Number Magic

18.1 INTRODUCTION

18.1.1 Goldbach’s Challenge

One of the most famous open problems in mathematics is the Goldbach conjecture:

Let \(g(n)\) denote the function which tells in how many ways we can write \(n\) as a sum of two primes. For example \(g(5)=2\), \(g(6)=1\) because \(5=2+3=3+2\) and \(g(6)=1\) because \(6+3+3\).

18.1.2 Goldbach Conjecture: Visualizing with Mathematica

Here is Mathematica code which allows to plot the comet, the graph of the function \(g\).

18.1.3 Cracking Goldbach with Calculus?

Why is this remarkable? It shows that computing the numbers \(f(n)\) could be done nicely using calculus by defining a function \(f\). Using Taylor’s theorem we can compute the entries \(g(n)\). The Goldbach conjecture is equivalent to \[\begin{aligned} g^{2 n}(0) \text { is nonzero for all } n \geq 1. \end{aligned}\] The only thing we would really need is to get a grip on the function \(f\). Unfortunately, nobody has seen how to write down the function \(f\) in terms of known functions. But it is not completely hopeless that there should not exist a modification \[f(x)=\sum_{p \text { prime }} a_{p} x^{p}\] with positive \(a_{p}\) such that \(f(x)\) is expressible using known functions. Also then, if \(g(x)=f(x)^{2}\) had positive even derivatives, Goldbach would follow.

18.2 SEMINAR

18.2.1 Calculus Hacks: Beyond Calculations

In this seminar, we see how calculus can help to compute things effectively and also hope to get insight into topics which are of more number theoretical nature. To find the cube root of \(10\) for example, we have \[10^{1 / 3} \sim 8^{1 / 3}+\frac{2}{3 \cdot 8^{2 / 3}}=2+\frac{2}{12}=2.1666 \ldots\] The actual value is \(2.15443\). We can also use linearization to find exact roots

18.2.2 Newton’s Method for Roots: A Powerful Tool

We could not mention the Newton method to find roots in class. It is a simple but effective iterative method. We can also do that to find roots. In order to find the cube root of \(9\) for example, we start with a first approximation like \(2\), then introduce the function \(f(x)=x^{3}-9\) for which we aim to find the root, then apply the Newton step \[T(x)=x-\frac{f(x)}{f^{\prime}(x)}.\] We have \(f^{\prime}(x)=3 x^{2}\) and so \(T(x)=x-(x^{3}-9) /(3 x^{2})\). This gives \(T(2)=25 / 12= 2.08333\). Already quite close to \(9^{1 / 3}=2.08008\).

18.2.3 Newton’s Method Takes a Complex Turn

There is an interesting story here when applying the Newton method in the complex plane. The function \(f(x)=x^{3}-9\) has exactly \(3\) roots in the complex plane. They are \(9^{1 / 3}\), \(9^{1 / 3} e^{i 2 \pi / 3}\) and \(9^{1 / 3} e^{i 4 \pi / 3}\). Check that these three numbers satisfy \(f(x)=0\)! Investigating the Newton method in the complex actually predated the Mandelbrot story. One can wonder what happens if you apply the Newton method with a given initial condition. The solution will end up in one of the three roots, but which ones? When drawing this, we see the Newton fractal. here is how you can plot the Newton fractal.¹

18.2.4 Geometric Series Demystified

In the exam you have proven \(1+3+3^{2}+\cdots+3^{n-1}=\left(3^{n}-1\right) / 2\). This is a special case of the geometric series formula \[1+a+a^{2}+\cdots+a^{n}=\frac{1-a^{n+1}}{1-a}.\] Of course, we could also prove this formula by induction. Better do it directly:

18.2.5 Convergence in Infinite Series

These were all finite sums but seeing the pattern allows us to take a limit and compute the infinite series:

18.2.6 Unlocking Taylor Series with Infinite Series

The Taylor series of a nice function is \(f(x)=\sum_{k=0}^{\infty} f^{(k)}(0) x^{k}\). Having just looked at C) can answer the trick question:

18.2.7 Deriving Logarithms with Series

How can you get from the last exercise the following identity?

18.2.8 Exploring the Series at -1

Now lets see what happens at \(x=-1\).

18.2.9 Number Theory Meets Calculus

How come that great number theorists like Leonard Euler or Godfrey Hardy were also masters in calculus? The reason is that many results of number theoretic nature have intimate relations with calculus. Lets look at the following problem:

Hint: compute first the Taylor series of \(f(x)=\arctan (x)\) using the Taylor series of \(1 /(1+x^{2})\) (the later is a geometric series), then evaluate \(f\) at \(x=1\).

18.2.10 Zeta Function and the Riemann Hypothesis

\[\operatorname{Li}_{s}(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{s}}=x+\frac{x^{2}}{2^{s}}+\frac{x^{3}}{3^{s}}+\cdots\] is called the poly logarithm function. For \(s=0\) it is Problem D, for \(s=1\) it is problem E. While in calculus, we might be more interested in the function as a function of \(x\), number theorists are more interested in the function as a function of \(s\) and \(s\) is complex. In the case \(x=1\), the function \(L i_{s}(x)\) is the Riemann zeta function \(\zeta(s)=\sum_{k=1}^{\infty} \frac{1}{k^{s}}\).

The Euler golden key relates \(\zeta\) with primes:

18.2.11 The Euler Golden Key Identity

First verify (maybe look at Problem C) that for a single prime \(p\) \[\frac{1}{1-\frac{1}{p^{s}}}=1+\frac{1}{p^{s}}+\frac{1}{p^{2 s}}+\frac{1}{p^{3 s}}+\cdots\] which is the sum over all \(\frac{1}{n^{s}}\), where \(n\) has only prime factors \(p\). Then look at the product of these for two primes \(p\), \(q\) and see that this is the sum over all \(\frac{1}{n^{s}}\) where \(n\) has only prime factors \(p\) and \(q\).

18.2.12 Exploring a Calculus-Based Equivalence to Goldbach

Lets come back to the topic of the introduction. Remember that the Goldbach conjecture tells that every even number larger than \(2\) is the sum of two primes. What is the relation with calculus? Define \(g(x)=(f(x))^{2}\) with \[f(x)=\sum_{p} \frac{x^{p}}{p !}=\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\frac{x^{7}}{7 !}+\cdots\] For the following, try to verify this carefully by showing both directions. If a statement \(A\), \(B\) are equivalent then this means that we have to show two things. We have to verify that \(A \implies B\) and \(B \implies A\).

EXERCISES