If you do not subscribe to the RSS feed for this blog, then you have by now stopped checking it for updates. I intend to post occasionally, but school is absorbing my mathematical thoughts, rather than writing.
Since I have no extra time, I have started a new blog, sister to this one. Whereas Orange Juice Files is primarily for mathematical (and nonmathematical) essays, Local Seasoning will be a repository of recipes and meals from my kitchen. Enjoy.
12 February 2008
08 January 2008
A composition law for discrete-time QM
One of the best ways to understand quantum mechanics — bear with me — is as a one-dimensional quantum field theory. No, it's not backwards, and we really should think of QM as inherently one-dimensional: there's one dimension of time. The configuration space of a particle is finite-dimensional; the size of the space of paths the particle could take — and Feynman says that a particle takes every possible path — is largely determined by number of "time" dimensions in the problem, since a path has a point in configuration space for every moment in time.
In any case, I'm going to think of it that way. And then I'm going to think about path integration: the transition amplitude between two configurations is some poorly-defined integral over the infinite-dimensional space of paths connecting those configurations. Rather than trying to compute in infinite dimensions, physicists since Feynman have formally expanded the integral asymptotically, and interpreted the coefficients combinatorially as diagrams. (And interpreted the diagrams as describing actual events, which is a matter for them physicists rather than for us mathematicians to discuss.)
These formal power series should — indeed, must, if the formalism is to make sense — satisfy a particular gluing axiom. But it seems that no one has gone to the trouble to verify that in fact they do; there is no proof available in the literature. In his fall-semester QFT class, Nicolai Reshetikhin suggested that someone try to fix this, and I volunteered. Silly me.
Long story short, after making progress on a related issue, and then avoiding the project for more than a month, I've worked the 0-dimensional analogue, where we approximate the time interval by a sequence of (equally-spaced) discrete points. If you would like to read about it, the paper is available here.
Update: The second, and likely final, installment is now available. It concludes with the gluing rules for arbitrary perturbative quantum field theories.
In any case, I'm going to think of it that way. And then I'm going to think about path integration: the transition amplitude between two configurations is some poorly-defined integral over the infinite-dimensional space of paths connecting those configurations. Rather than trying to compute in infinite dimensions, physicists since Feynman have formally expanded the integral asymptotically, and interpreted the coefficients combinatorially as diagrams. (And interpreted the diagrams as describing actual events, which is a matter for them physicists rather than for us mathematicians to discuss.)
These formal power series should — indeed, must, if the formalism is to make sense — satisfy a particular gluing axiom. But it seems that no one has gone to the trouble to verify that in fact they do; there is no proof available in the literature. In his fall-semester QFT class, Nicolai Reshetikhin suggested that someone try to fix this, and I volunteered. Silly me.
Long story short, after making progress on a related issue, and then avoiding the project for more than a month, I've worked the 0-dimensional analogue, where we approximate the time interval by a sequence of (equally-spaced) discrete points. If you would like to read about it, the paper is available here.
Update: The second, and likely final, installment is now available. It concludes with the gluing rules for arbitrary perturbative quantum field theories.
27 December 2007
On Presidential Primaries
One of the crimes of our current electoral system — the United States picks its President in about as undemocratic a method imaginable — is that the same four million people pick the candidates for both major parties every year.
Wikipedia lists the fifty states' and six non-state U.S. territories' populations, based on official estimates by the U.S. Census Bureau. In light of next month's high-profile primary elections, the numbers are remarkable.
Elections are very expensive: candidates for primaries must go door-to-door, and raise money for media buys. It would be highly undemocratic for California to be the first primary. Eventually, the nominee must win California, but s/he should not have to compete there until most of the candidates have been culled from the game, and fundraising is funneled to only a few folks who've had lots of free media. Indeed, if California was the first primary, you can be sure that only the candidates who come into the game with big names and lots of personal money would be competitive. (Perhaps, of course, this is better? I wouldn't mind a system that rewarded career politicians who worked their way to the top after long terms in the Senate, Cabinet, and state Governor mansions.) So we must, if we are to have some semblance of a meritocratic democracy, begin the nominating contest in small states.
With that said, the current choices of Iowa, New Hampshire, and South Carolina cannot be justified.
Iowa, with just shy of three million people, is not large. It has five members of the House of Representatives, and its population is largely white, pro-farm, and anti-immigrant. New Hampshire is legitimately small: it has only 1.3 million people and two representative in Congress. South Carolina, on the other hand, houses more than four million people, and is the only early primary with a sizable non-white population.
But Iowa is larger than twenty other states (and the District of Columbia), any of whom would make a reasonable first-in-the-nation primary. From smallest to largest, they are:
Wyoming, DC, Vermont, North Dakota, Alaska, South Dakota, Delaware, Montana, Rhode Island, Hawaii, New Hampshire, Maine, Idaho, Nebraska, West Virginia, New Mexico, Nevada, Utah, Kansas, Arkansas, and Mississippi.
These twenty one voting districts represent the full range of American demographics. DC is urban, Black, and overwhelmingly Democratic. Hawaii is majority Asian; New Mexico has a large Mexican immigrant population. Many of the small states are rural and conservative, while others, like Rhode Island and Delaware, easily represent the metropolitan North East. Vermont has a bizarre politics all to its own.
The 2008 election cycle will begin with the same four million people that have started the game for the last twenty years. Four million people whom the candidates, when they are in their most pandering mode, call "uniquely qualified" to pick a candidate. Are the rest of us stupid and uneducated? Are North Dakotans or Hawaiians somehow less able to consider the candidates' experiences and abilities? A democratic system should not enfranchise only those who are educated, or moneyed, or otherwise "qualified" to vote.
In a better system, a bipartisan, independent commission, taking input from major party leaders, would select the Primary order each cycle. Their selections should, by law or policy, begin always with a couple small states — say, two states from the twenty smallest. Criteria should include some complementarity: one liberal, urban population, with one rural conservative one, say. And, most importantly, their selections should be different from year to year. In one cycle, Vermont and Idaho start off; in another, Utah and Delaware.
There have been many complaints this year about the election starting too early. It has been an expensive year (see, for instance, the listings at opensecrets.org): Clinton and Obama have each raised close to one hundred million dollars and spend forty million of it; Romney has raised and spent around sixty million. The general election will probably cost each party around a billion. And it's exhausting, and a distraction from the important work of running the country. Oregon's Governor, for instance, is out in Iowa stumping for Clinton, rather than doing his gubernatorial work. Will the 2012 election begin similarly early? It wouldn't if the Primary committee waited to announce the order, and perhaps was empowered also to announce the dates.
A system like this — or, indeed, any proposal to reform the primaries — would most likely require, in order to go into effect, the commitment of both parties as well as Congressional involvement. That's a tough order, but a necessary one if we are going to have a truly democratic democracy.
Wikipedia lists the fifty states' and six non-state U.S. territories' populations, based on official estimates by the U.S. Census Bureau. In light of next month's high-profile primary elections, the numbers are remarkable.
Elections are very expensive: candidates for primaries must go door-to-door, and raise money for media buys. It would be highly undemocratic for California to be the first primary. Eventually, the nominee must win California, but s/he should not have to compete there until most of the candidates have been culled from the game, and fundraising is funneled to only a few folks who've had lots of free media. Indeed, if California was the first primary, you can be sure that only the candidates who come into the game with big names and lots of personal money would be competitive. (Perhaps, of course, this is better? I wouldn't mind a system that rewarded career politicians who worked their way to the top after long terms in the Senate, Cabinet, and state Governor mansions.) So we must, if we are to have some semblance of a meritocratic democracy, begin the nominating contest in small states.
With that said, the current choices of Iowa, New Hampshire, and South Carolina cannot be justified.
Iowa, with just shy of three million people, is not large. It has five members of the House of Representatives, and its population is largely white, pro-farm, and anti-immigrant. New Hampshire is legitimately small: it has only 1.3 million people and two representative in Congress. South Carolina, on the other hand, houses more than four million people, and is the only early primary with a sizable non-white population.
But Iowa is larger than twenty other states (and the District of Columbia), any of whom would make a reasonable first-in-the-nation primary. From smallest to largest, they are:
Wyoming, DC, Vermont, North Dakota, Alaska, South Dakota, Delaware, Montana, Rhode Island, Hawaii, New Hampshire, Maine, Idaho, Nebraska, West Virginia, New Mexico, Nevada, Utah, Kansas, Arkansas, and Mississippi.
These twenty one voting districts represent the full range of American demographics. DC is urban, Black, and overwhelmingly Democratic. Hawaii is majority Asian; New Mexico has a large Mexican immigrant population. Many of the small states are rural and conservative, while others, like Rhode Island and Delaware, easily represent the metropolitan North East. Vermont has a bizarre politics all to its own.
The 2008 election cycle will begin with the same four million people that have started the game for the last twenty years. Four million people whom the candidates, when they are in their most pandering mode, call "uniquely qualified" to pick a candidate. Are the rest of us stupid and uneducated? Are North Dakotans or Hawaiians somehow less able to consider the candidates' experiences and abilities? A democratic system should not enfranchise only those who are educated, or moneyed, or otherwise "qualified" to vote.
In a better system, a bipartisan, independent commission, taking input from major party leaders, would select the Primary order each cycle. Their selections should, by law or policy, begin always with a couple small states — say, two states from the twenty smallest. Criteria should include some complementarity: one liberal, urban population, with one rural conservative one, say. And, most importantly, their selections should be different from year to year. In one cycle, Vermont and Idaho start off; in another, Utah and Delaware.
There have been many complaints this year about the election starting too early. It has been an expensive year (see, for instance, the listings at opensecrets.org): Clinton and Obama have each raised close to one hundred million dollars and spend forty million of it; Romney has raised and spent around sixty million. The general election will probably cost each party around a billion. And it's exhausting, and a distraction from the important work of running the country. Oregon's Governor, for instance, is out in Iowa stumping for Clinton, rather than doing his gubernatorial work. Will the 2012 election begin similarly early? It wouldn't if the Primary committee waited to announce the order, and perhaps was empowered also to announce the dates.
A system like this — or, indeed, any proposal to reform the primaries — would most likely require, in order to go into effect, the commitment of both parties as well as Congressional involvement. That's a tough order, but a necessary one if we are going to have a truly democratic democracy.
13 December 2007
Linear Differential Equations
In the calculus class I'm TAing, we spent some time learning how "the method of undetermined coefficients" could be used to solve linear differential equations. I have never taken a first-year differential equations class, so although I'd solved many differential equations this way, I had never really though about such methods with any real theory. My goal in this entry is to describe the method and explain why it works, using more sophisticated ideas than in a first-year class, but still remaining very elementary. I had hoped to have this written and posted weeks ago; as it is, I'm writing it while proctoring the final exam for the class.
First, let me remind you of the set-up of the problem. We are trying to solve a non-homogeneous differential equation with constant coefficients:
$$ a_n y^{(n)} + \dots + a_1 y' + a_0 y = g(x) $$
I will assume that $a_n$ is not 0; then the left-hand side defines a linear operator $D$ of on the space of functions, with $n$-dimensional kernel.
We can diagonalize this operator by Fourier-transforming: if $y = e^{rx}$, then $D[y] = a(r) e^{rx}$, where $a(r)$ is the polynomial $a_n r^n + \dots + a_1 r + a_0$. If $a(r)$ has no repeated roots, then we can immediately read off a basis for the kernel as $e^{rx}$ for $r$ ranging over the $n$ roots. If there is a repeated root, then $a'(r)$ and $a(r)$ share a common factor; $a'(r)$ corresponds to the operator
$$ E[y] = a_n n y^{(n-1)} + \dots + a_1 $$
Then, since $\frac{d^k}{dx^k} [x y] = x y^{(k)} + k y^{(k-1)}$, we see that
$$ D[x y] = x D[y] + E[y] $$
so $r$ is a repeated root of $a(r)$ if and only if $e^{rx}$ and $x e^{rx}$ are zeros of $D$.
More generally, linear differential operators with constant coefficients satisfy a Leibniz rule:
$$ D[p(x) e^{rx}] = \left( p(x) a(r) + p'(x) a'(r) + \dots + p^{(n)}(x) a^{(n)}(r) \right) e^{rx} $$
for polynomial $p(x)$.
Thus, our ability to solve linear homogeneous differential equations with constant coefficients depends exactly on our ability to factor polynomials; for example, we can always solve second-order equations, by using the quadratic formula.
But, now, what if $g(x) \neq 0$? I will assume that, through whatever conniving factorization methods we use, we have found the entire kernel of $D$. Then our problem will be solved if we can find one solution to $D[y] = g(x)$; all other solutions are the space through this function parallel to the kernel. In calculus, we write this observation as "$y_g = y_c + y_p$" where g, c, and p stand for "general", "[c]homogenous", and "particular", respectively.
For a general $g$, we can continue to work in the Fourier basis: Fourier transform, sending $D[y] \mapsto a(r)$, then divide and integrate to transform back. This may miss some solutions at the poles, and is computationally difficult: integrating is as hard as factoring polynomials. For second-order equations, we can get to "just an integral" via an alternative method, by judiciously choosing how to represent functions in terms of the basis of solutions for $D[y]=0$.
But for many physical problems, $g(x)$ is an especially simple function (and, of course, it can always be Fourier-transformed into one).
In particular, let's say that $g(x)$ is, for example, a sum of products of exponential (and sinosoidal) and polynomial functions. I.e. let's say that $g(x)$ is a solution to some homogeneous linear constant-coefficient $C[g(x)] = 0$. Another way of saying this: let's say that the space spanned by all the derivatives $g$, $g'$, $g''$, etc., is finite-dimensional. If it has dimension $= m$, then $g^{(m)}(x)$ is a linear combination of lower-order derivatives, and thus I can find a minimal (order) $C$ of degree $m$ so that $C[g] = 0$. By the Leibniz rule, functions with this property form a ring. When I add two functions, the dimensions of their derivative spaces no more than add; when I multiply, the dimensions no worse than multiply. Indeed, by the earlier discussion, we have an exact description of such functions: they are precisely sums of products of $x^s e^{rx}$ ($s$ an non-negative integer).
In any case, let's say $g$ is of this form, i.e. we have $C$ (with degree $m$) minimal such that $C[g] = 0$, and first let's assume that the kernels of $C$ and $D$ do not intersect. Then $D$ acts as a one-to-one linear operator on finite-dimensional space $\ker C$, which by construction is spanned by the derivatives of $g$, and so must be onto. I.e. there is a unique point $y_p(x) = b_0 g(x) + b_1 g'(x) + \dots + b_{m-1} g^{(m-1)}(x) \in \ker C$ so that $D[y_p] = g$. Finding it requires only solving the system of linear equations in the $b_i$.
If, however, $\ker C$ and $\ker D$ intersect, then we will not generically be able to solve this system of linear equations. Because $C$ is minimal, $g$ is not a linear combination of fewer than $m$ of its derivatives; if $D$ sends (linear combinations of) some of those derivatives to 0, we will never be able to get a map onto $g$. Let me say this again. If $D$ does not act one-to-one on $\ker C$, but $g$ is in the range of this matrix, then $g$ is in a smaller-than-$m$-dimensional space closed under a differential operator; thus, there is a differential operator of lower-than-$m$ degree that annihilates $g$.
How can we then solve the equation? By the Leibniz rule, we observed earlier, $(xg)' = g + x g'$, and so
$$\frac{d^k}{dx^k}[x g(x)] = x g^{(k)}(x) + k g^{(k-1)}(x)$$
Then $C[xg]$ is a linear combination of derivatives of $g$; i.e. $C[xg] \in \ker C$. If we take the space spanned by the derivatives of $x g(x)$, it is one dimension larger than $\ker C$. We can repeat this trick ---
$$ \frac{d^k}{dx^k}[ x^p g(x) ] = \sum_{i=0}^k \frac{k!p!}{i!(k-i)!(p-k+i)!} x^{p-k+i} g^{(i)}(x) $$
--- and eventually get a space that's $n+m$ dimensional, containing, among other things, $g$, and closed under differentiation. The kernel of $D$ in this larger space is at most $n$ dimensional (since $n = \dim \ker D$ in the space of all functions), and so the range is $m$-dimensional, and must contain $g$: the system of linear equations is solvable.
Of course, we often can stop before getting all the way to $n$ extra dimensions. But so long as we are only interested in functions that are zeros of constant linear differential operators, then we can always solve differential equations. For example, every linear equation from a physics class, and almost every one in first-year calculus, is solvable with this method.
One final remark:
Composition of differential operators follows matrix multiplication, and hence yields differential operators. If $g$ satisfies $C[g] = 0$, and if we're trying to solve $D[y]=g$, then we might decide to solve more generally $CD[y] = 0$. The left-hand-side is $n+m$ dimensional, and if we're truly gifted at factoring polynomials, then we can solve it directly. Then the solutions to our original equation must be in this kernel.
First, let me remind you of the set-up of the problem. We are trying to solve a non-homogeneous differential equation with constant coefficients:
$$ a_n y^{(n)} + \dots + a_1 y' + a_0 y = g(x) $$
I will assume that $a_n$ is not 0; then the left-hand side defines a linear operator $D$ of on the space of functions, with $n$-dimensional kernel.
We can diagonalize this operator by Fourier-transforming: if $y = e^{rx}$, then $D[y] = a(r) e^{rx}$, where $a(r)$ is the polynomial $a_n r^n + \dots + a_1 r + a_0$. If $a(r)$ has no repeated roots, then we can immediately read off a basis for the kernel as $e^{rx}$ for $r$ ranging over the $n$ roots. If there is a repeated root, then $a'(r)$ and $a(r)$ share a common factor; $a'(r)$ corresponds to the operator
$$ E[y] = a_n n y^{(n-1)} + \dots + a_1 $$
Then, since $\frac{d^k}{dx^k} [x y] = x y^{(k)} + k y^{(k-1)}$, we see that
$$ D[x y] = x D[y] + E[y] $$
so $r$ is a repeated root of $a(r)$ if and only if $e^{rx}$ and $x e^{rx}$ are zeros of $D$.
More generally, linear differential operators with constant coefficients satisfy a Leibniz rule:
$$ D[p(x) e^{rx}] = \left( p(x) a(r) + p'(x) a'(r) + \dots + p^{(n)}(x) a^{(n)}(r) \right) e^{rx} $$
for polynomial $p(x)$.
Thus, our ability to solve linear homogeneous differential equations with constant coefficients depends exactly on our ability to factor polynomials; for example, we can always solve second-order equations, by using the quadratic formula.
But, now, what if $g(x) \neq 0$? I will assume that, through whatever conniving factorization methods we use, we have found the entire kernel of $D$. Then our problem will be solved if we can find one solution to $D[y] = g(x)$; all other solutions are the space through this function parallel to the kernel. In calculus, we write this observation as "$y_g = y_c + y_p$" where g, c, and p stand for "general", "[c]homogenous", and "particular", respectively.
For a general $g$, we can continue to work in the Fourier basis: Fourier transform, sending $D[y] \mapsto a(r)$, then divide and integrate to transform back. This may miss some solutions at the poles, and is computationally difficult: integrating is as hard as factoring polynomials. For second-order equations, we can get to "just an integral" via an alternative method, by judiciously choosing how to represent functions in terms of the basis of solutions for $D[y]=0$.
But for many physical problems, $g(x)$ is an especially simple function (and, of course, it can always be Fourier-transformed into one).
In particular, let's say that $g(x)$ is, for example, a sum of products of exponential (and sinosoidal) and polynomial functions. I.e. let's say that $g(x)$ is a solution to some homogeneous linear constant-coefficient $C[g(x)] = 0$. Another way of saying this: let's say that the space spanned by all the derivatives $g$, $g'$, $g''$, etc., is finite-dimensional. If it has dimension $= m$, then $g^{(m)}(x)$ is a linear combination of lower-order derivatives, and thus I can find a minimal (order) $C$ of degree $m$ so that $C[g] = 0$. By the Leibniz rule, functions with this property form a ring. When I add two functions, the dimensions of their derivative spaces no more than add; when I multiply, the dimensions no worse than multiply. Indeed, by the earlier discussion, we have an exact description of such functions: they are precisely sums of products of $x^s e^{rx}$ ($s$ an non-negative integer).
In any case, let's say $g$ is of this form, i.e. we have $C$ (with degree $m$) minimal such that $C[g] = 0$, and first let's assume that the kernels of $C$ and $D$ do not intersect. Then $D$ acts as a one-to-one linear operator on finite-dimensional space $\ker C$, which by construction is spanned by the derivatives of $g$, and so must be onto. I.e. there is a unique point $y_p(x) = b_0 g(x) + b_1 g'(x) + \dots + b_{m-1} g^{(m-1)}(x) \in \ker C$ so that $D[y_p] = g$. Finding it requires only solving the system of linear equations in the $b_i$.
If, however, $\ker C$ and $\ker D$ intersect, then we will not generically be able to solve this system of linear equations. Because $C$ is minimal, $g$ is not a linear combination of fewer than $m$ of its derivatives; if $D$ sends (linear combinations of) some of those derivatives to 0, we will never be able to get a map onto $g$. Let me say this again. If $D$ does not act one-to-one on $\ker C$, but $g$ is in the range of this matrix, then $g$ is in a smaller-than-$m$-dimensional space closed under a differential operator; thus, there is a differential operator of lower-than-$m$ degree that annihilates $g$.
How can we then solve the equation? By the Leibniz rule, we observed earlier, $(xg)' = g + x g'$, and so
$$\frac{d^k}{dx^k}[x g(x)] = x g^{(k)}(x) + k g^{(k-1)}(x)$$
Then $C[xg]$ is a linear combination of derivatives of $g$; i.e. $C[xg] \in \ker C$. If we take the space spanned by the derivatives of $x g(x)$, it is one dimension larger than $\ker C$. We can repeat this trick ---
$$ \frac{d^k}{dx^k}[ x^p g(x) ] = \sum_{i=0}^k \frac{k!p!}{i!(k-i)!(p-k+i)!} x^{p-k+i} g^{(i)}(x) $$
--- and eventually get a space that's $n+m$ dimensional, containing, among other things, $g$, and closed under differentiation. The kernel of $D$ in this larger space is at most $n$ dimensional (since $n = \dim \ker D$ in the space of all functions), and so the range is $m$-dimensional, and must contain $g$: the system of linear equations is solvable.
Of course, we often can stop before getting all the way to $n$ extra dimensions. But so long as we are only interested in functions that are zeros of constant linear differential operators, then we can always solve differential equations. For example, every linear equation from a physics class, and almost every one in first-year calculus, is solvable with this method.
One final remark:
Composition of differential operators follows matrix multiplication, and hence yields differential operators. If $g$ satisfies $C[g] = 0$, and if we're trying to solve $D[y]=g$, then we might decide to solve more generally $CD[y] = 0$. The left-hand-side is $n+m$ dimensional, and if we're truly gifted at factoring polynomials, then we can solve it directly. Then the solutions to our original equation must be in this kernel.
14 October 2007
Divergent Series take 1
The following talk is significantly too long. Some parenthetical remarks are easy enough to excise, but what else should I drop?
The talk is available here (pdf). I will give it on Thursday at "Many Cheerful Facts", a brown-bag student-organized talk series in which different graduate students present general-audience material: if you know the subject already, you won't learn anything in the talk. Someone bakes something tasty each week.
Abstract: Whereas modern physicists write down divergent series all the time, mathematicians through the ages have been variously terrified or only mildly scared of such sums. In this talk, I will survey the most important methods of summing divergent series, and make general vague remarks about them. I will quote many results, but will studiously avoid proving anything.
The talk is available here (pdf). I will give it on Thursday at "Many Cheerful Facts", a brown-bag student-organized talk series in which different graduate students present general-audience material: if you know the subject already, you won't learn anything in the talk. Someone bakes something tasty each week.
Abstract: Whereas modern physicists write down divergent series all the time, mathematicians through the ages have been variously terrified or only mildly scared of such sums. In this talk, I will survey the most important methods of summing divergent series, and make general vague remarks about them. I will quote many results, but will studiously avoid proving anything.
30 September 2007
Whole grains, it bears repeating, are tasty and nutritious. They cook easily, but many take a fair amount of time. Whole grains are processed and sold dried: before eating, they must be boiled in (potentially salted or flavored) water. Most grains should be combined with a prescribed amount of water in a pot with a well-fitting lid, brought to a boil, and simmered covered for a prescribed amount of time. Length of time is determined by the grain; amount of water should be just enough to have almost entirely evaporated off / been soaked up in that amount of time. If your pot does not have a well-fitting lid, you'll need to use more water. Some cooks prefer to soak their grains overnight, as this reduces cooking time. If you use too much water, boil uncovered for the last few minutes to evaporate off the excess. Do not stir your grains unless you want to develop the starches into a mushy mix. Grains hold their heat covered exceedingly well. To make a better seel,
What follows is a first-approximation of how much water and for how long for different grains. For details on grains' nutrition and substitutions, I refer you to The Cook's Thesaurus. For continual updates, check here.
What follows is a first-approximation of how much water and for how long for different grains. For details on grains' nutrition and substitutions, I refer you to The Cook's Thesaurus. For continual updates, check here.
| Grain Type | Amount of Water per cup grain | Cooking Time | Notes |
| Corn | Enough | 10 minutes | Corn can be steamed or boiled (or grilled or microwaved). |
| Oats, rolled | 1, and add more if starts to burn | 5 minutes, stirring (uncovered), or until desired consistency | A traditional breakfast cereal, cooked as a mush. I suggest cooking with raisins, a stick of cinnamon, and some maple syrup. Rolled oats have been steamed once, so cooks fast. |
| Quinoa | 1.5 | 10 minutes | Very fast, high protein. Rinsing first will reduce the slightly bitter flavor. |
| Rice, brown | 1.5 | 20 minutes, plus 30 minutes with heat turned off | Do not remove lid during the entire process. Just turn off the heat and let the rice continue to cook in the steam in the pot. |
| Rice, white, Persian style | 2, or enough to cover by 2 inches | 10 minutes uncovered, then 45 minutes covered | Boil rice, then drain, rinse in cold water, and drain again. Melt in a large saucepan 1 Tbsp butter per cup uncooked rice, and add rice and stir once to coat well. Cover and steam over very low heat. Bottom should be crispy and golden when done. |
| Wheat, berry | 2.5 | 1 hour | Good pasta substitute, especially with tomato sauce. Given the time involved in cooking, many suggest soaking first, or slow-cooking overnight. I haven't tried these techniques. |
| Wheat, bulgur | .75 | 7 minutes, plus 15 minutes with heat turned off | Do not remove lid during the entire process. Just turn off the heat and let the wheat continue to cook in the steam in the pot. Bulgur has been steel-cut, soaked, and baked, so cooks fast. For a tasty pilaf, sauté thin-sliced onion with two-inch pieces of vermicelli, then add bulgur. |
16 September 2007
Partial Fractions
I really am taking my own classes, and thinking about my own mathematics. But so far my classes have discussed supermathematics, which is cool but to which I have nothing so far to add, and classical (Lagrangian and Hamiltonian) mechanics, which I had intended to blog about last year. Perhaps I will some day write about such stuff; for now, I'd like to tell you about another topic we've been discussing in my calculus class.
Let's say I have a (proper) fraction m/n (in lowest terms). It's not a very simple expression: n most likely has lots of factors, and it would be nice to understand how much each factor contributes to the whole. For instance:
7/15 = 2/3 - 1/5
Can we always split up a number like this? At best, we could write a proper fraction as a sum of (proper) fractions with very small denominators.
Let's start by considering the case when n has two relatively prime factors: n = rs. We want to write
m/n = A/r + B/s.
Multiple both sides by n; we see that this is equivalent to solving the following (easy) Diophantine equation:
m = As + Br
This certainly has a solution. For instance, we can use Euclid's algorithm to write
1 = Xs + Yr
and then use A = mX and B = mY. Of course, this choice of A and B will generally be much larger than hoped-for. Never fear, though: we can always shift A and B simultaneously in opposite directions multiples of r and s. Thus we can assure that
0 < A < r
in which case
0 < As = m - Br < rs = n
so
-n < m-n < Br < m < n
and thus
-s < B < s.
Thus, we can factor the denominator into prime-power parts, and use induction. Going directly: we're looking for A, B, ..., C such that
m/(rs...t) = A/r + B/s + ... + C/t
If we multiply both sides by (s...t), this is
m/r = A(s...t)/r + integer
so we're looking for an A such that
m = A(s...t) (mod r).
Since A, r, and (s...t) are pairwise relatively prime, we can definitely do this, and we can be sure that
0 < A < r.
Doing this for each term yields
m/n = A/r + B/s + ... + C/t + integer
and this integer is definitely negative and no more (in absolute value) than the number of other summands, since each fraction is between 0 and 1. Thus we can subtract 1 from some of the summands to make the integer disappear.
11/60 = 1/4 + 1/3 + 3/5 - 1 = -3/4 + 1/3 + 3/5 = 1/4 - 2/3 + 3/5 = 1/4 + 1/3 - 2/5.
This last step, where we have to subtract, reminds us that this decomposition is not unique. It's close: we have two choices for each term, but of course making some choices constrains others.
If we're working with polynomials, on the other hand, we never have to subtract. The division works exactly as with integers, but now all inequalities should be written in terms of the degrees of the polynomials. By counting total degree, we see that the left-hand side has total degree less than 0, so that "+ integer" on the right-hand side must be 0. This is a proof that the partial-fractions decomposition of polynomial fractions is unique.
Or, rather, there's one more step in the partial-fractions decomposition. What I've written so far allows us to factor the denominator into prime powers, and write one fraction for each power. But we can go one step further: if q < p^d, then we can write
q/p^d = A_1/p + A_2/p^2 + ... + A_d/p^d
with 0 ≤ A_i < p for each i. This is, of course, trivial, and is how we traditionally write integers in terms of a certain "base":
q = q_1 p + A_1
q_1 = q_2 p + A_2
...
One more point bears stating. In the case when we're working with polynomials, and when we can completely factor the denominator into linear factors, partial-fractions decomposition becomes extremely easy, because dividing by a linear factor is trivial:
m(x) = (x-a)q(x) + m(a)
I.e. the remainder mod (x-a) is just the value of the polynomial at x=a. To get the quotients, synthetic division is very fast. This makes the last step trivial, and repeatedly dividing by p allows us to divide by p^d, so really we can do the initial steps quickly as well.
Let's say I have a (proper) fraction m/n (in lowest terms). It's not a very simple expression: n most likely has lots of factors, and it would be nice to understand how much each factor contributes to the whole. For instance:
7/15 = 2/3 - 1/5
Can we always split up a number like this? At best, we could write a proper fraction as a sum of (proper) fractions with very small denominators.
Let's start by considering the case when n has two relatively prime factors: n = rs. We want to write
m/n = A/r + B/s.
Multiple both sides by n; we see that this is equivalent to solving the following (easy) Diophantine equation:
m = As + Br
This certainly has a solution. For instance, we can use Euclid's algorithm to write
1 = Xs + Yr
and then use A = mX and B = mY. Of course, this choice of A and B will generally be much larger than hoped-for. Never fear, though: we can always shift A and B simultaneously in opposite directions multiples of r and s. Thus we can assure that
0 < A < r
in which case
0 < As = m - Br < rs = n
so
-n < m-n < Br < m < n
and thus
-s < B < s.
Thus, we can factor the denominator into prime-power parts, and use induction. Going directly: we're looking for A, B, ..., C such that
m/(rs...t) = A/r + B/s + ... + C/t
If we multiply both sides by (s...t), this is
m/r = A(s...t)/r + integer
so we're looking for an A such that
m = A(s...t) (mod r).
Since A, r, and (s...t) are pairwise relatively prime, we can definitely do this, and we can be sure that
0 < A < r.
Doing this for each term yields
m/n = A/r + B/s + ... + C/t + integer
and this integer is definitely negative and no more (in absolute value) than the number of other summands, since each fraction is between 0 and 1. Thus we can subtract 1 from some of the summands to make the integer disappear.
11/60 = 1/4 + 1/3 + 3/5 - 1 = -3/4 + 1/3 + 3/5 = 1/4 - 2/3 + 3/5 = 1/4 + 1/3 - 2/5.
This last step, where we have to subtract, reminds us that this decomposition is not unique. It's close: we have two choices for each term, but of course making some choices constrains others.
If we're working with polynomials, on the other hand, we never have to subtract. The division works exactly as with integers, but now all inequalities should be written in terms of the degrees of the polynomials. By counting total degree, we see that the left-hand side has total degree less than 0, so that "+ integer" on the right-hand side must be 0. This is a proof that the partial-fractions decomposition of polynomial fractions is unique.
Or, rather, there's one more step in the partial-fractions decomposition. What I've written so far allows us to factor the denominator into prime powers, and write one fraction for each power. But we can go one step further: if q < p^d, then we can write
q/p^d = A_1/p + A_2/p^2 + ... + A_d/p^d
with 0 ≤ A_i < p for each i. This is, of course, trivial, and is how we traditionally write integers in terms of a certain "base":
q = q_1 p + A_1
q_1 = q_2 p + A_2
...
One more point bears stating. In the case when we're working with polynomials, and when we can completely factor the denominator into linear factors, partial-fractions decomposition becomes extremely easy, because dividing by a linear factor is trivial:
m(x) = (x-a)q(x) + m(a)
I.e. the remainder mod (x-a) is just the value of the polynomial at x=a. To get the quotients, synthetic division is very fast. This makes the last step trivial, and repeatedly dividing by p allows us to divide by p^d, so really we can do the initial steps quickly as well.
Subscribe to:
Posts (Atom)
