Life of \pi

Is $\pi = \frac{22}{7}$?

There was an interesting problem that had come up during one of the JEE exams.

It asked us to integrate,
$$I = \int_{0}^{1} \frac{x^4 (1-x)^4}{1+x^2} dx$$
Use Polynomial Long Division,
$$I=={\displaystyle\int}\left(-\dfrac{4}{x^2+1}+x^6-4x^5+5x^4-4x^2+4\right)\mathrm{d}x$$

Use Linearity,
$$=-\class{steps-node}{\cssId{steps-node-1}{4}}{\displaystyle\int}\dfrac{1}{x^2+1}\,\mathrm{d}x+$$

$${\displaystyle\int}x^6\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-2}{4}}{\displaystyle\int}x^5\,\mathrm{d}x+\class{steps-node}{\cssId{steps-node-3}{5}}{\displaystyle\int}x^4\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-4}{4}}{\displaystyle\int}x^2\,\mathrm{d}x+\class{steps-node}{\cssId{steps-node-5}{4}}{\displaystyle\int}1\,\mathrm{d}x$$

$$I=-4\arctan\left(x\right)+\dfrac{x^7}{7}-\dfrac{2x^6}{3}+x^5-\dfrac{4x^3}{3}+4x$$

Substitute the values and we will get,
$$I = -\dfrac{7{\pi}-22}{7} = -(\pi - \frac{22}{7})$$
Which comes to,
$$\displaystyle I = \frac{22}{7} - \pi \tag*{(1)}$$

Since, Area cannot be negative, it is evident from $(1)$ that,
$$\frac{22}{7} \geq \pi$$


Also, $\pi \neq \frac{22}{7}$ is evident from,
$$\boxed{ e^{\pi i}+1 \neq e^{\frac{22}{7} i}+1}$$
Cheers!

Preparation of exam.

Preparation Notes - Hell lot of Math and Algorithms; very helpful.

Note : These notes are made during my preparation for my exams. They might be crude and may not be in their best form of presentation. Read at your own risk.

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Subject : IC 021 - Computational Techniques in Control Engineering

Professor : Dr. A Ramakalyan.

Difficulty : Runs to Heavens and goes around it $\infty$ times.

Books:
1. B N Datta’s Numerical Methods for Linear Control Systems. - referenced as [1]
2. Gilbert Strang’s Linear ALgebra and it’s applications. - referenced as [2]
3. S Kumaresen’s Linear Algebra - A Geometric Approach. - referenced as [3]

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1. Computing $e^\mathcal{A}$[1]

1. There are many ways of doing this, one of the very popular one is the Taylor Series method.
   We simply expand $e^\mathcal{A}$ as per the Taylor series.
   Hence, $e^\mathcal{A}$ can be written as,
   $$\boxed{e^\mathcal{A} = I + \mathcal{A} + \frac{\mathcal{A}^{2}}{2!} + \frac{\mathcal{A^3}}{3!} + ...}$$

   • The drawbacks to this method are that a large number of terms is needed for convergence, and even when convergence occurs, the answer can be totally wrong.

2. Hence, I will now get into the Pade’s Scaling and Squaring algorithm for this task.
   Suppose that, $f(x)$ is a power series presented as,
   $$f(x) = f_0 + f_1 x+ f_2 x^2+...$$
   Then the Pade’s Scaling and Squaring Algorithm, given an approximation of $f(x)$ as ,
   $$\boxed{f(x) \approx \frac{c(x)}{d(x)}= \frac{\sum_{k=0}^p c_{k} x^{k}}{\sum_{k=0}^q d_{k} x^{k}}}$$
   The $c_k$ and $d_k$ are chosen such that the terms containing $x^0, x^1, x^2,...,x^{p+q}$ are cancelled out in $f(x)d(x) - c(x)$. The order of Pade’s approximation is $p+q$. The ratio $(\frac{c(x)}{d(x)})$ is unique and the coefficients $c_0, c_1,...,c_p$ and $d_0, d_1,...,d_q$ always exist.
   
   Similarly, Pade’s approximation for $e^\mathcal{A}$ is given by,
   $$\boxed{R_{pq}(\mathcal{A}) = [D_{pq}(\mathcal{A})]^{-1}N_{pq}(\mathcal{A})}$$
   Where,
   $$\boxed{D_{pq}(\mathcal{A})=\sum_{k=0}^q\frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}(-\mathcal{A})^{j}}$$
   and
   $$\boxed{N_{pq}(\mathcal{A})=\sum_{k=0}^p\frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}(\mathcal{A})^{j}}$$

   • It can be shown (Exercise 5.16) that if p and q are large, or if A is a matrix having all its eigenvalues with negative real parts, then $D_{pq}(\mathcal{A})$ is nonsingular.
   • The difficulty of Round off errors in Pade’s method can be controlled by Scaling and squaring.
   • Since $e^\mathcal{A}$ is same as $(e^\frac{\mathcal{A}}{m})^m$, the idea is it to find out $m$ in terms of powers of $2$ so that $e^\frac{\mathcal{A}}{m}$ can be computed accurately and then find $(e^\frac{\mathcal{A}}{m})^m$ with repetitive squaring.
   • This method has favorable numerical properties when $p=q$.
     
     Let $m = 2^j$ be chosen such that $||A||_{\infty} \leq 2^{j-1}$, then Moler and Van Loan(1978) have shown that there exists an $E$ such that ,
     $$\boxed{[R_{pp}(A/2^j)]^{2^j} = e^{\mathcal{A} + E}},$$
     Where $||E||_{\infty} \leq \epsilon ||\mathcal{A}||_{\infty}$, with
     $$\epsilon = 2^{3-2p} (\frac{(p!)^2}{(2p)!(2p+1)!)}) \\ \boxed{\epsilon = \frac{8}{2^{2p}} (\frac{(p!)^2}{(2p)!(2p+1)!)})}$$
     Given an error tolerance of $\delta$ we should choose $p$ such that $||E||_{\infty} \leq \delta ||\mathcal{A}||_{\infty}$
     
     The Algorithm is as follows:
     Input : $A \in \mathcal{R}^{n \times n}$, $\delta > 0$
     Output : $F = e^{\mathcal{A} + E}$ with $||E||_{\infty} \leq \delta ||\mathcal{A}||_{\infty}$
     
     Step 1: Choose $j$ such that $||\mathcal{A}||_{\infty} \leq 2^{j-1}$, set $\mathcal{A} \equiv \mathcal{A}/2^j$.
     Step 2: Choose $p>0$ and smallest non-negative integer satisfying,
     $$\boxed{(\frac{8}{2^{2p}}) (\frac{(p!)^2}{(2p)!(2p+1)!)}) \leq \delta}.$$
     Step 3: Set $\boxed{D \equiv I, N \equiv I, Y \equiv I, c = 1}$
     Step 4: For $k = 1, 2, ..., p$ do
     $$\boxed{c \equiv c(p - k + 1)/[(2p - k + 1)k] \\ Y \equiv AY,\\ N \equiv N + cY , \\D = D+(-1)^{k}cY.}$$
     Step 5: Solve for $F$ : $\boxed{DF = N}$.
     Step 6: For $k = 1, 2, ..., j$ do,
     $$\boxed{F \equiv F^{2}}$$
   • The algorithm requires about $2(p + j + (1/3)n^3$ flops.
   • The algorithm is numerically stable when A is normal. When A is non-normal, an analysis of the stability property becomes difficult, because, in this case e a may grow before it decays during the squaring process; which is known as “hump” phenomenon.
     Example at Page 143 of [1]

3. We have another effective method called the Matrix decomposition method uses the Real Schur form to do so.
   
   Let $\mathcal{A}$ be transformed to a real Schur matric $R$ using an orthogonal similarity transformation:
   $$\boxed{P^{T}\mathcal{A}P = R}$$
   Then,
   $$\boxed{e^{\mathcal{A}}=Pe^{R}P^{T}}$$
   Note that $R$ is upper-triangular.
   
   The algorithm is:
   Step 1: Transform $\mathcal{A}$ to $R$ using the $QR$ iteration algorithm.
   $$\boxed{P^{T}\mathcal{A}P = R}$$
   Step 2: Compute $e^R = G = g_{ij}$:
   for $i = 1,...,n$ do,
   $$g_{ii} = e^{r_{ii}}$$
   end.
   
   for $k = 1, 2, ..., n-1$ do
   for $i = 1, 2, ..., n-k$ do
   set $j = i+k$
   $$\boxed{g_{ij}=\frac{1}{(r_{ii}-r_{jj})}\Big[r_{ij}(g_{ii}-g_{jj})+\sum_{p=i+1}^{j-1}(g_{ip}r_{pj}-r_{ip}g_{pj})\Big]}$$
   end
   end
   Step 3: Now we have $e^R$ then we compute $e^\mathcal{A}$

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2.Lyapunov Stability theorem: [1]

Lyapunov Stability theorem,
The system :
$$\boxed{\dot{x}(t)=\mathcal{A}x(t)}$$
is asymptotically stable $iff$, for any symmetric positive definite matrix $M$, there exists a unique symmetric positive definite matrix $X$ satisfying the equation:
$$\boxed{X\mathcal{A}+\mathcal{A}^{T}X=-M}$$

Kronecker Product and Lyapunov:
$$\boxed{\mathcal{A} \bigotimes B = \big[ a_{ij} [B] \big]}$$
For example,
if
$$\mathcal{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix}$$
and,
$$B = \begin{bmatrix} b_{11} &b_{12} &b_{13}\\ b_{21} &b_{22} &b_{23}\\ \end{bmatrix}$$
Then, the Kronecker product will look like,
$$\mathcal{A} \bigotimes B = \begin{bmatrix} a_{11}b_{11} &a_{11}b_{12} &a_{11}b_{13} &a_{12}b_{11} &a_{12}b_{12} &a_{12}b_{13} \\ a_{21}b_{21} &a_{21}b_{22} &a_{21}b_{23}&a_{22}b_{11} &a_{22}b_{12} &a_{22}b_{13} \\ \end{bmatrix}$$
It is evident that,
$$\boxed{\mathcal{A} \bigotimes B \neq B\bigotimes\mathcal{A}}$$
Now, we formulate the $X\mathcal{A}+\mathcal{A}^{T}X=M$ using the Kronecker product. Note that $M$ is negative semi-definite. $X$ is positive definite and symmetric.

Vectorisation of a matrix $\large \triangleq$ Stack all elements of
the matrix into one column.

Going directly to the final answer,
$$\boxed{X\mathcal{A}+\mathcal{A}^{T}X=M = [(A^{T} \bigotimes I )+( I\bigotimes A^{T})]\overrightarrow{X} = \overrightarrow{M}}$$
Verify it once.

If $\mathcal{A}$ is stable then $X$ has unique solution.


Schur method of Lyapunov:
Method proposed by Bartels and Stewart. It is based on the reduction of $A^{T}$ to $RSF$.

Step 1: Reduce the problem.
For,
$$X\mathcal{A}+\mathcal{A}^{T}X = C$$
is reduced to,
$$\boxed{YR^{T}+RY=\hat{C}}$$
Where,
$R = U^{T}A^{T}U$, $\hat{C}=U^{T}CU$ and $Y=U^{T}XU$.

Step 2: Solution of the reduced form.
Let,
$$Y = (y_1, y_2,...,y_n),$$
$$\hat{C}=(c_1,...,c_n)$$
and $$R=(r_{ij})$$
Assume that the columns $y_{k+1}$ through $y_{n}$ have been computed, and consider the following two cases,

Case 1: $r_{kk-1} = 0$. Then $y_k$ is determined by solving the quasi-triangular system,
$$\boxed{(R+r_{kk}I)y_k=c_k-\sum_{j=k+1}^nr_{kj}y_j}$$
If, in particular, R is upper triangular, that is there are no “Schur bumps” on the diagonal, then each $y_i$, $i=n, n - 1 ..... 2, 1$ can be obtained
by solving an $n \times n$ upper triangular system as follows:
$$\boxed{(R+r_{nn}I)y_n=c_n} \\ \boxed{(R+r_{n-1,n-1}I)y_{n-1}=c_{n-1}-r_{n-1,n}y_n}$$

Case 2: $r_{kk-1} \neq 0$ this shows a Schur Bump.
$$\boxed{R(y_{k-1}, y_k)+(y_{k-1}, y_k) \begin{bmatrix} r_{k-1,k-1} &r_{k,k-1}\\ r_{k-1,k} &r_{k,k}\\ \end{bmatrix}\\ =(c_{k-1},c_k)-\sum_{j=k+1}^n(r_{k-1,j}y_j, r_{kj}y_j)=(d_{k-1},d_k)}$$
Example at Page 274


If $\mathcal{A}$ doesn’t have Eigen Values in the Limit circle , we modify $\mathcal{A}$ as,
$$A_{1} = (I-A)^{-1}(I+A) \text{ and } Q_{1} = (I+A^{T}_{1})(I+A_{1})\\ \text { solve for A } \implies (I-A)A_{1}=(I+A)\\ \implies A_{1}-AA_{1}=I+A\\ \implies A_{1} - I = AA_{1}+A\\ \implies A(A_{1}+I) = A_{1}-I\\ \implies A=(A_{1}-I)(A_{1}+I)^{-1}$$
Use this $A$ in the Discrete Lyapunov which is given by,
$$\mathcal{L}_{D} : A^{T}PA-P=Q$$
The resulting equation is,
$$A^{T}_1P+PA_{1}=Q_{1} \text{ assume, } Q = -2I$$
You will get a similar equation to that of the normal Lyapunov equation.

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3. Modified QR algorithms:[4]

1. Modified Gram-Schmidt method:
   for
   $$k = 1:n\\ q^{T}_k = a^{T}_k$$
   end
   
   for
   $$k = 1:n\\ r_{kk} = (q^{T}_{k}q_k)^{1/2}\\ q^{T}_k= \frac{q^{T}_k}{r_{kk}}$$
   -for
   $$j = k+1:n\\ r_{kj}=a^{T}_{j}q_{k}\\ q^{T}_{j}=q^{T}_{j}-r_{kj}q^{T}_{k}$$
   -end
   end
   It requires $\boxed{O(\frac{8}{3}n^{3})}$ operations.
   Practice example from the above link

2. Original algorithm of QR decomposition:
   Aim is to write matrix $\mathcal{A}$ in terms of $\boxed{Q\times R}$.
   Here, $Q$ is an orthogonal matrix with orthonormal columns; and $R$ is an upper-triangular matrix.
   
   Now, we take an example,
   $$\mathcal{A}=\begin{bmatrix} 1 & 2 & 3\\ -1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix}$$
   We will first find $q_1$, $q_2$ and $q_3$ as the three orthonormal vectors using the GS-strategy.
   Now, we will define $a$, $b$ and $c$ as,
   $$a = \begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}, b=\begin{bmatrix} 2\\ 1\\ 1\\ 1 \end{bmatrix} \text{and}, c = \begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix}$$
   We define a partial basis as $\mathcal{X}$ which will be updated at each step.
   
   For the first step,
   we find $v_1$ as,
   $$\boxed{v_1 = \frac{a}{1} = \begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}} \text{now the partial basis is,} \mathcal{X} = \big[\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix} \big]$$
   similarly, we find $v_2$ as follows,
   $$v_2 = b - proj_{\mathcal{X}}\begin{bmatrix} 2\\ 1\\ 1\\ 1 \end{bmatrix}\\ = \begin{bmatrix} 2\\ 1\\ 1\\ 1 \end{bmatrix}-\frac{\begin{bmatrix} 2\\ 1\\ 1\\ 1 \end{bmatrix}.\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}}{\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}.\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}}\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix} = \begin{bmatrix} 2\\ 1\\ 1\\ 1 \end{bmatrix}-\frac{3}{4}\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}\\ \text{Now,} \boxed{4v_2=\begin{bmatrix} 8\\ 4\\ 4\\ 4 \end{bmatrix}-\begin{bmatrix} 3\\ -3\\ 3\\ 3 \end{bmatrix}\\ =\begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}}$$
   Now, we update $\mathcal{X}$,
   $$\mathcal{X} = \Big[\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix} \Big]$$
   Now, we find $v_3$ as follows,
   $$v_3 = \begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix} - proj_{\mathcal{X}}\begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix}\\ v_3 = \begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix}-\frac{\begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix}.\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}}{\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}.\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}}\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}-\frac{\begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix}.\begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}}{\begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}.\begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}}\begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}\\ v_3 = \begin{bmatrix} 3\\ 1\\ 1\\ 1 \end{bmatrix}-\frac{4}{4}\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}-\frac{24}{76}\begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}\\ \text{now},\\ \boxed{19v_3 = \begin{bmatrix} 8\\ -4\\ -6\\ -6 \end{bmatrix}}$$
   Hence, the partial basis is,
   $$\boxed{\mathcal{X} = \Big[\begin{bmatrix} 1\\ -1\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 5\\ 7\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 8\\ -4\\ -6\\ -6 \end{bmatrix}\Big]}$$

$$proj_{\overrightarrow{a}}{\overrightarrow{b}} = \frac{{\overrightarrow{a}}\cdot {\overrightarrow{b}}}{{\overrightarrow{a}}^2}{\overrightarrow{a}}$$

This partial basis is the orthogonal set, hence we now make it orthonoamal,
$$Q_{orthogonal} = \begin{bmatrix} 1 & 5 & 8\\ -1 & 7 & -2\\ 1 & 1 & -3\\ 1 & 1 & -3\\ \end{bmatrix}$$
Hence,
norm of first column, $$||n_1|| = \sqrt{1^2+(-1)^2+1^2+1^2} = 2$$
Similarly,
$$||n_2|| = \sqrt{76}$$
and
$$||n_2|| = \sqrt{38}$$
Hence, the orthonormal $Q$ is,
$$Q_{orthonormal}=\begin{bmatrix} 1/2 & 5/\sqrt{76} & 8/\sqrt{38} \\ -1/2 & 7/\sqrt{76} & -2/\sqrt{38}\\ 1/2 & 1/\sqrt{76} & -3/\sqrt{38}\\ 1/2 & 1/\sqrt{76} & -3/\sqrt{38}\\ \end{bmatrix}$$
Now, we have, $\mathcal{A}$ and $Q$; we now find $R$; which by theory has to be upper-triangular,
$$\mathcal{A}= Q \cdot R \\ Q^{T} \cdot \mathcal{A} = Q^{T} \cdot Q \cdot R\\ \text{NOTE:}\text{because } Q \text{ is orthonormal } \boxed{Q^{T} \cdot Q = I}.\\ \therefore \boxed{R = Q^{T} \cdot \mathcal{A}}$$
Make up the algorithm from the above steps.

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4. Condition Number:[5]

As a rule of thumb, if the condition number $\kappa (A)=10^{k}$, then you may lose up to $k$ digits of accuracy on top of what would be lost to the numerical method due to loss of precision from arithmetic methods.

A problem with a low condition number is said to be well-conditioned, while a problem with a high condition number is said to be ill-conditioned. The condition number is a property of the problem.

Condition number is mathematically defined as,
$$\kappa(A) = ||A||_{1}||A^{-1}||_{1}$$
We define the $1$-norm as follows,
For a matrix $M$
$$M = \begin{bmatrix} 2 & 3\\ 1 & -1 \end{bmatrix}$$
The $1$-norm is,
$$\text{Maximum of }(2+1,3+1) = 4.\\ \text{Hence, }||M||_{1} = 4.$$

So, for a given matrix,
$$P = \begin{bmatrix} 1 & 10^{4}\\ -1 & 2 \end{bmatrix}$$
We get,
$$P^{-1} = \frac{1}{2+10^{4}}\begin{bmatrix} 2 & -10^{4}\\ 1 & 1 \end{bmatrix}$$
Hence,
$$||P||_{1} = \text{maximum of }(1+1, 2+10^{2}) = 2+10^{4}$$
and,
$$||P^{-1}||_{1} = \frac{1}{2+10^{4}}\text{maximum of }(2+1, 10^{4}+1) = \\ \frac{1}{2+10^{4}}(1+10^{4})$$
Therefore,
$$\text{Condition number},\\ \kappa(P) = ||P||_{1}||P^{-1}||_{1} = 1+10^{4}$$
Thus, the matrix (or system) is ill-conditioned.

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The question paper that came for this exam was pretty sadistic.
Give it a try, if you can and tell me how awesome it feels. I will score around 30/50.

a FAILED mathematical approach into the search for the E.T.

Are we alone?

Scientists have been searching for another form of intelligent life for decades now. We have searched and have failed, we have the evidence and yet do not understand what they are.
Numerous Physicists, Astronomers, Conspiracy theorists and even common man ( :p ) have spent most or a significant time of their lives looking into the big blue sky and asking the question, “Am I alone?”
This sounds highly unlikely that we are the ONLY intelligent form of life in this huge universe.
I don’t care if the probability of any life in this universe, other than us, is $\rightarrow 0$, until and unless it is not $0$, I want to keep looking.
There have been so many things, let us take the Nazka lines of Peru. There are usually called Geoglyphs. These took a lot of time to come in light, Pedro Cieza de León mistook them as trail marks; some people mistook them as irrigation lines.
It was not until the days of flights and aeroplanes, when people saw these clearly; that is when they understood that these were not trail marks or anything. These were signs and drawings.
The one I like the most is as follows:



Now, what I ask is,
- How did they make these?
- Who did they make these for?
Unfortunately, these answers are not yet answered.

Some archeologists and geologists studied these lines and the results are shocking. These are not just lines with some nonsensical scribbling; these are made to last for a long time. Let alone the design; the methodology of design is scientific too. An excerpt from wikipedia is says the following,

On the ground, most of the lines are formed by a shallow trench with a
depth between 10 and 15 cm (4 and 6 in). Such trenches were made by
removing the reddish-brown iron oxide-coated pebbles that cover the
surface of the Nazca Desert. When this gravel is removed, the
light-colored clay earth which is exposed in the bottom of the trench
produces lines which contrast sharply in color and tone with the
surrounding land surface. This sublayer contains high amounts of lime
which, with the morning mist, hardens to form a protective layer that
shields the lines from winds, thereby preventing erosion.

Are these just drawings after all?
These are maps, some theorists say. Don’t believe them? have a look at the image below.




One of my favorite conspiracy theory is by Erich von Däniken, it is called the Cargo Cult

Erich von Däniken’s theory is the most famous approach to solve the
mystery of Nazca. He had the idea that long time ago visitors from
other stars visited the earth and naturally Nazca. At this place they
landed, during the landing stones was blown away by the power of
rocket propulsion. By approaching more the power was increasing and
the cleaned band broader. In this way the first trapezes emerged.
Later the Aliens disappeared and left confused people. Like in the
modern cargo cults they tried to call the Gods back by drawing lines,
figures and trapezes. Never Däniken said the formations was made by
Aliens. He discovered the GGF/Mandala/Zodiac and the mirror -
Formation and compares them with modern VASIS or PAPI-Signs.

To read more please visit : Conspiracy theories of Nazca Lines.

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Mathematics:

There is a mathematical formula by American Astronomer and Astrophysicist Frank Drake [3]. He is hugely regarded as the father of modern SETI(Search for extraterrestrial intelligence).
The criticism related to Drake’s equation is not about the correctness of the equation itself, but about the highly ambiguous estimations of the various variable used in the equation.
The equation is as follows,
$$N = R^* \times f_p \times n_e \times f_l \times f_i \times f_c \times L$$
Here,
$N$ = the number of civilizations in our galaxy with which communication might be possible (i.e. which are on our current past light cone);
$R*$ = the average rate of star formation in our galaxy
$f_p$ = the fraction of those stars that have planets
$n_e$ = the average number of planets that can potentially support life per star that has planets
$f_l$ = the fraction of planets that could support life that actually develop life at some point
$f_i$ = the fraction of planets with life that actually go on to develop intelligent life (civilizations)
$f_c$ = the fraction of civilizations that develop a technology that releases detectable signs of their existence into space
$L$ = the length of time for which such civilizations release detectable signals into space

For people wondering what is a light cone, here is a picture. This is
a gloried way of looking into the dynamics of time.


In special and general relativity, a light cone is the path that a flash of light, emanating from a single event (localized to a single point in space and a single moment in time) and traveling in all directions, would take through spacetime

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Criticism:

1. One major setback of this equation is that it takes no account of the cosmological developmental phases and time, the value of $N$ would heavily rely on the speed of cosmological development[1].
2. There is no derivation or methodology of how this equation came into being; which suggests that it is just a stab in the dark.
3. The factor $n_e$ determines habitability; how in the fuck’s world do you find that? Oxygen and Water makes Earth habitable for humans; and probably in some planet $\mathcal{X}$, Nitrogen and Alcohol makes it habitable [2].

---

Here is a beautiful image about Drake’s equation,

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Conclusion:

1. This is a very badly written equation; not because it is wrong. But, because it has no methodology involved. Looks like couple of fresh college graduates wrote this while having some really potent weed.
   2.Drake’s equation is therefore a Fermi Paradox.

The Fermi paradox or Fermi’s paradox, named after physicist Enrico
Fermi, is the apparent contradiction between the lack of evidence and
high probability estimates, e.g., those given by the Drake equation,
for the existence of extraterrestrial civilizations.

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An interesting coming from xkcd[3], it added another term $B_6$ which is the amount of bull shit you are gonna buy from the the drake equation.

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Cheers!

Euler - Riemann Zeta Function.

“Madam, I have just come from a country where people are hanged if
they talk.” ― Leonhard Euler

The Riemann Zeta function, denoted as $\zeta{(s)}$ is a function of a complex variable $s$ that analytically continues the sum of the Dirichlet Series, [1]
$$\LARGE \zeta{(s)}= \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
The entire blog will be divided into the following parts:

1. What is Analytic Continuation?
2. What is Dirichlet Series?
3. Recurrence relation between Bernoulli Numbers.
4. Relationship between Bernoulli, Riemann and Euler.

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1.

Before we talk about Analytic continuation, we will have to know what an Analytic function is.

There are multiple ways of defining the Analytic functions[2]. The one that I find most comfortable is given below:
$\underline{\text{Definition 1}}$: A function $f(z)$ is said to be analytic in a region $\mathcal{R}$ of the complex plane if $f(z)$ has a derivative at each point of $\mathcal{R}$ and if $f(z)$ is single valued.

For better understanding, I will extend an example,
$\underline{\text{Example 1}}$: Consider $f(z)$ = $2xy + i(x^{2}-y^{2})$, determine if the given function is analytic or not.
$\underline{\text{Solution}}$: We will use the Cauchy-Riemann equations.

For a given, $f(z) = f(x+iy) = u(x,y) + i v(x,y)$
Now we have the following definitions,
$$u_{x} = \frac{\partial{u}}{\partial{x}}, \\ u_{y} = \frac{\partial{u}}{\partial{y}}, \\ v_{x} = \frac{\partial{v}}{\partial{x}}, \\ v_{y} = \frac{\partial{v}}{\partial{y}}.$$
Now, according to the Cauchy-Riemann Equation, any $f(z)$ is an analytic function, iff,
$$u_{x}=v_{y}, \\ u_{y} = -v_{x}.$$

Now, for the given $f(z)$,
$$u(x,y) = 2xy, \\ v(x,y) = x^{2}-y^{2}.$$
Hence, the relations are,
$$u_{x} = \frac{\partial{(2xy)}}{\partial{x}} = 2y, \\ u_{y} = \frac{\partial{(2xy)}}{\partial{y}}=2x, \\ v_{x} = \frac{\partial{(x^{2}-y^{2})}}{\partial{x}}=2x, \\ v_{y} = \frac{\partial{(x^{2}-y^{2})}}{\partial{y}}=-2y.$$
It is clear that,
$$u_{x} \neq v_{y}, \\ u_{y} \neq -v_{x}.$$
Therefore, the given $f(z)$ is not an analytic function.

Analytic continuation is a pretty simple concept to understand really.
$\underline{\text{Definition 2}}$: Say $f(z)$ is an analytic function defined over a non-empty open subset $W$ of the complex plane $\mathcal{C}$. If $V$ is a larger open subset of $\mathcal{C}$, containing $W$, and $\mathcal{F}$ is an analytic function defined on $V$ such that,
$$\mathcal{F}(z) = f(z), \forall z \in V$$
In other words, Analytic continuation is the method of extending the domain of an analytic function. [3]

For some cool examples on Analytic continuation refer Virginia Tech’s paper here [4].

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2.

Dirichlet Series[5] is any series of general form,
$$\sum_{n=1}^{\infty}\frac{a_n}{n^{s}}$$
Now, with a slight modification,
$$a_n = 1 , \forall n \in \mathcal{R}$$
We get,
$$\sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
For a function defined as,
$$\zeta{(s)} = \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
is the Riemann Zeta function.

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3.

Definitions,[6]
$$s_{0}(n)=n,\\ s_{1}(n) = \frac{n(n-1)}{2},\\ s_{2}(n) = \frac{n(n-1)(2n-1)}{6}...$$
Therefore, $s_{p}(n)$ is given as,
$$s_{p}(n) = \frac{1}{p+1}n^{p+1}-\frac{1}{2}n^{p}+\frac{p}{12}n^{(p-1)}+0 \times n^{(p-2)}+...$$
More generally,
$$s_{p}(n)=\sum_{k=0}^{p} \frac{B_{k}}{k!} \frac{p!}{(p+1-k)!}n^{(p+1-k)}$$
for $n=1$,
$$0=\sum_{k=0}^{p} \frac{B_{k}}{k!} \frac{p!}{(p+1-k)!}$$
or equivalently,
$$B_p=-\frac{1}{p+1}\sum_{k=0}^{p-1} \binom{p+1}{k}$$
The above relation is symbolically written as,
$$(B+1)^{(p+1)}-B^{(p+1)}=0$$
On expansion, all $k$-th powers of $B$, $B^{k}$ must be written as $B_k$ and treated as Bernoulli Numbers.
The expansion is done on the basis of Binomial Theorem, which statest that,
$$\forall a,b \in \mathcal{Z} \ \text{and} \ n \in \mathcal{N}\\ \large (a+b)^{n}= \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$$
Therefore, for $n=5$,
$$(a+b)^{5} = \sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\\ \implies(a+b)^{5} = \binom{5}{0}a^{5-0}b^{0}+\binom{5}{1}a^{5-1}b^{1}+\binom{5}{2}a^{5-2}b^{2}+\\ \binom{5}{3}a^{5-3}b^{3}+\binom{5}{4}a^{5-4}b^{4}++\binom{5}{5}a^{5-5}b^{5}\\ \implies(a+b)^{5} = 1 \times a^{5}b^{0}+5\times a^{5-1}b^{1}+10\times a^{5-2}b^{2}+\\ 10\times a^{5-3}b^{3}+5\times a^{5-4}b^{4}+1\times a^{5-5}b^{5}\\ $$
Therefore, for $(B+1)^{(p+1)}-B^{(p+1)}=0$, after expanding for $p=4$ we have,
$$(B+1)^{(4+1)}-B^{(4+1)}=0 \\ \implies (B+1)^{5}-B^{5}=0 \\ \implies 1\times B_5+5\times B_4+10 \times B_3 + 10 \times B_2 + \\ 5 \times B_1+1\times B_0 =0\\ \implies 1\times 0+5\times B_4+10 \times 0+ 10 \times (\frac{1}{6}) + \\ 5 \times (\frac{-1}{2})+1\times 1 =0\\ \implies 5\times B_4+10 \times (\frac{1}{6}) + 5 \times (\frac{-1}{2})+1 =0\\ \implies B_4=-\frac{1}{30}\\ $$

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4.

Euler found a formula which easily defined the **even-numbered zeta**functions as follows[7]:
$$\zeta(2n)=(-1)^{(n+1)}\frac{{B_{2n}}{(2\pi)^{2n}}}{2(2n)!}$$
Interestingly, $B_{2n+1}=0 \ \ \forall n \geq 1$, so there is no counter-part for $\zeta(2n+1)$ and yet a value of $\zeta(3)$ exists. Well, that is beyond my scope of this blog.
Cheers!

Collatz Conjecture - a study.

God does not care about our mathematical difficulties. He integrates empirically. - Einstein.

In Mathematics, we often come across Conjectures. A conjecture is a conclusion or proposition based on incomplete information, for which no proof has been found[1].
Collatz Conjecture is named after Lothar Collatz [2]. The conjecture is also known as the $3n+1$ conjecture.
Now, let us see how it works, it is, in fact very simple, no $\int$, no $\sum$, no $\frac{d}{dt}$ etc.
For any given positive integer $\mathcal{n}$,
$$f(n)= \begin{cases} \frac{n}{2},& \text{if } n \ \text {is even}\\ 3n+1, & n \ \text{is odd} \end{cases}$$
If we keep repeating this process,
We use the Half Or Triple Plus One acronymed to $\mathcal{HOTP}$.
So, for any given positive number, say $15$, we do the following steps:

15 is odd $\implies$15 $\times$3+1=46$\implies$46 is even $\implies$ $\frac{46}{2} = 23$ $\implies$
23 is odd $\implies$23$\times$3+1=70$\implies$35$\implies$106$\implies$53$\implies$160$\implies$
80$\implies$40$\implies$20$\implies$10$\implies$5$\implies$16$\implies$8$\implies$4$\implies$2$\implies$$1$

You get the idea now. In this blogpost, I will explore various aspects of the Collatz Conjecture. A loose structure of the blog is going to be,

1. Collatz for Negative numbers.
2. Collatz for Fractions.
3. Collatz for Irrational Numbers.
4. Collatz for Prime numbers and pattern recognition (if any) in the number of steps it needs for different scenarios to reach to one.
5. What is the possible approach for the proof of collatz conjecture?
6. Difficult and beautiful.

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1.

I don’t understand why do they have the condition, $\mathcal{n} > 0$ in the conjecture. Let’s look at an example and see what happens when Collatz conjecture is applied to number, $\mathcal{n}$ when $\mathcal{n} < 0$.

For, $\mathcal{n}=-5$, let us do the collatz conjecture,

-5 is odd $\implies$ -5 $\times$ 3+1 $\implies$-14 is even $\implies$ -7 is odd $\implies$ -20 is even $\implies$ -10 is even $\implies$ -5 is odd $\implies$ -5 $\times$ 3+1 $\implies$-14 is even $\implies$ -7 is odd $\implies$ -20 is even $\implies$ -10 is even $\implies$ -5

So, there it is.
$\underline{Conclusion \ 1:}$ If we use the Collatz conjecture to negative numbers, we get oscillating values and they even eventually set to the starting number itself.

Modified Collatz Conjecture:
$$f(n)= \begin{cases} \frac{n}{2},& \text{if } n \ \text {is even}, n>0\\ 3n+1, & n \ \text{is odd}, n>0 \\ \text{comes back to starting integer}, & n < 0 \end{cases}$$

These are called “Cycles” that exist for negative numbers and probably also for positive numbers. There is no form of proof, that says that there is absolutely no positive number that gives rise to a similar situation.

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2.

For this section, we define something called Number-decimals.
Although, we already know that Even, Odd or Prime are only defined for whole numbers.
$\underline{Definition \ 1}$: Even-Number-decimal are defined as terminating fractions, say $\large \frac{p}{q}$ that are converted to decimals, say $0.00k$. Then we remove the decimal point from the number to get $k$.
(If you find this definition to be strange, please comment below and mail me with a possible modification)

Now, if $\large \frac{k}{2}=0$, then $\large 0.00k$ is an Even-Number-decimal and if $\large \frac{k}{2} \neq 0$, then then $\large 0.00k$ is an Odd-Number-decimal
So, it is pretty clear that the conjecture behaves the same in this scenario.

For example, take $\frac{1}{5}$, which when converted to decimals comes as $0.2$. Now, we run the algorithm on $2$ to get 1. Similarly for other fractions as well.


In the above approach, we considered fractions as decimal numbers, that is only how we can use the Collatz Conjecture on fractions; that too only if the decimal is terminating in nature. Fractions such as $\frac{22}{7}$ ($\pi$), $\frac{1}{6}$, $\frac{1}{3}$ etc. cannot be used in the Collatz conjecture for fractions.

---

3.

Collatz conjecture for irrational numbers is same as saying that we apply Collatz Conjecture to non-terminating decimals.
Just like $\pi$, as mentioned above cannot work for Collatz; irrational numbers won’t work too.
For revision purpose, an irrational number is defined as a number that cannot be determined as the ratio of two integers. For illustration, $0.5 = \frac{3}{2}$, however, $\pi =3.14159...= \frac{?}{?}$ , no exact way of saying this.

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4.

Collatz for primes will behave the same, that is my guess. Let’s have a look at it.


Let’s take all the primes from $1 \ \text{to} \ 20$, namely, $2, 3, 5, 7, 11, 13, 17, 19$.

We are sure that all of them will reach $1$ eventually. I want to analyze that, in how many steps do they reach $1$ and if there is any correlation.

For this, I wrote a small script in R. The code is as follows:

collatz_numbers <- function(n, list_col=c()) {
  if(n==1) return(c(list_col, 1));
  collatz(ifelse(n%%2==0, n/2, 3*n +1), c(list_col, n))} 

We shall define the number of steps taken by the number to reach $1$ as $\mathcal{m}$.

So,
collatz_number(2)=1 ( $\mathcal{m}$ = 1)
collatz_number(3)=3, 10, 5, 16, 8, 4, 2, 1 ($\mathcal{m}$ = 8)
collatz_number(5)=5, 16, 8, 4, 2, 1($\mathcal{m}$ = 6)
collatz_number(7)=7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1($\mathcal{m}$ = 16)
collatz_number(11) =11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ($\mathcal{m}$ = 15)
collatz_number(13)=13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ($\mathcal{m}$ = 10)
collatz_number(17)=17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1($\mathcal{m}$ = 13)
collatz_number(19)=19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1($\mathcal{m}$ = 21)
If we plot the numbers versus the number of steps it took to get to $1$, we get the following plot,

Which is an increasing series, as per the Wikipedia article[1], have a look at this [3] to have a more clear idea.
The picture below also agrees to our finding,

One interesting observation is how HUGE numbers appear in the sequence. For that, I thoughtlessly try numbers out on my function,
collatz_number(20) = 20 10 5 16 8 4 2 1.
collatz_number(21)=21 64 32 16 8 4 2 1
collatz_number(22)=22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
collatz_number(23)=23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
collatz_number(24)=24 12 6 3 10 5 16 8 4 2 1
collatz_number(25)=25 76 38 19 58 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
collatz_number(26)=26 13 40 20 10 5 16 8 4 2 1
collatz_number(27)=27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
collatz_number(28)=28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

All the largest numbers in the sequence are made bold. We can clearly see their chaotic nature, but there is never really any chaos, we just don’t look from a broad enough perspective.

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5.

I have failed pathetically, at finding a proper proof for the the conjecture. However, I have some points that could be the lines on which someone could work for proving the conjecture. I will mention them in bullet points,

• The first task is to show that for all positive numbers, there is not a single number that gives rise to a cycle (to know what is a “cycle”, read section 1).
• Understanding the behavior of steps with increasing or decreasing numbers.
• Understanding that $3n+1$ is usually the universal way of converting any number into an Even number, so I guess, he played the CLEVER trick there.
• Looking at even and odd as $1'\text{s}$ and $0'\text{s}$ and finding if they follow a particular pattern. Further, we can convert these Binary strings to Hexadecimal strings and see if there is a possibility of a Collatz conjecture in Alphabetical domain. [Original idea by Pragyaditya Das].

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6.

Collatz conjecture is very nice, if you haven’t seen it already, let me point it out to you.

No matter how many time we do $3\times$ and $+1$ to a system; if we semi-periodically keep pulling $2'\text{s}$ from the system; it will eventually lead to $1$.

Collatz conjecture is the simplest mathematical open-problem available; you can explain all your non-math, non-science or non-engineering friends about it; hell! they might even give a noob try to prove it even.

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Cheers! with a sad end…

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 :

---

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!