Convergence Of Infinite Products

Infinite Series (Sums)

At school we learned about infinite series.

$$  \sum_{n=1}^\infty a_n = a_1+ a_2 + a_3 + \ldots  $$

If the partial sum $\sum_{n=1}^{x} a_n$ tends to a limit as $x \rightarrow \infty$, we can say the series converges to a finite sum. There are many tests that can be used to see if a series converges or not, such as the popular ratio test:

$$ \lim_{n \rightarrow \infty} | \frac{a_{n+1}}{a_n} | <1 $$

If we think about the series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$, the ratio test gives us $|\frac{ 2^{n-1} }{ 2^{n} }| = \frac{1}{2}$, confirming the series is absolutely convergent.

Infinite Products

Infinite products don't seem to be discussed as much as series.

$$ \prod_{n=1}^\infty a_n = a_1 \times a_2 \times a_3 \times \ldots  $$

When do such infinite products converge? Let's think about this before jumping to the answer.

Thoughts

It seems intuitive that if the elements $a_n$ are getting larger, the product grows and diverges. The following example gets ever larger and doesn't converge. 

$$  2 \times 3 \times 4 \times 5 \times \ldots $$

Let's look at another example.

$$ 0.5 \times 0.55 \times 0.555 \times 0.5555 \times \ldots $$

Here the elements are getting larger, but are individually less than 1, so it isn't immediately obvious whether the product converges or not.

What if the product contains a zero? That sounds easy. A single zero would immediately make the overall product zero. 

But what about this product? 

$$ -2 \times -1 \times 0 \times 1 \times 2 \times \ldots $$

With the zero, we'd suggest the product is zero overall. But without that zero, the product gets larger and diverges. Is the single zero somehow 'powerful' enough to cancel the infinity?

Definition

A consensus has emerged around infinite products which may, or may not, seem natural.

If the partial product $\prod_{n=1}^{x} a_n$ converges to a non-zero limit $P$ as $x \rightarrow \infty$, we say the infinite product converges to $P$.

This makes sense but the fact that the limit needs to be non-zero sets infinite products apart from infinite series.

If a finite number of zeros $a_n$ can be removed from a product, and the remaining product converges, we say the infinite product converges to 0.

This is rather specific, and means that we can't say a product with infinitely many zero elements converges. For example, we can't say the following converges:

$$  1 \times 0 \times 1 \times 0 \times 1 \times 0 \times 1 \times \ldots $$

Let's look at some examples to bring these ideas to life.

Example 1

Does the following infinite product converge or diverge?

$$ \prod_{n=1}^{\infty}(1+\frac{1}{n}) $$

If we consider the partial sum, and look at at the first few elements of this product, we'll see a pattern.

$$ \begin{align} \prod_{n=1}^{N}(1+\frac{1}{n}) &=  \prod_{n=1}^{N}(\frac{n+1}{n}) \\ \\ &= \frac{\cancel{2}}{1} \times \frac{\cancel{3}}{\cancel{2}} \times \frac{4}{\cancel{3}} \times  \ldots \times \frac{N+1}{N} \\ \\ &= N + 1 \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to $\infty$, so the infinite product diverges. This isn't surprising when we look at each element and that they are all above 1.

Example 2

Let's now look at an example where each element is less than 1. Notice $n$ starts at 2, not 1. 

$$ \prod_{n=2}^{\infty}(1-\frac{1}{n}) $$

Again, if we consider the partial sum, and look at at the first few elements of this product, we'll see a pattern.

$$ \begin{align} \prod_{n=2}^{N}(1-\frac{1}{n}) &=  \prod_{n=2}^{N}(\frac{n-1}{n}) \\ \\ &= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{4} \times  \ldots \times \frac{N-1}{N} \\ \\ &= \frac{1}{N} \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to 0. Because this limit is not a non-zero limit, as required by the definitions above, we say the infinite product diverges. 

Turning Products Into Sums

We can convert a product into a sum using logarithms.

$$ \begin{align} \ln (\prod_{n=1}^\infty a_n) &= \ln( a_1 \times a_2 \times a_3 \times \ldots)  \\ \\ &= \sum_{n=1}^\infty \ln(a_n) \end{align}$$

We can say the infinite product converges if the sum $\sum_{n=1}^\infty \ln(a_n)$ converges. 

If that product tends to zero, this sum tends to $-\infty$, which is why we say the infinite product diverges to zero, not converges to zero.

A Different Test For Convergence

If we can rewrite the product elements as $(1 + a_n)$ where $a_n >0$ we can use a simpler convergence test.

$$ \begin{align} \prod_{n=1}^\infty (1 + a_n) &\leq  \prod_{n=1}^\infty e^{ a_n} \\ \\ &=   e^{\sum_{n=1}^\infty a_n} \end{align} $$

The inequality comes from $e^x=1 + x + O(x^2)$. We can make the following conclusion.

If $\sum_{n=1}^\infty a_n$ converges, then so does the product $\prod_{n=1}^\infty (1 + a_n)$, as long as $a_n >0$.

This test allows us to easily say the following infinite series converges:

$$ \prod_{n=1}^\infty (1 + \frac{1}{n^2}) $$

We can say this because we know $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.

Infinite Products Over Complex Numbers

There is an equivalent form of the previous test for complex $z_n$.

If $\sum_{n=1}^\infty |a_n|$ converges, then so does the product $\prod_{n=1}^\infty (1 + a_n)$, as long as $a_n \neq -1$.

Mathologer Video on Euler's Product Formula

The Mathologer youtube channel is a treasure trove of mathematics made fun and accessible. His video on the Euler product formula is particularly clear and worth watching.

He even includes some really nice and easily derived consequences of Euler's product formula, including:

• there are infinitely many primes because the harmonic series diverges

• there are infinitely many primes because $\pi$ is not a rational number

• probability that two randomly chosen integers are coprime is $\frac{6}{\pi^2}$

Euler's Golden Bridge

Euler was the first to find a connection between the world of primes and the world of ordinary counting numbers. Many insights about the primes have been revealed by travelling over this `golden bridge'.

Let's recreate Euler's discovery for ourselves. 

Video: [youtube]

Slides: [pdf]

Reimann Zeta Function

We know the harmonic series $\sum1/n$ diverges. We also know the series $\sum1/n^{2}$ converges. 

It's natural to ask for which values of $s$ the more general series, known as the Riemann Zeta function $\zeta(s)$, converges.

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

Series like $\sum1/n^{3}$ and $\sum1/n^{4}$ converge because each term is smaller than the corresponding one in $\sum1/n^{2}$. Less obvious is when $s<2$.

A separate blog post presents a short proof that $\zeta(s)$ converges for $s>1$. 

Sieving The Zeta Function

Let's write out the zeta function again, noting that $1^{s}=1$.

$$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\ldots$$

We can divide this series by $2^{s}$. 

$$\frac{1}{2^{s}}\zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\frac{1}{12^{s}}\ldots$$

These denominators are multiples of $2^{s}$. By subtracting these terms from $\zeta(s)$, we sieve out terms with these multiples of $2^{s}$. 

$$(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\ldots$$

This dividing and subtracting of infinite series is only valid because they are absolutely convergent for $s>1$.

Let's now divide this series by $3^{s}$.

$$\frac{1}{3^{s}}\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=\frac{1}{3^{s}}+\frac{1}{9^{s}}+\frac{1}{15^{s}}+\frac{1}{21^{s}}+\frac{1}{27^{s}}+\ldots$$

These denominators are all multiples of $3^{s}$, but not all multiples of $3^{s}$ are here. Some like $6^{s}$ and $12^{s}$ were removed in the previous step. Subtracting these from the previous series leaves terms with denominators that are not multiples of 2 or 3.

$$(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

We can't remove terms with multiples of $4^{s}$ because they were sieved out when we removed multiples of $2^{s}$. The next useful step is to remove multiples of $5^{s}$.

Repeating this process leaves terms with denominators that are not multiples of 2, 3 or 5.

$$(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

Again, we can't remove multiples of $6^{s}$ because they were already removed. If we repeat this several times, we'll see we can only remove multiples of successive primes. We'll also see that after each removal, the very first term 1 always survives. 

If we kept going, we'd end up with an infinite product on the left, and only 1 on the right.

$$\ldots\cdot(1-\frac{1}{11^{s}})\cdot(1-\frac{1}{7^{s}})(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1$$

We can rearrange this to isolate $\zeta(s)$. 

$$\zeta(s)=\prod_{p}(1-\frac{1}{p^{s}})^{-1}$$

The symbol $\prod$ means product, just like $\sum$ means sum.

Euler's Product Formula

We've arrived at Euler's product formula.

$$\boxed{\sum_{n}\frac{1}{n^{s}}=\prod_{p}(1-\frac{1}{p^{s}})^{-1}}$$

The product of $(1-\frac{1}{p^{s}})^{-1}$ over all primes $p$ is the sum of $\frac{1}{n^{s}}$ over all positive integers $n$, as long as we remember to keep $s>1$.

To say this result is amazing would not be an exaggeration. It reveals a deep connection between the primes and the ordinary counting numbers, a connection that doesn't appear too complicated at first sight. 

The Prime Number Theorem

We’ve just seen experimental evidence that $n/\ln(n)$ approximates $\pi(n)$ fairly well. Although the error itself grows as $n \rightarrow \infty$, the proportional error gets ever smaller.

Let’s write that out.

$$ \lim_{n \rightarrow \infty} \frac{\pi(n) - n/\ln(n)}{\pi(n)} = 0 $$

Rearranging this gives us the following.

$$ \lim_{n \rightarrow \infty} \frac{\pi(n)}{n/\ln(n)} = 1 $$

This says the ratio of $\pi(n)$ and the approximation $n/ ln(n)$ tends to 1 as $n \rightarrow \infty$. And this is precisely what the prime number theorem says.

$$ \pi(n) \sim n/ \ln(n) $$

The symbol $\sim$ says that both sides are asymptotically equivalent. For example, $f(n) \sim g(n)$ means $f(n)/g(n) = 1$ as $n \rightarrow \infty$.

Video: [youtube]

Slides: [pdf]

Asymptotic Equivalence

Some examples of asymptotic equivalence will help clarify its meaning.

If $f(x)=x^2 + x$ and $g(x)=x^2$, then $f \sim g$. Both $f$ and $g$ have the same dominant term $x^2$.

$$ \lim_{x\rightarrow\infty} \frac{x^2+x}{x^2} = \lim_{x\rightarrow\infty}1+\frac{1}{x}=1 $$

Swapping $f$ and $g$ doesn't break asymptotic equivalence, $g \sim f$.

$$ \lim_{x\rightarrow\infty} \frac{x^2}{x^2+x}=\lim_{x\rightarrow\infty}\frac{1}{1+\frac{1}{x}}=1 $$

However, if $f(x)=x^{3}$ and $g(x)=x^{2}$, then $f$ and $g$ are not asymptotically equivalent because the ratio $f/g$ tends to $x$, not 1. 

If we know that $f \sim g$ and $g \sim h$, then we can also say $f \sim h$. This property, called transitivity, is familiar from normal equality.

What About li(n)?

Gauss' second approximation $\text{li}(n)$ appeared to be a better approximation for $\pi(n)$. You'll find the prime number theorem is sometimes expressed using the logarithmic integral.

$$ \pi(n) \sim \text{li}(n) $$

Surely the prime number theorem must be about one of the approximations, not both? The only solution is for both approximations to be asymptotically equivalent. Let's see that this is indeed the case.

Let's set $f(n)=\frac{{n}}{\ln(n)}$ and $g(n)=\int_{0}^{n}\frac{1}{\ln(x)}dx$. 

To show $f \sim g$ we need to find the limit of $f(n)/g(n)$ as $n\rightarrow\infty$ and confirm it is 1. Sadly, both $f(n)$ and $g(n)$ become infinitely large as $n\rightarrow\infty$, so the ratio is undefined. 

When this happens, we usually try l'Hopital's rule as an alternative way to find the limit.

$$ \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)}=\lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} $$

It's fairly easy to work out $f'(n)=\frac{\ln(n)-1}{\ln^{2}(n)}$, and $g'(n)=\frac{1}{\ln(n)}$ pops out of the definition of $\text{li}(n)$. 

$$ \begin{align} \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} &= \lim_{n\rightarrow\infty}\frac{(\ln(n)-1)\ln(n)}{\ln^{2}(n)} \\ &= \lim_{n\rightarrow\infty}1-\frac{1}{\ln(n)} \\ &= 1 \end{align} $$

So the prime number theorem can refer to either of the two approximations, $n/ln(n)$ and $\text{li}(n)$, because they are asymptotically equivalent.

What Does The Prime Number Theorem Really Say?

The prime number theorem says that $\pi(n)$ grows in a way that is asymptotically equivalent to functions like $n/\ln(n)$ and $\text{li}(n)$. 

It doesn't say that these are the only or best functions for approximating $\pi(n)$, which leaves open the intriguing possibility of other functions that are even better than $\text{li}(n)$.

Bertrand's Postulate

The prime number theorem, even if it looks imprecise, is rather useful. Let's look at a particularly simple example.

In 1845 Bertrand proposed that there is at least one prime between a counting number and its double, $n<p<2n$.

Let's compare the number of primes up to $2x$, with the number of primes up to $x$.

$$ \frac{\pi(2x)}{\pi(x)} \sim \frac{2x}{\ln(2x)} \cdot \frac{\ln(x)}{x} \sim 2 $$

This shows that for sufficiently large $x$, between $x$ and $2x$, there are $x/\ln(x)$ primes. This is a stronger statement than Bertrand's postulate which merely suggests there is at least one prime.

There are proofs that don't require $x$ to be sufficiently large, but they're nowhere near as simple as this deduction from the prime number theorem.

Historical References For $\pi(n)$

Gauss, 1791

Gauss’ 1791 ‘Some Asymptotic Laws Of Number Theory’ can be found in volume 10 of his collected works. In it he presents his approximation for $\pi(n)$.

$$ \frac{a}{la} $$

Today, this would be written as $n/ \ln(n)$.

Gauss’ 1971 Some Asymptotic Laws Of Number Theory.

Source: http://resolver.sub.uni-goettingen.de/purl?PPN236018647

Legendre, 1797

Legendre in his first edition of ‘Essai Sur La Theorie Des Nombres’ presented his approximation.

$$ \frac{a}{A \text{log}(a)+B} $$

The logarithm is the natural $\ln(a)$. In his 1808 second edition he quantifies the constants.

$$ \frac{x}{\text{log}(x) - 1.08366} $$

Legendre’s 1797 Essai Sur La Theorie Des Nombres.

Source: https://gallica.bnf.fr/ark:/12148/btv1b8626880r/f55. image

Gauss, 1849

Gauss wrote a letter to astronomer Encke dated Decemer 24th 1849, in which he first presents an integral form of a prime counting function. He states this is based on work he started in 1792 or 1793.

Gauss uses the following expression.

$$ \int \frac{dn}{\text{log}n} $$

Today this would be written as the logarithmic integral function.

$$ \int_0^n \frac{1}{\ln(x)} dx $$

First page of Gauss’ 1849 letter to Encke.

Source: https://gauss.adw-goe.de/handle/gauss/199

Experimental Maths - The Distribution Of Primes

Given primes are so resistant to encoding into a simple generating formula, let’s take a detour and try a different approach, experimental mathematics.

Video: [youtube]

Slides: [pdf]

Number of Primes Up To A Number

We showed that primes don’t run out as we explore larger and larger numbers. We also showed they don’t thin out as quickly as the squares. So how quickly do they thin out?

One way to explore this is to keep a count of the number of primes as we progress along the whole numbers.

The expression $\pi(n)$ has become an abbreviation for ‘the number of primes up to, and including, n’. For example, $\pi(5) = 3$ because there are 3 primes up to, and including, 5. The next number 6 is not prime, so $\pi(6)$ remains 3. The use of the symbol $\pi$ can be confusing at first.

The chart below shows $\pi(n)$ for $n$ up to 100. A fairly smooth curve seems to be emerging. This is slightly unexpected because the primes appear to be randomly placed amongst the numbers. The curve suggests the primes are governed by some kind of constraint. It wouldn’t be too adventurous to say the curve looks logarithmic, like $\ln(n)$. (Click to enlarge)

$\pi(n)$ for $n$ from 1 to 100.

Rather Good Approximations for $\pi(n)$

Gauss, one the most prolific mathematicians in history, was the first to find an expression that approximates $\pi(n)$ fairly well. He was aged about 15 at the time.

$$ π(n) \approx \frac{n}{\ln(n)} $$

The expression is surprisingly simple. It is worth pondering on what hidden pattern in the primes is captured by the natural logarithm $\ln(n)$.

Just a year later, Gauss developed a different expression that approximates $\pi(n)$ even more closely.

$$ \pi(n) \approx \int_0^n \frac{1}{\ln(x)} dx $$

At first glance, this logarithmic integral function, shortened to $\text{li}(n)$, appears to be a continuous form of the first approximation. 

The chart below shows a comparison of these approximations with the actual $\pi(n)$ for $n$ all the way up to 10,000. It's clear the logarithmic integral function is much closer to the actual prime counts.

Comparing $\text{li}(n)$ and $n/\ln(n)$ with $\pi(n)$.

Proportional Error

Looking again at the previous chart, the prime counting approximation $n/\ln(n)$ appears to be diverging away from the true prime count $\pi(n)$ as $n$ gets larger. That is, the error appears to be getting ever larger. If we looked at the numbers, we’d also see $\text{li}(n)$ diverging away from $\pi(n)$ too. Does this mean the approximations become useless as $n$ gets larger?

The chart below paints a different picture. It shows the error as a proportion of $\pi(n)$. We can see this proportional error becomes smaller as $n$ grows to 10,000. It's also clear that $\text{li}(n)$ has a distinctly smaller proportional error than $n/\ln(n)$.

Proportional errors for $\text{li}(n)$ and $n/\ln(n)$.

There are 1229 primes amongst the first 10,000 whole numbers. The logarithmic integral gives us $\text{li}(10,000)=1246$. The error is just 17, and as a proportion of 1229, an impressively small 0.0138.

If we extended $n$ to even larger values, we'd find the proportional error would fall further towards zero. Perhaps these approximations are correct in the limit $n\rightarrow\infty$?

Prime Density

Let’s look again at those approximations and see if we can interpret their form. The following compares Gauss’ first approximation with a general expression for calculating the mass of a volume of stuff with a given average density.

$$ \begin{align} \text{mass} &= \text{density} \times \text{volume} \\ \\ \pi(n) &\approx \frac{1}{\ln(n)} \times n \end{align} $$

The comparison suggests that $1/ \ln(n)$ is the average density of primes. If true, this would be a remarkable insight into the primes.

We can apply a similar analogy to Gauss’ second approximation too. This time we compare it with another general expression for calculating mass where the density is not assumed to be constant throughout its volume.

$$ \begin{align} \text{mass} &= \int (\text{density}) dv \\ \\ \pi(n) &\approx \int_0^n \frac{1}{\ln(x)} dx \end{align} $$

Again, $1/\ln(x)$ emerges as a more locally accurate density of primes around a number $x$.

It was this density of primes around a number that the young Gauss first noticed as he studied the number of primes in successive ranges of whole numbers, 1-1000, 1001-2000, 2001-3000, and so on.

Nth Prime?

If the density of primes is $1/ \ln(n)$ then we can say average distance between primes is $\ln(n)$. We can then make the short leap to say the $n^{th}$ prime is approximately $n\ln(n)$.

Before we get too excited about having found a simple function for generating the $n^{th}$ prime, this expression is an approximation, based on an average, itself based on Gauss’ approximation for $\pi(n)$.

It’s still interesting to see how well this expression for the $n^{th}$ prime performs. The chart below shows the error in $n\ln(n)$ as a proportion of the actual $n^{th}$ prime. As $n$ increases to 10,000,000, the proportional error falls to approximately 0.1017.

Proportional error of $n\ln(n)$ against the $n^{th}$ prime.

Looking at the graph, it is tempting to conclude the proportional error in $n\ln(n)$ approaches 0.1 as $n \rightarrow \infty$. We should be cautious because the error could be falling to zero very very slowly.

Anecdotal evidence is not mathematical proof, no matter how compelling it looks.

For example, our experiments show $\text{li}(n)$ is always a bit higher than the true $\pi(n)$. If we extended $n$ from 10,000 to 10,000,000 we’d still find $\text{li}(n)$ was always higher. But in 1914 Littlewood proved that $\text{li}(n)$ can become lower than $\pi(n)$ infinitely many times. More recent proofs show this starts to happen somewhere between $10^{19}$ and a mind-blowingly large $10^{316}$.

Imperfect History

The question of who first developed an approximation for $\pi(n)$ is not perfectly clear. Gauss didn’t always publish his work, leaving us to reconstruct history from notes and letters.

In his 1797 book on number theory, Legendre first published a form $n/(A \ln(n) + B)$, which he updated in his 1808 second edition to $n/(\ln(n) − 1.08366)$.

However, in 1849 Gauss wrote a letter to astronomer, and former student, Encke telling him that he had, in ‘1792 or 1793’, developed the logarithmic integral approximation, which he wrote as $\int\frac{dn}{\text{log}n}$. His collected works also reveal that in 1791 he had written about the simpler approximation, $\frac{a}{la}$ as he wrote it.

A separate blog post presents reproductions of the relevant parts of these historical works.

Primes Aren't That Spread Out

We’ve seen there is no limit to the supply of primes. A good question to ask next is how frequently they occur.

One way to explore this is by looking at the sum of their inverses, or reciprocals.

Video: [youtube]

Slides: [pdf]

Infinite Sum Of Reciprocals

The counting numbers $1, 2, 3, 4, \ldots$ are spaced 1 apart. The sum of their inverses is called the harmonic series.

$$ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots $$

This series is known to diverge, that is, the sum is infinitely large. A separate blog post has an easy short proof.

The square numbers $1, 4, 9, 16, \ldots$ are spaced further apart than the counting numbers.

$$ 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \ldots $$

The sum of their inverses converges.  A separate blog post walks through Euler’s historic and rather adventurous proof showing it converges to $\frac{\pi^2}{6}$.

We can interpret this to mean the squares $n^2$ are so spread out that the terms in the series become small quickly enough for the sum not to become infinitely large.

It is natural to then ask the same question about the primes. Are they so spread out that the infinite sum of their inverses converges too?

Infinite Sum Of Prime Reciprocals

Let’s start by assuming the infinite series of prime reciprocals does in fact converge to a finite sum $S$.

$$ S = \sum_{n=1} \frac{1}{p_n} $$

Because $S$ is finite, and each term is smaller than the previous one, there must be a value of $k$ such that the infinite series after $\frac{1}{p_k}$ sums to less than 1. We can call this sum $x$. 

$$ x = \sum_{n=k+1} \frac{1}{p_n} <1 $$

Let’s build an infinite geometric series based on this $x$.

$$ G = x + x^2 + x^3 + x^4 + \ldots $$

This new series $G$ converges because the ratio between terms $x$ is less than 1.

Let’s think a little more carefully about the terms in $G$. Any term in $G$ will be of the form $\frac{1}{N}$ where $N$ has prime factors $p_{k+1}$ or larger. This is because $x$ was intentionally constructed with primes $p_{k+1}$ and larger.

Now consider a second series F where, in contrast to G, the terms are constructed from all the primes $p_k$ and smaller.

$$ F = \sum_{j=1} \frac{1}{1 + j \cdot (p_1 \cdot p_2 \cdot p_3 \ldots p_k)}$$

Between each term, only $j$ changes. Now let’s look more closely at the expression $1 + j \cdot (p_1 \cdot p_2 \cdot p_3 \ldots p_k)$. This has no prime factors from the range $p_1$ to $p_k$. Since all whole numbers have prime factors, its prime factors must be from the set $p_{k+1}$ and larger.

That means $F$ is a subseries of $G$. That is, the terms of $F$ appear in the terms of $G$.

Now, if we compare the terms of $F$ to the harmonic series, we can test whether $F$ diverges.

We do this with the limit comparison test, which tests what happens to the ratio of terms from each series as they extend to infinity. If the ratio is finite, the series either both converge, or both diverge.

$$ \lim_{j \rightarrow} \frac{1 + j \cdot (p_1 \cdot p_2 \cdot p_3 \ldots p_k)}{j} = p_1 \cdot p_2 \cdot p_3 \ldots p_k $$

The ratio is finite, and since the harmonic series diverges, so does $F$.

Since $F$ diverges, and is a subseries of $G$, then $G$ must also diverge. But we constructed $G$ to converge. This contradiction proves the initial assumption that the infinite series of prime reciprocals converges was wrong.

That $\sum \frac{1}{p_n}$ diverges is a little surprising because our intuition was that primes thin out rather rapidly.

Legendre’s Conjecture

The fact that $\sum \frac{1}{n^2}$ converges suggests the primes are not as sparse as the squares. This leads us to an interesting proposal attributed to Legendre, but actually first published by Desboves in 1855, that there is at least one prime number between two consecutive squares.

$$ n^2 < p < (n+1)^2 $$

This remains a deep mystery of mathematics. Nobody has been able to prove or disprove it.

$\sum \frac{1}{n^2}$ Converges

The sum of the reciprocals of the square numbers was a particularly difficult challenge, first posed around 1650, and later named the Basel problem.

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

Although there are more modern proofs, we will follow Euler’s original proof from 1734 because his methods were pretty audacious, and later influenced Riemann’s work on the prime number theorem.

Slides: [pdf]

Video: [youtube]

Taylor Series For sin(x)

We start with the familiar Taylor series for $\sin(x)$. 

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

Euler’s New Series For sin(x)

The polynomial $f(x) = (1− \frac{x}{a})(1+ \frac{x}{a})$ has factors $(1− \frac{x}{a})$ and $(1+ \frac{x}{a})$, and zeros at $+a$ and $−a$. We can shorten it to $f(x)=(1−\frac{x^2}{a^2})$.

Euler’s novel idea was to write $\sin(x)$ as a product of similar linear factors, which would lead him to a different series.

The zeros of $\sin(x)$ are at $0, \pm π, \pm 2π, \pm 3π, \ldots$ so the product of factors looks like the following.

$$ \sin(x) = A \cdot x \cdot (1 - \frac{x^2}{\pi^2}) \cdot (1 - \frac{x^2}{(2\pi)^2}) \cdot (1 - \frac{x^2}{(3\pi)^2}) \cdot \ldots  $$

The constant $A$ is 1 because we know $\frac{\sin(x)}{x} \rightarrow 1$ as $x \rightarrow 0$. Alternatively, taking the first derivative of both sides gives $A = 1$ when $x = 0$.

The second factor is $x$ and not $x^2$ because the zero of $\sin(x)$ at $x = 0$ has multiplicity 1.

Euler then expanded out the series.

$$ \sin(x) = x \cdot [ 1 - \frac{x^2}{\pi^2} ( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots) + X] $$

Inside the square brackets, the terms with powers of $x$ higher than 2 are contained in $X$.

Comparing The Two Series

The terms in Euler’s new series and the Taylor series must be equivalent because they both represent $\sin(x)$. Let’s pick out the $x^3$ terms from both series.

$$ \frac{x^3}{3!} = \frac{x^3}{\pi^2} ( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots ) $$

We can easily rearrange this to give us the desired infinite sum.

$$ \frac{\pi^2}{6} =  \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots  $$

Euler, aged 28, had solved the long standing Basel problem, not only proving the infinite series of squared reciprocals converged, but giving it an exact value.

$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

Rigour

Euler’s original proof was adventurous in expressing $\sin(x)$ as an infinite product of simple linear factors. It made intuitive sense, but at the time was not rigorously justified.

It was almost 100 years later when Weierstrass developed and proved a factorisation theorem that confirmed Euler’s leap was legitimate.

$\sum \frac{1}{n}$ Diverges

Infinite Series

Have a look at the following infinite series.

$$ 1 + 1 + 1 + 1 + \ldots $$

We can easily see this sum is infinitely large. The series diverges.

The following shows a different infinite series. Each term is half the size of the previous one.

$$ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots $$

We can intuitively see this series gets ever closer to 2. Many would simply say the sum is in fact 2. The series converges.

Video: [youtube]

Slides: [pdf]

Harmonic Series

Now let’s look at this infinite series, called the harmonic series. 

$$ S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}+ \ldots $$

Each term is smaller than the previous one, and so contributes an ever smaller amount to the sum. Perhaps surprisingly, the harmonic series doesn’t converge. The sum is infinitely large.

The following, rather fun, proof is based on Oresme’s which dates back to the early 1300s.

We start by grouping the terms in the series as follows.

$$ S = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + \ldots $$

The brackets will have $2, 4, 8, 16 \ldots$ terms inside them. Replacing each term in a group by its smallest member gives us the following new series.

$$ \begin{align} T &= 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + \ldots \\ \\ &= 1 + \frac{1}{2} + (\frac{1}{2}) + (\frac{1}{2}) + \ldots \end{align} $$

We can see straight away this series diverges.

Because we replaced terms in $S$ by smaller ones to make $T$, we can say $S > T$ .

And because $T$ diverges, so must the harmonic series $S$.

Primes Are Rather Elusive

If we listed all the counting numbers $1, 2, 3, 4, 5, \ldots$, excluded 1, and then crossed out all the multiples of $2, 3, 4, \ldots$ we’d be left with the primes. This sieving process emphasises that primes are defined more by what they are not, than by what they are.

If there was a simple pattern in the primes, we’d be able to encode it into a simple formula for generating them. For example, the triangle numbers $1, 3, 6, 10, 15, \ldots$ can be generated by the simple expression $\frac{1}{2}n(n + 1)$. The prime numbers, however, have resisted attempts by mathematicians over hundreds of years to find precise and simple patterns in them.

One of the first questions anyone enthusiastic about prime numbers asks is whether a polynomial can generate the $n^{th}$ prime. Polynomials are both simple and rather flexible, and it would be quite pleasing if one could generate primes.

Let’s prove that prime numbers are so elusive that no simple polynomial in $n$ can generate the $n^{th}$ prime.

slides: [pdf]

video: [youtube]

No Simple Polynomial Generates Only Primes

A polynomial in $n$ has the following general form, simple yet flexible.

$$ P(n) = a + bn + cn^2 + dn^3 +\ldots + \alpha n^{\beta} $$

By simple polynomial we mean the coefficients $a,b,c \ldots \alpha$ are whole numbers. Let’s also  say that $b, c, d, \ldots \alpha$ are not all zero. This way we exclude trivial polynomials like $P(n) = 7$ that only generate a single value no matter what $n$ is.

Let’s start our proof by assuming there is indeed a $P(n)$ that generates only primes, given a counting number $n$. When $n = 1$, it generates a prime, which we can call $p_1$.

$$ p_1 = P(1) = a + b + c + d + \ldots + \alpha $$

Now let’s try $n = (1 + p_1)$.

$$ P(1+p_1) = a + b(1+p_1) + c(1+p_1)^2  + d(1+p_1)^3 + \ldots $$

That looks complicated, but all we need to notice is that if we expand out all the terms, we’ll have two kinds, those with $p_1$ as a factor, and those without. We can collect together all those terms with factor $p_1$ and call them $p_1 \cdot X$.

$$ P(1+p_1) = (a + b + c + d + e + \ldots \alpha) + p_1 ·X $$

We then notice that $(a + b + c + d + e + \ldots \alpha)$ is actually $p_1$.

$$ \begin{align} P(1+p_1) &= p_1 + p_1 \cdot X \\ \\ &= p_1(1 + X) \end{align} $$

Since $X$ is a whole number, this is divisible by $p_1$. It shouldn’t be because $P(1 + p_1)$ is supposed to be a prime. This contradiction means the starting assumption that there is a polynomial $P(n)$ that generates only primes is wrong.

We’ve actually proved a stronger statement than we intended. We intended to prove that there is no polynomial $P(n)$ that generates the $n^{th}$ prime. We ended up proving that no simple polynomial $P(n)$ can generate only primes.

Polynomials With Rational Coefficients

Insisting on integer coefficients for polynomials might seem overly restrictive. Let's broaden our definition to allow rational coefficients of the form $\frac{s}{t}$ where $s$ and $t$ are integers. 

We again assume $P(n)$ does indeed generate only primes, and so $p_{1}=P(1)$ is prime. This time we'll consider $n = (1+k\cdot p_{1})$.

$$ \begin{align} P(1+k\cdot p_{1}) &= a+b(1+k\cdot p_{1})+c(1+k\cdot p_{1})^{2}+\ldots \\ &= p_{1}+k\cdot p_{1}\cdot X \\ &= p_{1}(1+k\cdot X) \end{align} $$

Here $X$ contains terms that are combinations of the rational coefficients $a,b,c\dots\alpha$ multiplied together. We can choose a $k$ which cancels all the denominators of the rational coefficients leaving $k\cdot X$ as an integer. The lowest common multiple of all the denominators is one way to do this.

Our proof by contradiction then continues as before because we've found an example of $P(n)$ that is not prime.

The primes really are rather elusive if even polynomials with rational coefficients can't generate only primes.

Primes Are The Building Blocks Of Numbers

We saw earlier that positive whole numbers have factors if they’re not a prime number. Let’s explore this a little further.

video: [youtube]

slides: [pdf]

Breaking A Number Into Its Factors

Let’s think about the number 12 and its factors. We can think of two combinations straight away.

$$ \begin{align} 12 &= 2 \times 6 \\ 12 &= 3 \times 4 \end{align} $$

Looking again at those factors we can see that 6 itself can be broken down into smaller factors 3 and 2. That 4 can also be broken down into factors 2 and 2.

$$ \begin{align}  12 = 2 \times (3 \times 2) \\ 12 = 3 \times (2 \times 2) \end{align} $$

We can’t break these smaller factors down any further, which means they’re prime numbers. Both combinations now look very similar. If we put those factors in order of size, we can see they are in fact exactly the same.

$$ \begin{align}12 = 2 \times 2 \times 3 \\ 12 = 2 \times 2 \times 3 \end{align} $$

Perhaps every number can be broken down into a list of prime factors that is unique to that number, much like DNA is unique to people. Let’s prove it.

Fundamental Theorem Of Arithmetic

We’ll split this proof into two steps.

• First we’ll show that any positive whole number can be broken down into a list of factors that are all prime.

• Second we’ll show this list of primes is unique to that number. 

Let’s imagine a number $N$ and write it out as a product of factors.

$$ N =f_1 \cdot f_2 \cdot f_3 \cdot \ldots \cdot f_m $$

We can look at each of these factors $f_i$ in turn. If a factor is not prime, we can break it down into smaller factors. For example, the factor $f_1$ might be broken down as $f_1 = g_1 \cdot g_2$. If a factor is prime, $f_2 = p_1$ for example, we leave it because we can’t break it into smaller factors.

$$ N =(g_1 \cdot g_2) \cdot p_1 \cdot (g_3 \cdot g_4 \cdot g_5) \cdot \ldots \cdot (g_x \cdot g_y) $$

If we keep repeating this process, all the factors will eventually be prime. How can we be so sure? Well, if any number in the list isn’t prime, we can apply the process again, breaking that number down into smaller factors. The only thing that stops us applying the process again is when all the factors are eventually prime.

The following picture shows an example of this iterative process applied to the number 720. (Click to enlarge the picture)

We can now write $N$ as a product of these primes.

$$ N =p_2 \cdot p_3 \cdot p_1 \cdot p_5 \cdot p_4 \cdot p_6 \cdot p_7 \cdot \ldots \cdot p_n $$

These primes won’t necessarily be in order of size. They may also repeat, for example $p_1$ might be the same as $p_7$. It doesn’t matter. We’ve shown that any positive whole number can be written as a product of primes.

Let’s now show that this list of primes is unique to that number $N$. For the moment, imagine this isn’t true and a number $N$ can be written as a product of two different lists of primes.

$$ \begin{align} N &= p_1 \cdot p_2 \cdot p_3 \cdot \ldots \cdot p_a \\ N &= q_1 \cdot q_2 \cdot q_3 \cdot \ldots \cdot q_a \cdot q_b \cdot q_c \cdot q_d \end{align} $$

These primes are not necessarily in order of size, and some might be repeated, so $p_2$ could be the same as $p_4$. Again, we won’t let that bother us. To keep our argument general, we’ll assume that the number of primes in the second list, $d$, is larger than the number of primes in the first list, $a$.

Now, we can see that $p_1$ is a factor of $N$. That means it must also be a factor of the second list. That means $p_1$ is one of the factors $q_i$. Because we we didn’t assume any order in these primes, let’s say it is $q_1$. That means we can divide both lists by $p_1 = q_1$.

$$ \cancel{p_1} \cdot p_2 \cdot p_3 \cdot \ldots \cdot p_a = \cancel{q_1} \cdot q_2 \cdot q_3 \cdot ...\cdot q_a \cdot q_b \cdot q_c \cdot q_d $$

We can apply the same logic again. The first list has a factor $p_2$ which means it must also be a factor of the second list. We can say that $p_2 = q_2$, and divide both lists by this factor.

$$ \cancel{p_1} \cdot \cancel{p_2} \cdot p_3 \cdot \ldots \cdot p_a = \cancel{q_1} \cdot \cancel{q_2} \cdot q_3 \cdot ...\cdot q_a \cdot q_b \cdot q_c \cdot q_d $$

We can keep doing this until all the factors in the first list have been matched up with factors in the second list. It doesn’t matter if a prime repeats, for example if $p_1$ is the same as $p_3$, the factors will still be matched correctly, in this case $p_1 = q_1$ and $p_3 = q_3$.

$$ \cancel{p_1} \cdot \cancel{p_2} \cdot \cancel{p_3} \cdot \ldots \cdot \cancel{p_a} = \cancel{q_1} \cdot \cancel{q_2} \cdot \cancel{q_3} \cdot ...\cdot \cancel{q_a} \cdot q_b \cdot q_c \cdot q_d $$

Let’s simplify the algebra.

$$ 1=q_b \cdot q_c \cdot q_d $$

What we’ve just shown is that if a number $N$ can be written as two separate lists of prime factors, their factors can be paired up as being equal, and if any are left over, they must equal 1. That is, the two lists are identical.

We’ve shown that any whole number $N$ can be decomposed into a list of prime factors, and this list of primes is unique to that number. This is rather profound, and is called the Fundamental Theorem of Algebra. 

How Many Primes Are There?

At first thought it might seem obvious that there is an unending supply of prime numbers.

If we think a little longer, a bit of doubt might intrude on our certainty. A small number like 6 has factors 2 and 3. There aren’t many more options to try as factors. Larger numbers like 720 have more numbers smaller than them, and that means more numbers which could be factors.

Let’s think about this another way. Every multiple of 2 is not a prime number, every multiple of 3 is not a prime number, every multiple of 4 is not a prime number, and so on. All these multiples are reducing the probability that a large number is prime.

We might be tempted to think that eventually prime numbers just fizzle out. Instead of relying on intuition, let’s decide the matter with rigorous mathematical proof.

Video: [youtube]

Slides: [pdf]

Proof There Are Infinitely Many Primes

A proof is not an intuition, nor is it a set of convincing examples. A proof is a watertight logical argument that leads to a conclusion we can’t argue with.

The proof that there is no limit to the number of primes is ancient and rather elegant, due to Euclid around 300 BC, and a nice one to have as our first example.

Let's start by assuming the number of primes is not endless but finite. If there are $n$ primes, we can list them. 

$$ p_1, p_2, p_3, p_4 \ldots pn $$

We can create a new number $x$ by multiplying all these primes together.

$$ x = p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot \ldots \cdot p_n $$

This $x$ is clearly not a prime number. It’s full of factors like $p_1$ , $p_3$ and $p_n$.

Let’s make another number $y$ in the same way, but this time we’ll also add 1.

$$ y = p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot \ldots \cdot p_n + 1 $$

Now $y$ could be a prime number, or it could not be a prime number.

These are the only two options for any positive whole number.

If $y$ is prime then we have a problem because we’ve just found a new prime number which isn’t part of the original finite set $p_1, p_2 \ldots p_n$. How do we know it’s not part of the original set? Well $y$ is bigger than any of the primes in the list because we created it by multiplying them all together, and adding 1 for good measure.

So perhaps $y$ is not a prime. In this case, it must have factors. And the factors must be one or more of the known primes $p_1, p_2 \ldots p_n$. That means $y$ can be divided by one of those primes $p_i$ exactly, leaving no remainder. Let’s write this out.

$$ \frac{y}{p_i} = \frac{p_1 \cdot p2_ \cdot p_3 \cdot p_4 \cdot ... \cdot p_n}{p_i} + \frac{1}{p_i} $$

The first part divides neatly without a remainder because $p_i$ is one of the primes $p_1, p_2 \ldots p_n$. The second part doesn’t divide neatly at all. That means $y$ can’t be divided by any of known primes. Which again suggests it is a new prime, not in the original list.

Both of these options point to the original list of primes being incomplete.

And that’s the proof. No finite list of primes can be a complete list of primes. So there are infinitely many primes.

A Common Misunderstanding

It is easy to think that $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot \ldots \cdot p_n + 1$ is a way of generating prime numbers. This is not correct. The proof only asks what the consequences are if $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot \ldots \cdot p_n + 1$ is prime, under the assumption that we have a limited list of primes $p_1, p_2, p_3, p_4 .\ldots p_n$.

We can prove that $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot \ldots \cdot p_n + 1$ is not always prime by finding just one counter-example. If we use prime numbers 2, 3, 5, 7, 11 and 13, we can see that $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1 = 30031$ which is not prime because $30031 = 59 \cdot 509$.

What Are Prime Numbers?

Let's start by looking at the most ordinary numbers we know, the counting numbers.

$$ 1, 2, 3, 4, 5, 6, 7, 8, \ldots $$

We became familiar with these numbers when we were just toddlers, counting apples in a bowl, for example.

Video: [youtube]

Slides: [pdf]

Multiplication

We soon learned to add and multiply these numbers. Many of us learned our times tables by heart. Almost without thinking we could recite multiplications like $2 \times 4 = 8$, and $5 \times 5 = 25$.

When we multiply 3 by 4, the answer is 12. This 12 is called a product, and the 3 and 4 are called factors.

If we pick any two numbers $a$ and $b$ and multiply them, the result is another number, which we can call $c$.

$$ a \times b = c $$

Because $a$ and $b$ are whole numbers, so is $c$.

An Innocent Question

Those factors $a$ and $b$ can be any counting number we feel like choosing. Does this freedom apply to $c$ as well?

Surely some combination of $a$ and $b$ can give us any number $c$ that we desire. Let’s try a couple of examples.

• If we want $c$ to be 12, we could choose $a = 3$ and $b = 4$. We could have chosen $a = 2$ and $b = 6$, and that would work too.

• If we want $c$ to be 100, we could choose $a = 2$ and $b = 50$. Another combination that works is $a = 10$ and $b = 10$.

What if we want $c$ to be 7?

If we try for a short while, we’ll find there doesn’t seem to be a combination of factors $a$ and $b$ that gives 7 as a product. In fact, if we try all the numbers in the range $2 \ldots 6$ we’ll see for ourselves there really is no combination that gives $a \times b = 7$.

What if we want $c$ to be 11? Again we’ll find no combination of whole number factors gives 11 as a product.

So the answer to our innocent-looking question is no, $c$ can’t be any whole number.

Numbers like 7 and 11 that don’t have whole number factors, are called prime numbers. Here are the first few.

$$ 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53, \ldots $$

In short, if we multiply two counting numbers, the answer is never a prime number.

What About 1?

You might have spotted that when we were trying to find factors of 7 we didn’t consider combinations like $a = 1$ and $b = 7$. That’s because we exclude 1 as a legitimate factor. Why? Because every number has 1 as a factor, and that’s not particularly interesting.

If we didn’t exclude 1, there would be no prime numbers because every number $c$ would have factors $a = 1$ and $b = c$.

Even worse, a number could have lots of factors as 1, which is also rather unhelpful. The number 12 could have an infinite number of factors.

$$ 12 = 4 × 3 × 1 × 1 × 1 × 1 × 1 × 1 × 1 \ldots $$

Negative Numbers?

Prime numbers were known about and discussed in ancient times, well before the idea of a negative number was accepted.

Over the hundreds of years since then, new ideas and insights were developed about prime numbers, and they were built on the original assumption that prime numbers could only be positive whole numbers.

Today almost all exploration of prime numbers continues under the same constraint that products, factors and primes are positive whole numbers greater than 1. This constraint really doesn’t limit the mysteries and surprises that prime numbers hold.

Apparent Randomness

Looking back at the list of prime numbers, there doesn't seem to be a pattern to them. Apart from never being even numbers, with the exception of 2, they seem to be fairly randomly located along the number line. 

For hundreds of years, mathematicians puzzled over the primes, attacking them with all sorts of exotic tools, trying to crack them open to reveal any elusive rules that govern their location. That endeavour continues to this day. 

Welcome!

This is a blog that will accompany our journey from the simple prime numbers to the foothills of the Riemann Hypothesis, one of the most important and still open challenges in mathematics.

The aim is to make the subjects we encounter along the way as accessible as possible. That means we wont' assume you are a university trained mathematician, but we will assume you studied maths at secondary school - so are familiar with concepts like convergence of series, calculus, and complex numbers.

Along the way we will encounter historic proofs such as the infinitude of the primes, techniques such as comparing discrete sums with continuous functions, important ideas such as the Prime Number Theorem, and we'll be using blogs, videos and code to explore and explain those ideas.