite_series_when_the_sum_of_all_positive_integ ers_is_a_small_negative.html) on Slate. Apparently, an infinite series of positive integers now sums to -1/12. Of course, that's -0.083333...

I have no clue as to what is going on here... Ammo for cocktail parties?

corner

2014-Jan-17, 10:05 PM

Phil Plait posted this bizarre article and video (http://www.slate.com/blogs/bad_astronomy/2014/01/17/infinite_series_when_the_sum_of_all_positive_integ ers_is_a_small_negative.html) on Slate. Apparently, an infinite series of positive integers now sums to -1/12. Of course, that's -0.083333...

I have no clue as to what is going on here... Ammo for cocktail parties?

Judging from the blog comment

Like the one you just saw. Let me say here, when I first saw it I thought it was an actual hoax, a joke (like being able to prove 1=2, an algebra trick most science/math dorks have seen). But it turns out that the conclusions they draw in that video are literally correct.

I'd say what is going on here is that a PhD-trained scientist has been duped.

Trebuchet

2014-Jan-18, 12:47 AM

I'd say what is going on here is that a PhD-trained scientist has been duped.

PZ Myers seems to agree (http://freethoughtblogs.com/pharyngula/2014/01/17/the-sum-of-all-natural-numbers-is-not-112/). I confess I haven't had the energy either to try to understand Phil's post, watch the video, nor to read the comments, especially given the utter disaster that it Slate's current commenting system.

ETA: Some commenters at PZ's seem to understand the idea. I've already had a glass of wine, it's getting dark, and I'm not even going to try!

Solfe

2014-Jan-18, 01:27 AM

I don't know... that looks more like a ratio of values between the two (or several) presented series.

My gut feeling is it is correct mathematically, but is portrayed as "infinity=-1/12" to the layman. It seems to be working a cycle of compared numbers and that cycle's average is -1/12, which is wildly different that "infinity=something or other".

I stink at math, so what do I know.

KlausH

2014-Jan-18, 01:27 AM

I'd say what is going on here is that a PhD-trained scientist has been duped.

I'd certainly agree with that.

This entire charade is based on the first "sum".

1-1+1-1+1-1+1... = 0.5.

That is of course nonsense.

That series simply does not have a final value and to set it to 0.5 is completely arbitrary.

It is alternating between 0 and 1 and that's all one can say about it.

So, this entire house of cards crumbles once that becomes clear.

No need to loose sleep over it.

corner

2014-Jan-18, 06:31 AM

PZ Myers seems to agree (http://freethoughtblogs.com/pharyngula/2014/01/17/the-sum-of-all-natural-numbers-is-not-112/).

I don't know who this PZ Myers person is, but he or she sounds quite sensible, at least based on this.

I confess I haven't had the energy either to try to understand Phil's post

That's just as well, unless "understand Phil's post" is interpreted as "understand what's wrong with it".

watch the video, nor to read the comments, especially given the utter disaster that it Slate's current commenting system.

ETA: Some commenters at PZ's seem to understand the idea. I've already had a glass of wine, it's getting dark, and I'm not even going to try!

KlausH summarises it nicely. It's off from the first step. They try to assign value to an infinite series which doesn't converge.

My gut feeling is it is correct mathematically

Not to be too harsh, but if that's your gut feeling, I've got to advise against trusting your gut.

Mathematically, it is something the dog threw up.

Infinite series are defined to have the value to which the partial sums converge. The one they start with doesn't converge. After assigning a value to that, they apply some mathematical results which are perfectly value when applied to converging infinite series, but which don't apply to infinite series which don't converge. Total rubbish.

The whole "shift and add" method does work on converging infinite series; in fact, you could even misapply it to the original series, to "prove" that this series sums to 1/2. It can be proven that it works on converging infinite series. If

a=\sum_{n=1}^{\infty}a_{n}

and

b=\sum_{n=1}^{\infty}b_{n},

then

a+b=\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right).

That result follows from the definition of a limit, and a few of the basic rules of arithmetic applied to finite expressions. Doesn't work when either of the original series fails to converge.

Ever seen the "proof" that

1+2+4+8+16+32+\ldots=-1?

It works (or more precisely, produces a nonsense answer) through the same error used to get this nonsense -1/12 result.

So this Phil Plait has something of a reputation as a debunker? If you ask me, when you post about this remarkable result at your blog, believing it to be correct, you lose your right to make fun of people who believe in Niburu …

corner

2014-Jan-18, 10:10 AM

Not sure why my earlier post isn't here. But, let's try a different series.

X=1+1+1+\ldots

Then

X=1+\underbrace{1+1+1+\ldots}_{=X}

So

X=1+X

and

0=1.

Proved. If you think it's wrong, and that zero is not equal to one, then point out which step in my derivation is wrong, and I'll point out where they use it in the video.

Trebuchet

2014-Jan-19, 12:34 AM

Phil has another post on it today. But what I really want to know is:

Does .99999... = 1.0?

(Runs and hides.)

corner

2014-Jan-19, 01:49 AM

Phil has another post on it today.

His new post is about analytic continuation. Take the series

1+x+x^{2}+x^{3}+x^{4}+\ldots

This series converges for -1<x<1, and has the value \frac{1}{1-x}. So the analytic continuation approach is to declare, the value of the series is therefore always \frac{1}{1-x}, even when x\geq1 or x\leq1, and the series diverges.

You can do that; you could define the value of a divergent series to be the value of the 19th element divided by the 12th element if you want. Using these alternate definitions, divergent series won't have all the same properties as convergent series. You could not necessarily add them term by term, which is incidentally exactly what they do in the video. Furthermore, different methods of analytic continuation, based on different convergent series, assign different values to the same divergent series. So different analytic continuation methods are not consistent with each other.

In my post above, I prove that 0=1 using only methods in the video. Are we believing this?

But what I really want to know is:

Does .99999... = 1.0?

(Runs and hides.)

Yes, that one converges. Is someone saying it doesn't?

Solfe

2014-Jan-19, 02:18 AM

phil has another post on it today. But what i really want to know is:

Does .99999... = 1.0?

(runs and hides.)

:) not it! :)

grapes

2014-Jan-19, 05:05 AM

Phil has another post on it today.

And, he strikes out part of the previous post, and adds this:

[Important Update/Correction (Jan. 19 at 2:00 UTC): After writing this post, I received a lot of objections about it. The critical one is that the mathematical method used in the video to solve the series does not actually work, since it violates some rules about what sorts of things you can do with series. You can read what I wrote below, but note what I wrote about the videos is incorrect. However, I wrote a follow-up post that should clear this all up, so please read that after reading the article below.]

Phil went off the rails years ago.

ETA:

Go down to the end of that second post ( http://www.slate.com/blogs/bad_astronomy/2014/01/18/follow_up_the_infinite_series_and_the_mind_blowing _result.html ), past this part:

Overall, a lot of what I wrote in the article is correct prima facie. A lot of it wasn’t. How this came to be makes me a bit red-faced as well as has me chuckling at myself.

I did talk a bit about the analytic continuation method, called it rigorous, and said it shows that the series can have a value of -1/12. But I made a couple of mistakes: one was not trusting my instincts, and the other was trusting them too much.

Hornblower

2014-Jan-19, 12:19 PM

Not sure why my earlier post isn't here. But, let's try a different series.

X=1+1+1+\ldots

Then

X=1+\underbrace{1+1+1+\ldots}_{=X}

So

X=1+X

and

0=1.

Proved. If you think it's wrong, and that zero is not equal to one, then point out which step in my derivation is wrong, and I'll point out where they use it in the video.

In my opinion your second line should have been:

X+1=1+\underbrace{1+1+1+\ldots}_{=X}

I was taught early on that to maintain the equality you must add the same thing to both sides.

slang

2014-Jan-19, 12:33 PM

Yes, that one converges. Is someone saying it doesn't?

This argument has led to very heated debate and, IIRC, infractions or suspensions. Let's not go there again.

corner

2014-Jan-19, 01:55 PM

In my opinion your second line should have been:

X+1=1+\underbrace{1+1+1+\ldots}_{=X}

I was taught early on that to maintain the equality you must add the same thing to both sides.

It should be exactly as it is, because I didn't add anything to either side. What I did was split off the first term of the series, and recognise that what remains is still equal to the original series, such series apparently having meaning in this strange world of alternative mathematics.

This was done in the video. If they're allowed to do it, I should be allowed as well.

This is a strange board, we seem to be allowed to discuss divergent series, but not convergent ones.

KlausH

2014-Jan-19, 02:00 PM

This is a strange board, we seem to be allowed to discuss divergent series, but not convergent ones.

This one is absolute genius!

I laughed long and hard and it is oh so true!

grapes

2014-Jan-19, 05:28 PM

This is a strange board, we seem to be allowed to discuss divergent series, but not convergent ones.

This one is absolute genius!

I laughed long and hard and it is oh so true!

Maybe. But at the same time, oh so false. Which is probably why it's funny. I think that follows (mumble)'s paradigm of humor.

We *can* discuss convergent and divergent sequences. The mod comment above was about a specific sequence that has generated acres of ill will, on this board as well as a host of other boards. There is nothing new to contribute to the discussion (you can find hours of reading on it here at this forum--at least one of the posters will probably present your favorite opinion of it), and it invariably devolves into ridiculousness.

Hlafordlaes

2014-Jan-19, 06:40 PM

Just chiming in to agree on that point. The old thread in question is happily lying dormant. Awakening it would be a Smaug-like event, all heat and little light, and loads of regets.

grapes

2014-Jan-19, 07:36 PM

Back to the topic at hand, let's look at a few functions.

f(x) = \frac{x^2 - 3 x + 2}{(x-1)}

g(x) = \frac{sin{(x-1)}}{(x-1)}

Both f and g have a division by zero when x=1, so they are undefined. Their graphs are nice smooth lines though, with a little hole in the line where x equals 1. That hole may as well be filled with the single point that makes their graphs into a continuous line, right? For f, we fill in the point (1,-1); for g, we fill in the point (1,1). Now, neither f or g are really defined at x=1, but we've managed to define new functions that match them everywhere but at x=1.

For f, that new function is just (x-2), as you can easily see. No matter which x you try, (x-2) matches the original f(x), except for x=1. And for x=1, it matches the point we added, (1,-1). Of course, the way I came up with (x-2) was just to factor the numerator and cancel the common factor. That means f(x) was just (x-2) "in disguise" all along.

For g, it's not as simple to find the continuation, but if you graph g(x) at wolframalpha.com, you'll indeed see that the point I chose makes the graph of g as continuous as (x-2). But having continued g, we can play the game of "formally" evaluating g at x=1. That is, we plug 1 into g(x) and see what expression we get that gives us the point (1,1): g(1) =1, so that would mean \frac{sin{(1-1)}}{(1-1)} = 1

In other words, we've just extended g, and the "formal" result is that 0/0 = 1

That's nonsense, but it does allow us to fill in the hole in g. Or rather, we can fill in the hole in g, but we can't really draw such conclusions from it. How am I so sure?

Look at f(x), and "formally" evaluate it at x=1. Then we get (1^2 - 3*1 + 2)/(1 - 1) = 0/0. But, we've filled the hole with (1,-1), so now our "formal" conclusion is 0/0 = -1, which contradicts what we got from working with g

grapes

2014-Jan-19, 08:54 PM

The series h(x) = 1 + x + x^2 + ... + x^(L-1) can be re-written as

h(x) = \sum_{n=0}^{(L-1)} x^n = \frac{x^{L} - 1}{x-1}

That x^{L} term in the formula on the right hand side (RHS) is the key to the behavior of the series. Notice, that that's the only part of the RHS that depends on L, the number of terms. We can let L get as large as we like, and the formula still works. When L is very large, and x is greater than 1, x^{L} will also be very large. When L is very large, and x is less than 1, x^{L} will be very small.

When we let L go to infinity, that x^{L} will get infinitely large too, unless x is less than 1.

There's a big "hole" there, where h(x) is infinite--in another word, undefined.

In other words, for that infinite sequence, if x is less than 1, h(x) = -1/(x-1) = 1/(1-x); otherwise it's undefined.

We can then use that definition of h to "fill the holes" where the sequence was undefined/infinite..

Then, instead of infinity, h(2) = 1/(1-2) = -1

Now h is defined everywhere except x=1, where it is still undefined, and it matches all the places where it was defined. But look at the original definition of h(2), which was 1 + 2 + 2^2 + 2^3 + ...

Our extension allows us to fill almost all the holes, but it leaves us with what *looks like* (that is, formally) -1 = 1 + 2 + 4 + 8 +...

And that's because we threw away the x^L term for all values of x, not just the ones where it went to zero.

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