Divergent Infinite Products

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tomtom2357
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Divergent Infinite Products

There are ways to give a value for infinite sums, so why not infinite products? Are there any methods for defining a value for 1*2*3*4*...?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

cyanyoshi
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Re: Divergent Infinite Products

I'd imagine the way to deal with an infinite product is to take its logarithm to turn it into an infinite sum of logarithms. When you're done analyzing it, just take the exponential of that series to recover the original problem. That's what I suspect at least; I haven't actually been required to work with infinite products in much depth.

Voekoevaka
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Re: Divergent Infinite Products

I like the idea of logarithms, but they're are ineficient to obtain a result here.

Maybe there's a way to distribute step by step and rearange the terms to obtain a series.
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Parralelex
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Re: Divergent Infinite Products

EDIT: Never mind, I made an error somewhere. Infinite products are hard
Last edited by Parralelex on Wed May 08, 2013 10:57 pm UTC, edited 1 time in total.
I put the "fun" in "mathematics".

And then I took it back out.

flownt
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Re: Divergent Infinite Products

I would define the limit of an infinite product the same way as is done for a series;
Let $a_n$ be a sequence of real numbers, and $p_n=\prod_{k=1}^n a_n$
then $\prod_{k=1}^\infty = \lim_{n\to\infty} p_n$
ie. the value of an infinite product is the limit of the partial products.

So if $\lim_{n\to\infty} |a_n| > 1$ then the infinite product can't converge,
if $\lim_{n\to\infty} |a_n| <1$ then the product must converge to 0,
the most interesting case is $\lim_{n\to\infty} |a_n|=1$.

It is quite clear that it can't converge or diverge unconditionally: $a_n=1$ makes the product converge to 1, while $a_n=-1$ makes the product diverge.
Let $a_1=2$ and for all $n>1$, $a_n=a_{n-1}^{1/2}$. Thm: $\prod_{k=1}^\infty a_k=4$.
To proof this I will first prove a lemma. Lemma $a_{n+1}=a_1^{1/2^n}$. Proof: Using induction. $a_1=a_1$ therefore the statement is true for $n=0$. Assume $n>1$ and for all $k<n$ the lemma holds. $a_{n+1}=a_n^{1/2}=(a_1^{1/2^{n-1}})^{1/2}=a_1^{1/2^n}$.

This means $p_n=\prod_{k=1}^n a_k = a_1^{1/2^{k-1}} = a_1^{\sum_{k=1}^n 1/2^{k-1} } = a_1^{2-1/2^{n-1} }$
In the limit this goes to $a_1^2$: $\lim_{n\to\infty} p_n=a_1^{2-\lim_{n\to\infty}1/2^{n-1}}=a_1^{2-0}=a_1^2=4$.

This shows a non constant converging product-series showing that infinite products can be given a value. Exercise: prove that this statement is equivalent to the exp-sum-log method mentioned by cyanyoshi.

@paralelex: rearrangement of terms is not valid (or proof that it is under your definition) unless your convergence is fast enough, compare http://en.wikipedia.org/wiki/absolute_convergence

Under my definition (and also under cyanyoshi's) the original product series doesn't converge.

Parralelex
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Re: Divergent Infinite Products

flownt wrote:I would define the limit of an infinite product the same way as is done for a series;
Let $a_n$ be a sequence of real numbers, and $p_n=\prod_{k=1}^n a_n$
then $\prod_{k=1}^\infty = \lim_{n\to\infty} p_n$
ie. the value of an infinite product is the limit of the partial products.

So if $\lim_{n\to\infty} |a_n| > 1$ then the infinite product can't converge,
if $\lim_{n\to\infty} |a_n| <1$ then the product must converge to 0,
the most interesting case is $\lim_{n\to\infty} |a_n|=1$.

It is quite clear that it can't converge or diverge unconditionally: $a_n=1$ makes the product converge to 1, while $a_n=-1$ makes the product diverge.
Let $a_1=2$ and for all $n>1$, $a_n=a_{n-1}^{1/2}$. Thm: $\prod_{k=1}^\infty a_k=4$.
To proof this I will first prove a lemma. Lemma $a_{n+1}=a_1^{1/2^n}$. Proof: Using induction. $a_1=a_1$ therefore the statement is true for $n=0$. Assume $n>1$ and for all $k<n$ the lemma holds. $a_{n+1}=a_n^{1/2}=(a_1^{1/2^{n-1}})^{1/2}=a_1^{1/2^n}$.

This means $p_n=\prod_{k=1}^n a_k = a_1^{1/2^{k-1}} = a_1^{\sum_{k=1}^n 1/2^{k-1} } = a_1^{2-1/2^{n-1} }$
In the limit this goes to $a_1^2$: $\lim_{n\to\infty} p_n=a_1^{2-\lim_{n\to\infty}1/2^{n-1}}=a_1^{2-0}=a_1^2=4$.

This shows a non constant converging product-series showing that infinite products can be given a value. Exercise: prove that this statement is equivalent to the exp-sum-log method mentioned by cyanyoshi.

@paralelex: rearrangement of terms is not valid (or proof that it is under your definition) unless your convergence is fast enough, compare http://en.wikipedia.org/wiki/absolute_convergence

Under my definition (and also under cyanyoshi's) the original product series doesn't converge.

We're not assuming that it converges, in fact it should be fairly obvious that it diverges. We're saying that IF it had a value, what would the value be?
I put the "fun" in "mathematics".

And then I took it back out.

Qaanol
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Re: Divergent Infinite Products

To give an example of cyanyoshi's approach, consider 2×4×16×256×65536×⋯×22^k×⋯

The base-2 log of that product is 1+2+4+8+16+⋯+2k+⋯

By the usual technique that equals -1, so the original product is 2-1
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Tirian
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Re: Divergent Infinite Products

Do logarithms work that way in the 2-adic world though? I mean, as a computer programmer I can wrap my mind around 1 + 2 + 4 + 8 + 16 + ... = -1, because that's how signed integers work if you squint and wave your hands. But the same intuition would lead me to wonder if 1 * 2 * 4 * 16 * 256 * ... = 0 in 2-adics, because the product would be kinda sorta just a 1 in an inaccessibly large digit place.

cyanyoshi
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Re: Divergent Infinite Products

So here's my (non-rigorous, sleep-deprived) attempt at this.
Let P = 2*3*4*...
ln(P) = ln(2) + ln(3) + ln(4) +...

The Riemann Zeta function is defined as
ζ(s) = 1 + 2^-s + 3^-s + 4^-s +... wherever it converges.
Thus it's derivative would be:
ζ'(s) = -ln(2)/2^s - ln(3)/3^s - ln(4)/4^s +...

By analytic continuation of this function,
ζ'(0) "=" -ln(2) - ln(3) - ln(4) +...
ζ'(0) = -ln(P)
P = exp(-ζ'(0))

...whatever that is.

Edit: It's apparently equal to (2π)^.5 or 2.506628.... Huh. So there you go, I guess.

tomtom2357
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Re: Divergent Infinite Products

Wow, good job! Never thought of differentiating the zeta function. I wonder how far this could extend, fairly certain it can work with series such as 1*3*5*7*..., and 2*4*6*8*..., but could it extend to 1*3*6*10*...?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

korona
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Re: Divergent Infinite Products

@Qaanol
What do you mean by usual technique? It is obvious that 1 + 2 + ... + 2^k = -1 mod 2^(k+1) but that statement does not make sense in a field of different characteristic.

@cyanyoshi
ln(2) + ln(3) + ... does not converge, as ln is an increasing function.

The post by flownt is the only meaningful answer to this question. It does not make sense to define a value for 2*3*4*... (neither does it make sense to define a value for 1+2+3+...)
There is no way to do that in a way that respects the usual topology and field structure of the reals so I could as well define 2*3*4*... = 42.

Qaanol
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Re: Divergent Infinite Products

korona wrote:@Qaanol
What do you mean by usual technique? It is obvious that 1 + 2 + ... + 2^k = -1 mod 2^(k+1) but that statement does not make sense in a field of different characteristic.

Or as follows:

S = 1 + 2 + 4 + 8 + ⋯
S = 1 + 2(1 + 2 + 4 + 8 + ⋯)
S = 1 + 2S
S = -1 (or S = ∞, which is why it is “nonstandard” to choose S=-1)

korona wrote:@cyanyoshi
ln(2) + ln(3) + ... does not converge, as ln is an increasing function.

Zeta regularization.

korona wrote:The post by flownt is the only meaningful answer to this question. It does not make sense to define a value for 2*3*4*... (neither does it make sense to define a value for 1+2+3+...)

It makes perfect sense. We are defining countable-addition and countable-multiplication in a different manner from the standard “it is defined as the limit of the partial sums/products”.

Call it a holistic approach. Call it nonstandard. Call it whatever you want, but it makes perfect sense. And frankly, to deny that it makes sense, rather suggests a rigidly dogmatic adherence to previous beliefs.

Sure, it makes sense to define countable-addition as the limit of partial sums, but it also makes sense to define it holistically, or p-adically, or any of a number of alternative ways.

korona wrote:There is no way to do that in a way that respects the usual topology and field structure of the reals so I could as well define 2*3*4*... = 42.

You can make any definition you want. The point it, zeta regularization has lots of nice properties that make it useful for things.
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korona
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Re: Divergent Infinite Products

Why are 2-adics the "usual technique"? Why not 3, 5, 7 or 11 adics?

Before we can talk about arguments like
S = 1 + 2 + 4 + 8 + ⋯
S = 1 + 2(1 + 2 + 4 + 8 + ⋯)
S = 1 + 2S
we have to define the RHS. Because the above equations look natural it would make sense to define countable addition in a way that they hold.
But I doubt that this will be possible for all countable sums of real numbers (not of p-adic numbers).

To define countable addition we need a map from (possibly a subset of) the space of all sequences L to the real numbers.
Let a_1, a_2, ... be a sequence. We want to define a_1 + a_2 + ...
Let's call the result A(a_1, a_2, ...)
As the result should behave like addition does, the map should at least be linear, i.e. A(u * a_1, u * a_2, ...) = u * A(a_1, a_2, ...)
and A(a_1 + b_1, a_2 + b_2, ...) = A(a_1, a_2, ...) + A(b_1, b_2, ...).
If only finitely many numbers of our sequence are non-zero, we want A(a_1, a_2, ...) to be the sum of those elements,
that implies that A is the canonical projection on the (proper) subspace spanned by the countable standard basis vectors. It also implies that the map is pairwise symmetric in this subspace.
We also want A(a_1, a_2, ...) = a_1 + A(a_2, a_3, ...), as that is required for your argument.
Is there any natural map that satisfies these axioms? As L has not even a countable dimension I guess that it will be very hard to construct such a map.

It makes sense to talk about regularizations (I did not know about zeta-regularizations, thank you) and p-adic numbers but then please state what you are talking about. And also state the limits of this technique, e.g. if does not make sense for non-rational numbers (as in the p-adic case) or loses properties that the finite addition of real numbers have.
Each post in this thread was talking about different non-equivalent concepts and was doing a lot of hand waving. It was not obvious that you were using a different definition than the log -> take limit of series -> exp definition.

Voekoevaka
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Re: Divergent Infinite Products

cyanyoshi wrote:So here's my (non-rigorous, sleep-deprived) attempt at this.
Let P = 2*3*4*...
ln(P) = ln(2) + ln(3) + ln(4) +...

The Riemann Zeta function is defined as
ζ(s) = 1 + 2^-s + 3^-s + 4^-s +... wherever it converges.
Thus it's derivative would be:
ζ'(s) = -ln(2)/2^s - ln(3)/3^s - ln(4)/4^s +...

By analytic continuation of this function,
ζ'(0) "=" -ln(2) - ln(3) - ln(4) +...
ζ'(0) = -ln(P)
P = exp(-ζ'(0))

...whatever that is.

Edit: It's apparently equal to (2π)^.5 or 2.506628.... Huh. So there you go, I guess.

Wonderful. Why did I bother with Bernoulli numbers whereas I could have stop with zeta function ?

tomtom2357 wrote:Wow, good job! Never thought of differentiating the zeta function. I wonder how far this could extend, fairly certain it can work with series such as 1*3*5*7*..., and 2*4*6*8*..., but could it extend to 1*3*6*10*...?

For 1*3*5*7*... and 2*4*6*8*..., I recommand the same method with Dirichlet Eta, Lambda and Beta function.

ζ(x)=∑((n)-x,n-1∈ℕ)
ζ'(0)=ln((2π)) ⇒ 1*2*3*4*5*6*7*...=(2π)

η(x)=∑((-1)n-1(n)-x,n-1∈ℕ)
η'(0)=2ln(2)ζ(0)-ζ'(0)=ln(1/2)-ln((2π))=ln(1/(2*(2π))) ⇒ 1/2*3/4*5/6*7/...=1/(2*(2π))

λ(x)=∑((2n+1)-x,n∈ℕ)
λ'(0)=ln(2)ζ(0)=-ln(2)/2=ln(2) ⇒ 1*3*5*7*9*11*...=2

β(x)=∑((-1)n(2n+1)-x,n∈ℕ)
β'(0)=ln(Γ²(1/4)/(2π2½)) ⇒ 1/3*5/7*9/11*...=Γ²(1/4)/(2π2½)

2*4*6*8*10*...=1*2*3*4*5*6*7*.../1*3*5*7*9*...=(2π)/2
I'm a dozenalist and a believer in Tau !

tomtom2357
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Re: Divergent Infinite Products

If countable addition is not commutative, then I'm not sure multiplication is, so I'm not sure you're calculation of 2*4*6*... is correct.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.