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Analytic Continuation and the Riemann Zeta Function

Analytic Continuation is a very important mathematical technique which allows us to extend the domain of functions.  It is essential in higher level mathematics and physics and leads to some remarkable results. For example, by using analytic continuation we can prove that the sum of the natural numbers (1 + 2 + 3 + ….) is -1/12. Results don’t get more surprising than that!

Analytic continuation concerns functions of the form:

f(z) where z is a complex number and f(z) is (complex) differentiable.

Remember complex numbers are of the form a + bi and can be thought of as coordinate points in an x,y axis.  For the purposes of this post we will only look at real values of z (real numbers are still a subset of complex numbers).

The idea of analytic continuation is to take an original function with a restricted domain, then to find another function which is the same within that restricted domain, but also is valid outside that domain.  This sounds very complicated – but let’s look at a couple of examples: $f(z) = \frac{(z+1)(z+2)}{(z+1)}$

This is a function which is defined for all values except for z = -1. When z = -1 we have zero on the denominator so the function doesn’t exist. However we can write a new function: $g(z) = (z+2)$

Now, g(z) = f(z) for all z when z is not -1, but g(z) also exists when z = -1. Therefore we can regard g(z) as the analytic continuation of f(z), and we have extended the domain of f(z) from all values except -1, to all values of z.

A more interesting example is the following: $f(z) = \sum_{n=0}^\infty z^{n}$

This is the infinite series: $1 + z + z^{2}+ ...$

This function is analytic (complex differentiable) only when   -1 < z< 1.  (Don’t worry about how this is calculated – though it is related to the domain of convergence). Therefore this is our restricted domain.

But we can notice that the sum of a geometric sequence formula allows us to calculate f(z) in a different way: $\sum_{n=0}^\infty z^{n} = \frac{1}{(1-z)}$

Here we have used the formula for summing a geometric, with the first term 1 and common ratio z.

Therefore we could write: $g(z) = \frac{1}{(1-z)}$

f(z) = g(z) when -1 < z< 1 , but g(z) is complex differentiable for all values except for z = 1 (when the denominator is 0).  Therefore g(z) is the analytic continuation of f(z) from  -1 < z< 1 to all values of z except z = 1. One example of analytic continuation that I’ve written about before is the Riemann Sphere. This extends by analytic continuation the complex plane into the complex plane plus infinity.

Another example is used in showing that the sum of natural numbers is -1/12. There are a few different methods to show this – some discussed previously here. I’m going to try and talk through another proof of this result. It’s a bit difficult, but try and understand the general method! The proof revolves around the Riemann Zeta function, (Riemann is pictured above). This is defined as: $\zeta(z)= \sum_{n=1}^{\infty}n^{-z}$

This can also be written as: $\zeta(z)=\frac{1}{1^{z}} +\frac{1}{2^{z}} +\frac{1}{3^{z}}..$

So, if we want to find the sum of 1 + 2 + 3 … then we need to substitute z = -1 into the above summation. However this formula for the zeta function is only valid for the domain z > 1, so we first need to extend the function through analytic continuation.

Through analytic continuation (where we extend the domain from z > 1 to all complex numbers apart from -1) we can rewrite the zeta function as: $\zeta(1-z)=2^{1-z}\pi^{-z}cos(0.5\pi z)\Gamma(z)\zeta(z)$

and substituting z = 2 into this formula, so that we end up with zeta(-1) we get: $\zeta(-1)=2^{-1}\pi^{-2}cos(\pi)\Gamma(2)\zeta(2)$

Now, $\zeta(2) = \frac{\pi^{2}}{6}$ $\Gamma(2) = 1$ $cos(\pi) = -1$

Therefore $\zeta(-1)=-\frac{1}{12}$

We have proved that 1 + 2 + 3 … = -1/12 !

If you enjoyed this post you might also like:

The Riemann Hypothesis Explained. What is the Riemann Hypothesis – and how solving it can win you \$1 million.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct.  How?

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