User mike stay - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:44:07Z http://mathoverflow.net/feeds/user/756 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114460/previous-work-on-this-generalization-of-continued-fractions Previous work on this generalization of continued fractions? Mike Stay 2012-11-25T22:27:09Z 2012-12-10T01:22:00Z <p>The 2x2 matrix representation of a continued fraction makes it clear that we're multiplying together a bunch of group elements. Inversion is essentially freely adjoining a generator to a Coxeter group; the usual notion of simple continued fractions comes from adding such a generator to $\tilde{I}_1$.</p> <p>Given a regular tiling of a two-dimensional space (whether spherical, Euclidean, or hyperbolic), we get a triangle group by barycentrically subdividing the regular polygons and considering the ways of reflecting the triangles into each other. The triangle group acts on the space in the obvious way.</p> <p>Say we pick the center of one of the original polygons as our origin, pick units so that the hypotenuse of the triangle has length 1, and pick one of the triangles with a vertex at the origin as the canonical one.</p> <p>We can act on the origin with an element of the triangle group and reflect it outside the unit circle, then geometrically invert the point and bring it back inside. Those points on the surface that one can reach in a finite number of such steps could be thought of as "rational".</p> <p>In the Euclidean and hyperbolic cases, we can also go the other way, since geometric inversion always takes a point inside the unit circle to a point outside of it. We can act on a point in the space with a group element and reflect it into the canonical triangle, then do geometric inversion to place it outside the unit circle. Those points that reach the origin in a finite number of moves could be called "rational"; those points with repeating continued fractions could be called "quadratic surds".</p> <p>Does anyone know of previous work on this idea? Does the generalization lead to any interesting number theory?</p> http://mathoverflow.net/questions/114687/information-about-mutant-leech-lattice-related-to-smallest-perfect-squared-square Information about mutant Leech lattice related to smallest perfect squared square Mike Stay 2012-11-27T18:23:42Z 2012-11-27T19:53:45Z <p>What happens if we follow the construction of the Leech lattice but replace the relation</p> <p>$\displaystyle \sum_{n=1}^{24} n^2 = 70^2$</p> <p>with the smallest perfect squared square? Explicitly, if we set up a dot product on $\mathbf{R}^{22}$</p> <p>$a \cdot b = a_1 b_1 + \ldots + a_{21} b_{21} - a_{22} b_{22}$</p> <p>and consider the lattice II${}_{21,1}$ whose coordinates are all integers or half integers and</p> <p>$a_1 + \ldots + a_{21} - a_{22}$</p> <p>is even, then the lattice contains the vector</p> <p>$v = (2, 4, 6, 7, 8, 9, 11, 15, 16, 17, 18, 19, 24, 25, 27, 29, 33, 35, 37, 42, 50, 112).$</p> <p>(This vector comes from the smallest perfect squared square.)</p> <p>Let $v^⊥ =${ $a \in \mbox{II}_{21,1}$ | $a\cdot v = 0$ }; then $v$ is in $v^⊥$. The lattice $v^⊥/v$ is like a strange cousin of the Leech lattice. What's known about it?</p> http://mathoverflow.net/questions/89086/examples-of-algorithms-that-came-from-category-theory Examples of algorithms that came from category theory? Mike Stay 2012-02-21T07:07:58Z 2012-02-22T20:42:13Z <p><a href="http://cseweb.ucsd.edu/~rtate/publications/proofgen/" rel="nofollow">Generating Compiler Optimizations from Proofs</a> is a wonderful paper. The authors say that they were faced with the problem, got stuck, then tried reasoning about it using category theory. They took the obvious tack, isolated the new idea, designed the abstract algorithm, and applied it to their specific case.</p> <p>What other examples like this do you know of, where category theory clearly had a role in producing a nontrivial algorithm?</p> http://mathoverflow.net/questions/70767/references-to-using-profunctors-in-program-analysis References to using profunctors in program analysis? Mike Stay 2011-07-19T19:13:43Z 2011-07-20T14:57:58Z <p>Profunctors from a category to itself seem like they'd be useful in representing the result of a program analysis; I can imagine a profunctor that given some information about a function it tells you what information you can derive about the composition of that function with something else.</p> <p>Please post references below, if you have any. Thanks!</p> http://mathoverflow.net/questions/68424/in-what-sense-do-the-categorical-trace-and-coend-count-fixed-points In what sense do the categorical trace and coend count fixed points? Mike Stay 2011-06-21T19:30:45Z 2011-06-22T22:54:09Z <p>According to the nlab, the categorical trace of a 1-endomorphism $F:C\to C$ in a 2-category is the set hom$(1_C, F)$ of global elements of $F$. If $F$ is a functor in the 2-category Cat, the categorical trace is a set of natural transformations that assign to each object of $C$ a coalgebra of $F$ such that the obvious square commutes.</p> <p>Any functor can be considered a special kind of profunctor; given an endofunctor, we can compute the coend of the corresponding profunctor.</p> <p>Both of these concepts are generalizations of the trace, which for a function counts the number of fixpoints. In what sense do these "count" the fixpoints of a functor? I don't see how the categorical trace of a functor relates to fixpoints at all.</p> <p>Also, does the notion of what constitutes a fixpoint change? The coend, in particular, seems like it might count an object $c$ as a fixpoint of $F$ if it's in the same <em>endomorphism</em> class rather than the same <em>isomorphism</em> class as $Fc$.</p> http://mathoverflow.net/questions/68424/in-what-sense-do-the-categorical-trace-and-coend-count-fixed-points/68564#68564 Answer by Mike Stay for In what sense do the categorical trace and coend count fixed points? Mike Stay 2011-06-22T22:54:09Z 2011-06-22T22:54:09Z <p>Simon Willerton explains it all very well here: <a href="http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf" rel="nofollow">http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf</a></p> http://mathoverflow.net/questions/68424/in-what-sense-do-the-categorical-trace-and-coend-count-fixed-points/68537#68537 Answer by Mike Stay for In what sense do the categorical trace and coend count fixed points? Mike Stay 2011-06-22T17:21:31Z 2011-06-22T17:21:31Z <p>Here's a partial answer: in the case of an endofunctor $F$ on a discrete category $C$ (i.e. $F$ is a function), the coend of $F$ gives the <em>set</em> of fixpoints rather than the <em>number</em>: A profunctor $F:C \not\to C$ adds extra morphisms to $C$ so that the result is still a category. I'll say these morphisms are "in $F$". The coend of $F$ is the set of endomorphisms in $F$ mod conjugation by the morphisms in $C$; since the morphisms of $C$ are all identities, we just get the set of endomorphisms in $F$, i.e. fixed points of $F$.</p> <p>The categorical trace doesn't reduce to anything useful in the case of a discrete category. A natural transformation $\alpha:1_C \Rightarrow F$ chooses for each $c \in C$ a morphism $\alpha_c:c \to Fc$ in $C$. Since we're assuming all the morphisms in $C$ are identities, $\alpha_c$ can't exist unless $Fc = c$. So it looks to me like the set hom$(1_C, F)$ is empty unless $F$ is the identity functor on $C$, in which case it's the terminal set.</p> http://mathoverflow.net/questions/47644/is-there-some-way-to-see-a-hilbert-space-as-a-c-enriched-category Is there some way to see a Hilbert space as a C-enriched category? Mike Stay 2010-11-29T05:30:13Z 2011-06-13T22:57:05Z <p>The inner product of vectors in a Hilbert space has many properties in common with a hom functor. I know that one can make a projectivized Hilbert space into a metric space with the Fubini-Study metric and then follow Lawvere in considering it a $[0,\infty)$-enriched category; it's particularly nice because the distance between two vectors is a function of the inner product.</p> <p>Is there a way of defining a monoidal structure on $\mathbb{C}$ such that the inner product is the hom so that we can think of a (perhaps projectivized) Hilbert space as $\mathbb{C}$-enriched?</p> http://mathoverflow.net/questions/66709/generator-of-translation-for-the-hyperbolic-plane Generator of translation for the hyperbolic plane? Mike Stay 2011-06-02T05:30:11Z 2011-06-02T06:05:34Z <p>What is the generator of translation in the Beltrami-Klein model of the hyperbolic plane?</p> http://mathoverflow.net/questions/54277/do-plain-functors-out-of-monoidal-categories-factor-into-a-nontrivial-monoidal-pa Do plain functors out of monoidal categories factor into a nontrivial monoidal part and a plain part? Mike Stay 2011-02-04T01:32:27Z 2011-02-04T03:44:57Z <p>Given symmetric monoidal closed categories $C, D$, a symmetric monoidal (not closed!) functor $F:C\to D$ factors into two parts: </p> <ul> <li><p>the first is a symmetric monoidal <strong>closed</strong> functor from $C$ to a "halfway house" $C'$, followed by </p></li> <li><p>a symmetric monoidal functor from $C'$ to $D$.</p></li> </ul> <p>To get $C'$, you do two changes of base: </p> <ul> <li><p>since $C$ is closed, consider it a $C$-enriched category and then apply $F$ to its hom objects to get a $D$-enriched category, and then </p></li> <li><p>apply the "points" functor $\mbox{hom}(1, -):D \to \mbox{Set}$ to get a plain category.</p></li> </ul> <p>Does something similar always happen when a functor fails to preserve the structure of the source and target categories? In particular, does a plain functor between monoidal categories factor into a nontrivial monoidal functor followed by another plain functor?</p> http://mathoverflow.net/questions/50832/references-for-promorphisms-of-profunctors References for promorphisms of profunctors? Mike Stay 2010-12-31T21:33:09Z 2010-12-31T21:33:09Z <p>We can interpret a profunctor $F:C^{\mbox{op}}\times D \to \mbox{Set}$ between small categories as adjoining some morphisms to the category $C \cup D$ to get a new category $\tilde{F}$. Then a natural transformation between profunctors $F, G$ can be seen as a functor between $\tilde{F}, \tilde{G}$. We can also look at profunctors between $\tilde{F}, \tilde{G}$, but I don't know what such things are called--searching Google for "pronatural transformation" gives no hits, and "natural pro-transformation" gives a single hit, S. Yokura's paper. Nothing for "cocontinuous natural transformation" or "natural cocontinuous transformation" either.</p> <p>Are these studied under some other name?</p> http://mathoverflow.net/questions/37382/what-is-the-proper-name-for-compact-closed-multiplicative-intuitionistic-linear What is the proper name for "compact closed" multiplicative intuitionistic linear logic? Mike Stay 2010-09-01T14:21:05Z 2010-12-06T19:56:44Z <p>Multiplicative intuitionistic linear logic (MILL) has only multiplicative conjunction $\otimes$ and linear implication $\multimap$ as connectives. It has models in symmetric monoidal closed categories. </p> <p>Compact closed categories are symmetric monoidal closed categories in which every object $A$ has a dual $A^*$ and $A \multimap B \cong A^* \otimes B$. Thought of as a resource, $A^*$ is a debt, owing someone an $A$. Is there a special name for MILL when these conditions hold?</p> http://mathoverflow.net/questions/47301/which-linear-transformations-between-f-d-hilbert-spaces-contract-the-inner-produ Which linear transformations between f.d. Hilbert spaces contract the inner product? Mike Stay 2010-11-25T05:49:38Z 2010-11-25T13:00:41Z <p>Given two finite-dimensional Hilbert spaces $U, V,$ a linear transformation $T:U\to V$ contracts the inner product if for all $x,y \in U,$ $$\langle x,y \rangle_U \ge \langle Tx, Ty\rangle_V.$$ All unitary transformations satisfy this criterion; is there a larger class of linear transformations that do?</p> http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid Is there a sensible way to enrich over SymMonCat such that id_X is not the monoidal unit? Mike Stay 2010-11-09T15:45:39Z 2010-11-12T16:23:19Z <p><strong>SymMonCat</strong> is the cartesian 2-category of symmetric monoidal categories, braided monoidal functors, and monoidal natural transformations. The terminal symmetric monoidal category <strong>1</strong> has one object $I$ and $I \otimes I = I$.</p> <p>A category enriched over a monoidal category $V$ assigns to each pair of objects $X, Y$ an object hom$(X,Y)$ in $V$ and to each object $X$ a morphism $id_X:I \to \mbox{hom}(X,X)$ in $V$.</p> <p>When $V =$ <strong>SymMonCat</strong>, the morphism $id_X:1 \to \mbox{hom}(X,X)$ is a braided monoidal functor; since monoidal functors preserve the monoidal unit and tensor product, it must map the unit $I$ in <strong>1</strong> to the unit $I$ in hom$(X,X)$.</p> <p>Is there a different way of enriching over <strong>SymMonCat</strong> such that $id_X$ does not pick out the monoidal unit (other than considering it a subcategory of <strong>Cat</strong>)? </p> http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid/45705#45705 Answer by Mike Stay for Is there a sensible way to enrich over SymMonCat such that id_X is not the monoidal unit? Mike Stay 2010-11-11T14:52:48Z 2010-11-11T14:52:48Z <p>Yes, the tensor product I want is the one described by <a href="http://arxiv.org/abs/0711.0324" rel="nofollow">Vincent Schmitt</a>; it satisfies the universal property that any other braided monoidal bifunctor is naturally isomorphic to it.</p> http://mathoverflow.net/questions/33230/matrices-whose-exponential-is-stochastic Matrices whose exponential is stochastic Mike Stay 2010-07-24T21:21:26Z 2010-07-25T06:54:53Z <p>The complex matrix exponential of a Hermitian matrix is unitary: <code>$e^{-iH} = U$</code>. Is there a name or a characterization for matrices Q whose real exponential is stochastic: <code>$e^{-Q} = S$</code>?</p> http://mathoverflow.net/questions/23476/is-this-related-to-the-j-function Is this related to the j-function? Mike Stay 2010-05-04T20:47:15Z 2010-05-04T22:05:42Z <p>I was browsing in Plouffe's inverter and found that $\frac{\exp(\pi)-\ln(3)}{\ln(2)}$ is very nearly $\frac{159}{5}.$ The continued fraction is</p> <p>[31, 1, 4, 12029125, ...].</p> <p>Is this the same magic as $\exp(\pi \sqrt{163})$?</p> http://mathoverflow.net/questions/21358/does-every-cocontinuous-functor-between-categories-of-presheaves-on-small-categor Does every cocontinuous functor between categories of presheaves on small categories have a right adjoint? Mike Stay 2010-04-14T16:59:41Z 2010-04-14T22:15:41Z <p>Let <em>C, E</em> be small categories, let <em>C&#x0302;</em> = Set<sup><em>C</em><sup>op</sup></sup>, and let <em>F:C&#x0302; &rarr; E&#x0302;</em> be cocontinuous. I think <em>F</em> will always have a right adjoint when <em>C, E</em> are small, but not necessarily if they're large. Is that right?</p> http://mathoverflow.net/questions/20580/coend-computation-continued Coend computation continued Mike Stay 2010-04-07T01:32:05Z 2010-04-07T04:41:37Z <p>This is a follow-up question to this <a href="http://mathoverflow.net/questions/20445/coend-computation" rel="nofollow">coend computation</a>. Here's an attempt at a slightly simpler computation:</p> <blockquote> <p><code>$\int^{a \in A} \mbox{hom_A(a,a)}$</code></p> </blockquote> <p>This should be similar to the trace operator. In attempting to follow the derivation</p> <blockquote> <p><code>$\begin{array}{l}\mbox{Set}(\int^{b \in B}\mbox{hom}(a, b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b), \mbox{Set}(F(b), S)) \\ \cong \mbox{Nat}(\mbox{hom}(a,-), \mbox{Set}(F(-), S) \\ \cong \mbox{Set}(F(a), S),\end{array}$</code> </p> </blockquote> <p>I get</p> <blockquote> <p><code>$\begin{array}{l}\mbox{Set}(\int^{a \in A} \mbox{hom}_A(a,a), S) \\ \cong \int_{a \in A} \mbox{Set}(\mbox{hom}_A(a,a), S) \\ \cong \mbox{Nat}(\mbox{hom}_A(-,-), S)\end{array}$</code></p> </blockquote> <p>So here I guess we have the set of natural transformations from the hom functor to the constant functor $S$. For any first parameter $a$, we have the set of natural transformations from hom$(a,-)$ to $S(a,-)$, which by Yoneda's lemma is isomorphic to $S(a,a) = S$. So I think it goes</p> <blockquote> <p><code>$\begin{array}{l}\cong \displaystyle \prod_a \mbox{Nat}(\mbox{hom}_A(a,-), S(a,-)) \\ \cong \prod_a S\\ \cong S^{\mbox{Ob}(A)} \\ \cong \mbox{Set}(\mbox{Ob}(A), S).\end{array}$</code></p> </blockquote> <p>So <code>$\int^{a \in A} \mbox{hom_A(a,a)} \cong \mbox{Ob}(A).$</code> Is that right?</p> http://mathoverflow.net/questions/20445/coend-computation Coend computation Mike Stay 2010-04-06T00:08:07Z 2010-04-06T00:34:51Z <p>Let </p> <blockquote> <p>$F:A^{\mbox{op}} \to \mbox{Set}$</p> </blockquote> <p>and define </p> <blockquote> <p><code>$G_a:A\times A^{\mbox{op}} \to \mbox{Set}$</code></p> <p><code>$G_a(b,c) = \mbox{hom}(a,b) \times F(c)$</code>. </p> </blockquote> <p>I <i>think</i> the coend of <code>$G_a$</code>, </p> <blockquote> <p><code>$\int^AG_a$</code>, </p> </blockquote> <p>ought to be $F(a)$--it's certainly true when A is discrete, since then hom is a delta function. But my colimit-fu isn't good enough to actually compute the thing and verify it's true. Can someone walk me through the computation, please?</p> http://mathoverflow.net/questions/20203/where-do-i-turn-for-help-with-generating-functions/20446#20446 Answer by Mike Stay for Where do I turn for help with generating functions? Mike Stay 2010-04-06T00:15:07Z 2010-04-06T00:15:07Z <p>Also, be sure there isn't another easier way of doing the same thing. Download Wilf's book <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">Generatingfunctionology</a> and check to see that someone else hasn't already solved the problem!</p> http://mathoverflow.net/questions/1083/do-good-math-jokes-exist/1239#1239 Answer by Mike Stay for Do good math jokes exist? Mike Stay 2009-10-19T17:19:06Z 2009-10-19T17:19:06Z <p>Here are a few of my own inventions:</p> <p>Old Macdonald had a form; e<sub>i</sub> /\ e<sub>i</sub> = 0</p> <p>Save the environment: use continuation passing style!</p> <p>What shape of pasta takes the least time to eat? Brachistochroni!</p> <p>You might be a mathematician if you think fog is a composition.</p> <p>The Yoda embedding, contravariant it is.</p> <p>How are Goethe's Faust novels like isomorphisms of sets? Dey're de monic epics.</p> <p>I'm kind of in two minds about this whole Schroedinger's cat thing...</p> <p>qwhine, n. self-recrimination</p> <p>recursive: (λ damn. damn (damn)) (λ damn. damn (damn))</p> <p>Coeschatology: the study of the beginning of times. The coend is ming!</p> http://mathoverflow.net/questions/89086/examples-of-algorithms-that-came-from-category-theory/89087#89087 Comment by Mike Stay Mike Stay 2012-02-21T23:12:28Z 2012-02-21T23:12:28Z Thanks! The parser example is more in line with what I'm looking for. I'm trying to explain the merits of category theory to users of imperative, stateful, dynamically-typed programming languages; their initial impression of monads is a hard-to-understand kludge that's only necessary because of a silly insistence on functional purity. While I'm a big fan of algebraic data types, functional programming, and monads, I'd like applications outside of type theory. http://mathoverflow.net/questions/59892/co-ends-as-a-trace-operation-on-profunctors/59897#59897 Comment by Mike Stay Mike Stay 2011-06-27T23:32:30Z 2011-06-27T23:32:30Z Simon Willerton has a nice set of slides on the subject: <a href="http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf" rel="nofollow">simonwillerton.staff.shef.ac.uk/ftp/&hellip;</a> http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid/45515#45515 Comment by Mike Stay Mike Stay 2010-11-12T16:18:39Z 2010-11-12T16:18:39Z Thanks to both of you! I voted up Mike's comment, but since it wasn't given as an &quot;answer&quot;, I'm accepting Chris' answer. http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid Comment by Mike Stay Mike Stay 2010-11-10T19:32:08Z 2010-11-10T19:32:08Z Yes, that sounds like just what I want. Is that the same tensor product described by Vincent Schmitt ( <a href="http://arxiv.org/abs/0711.0324" rel="nofollow">arxiv.org/abs/0711.0324</a> )? http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid/45576#45576 Comment by Mike Stay Mike Stay 2010-11-10T19:19:41Z 2010-11-10T19:19:41Z I've discovered that I <i>don't</i> want the cartesian product, for the same reason that a one-object category enriched over (<b>Ab</b>, $\times$, 1) is just an abelian group, not a ring. You have to enrich over (<b>Ab</b>, $\otimes$, $\mathbb{Z}$) to get a ring. So my question has become how to define $\otimes$ properly on <b>SymMonCat</b> and what its monoidal unit is. http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid/45452#45452 Comment by Mike Stay Mike Stay 2010-11-09T22:10:38Z 2010-11-09T22:10:38Z Just thought of at least one place I'm going wrong: the monoidal unit in (Ab, $\otimes$) is $\mathbb{Z}$, not the trivial group. Is there a similar tensor product on SymMonCat? http://mathoverflow.net/questions/45449/is-there-a-sensible-way-to-enrich-over-symmoncat-such-that-id-x-is-not-the-monoid/45452#45452 Comment by Mike Stay Mike Stay 2010-11-09T21:27:40Z 2010-11-09T21:27:40Z Thinking about it some more, I guess I have the same question about Ab-enriched categories, aka ringoids. A one-object Ab-enriched category is a ring; multiplication is composition and addition comes from the abelian group structure. Every object $X$ comes with a group homomorphism from the trivial group to hom$(X,X)$; this preserves the unit and products, so $id_X$ has to pick out the <i>additive</i> identity, not the <i>multiplicative</i> identity. Where am I going wrong? http://mathoverflow.net/questions/33230/matrices-whose-exponential-is-stochastic/33247#33247 Comment by Mike Stay Mike Stay 2010-07-25T16:01:54Z 2010-07-25T16:01:54Z Thanks! When Googling &quot;infinitesimally stochastic&quot;, the first hit with a definition seemed to be this one: <a href="http://arxiv.org/abs/1002.4773" rel="nofollow">arxiv.org/abs/1002.4773</a> http://mathoverflow.net/questions/20580/coend-computation-continued/20586#20586 Comment by Mike Stay Mike Stay 2010-04-07T05:03:44Z 2010-04-07T05:03:44Z Yeah, I clearly got that wrong. At best I've got a dinatural transformation from a constant functor to hom that picks out the identity. http://mathoverflow.net/questions/20580/coend-computation-continued Comment by Mike Stay Mike Stay 2010-04-07T04:42:08Z 2010-04-07T04:42:08Z I copied over the answer from the previous question; hopefully it's clearer now. http://mathoverflow.net/questions/20445/coend-computation/20451#20451 Comment by Mike Stay Mike Stay 2010-04-07T01:32:17Z 2010-04-07T01:32:17Z I've asked a follow-up question here: <a href="http://mathoverflow.net/questions/20580/coend-computation-continued" rel="nofollow" title="coend computation continued">mathoverflow.net/questions/20580/&hellip;</a> http://mathoverflow.net/questions/20203/where-do-i-turn-for-help-with-generating-functions/20446#20446 Comment by Mike Stay Mike Stay 2010-04-06T19:59:43Z 2010-04-06T19:59:43Z Oops--didn't mean to be redundant! http://mathoverflow.net/questions/20445/coend-computation/20451#20451 Comment by Mike Stay Mike Stay 2010-04-06T18:30:23Z 2010-04-06T18:30:23Z Thanks! The third step is really the one I need to understand, so I'll think about that for a while and come back if I need more help.