4 articles tagged “mathematics” — software, engineering, and related writing by idmontie.

Mathematics blog posts - idmontie's Portfolio

In the previous article, we discussed the geometric series as an example of an infinite series that
has a closed form solution. We approached this as the inifinite series acting as the Taylor series
of that closed form solution:

$$
S(x) = 1 + x + x^2 + x^3 + ... = \frac{1}{1-x}
$$

What we skipped over however is that this is only true for the domain $|x| < 1$. If we try to evaluate this series for $x = 1$ we get:

$$
S(1) = 1 + 1 + 1 + 1 + ... = \infty
$$

And any value larger than 1 or less than -1 will result in a divergent series. The infinite series is only
convergent for values between -1 and 1, but the closed form solution is value for all real numbers except for $x = 1$.

Can we apply this domain expansion to other inifinite series? Let's take a look at the zeta function.

## The Riemann Zeta Function

Is it possible that the domain of the zeta function

$$
\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots
$$

is larger than just all numbers greater than 1?

Euler discussed in "Remarques sur un beau rapport entre les series des puissances tant directes que reciproques" a related function ot the zeta function, where the terms of the zeta function alternate in sign. He defined the series as:

$$
1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \dots
$$

which we write in modern notation as the Alternating Zeta Function (or the Dirichlet Eta Function):

$$
\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = 1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \dots
$$

We note that this series is convergent for all $\Re(s) > 0$ without proof.

We can express the eta function in terms of the zeta function:

$$
\begin{align*}
\eta(s) & = 1 =- \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \dots \\
& = 1 + \frac{1}{2^s} + 2\frac{1}{2^s} + \frac{1}{3^s} - 2\frac{1}{4^s} + \dots \\
& = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots - 2\left(\frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \dots\right) \\
& = \sum_{n=1}^{\infty} \frac{1}{n^s} - 2\sum_{n=1}^{\infty} \frac{1}{2n^s} \\
& = \zeta(s) - 2^{1-s}\zeta(s) \\
& = \zeta(s)(1 - 2^{1-s})
\end{align*}
$$

Rearranging the terms to write the zeta function in terms of the eta function, we get:

$$
\boxed{\zeta(s) = \frac{1}{1 - 2^{1-s}} \eta(s) \text{ for }
\Re(s) > 0 \text{ and } 1 - 2^{1-s} \neq 0.}
$$

The above equation extends the zeta function to the half-plane $\Re(s) > 0$. We have to remember though that

$$
1 - 2^{1-s} = 0
$$

has infinitely many solutions in the complex plane and not just the real solution of $s = 1$. For example, we have the complex solutions of $s = 1 + 2\pi i k$ for any integer $k$.

## The Critical Strip

This definition of the zeta function in terms of the eta function extends the domain of the function into what is known as the critical strip $0 < \Re(s) < 1$ of the complex plane. Since the leading term $1/(1 - 2^{1-s})$ is never zero in this domain, the zeta function and the eta function have the same zeros in this region.

In later articles we will show that Riemann's zeta function can be extended to the entire complex plane, and we will go into how the critical strip plays a role in the Riemann hypothesis and its impact on the Prime Number Theorem.

Extending the Domain of the Zeta Function

When we last looked at the [On the Addition of Fractions](/blog/post/2025-03-16-on-the-addition-of-fractions), I covered a proof
that the harmonic series:

$$
\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots
$$

divereses. The obvious next question is do similar sums converge or diverge? In 1650, Petri Mengoli proposed the
following problem in his "Novae quadraturae arithmeticae" series:

> Find the sum of the series:
>
> $$
> \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots
> $$

We can calculate some of the partial sums to see how quickly it grows:

$$
\begin{align*}
\frac{1}{1^2} & = 1 \\
\frac{1}{1^2} + \frac{1}{2^2} & = 1.25 \\
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} & = 1.361111111111111 \\
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} & = 1.423611111111111 \\
\end{align*}
$$

It is clear that the sum grows, but if we sum the first $n$ terms, we get:

| First $n$ terms | Sum |
| --------------- | ----------------- |
| 10 | 1.549767731166541 |
| 100 | 1.634983900184893 |
| 1000 | 1.643934566681561 |
| 10000 | 1.644834071848065 |
| 100000 | 1.644923078134810 |
| 1000000 | 1.644932078134810 |

The sum seems to not grow indefinitely like the harmonic series, and seems to converge to some value. When summing
the terms individually, we find that it converges extremely slowly. By Euler's time in 1734, it was already known that
ths sum converged to a value, but there was no known closed-form expression for the sum. A closed-form expression
is one that can be written as a finite number of operations. The open question of whether there was a closed form for the sum, became known as the Basel problem in honor of Euler's home city since he was the one who discovered the solution to the problem.

Let's cover a few tricks that Euelr would have already known from his many other discoveries and mathematical research. First, we know that some infinite series have closed forms, in particular, we know that the geometric series has a closed form:

$$
S = \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty} x^n
$$

We can verify this using another technique known in Euler's time, called Taylor series expansion:

$$
f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots
$$

Let $a = 0$ and $f(x) = 1 / (1 - x)$, then we see:

$$
\begin{align*}
f'(x) = \frac{1}{(1-x)^2} \\
f''(x) = \frac{2}{(1-x)^3} \\
f'''(x) = \frac{6}{(1-x)^4} \\
\vdots
\end{align*}
$$

And evaluating these derivatives at $x = 0$, we get:

$$
\begin{align*}
f'(0) = 1 \\
f''(0) = 2 \\
f'''(0) = 6 \\
\vdots
\end{align*}
$$

So the Taylor series expansion of $1 / (1 - x)$ around $x = 0$ is:

$$
\boxed{\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty} x^n}
$$

Euler's approach in his 1734 letter was similar to this techique, but insted of looking at $1 / (1 - x)$, he looked at the
$\sin$ function instead. The Taylor series expansion of $\sin(x)$ would have been known to Euler and is easily derived from
the Taylor series definition we provided above. The expansion is:

$$
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
$$

We will divide each side by $x$ to make a step later in this post easier to follow:

$$
\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots
$$

So far, we have focused on Taylor series expansion, where we change how we represent a function in terms of a series of
powers of $x$. Another technique known at the time was thinking of polynomials as being defined by their roots. If we
have a n-degree polynomial:

$$
p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
$$

Then we can write it in terms of its roots $r_1, r_2, \cdots, r_n$:

$$
p(x) = a_n (x - r_1) (x - r_2) \cdots (x - r_n)
$$

Where $r_1, r_2, \cdots, r_n$ are the roots of the polynomial.

This begs the question: can we use a similar technique with other functions? Euler seemed to think that since $\sin$ could
be written as an infinite polynomial, then we could use a similar technique to write $\sin$ in terms of its roots. This technique would not be formally proven until centuries later by Karl Weierstrass. Without Weierstrass's work, we will have to consider Euler's argument as less than rigorous.

Euler made the leap that if a polynomial function is equal to the product of its roots, then he could define $sin(x) / x$ as an infinite product:

$$
\frac{\sin(x)}{x} = (1 - \frac{x}{\pi})(1 + \frac{x}{\pi})(1 - \frac{x}{2\pi})(1 + \frac{x}{2\pi})(1 - \frac{x}{3\pi})(1 + \frac{x}{3\pi}) \cdots
$$

If we begin to expand this product, multiplying the first term across the rest, we get:

$$
\frac{\sin(x)}{x} = 1 - \frac{x^2}{\pi^2} - \frac{x^2}{4\pi^2} - \frac{x^2}{9\pi^2} + \cdots
$$

And we collect the terms with $x^2$ together, we get:

$$
\frac{\sin(x)}{x} = 1 - x^2(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots) - \cdots
$$

Now we can compare the coefficient of $x^2$ in this representation to the coefficient of $x^2$ in the Taylor series expansion of $\sin(x) / x$:

$$
\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots = \frac{1}{3!}
$$

And multiplying both sides by $\pi^2$ we get:

$$
\boxed{1 + \frac{1}{4} + \frac{1}{9} + \cdots = \frac{\pi^2}{6}}
$$

Euler showed that the Basel problem has the closed form of $\pi^2 / 6$.

Euler would go on to make his proof more rigorous in letters in 1741, 1745, and 1755, but the core idea is the same. He continued to investigate similar sums with powers up to $n=10$ and found that the values were all rational multiples of
powers of $\pi$.

Later, a more formal equation for the values of positive even integers was discovered and is given by the formula:

$$
\zeta(2n) = \sum_{k=1}^{\infty} \frac{1}{k^{2n}} = \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n}}{2(2n)!}
$$

Where $B_{2n}$ are the Bernoulli numbers.

The general form of the function that Euler studied is known as the zeta-function and is defined as:


$$
\zeta(s) = \sum_{k=1}^{\infty} \frac{1}{k^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots
$$

This function is very well studied and has many interesting properties relating to prime numbers and the distribution of primes. Even with all of the research done on this function, there are still two still open questions related to the zeta function:

$$
\boxed{\text{Is Apéry's constant $\zeta(3)$ transcendental?}}
$$

and

$$
\boxed{\text{Are all odd values of the zeta function transcendental?}}
$$

Where a transcendental number is a number that is not algebraic, meaning it is not a solution to any polynomial equation with rational coefficients.

The Basel problem is a simple problem that can be solved using a few tricks that Euler already knew from his many other discoveries and mathematical research. It is a beautiful example of how mathematics can be used to solve problems and how different techniques can be used to solve the same problem.

The Basel Problem

In my previous post, ["On the Addition of Fractions, by Petri Mengoli"](/blog/post/2025-03-16-on-the-addition-of-fractions), I
wrote that the following inequality holds, with no proof:

$$
\frac{1}{1-n} + \frac{1}{n} + \frac{1}{n + 1} > \frac{3}{n}
$$

If we don't assume a specific value for the numerator of the right hand side of the
inequality, we can derive the inequality by starting with

$$
\frac{1}{1-n} + \frac{1}{n} + \frac{1}{n + 1} > \frac{a}{n}.
$$

Multiply through by $n$:

$$
\frac{n}{1-n} + \frac{n}{n + 1} + 1 > a.
$$

Set a common denominator for the left hand side:

$$
\frac{n(n+1) + n(n-1)}{(n-1)(n+1)} > a - 1
$$

Expanding the expression

$$
\frac{n^2 + n + n^2 - n}{n^2 - 1} > a - 1
$$

And simplifying

$$
\frac{2n^2}{n^2 - 1} > a - 1
$$

We notice that when $|n| > 1$, the left hand side of the expression is always positive.

If we let $n\to\infty$, then we see that

$$
\lim_{n\to\infty} \frac{2n^2}{n^2 - 1} \overset{\mathrm{H}}{=} \lim_{n\to\infty} \frac{4n}{2n} = 2
$$

through the use of L'Hospital's rule. We find that as $n\to\infty$, the value of n is always positive and a value greater than 2.

If we slightly re-arrange, we find the following:

$$
\frac{2n^2}{n^2 - 1} + 1 > a
$$

And now knowing that the left hand side approaches 3 from above as $n\to\infty$, we know
that $a$ can be set to $3$ as the largest value that still satisfies the expression.

Hence, we find that the original inequality posited holds.

Petri Mengoli Fraction Inequality

I was researching the origin of the Basel problem to write a short blog post about
how Euler approached the problem (only to get beaten by <a href="https://www.youtube.com/watch?v=jmtiWGnj5os">anime characters talking
about it</a>). I however wanted to
dive deeper into the context of why Euler wanted to even approach this problem,
so I investigated the work of Petri Mengoli.
The publication of his work &quot;On the Addition of Fractions&quot; is available via the
<a href="https://archive.org/details/bub_gb_PrKgVx1LcUUC/mode/1up">Internet Archive in its original Latin</a>.
Petri Mengoli&#x27;s work focuses on the harmonic series in the preface and shows
that it diverges. The harmonic series is written in modern notation as:
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><munderover><mo>∑</mo><mrow><mi>n</mi><mo>=</mo><mn>1</mn></mrow><mi mathvariant="normal">∞</mi></munderover><mfrac><mn>1</mn><mi>n</mi></mfrac><mo>=</mo><mn>1</mn><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><mo>+</mo><mo>…</mo></mrow><annotation encoding="application/x-tex">\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots</annotation></semantics></math>n=1∑∞​n1​=1+21​+31​+41​+…
His proof relies on finding groupings of fractions that sum greater to one
(which his paper refers to as &quot;unity&quot;). First, look at the first 3 terms after
&quot;unity&quot;: <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mi mathvariant="normal">/</mi><mn>2</mn><mo>+</mo><mn>1</mn><mi mathvariant="normal">/</mi><mn>3</mn><mo>+</mo><mn>1</mn><mi mathvariant="normal">/</mi><mn>4</mn></mrow><annotation encoding="application/x-tex">1/2 + 1/3 + 1/4</annotation></semantics></math>1/2+1/3+1/4. When added, we see that when we set common denominators:
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>6</mn><mi mathvariant="normal">/</mi><mn>12</mn><mo>+</mo><mn>4</mn><mi mathvariant="normal">/</mi><mn>12</mn><mo>+</mo><mn>3</mn><mi mathvariant="normal">/</mi><mn>12</mn><mo>=</mo><mn>13</mn><mi mathvariant="normal">/</mi><mn>12</mn><mo>&gt;</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">6/12 + 4/12 + 3/12 = 13 / 12 &gt; 1</annotation></semantics></math>6/12+4/12+3/12=13/12&gt;1
And in fact, for any three terms:
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mn>1</mn><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></mfrac><mo>+</mo><mfrac><mn>1</mn><mi>n</mi></mfrac><mo>+</mo><mfrac><mn>1</mn><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></mfrac><mo>&gt;</mo><mfrac><mn>3</mn><mi>n</mi></mfrac></mrow><annotation encoding="application/x-tex">\frac{1}{n - 1} + \frac{1}{n} + \frac{1}{n+1} &gt; \frac{3}{n}</annotation></semantics></math>n−11​+n1​+n+11​&gt;n3​
Next Mengoli shows that there is another group of fractions that adds up to
greater than one, and another set of fractions after that set that also adds up
to one. He demonstrates that for any given number, there is a finite set of
fractions that can be added to get a value greater than that. Therefore the
series must diverge.
His approach here is inspired by Archimedes&#x27; &quot;Quadrature of the Parabola&quot;,
and focuses on using the &quot;method of exhaustion&quot;, rather than manipulating
infinite series or dealing with limits.
From this work, comes the original quote of what became known as the Basel
problem:
<blockquote class="border-l-4 border-gray-300 pl-4">
Ab huius fractionum dispositionis contemplatione feliciter
expeditus, ad aliam progrediebar dispositionem, in qua
singulæ unitates numeris quadratis denominantur.
Hæc speculatio fructus quidem laboris rependit, nondum tamen
effecta est solvendo, sed ingenii ditioris postulat adminiculum,
ut præcisam dispositionis, quam mihimetipst proposui, summam
valeat reportare.
</blockquote>
Roughly translating to:
<blockquote class="border-l-4 border-gray-300 pl-4">
Having successfully concluded my contemplation of this arrangement
of fractions, I proceeded to another arrangement, in which
each unit is denominated by square numbers.
This speculation indeed rewards labor with fruit, but it has not
yet been completed in solving and demands the assistance of
greater ingenuity so that the precise sum of the arrangement
I proposed to myself may be recovered.
</blockquote>
This is the work that perhaps sparked the inspiration for Euler&#x27;s proof that the
sum of the reciprocal of the squares adds up to <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>π</mi><mn>2</mn></msup><mi mathvariant="normal">/</mi><mn>6</mn></mrow><annotation encoding="application/x-tex">\pi^2 /6</annotation></semantics></math>π2/6 . Euler required much
more advanced mathematics and insight in order to come up with this equality, and
he revisited the sum in multiple letters with slightly different arguments to
validate his work.
For a full write up on this, see my self-published work here:
<a class="block flex-1 overflow-visible rounded-lg bg-gradient-to-br from-slate-100 to-slate-100 no-underline shadow-lg dark:from-slate-800/60 dark:to-slate-700/60" href="/docs/on-the-addition-of-fractions.pdf"><div class="p-6"><h4 class="!m-0 !p-0 !text-base font-bold !no-underline">On the addition of fractions.pdf</h4></div></a>

I was researching the origin of the Basel problem to write a short blog post about
how Euler approached the problem (only to get beaten by [anime characters talking
about it](https://www.youtube.com/watch?v=jmtiWGnj5os)). I however wanted to
dive deeper into the context of why Euler wanted to even approach this problem,
so I investigated the work of Petri Mengoli.

The publication of his work "On the Addition of Fractions" is available via the
[Internet Archive in its original Latin](https://archive.org/details/bub_gb_PrKgVx1LcUUC/mode/1up).
Petri Mengoli's work focuses on the harmonic series in the preface and shows
that it diverges. The harmonic series is written in modern notation as:

$$
\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots
$$

His proof relies on finding groupings of fractions that sum greater to one
(which his paper refers to as "unity"). First, look at the first 3 terms after
"unity": $1/2 + 1/3 + 1/4$. When added, we see that when we set common denominators:

$$
6/12 + 4/12 + 3/12 = 13 / 12 > 1
$$

And in fact, for any three terms:

$$
\frac{1}{n - 1} + \frac{1}{n} + \frac{1}{n+1} > \frac{3}{n}
$$

Next Mengoli shows that there is another group of fractions that adds up to
greater than one, and another set of fractions after that set that also adds up
to one. He demonstrates that for any given number, there is a finite set of
fractions that can be added to get a value greater than that. Therefore the
series must diverge.

His approach here is inspired by Archimedes' "Quadrature of the Parabola",
and focuses on using the "method of exhaustion", rather than manipulating
infinite series or dealing with limits.

From this work, comes the original quote of what became known as the Basel
problem:

> Ab huius fractionum dispositionis contemplatione feliciter
expeditus, ad aliam progrediebar dispositionem, in qua
singulæ unitates numeris quadratis denominantur.
Hæc speculatio fructus quidem laboris rependit, nondum tamen
effecta est solvendo, sed ingenii ditioris postulat adminiculum,
ut præcisam dispositionis, quam mihimetipst proposui, summam
valeat reportare.
>

Roughly translating to:

> Having successfully concluded my contemplation of this arrangement
of fractions, I proceeded to another arrangement, in which
each unit is denominated by square numbers.
This speculation indeed rewards labor with fruit, but it has not
yet been completed in solving and demands the assistance of
greater ingenuity so that the precise sum of the arrangement
I proposed to myself may be recovered.
>

This is the work that perhaps sparked the inspiration for Euler's proof that the
sum of the reciprocal of the squares adds up to $\pi^2 /6$ . Euler required much
more advanced mathematics and insight in order to come up with this equality, and
he revisited the sum in multiple letters with slightly different arguments to
validate his work.

For a full write up on this, see my self-published work here:

<DownloadCard
 href="/docs/on-the-addition-of-fractions.pdf"
 title="On the addition of fractions.pdf"
/>