# A simple way to look at the Riemann Zeta Function

The Riemann hypothesis. Walk up to any mathematician and ask him the one problem sought after for a solution, and he would say that’s the one. For several reasons, it’s the holy grail of modern maths, the coupe de grace of centuries of mathematical advancement. It’s also unfortunately intractable. To understand the statement itself (not the as yet ungotten solution), one needs training in multiple disciplines of mathematics.

There are resources which go through the long history of the problem, and wonderful insights that some really good mathematicians came up with. Other places can be found that explain its importance and its relationship to the distribution of primes. Here we just try to explain the problem in a (novel) way without having to go through a pesky and unintuitive trick called analytic continuation. Beginning with a few simple postulates, we can reach the famous and crazy identity:

1 + 2 + 3 + 4 + … = -1/12

Let’s start off with the definition of the Riemann Zeta function. It begins in a deceptively innocent way. It starts off by adding all the numbers known to mankind.

S = 1 + 2 + 3 + 4 + …

And mathematicians being fond of generalisations, we raise each term to a power.

S = 1ˢ + 2ˢ + 3ˢ + 4ˢ + …

And further, mathematicians being mathematicians, we replace the s by -s and get the Riemann Zeta function

ζ(s) = 1⁻ˢ + 2⁻ˢ + 3⁻ˢ + 4⁻ˢ + …

The above sum is valid where Re(s)> 1. For Re(s)< 1, the sum diverges, i.e, goes off to ∞. This is solid mathematics, and to find a ‘value’ for Re(s) < 1, we use a trick called analytic continuation. All analytic continuation does is ‘extend’ the function keeping everything ‘smooth’ and assigns a value for all s. It looks somewhat like:

The function to the left of the line x = 1 is extended in a unique way. However, were we to reexamine our basic ideas of distance, we can reach the same results in a more intuitive manner. Thus far in our computation, we declare that the value of the Riemann Zeta function is given by the ‘final’ distance of the infinite sum from the x axis. Herein lies the assumption that the x axis gives us a say in the outcome. But why grant only the x axis this privilege? This leads us to the postulate:

All (rotated) axes are equal. There is nothing special about the x-axis.

Using this, we can instantly prove the standard identity

ζ(0) = 1 + 1 + 1 + 1 + … = -1/2

How do we do so? Consider the graph of partial sums, where each point (x, y) represents sum of first x terms:

If one were to travel along the green line given above, one would testify that the sequence of sums oscillates about 0. Travelling along any other line, including the x axis, we would say the sum diverges. Thus we put our faith in the one line that gives us a non-infinite answer. We say that the Riemann sum can be replaced by the line, because it oscillates about the line with an average value 0.

Now how to ascribe a value to this line? Thats simple, this line is the same as moving the x axis down by 1/2 and rotating it by 45˚. But as all rotated axes are equal, the value for this line is how much we moved relative to the origin, which is -1/2! Thus

1 + 1 + 1 + 1 + … = -1/2

We found a notable result, that the ‘effective’ value of any line is given by its intercept on the y axis. This powerful result is a consequence of assuming rotational symmetry.

Similarly, one can show

1 - 2 + 3 - 4 + … = 1/4

The two broken purple lines above meet on the y-axis at 1/4.

Now let’s try to extend this ‘effective’ value to functions other than straight lines. We finally reach the famous series

ζ(-1) = 1 + 2 + 3 + 4 + …

To find a function about which the sequence of partial sums oscillates about 0, we again plot the partial sums. Now as the average viewed from the function is 0, area swept by the function over the unit is same as area swept by the partial sum over the unit. As an equation,

Using 1 + 2 + 3 … n = n(n + 1) / 2, and knowing that f(x) is quadratic, we get

f(x) = x²/2 - 1/6

All that remains is finding the effective value of this function. To do this, we use the following properties:

1. Effective value is determined strictly by the y-intercept, as all other values of the function at x != 0 would not satisfy rotational symmetry. This is similar to that of straight lines.
2. Drawing a tangent line at any point on the function, the y intercept of the tangent would always be twice the y intercept of the function as seen from the point (as it’s a parabola). As this is true at every point, the function turns out to be a squeezing operator, that is, it takes the value expected by the straight line, and squeezes it by half.
3. Thus the effective value of the function f(x) = x²/2 - 1/6 is its y intercept divided by 2, giving -1/12. So

ζ(-1) = 1 + 2 + 3 + 4 + … = -1/12

In general, the effective value of a polynomial p(x) = axⁿ + bxⁿ⁻¹ + … u should be given by u/n, for n ≥ 1. Extension to complex exponents and coefficients shouldn’t break anything.

This technique does not work for polynomials whose order is less than 1. A different approach is required here, which necessitates another post to explain in its entirety. However that analysis turns out to be fruitful as we can extend the Riemann Zeta function into the critical strip (0 < Re(s) < 1) and prove that

ζ(s) = 1⁻ˢ + 2⁻ˢ + 3⁻ˢ + 4⁻ˢ + … n⁻ˢ - ((n+1)²⁻ˢ - n²⁻ˢ) / ((2-s)(1-s))

where n → ∞, Re(s) ≥ 0