Lill’s Method, Prime Numbers and Tangent of Sum of Angles

In this post I want to explore again the property discussed in my paper “Lill’s Method and the Sum of Arctangents”. I’ll apply the property to this question: If tan(θ₁)=2, tan(θ₂)=3,tan(θ₃)=5,…,tan(θ_n)=n-th prime number, then what is tan(θ₁ + θ₂ +… θ_n)?

The question can be easily solved with a calculator. We’ll see that the answer is always a rational number, so I want to obtain the fraction that is equal to the tangent of sum of angles. We’ll also see that the answer can be obtained using the coefficients of polynomial equations and that Lill’s method can provide a visual representation of the answer.

Polynomial Equations and Lill’s Method

In Lill’s method, each real root x has a corresponding angle θ such that x=-tan(θ). Because of the negative sign before the tangent, it’s more convenient to consider the Polynomials that have -tan(θ₁)=-2, -tan(θ₂)=-3,-tan(θ₃)=-5,…,-tan(θ_n)=-n-th prime number as roots when we’ll do the Lill representation. These polynomials are of the form (x+2)(x+3)(x+5)…(x+p_n).

If P(x)=xⁿ+a_n-1x^n-1+…+a₁x+a₀ , then P₀P₁ represents the coefficient 1 (coefficient of xⁿ ), P₁P₂ =a_n-1 ,… P_nP_n+1 =a₀ . Also tan( P₁P₀P_n+1)= tan(θ₁ + θ₂ +… θ_n). In the next section we’ll see the visual representation using Lill’s method.

Algebraically, the answer can be obtained using the following formula, where a_n =1:

Sequence

By definition tan(θ₁)=2, so no representation is required. To obtain tan(θ₁ + θ₂ ) let’s consider the polynomial (x+2)(x+3)= x²+5x+6. We can see that tan( P₁P₀P₃)=5/(1-6)=5/-5=-1.

To obtain tan(θ₁ + θ₂ +θ₃ ), let’s consider the polynomial (x+2)(x+3)(x+5)=x³ + 10 x² + 31x + 30. tan( P₁P₀P₄)=(10-30)/(1-31)=-20/-30=2/3.

As the degree increases, we’ll get bigger and bigger spirals rotating counterclockwise. For bigger degrees is easier to just use the algebraic equation mentioned in the previous section. So to obtain tan(θ₁ + θ₂ +θ₃ +θ₄ ), we’ll consider the polynomial (x+2)(x+3)(x+5)(x+7)=x⁴ + 17 x³ + 101 x² + 247 x + 210. tan( P₁P₀P₅)=(17-247)/(1-101+210)=-230/110=-23/11.

Using the same method we get tan(θ₁ + θ₂ +θ₃ +θ₄ +θ₅ )=49/132, tan(θ₁ + θ₂ +θ₃ +θ₄ +θ₅ +θ₆ )=-353/101 and tan(θ₁ + θ₂ +θ₃ +θ₄ +θ₅ +θ₆ +θ₇ )=682/3051.

Final Notes

We can see that tan(θ₁) and tan(θ₁ + θ₂ ) are integers. Are there tan(θ₁ + θ₂ +… θ_n) equal to integers for n>2? Are there tan(θ₁ + θ₂ +… θ_n) that have the denominator or the numerator equal to 0?

The method used in this post can be used to obtain the OEIS sequences A180657, A220447, A105750 and A105751 if we let tan(θ₁)=1, tan(θ₂)=2,tan(θ₃)=3,…,tan(θ_n)=n.

In my paper “Representing Polynomials with Complex Coefficients using Lill’s Method” I mentioned that sum of angles property can be extended even to polynomials with complex roots (I might add complex roots and complex coefficients). A more fitting name for the property is probably “Lill’s sum of root angles property”.