*d*can be written as a series in the form

where each

*a*

_{n}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

*n*th 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*) + 3

*f*

*(*

^{(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*) + 10

*f*

*(*

^{(2)}*g*(

*x*))

*g*

*(*

^{(2)}*x*)

*g*

*(*

^{(3)}*x*) + 5

*f*

*(*

^{(2)}*g*(

*x*))

*g*

*(*

^{(1)}*x*)

*g*

*(*

^{(4)}*x*) + 15

*f*

*(*

^{(3)}*g*(

*x*))

*g*

*(*

^{(1)}*x*)

*g*

*(*

^{(2)}*x*)

^{2}+ 10

*f*

*(*

^{(3)}*g*(

*x*))

*g*

*(*

^{(1)}*x*)

^{2}

*g*

*(*

^{(3)}*x*) + 10

*f*

*(*

^{(4)}*g*(

*x*))

*g*

*(*

^{(1)}*x*)

^{3}

*g*

*(*

^{(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 / 2

^{1}

1 / 2

^{4}

1 / 2

^{4}

43 / 2

^{8}

223 / 2

^{8}

60623 / 2

^{13}

764783 / 2

^{13}

107351407 / 2

^{16}

2499928867 / 2

^{16}

596767688063 / 2

^{19}

22200786516383 / 2

^{19}

64470807442488761 / 2

^{25}

3504534741776035061 / 2

^{25}

3597207408242668198973 / 2

^{29}

268918457620309807441853 / 2

^{29}

185388032403184965693274807 / 2

^{32}

18241991360742724891839902347 / 2

^{32}

16262449142923535035348829394713 / 2

^{35}

2037403289344551020141524968209753 / 2

^{35}

4561840680589672832866190269667090909 / 2

^{39}

708949733527461872865880740295881887369 / 2

^{39}

15579885703040482381715272205722723759995121 / 2

^{46}

2941124283010451165887914573127749096121128721 / 2

^{46}

4862902621226153133515412278839571188070361725789 / 2

^{49}

1096038034387794685488112001145283295041059426870329 / 2

^{49}

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 64**570 711495 ...

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