Euler's Formula

Graphic created using Gemini

Courtesy of Gemini:

Often cited as the most beautiful theorem in mathematics, Euler’s identity (\(e^{i\pi} + 1 = 0\)) achieves its fame by elegantly linking the most fundamental constants and operations in a single, concise equation.

Here is the history behind its discovery and the mathematical proof of why it works.

The History Behind the Identity

The equation is a specific case of Euler’s formula, which states that for any real number \(x\):

\[e^{ix} = \cos(x) + i\sin(x)\]

While the identity bears the name of the brilliant Swiss mathematician Leonhard Euler, the groundwork was laid slightly earlier. In 1714, the English mathematician Roger Cotes discovered a precursor to this relationship, expressing it in terms of natural logarithms: \(\ln(\cos x + i\sin x) = ix\).

However, it was Euler who, in 1748, formalized the mathematics into the modern exponential form we use today. He published it in his foundational text Introductio in analysin infinitorum (Introduction to the Analysis of the Infinite).

The identity is celebrated because it connects five foundational mathematical constants from entirely different disciplines:

• \(0\): The additive identity (the foundation of arithmetic).
• \(1\): The multiplicative identity (the foundation of counting).
• \(\pi\): The fundamental constant of geometry (the ratio of a circle's circumference to its diameter).
• \(e\): The fundamental constant of growth and continuous change (the base of natural logarithms).
• \(i\): The imaginary unit (\(\sqrt{-1}\)), which unlocks complex algebra.

It also seamlessly utilizes the three most basic mathematical operations: addition, multiplication, and exponentiation.

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Why the Equation Holds True

To prove why this works, we must look to Taylor series (specifically, Maclaurin series expansions). A Maclaurin series allows us to express complex functions as an infinite sum of polynomial terms.

The series expansion for the exponential function \(e^x\) is:

\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots\]

If we substitute \(x = i\theta\) into this series, we get:

\[e^{i\theta} = 1 + (i\theta) + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \dots\]

Next, we evaluate the powers of \(i\). By definition, \(i = \sqrt{-1}\), which creates a repeating pattern:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)
• \(i^5 = i\)

Substituting these evaluated powers back into our series gives:

\[e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \dots\]

Now, we group the real terms (those without an \(i\)) and the imaginary terms (those multiplied by \(i\)) together:

\[e^{i\theta} = \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots \right) + i\left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \right)\]

This is where Euler's brilliant connection becomes visible. These two grouped polynomial expansions perfectly match the standard Maclaurin series expansions for the trigonometric functions cosine and sine:

\[\cos(\theta) = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots\]

\[\sin(\theta) = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots\]

By substituting \(\cos(\theta)\) and \(\sin(\theta)\) back into our grouped equation, we arrive at Euler's formula:

\[e^{i\theta} = \cos(\theta) + i\sin(\theta)\]

Reaching the Final Identity

To get Euler's Identity, we evaluate the formula at \(\theta = \pi\) radians (which corresponds to \(180\) degrees on the unit circle):

\[e^{i\pi} = \cos(\pi) + i\sin(\pi)\]

Looking at a standard trigonometric unit circle, we know that:

• \(\cos(\pi) = -1\)
• \(\sin(\pi) = 0\)

Substituting these values in:

\[e^{i\pi} = -1 + 0i\]

\[e^{i\pi} = -1\]

By simply adding \(1\) to both sides, we arrive at the final state of the identity, linking the five constants perfectly together:

\[e^{i\pi} + 1 = 0\]

Here is a Gemini link to information about the mathematicians Taylor and MacLaurin.

Palindromic Day 28182

Palindromic properties of 28182 (showing only sequence members up to 40000):

A098834: palindromic Smith numbers.

4, 22, 121, 202, 454, 535, 636, 666, 1111, 1881, 3663, 7227, 7447, 9229, 10201, 17271, 22522, 24142, 28182, 33633, 38283

A Smith number is a composite number where the sum of its digits equals the sum of the digits of its prime factors. For 28182:$$ \begin{align} 28182 &\rightarrow 2+8+1+8+2 = 21 \\ 28182 &= 2 \times 3 \times 7 \times 11 \times 61\\ &\rightarrow 2 + 3 + 7 + 1+1+6 +1 =21 \end{align}$$

A046395: palindromes that are the product of 5 distinct primes.

6006, 8778, 20202, 28182

Here \(28182 = 2 \times 3 \times 7 \times 11 \times 61 \)

A099052: all palindromes of length > 1 in the decimal expansion of \(e\).

\(e\) = 2.71828182845904523536028747135266 ...

A045571: numbers that are palindromic, divisible by 11 and have an odd number of digits.

121, 242, 363, 484, 616, 737, 858, 979, 10901, 11011, 12221, 13431, 14641, 15851, 17171, 18381, 19591, 20702, 21912, 22022, 23232, 24442, 25652, 26862, 28182, 29392, 30503, 31713, 32923, 33033, 34243, 35453, 36663, 37873, 39193

All the palindromic numbers with an even number of digits are divisible by 11. The number of palindromic numbers with \(2k+1\) digits that are divisible by 11 is \((10^{k+1} + (-1)^k)/11\), and their asymptotic relative density within the set of all palindromic numbers with an odd number of digits is 1/11 (from OEIS comments).

A113838: palindromes sandwiched between twin primes.

4, 6, 282, 828, 858, 2112, 21012, 21612, 23832, 26262, 26862, 28182

Here of course the twin primes are 28181 and 28183.

A032751: palindromic Super-3 Numbers.

4554, 6776, 17471, 22322, 22722, 28182

Super-3 numbers \(n\) are of the form \(3 \times n^3 \) and contain three consecutive 3's.

Here \(3 \times 28182^3 = 67148557\textbf{333}704\)

Simple Permutations

One of the properties associated with my diurnal age today (28146) is that it's a member of OEIS A111111:

A111111: number of simple permutations of degree \(n\).

The initial members of this sequence beginning with \(n=1\) up to \(n=9\) are 1, 2, 0, 2, 6, 46, 338, 2926 and 28146.

I did not understand the explanation in the OEIS comments about what a simple permutation of degree \(n\) was so I turned to Gemini for help and its response forms the content of this blog post. I got NotebookLM to create a video overview of the post and then uploaded it to YouTube.

Understanding Simple Permutations

To understand OEIS A111111, it helps to think of the problem in terms of finding "hidden chunks" or "blocks" within a sequence of numbers.

The core rule is this: a permutation is simple if you cannot find any contiguous sub-sequence of length 2 or more (but less than the total length) that contains a continuous range of numbers, regardless of the order those numbers appear in.

Deconstructing the Example: 2647513

The comment in the sequence notes that the permutation 2647513 is not simple because of the interval of indices [2..5]. Let's map the positions (indices) to their actual values.

• Position 1: 2
• Position 2: 6
• Position 3: 4
• Position 4: 7
• Position 5: 5
• Position 6: 1
• Position 7: 3

If we look strictly at positions 2 through 5, the numbers present are 6, 4, 7, and 5. If we sort that subset of numbers, we get 4, 5, 6, 7. This forms a perfect, unbroken mathematical sequence (an interval). Because we found a contiguous block of positions that maps to a contiguous block of values, and this block is smaller than the entire sequence but larger than a single number, the permutation fails the test. It is not simple.

Why is n=3 exactly 0?

Looking at the sequence (1, 2, 0, 2, 6, 46...), the third term is 0. This means there are absolutely zero simple permutations of length 3. We can prove this by testing all six possible permutations of the numbers 1, 2, and 3:

• 123: Positions 1-2 contain {1, 2}. (Fails)
• 132: Positions 2-3 contain {3, 2}, which sorts to {2, 3}. (Fails)
• 213: Positions 1-2 contain {2, 1}, which sorts to {1, 2}. (Fails)
• 231: Positions 1-2 contain {2, 3}. (Fails)
• 312: Positions 2-3 contain {1, 2}. (Fails)
• 321: Positions 1-2 contain {3, 2}, which sorts to {2, 3}. (Fails)

Every single permutation of length 3 contains an embedded chunk of length 2 that forms a continuous range. Therefore, none of them are simple.

The n=4 Case: The Two Survivors

The sequence tells us there are exactly 2 simple permutations for n=4. Out of the 24 possible combinations, only two survive the test: 2413 and 3142.

Let's test 2413 to see why it passes:

Check all chunks of length 2:

• Positions 1-2: {2, 4} — Gap between 2 and 4.
• Positions 2-3: {4, 1} — Gap between 1 and 4.
• Positions 3-4: {1, 3} — Gap between 1 and 3.

Check all chunks of length 3:

• Positions 1-3: {2, 4, 1} — Sorts to {1, 2, 4}. Missing the 3.
• Positions 2-4: {4, 1, 3} — Sorts to {1, 3, 4}. Missing the 2.

Because no sub-sequence forms an unbroken chain of numbers, 2413 is a simple permutation. The same logic applies to 3142, making them the only two simple permutations of degree 4.

I then asked Gemini the following:

So in the case of n=9 we are looking at all permutations of the digits 123456789 and finding that there are 28146 simple permutations? Is that correct? Also, can you explain what is meant by a permutation of degree n. Is there a formula for calculating the number of simple permutations of degree n? 

The n=9 Case

For n=9, you are looking at all possible arrangements of the digits 1 through 9. There are 362880 total possible permutations for those nine digits (which is 9 factorial). If you were to test every single one of those 362880 arrangements for hidden consecutive blocks, you would find that exactly 28146 of them are simple permutations.

What is a "Permutation of Degree n"?

In combinatorial mathematics, the "degree" simply refers to the total number of elements in the set you are arranging.

• A permutation of degree 4 means you are finding the arrangements of a set containing exactly 4 distinct items (usually the integers 1, 2, 3, and 4).
• A permutation of degree n just means you are arranging the integers from 1 up to n.

It is just a formal way of stating the length of the string of numbers you are working with.

Is There a Formula?

There is no simple, single-line closed-form algebraic formula to instantly calculate the exact number of simple permutations for a specific degree. You cannot just plug n into a basic equation to get the answer.

Instead, mathematicians calculate exact values using generating functions and recurrence relations. The foundational logic is based on substitution decomposition: the idea that every standard permutation can be uniquely constructed by taking a simple permutation and "inflating" its individual points with other, smaller permutations. By knowing the total number of permutations (n!), mathematicians can work backward recursively to strip away the complex ones and isolate the count of simple ones.

However, there is a very elegant asymptotic formula for this sequence.

As the degree n becomes very large, the number of simple permutations, denoted as \( S_n\), can be approximated by:$$ S_n \approx \frac{n!}{e^2}$$Here, e is Euler's number. Because e² is approximately 7.389056, this formula reveals a fascinating property of permutations: as the length of the number string approaches infinity, the probability that a randomly generated permutation will be simple approaches 1 / e², or roughly 13.53%.

So, while calculating the exact integer for a specific degree requires heavy recursive computation, calculating the approximate scale of simple permutations for massive degrees is surprisingly straightforward.

I asked Gemini to give an example for the case of \(n=9\):

An Example for \(n=9\)

Here is one of the 28146 simple permutations of the digits 1 through 9:

4 1 7 3 9 6 2 8 5

To verify that this permutation is simple, we can apply the same logic used previously. We must ensure that no subset of contiguous positions (from length 2 up to length 8) contains a continuous range of values.

Why This Example Works

The easiest way to prove it is simple is to test the difference between the highest and lowest values within various blocks. If the difference between the maximum and minimum values in a block equals the length of the block minus 1, it forms a continuous interval.

• Testing Length 2 Blocks: Let's look at the adjacent pairs: {4, 1}, {1, 7}, {7, 3}, {3, 9}, {9, 6}, {6, 2}, {2, 8}, and {8, 5}. None of these pairs are consecutive numbers. The smallest gap is 3 (between 1 and 4, 6 and 9, 5 and 8). Because no pair has a difference of exactly 1, there are no length 2 intervals.
• Testing Length 4 Blocks: If we take the first four positions (4, 1, 7, 3), the highest value is 7 and the lowest is 1. The span is 6, which is much larger than the block length of 4. If we take the last four positions (6, 2, 8, 5), the highest is 8 and the lowest is 2. The span is 6. Because the numbers are scattered, they do not pack tightly enough to form an interval.
• Testing Length 8 Blocks: If we take the first eight numbers (4, 1, 7, 3, 9, 6, 2, 8), the values range from 1 to 9. The only missing number is 5. Because 5 is missing from the middle of the set, the remaining numbers cannot form an unbroken sequence. The same applies to the last eight numbers (1, 7, 3, 9, 6, 2, 8, 5), which are missing the 4.

Because this "scattering" holds true for every possible block size between 2 and 8, the permutation strictly avoids all intervals and perfectly satisfies the definition of a simple permutation.

I asked Gemini the following: 

Of what use are simple permutations? Are there any practical applications or is the concept purely theoretical.

While simple permutations originated as a purely theoretical concept in combinatorial mathematics, they have proven to be surprisingly useful in several practical, real-world applications. Because mathematics is the language of structure, these "irreducible" sequences provide the foundational logic for any field that needs to analyze, dismantle, or scramble data.

Here is a breakdown of how simple permutations bridge the gap between pure theory and practical application.

The "Prime Numbers" of Permutations (Theoretical Foundation)

Before looking at applied fields, it is helpful to understand their role in pure mathematics. Just as every integer can be factored into a unique set of prime numbers, every single permutation can be uniquely broken down into a "tree" of simple permutations through a process called substitution decomposition. In this sense, simple permutations are the fundamental, indivisible building blocks of all mathematical sequences.

Bioinformatics and Evolutionary Genomics

One of the most robust practical applications of this concept is in computational biology, specifically regarding genome rearrangement.

When scientists compare the DNA sequences of different species, they look at how chunks of genes have moved around over millions of years of evolution. A sequence of genes in a fruit fly might be a permuted version of the same genes in a mosquito. If a block of genes moves together as a single, unbroken unit during evolution, it forms an "interval" (meaning the permutation of that genome is not simple).

Simple permutations represent the extreme cases of genetic mutation: complete and total structural rearrangement where no contiguous blocks of genes have been preserved intact. By programming algorithms to search for and isolate these simple permutation structures, geneticists can map exactly where and how evolutionary divergence occurred.

Computer Science and Algorithm Optimization

In computational analysis, particularly when writing custom data analysis scripts in Python or SageMath, efficiency is critical. Because every complex sequence can be mathematically decomposed into a structural tree of simple permutations, computer scientists use these trees to drastically speed up sorting and search algorithms.

Problems involving "permutation pattern matching" that would normally take an impossible amount of exponential computing time can often be reduced to manageable polynomial time. The algorithm simply breaks the massive dataset down into its simple permutation components first, and then solves the problem on those smaller, irreducible blocks.

Cryptography and Data Diffusion

A secure cryptographic cipher relies on two primary principles: confusion and diffusion. Diffusion requires that a small, localized change in the input data is spread completely across the entire output data.

Because the strict definition of a simple permutation means it perfectly avoids forming neat, predictable blocks of adjacent numbers, it represents a state of maximum "scattering." This mathematical framework helps cryptographers design the substitution-permutation networks (the architecture behind modern encryption standards like AES) that resist pattern analysis and keep digital communications secure.

While you may never see a simple permutation explicitly named in a piece of commercial software, the mathematical rules governing them are silently working behind the scenes to optimize data, track genetic history, and secure information.

From Polynomial to Linear Recurrence

I noticed that the number associated with my diurnal age today (28080) is a member of OEIS  A144459 formed from a quadratic polynomial:$$ \text{a}(n) = (3n+1)(5n+1) $$The initial members are:

1, 24, 77, 160, 273, 416, 589, 792, 1025, 1288, 1581, 1904, 2257, 2640, 3053, 3496, 3969, 4472, 5005, 5568, 6161, 6784, 7437, 8120, 8833, 9576, 10349, 11152, 11985, 12848, 13741, 14664, 15617, 16600, 17613, 18656, 19729, 20832, 21965, 23128, 24321, 25544, 26797, 28080

What interested me was that the sequence members could also be generated from a linear recurrence:$$ \begin{align} \text{a}(n) = 3\text{a}(n-1) -3\text{a}(n-2) +\text{a}(n-3) \\ \text{with } \text{a}(0)=1, \text{a}(1)=24 \text{ and } \text{a}(2)=77 \end{align}$$How do we get from \( \text{a}(n) = (3n+1)(5n+1) \text{ to } \text{a}(n-1) -3\text{a}(n-2) +\text{a}(n-3) \)? Gemini's response was a follows:

Deriving the Linear Recurrence for OEIS A144459

To derive the linear recurrence \(a(n) = 3a(n-1) - 3a(n-2) + a(n-3)\) from the explicit formula \(a(n) = (3n+1)(5n+1)\), we utilize the Method of Finite Differences.

This method relies on a fundamental rule of sequence algebra: Any sequence defined by a polynomial of degree \(d\) satisfies a linear recurrence relation of order \(d+1\) with binomial coefficients.

Step 1: Determine the Degree of the Polynomial

First, expand the explicit formula to see the structure of the polynomial:

$$ a(n) = (3n+1)(5n+1) $$ $$ a(n) = 15n^2 + 3n + 5n + 1 $$ $$ a(n) = 15n^2 + 8n + 1 $$

Since the highest power of \(n\) is 2, this is a quadratic polynomial (degree \(d=2\)).

Step 2: Apply the Difference Operator Rule

In the calculus of finite differences, we use the difference operator \(\Delta\). The rule states that if \(a(n)\) is a polynomial of degree 2:

• The first differences are linear (degree 1).
• The second differences are constant (degree 0).
• The third differences are exactly zero.

Mathematically, this means:

$$ \Delta^3 a(n) = 0 $$

Step 3: Expand using Shift Operators

We express the difference operator \(\Delta\) in terms of the shift operator \(E\), where \(E a(n) = a(n+1)\). We know that \(\Delta = E - 1\).

Since the third difference is zero, we substitute \((E-1)\) for \(\Delta\):

$$ (E - 1)^3 a(n) = 0 $$

Now, we expand the cubic term using standard binomial expansion:

$$ (E^3 - 3E^2 + 3E - 1) a(n) = 0 $$

Step 4: Apply the Operators to the Sequence

Distribute the operator \(a(n)\) across the expanded terms. Recall that \(E^k a(n) = a(n+k)\):

$$ a(n+3) - 3a(n+2) + 3a(n+1) - a(n) = 0 $$

To match the standard recurrence form (where the highest term is \(a(n)\)), we shift the indices down by 3 (replace \(n\) with \(n-3\)):

$$ a(n) - 3a(n-1) + 3a(n-2) - a(n-3) = 0 $$

Step 5: Rearrange to the Final Recurrence

Move the negative terms to the right side to isolate \(a(n)\):

$$ a(n) = 3a(n-1) - 3a(n-2) + a(n-3) $$

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Verification

We verify this by checking the initial values given for A144459.

Explicit Formula: \(a(n) = 15n^2 + 8n + 1\)

• \(a(0) = 1\), \(a(1) = 24\), \(a(2) = 77\)

Recurrence Check for \(a(3)\):

$$ a(3) = 3(77) - 3(24) + 1 $$ $$ a(3) = 231 - 72 + 1 $$ $$ a(3) = 160 $$

This matches the explicit calculation \(15(3)^2 + 8(3) + 1 = 160\).

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Here is another example just to reinforce what's been covered earlier.

1. The Transformation Process

To convert the polynomial closed form into a linear recurrence relation, we use the method of finite differences. Since the polynomial \( a_n = 2n^2 + 6n + 1 \) is of degree \( d=2 \), the sequence has a constant second difference and a zero third difference.

Step 1: Set up the Difference Operator

We define the shift operator \( E \) such that \( E a_n = a_{n+1} \). The difference operator is \( \Delta = E - 1 \). Because the degree is 2, the third difference is zero: \[ \Delta^3 a_n = 0 \]

Step 2: Expand the Operator

Substitute \( \Delta = E - 1 \) and expand the cubic expression: \[ (E - 1)^3 a_n = 0 \] \[ (E^3 - 3E^2 + 3E - 1) a_n = 0 \]

Step 3: Apply the Shift Operator

Distributing the operator to \( a_n \) gives the recurrence relation in terms of forward shifts: \[ a_{n+3} - 3a_{n+2} + 3a_{n+1} - a_n = 0 \]

Step 4: Final Linear Recurrence

To solve for the current term based on previous terms, we shift the indices down by 3 (replace \( n+3 \) with \( n \)): \[ \boxed{ a_n = 3a_{n-1} - 3a_{n-2} + a_{n-3} } \]

Initial Conditions:

We calculate the first three seeds using the original polynomial \( 2n^2 + 6n + 1 \):

• \( a_0 = 1 \)
• \( a_1 = 9 \)
• \( a_2 = 21 \)

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2. Comparison and Verification

The table below compares the values generated by the closed form polynomial against the derived linear recurrence.

\[ \begin{array}{|c|c|c|c|} \hline \textbf{n} & \textbf{Closed Form} & \textbf{Recurrence} & \textbf{Match} \\ & (2n^2 + 6n + 1) & (3a_{n-1} - 3a_{n-2} + a_{n-3}) & \\ \hline 0 & 1 & 1 & \text{True} \\ \hline 1 & 9 & 9 & \text{True} \\ \hline 2 & 21 & 21 & \text{True} \\ \hline 3 & 37 & 37 & \text{True} \\ \hline 4 & 57 & 57 & \text{True} \\ \hline 5 & 81 & 81 & \text{True} \\ \hline 6 & 109 & 109 & \text{True} \\ \hline 7 & 141 & 141 & \text{True} \\ \hline 8 & 177 & 177 & \text{True} \\ \hline 9 & 217 & 217 & \text{True} \\ \hline \end{array} \]

An Interesting Limit

Here is a report generated by Gemini Pro based on the content of this video. I found the video moved a little too quickly at times and the author read out every mathematical expression symbol by symbol which wasn't really necessary and it didn't aid comprehension. It's useful to be able to take the content of the video and put into a report that can be read at your leisure. If you want to tweak the LaTeX code, you can do so easily and I've done that below for a few of expressions.

The Limit of the nth Root: An Exploration of Euler's Sequence

At first glance, the limit of the sequence involving factorials and powers can seem intuitive, but as with many problems in analysis, the "obvious" answer can be misleading. This report dismantles the problem of finding the limit of the nth root of n factorial divided by n to the power of n. Following the path likely walked by Leonhard Euler, we move from discrete sequences to continuous calculus to find a beautiful truth connecting factorials, limits, and Euler’s number, e.

1. The Sequence and Intuition

We begin by defining our sequence \( a_n \):

$$ a_n = \frac{n!}{n^n} $$

Intuitively, the denominator \( n^n \) grows much faster than the numerator \( n! \), suggesting that the terms drop toward zero. To prove this rigorously, we employ the Ratio Test. We investigate the ratio of consecutive terms:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} $$

Simplifying the factorials (\( (n+1)! = (n+1)n! \)) and regrouping terms gives:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left( \frac{n}{n+1} \right)^n $$

This expression can be rewritten to reveal a fundamental definition of Euler's number:

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{\left( \frac{n+1}{n} \right)^n} = \lim_{n \to \infty} \frac{1}{\left( 1 + \frac{1}{n} \right)^n} = \frac{1}{e} $$

Since \( \dfrac{1}{e} \approx 0.3679 \) is strictly less than 1, the Ratio Test confirms that the sequence converges to 0.

2. The Main Problem: The nth Root

Establishing that the base sequence goes to 0 is just the preamble. The true challenge is finding the limit of the nth root of this sequence:

$$ L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} $$

Since the base approaches 0 and the exponent \( \frac{1}{n} \) also approaches 0, we are facing the indeterminate form \( 0^0 \). To resolve this, we use the natural logarithm to transform the product into a sum, bridging the gap between discrete algebra and integral calculus.

3. From Logarithms to Riemann Sums

Let \( y = \sqrt[ \uproot{4} n]{a_n} \). Taking the natural log of the limit allows us to work with sums:

$$ \ln(L) = \lim_{n \to \infty} \ln \left( \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} \right) = \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) $$

Using logarithm rules, we expand the term inside:

$$ \ln \left( \frac{n!}{n^n} \right) = \ln(n!) - \ln(n^n) = \sum_{k=1}^{n} \ln(k) - n \ln(n) $$

We can cleverly rewrite \( n \ln(n) \) as a sum of \( n \) identical terms: \( \displaystyle \sum_{k=1}^{n} \ln(n) \). Substituting this back into our limit equation gives:

$$ \ln(L) = \lim_{n \to \infty} \frac{1}{n} \left( \sum_{k=1}^{n} \ln(k) - \sum_{k=1}^{n} \ln(n) \right) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \ln(k) - \ln(n) \right) $$

Combining the logs, we arrive at a recognizable form:

$$ \ln(L) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} \right) $$

This structure—a sum of function values multiplied by a small width \( \frac{1}{n} \) is precisely the definition of a Riemann Sum on the interval \([0, 1]\). As \( n \to \infty \), the sum converges to a definite integral:

$$ \ln(L) = \int_{0}^{1} \ln(x) \, dx $$

4. Evaluating the Integral

We now evaluate the improper integral of \( \ln(x) \) from 0 to 1. Using integration by parts (where \( u = \ln(x) \) and \( dv = dx \)), we get:

$$ \int \ln(x) \, dx = x \ln(x) - x $$

We evaluate the definite integral by taking the limit as the lower bound approaches 0:

$$ \Bigl[ x \ln(x) - x \Bigr]_0^1 = (1 \ln(1) - 1) - \lim_{a \to 0^+} (a \ln(a) - a) $$

We know \( \ln(1) = 0 \), so the first term is \( -1 \). For the limit \( \lim_{a \to 0^+} a \ln(a) \), we use L'Hôpital's rule, which shows that this term vanishes to 0. Thus:

$$ \ln(L) = -1 - 0 = -1 $$

Conclusion

We have found that the natural logarithm of the limit is \( -1 \). To find the original limit \( L \), we simply exponentiate both sides:

$$ L = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}}=e^{-1} = \frac{1}{e} $$

We have proven that while the sequence \( \dfrac{n!}{n^n} \) collapses to zero, its geometric mean—the nth root—stabilizes at exactly 1 over e. This problem serves as an elegant bridge connecting basic sequence properties, the definition of e, and the power of integral calculus.

Primes Within Transcendental Numbers

One of the properties of the number associated with my diurnal age today, 27943, is that it is a prime formed by merging five successive digits in the decimal expansion of \(e \). Primes with this property form OEIS A104846. This got me thinking about whether all possible primes could be expressed by merging successive digits in the decimal expansions of all transcendental numbers like \(e\) and \( \pi \).  

However, as Grok explained:

No, it cannot be proven that the decimal expansions of transcendental numbers like \( \pi \) or \(e\) contain representations of all prime numbers through progressive concatenations of consecutive digits. This interpretation hinges on whether every possible finite sequence of digits—including the decimal representations of all primes—appears as a substring somewhere in their infinite, non-repeating expansions. While this would be true if \( \pi \) and \(e\) are normal numbers (where every digit sequence of length \(k\) appears with equal frequency \(1/10^k\) in the limit), the normality of \( \pi \) and \(e\) remains an open problem in mathematics. 

If numbers like \( \pi \) and \(e\) could be proven to be normal then not only would every prime occur but it would occur an infinite number of times. It's interesting to look at 27943 and see how many times it appears in the first million digits of \(e\). Here are the initial positions for the "2" (permalink):

166, 46606, 217029, 234391, 248347, 258504, 483659, 832027, 941722

For \( \pi \), it takes a while before the first occurrence but afterwards the occurrences of 27943 are more frequent (permalink):

94943, 146738, 216793, 253032, 283291, 290571, 302660, 350882, 358637, 520562, 632389, 661691, 715324, 805161, 843112, 890626

I wondered if there was any difference between algebraic numbers like \( \phi \), the golden ratio, and transcendental numbers in regards to being able to represent all primes as concatenations of successive digits. Here's was Grok's response:

No, it cannot be proven that the decimal expansions of algebraic irrational numbers like the golden ratio (\( \phi  \approx 1.6180339887 \dots \)) contain representations of all prime numbers through progressive concatenations of consecutive digits, for essentially the same reasons as with transcendentals like \( \pi \) and \(e\). Their normality in base 10 remains unproven. 

Just as with π and e, the golden ratio's decimal expansion is infinite and non-repeating (since it's irrational), but this doesn't guarantee it contains all sequences. Empirical checks of millions of digits suggest randomness consistent with normality, but no proof exists—not even that all digits 0-9 appear infinitely often. It's widely conjectured that φ (a quadratic irrational algebraic number) is normal in base 10, mirroring beliefs about π and e. Proving this would be a major result in number theory.

There's a specific conjecture (dating back to Borel in 1909, refined later) that every irrational algebraic number is absolutely normal (normal in every integer base \(\geq 2)\). This includes \( \phi, \sqrt{2}, \sqrt{3} \), and others. No counterexamples exist, but crucially, no irrational algebraic has ever been proven normal in any base. In contrast, for transcendentals, normality isn't conjectured for all—only for certain "natural" ones like \( \phi \) and \(e\)—while many non-normal transcendentals are known (e.g., Liouville numbers designed to lack certain digits). 

Proven examples: We have constructed transcendentals proven to be normal, such as Champernowne's constant ( 0.123456789101112 ...)  or the Copeland–Erdős constant (formed by concatenating primes: 0.235711131719... ). These do contain every finite digit sequence infinitely often, including all primes. No such proven-normal algebraic irrationals exist, partly because algebraics satisfy polynomial equations, which might impose subtle constraints on their expansions (though the conjecture suggests not). 

In summary, algebraic irrationals like \( \phi \) aren't fundamentally "different" in the sense that the core question—whether all primes appear as substrings—remains unproven and conjectured true, just like for \( \phi\) and \(e\). The main distinction lies in the broader landscape: Algebraics have a stronger universal conjecture without exceptions or proven cases, while transcendentals show a mix of proven normal and non-normal examples. If the conjectures hold, then yes, \( \phi \)'s expansion would contain all primes, but we can't prove it yet.

So at what positions does the prime 27943 appear in the decimal expansion of \( \phi \). Here are the locations (permalink):

33280, 127164, 129935, 529433, 597796, 795399, 971620

Farey Fractions

The number associated with my diurnal age today is 27634 and one its properties qualifies for membership in OEIS A119015:

A119015: denominators of "Farey fraction" approximations to \(e\).

I couldn't understand the explanation in the OEIS comments about how these fractions were formed but an investigation of the number in this blog revealed that I'd already dealt with these types of fractions as applied to \(e\) before in a post titled The Mediant on the 5th November 2022. The explanation in that post is much easier to understand but in it I'd only considered the fractions themselves and not their numerators and denominators considered separately, as part of different sequences. The denominator sequence is linked above and here is the corresponding sequence for the numerators:

A119014: numerators of "Farey fraction" approximations to \(e\).

Here are the fractions again (permalink):

5/2, 8/3, 11/4, 19/7, 30/11, 49/18, 68/25, 87/32, 106/39, 193/71, 299/110, 492/181, 685/252, 878/323, 1071/394, 1264/465, 1457/536, 2721/1001, 4178/1537, 6899/2538, 9620/3539, 12341/4540, 15062/5541, 17783/6542, 20504/7543, 23225/8544, 25946/9545, 49171/18089, 75117/27634, 124288/45723, 173459/63812, 222630/81901, 271801/99990, 320972/118079, 370143/136168, 419314/154257, 468485/172346, 517656/190435, 566827/208524, 1084483/398959

Here are the Farey fraction denominators (permalink):

2, 3, 4, 7, 11, 18, 25, 32, 39, 71, 110, 181, 252, 323, 394, 465, 536, 1001, 1537, 2538, 3539, 4540, 5541, 6542, 7543, 8544, 9545, 18089, 27634, 45723, 63812, 81901, 99990, 118079, 136168, 154257, 172346, 190435, 208524, 398959

Here are the Farey fraction numerators (permalink):

5, 8, 11, 19, 30, 49, 68, 87, 106, 193, 299, 492, 685, 878, 1071, 1264, 1457, 2721, 4178, 6899, 9620, 12341, 15062, 17783, 20504, 23225, 25946, 49171, 75117, 124288, 173459, 222630, 271801, 320972, 370143, 419314, 468485, 517656, 566827, 1084483

There are OEIS entries for the numerators and denominators of the Farey fraction approximations to most of the standard constants e.g. \( \pi \). Here are the Farey fractions together with their numerators and denominators for \( \pi \):

The progressive list of approximating Farey fractions is given by (permalink):

[7/2, 10/3, 13/4, 16/5, 19/6, 22/7, 25/8, 47/15, 69/22, 91/29, 113/36, 135/43, 157/50, 179/57, 201/64, 223/71, 245/78, 267/85, 289/92, 311/99, 333/106, 355/113, 688/219, 1043/332, 1398/445, 1753/558, 2108/671, 2463/784, 2818/897, 3173/1010, 3528/1123, 3883/1236, 4238/1349, 4593/1462, 4948/1575, 5303/1688, 5658/1801, 6013/1914, 6368/2027, 6723/2140, 7078/2253, 7433/2366, 7788/2479, 8143/2592, 8498/2705, 8853/2818, 9208/2931, 9563/3044, 9918/3157, 10273/3270, 10628/3383, 10983/3496, 11338/3609, 11693/3722, 12048/3835, 12403/3948, 12758/4061, 13113/4174, 13468/4287, 13823/4400, 14178/4513, 14533/4626, 14888/4739, 15243/4852, 15598/4965, 15953/5078, 16308/5191, 16663/5304, 17018/5417, 17373/5530, 17728/5643, 18083/5756, 18438/5869, 18793/5982, 19148/6095, 19503/6208, 19858/6321, 20213/6434, 20568/6547, 20923/6660, 21278/6773, 21633/6886, 21988/6999, 22343/7112, 22698/7225, 23053/7338, 23408/7451, 23763/7564, 24118/7677, 24473/7790, 24828/7903, 25183/8016, 25538/8129, 25893/8242, 26248/8355, 26603/8468, 26958/8581, 27313/8694, 27668/8807, 28023/8920, 28378/9033, 28733/9146, 29088/9259, 29443/9372, 29798/9485, 30153/9598, 30508/9711, 30863/9824, 31218/9937, 31573/10050, 31928/10163, 32283/10276, 32638/10389, 32993/10502, 33348/10615, 33703/10728, 34058/10841, 34413/10954, 34768/11067, 35123/11180, 35478/11293, 35833/11406, 36188/11519, 36543/11632, 36898/11745, 37253/11858, 37608/11971, 37963/12084, 38318/12197, 38673/12310, 39028/12423, 39383/12536, 39738/12649, 40093/12762, 40448/12875, 40803/12988, 41158/13101, 41513/13214, 41868/13327, 42223/13440, 42578/13553, 42933/13666, 43288/13779, 43643/13892, 43998/14005, 44353/14118, 44708/14231, 45063/14344, 45418/14457, 45773/14570, 46128/14683, 46483/14796, 46838/14909, 47193/15022, 47548/15135, 47903/15248, 48258/15361, 48613/15474, 48968/15587, 49323/15700, 49678/15813, 50033/15926, 50388/16039, 50743/16152, 51098/16265, 51453/16378, 51808/16491, 52163/16604, 52518/16717, 52873/16830, 53228/16943, 53583/17056, 53938/17169, 54293/17282, 54648/17395, 55003/17508, 55358/17621, 55713/17734, 56068/17847, 56423/17960, 56778/18073, 57133/18186, 57488/18299, 57843/18412, 58198/18525, 58553/18638, 58908/18751, 59263/18864, 59618/18977, 59973/19090, 60328/19203, 60683/19316, 61038/19429, 61393/19542, 61748/19655, 62103/19768, 62458/19881, 62813/19994, 63168/20107, 63523/20220, 63878/20333, 64233/20446, 64588/20559, 64943/20672, 65298/20785, 65653/20898, 66008/21011, 66363/21124, 66718/21237, 67073/21350, 67428/21463, 67783/21576, 68138/21689, 68493/21802, 68848/21915, 69203/22028, 69558/22141, 69913/22254, 70268/22367, 70623/22480, 70978/22593, 71333/22706, 71688/22819, 72043/22932, 72398/23045, 72753/23158, 73108/23271, 73463/23384, 73818/23497, 74173/23610, 74528/23723, 74883/23836, 75238/23949, 75593/24062, 75948/24175, 76303/24288, 76658/24401, 77013/24514, 77368/24627, 77723/24740, 78078/24853, 78433/24966, 78788/25079, 79143/25192, 79498/25305, 79853/25418, 80208/25531, 80563/25644, 80918/25757, 81273/25870, 81628/25983, 81983/26096, 82338/26209, 82693/26322, 83048/26435, 83403/26548, 83758/26661, 84113/26774, 84468/26887, 84823/27000, 85178/27113, 85533/27226, 85888/27339, 86243/27452, 86598/27565, 86953/27678, 87308/27791, 87663/27904, 88018/28017, 88373/28130, 88728/28243, 89083/28356, 89438/28469, 89793/28582, 90148/28695, 90503/28808, 90858/28921, 91213/29034, 91568/29147, 91923/29260, 92278/29373, 92633/29486, 92988/29599, 93343/29712, 93698/29825, 94053/29938, 94408/30051, 94763/30164, 95118/30277, 95473/30390, 95828/30503, 96183/30616, 96538/30729, 96893/30842, 97248/30955, 97603/31068, 97958/31181, 98313/31294, 98668/31407, 99023/31520, 99378/31633, 99733/31746, 100088/31859, 100443/31972, 100798/32085, 101153/32198, 101508/32311, 101863/32424, 102218/32537, 102573/32650, 102928/32763, 103283/32876, 103638/32989, 103993/33102, 104348/33215, 208341/66317, 312689/99532, 521030/165849, 833719/265381, 1146408/364913, 1980127/630294, 3126535/995207, 4272943/1360120]

Here are the Farey fraction numerators for \( \pi \) (permalink):

7, 10, 13, 16, 19, 22, 25, 47, 69, 91, 113, 135, 157, 179, 201, 223, 245, 267, 289, 311, 333, 355, 688, 1043, 1398, 1753, 2108, 2463, 2818, 3173, 3528, 3883, 4238, 4593, 4948, 5303, 5658, 6013, 6368, 6723, 7078, 7433, 7788, 8143, 8498, 8853, 9208, 9563, 9918, 10273, 10628, 10983, 11338, 11693, 12048, 12403, 12758, 13113, 13468, 13823, 14178, 14533, 14888, 15243, 15598, 15953, 16308, 16663, 17018, 17373, 17728, 18083, 18438, 18793, 19148, 19503, 19858, 20213, 20568, 20923, 21278, 21633, 21988, 22343, 22698, 23053, 23408, 23763, 24118, 24473, 24828, 25183, 25538, 25893, 26248, 26603, 26958, 27313, 27668, 28023, 28378, 28733, 29088, 29443, 29798, 30153, 30508, 30863, 31218, 31573, 31928, 32283, 32638, 32993, 33348, 33703, 34058, 34413, 34768, 35123, 35478, 35833, 36188, 36543, 36898, 37253, 37608, 37963, 38318, 38673, 39028, 39383, 39738, 40093, 40448, 40803, 41158, 41513, 41868, 42223, 42578, 42933, 43288, 43643, 43998, 44353, 44708, 45063, 45418, 45773, 46128, 46483, 46838, 47193, 47548, 47903, 48258, 48613, 48968, 49323, 49678, 50033, 50388, 50743, 51098, 51453, 51808, 52163, 52518, 52873, 53228, 53583, 53938, 54293, 54648, 55003, 55358, 55713, 56068, 56423, 56778, 57133, 57488, 57843, 58198, 58553, 58908, 59263, 59618, 59973, 60328, 60683, 61038, 61393, 61748, 62103, 62458, 62813, 63168, 63523, 63878, 64233, 64588, 64943, 65298, 65653, 66008, 66363, 66718, 67073, 67428, 67783, 68138, 68493, 68848, 69203, 69558, 69913, 70268, 70623, 70978, 71333, 71688, 72043, 72398, 72753, 73108, 73463, 73818, 74173, 74528, 74883, 75238, 75593, 75948, 76303, 76658, 77013, 77368, 77723, 78078, 78433, 78788, 79143, 79498, 79853, 80208, 80563, 80918, 81273, 81628, 81983, 82338, 82693, 83048, 83403, 83758, 84113, 84468, 84823, 85178, 85533, 85888, 86243, 86598, 86953, 87308, 87663, 88018, 88373, 88728, 89083, 89438, 89793, 90148, 90503, 90858, 91213, 91568, 91923, 92278, 92633, 92988, 93343, 93698, 94053, 94408, 94763, 95118, 95473, 95828, 96183, 96538, 96893, 97248, 97603, 97958, 98313, 98668, 99023, 99378, 99733, 100088, 100443, 100798, 101153, 101508, 101863, 102218, 102573, 102928, 103283, 103638, 103993, 104348, 208341, 312689, 521030, 833719, 1146408, 1980127, 3126535, 4272943 (OEIS A097545)

Here are the Farey fraction denominators for \( \pi \) (permalink):

2, 3, 4, 5, 6, 7, 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99, 106, 113, 219, 332, 445, 558, 671, 784, 897, 1010, 1123, 1236, 1349, 1462, 1575, 1688, 1801, 1914, 2027, 2140, 2253, 2366, 2479, 2592, 2705, 2818, 2931, 3044, 3157, 3270, 3383, 3496, 3609, 3722, 3835, 3948, 4061, 4174, 4287, 4400, 4513, 4626, 4739, 4852, 4965, 5078, 5191, 5304, 5417, 5530, 5643, 5756, 5869, 5982, 6095, 6208, 6321, 6434, 6547, 6660, 6773, 6886, 6999, 7112, 7225, 7338, 7451, 7564, 7677, 7790, 7903, 8016, 8129, 8242, 8355, 8468, 8581, 8694, 8807, 8920, 9033, 9146, 9259, 9372, 9485, 9598, 9711, 9824, 9937, 10050, 10163, 10276, 10389, 10502, 10615, 10728, 10841, 10954, 11067, 11180, 11293, 11406, 11519, 11632, 11745, 11858, 11971, 12084, 12197, 12310, 12423, 12536, 12649, 12762, 12875, 12988, 13101, 13214, 13327, 13440, 13553, 13666, 13779, 13892, 14005, 14118, 14231, 14344, 14457, 14570, 14683, 14796, 14909, 15022, 15135, 15248, 15361, 15474, 15587, 15700, 15813, 15926, 16039, 16152, 16265, 16378, 16491, 16604, 16717, 16830, 16943, 17056, 17169, 17282, 17395, 17508, 17621, 17734, 17847, 17960, 18073, 18186, 18299, 18412, 18525, 18638, 18751, 18864, 18977, 19090, 19203, 19316, 19429, 19542, 19655, 19768, 19881, 19994, 20107, 20220, 20333, 20446, 20559, 20672, 20785, 20898, 21011, 21124, 21237, 21350, 21463, 21576, 21689, 21802, 21915, 22028, 22141, 22254, 22367, 22480, 22593, 22706, 22819, 22932, 23045, 23158, 23271, 23384, 23497, 23610, 23723, 23836, 23949, 24062, 24175, 24288, 24401, 24514, 24627, 24740, 24853, 24966, 25079, 25192, 25305, 25418, 25531, 25644, 25757, 25870, 25983, 26096, 26209, 26322, 26435, 26548, 26661, 26774, 26887, 27000, 27113, 27226, 27339, 27452, 27565, 27678, 27791, 27904, 28017, 28130, 28243, 28356, 28469, 28582, 28695, 28808, 28921, 29034, 29147, 29260, 29373, 29486, 29599, 29712, 29825, 29938, 30051, 30164, 30277, 30390, 30503, 30616, 30729, 30842, 30955, 31068, 31181, 31294, 31407, 31520, 31633, 31746, 31859, 31972, 32085, 32198, 32311, 32424, 32537, 32650, 32763, 32876, 32989, 33102, 33215, 66317, 99532, 165849, 265381, 364913, 630294, 995207, 1360120 (OEIS A097546)

The progressive list of approximating Farey fractions for \( \sqrt{2} \) is given by (permalink):

3/2, 4/3, 7/5, 10/7, 17/12, 24/17, 41/29, 58/41, 99/70, 140/99, 239/169, 338/239, 577/408, 816/577, 1393/985, 1970/1393, 3363/2378, 4756/3363, 8119/5741, 11482/8119, 19601/13860, 27720/19601, 47321/33461, 66922/47321, 114243/80782, 161564/114243, 275807/195025, 390050/275807, 665857/470832, 941664/665857, 1607521/1136689

Here are the Farey fraction numerators for \( \sqrt{2} \) (permalink):

3, 4, 7, 10, 17, 24, 41, 58, 99, 140, 239, 338, 577, 816, 1393, 1970, 3363, 4756, 8119, 11482, 19601, 27720, 47321, 66922, 114243, 161564, 275807, 390050, 665857, 941664, 1607521 (OEIS A119016)

Here are Farey fraction denominators for \( \sqrt{2} \) (permalink):

2, 3, 5, 7, 12, 17, 29, 41, 70, 99, 169, 239, 408, 577, 985, 1393, 2378, 3363, 5741, 8119, 13860, 19601, 33461, 47321, 80782, 114243, 195025, 275807, 470832, 665857, 1136689 (OEIS A002965)

Other terms that are used in this context are "interleave denominators" and "interleave numerators". The Farey fractions for the golden ratio are the successive ratios of Fibonacci numbers (higher term as numerator and lower term as denominator). Permalink.

Now what about the mathematician to whom these kinds of fractions owe their name. Well, he wasn't really a mathematician but rather a geologist, although he does have an entry in MacTutor biographies of mathematicians. Here is a link. Some details follow:

Born on the 24th of September 1766 in Woburn, Bedfordshire, England

Died on the 6th January 1826 in London, England

Summary: John Farey was an Engish geologist, noted as a mathematician for the Farey sequence which is a listing of the rationals.

Some Interesting Constants

There was an interesting tweet by Cliff Pickover today that is shown below as Figure 1.

Figure 1

The URL shown in the tweet leads to the Wikipedia article about it. First and formost however, who was Gelfond? As usual the MacTutor site at St.Andrews provides a brief biography.

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Born: 24th October 1906 in St Petersburg, Russia

Died: 7th November 1968 in Moscow, Russia. 

Summary

Gelfond developed basic techniques in the study of transcendental numbers.

Biography 

Aleksandr Osipovich Gelfond's father was Osip Isaacovich Gelfond who was a physician who also had an interest in philosophy. Gelfond entered Faculty of Physics and Mathematics at Moscow State University in 1924 and completed his undergraduate studies in 1927. He then began research under the supervision of Aleksandr Khinchin and Vyacheslaw Stepanov and completed his postgraduate studies in 1930.

During 1929-30 he taught mathematics at Moscow Technological College but already he had published some important papers: The arithmetic properties of entire functions (1929); Transcendental numbers (1929); and An outline of the history and the present state of the theory of transcendental numbers (1930). The second of these 1929 papers contained the lecture which Gelfond gave to the First All-Union Mathematics Congress held in Kharkov in 1930. These papers by Gelfond represent a major step forward in the study of transcendental numbers. The first of the papers examines the growth of an entire function which assumes integer values for integer arguments. In the second of the 1929 papers Gelfond applied this result to prove that certain numbers are transcendental, so solving a special case of Hilbert's Seventh Problem. We explain some of these ideas below.

In an article he wrote, Gelfond describes the four month visit which he made in 1930 to Germany where he spent time at both Berlin and Göttingen. He was particularly influenced by Hilbert, Siegel and Landau during his visit. After his return to Russia, Gelfond taught mathematics from 1931 at Moscow State University where he held chairs of analysis, theory of numbers and the history of mathematics. From 1933 he also worked in the Mathematical Institute of the Russian Academy of Sciences.

Gelfond developed basic techniques in the study of transcendental numbers, that is numbers that are not the solution of an algebraic equation with rational coefficients. In addition to his important work in the number theory of transcendental numbers, Gelfond made significant contributions to the theory of interpolation and the approximation of functions of a complex variable. He also contributed to the study of differential and integral equations and to the history of mathematics.

In 1934 he proved a special case of his conjecture namely that \(a^x\) is transcendental if \(a\) is algebraic (\(a \neq 0,1 \)) and \(x\) is an irrational, algebraic number. This result is now known as Gelfond's theorem and solved Problem 7 of the list of Hilbert problems. It was solved independently by Schneider. In 1966 Alan Baker proved Gelfond's Conjecture in general. Gelfond's papers in 1933 and 1934, which include his remarkable achievement, are:  

• Gram determinants for stationary series (written jointly with Khinchin) (1933) 

• A necessary and sufficient criterion for the transcendence of a number (1933) 

• Functions that take integer values at the points of a geometric progression (1933) 

• On the seventh problem of D Hilbert (1934) 

Gelfond addressed the Second All-Union Mathematics Congress in Leningrad in 1934) on Transcendental numbers.

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The decimal expansion of Gelfond's Constant is:$$e^{\pi} = 23.14069263277926900572 \dots$$The Wikipedia article states that \(e^{\pi} \) also appears in the volumes of hyperspheres. The volume of an \(n\)-sphere with radius \(R\) is given by:$$V_n(R) = \frac{\pi^{n/2} R^n}{\Gamma (n/2+1)}$$where \(R\) is the gamma function. If we take \(R=1\) we get:$$V_n(1) = \frac{\pi^{n/2} }{\Gamma (n/2+1)}$$Any even dimensional \(2n\)-sphere now gives:$$V_{2n}(1) = \frac{\pi^{n} }{\Gamma (n+1)}$$Summing up all even-dimensional unit sphere volumes and utilizing the series expansion of the exponential function gives:$$ V_{2n}(1) = \sum_{0}^{\infty} \frac{\pi^{n}}{n!}=e^{\pi}$$The Wikipedia article also looks at some related functions and the first of them is Ramanujan's Constant:$$ \begin{align} e^{\pi \sqrt{163}} &= 262537412640768743.99999999999925007259 \dots \\ &\approx 640320^3 + 744 \text{ to within one trillionth}\end{align}$$The article goes on to say that:

This is an application of Heegner numbers, where 163 is the Heegner number in question. This number was discovered in 1859 by the mathematician Charles Hermite. In a 1975 April Fool article in Scientific American magazine, "Mathematical Games" columnist Martin Gardner made the hoax claim that the number was in fact an integer, and that the Indian mathematical genius Srinivasa Ramanujan had predicted it—hence its name. Ramanujan's constant is also a transcendental number.

There is also the transcendental constant \(e^{\pi}-\pi \) which approximates to:$$e^{\pi}-\pi =19.99909997918947576726 \dots$$The constant \(i^i\) can be evaluated as follows:$$ \begin{align} i^i &= (e^{i\pi /2})^i\\ &= e^{-\pi/2} \\&=(e^{\pi})^{-1/2}\\ &= 0.20787957635076190854 \dots \end{align}$$As for \( \pi^e \), it is not known whether it is transcendental or not.

Following the link above for Charles Hermite yields the following biography. I've included it in full because I believe it's important to promote awareness of the lives of famous mathematicians to whom we owe so much.

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Born: 24 December 1822 in Dieuze, Lorraine, France

Died: 14 January 1901 in Paris, France 

Summary

Charles Hermite's work in the theory of functions includes the application of elliptic functions to the quintic equation. He published the first proof that e is a transcendental number.

Biography

Charles Hermite's father was Ferdinand Hermite and his mother was Madeleine Lallemand. Ferdinand Hermite was a trained engineer and he worked in this capacity in a salt mine near Dieuse. After he married Madeleine he joined in the draper's trade in which her family were involved. However he was an artistic man who always wanted to pursue art as a career. He had his wife look after the draper's business and he took up art. Charles was the sixth of his parents seven children and when he was about seven years old his parents left Dieuse and went to live in Nancy to where the business had moved.

Education was not a high priority for Charles's parents but despite not taking too much personal interest in their children's education, nevertheless they did provide them with good schooling. Charles was something of a worry to his parents for he had a defect in his right foot which meant that he moved around only with difficulty. It was clear that this would present him with problems in finding a career. However he had a happy disposition and bore his disability with a cheerful smile.

Charles attended the Collège de Nancy, then went to Paris where he attended the Collège Henri. In 1840-41 he studied at the Collège Louis-le-Grand where some fifteen years earlier Galois had studied. In fact he was taught mathematics there by Louis Richard who had taught Galois. In some ways Hermite was similar to Galois for he preferred to read papers by Euler, Gauss and Lagrange rather than work for his formal examinations.

If Hermite neglected the studies that he should have concentrated on, he was showing remarkable research ability publishing two papers while at Louis-le-Grand. Also like Galois he was attracted by the problem of solving algebraic equations and one of the two papers attempted to show that the quintic cannot be solved in radicals. That he was unfamiliar with Galois's contributions, despite being at the same school, is not at all surprising since the mathematical community were completely unaware of them at this time. However he might reasonably have known of the contributions of Ruffini and Abel to this question, but apparently he did not.

Again like Galois, Hermite wanted to study at the École Polytechnique and he took a year preparing for the examinations. He was tutored by Catalan in 1841-42 and certainly Hermite fared better than Galois had done for he passed. However it was not a glorious pass for he only attained sixty-eighth place in the ordered list. After one year at the École Polytechnique Hermite was refused the right to continue his studies because of his disability. Clearly this was an unfair decision and some important people were prepared to take up his case and fight for him to have the right to continue as a student at the École Polytechnique. The decision was reversed so that he could continue his studies but strict conditions were imposed. Hermite did not find these conditions acceptable and decided that he would not graduate from the École Polytechnique.

Hermite made friends with important mathematicians at this time and frequently visited Joseph Bertrand. On a personal note this was highly significant for he would marry Joseph Bertrand's sister. More significantly from a mathematical point of view he began corresponding with Jacobi and, despite not shining in his formal education, he was already producing research which was ranking as a leading world-class mathematician. The letters he exchanged with Jacobi show that Hermite had discovered some differential equations satisfied by theta-functions and he was using Fourier series to study them. He had found general solutions to the equations in terms of theta-functions. Hermite may have still been an undergraduate but it is likely that his ideas from around 1843 helped Liouville to his important 1844 results which include the result now known as Liouville's theorem.

After spending five years working towards his degree he took and passed the examinations for the baccalauréat and licence which he was awarded in 1847. In the following year he was appointed to the École Polytechnique, the institution which had tried to prevent him continuing his studies some four years earlier; he was appointed répétiteur and admissions examiner.

Hermite made important contributions to number theory and algebra, orthogonal polynomials, and elliptic functions. He discovered his most significant mathematical results over the ten years following his appointment to the École Polytechnique. In 1848 he proved that doubly periodic functions can be represented as quotients of periodic entire functions. In 1849 Hermite submitted a memoir to the Académie des Sciences which applied Cauchy's residue techniques to doubly periodic functions. Sturm and Cauchy gave a good report on this memoir in 1851 but a priority dispute with Liouville seems to have prevented its publication.

Another topic on which Hermite worked and made important contributions was the theory of quadratic forms. This led him to study invariant theory and he found a reciprocity law relating to binary forms. With his understanding of quadratic forms and invariant theory he created a theory of transformations in 1855. His results on this topic provided connections between number theory, theta functions, and the transformations of abelian functions.

On 14 July 1856 Hermite was elected to the Académie des Sciences. However, despite this achievement, 1856 was a bad year for Hermite for he contracted smallpox. It was Cauchy who, with his strong religious conviction, helped Hermite through the crisis. This had a profound effect on Hermite who, under Cauchy's influence, turned to the Roman Catholic religion. Cauchy was also a very staunch royalist and Hermite was influenced by him to also become a royalist. We made comparisons with Galois earlier on in this article, but with royalist views, Hermite was now completely opposed to the views which the staunch republican Galois had held.

The next mathematical result by Hermite which we must mention is one for which he is rightly famous. Although an algebraic equation of the fifth degree cannot be solved in radicals, a result which was proved by Ruffini and Abel, Hermite showed in 1858 that an algebraic equation of the fifth degree could be solved using elliptic functions. He applied these results to number theory, in particular to class number relations of quadratic forms.

In 1862 Hermite was appointed maître de conférence at the École Polytechnique, a position which had been specially created for him. In the following year he became an examiner there. The year 1869 saw him become a professor when he succeeded Duhamel as professor of analysis both at the École Polytechnique and at the Sorbonne. Hermite resigned his chair at the École Polytechnique in 1876 but continued to hold the chair at the Sorbonne until he retired in 1897. In the 1890s Hermite became much less interested in the new results found by the mathematicians of the next generation.

The 1870s saw Hermite return to problems which had interested him earlier in his career such as problems concerning approximation and interpolation. In 1873 Hermite published the first proof that e is a transcendental number. This is another result for which he is rightly famous. Using method's similar to those of Hermite, Lindemann established in 1882 that π was also transcendental. Many historians of science regret that Hermite, despite doing most of the hard work, failed to use it to prove the result on which would have brought him fame outside the world of mathematics. Hermite is now best known for a number of mathematical entities that bear his name: Hermite polynomials, Hermite's differential equation, Hermite's formula of interpolation and Hermitian matrices.

For Hermite certain areas of mathematics were much more interesting than other areas. Hadamard, who unlike his teacher Hermite worked in all areas of mathematics, spoke of Hermite's dislike for geometry:

[Hermite] had a kind of positive hatred of geometry and once curiously reproached me with having made a geometrical memoir.

Hermite's great love was for analysis and, not surprisingly, he had a great respect for Weierstrass. When Mittag-Leffler arrived in Paris to study with him, Hermite greeted him warmly but said:

You have made a mistake, sir, you should follow Weierstrass's course in Berlin. He is the master of us all.

Poincaré is almost certainly the best known of Hermite's students. He once suggested that Hermite's mind did not proceed in logical fashion. He wrote:

But to call Hermite a logician! Nothing can appear to me more contrary to the truth. Methods always seemed to be born in his mind in some mysterious way.

Hadamard like Poincaré was very interested in the way that mathematics was discovered. He also had this to say about the way that Hermite made his discoveries:

Hermite used to observe [that biology] may be a most useful study even for mathematicians, as hidden and eventually fruitful analogies may appear between processes in both kinds of studies.

Hadamard had great respect for Hermite as a teacher. He said:

I do not think that those who never listened to him can realise how magnificent Hermite's teaching was, overflowing with enthusiasm for science, which seemed to come to life in his voice and whose beauty he never failed to communicate to us, since he felt it so much himself to the very depth of his being.

[Hermite] was making a deep impression on us, not only with his methods and those of Weierstrass, but also with his enthusiasm and love of science; in our brief but fruitful conversations, Hermite loved to direct to me remarks such as: "He who strays from the paths traced by providence crashes." These were the words of a profoundly religious man, but an atheist like me understood them very well, especially when he added at other times: "In mathematics, our role is more of servant than of master." It goes without saying that gradually, as years and my scientific work unfolded, I came to understand more and more deeply the aptness and scope of his words.

Cross, reviewing where 125 letters from Hermite to Mittag-Leffler are reproduced, writes:

So there stands revealed one of the most engaging and influential men in Parisian and French mathematics in the second half of the 19th century, one might even say the central character for the period in which he published, 1842-1901. What radiates from the text is [Hermite's] humility, his Catholicism, his concern for his (very extended) family, his willingness to fight for colleagues whose merit he discerns, and his devotion to family, merit, and principle rather than simple influence.

In terms of his family life Hermite had married Louise Bertrand, Joseph Bertrand's sister. One of their two daughters married Émile Picard. Struik writes:

Hermite lived a retired life, with his family. His working hours were devoted to mathematical research and teaching. His outlook on mathematics was realistic in the Platonic sense: a mathematician, like a naturalist, discovers an outside world, in his case a world of ideas. Hermite, therefore, disliked Cantor's world, in which a new mathematical world was created.

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