avatarKeith McNulty

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A Beautiful Expression for Pi

Proving a surprising and pleasing mathematical result

Today I am going to prove a beautiful and somewhat surprising expression for π. The expression that I will prove is the following:

where the denominator is an infinite product of nested surds.

Step 1 of my proof will establish an important trigonometric identity, which we will use in Step 2 by taking limits and applying the result to π.

Proving an important trigonometric identity

Our expression for π can be derived from the following trigonometric identity:

for all positive integers n and where ⍺ is not an integer multiple of π.

We are going to use classic induction to prove this. Remember that you can prove a result for all positive integers n if and only if you can prove it is true for n = 1 (the induction start), and you can prove that if it is true for n = k then it is true for n = k + 1 (the induction step).

To prove the induction start, let’s use the standard trigonometric identity sin(θ+ϕ) = sinθcosϕ + cosθsinϕ. If θ = ϕ = ⍺/2, we have sin⍺ = 2sin(⍺/2)cos(⍺/2). Dividing both sides by cos(⍺/2) gives the result for n = 1.

Now let’s assume the result is true for n = k. We can use the same standard trigonometric identity to derive the result for n = k + 1, as follows:

Hence the result is true for all positive integers n.

Using this to prove our beautiful expression for π

Now we make the observation that as an angle θ approaches zero, the value of (sinθ)/θ approaches 1. From this, with a little simple manipulation, we can say that

So applying limits to both sides of our trigonometric identity from the last section, we can conclude the following:

Now we are going to use another well known trigonometric identity, which is that cos(θ+ϕ) = cosθcosϕ -sinθsinϕ. Again, with θ = ϕ = ⍺/2, we can say that cos⍺ = cos²(⍺/2)-sin²(⍺/2) = 2cos²(⍺/2)-1 by Pythagoras’ Theorem. Rearranging this, we have:

And so we can rewrite our previous expression for ⍺ as:

Now taking ⍺ = π/2, we know that sin(π/2) = 1 and cos(π/4) = √(1/2), hence:

and by multiplying both sides by 2 (or dividing both sides by 1/2), we get our final result.

Verifying using R

In this R code, I verify this result by writing a function that calculates the nested surd expression for π to n nested surds, and then I calculate the value generated at n = 2, n = 5, n = 10 and n = 100. We can see that this expression does indeed approximate π.

calc_pi <- function (n) {
  multiplier <- sqrt(0.5 + 0.5*sqrt(0.5)) 
  val <- 0.5*sqrt(0.5)*multiplier
  for (i in 1:n) {
    multiplier <- sqrt(0.5 + 0.5*multiplier)
    val <- val*multiplier
  }
  1/val
}
for (k in c(2, 5, 10, 100)) {
  print(calc_pi(k))
}
[1] 3.136548
[1] 3.141514
[1] 3.141593
[1] 3.141593

What did you think of this result? Can you relate it to other things you know about π? Feel free to comment.

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