National Research Council. "21. The Error Term." Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. Washington, DC: The National Academies Press, 2003. 1. Print.
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Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics
mial is all complex numbers. A polynomial never “equals infinity.” There is no argument z for which its value just can’t be calculated. Calculating the value of a polynomial function for any given argument just involves raising the argument to natural-number powers, multiplying it by numbers, and adding the results together. You can do that with any number.
Functions whose domain is all complex numbers and which are decently well behaved (there is a precise mathematical definition of that!) are called entire functions. All polynomials are entire functions, so is the exponential function. Those rational functions I showed in Chapter 17.ii, however, are not entire functions, since their denominators can be zero. The log function is not an entire function, either: it has no value at argument zero. Riemann’s zeta function, likewise, has no value at argument 1, and so is not an entire function.
An entire function might have no zeros at all (like the exponential function: ez = 0 is never true), or it might have many (like a polynomial: 4 and 7 are the zeros of z2 – 11z + 28), or it might have infinitely many (like the sine function, which is zero at every integer multiple of π ).127 Now since polynomials can be rewritten in terms of their zeros, can all entire functions that have zeros be rewritten this way? Suppose I have some entire function, call it F, that can be defined by an infinite sum, F(z) = a + bz + cz2 + dz3 + …. And suppose I happen to know that this function has infinitely many zeros; call them ρ , σ , τ …. Can I rewrite this function in terms of its zeros, as an infinite product, F(x) = a (1 – z / ρ )(1 – z / σ )(1 – z / τ )…? As if the infinite sum were a sort of super-polynomial?
The answer is that under certain conditions, yes, you can. When you can do it, it’s often a very handy thing. This, for example, is how Euler solved the Basel problem, by applying this reasoning to the sine function.
How does this help us with the zeta function, which unfortunately is not an entire function? Well, as part of that complicated