For the discussion of math. Duh.

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Spambot5546
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I couldn't fine anything in this forum for math that is, well, bad.

Probably the best known example is the 1=2 proof: Most of us know the reason this is bullshit: the division by 0 in the transition from step 4 to 5.

The one that inspired me to start this thread was a video that isn't particularly new, but which I only recently saw, claiming that the sum of the natural numbers is -1/12.

Now, my knowledge of math only goes as far as what it takes to get an undergrad degree, but I know a gaussian sum when I see one, so I know that isn't accurate. I couldn't bring myself to watch the whole video, but I made it as far as one fallacy. The guy in the video claims

1-1+1-1+1-1... = 1/2

A claim he justifies by saying that if you end the series at an odd iteration the answer is one and on an even number the answer is 0, so the correct answer is to take the average of the two. Having doubts I represented the series as a sum of (-1)^n-1 and let Wolfram Alpha confirm that the series does not converge. So yeah, BS.
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alessandro95
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there is a good reason to say that 1+2+3+4...=-1/12, because that's what the analytic continuation of the zeta function says, but you could obtain different results with similar methods so there isn't really a "right" result, because the series is clearly divergent, but there are some context in which it makes sense to say that this series sums to -1/12
The primary reason Bourbaki stopped writing books was the realization that Lang was one single person.

Cleverbeans
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S = 1-1+1-1+1-1+....
S = 1 - (1-1+1-1+...)
S = 1 - S
2S = 1
S = 1/2

Rigor is for suckers, long live unjustified algebraic manipulation! "Labor is prior to, and independent of, capital. Capital is only the fruit of labor, and could never have existed if labor had not first existed. Labor is the superior of capital, and deserves much the higher consideration." - Abraham Lincoln

Tirian
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As much as that video has always set my teeth on edge, it is fascinating how naturally the result that 1 + 2 + 3 + ... = -1/12 flows from the hypothesis that 1 - 1 + 1 - 1 + ... = 1/2. I just wish they would stop patting themselves on the brain long enough to mention that this is a great illustration why 1 - 1 + 1 - 1 + ... is NOT equal to 1/2 or any other number in standard analysis.

Twistar
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It's not equality. At least it's not equality in the conventional sense. Typically when we write something like: 1+1/2+1/4+1/8+....=2 We mean the infinite sum on the left side of equality CONVERGES to the value on the right side (meaning the sequences of partial sums converges to the value). When they write 1-1+1-1+1...=1/2 they are NOT implying that the infinite sum on the left CONVERGES (in the usual sense) to the value on the right. No one is saying that. It turns out, instead, that the sum on the left hand sign is Cesaro summable and the when you calculate the cesaro sum you get 1/2. This isn't convergence in the usual sense but it IS convergence in SOME sense and this why people choose to use this sort of thing.

cyanyoshi
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Using the ordinary definition of a "sum" as a limit of partial sums, just like we were taught in calculus class, then Grandi's sum does not converge. If you were in school, you are free to stop there, and you will be patted on the back for doing so. But that's a little boring.

There are other, more widely applicable definitions of the "sum" of a series that can be used to assign a value to a series, such as Cesàro summation that keep some important properties that are expected of an infinite sum, and have the advantage that they give a meaningful value to some divergent series. I admit that the Numberphile video in the blog is rather sloppy in the way they approached the whole issue. I would recommend watching this follow-up video also that they put together that gets closer to what mathematicians mean by the "sum" of a divergent series.
Last edited by cyanyoshi on Tue May 13, 2014 5:18 am UTC, edited 1 time in total.

Moole
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Really, they're just working with some function f that replaces the sum - so it takes a sequence and spits out a real number representing the sum - that is assumed to have the following properties:

1. f(S) for a convergent sequence S agrees with the sum.
2. If S' is the same as sequence S, except with zeros inserted, f(S') = f(S).
3. If C is the term-wise sum of sequences A and B, then f(C) = f(A) + f(B)

and managing to the property that f(1,2,3,4,...) = -1/12 given only the above rules. But, the second rule is contradictory as one might note that the sum of the sequence {1,-1,0,1,-1,0,...} repeating would have to equal f(1,-1,1,-1,...) = 1/2 repeating, but if we sum the former with its shifts {0,1,-1,0,1,-1,0,...} and {0,0,1,-1,0,1,-1,0...} we get {1,0,0,0,0,0,0,...} which has f(1,0,0,...) = 1, so 3f(1,-1,0,1,-1,0)... = 1 and f(1,-1,0,1,-1,0) = 1/3, not 1/2 like we wish. We would have to replace the second rule with "with zeros inserted every other term in the sequence or prepended to the sequence" to even have a chance at such an f being well-defined, at which point the rules begin to seem a bit arbitrary (not to mention that it's still not obvious whether such an f exists), though such rules would suffice for their argument and certainly imply f(1,2,3,4,...) = -1/12... if not also other values.
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

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I've always liked the proof by induction that all horses are white:

Theorem: For any (finite) set of horses, all of the horses are the same color.

We prove this by induction on n, the size of the set of horses.

Base case: n=1, obviously since there's only one horse in the set it has just one color.

Inductive step: Suppose that all sets of n-1 horses are of the same color, and consider a set of size n. Removing one horse, call it h, we obtain a set of size n-1, so by the inductive hypothesis all horses except h have the same color. But putting h back and removing a different horse gives us another set of size n-1 including h, so the inductive hypothesis shows that h has the same color as the other horses, so all the horses in the full set have the same color.

Thus, all horses in the world have the same color. Since I've seen a white horse, all horses are white.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

Twistar
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Joined: Sat May 09, 2009 11:39 pm UTC

Cool. Is this the issue:
Spoiler:
Is the problem here that the proof of the inductive step fails for n=2? i.e. the horses all being the same color doesn't follow from the above paragraph because the set of "all other horses" two which the two horses are being compared is empty?

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Twistar:
Spoiler:
You are correct, the induction step fails at n=2. The language gets just a bit vague at that point (of course), so it's tricky to say exactly which statement is false. You could interpret "h has the same color as the other horses" to mean all horses besides the one that was removed, which is technically true since that set is empty, but then you can't conclude that h and the other removed horse are the same color just because they are both the same color as "all the other horses".

This fake proof is kind of funny in that it works best on people familiar with math; in my experience if you try to explain it to someone who doesn't already understand proof by induction intuitively they find the error right away.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

DR6
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Moole wrote:Really, they're just working with some function f that replaces the sum - so it takes a sequence and spits out a real number representing the sum - that is assumed to have the following properties:

1. f(S) for a convergent sequence S agrees with the sum.
2. If S' is the same as sequence S, except with zeros inserted, f(S') = f(S).
3. If C is the term-wise sum of sequences A and B, then f(C) = f(A) + f(B)

and managing to the property that f(1,2,3,4,...) = -1/12 given only the above rules. But, the second rule is contradictory as one might note that the sum of the sequence {1,-1,0,1,-1,0,...} repeating would have to equal f(1,-1,1,-1,...) = 1/2 repeating, but if we sum the former with its shifts {0,1,-1,0,1,-1,0,...} and {0,0,1,-1,0,1,-1,0...} we get {1,0,0,0,0,0,0,...} which has f(1,0,0,...) = 1, so 3f(1,-1,0,1,-1,0)... = 1 and f(1,-1,0,1,-1,0) = 1/3, not 1/2 like we wish. We would have to replace the second rule with "with zeros inserted every other term in the sequence or prepended to the sequence" to even have a chance at such an f being well-defined, at which point the rules begin to seem a bit arbitrary (not to mention that it's still not obvious whether such an f exists), though such rules would suffice for their argument and certainly imply f(1,2,3,4,...) = -1/12... if not also other values.

Also, if 3 holds then f can't be anything but the sum... or I think so.

Proof: given an infinite countable sequence A, we construct a family of sequences A' where A'(n) is equal to the nth term of A at the nth term and zero otherwise. That is, from:

1,2,3,4,5,6...
we make
1,0,0,0...
0,2,0,0,0....
0,0,3,0....

and so on.

Then, because of 3, f(A) = sum from i=0 to inf of A'(i).... so yeah, QED.

Moole
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DR6 wrote:
Moole wrote:Really, they're just working with some function f that replaces the sum - so it takes a sequence and spits out a real number representing the sum - that is assumed to have the following properties:

1. f(S) for a convergent sequence S agrees with the sum.
2. If S' is the same as sequence S, except with zeros inserted, f(S') = f(S).
3. If C is the term-wise sum of sequences A and B, then f(C) = f(A) + f(B)

and managing to the property that f(1,2,3,4,...) = -1/12 given only the above rules. But, the second rule is contradictory as one might note that the sum of the sequence {1,-1,0,1,-1,0,...} repeating would have to equal f(1,-1,1,-1,...) = 1/2 repeating, but if we sum the former with its shifts {0,1,-1,0,1,-1,0,...} and {0,0,1,-1,0,1,-1,0...} we get {1,0,0,0,0,0,0,...} which has f(1,0,0,...) = 1, so 3f(1,-1,0,1,-1,0)... = 1 and f(1,-1,0,1,-1,0) = 1/3, not 1/2 like we wish. We would have to replace the second rule with "with zeros inserted every other term in the sequence or prepended to the sequence" to even have a chance at such an f being well-defined, at which point the rules begin to seem a bit arbitrary (not to mention that it's still not obvious whether such an f exists), though such rules would suffice for their argument and certainly imply f(1,2,3,4,...) = -1/12... if not also other values.

Also, if 3 holds then f can't be anything but the sum... or I think so.

Proof: given an infinite countable sequence A, we construct a family of sequences A' where A'(n) is equal to the nth term of A at the nth term and zero otherwise. That is, from:

1,2,3,4,5,6...
we make
1,0,0,0...
0,2,0,0,0....
0,0,3,0....

and so on.

Then, because of 3, f(A) = sum from i=0 to inf of A'(i).... so yeah, QED.

That assumes countable additivity; the third condition only makes a statement about finite sums. If we wanted countable addivity, we'd want something more like

f(sum A'(i)) = f(f(A'(1)),f(A'(2)),f(A'(3)),...)

where we use another application of f instead of an infinite sum, since f was supposed to replace such sums.

You can satisfy the first and third condition pretty easily for an f other than the normal sum; for instance, the Césaro sum, which basically averages together the partial sums of a series satisfies both, but is defined on some divergent series (i.e. 1-1+1-1+1-1 ... has the sum of 1/2 under that). I mean, those two basically say:

f is linear
On a linear subspace (the convergent series), f takes on specified values.

And, by assigning values in a quotient space of the space of real sequences quotiented by the space of convergent sequences, one could in fact generate function satisfying both conditions one and three on the entire space of real sequences. Really, it's the second condition that's interesting, since it takes a vector space and gives it more structure. The condition of f(sum A'(i)) = f(f(A'(1)),f(A'(2)),f(A'(3)),...) could also be interesting, perhaps as a substitute for the second condition, since it takes f and gives it more structure than any linear map.
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

DR6
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Moole wrote:
DR6 wrote:
Also, if 3 holds then f can't be anything but the sum... or I think so.

Proof: given an infinite countable sequence A, we construct a family of sequences A' where A'(n) is equal to the nth term of A at the nth term and zero otherwise. That is, from:

1,2,3,4,5,6...
we make
1,0,0,0...
0,2,0,0,0....
0,0,3,0....

and so on.

Then, because of 3, f(A) = sum from i=0 to inf of A'(i).... so yeah, QED.

That assumes countable additivity; the third condition only makes a statement about finite sums. If we wanted countable addivity, we'd want something more like

f(sum A'(i)) = f(f(A'(1)),f(A'(2)),f(A'(3)),...)

where we use another application of f instead of an infinite sum, since f was supposed to replace such sums.

You can satisfy the first and third condition pretty easily for an f other than the normal sum; for instance, the Césaro sum, which basically averages together the partial sums of a series satisfies both, but is defined on some divergent series (i.e. 1-1+1-1+1-1 ... has the sum of 1/2 under that). I mean, those two basically say:

f is linear
On a linear subspace (the convergent series), f takes on specified values.

And, by assigning values in a quotient space of the space of real sequences quotiented by the space of convergent sequences, one could in fact generate function satisfying both conditions one and three on the entire space of real sequences. Really, it's the second condition that's interesting, since it takes a vector space and gives it more structure. The condition of f(sum A'(i)) = f(f(A'(1)),f(A'(2)),f(A'(3)),...) could also be interesting, perhaps as a substitute for the second condition, since it takes f and gives it more structure than any linear map.

Oh, ok. I thought countable additivity would follow from finite additivity, and it would be uncountable additivity giving problem. Seems to not be the case.

jestingrabbit
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Uncountable additivity is something that you'll really struggle to make sense of no matter the setting.
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arbiteroftruth
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jestingrabbit wrote:Uncountable additivity is something that you'll really struggle to make sense of no matter the setting.

Really? Isn't uncountable additivity just integration when you interpret "dx" as a literal infinitesimal hyperreal?

mathmannix
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Okay, so
S = 1+2+3+4+5+6+7+ ... = -1/12.

Let S3 = 1+3+5+7+9+11+13+15+ ...

S - S3 = 2+4+6+8+10+12+14+...
= 2*(1+2+3+4+5+6+7+...)
= 2*(-1/12)
= -1/6.

Therefore, S3 = S - (-1/6)
S3 = (-1/12) - (-1/6)
S3 = 1/12.

Now, notice this:
1=1 (= 1²).
1+3 = 4 (= 2²).
1+3+5 = 9 (=3²).
1+3+5+7 = 16 (=4²).
These are the perfect squares.

What is the largest perfect square?
1+3+5+7+9+11+13+15+... = S3 = 1/12.

(It follows that the largest integer is 1/√12.)
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jestingrabbit
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arbiteroftruth wrote:
jestingrabbit wrote:Uncountable additivity is something that you'll really struggle to make sense of no matter the setting.

Really? Isn't uncountable additivity just integration when you interpret "dx" as a literal infinitesimal hyperreal?

If you need to start by multiplying all your summands by an infinitesimal, then I think I've proven my point.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

lalop
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I'm a little bit rusty, so sorry if this is a trivial question, but is it known for sure that there are uncountable hyperrnaturals (i.e. {1,...,N} is uncountable, as an index set for an uncountable sum), and, well, in what sense, since that set is finite in the hyperreals. Does this mean there are "countable hypernaturals" as well?

snowyowl
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I don't know hypernaturals, but there is such a thing as an uncountable ordinal.
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mathmannix
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Imagine a natural number bigger than you can imagine anyone or any computer counting to. It's uncountable!

I mean, imagine a prime number Q1, which is otherwise uninteresting (not a power of two minus one or anything like that), whose index in the list of primes is larger than the theoretically fastest computer could count to in the length of time from the beginning to end of time (assuming finite time). Then imagine an otherwise-uninteresting prime number Q2 with Q12+1 digits. I'd say that number is safely uncountable, but it is still a positive integer, a natural number.
I hear velociraptor tastes like chicken.

alessandro95
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There is a precise and well defined meaning of "countable" and "uncountable" in mathematics, you cannot use this terms as you wish, the set of natural numbers is countable and so is any of its subsets, doesn't matter how big or wether it can be computed with modern technologies in a useful time
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Schrollini
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mathmannix wrote:I mean, imagine a prime number Q1, which is otherwise uninteresting (not a power of two minus one or anything like that)

But there is no such number, since all positive integers are interesting.

Proof: Suppose there are uninteresting positive integers. Then there is a smallest uninteresting positive integer. Hey, that's interesting.

Spoiler:
:D
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DR6
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mathmannix wrote:Imagine a natural number bigger than you can imagine anyone or any computer counting to. It's uncountable!

I mean, imagine a prime number Q1, which is otherwise uninteresting (not a power of two minus one or anything like that), whose index in the list of primes is larger than the theoretically fastest computer could count to in the length of time from the beginning to end of time (assuming finite time). Then imagine an otherwise-uninteresting prime number Q2 with Q12+1 digits. I'd say that number is safely uncountable, but it is still a positive integer, a natural number.

"Countable" and "uncountable" are words used to describe sets, not numbers.

Nautilus
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A very relevant blog.

(A little melodramatic at times, but pretty interesting!)
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jestingrabbit
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lalop wrote:I'm a little bit rusty, so sorry if this is a trivial question, but is it known for sure that there are uncountable hyperrnaturals (i.e. {1,...,N} is uncountable, as an index set for an uncountable sum), and, well, in what sense, since that set is finite in the hyperreals. Does this mean there are "countable hypernaturals" as well?

I'm trying to pick this apart. By "the hypernaturals" we usually mean the set of sequences of natural numbers, factored by a particular equivalence relation. I believe that there are an uncountable infinity of such things, with the equivalence classes of {floor(alpha*n)| n\in N}, where alpha in R+, demonstrating this fact: any two such sequences can agree on only a finite number of places for different alpha and beta, and because the nonprincipal ultrafilter must be co-finite, all these sequences must fit into different equivalence classes, and so there must be an uncountable infinity of them.

The rest doesn't really gel. Elements can't be countable, and I'm not sure what you mean by "{1,...,N} is uncountable".

Anyway, trying to make sense of uncountable sums by using hypernaturals to index them is a fools errand imo. It just doesn't seem like a coherent endeavour.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

brenok
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Nautilus wrote:A very relevant blog.

(A little melodramatic at times, but pretty interesting!)

I think this post in particular might also be relevant

Flumble
Yes Man
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Spambot5546 wrote:The one that inspired me to start this thread was a video that isn't particularly new, but which I only recently saw, claiming that the sum of the natural numbers is -1/12.

While that derivation isn't rigorous, as multiple (possible) flaws have been pointed out in posts and links here, the Euler-Riemann ζ does state that 1+2+3+4... equals ζ(-1)=-1/12 via analytic continuation.

I do like to know what premises are needed for ζ(-1) to hold, as I'm not that much into math.

PM 2Ring
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A simpler example than the ζ series is the sum of a geometric progression.
Let S(x) = 1 + x + x² + x³ + x⁴ + ...
This series converges if |x| < 1, and it's easy to show that if |x| < 1 then
S(x) = 1 / (1 - x)

Spoiler:
Let S = 1 + x + x² + x³ + x⁴ + ... + xn
xS = x + x² + x³ + x⁴ + ... + xn+1
S - xS = 1 - xn+1
S(1 - x) = 1 - xn
S = (1 - xn) /(1 - x)
As n → ∞, xn → 0 iff |x| < 1
So S(x) = 1 / (1 - x)

That formula for S(x) is undefined for x = 1, but it gives negative values for x > 1.
Eg, S(2) = 1/ (1 - 2) = -1.
We don't usually say that 1 + 2 + 2² + 2³ + 2⁴ + ... = -1 , although there are occasions where it is useful to assign that measure to that divergent series.

onoresrts63
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Spambot5546 wrote:I couldn't fine anything in this forum for math that is, well, bad.

Probably the best known example is the 1=2 proof: Most of us know the reason this is bullshit: the division by 0 in the transition from step 4 to 5.

The one that inspired me to start this thread was a video that isn't particularly new, but which I only recently saw, claiming that the sum of the natural numbers is -1/12.

Now, my knowledge of math only goes as far as what it takes to get an undergrad degree, but I know a gaussian sum when I see one, so I know that isn't accurate. I couldn't bring myself to watch the whole video, but I made it as far as one fallacy. The guy in the video claims

1-1+1-1+1-1... = 1/2

A claim he justifies by saying that if you end the series at an odd iteration the answer is one and on an even number the answer is 0, so the correct answer is to take the average of the two. Having doubts I represented the series as a sum of (-1)^n-1 and let Wolfram Alpha confirm that the series does not converge. So yeah, BS.

so the correct answer is to take the average of the two? That logic looks like its from a 3-year old. Math is way more beautiful than that kind of logic. Believe Wolfram.

snowyowl
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There's several lines of reasoning that make more sense. For example, the infinite sum 1 + s1 + s2 + s3 + s4 + ... converges to 1/(1-s) for all |s|<1 (even if s is complex) and diverges otherwise. Now, look at the behaviour of this function near s=-1. As s tends to -1, the value of 1/(1-s) tends to 1/2.

Edit: Wait, someone already said this. Never mind.
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mike-l
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onoresrts63 wrote:so the correct answer is to take the average of the two? That logic looks like its from a 3-year old. Math is way more beautiful than that kind of logic. Believe Wolfram.

Actually that's just the Cesàro summation.

Sometimes I feel like the folks who get really upset about 1+2+3+.... = -1/12 are those who know just enough math to be completely uninteresting to talk to. They're completely correct that the sum diverges (using the definition of diverge you learn in calc 1), but limiting yourself to only thinking in terms of Calc 1 is missing out on tons of 'way more beautiful' ideas. It's not unlike someone saying that 3 doesn't divide into 4. They're correct in their grade 2 definition of division, but they're missing out on the wonderful world of fractions.

Edit: the common way to rigorously talk about 1+2+3+... Is http://en.m.wikipedia.org/wiki/Zeta_fun ... larization which indeed gives the value -1/12. And this crops up in all sorts of beautiful things ranging from the Prime Number Theorem to Bosonic String Theory
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PM 2Ring
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And if you find Cesàro summation too simple for your tastes there's always Ramanujan summation, which does groovy things with Bernoulli numbers.

Actually, Ramanujan summation is used in Zeta regularization, although it's not explicitly mentioned in mike-l's link.

Spambot5546
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mike-l wrote:Sometimes I feel like the folks who get really upset about 1+2+3+.... = -1/12 are those who know just enough math to be completely uninteresting to talk to.

I have a BA in math and had never heard of a Cesaro summation until this thread. I think you're overestimating how common knowledge they are.
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Because it is bitter,
And because it is my heart."

mike-l
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Spambot5546 wrote:
mike-l wrote:Sometimes I feel like the folks who get really upset about 1+2+3+.... = -1/12 are those who know just enough math to be completely uninteresting to talk to.

I have a BA in math and had never heard of a Cesaro summation until this thread. I think you're overestimating how common knowledge they are.

I never said anything about common knowledge.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

Eebster the Great
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DR6 wrote:
mathmannix wrote:Imagine a natural number bigger than you can imagine anyone or any computer counting to. It's uncountable!

I mean, imagine a prime number Q1, which is otherwise uninteresting (not a power of two minus one or anything like that), whose index in the list of primes is larger than the theoretically fastest computer could count to in the length of time from the beginning to end of time (assuming finite time). Then imagine an otherwise-uninteresting prime number Q2 with Q12+1 digits. I'd say that number is safely uncountable, but it is still a positive integer, a natural number.

"Countable" and "uncountable" are words used to describe sets, not numbers.

It depends on what you mean by "number." The cardinality of an uncountable set is an uncountable cardinal number.

Of course, countability doesn't have anything to do with literally being able to "count to" a number, since then even countably infinite numbers would be "uncountable."

Spambot5546
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mike-l wrote:
Spambot5546 wrote:
mike-l wrote:Sometimes I feel like the folks who get really upset about 1+2+3+.... = -1/12 are those who know just enough math to be completely uninteresting to talk to.

I have a BA in math and had never heard of a Cesaro summation until this thread. I think you're overestimating how common knowledge they are.

I never said anything about common knowledge.

You're talking about the principal like anyone who doesn't know about it doesn't know much math. That might be true at a post graduate level, but one can know a lot of interesting math without ever having learned this particular esoteric version of summation.
"It is bitter – bitter", he answered,
"But I like it
Because it is bitter,
And because it is my heart."

Eebster the Great
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Cesaro summation is somewhat "esoteric" I guess, but zeta-function regularization is not.

PM 2Ring
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And anyone who's had some exposure to complex analysis ought to be at least vaguely familiar with the concept of analytic continuation.

Spambot5546
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Did you guys seriously get to take complex analysis classes as undergrads? My school considered real analysis a 400 level, and the only class dealing with the complex plane was Complex Variables, the 300 level that introduced the concept. Did I go to a shitty university? :'(
"It is bitter – bitter", he answered,
"But I like it
Because it is bitter,
And because it is my heart."

Eebster the Great
Posts: 3460
Joined: Mon Nov 10, 2008 12:58 am UTC
Location: Cleveland, Ohio