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The solutions are in! Last week’s POW-9 is a problem for which you need a bit of insight to solve at all (even with calculator or computer), and then a bit more to solve elegantly without machine aid.

Solution by Philipp Lampe, University of Bonn: The answer is n = 65536. The decimal expansion of 1/65537 is eventually periodic with (minimum) period length n. Suppose it takes the form

\displaystyle s \cdot 10^{-k} + t \cdot 10^{-(k+n)} + t \cdot 10^{-(k + 2n)} + t \cdot 10^{-(k + 3n)} + \ldots

\displaystyle = \frac{s}{10^k} + \frac{t}{10^k (10^n - 1)}

for integers 0 \leq s < 10^k and 0 \leq t < 10^n. Then 65537| 10^k (10^n - 1), and therefore 65537|10^n - 1; equivalently 10^n \equiv 1 \mod 65537. The minimum m such that 10^m \equiv 1 \mod 65537 (i.e., the order of 10 modulo 65537) must then divide n. On the other hand, from 65537u = 10^m - 1, we derive

\displaystyle \frac1{65537} = \frac{u}{10^m - 1} = \frac{u}{10^m} + \frac{u}{10^{2m}} + \ldots

Thus by minimality of n, we conclude m = n, and that the decimal expansion is periodic (not merely eventually periodic).

It is well known that p = 65537 = 2^{16} + 1 is prime (it is the largest known Fermat prime), so by Fermat’s little theorem, the order of 10 modulo p divides \phi(p) = p - 1 = 2^{16} (where \phi is the Euler phi-function), and so it must be a power of 2, at most 2^{16}. That the order is in fact 2^{16} can be checked either by brute force (by repeated squaring and reduction modulo 65537, best carried out with calculator or computer), or by the following number-theoretic argument.

Since p is prime, the multiplicative group of nonzero elements of the field \mathbb{Z}_p is cyclic (is generated by a single element, say a). Hence 10 \equiv a^i \mod p where 1 \leq i < 65536, and 10 itself is a generator (has order 65536) iff i is prime to 65536, i.e., iff i is odd. This occurs iff 10 is a non-square modulo p. Defining the Jacobi symbol by the usual formula

\displaystyle (\frac{a^i}{p}) = (-1)^i

(so that (\frac{x}{p}) = 1 iff x is a square modulo p), we have

\displaystyle (\frac{10}{p}) = (\frac{2}{p}) \cdot (\frac{5}{p})

The second factor is, by the law of quadratic reciprocity (and the fact p \equiv 1 \pmod 4),

\displaystyle (\frac{5}{p}) = (\frac{p}{5}) = (\frac{2}{5}) = -1.

Also, by the “second supplement“, we compute (\frac{2}{p}) = (-1)^{(p^2 - 1)/8} = 1. Hence (\frac{10}{p}) = -1: 10 is a non-square as desired. This completes the proof.

Remarks:The argument in the first paragraph of Philipp’s solution proves for 1/65537 a fact which holds generally for any reduced fraction 0 < q/p < 1 where p is prime to 10: that the decimal expansion is actually periodic, not just eventually periodic. Henry Wilton also took the time and trouble to argue this point. I’m not sure how widely known that is, so I was glad the point was made; of course the main thing we need from that paragraph is the general fact that the period length of q/p equals the order of 10 modulo p.

A number of solvers computed the order of 10 by brute force, which I accepted. The use of quadratic reciprocity is slick but gets into rather deeper waters. As the Wikipedia article explains, this law was known to (or at least conjectured by) Euler and Legendre, but resisted a complete proof until Gauss found one after an intensive effort; it was one of his proudest achievements, and a theorem he returned to over the decades, giving over half a dozen more proofs in his life. Now more than 200 proofs are known. One proof that I like was given by Zolotarev, which is given a nice account in an appendix to John Conway’s very entertaining book, The (Sensual) Quadratic Form. Arin Chaudhuri pointed out to me that it’s also given as an exercise in Knuth’s The Art of Computer Programming (Vol. I, p. 45, exercise 47).

The “second supplement” was, according to the Wikipedia article, first proved by Euler. Its use here can be circumvented by observing that 2 cannot have order p -1 = 65536 since already 2^{16} = -1 \mod p, so that by Lampe’s argument, 2 is a square mod p, i.e. (\frac{2}{p}) = 1.

Quadratic reciprocity is no isolated curiosity: most of the deeper, more conceptual treatments involve some nontrivial algebraic number theory, and the efforts to formulate reciprocity laws for higher powers had a lot to do with the creation of class field theory, culminating in Artin reciprocity, one of the crowning achievements of early 20th century mathematics.

Also solved by Arin Chaudhuri, Ashwin Kumar, Vladimir Nesov, Américo Tavares, Nathan Williams, and Henry Wilton. Thanks to all who wrote in!

In my post, I made a comment in passing that continued fractions have applications to knot theory. Now A Neighborhood of Infinity has a series of posts on the very topic I had in mind, namely the theory of rational tangles. Much of the theory is due to the great John Horton Conway.

Things have been a little quiet lately on this blog; I was away for a bit on vacation, and I’m still catching up on some things, and I know Vishal has been busy. Hopefully we’ll have more posts soon.

In any case, it’s high time for another Problem of the Week! This one is undoubtedly classical in nature, but may require a famously nontrivial result to solve quickly by hand:

Find the length of the period of the repeating decimal representation of \displaystyle \frac1{65537}.

Please submit solutions to topological[dot]musings[At]gmail[dot]com by Friday, August 29, 11:59 pm (UTC); do not submit solutions in Comments. Everyone with a correct solution will be inducted into our Hall of Fame! We look forward to your response.

One of the rare pleasures of doing mathematics — not necessarily high-powered research-level mathematics, but casual fun stuff too — is finally getting an answer to a question tucked away at the back of one’s mind for years and years, sometimes decades. Let me give an example: ever since I was pretty young (early teens), I’ve loved continued fractions; they are a marvelous way of representing numbers, with all sorts of connections to non-trivial mathematics [analysis, number theory (both algebraic and transcendental!), dynamical systems, knot theory, …]. And ever since I’ve heard of continued fractions, there’s one little factoid which I have frequently seen mentioned but which is hardly ever proved in the classic texts, at least not in the ones I looked at: the beautiful continued fraction representation for e = 2.7182818....

[Admittedly, most of my past searches were done in the pre-Google era — today it’s not that hard to find proofs online.]

This continued fraction was apparently “proved” by Euler way back when (1731); I once searched for a proof in his Collected Works, but for some reason didn’t find it; perhaps I just got lost in the forest. Sometimes I would ask people for a proof; the responses I got were generally along the lines of “isn’t that trivial?” or “I think I can prove that”. But talk is cheap, and I never did get no satisfaction. That is, until a few (maybe five) years ago, when by accident I spotted a proof buried in Volume 2 of Knuth’s The Art of Computer Programming. Huge rush of relief! So, if any of you have been bothered by this yourselves, maybe this is your lucky day.

I’m sure most of you know what I’m talking about. To get the (regular) continued fraction for a number, just iterate the following steps: write down the integer part, subtract it, take the reciprocal. Lather, rinse, repeat. For example, the sequence of integer parts you get for \sqrt{2} is 1, 2, 2, 2, … — this means

\sqrt{2} = 1 + 1/(2 + 1/(2 + 1/(2 + ...))),

giving the continued fraction representation for \sqrt{2}. Ignoring questions of convergence, this equation should be “obvious”, because it says that the continued fraction you get for \sqrt{2} - 1 equals the reciprocal of the continued fraction for \sqrt{2} + 1.

Before launching in on e, let me briefly recall a few well-known facts about continued fractions:

  • Every rational number has a continued fraction representation of finite length. The continued fraction expresses what happens when one iterates the Euclidean division algorithm.

For example, the integer parts appearing in the continued fraction for 37/14:

37/14 = \mathbf{2} + 1/(\mathbf{1} + 1/(\mathbf{1} + 1/(\mathbf{1} + 1/\mathbf{4})))

duplicate the successive quotients one gets by using the division algorithm to compute \mbox{gcd}(37, 14):

37 = \mathbf{2} \cdot 14 + 9

14 = \mathbf{1} \cdot 9 + 5

9 = \mathbf{1} \cdot 5 + 4

5 = \mathbf{1} \cdot 4 + 1

4 = \mathbf{4} \cdot 1 + 0

  • A number has an infinite continued fraction if and only if it is irrational. Let I denote the space of irrationals between 0 and 1 (as a subspace of \mathbb{R}). The continued fraction representation (mapping an irrational x \in I to the corresponding infinite sequence of integer parts a_1, a_2 \ldots in its continued fraction representation x = \left[0, a_1, a_2, \ldots \right]) gives a homeomorphism I \to \mathbb{N}^{\mathbb{N}}, where \mathbb{N}^{\mathbb{N}} carries a topology as product of countably many copies of the discrete space \mathbb{N}.

In particular, the shift map \displaystyle \sigma: \mathbb{N}^{\mathbb{N}} \to  \mathbb{N}^{\mathbb{N}}, defined by (\sigma(f))(n) = f(n+1), corresponds to the map \tau: I \to I defined by \tau(x) = 1/x \pmod 1. The behavior of \tau is a paragon, an exemplary model, of chaos:

  1. There is a dense set of periodic points of \tau. These are quadratic surds like \sqrt{2} - 1: elements of I that are fixed points of fractional linear transformations \displaystyle x \mapsto \frac{ax + b}{cx + d} (for integral a, b, c, d and |ad - bc| = 1).
  2. The transformation \tau is topologically mixing.
  3. There is sensitive dependence on initial conditions.

For some reason, I find it fun to observe this sensitive dependence using an ordinary calculator. Try calculating something like the golden mean \displaystyle \frac{\sqrt{5} - 1}{2}, and hit it with \tau over and over and watch the parade of integer parts go by (a long succession of 1’s until the precision of the arithmetic finally breaks down and the behavior looks random, chaotic). For me this activity is about as enjoyable as popping bubble wrap.

  • Remark: One can say rather more in addition to the topological mixing property. Specifically, consider the measure \displaystyle \mu(E) := \int_E \frac1{1 + x} dx on I, where \mu(I) = \log 2. It may be shown that \tau: I \to I is a measure-preserving transformation; much more significantly, \tau is an ergodic transformation on the measure space. It then follows from Birkhoff’s ergodicity theorem that whenever f: I \to \mathbb{R} is integrable, the time averages \displaystyle \frac1{n} \sum_{k=0}^{n-1} f(\tau^k(x)) approach the space average \displaystyle \frac1{\log 2}\int_I f(u) \frac{du}{1 + u} for almost all x \in I. Applying this fact to f([0, a_1, a_2, \ldots]) = \log(a_1), it follows that for almost all irrationals x \in I, the geometric mean of the integer parts a_1, a_2, \ldots a_n approaches a constant, Khinchin’s constant K = 2.685452.... A fantastic theorem!

Anyway, I digress. You are probably waiting to hear about the continued fraction representation of e, which is \left[2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, \ldots \right]:

e = 2 + 1/(1 + 1/(2 + 1/(1 + 1/(1 + 1/(4 + \ldots)))))

Cute little sequence, except for the bump at the beginning where there’s a 2 instead of a 1. One thing I learned from Knuth is that the bump is smoothed away by writing it in a slightly different way,

e = \left[1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots \right],

involving triads 1, 2n, 1, where n = 0, 1, 2, \ldots.

Anyway, how to prove this fact? I’ll sketch two proofs. The first is the one I found in Knuth (loc. cit., p. 375, exercise 16; see also pp. 650-651), and I imagine it is close in spirit to how Euler found it. The second is from a lovely article of Henry Cohn which appeared in the American Mathematical Monthly (Vol. 116 [2006], pp. 57-62), and is connected with Hermite’s proof of the transcendence of e.

PROOF 1 (sketch)

Two functions which Euler must have been very fond of are the tangent function and its cousin the hyperbolic tangent function,

\displaystyle \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}},

related by the equation \tan(ix) = i\tanh(x). These functions crop up a lot in his investigations. For example, he knew that their Taylor expansions are connected with Bernoulli numbers, e.g.,

\displaystyle \tanh(x) = \sum_{n \geq 1} \frac{B_{2n}4^n (4^n - 1)}{(2n)!} x^{2n-1}.

The Taylor coefficients T_n = E_{2n-1} where \displaystyle \tan(x) = \sum_{n \geq 1} \frac{E_{2n-1} x^{2n-1}}{(2n-1)!} are integers called tangent numbers; they are the numbers 1, 2, 16, … which appear along the right edge of the triangle

1

0, 1

1, 1, 0

0, 1, 2, 2

5, 5, 4, 2, 0

0, 5, 10, 14, 16, 16

where each row is gotten by taking partial sums from the preceding row, moving alternately left-to-right and right-to-left. The numbers 1, 1, 5, … which appear along the left edge are called secant numbers S_n, the Taylor coefficients of the secant function. Putting E_{2n} = S_n, the secant and tangent numbers E_n together are called Euler numbers, and enjoy some interesting combinatorics: E_n counts the number of “zig-zag permutations” k \mapsto a_k of \{1, 2, \ldots, n\}, where a_1 > a_2 < a_3 > \ldots. For more on this, see Stanley’s Enumerative Combinatorics (Volume I), p. 149, and also Conway and Guy’s The Book of Numbers, pp. 110-111; I also once gave a brief account of the combinatorics of the E_n in terms the generating function \sec(x) + \tan(x), over here.

Euler also discovered a lovely continued fraction representation,

\displaystyle \tanh(x) = 1/(x^{-1} + 1/(3x^{-1} + 1/(5x^{-1} + \ldots ))),

as a by-product of a larger investigation into continued fractions for solutions to the general Riccati equation. Let’s imagine how he might have found this continued fraction. Since both sides of the equation are odd functions, we may as well consider just x > 0, where 0 < \tanh(x) < 1. Thus the integer part is 0; subtract the integer part and take the reciprocal, and see what happens.

The MacLaurin series for \tanh(x) is x - x^3/3 + \ldots = x(1 - x^2/3 + \ldots); its reciprocal has a pole at 0 of residue 1, so

\displaystyle \frac1{\tanh(x)} - \frac1{x} = f_1(x)

gives a function f_1(x) = x/3 + \ldots which is odd and analytic near 0. Now repeat: reciprocating f_1(x), we get a simple pole at 0 of residue 3, and

\displaystyle \frac1{f_1(x)} - \frac{3}{x} = f_2(x)

gives a function f_2(x) which is odd and analytic near 0, and one may check by hand that its MacLaurin series begins as x/5 + \ldots.

The pattern continues by a simple induction. Recursively define (for x > 0)

\displaystyle f_{n+1}(x) = \frac1{f_n(x)} - (2n+1)x^{-1}

It turns out (lemma 1 below) that each f_n(x) is odd and analytic near 0, and then it becomes plausible that the continued fraction for \tanh(x) above is correct: we have

\displaystyle \tanh(x) = 1/(x^{-1} + 1/(3x^{-1} + \ldots 1/((2n-1)x^{-1} + f_n(x)) \ldots ))

Indeed, assuming the fact that f_n(x) is uniformly bounded over n, these expressions converge as n \to \infty, so that the continued fraction expression for \tanh(x) is correct.

Lemma 1: Each f_n(x) (as recursively defined above) is odd and analytic near 0, and satisfies the differential equation

\displaystyle f_{n}'(x) = 1 - f_n(x)^2 - 2n x^{-1}f_n(x).

Proof: By induction. In the case n = 0, we have that f_0(x) = \tanh(x) is analytic and

f_{0}'(x) = 1/\cosh^2(x) = 1 - f_{0}^2(x).

Assuming the conditions hold when n = k, and writing

f_k(x) = a_{k1}x + a_{k3}x^3 + \ldots

we easily calculate from the differential equation that a_{k1} = 1/(2k+1). It follows that

\displaystyle f_{k+1}(x) := \frac1{f_k(x)} - (2k+1)x^{-1}

is indeed analytic in a neighborhood of 0. The verification of the differential equation (as inductive step) for the case n = k+1 is routine and left to the reader. \Box

  • Remark: The proof that the continued fraction above indeed converges to \tanh(x) is too involved to give in detail here; I’ll just refer to notes that Knuth gives in the answers to his exercises. Basically, for each x in the range (-\pi/2, \pi/2), he gets a uniform bound |f_n(x)| \leq \tan|x| for all n, and notes that as a result convergence of the continued fraction is then easy to prove for such x (good enough for us, as we’ll be taking x = 1/2). He goes on to say, somewhat telegraphically for my taste, “Careful study of this argument reveals that the power series for f_n(z) actually converges for |z| < \pi \sqrt{2n+1}/2; therefore the singularities of f_n(z) get farther and farther away from the origin as n grows, and the continued fraction actually represents \tanh(z) throughout the complex plane.” [Emphasis his] Hmm…

Assuming the continued fraction representation for \tanh(x), let’s tackle e. From the continued fraction we get for instance

\displaystyle \tanh(1/2) = \frac{e^{1/2} - e^{-1/2}}{e^{1/2} + e^{-1/2}} = 1/(2 + 1/(6 + 1/(10 + \ldots)))

Taking reciprocals and manipulating,

\displaystyle 1 + \frac{2}{e - 1} = 2 + 1/(6 + 1/(10 + 1/(14 + \ldots ))).

Theorem 1: e-1 = \left[ 1, 1, 2, 1, 1, 4, 1, 1, 6, \ldots \right].

Proof: By the last displayed equation, it suffices to show

2\left[ 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, \ldots \right] = \left[ 1, 6, 10, 14, \ldots \right].

This follows from a recursive algorithm for multiplying a continued fraction by 2, due to Hurwitz (Knuth, loc. cit., p. 375, exercise 14):

Lemma 2: 2\left[ 0, 2a + 1, b, c, \ldots \right] = \left[ 0, a, 1, 1 + 2\left[ 0, b-1, c, \ldots \right] \right], and

2\left[0, 2a, b, c, \ldots \right] = \left[ 0, a, 2\left[ b, c, \ldots \right] \right] .

I won’t bother proving this; instead I’ll just run through a few cycles to see how it applies to theorem 1:

2\left[ 0, 1, 1, 2, 1, 1, 4, \ldots \right] = \left[ 0, 0, 1, 1 + 2\left[ 0, 0, 2, 1, 1, 4\ldots \right] \right]

= \left[ 1, 1 + 2\left[ 2, 1, 1, 4, \ldots \right] \right]

= \left[ 1, 1 + 4 + 2\left[ 0, 1, 1, 4, \ldots \right] \right]

= \left[ 1, 5 + \left[ 0, 0, 1, 1 + 2\left[ 0, 0, 4, \ldots \right] \right] \right]

= \left[ 1, 6 + \left[ 0, 1 + 2\left[ 4, 1, 1, 6, \ldots \right] \right] \right]

= \left[ 1, 6, 1 + 2\left[ 4, 1, 1, 6, \ldots \right] \right]

= \left[ 1, 6, 1 + 8 + 2\left[ 0, 1, 1, 6, \ldots \right] \right]

and so on. Continuing this procedure, we get \left[ 1, 6, 10, 14, \ldots \right], which finishes the proof of theorem 1. \Box

PROOF 2

I turn now to the second proof (by Cohn, loc. cit.), which I find rather more satisfying. It’s based on Padé approximants, which are “best fit” rational function approximations to a given analytic function, much as the rational approximants \displaystyle \frac{p_n}{q_n} = \left[ a_0, a_1, \ldots, a_n \right] provide “best fits” to a given real number x = \left[ a_0, a_1, \ldots \right]. (By “best fit”, I mean a sample theorem like: among all rational numbers whose denominator is bounded above in magnitude by q_n, the approximant \displaystyle \frac{p_n}{q_n} comes closest to x.)

Definition: Let f be a function analytic in a neighborhood of 0. The Padé approximant to f of order (m, n), denoted r_{m, n}(x), is the (unique) rational function \displaystyle \frac{p(x)}{q(x)} such that \mbox{deg}(p) = m, \mbox{deg}(q) = n, and the MacLaurin coefficients of r_{m, n} agree with those of f up to degree m + n. \Box

This agreement of MacLaurin coefficients is equivalent to the condition that the function

\displaystyle \frac{q(x)f(x) - p(x)}{x^{m+n+1}}

is analytic around 0. Here, we will be interested in Padé approximants to f(x) = e^x.

In general, Padé approximants may be computed by (tedious) linear algebra, but in the present case Hermite found a clever integration trick which gets the job done:

Proposition 1: Let r be a polynomial of degree k. Then there are polynomials p, q of degree at most k such that

\displaystyle \int_{0}^1 r(t)e^{xt} dt = \frac{q(x)e^x - p(x)}{x^{k+1}}

Explicitly,

p(x) = r(0)x^k - r'(0)x^{k-1} + r''(0)x^{k-2} - \ldots

q(x) = r(1)x^k - r'(1)x^{k-1} + r''(1)x^{k-2} - \ldots

Proof: Integration by parts yields

\displaystyle \int_{0}^1 r(t)e^{xt} dt = \frac{r(1)e^x - r(0)}{x} - \frac1{x} \int_{0}^1 r'(t)e^{xt} dt

and the general result follows by induction. \Box

It is clear that the integral of proposition 1 defines a function analytic in x. Taking k = m + n, this means we can read off the Padé approximant r_{m, n}(x) = p(x)/q(x) to e^x from the formulas for p, q in proposition 1, provided that the polynomial r(t) [of degree m + n] is chosen so that \mbox{deg}(p) = m, \mbox{deg}(q) = n. Looking at these formulas, all we have to do is choose r to have a zero of order n at t = 0, and a zero of order m at t = 1. Therefore r(t) = t^n (t-1)^m fits the bill.

Notice also we can adjust r by any constant multiple; the numerator and denominator p(x), q(x) are adjusted by the same constant multiples, which cancel each other in the Padé approximant r_{m, n}(x) = p(x)/q(x).

Taking r(t) = t^n (t-1)^m in proposition 1, we then infer

\displaystyle \int_{0}^1 t^n (t-1)^m e^t dt = q(1)e - p(1).

Notice that this integral is small when m, n are large. This means that r_{m, n}(1) = p(1)/q(1) will be close to e (see the following remark), and it turns out that by choosing m, n appropriately, the values r_{m, n}(1) coincide exactly with rational approximants coming from the continued fraction for e.

  • Remark: Note that for the choice r(t) =t^n (t-1)^m, the values p(1), q(1) derived from proposition 1 are manifestly integral, and q(1) \neq 0. [In particular, |q(1)| \geq 1, justifying the claim that |p(1)/q(1) - e| is small if q(1)e - p(1) is.] In fact, p(1), q(1) may be much larger than necessary; e.g., they may have a common factor, so that the fraction p(1)/q(1) is unreduced. This ties in with how we adjust r(t) by a constant factor, as in theorem 2 below.

For n \geq 0, let p_n/q_n denote the n^{th} rational approximant arising from the infinite continued fraction

\left[ a_0, a_1, a_2, \ldots \right] = \left[ 1, 0, 1, 1, 2, 1, 1, 4, 1, \dots \right],

where \mbox{gcd}(p_n, q_n) = 1. From standard theory of continued fractions, we have the following recursive rule for computing the integers p_n, q_n from the a_n: p_0 = a_0, q_0 = 1, p_{-1} := 1, q_{-1} := 0, and

p_{n+1} = a_{n+1} p_n + p_{n-1} \qquad q_{n+1} = a_{n+1} q_n + q_{n-1}.

Explicitly, a_{3n} = a_{3n+2} = 1 and a_{3n+1} = 2n, so

  1. p_{3n} = p_{3n-1} + p_{3n-2} \qquad q_{3n} = q_{3n-1} + q_{3n-2}
  2. p_{3n+1} = 2n p_{3n} + p_{3n-1} \qquad q_{3n+1} = 2n q_{3n} + q_{3n-1}
  3. p_{3n+2} = p_{3n+1} + p_{3n} \qquad q_{3n+2} = q_{3n+1} + q_{3n}

[Note: p_1 = 1 and q_1 = 0, so p_1/q_1 is infinite, but that won’t matter below.]

Theorem 2: Define, for n \geq 0,

\displaystyle A_n := \int_{0}^1 \frac{t^n (t-1)^n}{n!} e^t dt

\displaystyle B_n := \int_{0}^1 \frac{t^{n+1} (t-1)^n}{n!} e^t dt

\displaystyle C_n := \int_{0}^1 \frac{t^n (t-1)^{n+1}}{n!} e^t dt

Then A_n = q_{3n}e - p_{3n}, B_n = p_{3n+1} - q_{3n+1}e, and C_n = p_{3n+2} - q_{3n+2}e.

Proof: It is easy to see A_0 = e - 1, B_0 = 1, and C_0 = 2 - e. In view of the recursive relations for the p_n, q_n above, it suffices to show

  • A_n = -B_{n-1} - C_{n-1}
  • B_n = -2n A_{n-1} + C_{n-1}
  • C_n = B_n - A_n

The last relation is trivial. The first relation A_n + B_{n-1} + C_{n-1} = 0 follows by integrating both sides of the identity

\displaystyle \frac{t^n (t-1)^n e^t}{n!} + \frac{t^n (t-1)^{n-1} e^t}{(n-1)!} + \frac{t^{n-1} (t-1)^n e^t}{(n-1)!} = \frac{d}{dt} \frac{t^n (t-1)^n e^t}{n!}

over the interval \left[ 0, 1 \right]. The second relation B_n + 2n A_{n-1} - C_{n-1} = 0 follows by integrating both sides of the identity

\displaystyle \frac{t^{n+1} (t-1)^n e^t}{n!} + 2n\frac{t^{n-1} (t-1)^{n-1} e^t}{(n-1)!} - \frac{t^{n-1} (t-1)^n e^t}{(n-1)!} = \frac{d}{dt} \frac{t^n (t-1)^{n+1} e^t}{n!}

which we leave to the reader to check. This completes the proof. \Box

Theorem 2 immediately implies that

e = \left[ 1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots \right];

indeed, the rational approximants p_k/q_k to the right-hand side have the property that q_k |p_k/q_k - e| is one of |A_n|, |B_n|, or |C_n| (for k = 3n, 3n+1, 3n+2 respectively), and looking at their integral expressions, these quantities approach 0 very rapidly.

This in turn means, since the denominators q_k grow rapidly with k, that the rational approximants p_k/q_k approach einsanely” rapidly, and this in itself can be used as the basis of a proof that e is transcendental (Roth’s theorem). To give some quantitative inkling of just “how rapidly”: Knuth in his notes gives estimates on how close the approximant

1/(x^{-1} + 1/(3x^{-1} + \ldots + 1/(2n+1)x^{-1} \ldots ))

is to the function \tanh(x). It’s something on the order of \displaystyle \frac{(n!)^2}{(2n)!(2n+1)!} (loc. cit., p. 651).

  • Remark: Quoting Roth’s theorem in support of a theorem of Hermite is admittedly anachronistic. However, the Padé approximants and their integral representations used here did play an implicit role in Hermite’s proof of the transcendence of e; in fact, Padé was a student of Hermite. See Cohn’s article for further references to this topic.

[Wow, another long post. I wonder if anyone will read the whole thing!]

[Edits in response to the comment below by Henry Cohn.]

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