Solving Van Roomen's Problem: Step Two

In my previous blog, I showed how Van Roomen's problem could be reduced to a summation formula:

(45)x - (3,795)x³ + (95,634)x⁵ - (1,138,500)x⁷ + (7,811,375)x⁹ - (34,512,075)x¹¹ + (105,306,075)x¹³ - (232,676,280)x¹⁵ + (384,942,375)x¹⁷ - (488,494,125)x¹⁹ + (483,841,800)x²¹ - (378,658,800)x²³ + (236,030,652)x²⁵ - (117,679,100)x²⁷ + (46,955,700)x²⁹ - (14,945,040)x³¹ + (3,764,565)x³³ - (740,259)x³⁵ + (111,150)x³⁷ - (12,300)x³⁹ + (945)x⁴¹ - (45)x⁴³ + x⁴⁵ =



where n = 45.

In today's blog, I will show how a recurrence relation can be introduced to simplify the problem further.

The key insight comes to generalizing the above summation formula. For example, let's define:

F_n(x) =



It turns out that it is possible to show that for any n ≥ 1:

2cos(nα) = F_n(2cosα)

For any odd n ≥ 1:

2 sin(nα) = (-1)^(n-1)/2*F_n(2sinα)

François Viète knew both of these identities and he would use them to find his solutions to Van Roomen's problem. In today's blog, I will show the proof for the recurrent relation and in my next blog, I will show how this recurrence relation leads to the trigonometric identities above.

Theorem 1: Recurrence Formula

If:

F_n(x) =



Then:

F_n(x) = x*F_n-1(x) + F_n-2(x)

Proof:

(1) F₁(x) = (-1)⁰*(1/1)*[(1!)/(0!1!)]x^1-0 = x

(2) F₂(x) = (-1)⁰*(2/2)*[(2!)/(0!2!)]x^2-0 + (-1)¹*(2/1)*[(1!)/(1!0!)]x^2-2 = x² - 2

(3) F₃(x) = (-1)⁰*(3/3)*[(3!)/(0!3!)]x^3-0 + (-1)¹*(3/2)*[(2!)/(1!1!)]x^3-2 = x³ - 3x

(4) We can see that x*F₂(x) - F₁(x) = x*(x² - 2) - x = x³ - 3x.

(5) So, we can assume that the recurrence formula is true up to some integer n-1 where n ≥ 4, that is:

F_n-1(x) = x*F_n-2(x) - F_n-3(x)

(6) Now, x*F_n-1(x) =



=


(7) Likewise, F_n-2(x) =



=



=



(8) So, we see that: x*F_n-1(x) - F_n-2(x) =





where if n is odd,

C = 0

and if n is even,

C =

-(-1)^{(n/2 - 1)}[(n-2)/(n-1-n/2)][(n-1-n/2)!/(n/2-1)!(n-1-n/2-(n/2-1))!]x^n-2(n/2) =

= (-1)^(n/2)[(n-2)/(n/2-1)][(n/2-1)!/(n/2 -1)!(0!)]x⁰ =

= (-1)^(n/2)[(n-2)]/[(n/2-1)](1)(1) = (-1)^(n/2)(n-2)*(2)/(n-2) =

= (-1)^(n/2)*2

(9) So, x*F_n-1(x) - F_n-2(x) - C - x_n =



(10) Focusing solely on the middle part, we see that:
















(11) If n is odd:

F_n(x) =



(12) If n is even:

F_n(x) =



(13) So, either way we have:

F_n(x) = x*F_n-1(x) - F_n-2(x).

QED