Friday, January 7, 2011

Kaplan's Infinite Series for d

Kaplan discovered that d can be written as a series in the form


where each an is rational. The first few terms of this series are:



Although there is no obvious pattern to these terms, they can be calculated from the Taylor series of the inverse of a certain function.

Let f(x) = x − cos(x) and let g(x) be the inverse of f(x). If follows that:

f(π/2) = π/2
g(π/2) = π/2
f(d) = 0
g(0) = d

We will construct a Taylor series for g(x) at π/2, and let x = 0 to obtain g(0) = d. The general form of a Taylor series of a function g(x) at c is


Where g(n)(x) is the nth derivative of g(x). Since x = 0, and c = π/2, the expression becomes


The values of g(n)(π/2) can be found using values of f(n)(π/2) and the chain rule. Successive derivatives of f(x) at π/2 are easy to find, and in fact follow a pattern: 2, 0, −1, 0, 1, 0, −1, 0, 1, 0, ...

To find g(1)(x), use the equality f(g(x)) = x. By taking the derivatives of both sides, we obtain

f(1)(g(x))g(1)(x) = 1
g(1)(x) = 1 / f(1)(g(x)).

Therefore g(1)(π/2) = 1/2.

With a little more work we can find g(2)(x) and g(3)(x). By taking the derivatives of f(1)(g(x))g(1)(x) = 1, we obtain

f(1)(g(x))g(2)(x) + f(2)(g(x))g(1)(x)2 = 0

And it follows that g(2)(π/2) = 0. Taking the derivatives again, we obtain:

f(1)(g(x))g(3)(x) + 3f(2)(g(x))g(1)(x)g(2)(x) + f(3)(g(x))g(1)(x)3 = 0 and g(3)(π/2) = 1/16.

Further derivatives become cumbersome quickly. For example the fifth derivative of f(g(x)) is

f(1)(g(x))g(5)(x) + 10f(2)(g(x))g(2)(x)g(3)(x) + 5f(2)(g(x))g(1)(x)g(4)(x) + 15f(3)(g(x))g(1)(x)g(2)(x)2 + 10f(3)(g(x))g(1)(x)2g(3)(x) + 10f(4)(g(x))g(1)(x)3g(2)(x) + f(5)(g(x))g(1)(x)5

Fortunately the derivatives can be found with Faà di Bruno's formula and the process can be automated. It follows from Faà di Bruno's formula that g(n)(π/2) = 0 for all even n > 0. It also turns out that the value of each odd n is a rational number with a power of 2 in the denominator.

Here are the first 25 values of g(n)(π/2) for odd n:

1 / 21
1 / 24
1 / 24
43 / 28
223 / 28
60623 / 213
764783 / 213
107351407 / 216
2499928867 / 216
596767688063 / 219
22200786516383 / 219
64470807442488761 / 225
3504534741776035061 / 225
3597207408242668198973 / 229
268918457620309807441853 / 229
185388032403184965693274807 / 232
18241991360742724891839902347 / 232
16262449142923535035348829394713 / 235
2037403289344551020141524968209753 / 235
4561840680589672832866190269667090909 / 239
708949733527461872865880740295881887369 / 239
15579885703040482381715272205722723759995121 / 246
2941124283010451165887914573127749096121128721 / 246
4862902621226153133515412278839571188070361725789 / 249
1096038034387794685488112001145283295041059426870329 / 249

The first 25 terms of the series can be calculated from these values. For each odd n, divide g(n)(π/2) by n!, and multiply by (–π/2)n. The sum of these first terms yields a good approximation of d, accurate to 17 digits.

0.73908 51332 15160 64570 711495 ...

Reference

Kaplan, S. R. "The Dottie Number." Mathematics Magazine Vol. 80, No. 1, February 2007. p 73-74.

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