The Equation for the Limit Point of a Golden Spiral
The purpose of this paper is to show how to compute the limit point of a Golden Spiral,
which is the point that the Golden Spiral converges to.
This monograph assumes that you are familiar with Golden Rectangles, so if you are not,
more information can be found at the Wikipedia article at:
http://en.wikipedia.org/wiki/Golden_rectangle.
Consider a Golden Rectangle whose upper left corner has coordinates (0,0), and whose
short side is 1. Then the longer side must be  (pronounced Phi), where  is the Golden
Ratio which is:
  ( 5  1) / 2 = 1.618…
One special property of a Golden Rectangle is that if you make a square using the short
side (see S1 below), the remaining part of the large Golden Rectangle is itself a (smaller)
Golden Rectangle. This process can be continued indefinitely. If you then create an arc
for each square, you get a Golden Spiral.
To see an Applet that let’s you create a Golden Spiral, go to:
http://www.utdallas.edu/~tfarage/goldenRectangleApp/build/GoldenSpiralJApplet.html.
The next page has a part of a Golden Spiral obtained from Mathematica:
Here’s part of a Golden Spiral obtained from Mathematica:
squares
12
drawing step
0.8
autozoom
zoom
face color gray colors
As you can see the Golden Spiral converges to a point. What are the coordinates of this
point?
First, let’s figure out the lengths of the sides of the squares. We’re assuming that the the
side length of S1 is 1. Now each subsequent square has a side that is 1/  times the length
of the side of the previous square. This is because in any Golden Rectangle the longer
side is  times the shorter side, so conversely, the shorter side is 1/  times the longer
side. Thus the sequence of the lengths of each square’s side is:
1, (1/  ), (1/  )2, (1/  )3, (1/  )4, (1/  )5, (1/  )6, etc.
These square sides get small quickly. For instance, (1/  )10 is 0.0081, which is less than
one hundredth of the original square’s length. These lengths are said to be decreasing
exponentially.
Now let determine the X coordinates of the points of the arcs. We’re assuming that the
upper left points has coordinates (0,0), that the side length of S1 is 1, and that X values
increase as you go to the right and that Y values increase as you go down (similar to pixel
coordinates on a computer screen). So we have that X[0] = 0, and looking at the end of
the first arc we see that X[1] = 1. Now an interesting pattern emerges when you just
consider whether the next X coordinates is to the right or to the left of the previous X
coordinate. X[1], the end point of the arc of S1 is to the right of X[0]. X[2], the end point
of the arc of S2 is to the right of X[2]. But X[3], the end point of the arc of S3 is to the left
of X[2]. And X[4], the end point of the arc of S4 is also to the left of X[3]. This pattern
continues, so we get this pattern for the direction of the X coordinates:
Right, right, left, left, right, right, left, left, etc.
Since going to the right means an X value increases, and going to the left means an X
value decreases, this means that a right means to add and a left means to subtract.
Now combining this information with the square lengths gives us the X coordinates:
X[0] = 0 (because this is given)
X[1] = 1
X[2] = 1 + (1/  )
X[3] = 1 + (1/  ) - (1/  )2
X[4] = 1 + (1/  ) - (1/  )2 - (1/  )3
X[5] = 1 + (1/  ) - (1/  )2 - (1/  )3 + (1/  )4
X[6] = 1 + (1/  ) - (1/  )2 - (1/  )3 + (1/  )4 + (1/  )5
X[7] = 1 + (1/  ) - (1/  )2 - (1/  )3 + (1/  )4 + (1/  )5 - (1/  )6
X[8] = 1 + (1/  ) - (1/  )2 - (1/  )3 + (1/  )4 + (1/  )5 - (1/  )6 - (1/  )7
Such a nice pattern!
But how do we determine the limit of this sequence? That is, what is: lim X [n] ?
n
Let’s call this limit Lx, so that:
Lx = 1 + (1/  ) - (1/  )2 - (1/  )3 + (1/  )4 + (1/  )5 - (1/  )6 - (1/  )7 + (1/  )8 + …
The double plus and minus signs make this a bit more difficult than usual, but if we group
the terms differently, things get easier. So let’s group the even powered terms and then
the odd powered terms like this:
Lx = 1 - (1/  )2 + (1/  )4 - (1/  )6 + (1/  )8 + …
(1/  ) - (1/  )3 + (1/  )5 - (1/  )7 - …
Each of these two sequences is now fairly easily computable. Let’s call the first sequence
E (for the even powers) and then second sequence D (for odd powers), so that:
Lx = E + D where
E = 1 - (1/  )2 + (1/  )4 - (1/  )6 + (1/  )8 + … and
D = (1/  ) - (1/  )3 + (1/  )5 - (1/  )7 - …
To compute E multiply both sides of the equation for E by (1/  )2. This gives:
E(1/  )2 = (1/  )2 - (1/  )4 + (1/  )6 - (1/  )8 - …
Now add this new equation to the original equation for E and notice that all the terms on
the right hand side cancel except for the 1! So we get:
E + E(1/  )2 = 1
Let’s get rid of the fraction by multiplying by  2 to get
E 2 + E =  2
Factoring E from the left side gives:
E(  2 + 1) =  2, and so we finally get that:
E =  2 / (  2 + 1)
Fortunately, finding D is similar, because we again just multiply both sides of the D
equation by (1/  )2, just as for the equation for E. We’ll let the reader work out the
details to find that:
D =  / (  2 + 1)
Finally, we can get our limit Lx = E + D =  2 / (  2 + 1) +  / (  2 + 1). This given us:
Lx = (  2 +  ) / (  2 + 1)
Let’s get an actual number here by plugging in the value for  given above, 1.618….
Lx = 1.1708204…
How about the limit of the Y coordinate? The computation is very similar. Looking at
the Golden Spiral we see that:
Y[0] = 0 (because this is given)
Y[1] = 1
Y[2] = 1 - (1/  )
Y[3] = 1 - (1/  ) - (1/  )2
Y[4] = 1 - (1/  ) - (1/  )2 + (1/  )3
Y[5] = 1 - (1/  ) - (1/  )2 + (1/  )3 + (1/  )4
Y[6] = 1 - (1/  ) - (1/  )2 + (1/  )3 + (1/  )4 - (1/  )5
Y[7] = 1 - (1/  ) - (1/  )2 + (1/  )3 + (1/  )4 - (1/  )5 + (1/  )6
Y[8] = 1 - (1/  ) - (1/  )2 + (1/  )3 + (1/  )4 - (1/  )5 + (1/  )6 + (1/  )7
These are very similar to the X coordinate values except the signs are shifted. Again
we’ll let the reader go through the process we went through to compute the limit of the X
coordinate. If done correctly the limit for the Y coordinate will be found to be:
Ly = (  2 -  ) / (  2 + 1)
Note that the only difference between this and Lx is in the minus sign.
Let’s get an actual number for the limit of the Y coordinate here by plugging in the value
for  again:
Ly = 0.2763932…
And there we have it: to the 4th decimal place, the coordinate of the point that the Golden
Spiral converges to is:
( 1.1708, 0.2764 )
But remember, you’ll never get there.
Tim Farage can be emailed at tfarage@hotmail.com. A mini-biography is available at
http://www.utdallas.edu/~tfarage/bio.html.