jessecmckeown

[jessecmckeown]

To tidily quarter a given angle in place, draw two circles with a common radius

[jessecmckeown]

there, that's better....

[jessecmckeown]

The same configuration, highlighted in two different ways. Puzzle: How Many Desargues-Homologous pairs in one diagram?

[jessecmckeown]

If this is happening in salt water, and the drinks are alcoholic, they will also float away; just turn the cups upside-down! (Where are you keeping the ice, anyway???)

i had a dream i worked in an underwater restaurant and people kept ordering ice in their drinks and then getting mad at me when it would float away. and i’d tell them beforehand that the ice would float away & they’d be like lol no that’s not how it works just give me the ice. I’m fighting customer service battles never seen before

[jessecmckeown]

It's Not $\Pi$ day

I've not had enough time to do maths properly for the past... well, quite a long while, actually, but in particular I got myself a REAL JOB in INDUSTRY, in fact as a welder, which takes up twelve or thirteen hours most days (ten of them are spent Working)... Anyways, I'd Like To Let the Tau Fans Know Something about Circles in Industry.

Nobody measures the radius of an Industrial Circle. Never. Radii are not a Thing. It's Diameters, if possible, that we measure, and if a circle is Unfinished, we compare it with a template. At least, most of the time. I imagine a Machinist working at a Lathe may well consider truly radial measurements, but we don't have any genuine lathes in our plant, and that's Just Fine.

And the Reason that no-one measures the Radius of an Industrial Circle is because it's basically impossible. You see, to measure the radius, you have to locate the center. This is, of course, doable in principle, but pretty much a waste of time, because the center is located at the intersection of DIAMETERS. And if you have a diameter, you may as well measure THAT.

Some Caveats: the principal method of producing one of our Industrial Circles is by rolling (that is, incrementally locally bending) some material beyond its elastic/plastic transition. Sometimes it's feasible to do this end-to-end and then check: have I closed a circle? sometimes this is not feasible, and instead the ends of the material to be bent circlewise must be first matched to a template. But there are other ways --- with a lathe, for instance, one may cut both external and internal circles. One will have a decent estimate of the radial location of the cutter, but you'll still measure a diameter. But SOMETIMES, particularly in preparing the templates we have mentioned (sometimes to cut out holes to admit other industrial circles passage), we will draw a circle using a compass, and then cut along the lign. This is the ONLY case in which the radius is more easily known than the diameter, and it's when we LEAST care about the circumference produced. For the compassed circles we most use, the center is promptly thrown away, because it would make the template too large and unwieldy.

That is all, you may now return to your what-have-you.

[jessecmckeown]

Whence Fabulous Faulhaber?

should I promise not to make this a habit?

Dear Mr. Haran, I'm grateful to you and correspondent Eischen (and Conway) for putting the name of Faulhaber on a calculation which heretofore I'd only known quoted, without attribution, by Heinrich Dörrie in Triumf der Mathematik---(which I've only read in translation). However, I'm most frustrated that neither Dörrie nor Eischen give any satisfying motivation for why the postulate should work.

For bystanders still catching up, this postulate is that if one defines a sequence of numbers $B_k$ "by expanding" $$(B-1)^{k+1} = B^{k+1}$$ and transcribing exponents to subscripts... one finds that the differences $$ (n + B)^{k+1} - B^{k+1} $$ similarly treated are equal to cumulative power sums, $$(k+1) \sum_{j \leq n} j^k$$

So the calculation is doable. My Beef is Dilemmimorphic: Either the notational abuse of $(n + B)^k$ suggests that $B$ should be Some Kind Of Linear Operator, in which case what is it? Or else there's an Amazing Coincidence being Overlooked!

It's a comparative Triviality that the power sums $\sum_1^N n^k$ should be polynomials of $N$, and that the leading term be $\frac{1}{k+1} N^{k+1}$ , so indeed it is perfectly reasonable to consider coefficients $B_{k,j}$ defined by $$ \sum_1^N n^k = \frac{1}{k+1} \sum \binom{k+1}{j} B_{k,j} N^{k+1-j} $$ BUT WHY SHOULD WE ASSUME that in fact $B_{k,j}$ depends only on $j$? That's STAGE MAGIC, and the fact that indeed it somehow works does not explain "where it comes from" (Eischen's favourite phrase on the matter).

So, in my customary way of starting with the actual problem and throwing at it what seems to me the minimum of thought, let's first explicate that "comparative triviality": the sequence of polynomials $p_k(j) = \binom{j+k}{j}$ are integral generators for the Integral-valued polynomials, and are recursively definable as iterated cumulative sums of the constant polynomial $p_0 \equiv 1$: $$\binom{j+k+1}{j} = \binom{j+k}{j} + \binom{j+k}{j-1}$$. Hence, cumulative sums of any polynomial, written in the binomial basis, can be obtained just by incrementing: $$\sum_{j=1}^N \sum a_n p_n(j) = \sum a_n p_{n+1}(N)$$

Next, cumulative sums are themselves defined by induction: $"\sum_{j=1}^0" P(j) = 0$ and $\sum_{j=1}^{N+1} P(j) = P(N+1) + \sum_{j=1}^N P(j)$, or said differently, by the Difference equation $$ SP(N+1) - SP(N) = P(N+1).$$ In other words we are trying to solve the Difference Equations $$ S_k(N) - S_k(N-1) = N^k,$$ but in the basis of Monomials $N^j$ instead of Binomials $p_j(N)$.

The binomial theorem, $$ (x+y)^k = \sum \binom{k}{j} x^{k-j} y^j $$ makes the Taylor-MacLaurin formula a Theorem for polynomials $$ (x+y)^k = \sum y^j \frac{1}{j!} \frac{d^j}{dx^j} x^k $$ which is fruitfully abbreviated $$ P(x+y) = e^{y\\, d/dx} P(x) $$ the Backwards Difference, then, is similarly $$ P(x) - P(x-1) = (1 - e^{- d/dx}) P(x) $$

Shall we say, The kernel of the Backward Difference is reasonably well understood? The differential operator is the retract of the Integral operator $\int_0$, so the Taylor-MacLaurin formula provides us also a section for the Forward Difference operator, $$ 1-e^{-x} = \frac{d}{dx} + A\frac{d^2}{dx^2} $$ where, for now, the main point is that the unbounded-degree differential operator $A$ commutes with $d/dx$, so that, for example $$ (1 - e^{-d/dx}) \left(\int_0 \sim dx - A + A^2 \frac{d}{dx} - A^3\frac{d^2}{dx^2} + - \cdots \right) P(x) = P(x)$$

Of course, there are various paths to the power series, other than via expansion of the powers of $A$, but there is a (Laurent) power series $$ \frac{1}{1-e^{-t}} = \frac{1}{2}\coth(\frac{t}{2})+\frac{1}{2} = \frac{1}{t} + \sum \frac{B_j}{j!} t^{j-1} $$ where $B_j$ are the faBulous Bernoulli numbers.

In any case, applied to simple powers, $$ \left( \int_0 \sim dx + \frac{1}{2} + \sum_{j=2}^{\infty} \frac{B_j}{j!} \frac{d^{j-1}}{dx^{j-1}} \right) x^k = \frac{1}{k+1} x^{k+1} + \sum_{j=1}^{k} \frac{k!}{j!(k-j+1)!} x^{k-j+1} B_j \\\\ {} = \frac{1}{k+1} \sum_{j=0}^{k} \binom{k+1}{j} B_j x^{k+1-j} $$ Finally, the power sum polynomials $S_k$ vanish both at zero (formally an empty sum) and at $-1$ (since $S_k(0) - S_k(-1) = 0^k$), so that in particular, $$ \sum_{j=0}^k \binom{k+1}{j} B_j (-1)^{k-j} = 0$$ THAT'S WHERE THIS IS COMING FROM.

[jessecmckeown]

Silly Question: What Is the Binomial Theorem?

by which I mean: it's obviously not "Pascal's" "Triangle"; That is: it's not just the fact that (in commutative algebra) there are "binomial coefficients"; nor even that, for reasons of applicable combinatorics, the binomial coefficients get to be called "en-choose-kay". If anything is The Binomial Theorem, it's the coincidence of the binomial coefficients and certain fractions involving factorials. And while I'm as happy as the next fellow to agree that the number of subsets of a given size out of a set of the given size is equal to the number of permutations of the whole set modulo the permutations that fix the prefix/suffix partition at a fixed index... there are still more ways to interpret that equation than "precommutative terms in an algebraic expression, modulo the commutativity relations". That is, once we've decided that there ARE Binomial Coefficients, and that they are Integers, we can choose any argument that gives us those integers, even if it doesn't look like it need give integers. The underlying combinatorics, even, may be identitical, but where we apply them doesn't have to be "the terms of an algebraic expression". In other words, we can read the Binomial Theorem as saying $$ \frac{(a+b)^n}{n!} = \sum_{k+l=n} \frac{a^k}{k!}\frac{b^l}{l!} $$ just as well as anything else; and we can construe the left hand side as the volume of an $n$-simplex with axes $(a+b)$, and the right hand side similarly as a sum of suitable products of simplices. So I wrote a p3/java sketch to see what that looks like. Incidentally, I've often told myself that I write code that is basically terrible to maintain. Can't seem to break those habits...

Oh. You wanted an argument. Yes. Of Course. The volume of a Euclidean $k$-simplex containing an orthonormal basis among its edges is $1/k!$, because a $k$-dimensional cube can be subdivided into $k!$ of them with lower-dimensional overlaps, because an indexed set of $k$ real numbers also has a natural partial-ordering --- $\leq$. Generalizing to a $k$-simplex with $k$ orthogonal edges of lengths $a$, $b$, or $(a+b)$ are straightforward enough. Partitioning! Let's focus on an interior point of an orthogonal $n$-simplex with edges of length $(a+b)$; that is, $\leq$ is a total order on its coordinates. We might as well be looking at $$ 0 \lt x_1 \lt x_2 \lt \dots \lt x_n \lt (a+b) $$ But there are furthermore $n+1$ possible comparisons of each $x_j$ with $a$! Videlicet, $$ 0 \lt x_1 \lt \dots \lt x_k \lt a \lt x_{k+1} \lt \dots \lt x_n \lt a+b $$ That is, comparing coordinates with $a$ gives a natural partition of the simplex into products of simplices: an orthogonal $k$-simplex of size $a$ and an orthogonal $(n-k)$-simplex of size $b$, hurrah! It's not meant to be a profound; but last year I just wanted to show some high-school students that the binomial theorem could be really geometrical, as well as algebraic.

[jessecmckeown]

Theta the Magical

So, these Elliptic Curve things --- It's become very natural, post-Asteroids™ to think of "Torus" as $\mathbb{C}/\mathbb{Z}^2$, and so a function on an elliptic curve is "the same as" a doubly-periodic function on $\mathbb{C}$; it breaks some of the symmetry of this picture, but there is something marvelous that happens when one views the Torus instead as the quotient of a Cyllinder and --- because $\exp$ teaches us how to view $\mathbb{C}^\times$ as a cyllinder --- considers functions on a Torus as functions on $\mathbb{C}^\times$ with a discrete scaling-invariance. Let $ 0 \lt |q| \lt 1 $, and for a warm-up, verify that the product $$ R_q(z) = \prod_{n\in\mathbb{N}} (1 + q^{2n+1} z) $$ is absolutely convergent, with zeros $-\frac{1}{q^{2n+1}}$, and has a nice scaling property: $$ R_q(z/q^2) = \frac{q+z}{q} R_q(z) $$ Check that similarly, $$ R_q(\frac{q^2}{z}) = \frac{z}{q+z} R_q(\frac{1}{z}) $$ so that the product $$ \Theta(q,z) = R_q(z) R_q(1/z) $$ has an even nicer scaling property, $$ \Theta(q,z/q^2) = \frac{z}{q} \Theta(q,z) $$ and the fraction, even better: $$\Psi(q,z) = \frac{\Theta(q,z)}{\Theta(q,-z)} = - \frac{\Theta(q,q^2z)}{\Theta(q,-q^2z)} = \Psi(q,z/q^4) $$ Furthermore, for straight-forward reasons, $\Theta(q,z)$ has critical points at $z=\pm 1$ so that the fraction $\Psi(q,z)$ has critical points at even powers of $q$, $z = q^{2m}$ All of the preceding should be Routine. As the title of this post is meant to suggest, what we've called $\Theta$ is more like the star of this show than is $\Psi$, although it's not really the famous $\theta$... we'll get to $\theta$ in a minute. And the genuine star $\theta$ will give us the same $\Psi$, in some sense. The way $\Theta$ is defined, it should also have a very good Laurent Series Expansion, but (here's the trick) for now we're only going to worry about the second variable: $$ \Theta(q,z) = \sum_{n\in\mathbb{Z}} a_n(q) z^n $$ because the scaling property from above, $$ \sum_{n\in\mathbb{Z}} a_n(q) (z/q^2)^n = \frac{z}{q} \sum_{n\in\mathbb{Z}} a_n(q) z^n $$ is telling us that $$ q^{2n-1} a_{n-1}(q) = a_n(q) $$ so that, inductively, $$ a_n(q) = q^{n^2} a_0(q) .$$ It follows that $\Psi(q,z)$ can alternatively be written $$ \Psi(q,z) = \frac{\sum_{n\in\mathbb{Z}} q^{n^2} z^n }{\sum_{n\in\mathbb{Z}} (-1)^n q^{n^2} z^n }$$ and the first bit of magic is this: we really have no right to expect that any Laurent series defined by a nice-looking product should be so very sparse in any of its variables, but only square powers of $q$ need appear in this fraction-of-series representation of $\Psi$. The Really Magical $\theta$ then is defined to be this sparse series $$\theta(q,z) = \sum_{n\in\mathbb{Z}} q^{n^2} z^n $$ But that's just a hint of the Magic!

The basic Critical Points of $\theta(q,z)$, $z=\pm 1$, tells us we should particularly consider the series, also called $\theta(q) = \sum_{n\in\mathbb{Z}} q^{n^2} $, as well as the particular product $$ \Theta(q,1) = \prod (1+q^{2n+1})^2 $$ There is a Curious Relation one may exploit: $$ \Theta(q,1) \Theta(q,-1) = \prod_\mathbb{N} (1+q^{2n+1})^2(1 - q^{2n+1})^2 = \prod_\mathbb{N} (1-(q^2)^{2n+1})^2 = \Theta(q^2,-1) .$$ If nothing else, this suggests we should also consider the product $$ \theta(q) \theta(-q) = \left(\sum_\mathbb{Z} q^{n^2}\right) \left(\sum_\mathbb{Z} (-q)^{n^2}\right) $$ about which, some observations: if $m+n$ is odd, then we have some cancellation, or more precisely: $$q^{m^2}(-q)^{n^2} + q^{n^2}(-q)^{m^2} = ((-1)^n+(-1)^m) q^{m^2} q^{n^2} $$ which means that only terms with $m$ and $n$ of the same parity contribute; and $m$ and $n$ being of the same parity means exactly that there is a unique solution $m = a + b; n= a-b$: $$ \theta(q) \theta(-q) = \sum_{\mathbb{Z}^2} (-1)^{a-b} q^{(a+b)^2} q^{(a-b)^2} = \sum_{\mathbb{Z}^2} (-q^2)^{a^2} (-q^2)^{b^2} = (\theta(-q^2))^2 .$$ Significant "Hmmm...", Number 1. Of course, the terms with $m=n$ only turn up once each, as $(-1)^m q^{2m^2}$; but $m=n$ is a special case of $m\equiv_2 n$ Moving on, comparing the two relations by means of $a_0(q) \theta(q) = \Theta(q,1)$, $$ \theta(-q^2)^2 = \theta(q)\theta(-q) = \frac{1}{a_0(q)^2} \Theta(q,1)\Theta(q,-1) = \frac{1}{a_0(q)^2} \Theta(q^2,-1) = \frac{a_0(q^2)}{a_0(q)^2} \theta(-q^2) $$ or $$ \frac{1}{a_0(q)^2} = \frac{1}{a_0(q^2)} \theta(-q^2) = \frac{1}{a_0(q^2)^2} \Theta(q^2,-1) $$ From here, it's easy to check that $\lim_{q\to 0} a_0(q) = 1,$ and so $$ \frac{1}{a_0(q)} = \prod_{n\in\mathbb{N}} (1 - q^{2(n+1)}) $$ or, more famously known as the Jacobi Triple Product Identity, $$ \prod_{n=1}^\infty (1-q^{2n}) (1-q^{2n-1}z)(1-q^{2n-1}z^{-1}) = \sum_{n\in\mathbb{Z}} q^{n^2} (-z)^n $$ ... talk about Excessively Sparse $q$-Series! There must, of course, be a nice pictographable combinatorial argument for this analytic result. Ask Professor Polster. We've had one Significant "Hmmm...", wherein we found ourselves looking at $\theta(-q^2)^2 = \sum_{\mathbb{Z}^2} (-1)^{(a+b)} q^{2(a^2+b^2)}$; Those signums, of course, look a tad irksome; it'd be nice if we could drop them, but keep the cancellation; but that's Not Too Bad Either. The cancellation we had was from $(-1)^m + (-1)^n$, but we get the same zeroness with $1+(-1)^{m+n}$, in which the nonzero terms are all $2$. In Short: $$ 2 \sum_{a,b:\mathbb{Z}} q^{2(a^2+b^2)} = \sum_{m,n:\mathbb{Z}} (1 + (-1)^{m+n}) q^{m^2+n^2} = \theta(q)^2 + \theta(-q)^2 $$ and THIS (Significant "Hmmm..." Number 2) is the Start of the Real Magic of Theta the Magical: $$ \theta(q^2)^2 = \frac{\theta(q)^2+\theta(-q)^2}{2} \qquad \theta(-q^2)^2 = \sqrt{\theta(q)^2\theta(-q)^2} $$ $$ \operatorname{AGM}\big(\theta(q)^2,\theta(-q)^2\big) = 1 $$

[jessecmckeown]

from the MIT AI Lab memo "HAKMEM"...

[jessecmckeown]

Landen Pt. 5: the notorious stuff

Let us Name some Maps. "The" "Landen" transform, $v\mapsto \frac{1}{\sqrt{k}} \frac{2v}{1+v^2} $ is a double-cover of the elliptic curve $y^2=\sqrt{(k-x^2)(1-kx^2)}$ by the elliptic curve $u^2=\sqrt{(l-v^2)(1-lv^2)}$, where $$ l = L(k) = \frac{k^2}{(1+\sqrt{1-k^2})^2} ;$$ which defines $L$, one of our Important Named Maps. Similarly, the transposition $v \mapsto \frac{1-v}{1+v}$ is an isomorphism of whatever elliptic curve $y^2=\sqrt{(k-x^2)(1-kx^2)}$ and $u^2=\sqrt{(r-v^2)(1-rv^2)}$ with $$ r = T(k) = \left(\frac{1-\sqrt{k}}{1+\sqrt{k}}\right)^2 $$ which defines $T$, another Important Named Map. About two and half Pt.s ago, we pointed out that $L(v)^2 + L(T(v))^2 = 1$, a Curious Circumstance Indeed.

Among Elliptic Integrals $$ \int_a^b \frac{Q(x)}{\sqrt{P(x)}} dx $$ the Periods get a lot of attention, which correspond (roughly) to choosing $a,b$ to be (adjacent) roots of $P$. It doesn't look like it, but the [quarter] periods commonly known as $E(k)$ and $K(k)$ are, in the arrangement of things from Pt 4, easily written $$ E(k) = \int_0^{\sqrt{k}} \sqrt{\frac{1-kx^2}{k-x^2}} dx \\\\ K(k) = \int_0^{\sqrt{k}} \frac{1}{\sqrt{(1-kx^2)(k-x^2)}} dx $$ and the calculation we described but didn't do also gives us the lovely relations $$ E(k) = \frac{2}{1+l} E(l) - (1-l) K(l) \\\\ K(k) = (1+l) K(l) $$ where $$ k = \frac{2\sqrt{l}}{1+l} $$ Nice as those relations are, it's even nicer to write them like this: $$ E(l) = \frac{1+l}{2} E\left(\frac{2\sqrt{l}}{1+l}\right) + \frac{1-l}{2} K\left(\frac{2\sqrt{l}}{1+l}\right)$$ $$ K(\sqrt{1-l^2}) = \frac{2}{1+l}K\left(\frac{1-l}{1+l}\right) $$ which, on multiplying, gives us the Curious Equation $$ E(l)K(\sqrt{1-l^2}) = E\left(\frac{2\sqrt{l}}{1+l}\right) K\left(\frac{1-l}{1+l}\right) + \frac{1-l}{1+l} K\left(\frac{2\sqrt{l}}{1+l}\right) K\left(\frac{1-l}{1+l}\right) $$ or, in terms of our Named Maps, $$ E(L k) K(L T k) = E(k) K(L L T k) +(LL Tk) K(k) K( LL T k) $$ about which we note: the elliptic products on both sides involve complementary arguments. Continuing, this arrangement also makes it easier to deduce the (equivalent) $$ E(LLTk) K(k) = E(LTk)K(Lk) + (L k) K(LTk)K(Lk) $$ which, added to the original, also gives us $$ E(Lk)K(LTk) + K(Lk)E(LTk) + (Lk)K(LTk)K(Lk) \\\\ = \\\\ E(LLTk)K(k) + K(LLTk)E(k) + (LLTk) K(k)K(LLTk) $$ From here, it is a simple "Exercise" to check that the very symmetric expression $$ E(Lk)K(LTk) + K(Lk)E(LTk) - K(Lk)K(LTk) $$ is invariant under replacement of $k$ with $L k$. At this point, the following should be easy calculations: $$ E(0)=K(0)=\pi/2 $$ $$ E(1) = 1 $$ $$ K(k) - E(k) \lt k^2 K(k) $$ $$ K(1 - k) \underset{k\to 0}{\sim} \log\frac{1}{k} $$ $$ \lim_{k\to 0} \left[ E(k)K(\sqrt{1-k^2}) + E(\sqrt{1-k^2})K(k) - K(\sqrt{1-k^2})K(k)\right] = \frac{\pi}{2} $$ BUT! Since $L^n k$ tends to $0$, and since the subject of the limit is invariant under $k\mapsto L k$, it follows that $$ E(k)K(\sqrt{1-k^2}) + E(\sqrt{1-k^2})K(k) - K(\sqrt{1-k^2})K(k) = \frac{\pi}{2} $$ for all $k$. Other folk derive that lovely equation by differentiating with respect to $k$... other smart folk. I don't think I could have made that approach look any prettier than this one here.

[jessecmckeown]

Landen Pt 4. More Integrands!

It's easier with a well-understood Computer Algebra Assistant, but, well... But that's very Woostery of me, starting in the wrong place. One develops a conviction that the simple maps representing double-covers of elliptic curves are "the right" way to think about Landen Transforms, and then you get lost looking for "the right" way to understand the next trick in these terms. It's easy to get lost. If we choose to norrmalize our Elliptic Curves in terms of the modulus $k$ as defined by $y^2 = (k-x^2)(1-kx^2)$, then writing in terms of $x$ there are roughly two globally-available double-covering maps... one could say there's only one, up to twists and re-orientations, but ... I digress. There's the Möbius-translated squaring map $$ x\mapsto \frac{x^2 - \lambda}{1-\lambda x^2} \tag{Sq1}$$ and there is the conjugate of this map by the transposition $u\mapsto \frac{1-u}{1+u}$, which one calculates is $$ x\mapsto \rho \frac{2x}{1+x^2} \tag{Sq2} $$ The particular relations between $\rho,\lambda,k$ will depend on whether one is mapping from or to the curve with modulus $k$. In one sense, it does not matter whether we now use $\mathrm{Sq1}$ or $\mathrm{Sq2}$; however, the algebra is much simpler using $\mathrm{Sq2}$, or specifically, substitutions $$ x = \frac{1}{\sqrt{k}} \frac{2u}{1+u^2} $$ which is a double cover mapping to a curve with modulus $k$, from one with modulus... something that you can work out if you really must. Well, we already know what happens to our favourite measure, $$ \frac{dx}{\sqrt{(k-x^2)(1-kx^2)}} $$ when pulling-back along this substitution, $$ \frac{dx}{\sqrt{(k-x^2)(1-kx^2)}} = \frac{du}{\sqrt{k^2u^2-(4-2k^2)u^2+k^2}} $$ (if you ask maxima to do this substitution, it'll ALMOST tell you that, except that it'll mention an extra pair of branch points at $u=\pm 1$ ... which I believe I've mentioned before...) So it's maybe a Good Idea to see what happens under this substitution to other elliptic integrands. How about this one: $$ \sqrt{\frac{1-kx^2}{k-x^2}} dx = \frac{1-kx^2}{\sqrt{(k-x^2)(1-kx^2)}} dx ?$$ They call that one "second kind". And the Answer, up to some Branching Considerations: $$ \frac{2(1-u^2)^2}{(1+u^2)^2\sqrt{(k^2u^2-(4-2k^2)u^2+k^2)}} $$ Hm. That may not look like Progress, but in fact It Is! The first opacity to overcome here is that the fraction we have acquired, $$\frac{2(1-u^2)^2}{(1+u^2)^2} $$ ... it really wants to be a Polynomial... let's do Partial Fraction decomposition. $$ \frac{2(1-u^2)^2}{(1+u^2)^2} = \frac{8}{(1+u^2)^2} - \frac{8}{1-u^2} + 2 $$ The second is: the conceit of this game is that we're now allowed to Integrate $\frac{1}{\sqrt{P(x)}}$. We also know how to differentiate. A perfectly good thing to Differentiate is $$ \sqrt{P(x)} $$! You can probably tell that this doesn't immediately help, so instead, differentiate products $$ \frac{x^m}{Q(x)} \sqrt{P(x)} $$ and in particular you find $$ \frac{d}{du} \frac{u\sqrt{k^2u^4-(4-2k^2)u^2+k^2}}{1+u^2} = \left(\frac{8}{(1+u^2)^2} -\frac{8}{1+u^2} +k^2 u^2+k^2\right)\frac{1}{\sqrt{k^2u^4-(4-2k^2)u^2+k^2}} $$ and this is great! The derivative and the transformed Second Kind Integrand have exactly proportional singularities, at least at $\pm i$, where they're new. Or, to put it differently, Landen-transforming the Second Kind Integrand gives a sum of another Second-Kind integrand, a First-kind integrand, and a derivative (whose integral we know trivially).

[jessecmckeown]

Landen Pt 3: double covers and duality ... and ...

... this one feels a little more I-don't-know-where-to-stop-ish... I hope you don't mind.

Any way, $\RR$eal elliptic curves (i.e. with $\RR$eal coefficients) have $\RR$eal duals, and while there isn't a "double one direction" map of one elliptic curve to itself, there's an elliptic curve that does receive such a map. Let's invent some confusing notation, and call the primary form of any given $\RR$eal Elliptic curve $E_{a,b}$ where $a$ and $b$ are suggestive names. The dual of $E_{a,b}$ will be $E_{b,a}$ and there's a double-covering map in one direction $E_{a,b} \to E_{a,2b} \sim E_{a/2,b}$ and a conjugate double-covering map in the other direction $E_{a,b} \to E_{2a,b} \sim E_{a,b/2}$. Let's say that $E_{a,b} \to E_{a,2b}$ is our "Squaring" map. Then the usual way to describe the Landen transforms is the other one; and so we want to describe the main parameter, call it $l$, of $E_{2a,b}$. But we know that the main parameters $l$ of $E_{2a,b}$, and $k$ of $E_{2a,2b}$ are related by $$ l^2 = \frac{4k}{(1+k)^2} $$ or equivalently $$ k = \frac{1-\sqrt{1-l^2}}{1+\sqrt{1-l^2}} = \frac{l^2}{(1+\sqrt{1-l^2})^2} $$ AND! $E_{a,b}$ and $E_{2a,2b}$ are isomorphic! So $$ l = \frac{2\sqrt{k}}{1+k} $$ is the relation we want anyways! The geometrically-sensitive among us will have noticed some Suspicious Fractions turning up ... well, all over the place. It really shouldn't be a surprise... a peripherally-centered inversion maps a circle onto a line... Well, anyways. With all the transformations to choose from and compose, it's easy to get lost. One would be excused for not immediately asking how to compare $E_{2a,b}$ and $E_{2b,a}$; we've just calculated the main parameter --- call it $k_{2a,b}$ --- for $E_{2a,b}$; From Fiddly Constants, we can also calulate $$ k_{2b,a} = \frac{2 \frac{1-\sqrt{k}}{1+\sqrt k} }{1 + \left(\frac{1-\sqrt{k}}{1+\sqrt{k}}\right)^2} = \left( \frac{1 - \frac{k}{1+\sqrt{1-k^2}}}{1 + \frac{k}{1+\sqrt{1-k^2}}}\right)^2 = \frac{1-k}{1+k}$$ from which the geometrically-sensitive among us will immediately remember $$ k_{2b,a}^2 + k_{2a,b}^2 = 1 $$ In the Elliptic Integration Liturature, this relationship is called complementarity, and would sometimes written $k_{2a,b} = k'_{2b,a}$, except that nobody else puts those subscripts there. Since we can use whatever values of $a$ we like, we can in fact write the same result a suggestive bunch of times: $$ k_{b,2^{n-1} a}^2 + k_{2^{n+1}a,b}^2 = 1 .$$

[jessecmckeown]

Landen Transforms Pt 2½: fiddly constants

It ends up being quite irksome to solve for $l$ in $k = \frac{2\sqrt{l}}{1+l}$ and then work out the full substitution for comparing the measures $$ \frac{du}{\sqrt{u(1-u)(1-k^2u)}} \quad\mathrm{vs}\quad \frac{dt}{\sqrt{(1-t^2)(1-l^2t^2)}} ; $$ so instead it's a good exercise to work the other way around, substituting $ t = \frac{u+1}{lu+1}\frac{l+1}{2} $, with which one calculates $$ \frac{dt}{\sqrt{(1-t^2)(1-l^2t^t)}} = \frac{du}{(1+l)\sqrt{u(1-u)(1-\frac{4l}{(1+l)^2} u)}} $$ In other words, $$ \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} \\\\ {} = \frac{du}{2\sqrt{u(1-u)(1-k^2u)}} = \frac{(1+l)}{2} \frac{dt}{\sqrt{(1-t^2)(1-l^2t^2)}} $$

Another useful fact about elliptic curves coming from Real Equations is that they have dual elliptic curves, with reciprocal aspect ratios, as it were. This means there's another tricky Möbius transform to pin down --- it's much easier if instead of taking the form $(1-t^2)(1-k^2t^2)$ as normal, we work with $(k-t^2)(1-kt^2)$, which has the same cross-ratio; then all cases of transposition are covered by the Möbius transposition $t\mapsto \frac{1-t}{1+t}$. One pushes through the appropriate calculus and gets an equivalence of measures $$ \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}} = \frac{2}{(1+\sqrt{k})^2} \frac{du}{\sqrt{-(1-u^2)\left(1-\left(\frac{1-\sqrt {k}}{1+\sqrt {k}}\right)^4u^2\right)}}$$

Radicand Main Parameter Scaling factor Normalized $(1-t^2)(1-k^2 t^2)$ $k$ $1$ Transposed $(1-t^2)(1- \left(\frac{1-\sqrt{k}}{1+\sqrt{k}}\right)^4 t^2)$ $\left(\frac{1-\sqrt{k}}{1+\sqrt{k}}\right)^2$ $\frac{-2i}{(1+\sqrt{k})^2}$ Rotated $t(1-t)(1- \frac{4k}{(1+k)^2} t^2)$ $k$ $\frac{1}{1+k}$ Squared a double-cover $t(1-t)(1-k^2 t)$ $ \frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}=\frac{k^2}{(1+\sqrt{1-k^2})^2}$ $\frac{1}{2}$

[jessecmckeown]

Landen Transforms Pt 2: remembering facts about Elliptic Curves

Since elliptic integrals are essentially integrals against the Haar Measure $dx/y$ on an elliptic curve $y^2 = P(x)$ (with the usual remarks about the degree and discriminant of $P$) --- or, in $\mathbb{C}$omplexland, integrals pulled-back or pushed-forward along the Exponential Map $\mathbb{C} \sim TE \to E$, depending on how you think of it, it is helpful to remember that elliptic curves, over $\mathbb{C}$, are, topologically and group-strucutrally, tori. So, it may be helpful to think of maps of tori when thinking about Elliptic Curves and Elliptic Integrals too. Of course there are the various projections, $\mathbb{T}^2\to \sph^1$, but of particular interest (and more accessible, in $\mathbb{C}$) are the finite covering maps. The particular elliptic curves with which Landen transformations are first used are represented by the equations $y^2 = (1-x^2)(1-k^2x^2)$, usually with $0 \lt k \lt 1$. For our purposes it will be useful to model our Elliptic curve as a branched cover of two sheets over $\mathbb{C}$, with the particular branching structure of the line segment between $-1$ and $1$, and the real half-lines above $1/k$ and below $-1/k$ --- construed as one branch cut passing through the point at infinity. The reason this choice of branch cuts is useful, even though it runs through the usual domain of interest in the integral $\int dx/\sqrt{P(x)}$, is that composing with The squaring map, $x\mapsto x^2$, gives a four-sheeted branched cover of $\mathbb{C}$ with just three branch-points at $0,1,1/k^2$, and in particular the squaring map folds our branch cuts onto reasonable branch cuts. If we initially branch-cut where $P(x) \lt 0$, we have to introduce a new cut to describe the branching structure, and in both sheets, too. It might be easier to see what's up by trying the substitution $u = x^2$: $ du = 2 x dx $ or $ dx = du / 2\sqrt{u} $, so $$ \frac{dx}{\sqrt{(1-x^2)(1-k^2 x^2)}} = \frac{du}{2\sqrt{u (1-u) (1-k^2 u)}} $$ To put it in words again, "square $x$" gives a double covering from an elliptic curve with cross-ratio $\mu_{-1/k,-1,1}(\frac{1}{k}) = \frac{4k}{(1-k)^2}$ to an elliptic curve with cross-ratio $$\mu_{\infty,0,1}(\frac{1}{k^2}) = \frac{1/k^2-\infty}{1/k^2-1}\frac{0-1}{0-\infty} = \frac{k^2}{1-k^2} $$ Now a first proper hint of the Landen Transforms: The cross-ratio of $(-1/l,-1,1,1/l)$ is $\frac{4l}{(1-l)^2}$, so that restoring the cubic curve $y^2 = u(1-u)(1-k^2 u)$ by a Möbius transform to the form $y^2 = (1-x^2)(1-l^2 u^2)$ with $0 \lt l \lt 1$, we want $$ \frac{4l}{(1-l)^2} = \frac{k^2}{1-k^2} $$ or $$ k = \frac{2\sqrt{l}}{1+l}$$

[jessecmckeown]

Landen transforms, Pt1 : Möbius transforms and Cross-Ratios

Given two ordered triples of distinct complex numbers, there is a unique Möbius transform mapping one to the other in order; it's particularly easy to verify this for mapping a random triple $(x_0,x_1,x_\infty)$ to $(0,1,\infty)$, because that immediately tells you the transform is $$ \mu_x(z) = \frac{z-x_0}{z-x_\infty} \frac{x_1-x_\infty}{x_1-x_0} .$$ It then follows by a simple observation that two ordered quartuples $(x_0,x_\infty,x_1,z)$ and $(y_0,y_\infty,y_1,w)$ are connected, in order, by a Möbius transform iff $\mu_x(z) = \mu_y(w)$, because the transformation has to be $\mu_y^{-1}\circ\mu_x$ for the first three constraints to work. This quantity $\mu_x(z)$ is known as "the" cross-ratio of the quartuple $(x_*,z)$; the action of $3!$ on any list of three things induces, by permutations of $(0,\infty,1)$, a representation of $3!$ in the Möbius group, with the transposition $(0\infty)$ represented by reciprocal $z\mapsto \frac{1}{z}$ and the transposition $(01)$ represented by unit complementation $z\mapsto 1-z$. Of course, a cross ratio $\mu_x(z)$ relates four things, and so it is useful to note that cyclically permuted $(z,0,1,\infty)$ has cross ratio $$ \frac{\infty-z}{\infty-1}\frac{0-1}{0-z} = \frac{1}{z} $$ so that the action of $4!$ on $(0,\infty,1,z)$ descends through a classic map $4! \to 3!$. In the light of the Möbius-stability of the subquintic elliptic integrals, Cross-ratios are of interest to the Elliptic Integrator on account of the way nonsingular quartic polynomials tend to have four roots --- as we have seen, we can pretend that a cubic polynomial has an "extra" root at infinity --- at any rate, has singularity one degree lower at infinity than a quartic would, which is nearly the same thing. When discussing Landen transforms, one seems to particularly like the measures $$ \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} $$ and correspondingly the quartics with symmetrically-arranged roots $$ \{ \pm 1, \pm \frac{1}{k} \} $$ It is good to check therefore that, taken for instance in the order $$ ( -1/k , -1 , 1, 1/k ) $$ such a quartuple has cross-ratio $$ \left(\frac{1/k+1/k}{1/k-1}\right)\left(\frac{-1-1}{-1+1/k}\right)=\frac{4k}{(k-1)^2} $$ so that there are about two mutually-reciprocal ways to arrange the roots of any such polynomial in any given order. ($k$ may be imaginary, if we take our roots in the Wrong Order) Our earlier calculation also gives the cycled tuple $(-1,1,1/k,-1/k)$ the reciprocal cross-ratio, $ \frac{(k-1)^2}{4k} $. And we'll sometimes refer to the cross-ratio of an ordered 4-tuple as the cross ratio "of" the related elliptic curve, although this will be slightly ambiguous --- after all, there are six cross-ratios of four points, among their 24 possible orderings --- but since they are related by the Möbius representation of $S_3$, any one determines all the others.

[jessecmckeown]

Landen Transform Pt 0: Why those integrals

"Elliptic integral of the first kind"... why is it the "first" kind? It's not as though there's a well-ordering of "kinds". But it is, for a couple of reasons, a very natural family of integrals. Reason one: let $P$ be a nonsingular polynomial of degree three or four; the measure $$ \frac{dx}{\sqrt{P(x)}} $$ under a Möbius change-of-variables $x = \frac{A u + B}{a u + b}$ pulls-back as $$ \frac{du}{\sqrt{Q(u)}} $$ where again $Q$ is a nonsingular polynomial of degree three or four. And indeed the degees of $P$ and $Q$ can differ, as happens when $P$ is quartic and $A/a$ is a root of $P$. The second reason for considering such a measure as $\frac{dx}{\sqrt{P(x)}}$ is plainest when $P$ is a nonsingular cubic; because in this setting, the relation $y^2 = P(x)$ describes an Elliptic curve in a simple way--- others can rehearse the group theory side of these things, but on a nonsingular plane cubic, colinearity defines the ternary relation $f + g + h = 0$ --- one just has to pick a $0$ to pin down the group structure; and our favourite choice of $0$ is "the point at infinity", at least when it happens to be an inflection point. And the particular plane cubics $y^2 = P(x)$ are symmetrical, so that group-structure-opposites correspond to reversing $y$; and since "small group element" means "near the point at infinity", one calculates that $(x,y)+_E\varepsilon = (x+y\varepsilon,y + y \varepsilon \frac{dy}{dx})$, or in other words that $dx/y$ is a Haar Measure on the Elliptic Curve.

The related plane quartics are also decent elliptic curves, but it does seem a bit trickier to motivate their group structure from the get-go.

[jessecmckeown]

It's potentially even worse than this, however! The question basically asks us to observe that $$ x^6 - 4 x^5 + 5 x^4 - 4x^2 + 4 - \frac{4}{1+x^2} = \frac{x^4(1-x)^4}{1+x^2}$$ with the point being that the polynomial bit is easy to integrate; but reading it the other way around, it's also telling us that $$ \frac{1}{1+x^2} = 1 - x^2 + \frac{5}{4} x^4 + x^5 - \frac{1}{4}x^6 - \frac{1}{4} \frac{x^4(1-x)^4}{1+x^2}$$ Or in other words, that the error of estimating $$ \frac{x^4(1-x)^4}{1+x^2} \sim x^4(1-x)^4\left(1-x^2+\frac{5}{4}x^4+x^5-\frac{1}{4}x^6\right) $$ is $$ \frac{1}{4} \frac{x^8 (1-x)^8}{1+x^2} $$ of which the integral (between zero and one) is obviously less than $B(9,9)/4 = \frac{1}{875160}$ (oh joy! recursion!) So the Really Sneaky Thing To Do is: $$ \frac{22}{7} - \pi = B(5,5) - B(5,7) + \frac{5}{4}B(5,9) + B(5,10) - \frac{1}{4} B(5,11) + \frac{\varepsilon}{875160} $$ which eventually tells us $$ 3.1415917\lt \pi \lt 3.1415928 $$

Improving on a self-spoiling problem

Some while ago, so the internet tells us, some editor decided that the definite integral $$ \int_0^1 \frac{x^4(1-x)^4}{1+x^2} dx = \frac{22}{7}-\pi $$ would make a cute exam problem. The internet meme picture suggests further that “conclude that $\pi \lt 22/7$” was a good note to add. What if we could do better? The numerator in that integrand fraction, $$ x^4(1-x)^4 $$ happens to be the integrand of what has become known as the Beta function (capital Beta, and so written $B$), $B(5,5) = \frac{4! 4!}{9!}$. This integrand is very small, and is concentrated around its critical point $x=½$, where $\frac{1}{1+x^2}$ can be neatly approximated $\frac{1}{1+x^2} = \frac{28}{25} - \frac{16x}{25} + O(x-½)^3$. Go ahead and estimate the error, if you like. This means we can approximate the actual integral $$ \int_0^1 \frac{x^4(1-x)^4}{1+x^2} dx \sim \frac{28}{25} B(5,5) - \frac{16}{25} B(5,6) \sim 0.001269 $$ so that in fact the exam hint gives us $$ \pi \sim 3.142857 - 0.001269 = 3.141587 $$