Bernoulli Numbers - Explanation

Augusta Ada King-Noel, Countess of Lovelace (10 December 1815 – 27 November 1852) was an English mathematician and writer, chiefly known for her work on Charles Babbage’s proposed mechanical general-purpose computer, the Analytical Engine. [1].
She is widely regarded as the first computer programmer. She wrote an algorithm to calculate the Bernoulli numbers, for more please visit the previous blog here.
So, I thought to study and analyze Bernoulli numbers. $\mathcal Crazy \ math \ ahead!$
$B_0 = 1; B_1= \pm \frac{1}{2}; B_2 = \frac{1}{6}; B_3 = 0; B_4=-\frac{1}{30}; B_5 = 0...$

Approach 1 :

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Let us begin,
Definition: The Bernoulli numbers are defined as the co-efficients of the power series of the expansion of $\large \frac{t}{e^{t} -1}$. For $m \geq 0$, we define $B_m$ so that,
$$\large \frac{t}{e^{t} -1} = \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m}-(1)$$


Let us apply some mathematical rigor into it and see what happens.
We have,
$$\large \frac{t}{e^{t} -1} = \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m}\\ \implies\large (e^{t} -1) \frac{t}{e^{t} -1} = (e^{t} -1)\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} \\ \implies\large {t}= (e^{t} -1)\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} -(2)$$
Now, we know the McLaurin Series is,
$f(t) = f(0) + f'(0) + \frac{f""(0)}{2!}t^{2} + \frac{f^{(3)}(0)}{3!}t^{3}+...+\frac{f^{(n)} }{n!}t^{n}$
Let, $f(t) = e^{t}$, to find its McLaurin coefficients we must evaluate $\large(\frac{d^{k}}{dt^{k}}f(t))|_{t=0}$ for all $k$ = $0, 1, 2, ...$

By substitution, the McLaurin Series expansion of $(e^{t})$ is,
$$\large \sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}(t-0)^{k} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large e^{t} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large e^{t} = 1 + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large (e^{t} -1) = \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large (e^{t} -1) = \sum_{k=1}^{\infty}\frac{t^{k}}{k!} - (3)$$

Using $(3)$ in $(2)$, we get,
$$\implies\large {t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large {t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large \frac{t}{t}= \frac{1}{t}\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large \frac{t}{t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k-1}}{k!} \\ \implies\large 1= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{(k+1)!} - (4)$$
Let’s see which coefficient $a_n$ has the $t^n$ in the expanded right-hand part of this equation for some $n > 0$ (we expect this coefficient to be a zero). $t^n$ may appear if $t^m$ from the first sum multiplies the $t^{n-m}$ from the second one for some $0\le m\le n$. Thus,
$$\large a_n = \sum_{m=0}^{n}\frac{B_{n-k}}{m!(n-m+1)!}\\ \large = \sum_{m=0}^{n} {n+1\choose m} \frac{B_{m}}{(n+1)!}\\ \large = \sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} + \\\binom{n+1}{n}\frac{B_n}{(n+1)!} = 0$$
for any $n > 0$. Thus
$$\large (n+1)B_n = -\sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!}$$
Now for $n = 0$ we have from $(1)$
$$\large \frac{B_0}{0!}\frac{1}{1!} = 1$$
and thus
$$B_0 = 1$$

Approach 2 :

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We know,
$$\large \sum_{i=1}^{n} (i) = 1 + 2 + 3 + ... = \frac{n(n+1)}{2},\\ \large \sum_{i=1}^{n} (i^{2}) = 1^{2} + 2^{2} + 3^{2} + ... = \frac{n(n+1)(2n+1)}{6},\\ \sum_{i=1}^{n} (i^{3}) = 1^{3} + 2^{3} + 3^{3} + ...=\frac{n^{2}(n+1)^{2}}{4},\\ \sum_{i=1}^{n} (i^{4}) = 1^{4} + 2^{4} + 3^{4} + ...=\frac{n(n+1)(2n+1)(3n^{2}+3n-1}{30}.$$
and so on…

Then, how about,
$$\large \sum_{i=1}^{\infty}(i^{p})=?$$
Mathematicians have always been fascinated with such classic general formulae. So was JohannFaulhaber.
Let, $$S_p(n) = \sum_{i=1}^{\infty}(i^{p}) \\ \forall p \geq 0$$
Define the following exponential generating function with (initially) indeterminate $z$
$G(z,n) = \sum_{p=0}^{\infty } S_{p}(n) \frac {1}{p!}z^{p}$
We find
$$G(z,n)=\sum _{p=0}^{\infty }\sum _{k=1}^{n}\frac {1}{p!}((kz)^{p} )\\ =\sum _{k=1}^{n}e^{kz} \\ =e^{z}.\frac {1-e^{nz}}{1-e^{z}} \\ ={\frac {1-e^{nz}}{e^{-z}-1}}. \\ $$

$$G(z,n) =\sum_{p=0}^{\infty} \sum_{k=1}^{n} \frac{1}{p!}(kz)^p\\ =\sum_{k=1}^{n}e^{kz}=e^{z}.\frac{1-e^{nz}}{1-e^{z}},\\ = \frac{1-e^{nz}}{e^{-z}-1}.$$
This is an entire function in $z$ so that $z$ can be taken to be any complex number.
We next recall the exponential generating function for the Bernoulli polynomials $B_{j}(x) B_j(x)$
$${\frac {ze^{zx}}{e^{z}-1}}=\sum _{j=0}^{\infty }B_{j}(x)\frac{z^{j}}{j!}$$
where $B_{j}=B_{j}(0)$ denotes the Bernoulli number (with the convention $B_{1}=\frac {1}{2}; B_{1}=-\frac{1}{2}$). We obtain the Faulhaber formula by expanding the generating function as follows:
$$G(z,n) = \sum_{j=0}^{\infty}B_j \frac{(-z)^{j-1}}{j!}(-\sum_{l=1}^{\infty}\frac{(nz)^l}{l!}\\ $$
Solving it is again very complex, so, finally we get,
$$\large \sum_{k=1}^{n}i^{p}=\frac{1}{p+1}\sum_{j=0}^{p}(-1)^{j}\binom{p+1}{j}B_jn^{p+1-j}$$
Note that, $B_j = 0$, $\forall$ odd $j>1$; that is why Faulhaber defines $B_{1} = \frac{1}{2}$.

Verification:

Let us consider the following values,
$$p = 2 \\ $$
$$L.H.S = \large \sum_{k=1}^{n}i^{p} \\ = 1^{2} + 2^{2} + 3^{2} +...+n^{2}= \frac{n(n+1)(2n+1)}{6}.$$
$$R.H.S = \frac{1}{2+1}\sum_{j=0}^{2}(-1)^{j}\binom{2+1}{j}B_jn^{2+1-j}\\ =\frac{1}{2+1}[(-1)^{0}\binom{3}{0}B_0n^{3-0}+(-1)^{1}\binom{3}{1}B_1n^{3-1}+\\(-1)^{2}\binom{3}{2}B_2n^{3-2}]\\ =\frac{1}{3}[1 \cdot 1\cdot B_0 \cdot n^{3}+(-1) \cdot 3\cdot B_1 \cdot n^{2}+\\1 \cdot 3 \cdot B_2 \cdot n^{1}]\\ $$
We put, $B_0=1$ , $B_1=\frac{1}{2}$ and $B_2=\frac{1}{6}$ and see that $LHS$ = $RHS$.

Hence, Proved.

A Byronic Hero, Harry Potter and Bernoulli Numbers.

George Gordon Byron (22-01-1788 to 19-4-1824), is commonly known as Lord Byron, was a British Poet, a Politician and one of the brightest figures of the Romantic Movement. He is very easily regarded as one of the greatest British Poets of all times.
Considered to be the first modern-style celebrity. His image and thought of the Byronic Hero fascinated the public, his peers and his followers from the Literature community and from other communities as well. [3]
The first Byronic Hero, was probably Bryon himself. His wife Annabella coined the common, uncommon term “Byromania” to closely explain and depict the clamor around him. He was the classic depiction of the modern day Rockstar. He was fond of self-promotion and was supremely self-aware. All his portraits were drawn with a mindset of advertising him as a “Man of Action” and not as a poet or a romanticist. [2]
Lord Byron was a bisexual; however, the claims are yet to confirmed. Of course, the confirmation is not possible due to the heavy wave of suppression it has faced for decades. [1]
The figure, Byronic Hero, infuses much of Lord Byron’s work; in fact, many consider Byron, himself to be a perfect example of the characteristics of the Byronic Hero.
Popular study points that Lord Byron’s Byronic Hero was influenced by John Milton’s famous piece Satan from Paradise Lost. [4]
An excerpt is as follows:

In shape and gesture proudly eminent,
Stood like a tower. His form had yet not lost
All her original brightness, nor appeared
Less then Archangel ruined, and the excess
Of glory obscured: as when the sun new-risen
Looks through the horizontal misty air
Shorn of his beams, or from behind the moon,
In dim eclipse, disastrous twilight sheds
On half the nations, and with fear of change
Perplexes monarchs. Darkened so, yet shone
Above them all the archangel; but his face
Deep scars of thunder had intrenched, and care
Sat on his faded cheek, but under brows
Of dauntless courage, and considerate pride
Waiting revenge. Cruel his eye, but cast
Signs of remorse and passion, to behold
The fellows of his crime, the followers rather
(Far other once beheld in bliss), condemned
Forever now to have their lot in pain.

The Byronic Hero is a variant of the Romantic Hero [5]. Although there are traits and characteristics that exemplify the type, both Byron’s own persona as well as characters from his writings are considered to provide defining features.
The Byronic Hero first made it’s public appearance with Lord Byron’s semi-autobiographical epic narrative poem Childe Harold’s Pilgrimage, can be read for free on Project Gutenberg here Childe Harold’s Pilgrimage.
Thomas Babington Macaulay, popularly known Lord Macaulay, a British Historian and critic described the character as “a man proud, moody, cynical, with defiance on his brow, and misery in his heart, a scorner of his kind, implacable in revenge, yet capable of deep and strong affection“.
Byronic Hero was the instrument to many Romantic Classics and Gothic Love pieces of the 19th Century. One of the noteworthy one is Emily Brontë’s Wuthering Heights. The notorious, yet loved, Heathcliff is a classic depiction of a Byronic Hero.
A picture depicting the salient features of Heathcliff’s characteristics is as follows:

In American Renaissance, writers like Poe and Hawthrone took active reference in their work from the Byronic Hero.
In fact, Professor Severus Snape, of the Harry Potter Series, played by Late Alan Rickman is a Byronic Hero. He is dark, he is arrogant; yet demonstrated supreme love and sacrifice.

In this blog, as we are talking of Lord Byron; I don’t want to miss the chance to show my admiration for Ada Lovelace, Lord Byron’s daughter. Who is widely regarded as the first computer programmer. She was the first person to actually recognize the capabilities of Charles Babbage’s Analytical Engine, beyond pure mathematical tasks.
Charles Babbage’s seminar at University of Turin in 1840 on his Analytical Engine was recorded by Luigi Menabrea, a young Italian Engineer and future Prime Minister of the country. The language of recording was French. That is when Charles Wheatstone, the founder of the Wheatstone Bridge, asked Ada to translate that. She not only translated the paper, but also added detailed notes to it. She broke-down the lecture into parts from $A$ to $G$. In section $G$, she explained the method of Bernoulli’s number using the Analytical Engine.
The following picture shows her algorithm:

For people wondering what are Bernoulli’s numbers, let me provide a very simple explanation,
Benoulli’s numbers are defined from the power series expansion of $\large \frac{t}{e^{t}-1}$; for integers $m \geq 1$ we write $B_m$ so that,
$\large \frac{t}{e^{t}-1} = \sum_{m=0}^{\infty} \frac{B_m}{m!}t^{m}$
Multiplying both sides by $(e^{t} -1)$; will give,
$B_0=1$,
$B_m= -\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}B_k$.
Very confusing, right? To me too. Some rigorous math has to go in to understand this; and Lovelace wrote an algorithm for this when women weren’t even allowed to attend school.
Cheers!