Calculating Square Root of 2

I had so much fun deriving that series which converges to pi the other day that I thought it’d be fun to occasionally whip up such series for other irrational numbers. Today I’ll do \sqrt{2}.

There are many different series that would do the job, the one I’ll derive is found by expanding \sqrt{x} around x=1 as a power series.

Indeed, for x satsfying |x-1| < 1 we have

\begin{array}{ccl} \sqrt{x} & = & 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3-\frac{5}{128}(x-1)^4+ ... \\ & = & 1+ \frac{1}{2}\sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}(x-1)^{n+1}\end{array}.

This series is guaranteed to converge for all 0<x<2, and to diverge for all x<0 and x>2.   The number 2 itself sits right on the boundary of the circle of convergence, so although we would like to say

\sqrt{2} = 1 + \sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)},

we first need to show that the series is convergent. Consider the sequence a_n = \frac{(2n)!}{4^n(n!)^2(n+1)}. This sequence is monotonically decreasing, and we will show that for all n>0, a_n < \frac{1}{n}. Firstly,

a_1 = \frac{2!}{4\cdot 2} = \frac{1}{4} < 1.

We now assume that a_k < \frac{1}{k} for some integer k and show that a_{k+1} < \frac{1}{k+1}. This is true, since

\begin{array}{ccl}\frac{(2k+2)!}{4^{k+1}((k+1)!)^2(k+2)} & = & \frac{(2k+1)(2k+2)}{4(k+1)(k+2)}\frac{(2k)!}{4^k(k!)^2(k+1)} \\ &=& \frac{(2k+1)(2k+2)}{4(k+1)(k+2)}a_k\\& < & \frac{(2k+1)(2k+2)}{4k(k+1)(k+2)}\\ & = & \frac{1}{4}(\frac{1}{k}+\frac{3}{k+2}) \\ & \leq & \frac{1}{k+1}\end{array},

where the last inequality holds for k\geq1. Thus a_{k} < \frac{1}{k} implies a_{k+1}<\frac{1}{k+1}, and so by induction a_n < \frac{1}{n} for all n.

To show convergence of our series, we note we have shown enough to invoke the alternating series test. So it is true that

1 + \sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}

converges, and by continuity  it must equal \sqrt{2}. Truncating the series at n=10,000,000 gives

\sqrt{2}\approx 1.4142135623775118

which is correct up 11 decimal places.

Mathematical Proof that God EXISTS!!

Trigger Warning: This post takes the position that Santa Claus doesn’t exist.

I heard this one from a friend of a friend, it’s a good one. It goes like this:

Consider the logical connective associated with implication, the material conditional as it is sometimes called. It basically captures the intuition behind the “if blah is true then bleep is true”. So if a and b are two truth values (either true or false) and we write the material conditional as “if a then b“, then we can write out the four possible outputs:

  1. “if false then false” is true (both are false, implication still makes sense).
  2. “if false then true” is true (remember a implies b, but b can be true whether or not a is).
  3. “if true then false” is false (a implies b, so this combination is not allowed).
  4. “if true then true” is true (a forces b to be true).

The material conditional evaluates to either true or false, depending on whether the truth values are consistent with the implication. Now we can consider a to represent the proposition “God exists”. Let b stand for “Santa Claus exists”. We can now ask ourselves, is a true? Well, clearly b is false. What about the implication “if a then b“? I think everyone would agree that the existence of God in no way implies the existence of Santa, so we can say “if God exists then Santa exists” is false. We now check out the above list and observe that case 3 uniquely matches our situation; the whole implication is false and b is false. From this we conclude that a must be true, and therefore God exists.
QED