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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeDec 4th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexA little bit of a page at [[infinite product]], mainly just a definition, which leads me to ask a question: how do we define convergence of an infinite product in constructive mathematics? The definition seems to depend on decidability of $=0$.

   A little bit of a page at infinite product, mainly just a definition, which leads me to ask a question: how do we define convergence of an infinite product in constructive mathematics? The definition seems to depend on decidability of $=0$.

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeDec 4th 2014
   • (edited Dec 4th 2014)
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexIs there any reason you'd not want to use the definition using sums of logarithms, as Wikipedia does? Though I suppose that only makes sense for real numbers as it stands. To make it work for complex numbers, you'd have to pick a branch cut avoiding all terms of the sequence. Then I suppose the constructive domain of the logarithm makes itself felt. Certainly if all the terms are apart from 0 (or have absolute value apart from 0) it's ok.

   Is there any reason you’d not want to use the definition using sums of logarithms, as Wikipedia does? Though I suppose that only makes sense for real numbers as it stands. To make it work for complex numbers, you’d have to pick a branch cut avoiding all terms of the sequence. Then I suppose the constructive domain of the logarithm makes itself felt. Certainly if all the terms are apart from 0 (or have absolute value apart from 0) it’s ok.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeDec 4th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexYes, using logarithms makes perfect sense for complex numbers, although you do want to be sure that the branch you choose agrees with the principal branch along the positive real line (otherwise $\log 1 \neq 0$ and your series is never going to converge). But for products that involve some terms that might be zero, the classical thing to do is to ignore those terms when considering convergence but allow the value of the product to then be zero; but constructively, how can we decide which terms to ignore?

   Yes, using logarithms makes perfect sense for complex numbers, although you do want to be sure that the branch you choose agrees with the principal branch along the positive real line (otherwise $\log 1 \neq 0$ and your series is never going to converge). But for products that involve some terms that might be zero, the classical thing to do is to ignore those terms when considering convergence but allow the value of the product to then be zero; but constructively, how can we decide which terms to ignore?

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeDec 4th 2014
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   Author: DavidRoberts
   Format: MarkdownItexWhat sort of real do you want it to converge to? Perhaps if we don't have a proof that all the terms are apart from zero, we cannot prove it converges (or converges to something apart from zero).

   What sort of real do you want it to converge to? Perhaps if we don’t have a proof that all the terms are apart from zero, we cannot prove it converges (or converges to something apart from zero).

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeDec 4th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexHere's a more concrete situation. Suppose I know that $a_1, a_2, \dots$ are all apart from 0 and that $\prod_{k=1}^\infty a_k$ converges (in the obvious sense of a limit of partial products) to some limit $L$ that is also apart from 0. I would like to be able to prove that $\prod_{k=0}^\infty a_k$ converges to the product $a_0 L$. With the classical definition, I have to divide into cases: if $a_0=0$, then $\prod_{k=1}^\infty a_k$ converges to $L$ by ignoring $a_0$, and then $\prod_{k=0}^\infty a_k$ converges to 0; whereas if $a_0\neq 0$, then $\prod_{k=0}^\infty a_k$ converges to $a_0 L$ directly (as a limit of partial products). Is there a constructive definition of infinite products that would allow me to prove this fact? It sure seems like $\prod_{k=0}^\infty a_k = a_0 \cdot \prod_{k=1}^\infty a_k$ shouldn't depend on knowing that $a_0$ is either 0 or apart from 0!

   Here’s a more concrete situation. Suppose I know that $a_1, a_2, \dots$ are all apart from 0 and that $\prod_{k=1}^\infty a_k$ converges (in the obvious sense of a limit of partial products) to some limit $L$ that is also apart from 0. I would like to be able to prove that $\prod_{k=0}^\infty a_k$ converges to the product $a_0 L$. With the classical definition, I have to divide into cases: if $a_0=0$, then $\prod_{k=1}^\infty a_k$ converges to $L$ by ignoring $a_0$, and then $\prod_{k=0}^\infty a_k$ converges to 0; whereas if $a_0\neq 0$, then $\prod_{k=0}^\infty a_k$ converges to $a_0 L$ directly (as a limit of partial products). Is there a constructive definition of infinite products that would allow me to prove this fact? It sure seems like $\prod_{k=0}^\infty a_k = a_0 \cdot \prod_{k=1}^\infty a_k$ shouldn’t depend on knowing that $a_0$ is either 0 or apart from 0!

6. • CommentRowNumber6.
   • CommentAuthorSridharRamesh
   • CommentTimeDec 4th 2014
   • (edited Dec 4th 2014)
   • PermaLink
   Author: SridharRamesh
   Format: MarkdownItexLike David, I don't understand what's wrong with taking the infinite product to simply be the limit of partial products, calling it "convergent" if you like when this result is nonzero, and considering a result of zero to be a form of "divergence". This is precisely what you would get by thinking about infinite summation in the ordinary way in logarithm world, considering zero to have infinite (and thus divergent) logarithm. [I'll also note that, if one does want to _explicitly_ phrase things in terms of logarithms, we needn't worry about branch cuts at all; simply take logarithms of nonzero complex numbers to be elements of the quotient group of the complex numbers by the subgroup generated by $2\pi i$, with obvious corresponding notion of infinite summation. There's no reason to choose representatives from this quotient.]

   Like David, I don’t understand what’s wrong with taking the infinite product to simply be the limit of partial products, calling it “convergent” if you like when this result is nonzero, and considering a result of zero to be a form of “divergence”. This is precisely what you would get by thinking about infinite summation in the ordinary way in logarithm world, considering zero to have infinite (and thus divergent) logarithm.

   [I’ll also note that, if one does want to explicitly phrase things in terms of logarithms, we needn’t worry about branch cuts at all; simply take logarithms of nonzero complex numbers to be elements of the quotient group of the complex numbers by the subgroup generated by $2\pi i$, with obvious corresponding notion of infinite summation. There’s no reason to choose representatives from this quotient.]

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeDec 4th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThe point is that the product $0\cdot 1 \cdot 1\cdot 1\cdots$ should *converge* to 0, while the product $1 \cdot \frac1 2 \cdot \frac1 3 \cdot \frac1 4 \cdots $ should *diverge* ("to 0"), but in both cases the partial products have a limit of 0. The reason we want the first one to converge is because we want to define analytic functions by infinite products and allow those functions to have zeros (that being the point of the Weierstrass product theorem). How do you distinguish those cases constructively?

   The point is that the product $0\cdot 1 \cdot 1\cdot 1\cdots$ should converge to 0, while the product $1 \cdot \frac1 2 \cdot \frac1 3 \cdot \frac1 4 \cdots$ should diverge (“to 0”), but in both cases the partial products have a limit of 0. The reason we want the first one to converge is because we want to define analytic functions by infinite products and allow those functions to have zeros (that being the point of the Weierstrass product theorem). How do you distinguish those cases constructively?

8. • CommentRowNumber8.
   • CommentAuthorSridharRamesh
   • CommentTimeDec 4th 2014
   • (edited Dec 5th 2014)
   • PermaLink
   Author: SridharRamesh
   Format: MarkdownItexI see. In that case, you could say a sequence converges if, for sufficiently large n, tossing out its first n factors yields a nonzero limiting product. The value converged to will, of course, be the limiting product when tossing out no factors. This is equivalent to the definition given in the article, but makes it easy to constructively prove what was desired: convergence is now manifestly a tail-property, so that $\prod_{k = 0}^{\infty} a_k$ converges just in case $\prod_{k = 1}^{\infty} a_k$ converges. And, for convergent sequences, the value converged to is given by the usual limit, so that we furthermore have that $\prod_{k = 0}^{\infty} a_k = a_0 \prod_{k = 1}^{\infty} a_k$.

   I see. In that case, you could say a sequence converges if, for sufficiently large n, tossing out its first n factors yields a nonzero limiting product. The value converged to will, of course, be the limiting product when tossing out no factors. This is equivalent to the definition given in the article, but makes it easy to constructively prove what was desired: convergence is now manifestly a tail-property, so that $\prod_{k = 0}^{\infty} a_k$ converges just in case $\prod_{k = 1}^{\infty} a_k$ converges. And, for convergent sequences, the value converged to is given by the usual limit, so that we furthermore have that $\prod_{k = 0}^{\infty} a_k = a_0 \prod_{k = 1}^{\infty} a_k$.

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeDec 5th 2014
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   Author: Mike Shulman
   Format: MarkdownItexThanks! Biking home today I came to more or less the same conclusion: my idea was to say that the product converges if (1) there is a cofinite subset of $\mathbb{N}$ on which $a_k$ is apart from 0, and (2) the series of logarithms converges when restricted to that subset. I guess that's probably about the same -- I assume the logarithm is only defined constructively on numbers apart from 0. (I feel like if I were a real constructive analyst, I would have thought of this immediately... (-: )

   Thanks! Biking home today I came to more or less the same conclusion: my idea was to say that the product converges if (1) there is a cofinite subset of $\mathbb{N}$ on which $a_k$ is apart from 0, and (2) the series of logarithms converges when restricted to that subset. I guess that’s probably about the same – I assume the logarithm is only defined constructively on numbers apart from 0.

   (I feel like if I were a real constructive analyst, I would have thought of this immediately… (-: )

10. • CommentRowNumber10.
    • CommentAuthorNikolajK
    • CommentTimeApr 14th 2015
    • (edited Apr 14th 2015)
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    Author: NikolajK
    Format: MarkdownItexIs there a name for the operation $\lim_{N\to\infty}\prod_{k=0}^N a_k(N)$, akin to the definition of the infinite product, but where one extends the definition as such that the terms being multiplied may be functions of the upper bound $N$? For example, in the calculation of some path integrals one may use $\lim_{N\to\infty} \prod_{n=0}^{N-1} \left(1+\frac{x_n}{N}\right)= \exp\left(\lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}x_n\right)$

    Is there a name for the operation

    $\lim_{N\to\infty}\prod_{k=0}^N a_k(N)$,

    akin to the definition of the infinite product, but where one extends the definition as such that the terms being multiplied may be functions of the upper bound $N$? For example, in the calculation of some path integrals one may use

    $\lim_{N\to\infty} \prod_{n=0}^{N-1} \left(1+\frac{x_n}{N}\right)= \exp\left(\lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}x_n\right)$

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeApr 15th 2015
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    Author: Urs
    Format: MarkdownItexProbably there is no special terminology for this case, but of course it's hard to exclude that somebody somewhere proposed something. But I think it's a good point that infinite products of this more general kind should be mentioned in the entry, as they are important.

    Probably there is no special terminology for this case, but of course it’s hard to exclude that somebody somewhere proposed something. But I think it’s a good point that infinite products of this more general kind should be mentioned in the entry, as they are important.

12. • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeApr 15th 2015
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    Author: Mike Shulman
    Format: MarkdownItexWhile it's certainly an important notion, I'm dubious about calling it an "infinite product" of any sort.

    While it’s certainly an important notion, I’m dubious about calling it an “infinite product” of any sort.

13. • CommentRowNumber13.
    • CommentAuthorNikolajK
    • CommentTimeApr 16th 2015
    • (edited Apr 16th 2015)
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    Author: NikolajK
    Format: MarkdownItex@Mike: I now found a reference for a large class of similar constructions of the form $\lim_{N\to \infty}\sum_{k=0}^{N} a_k(N)$. They denote it by a stylized $\int$ for sum, however I’m dubious about calling it an “infinite sum” of any sort. :) edit: Okay, I now found a reference to the idea, the above is akin to a [Type II Volterra product integral](http://en.wikipedia.org/wiki/Product_integral). Oh, it has its own [calculus](http://en.wikipedia.org/wiki/Multiplicative_calculus#Multiplicative_derivatives) with fundamental theorem and all.

    @Mike: I now found a reference for a large class of similar constructions of the form $\lim_{N\to \infty}\sum_{k=0}^{N} a_k(N)$. They denote it by a stylized $\int$ for sum, however I’m dubious about calling it an “infinite sum” of any sort. :)

    edit: Okay, I now found a reference to the idea, the above is akin to a Type II Volterra product integral. Oh, it has its own calculus with fundamental theorem and all.

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeApr 16th 2015
    • (edited Apr 16th 2015)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI don't think that not mentioning something that people will tend to expect just because on strict erudite grounds they may be wrong to expect it is a good idea for a wiki like the $n$Lab. A little googling demostrates that path integral textbooks will easily speak of "infinite products" for the thing in #10, and I don't find it surprising.

    I don’t think that not mentioning something that people will tend to expect just because on strict erudite grounds they may be wrong to expect it is a good idea for a wiki like the $n$Lab.

    A little googling demostrates that path integral textbooks will easily speak of “infinite products” for the thing in #10, and I don’t find it surprising.

15. • CommentRowNumber15.
    • CommentAuthorNikolajK
    • CommentTimeApr 16th 2015
    • (edited Apr 16th 2015)
    • PermaLink
    Author: NikolajK
    Format: MarkdownItexI'm thinking about the topic because by a telescoping argument I "discovered" the formula of how to truncate a product _additively_ at any mid point $M$, actually turning them to sums with finitely computable summands. Namely $\prod_{k=K}^\infty b_k = \prod_{k=K}^{M-1}b_k+\sum_{n=M}^\infty(b_n-1)\,\prod_{k=K}^{n-1}b_k$ I surprised some people on MathOverflow/StackExchange with idenitities like $\prod_{k=0}^\infty \left( 1 + \frac {1} {k!} \right) = 7 + \sum_{n=4}^\infty \frac {1} {n!}\,\prod_{k=0}^{n-1} \left( 1 + \frac {1} {k!} \right)$, $\prod_{k=1}^\infty \left( 1 + \frac {1} {k^k} \right) = \frac {70} {3^3} + \sum_{n=4}^\infty \frac {1} {n^n}\,\prod_{k=1}^{n-1} \left( 1 + \frac {1} {k!} \right)$. This implies accurate approximations that are easy to reason about, such as $\prod_{k=1}^\infty \left( 1 + \frac {1} {k^k} \right) \approx \frac {70} {3^3} \left( 1+ \sum_{n=4}^\infty \frac {1} {n^n}\right)$. and I'm wondering who knew it before. My derivation uses an infinite upper bound, but that might not be necessary. I also want to find the $N$ dependent version of that tool.

    I’m thinking about the topic because by a telescoping argument I “discovered” the formula of how to truncate a product additively at any mid point $M$, actually turning them to sums with finitely computable summands. Namely

    $\prod_{k=K}^\infty b_k = \prod_{k=K}^{M-1}b_k+\sum_{n=M}^\infty(b_n-1)\,\prod_{k=K}^{n-1}b_k$

    I surprised some people on MathOverflow/StackExchange with idenitities like

    $\prod_{k=0}^\infty \left( 1 + \frac {1} {k!} \right) = 7 + \sum_{n=4}^\infty \frac {1} {n!}\,\prod_{k=0}^{n-1} \left( 1 + \frac {1} {k!} \right)$,

    $\prod_{k=1}^\infty \left( 1 + \frac {1} {k^k} \right) = \frac {70} {3^3} + \sum_{n=4}^\infty \frac {1} {n^n}\,\prod_{k=1}^{n-1} \left( 1 + \frac {1} {k!} \right)$.

    This implies accurate approximations that are easy to reason about, such as $\prod_{k=1}^\infty \left( 1 + \frac {1} {k^k} \right) \approx \frac {70} {3^3} \left( 1+ \sum_{n=4}^\infty \frac {1} {n^n}\right)$.

    and I’m wondering who knew it before. My derivation uses an infinite upper bound, but that might not be necessary. I also want to find the $N$ dependent version of that tool.

16. • CommentRowNumber16.
    • CommentAuthorMike Shulman
    • CommentTimeApr 16th 2015
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex@Nicolai #13: is that supposed to be a snarky argument that because integrals are related to sums, this is related to products? I don't dispute that it's related to products, but I think you've actually made my case, because integrals are not *called* "infinite products"; they have their own name "integral" and their own symbol $\int$ which is, as you say, a stylized sort of "S" for sum. It appears from your reference that the same is true here; these things are called "product integrals"? I'm not against mentioning this thing on the page [[infinite product]], I just don't think it should be called an "infinite product".

    @Nicolai #13: is that supposed to be a snarky argument that because integrals are related to sums, this is related to products? I don’t dispute that it’s related to products, but I think you’ve actually made my case, because integrals are not called “infinite products”; they have their own name “integral” and their own symbol $\int$ which is, as you say, a stylized sort of “S” for sum. It appears from your reference that the same is true here; these things are called “product integrals”? I’m not against mentioning this thing on the page infinite product, I just don’t think it should be called an “infinite product”.

17. • CommentRowNumber17.
    • CommentAuthorNikolajK
    • CommentTimeApr 16th 2015
    • PermaLink
    Author: NikolajK
    Format: MarkdownItex@Myke: I didn't want to sound snarky, sorry if it came across that way. In fact I never even proposed to call them infinite product, I was just pointing out that it's not so far fetched.

    @Myke: I didn’t want to sound snarky, sorry if it came across that way. In fact I never even proposed to call them infinite product, I was just pointing out that it’s not so far fetched.

18. • CommentRowNumber18.
    • CommentAuthorMike Shulman
    • CommentTimeApr 16th 2015
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexOk, sorry if I misinterpreted. (-: Feel free to create a page about product integrals and add a link to it from [[infinite product]]!

    Ok, sorry if I misinterpreted. (-: Feel free to create a page about product integrals and add a link to it from infinite product!

19. • CommentRowNumber19.
    • CommentAuthorNikolajK
    • CommentTimeApr 16th 2015
    • (edited Apr 16th 2015)
    • PermaLink
    Author: NikolajK
    Format: MarkdownItexYeah I might do that once I'm done with my quest of uniting the tool. Not sure about nPOV here, though. --- The theorems on that Wikipedia page tell you how product integral translate to sums, e.g. $\prod_a^b (1+f(x)\,dx) =\exp\left(\int_a^b f(x) \, dx\right)$, (where we know how to expand the exponential as a sum) and the formula in #15 I deviced above also translates an actual product to a sum. In fact the "infinite product is really an infinite sum of terms involving finite products" is akin to the Dyson series $|\Psi(t)\rangle=\sum_{n=0}^\infty {(-i)^n\over n!}\left(\prod_{k=1}^n \int_{t_0}^t dt_k\right) \mathcal{T}\left\{\prod_{k=1}^n e^{iH_0 t_k}Ve^{-iH_0 t_k}\right \}|\Psi(t_0)\rangle.$ It makes me question if there are limits of products at all, or if they all have some additive normal form. (Might relate to expansion and infinite factorizaiton of holomorphic objects.) I played with the product integral in Mathematica and it has a hard time evaluating stuff. Product integrating $3x^2$ takes alomst a second (=long) and gives $\exp(326)$ or something (as the formula claims). What's notable is that many of the product integrals evaluate to terms involving Pochhammer symbols. Now there are some notable product->sum translations involving those. One is lead to looking up classical formulas such as Eulers $\phi(q):=\prod_{k=1}^\infty (1-q^k)$ $=\sum_{n=-\infty}^\infty (-1)^n q^{(3n^2-n)/2}$ I tried to see how my truncation process relates e.g. to the Jacobi triple product formula, but it doesn't at all, the terms are sorted completely different.

    Yeah I might do that once I’m done with my quest of uniting the tool. Not sure about nPOV here, though.

    ---

    The theorems on that Wikipedia page tell you how product integral translate to sums, e.g.

    $\prod_a^b (1+f(x)\,dx) =\exp\left(\int_a^b f(x) \, dx\right)$,

    (where we know how to expand the exponential as a sum)

    and the formula in #15 I deviced above also translates an actual product to a sum. In fact the “infinite product is really an infinite sum of terms involving finite products” is akin to the Dyson series

    $|\Psi(t)\rangle=\sum_{n=0}^\infty {(-i)^n\over n!}\left(\prod_{k=1}^n \int_{t_0}^t dt_k\right) \mathcal{T}\left\{\prod_{k=1}^n e^{iH_0 t_k}Ve^{-iH_0 t_k}\right \}|\Psi(t_0)\rangle.$

    It makes me question if there are limits of products at all, or if they all have some additive normal form. (Might relate to expansion and infinite factorizaiton of holomorphic objects.)

    I played with the product integral in Mathematica and it has a hard time evaluating stuff. Product integrating $3x^2$ takes alomst a second (=long) and gives $\exp(326)$ or something (as the formula claims). What’s notable is that many of the product integrals evaluate to terms involving Pochhammer symbols. Now there are some notable product->sum translations involving those. One is lead to looking up classical formulas such as Eulers

    $\phi(q):=\prod_{k=1}^\infty (1-q^k)$

    $=\sum_{n=-\infty}^\infty (-1)^n q^{(3n^2-n)/2}$

    I tried to see how my truncation process relates e.g. to the Jacobi triple product formula, but it doesn’t at all, the terms are sorted completely different.