The portion of a tree, shown in red at right, from the base to LLLL... is a spiral.
Another spiral, the one which motivates this digression, is shown in blue and is formed by the segments on the tip-set

Definition-- Given a ratio r<1 and an angle theta, we will define a left-hand spiral by starting with a base segment and adjoining further segments, each new length is r times the previous length and each new direction is angle theta to the left of the previous direction.
The definition makes it clear that the spiral is (almost) self-similar (the base segment is missed in the image) by a move involving rotation through angle theta and a shrink by factor r. (We could make it truly self-similar by also adding longer and longer segments at the other end, ie, a doubly infinite construction.)

The center of the rotation is at the tip of the spiral. In the second example above (the blue spiral), the center is the point RLLLL.... If we call the center C and call the segments

[P₀, P₁], [P₁, P₂], [P₁, P₂], … [P_i, P_i+1], …

then the triangles CP_iP_i+1 are all similar, with equal angle measures theta at the vertex C.

The x-coordinate of the vertex P_N on this tip-set spiral is given by the formula

      

If we extend to the infinite sum, the expression       

gives the x-coordinate of the center C of the spiral. We found, by manipulating Fourier series, the function represented by the sum

Let us derive this formula directly from the geometry of the blue spiral.

DIRECT DERIVATION OF THE CLOSED FORM

The picture above has been cut down, rescaled and relabeled for simplicity: P₀ = P, P₁ = Q, P₂ = R,
segment labels are lengths.

We have similar triangles PQC and QRC, the angles CPQ and CQR both have measure phi, while angles PCQ, QCR, RQA all have measure theta.

Hence angle CQA has measure phi+theta.
Finally, triangles CAQ and CAP are a right triangles.

We need to express the distance b+1 as a function of r and theta . We see that

The law of sines tells us that

We substitute these into the formula for the sine of a sum of two angles:

Remove the factor       to get        

We need something to substitute for a² so we apply the Pythagorean theorem to the two right triangles:    .

We subtract, and get    .

Then the equation above becomes     .

Now we clear the fraction and solve for b:

,    

so         and     .

The extra factor r² in the original formula above is required to make the construction fit into one half of the tree.