January 11, 2012June 11, 2020 by Nirupam Biswas

Digit Math Application: Proving the correctness of shortcut method to squaring numbers ending with 5

Please read Digit Math: Introduction before you continue.

The problem

Someday, somewhere I came to know that any number which ends with the digit 5 can be easily squared. The trick can be easily demonstrate using an example. Suppose we want to find the square of 25.

Trick is to take the number before 5 (which will be 2 here), add one to it (2 + 1 = 3) and then multiply them together (2 x 3 = 6). Now the final answer would be the product followed by the number 25, i.e. 625 in this case.

Now let’s try it out yo find $215^2$.

$$
\begin{align}
215^2 &= (21 \times (21 + 1)) \omega 25\\
&= (21 \times 22) \omega 25\\
&= 462\omega25\\
&= 46225\tag{Answer}
\end{align}
$$

This always seemed to work out very well. The problem was, can I trust this trick? Will this always hold true? I didn’t have answers to those questions, until I proved it myself using Digit Math. Good news is that this trick will always hold true.

The proof

Let the number be $x = a\omega 5$. $a$ can have any number of digits.

Case 1:

$a$ has exactly one digit. So, $]a[ = ]5[ = 1$.

$$
\begin{align}
\therefore (a\omega 5)^2 &= aa\omega (5a+5a) \omega 5.5\tag{Using Bimultiplication formula}\\
&= a^2 \omega 10a \omega \underline{2}5\\
&= a^2 \omega (a\omega0) \omega \underline{2}5\\
&= a^2 \omega (a\omega2) \omega 5\\
&= (a^2 + a) \omega 2 \omega 5\\
&= \big(a(a+1)\big) \omega 25\tag{Proved}
\end{align}
$$

Case 2:

$a$ has more than one digits. So, $]a[ > (]5[ = 1)$.

But to apply Bimultiplication formula $a$ must have the same number of digits in $5$, which is obviously not the case here. So, we will use one trick. We will pad $5$ with some number of zeroes on the right, so that, $]a[ = ]5\omega c[$, where, $c$ is all zeroes and $]c[ = ]a[ – 1$. So, if $a = 123 \Rightarrow c = 00$.

$$
\begin{align}
\therefore (a\omega 5)^2 &= \big(a \omega ]5 \omega c[^{]a[}\big)^2\\
&= a^2 \omega \big(a(5\omega c) + a(5\omega c)\big) \omega (5\omega c)^2\tag{Using Bimultiplication}\\
&= a^2 \omega 2a(5\omega c) \omega (5\omega c)^2\\
&= a^2 \omega (10a \omega 2ac) \omega (5\omega c)^2\\
&= a^2 \omega (10a \omega c) \omega (5\omega c)^2\tag{Since, c is all zeroes}\\
&= a^2 \omega (a \omega 0 \omega c) \omega (5\omega c)^2\\
&= a^2 \omega (a \omega 0 \omega c) \omega (25 \omega 10c \omega c^2)\\
&= a^2 \omega (a \omega 0 \omega c) \omega (25 \omega c \omega c)\tag{1}
\end{align}
$$

Since each digit group must have $]a[$ digits, so let us move one zero from the middle $c$ in $25 \omega c \omega c$ to the rightmost $c$. So, now that group becomes $25\omega d\omega e$, where $]d[ = ]c[ – 1$ and $]e[ = ]c[ + 1$.

$$
\begin{align}
\therefore (1) &= a^2 \omega \big( (a \omega 0 \omega c) + (25 \omega d)\big) \omega e\\
&= a^2 \omega \big( a \omega (0+25) \omega (c+d) \big) \omega e\\
&= a^2 \omega ( a \omega 25 \omega d ) \omega e\tag{2}
\end{align}
$$

Now,

$$
\begin{align}
]25 \omega d[ &= ]25[ + ]d[\\
&= 2 + (]c[ – 1)\\
&= 2 + \big((]a[ – 1) – 1\big)\\
&= ]a[
\end{align}
$$

So, in the group $a \omega 25 \omega d$, $a$ is excess.

$$
\begin{align}
\therefore (2) &= (a^2 + a) \omega (25 \omega d) \omega e\\
&= a(a+1) \omega 25 \omega d \omega e\\
&= a(a+1) \omega 25 \omega c \omega c\tag{Shifting a zero from e to d}\\
\end{align}
$$

So finally,

$$
\begin{align}
&(a\omega 5 \omega c)^2 = a(a+1) \omega 25 \omega c \omega c\\
&\Rightarrow \big((a\omega 5) \times 10^c\big)^2 = \big(a(a+1) \omega 25 \omega c\big) \times 10^c\\
&\Rightarrow (a\omega 5)^2 \times 10^{2c} = \big(a(a+1) \omega 25\big) \times 10^{2c}\\
&\Rightarrow (a\omega 5)^2 = a(a+1) \omega 25\tag{Proved}
\end{align}
$$

So, we see that this trick is applicable for all kinds of whole numbers that end with 5.

Check out other applications of Digit Math

Link to list of other applications of Digit Math.

January 11, 2012June 11, 2020 by Nirupam Biswas

Digit Math Application: Proving that multiplying with 10**n puts n zeros at the end

Please read Digit Math: Introduction before you continue.

The problem

This is a very fundamental concept that we were taught when we were in junior schools. Now, think of it, what it says. If you add $x$ ten times then you will get $x\omega0$. If you add $x$ hundred times then you will get $x\omega00$; and so on. How do we know that this will always be true? We know it since we have never seen one violation of it, but, anyway I will try to prove it know to rest the uneasy souls.

The proof

Before I continue with this proof, I must make sure we understand another empirical rule.

When any number is multiplied by zero then the result is always zero. This is not hard to see why. When you multiply $x$ by $2$ then it is like adding $x$ twice. When you multiply $x$ by $1$ then it is like adding $x$ once. When you multiply $x$ by $0$ then it is like adding $x$ zero times, which in other words is that we never added $x$ in the first place so $x$ never existed, so we had nothing, and that nothing is zero.

Back to proof.

Let the number $a\omega b$ be multiplied by $1\omega d$, where $d$ is all all zeroes, $n$ times. To use Bimultiplication forumula we need to make sure that $b$ too has $n$ digits. We can always partition a number such that $b$ will always have $n$ digits. For example if $d = 000$ and $a\omega b = 2$ then we make $a = 000$ and $b = 002$.

So,

$$
\begin{align}
& a \omega b \times 10^d\\
&= a \omega b \times 1 \omega d\\
&= a.1 \omega (a.d + b.1) \omega b.d\tag{By Bimultiplication}\\
&= a \omega (d + b) \omega d\tag{Since d is all zeroes}\\
&= a \omega b \omega d\tag{Since d is all zeroes}
\end{align}
$$

So, in the above equations we see that multiplying a number by $10^d$ will give us the same number, but followed with $d$ zeroes. (Proved)

Check out other applications of Digit Math

Link to list of other applications of Digit Math.

Bertrand Russell (Famous Mathematician and Logician)

January 11, 2012August 17, 2021 by Nirupam Biswas

Digit Math: Introduction

Digit Math, Featured, Math, My Mind, My Work, Regulars
5 Comments

The person in the above image is Bertrand Russell. Famous Mathematician and Logician. An excellent graphic novel, Logicomix, is based around his life. That is a must read.

First thing first. Even if you are not a Math demi-god, it is fine, Digit Math is still for you. In fact my academic track record shows that I barely survived Mathematics! I invented this thing when I was in standard eleventh while trying to create some shortcut formula, because I kept forgetting long formulae and was very slow in arithmetic. Digit Math is a slightly different approach to classical Math. To understand this all you need to know is how to add and multiply, and a little bit of logic. That is it.

What is Digit Math?

Numbers are group of digits. Classical Mathematical operators are not meant to deal with few digits of a number. Digit Math allows that. You can interact with a single digit or a group of digits in a number. You might wonder, what is the use? Well, I would say, for fun. Like any good puzzle, this too forces you to think differently. It is not all pointless fun though.

Can you prove that any number which ends with 5 must be divisible by 5? Well try that using classical Math. I have used Digit Math to prove the same. There some more empirical facts which I have proved using Digit Math. One of that is proving that multiplying any number by $10^n$ will indeed give us that number but followed by $n$ zeroes. This might look obvious to you, but think for a moment, why does adding a number like $6$, $100$ times is guaranteed to yield the exact number $600$.

Before we begin…

Before we begin I must list out the notations and symbols I would be using. They were introduced specifically for this, since I don’t know of any existing symbols which convey the same information.

Symbols and Operators

I would be using these symbols throughout the articles on Digit Math, so pay attention to them.

$\omega$ (With-operator)

This is the main operator, which delimits the digits or groups of digits. I call this the ‘with-operator’. So, 251 can be written as $2\omega51$ (spoken as two-with-fifty-one) or $25\omega1$ (spoken as twenty-five-with-one) or $2\omega5\omega1$ (spoken as two-with-five-with-one) or even $0\omega251$. They all are the same, only the digit partitioning is different. How you partition depends on the problem at hand.

$]x[$ (Digit Group operator)

This is used in multiple ways. Its primary use is to group digits into a digits group. While in $2\omega51$ it is quite clear that 2 and 51 are different digit groups of one and two digits respectively, but sometimes we need to convey more information about a digit group.

$]x[$ – This is equal to the number of digits in x, when used on its own. So, $]25[ = 2$.
$]x[^y$ – This means that x should have y digits. Note that I said ‘should’ have. So, it is possible that while solving the problem, at intermediate steps we can have less or more digits. Eventually, the final result must have exactly y digits. This kind of notation is usually used when actual value of x is not known and x is part of a bigger number. So, $2\omega]x[^2$ means that this is a number which starts with 2 but ends with two digits.
Underline notation – Take the example of $2\omega]345[^2$. Here 3 is the extra digit in this digit group which has the maximum capacity of two digits. In this case since all the digits are known, so here we can use a short-hand notation of underlining the excess digit in the group. So, $2\omega]345[^2 = 2\omega\underline{3}45$. (Note that 3 is underlined.)

$\omega\sum$ (Digits summation operator)

So,$$\omega\sum_{i=1}^3a_i = a_3{\omega}a_2{\omega}a_1$$

Note that $a_1$ is the rightmost digit and $a_3$ the leftmost. Maintaining the sequence in digitized equations are of massive importance.

Precedence of With-operator

$\omega$ has lower precedence than multiply and division. So the full operator precedence would be:

B – Bracket
O – Orders (powers, roots, etc.)
M – Multiply
D – Divide
W – With ($\omega$)
A – Add
S – Subtract

Properties of With-operator

This is not commutative. This is because, $2\omega5 = 25 \neq 52 = 5\omega2$.
This is associative, but only when none of the digit groups have excess digits. That is, less number of digits than required is tolerated. So, $a\omega(b{\omega}c) = (a{\omega}b){\omega}c$.
This is distributive, but only when all digit groups have the exact number of digits. So, less digits is not tolerated here. So, $n \times (a{\omega}b) = (na)\omega(nb)$.

Some Important Formulae

General formula of transformation

This provides the relation between a normal equation and a digitized equation.

So,$$\omega\sum_{i=1}^{n}a_i = \sum_{i=1}^{n}a_i \times {10}^{n-i}$$

Provided $a_i$ has one digit only (i.e. $]a_i[ = 1$). This is not hard to see, how the above is true.

Bi-multiplication formula

This formula is vital. This describes how the groups of digits behave when they are multiplied. This starts with how we usually manually multiply two numbers.

$$
\begin{align}
x\omega y&\\
\underline{\times\hspace1em a\omega b}&\\
bx\omega by&\\
ax \omega ay\hspace1.5em&\\
\overline{ax\omega (ay+bx)\omega by}&
\end{align}
$$

So,

$$\underline{x \omega y \times a \omega b = ax \omega (ay+bx) \omega by}\tag{bi-multiplication formula}$$

The above relation is pretty easy to understand, but it does say how many digits should be in each group (digits separated by $\omega$). To understand this, try multiplying 12 by 34, the manual way. This would be like

$$
\begin{align}
1\omega 2&\\
\underline{\times\hspace1em 3\omega 4}&\\
4\omega 8&\\
3 \omega 6\hspace1.5em&\\
\overline{3\omega (6+4)\omega 8}&\\
\Rightarrow 3\omega 10\omega 8&\\
\Rightarrow 4\omega 0\omega 8&\tag{adding carry over 1 from center to left group}\\
\Rightarrow 408&
\end{align}
$$

Now lets try multiplying 111 by 123.

$$
\begin{align}
01\omega 11&\tag{prefixing with 0 to equate digit counts}\\
\underline{\times\hspace1em 01\omega 23}&\\
23\omega 253&\\
1 \omega 11\hspace1.5em&\\
\overline{1\omega (11+23)\omega 253}&\\
\Rightarrow 1\omega (11+23+2)\omega 53&\tag{2 is the carry over from 253}\\
\Rightarrow 1\omega 36\omega 53&\\
\Rightarrow 13653&
\end{align}
$$

So you can observe from the above scenarios that:-

$$
]a\times b[ = n\hspace1em\ldots\text{where ]a[ = ]b[ = n}
$$

Similarly,

$$
]a + b[ = n\hspace1em\ldots\text{where ]a[ = ]b[ = n}
$$

Any excess digits should be carried over to the digit group on the left and added there.

So,

$$
\begin{align}
1&\\
\underline{+\hspace1em9}&\\
0\omega\underline{1}0&\\
\Rightarrow (0+1)\omega0\\
\Rightarrow 1\omega0\\
\Rightarrow 10
\end{align}
$$

If $]a[ \neq ]b[$ then I am not sure what should be $]a \times b[$. So, we need to make sure that all digit groups have equal number of digits. This is restrictive but fortunately this does not cause problem as there is always a way around.

One more example. Let’s multiply 98 by 76.

$$
\begin{align}
9\omega 8&\\
\underline{\times\hspace1em 7\omega 6}&\\
54\omega 48&\\
63 \omega 56\hspace1.5em&\\
\overline{63\omega (56+54)\omega 48}&\\
\Rightarrow 63\omega 110\omega 48&\\
\Rightarrow 63\omega (110+4)\omega 8&\tag{since max size of each group is 1}\\
\Rightarrow 63\omega 114\omega 8&\\
\Rightarrow (63+11)\omega 4\omega 8&\\
\Rightarrow 74\omega4\omega8&\\
\Rightarrow 7448&
\end{align}
$$

Rules for using the bi-multiplication formula

You have already witness many of the rules, but let me summarize them for clearly. For the following assume that we are multiplying $a\omega b$ by $x\omega y$.

Make two fragments of both the numbers, by placing $\omega$ at the positions suitable for you. Remember, the number of digits in all the four fragments should be equal. That is, $]a[ = ]b[ = ]x[ = ]y[$.Partitioning $255$ as $2\omega55$ is perfectly fine, since $]02[ = ]55[ = 2$.
If the number of digits in $by$ or $ay+bx$ terms are less than the required number then insert zeroes in front of them to get the exact number of digits.
If the number of digits in $by$ or $ay+bx$ terms are more than the required number then remove the excess digits from the front (left) of them and add these excess digits to the group on left. Note, that shifting of excess digits (carry over) should started from the rightmost group and then progress towards left sequentially.
When all the above conditions are met then you are good to remove the $\omega$ (with-operator).

So, finally the bi-multiplication formula can be precisely expressed as,

$$\boxed{x \omega y \times a \omega b = ax\hspace2pt\omega\hspace2pt]ax+by[^n\hspace2pt\omega\hspace2pt]by[^n}\hspace1em\ldots\text{where ]a[ = ]b[ = ]x[ = ]y[ = n}$$

Squaring formula

This is directly derivable from bi-multiplication formula.

$$
\begin{align}
(a \omega b)^2&\\
\Rightarrow& a \omega b \times a \omega b\\
\Rightarrow& \boxed{a^2\omega ]2ab[^n \omega ]b^2[^n}\hspace1em\ldots\text{where ]a[ = ]b[ = n}
\end{align}
$$

General Powering Formula

For now I will provide only the formula. The proof is provided later in this article. The way I derived it was to I manually find the values of $(a\omega b)^2$, $(a\omega b)^3$, $(a\omega b)^4$ and so on. I found a pattern in all these and from there I got this formula. It was later when it struck me as to how to prove it.

$$
\boxed{(a\omega b)^n = \omega\sum_{r=0}^n\hspace1em{]^nC_r \times a^{(n-r)} \times b^r[}^d}\hspace1em\ldots\text{where ]a[ = ]b[ = d}
$$

Notice that the above equation looks very similar to Binomial equation. In fact I found this to be true for many equations. If we replace $+$ by $\omega$ then we end up with a ‘digitized’ version of that equation.

Now let me demonstrate how to use this equation. Let’s find cube of 99, i.e. $99^3$.

$$
\begin{align}
&(9\omega9)^3\\
&= ]^3C_0 \times 9^3 \times 9^0[^1 \omega ]^3C_1 \times 9^2 \times 9^1[^1\\
&= 9^3 \omega 3\times9^3 \omega 3\times9^3 \omega 9^3\tag{for brevity removing ].[ operator}\\
&= 729 \omega \underline{218}7 \omega \underline{218}7 \omega \underline{72}9\\
&= 729 \omega \underline{218}7 \omega \underline{225}9 \omega 9\tag{adding carry overs}\\
&= 729 \omega \underline{241}2 \omega 9 \omega 9\\
&= 970 \omega 2 \omega 9 \omega 9\\
&= 970299
\end{align}
$$

I didn’t underline the excess digits in leftmost group, since that is anyway not going to affect the result. Now let us find $241^4$.

$$
\begin{align}
&(02\omega41)^4\\
&= \omega\sum_{r=0}^4\hspace1em]^4C_r \times 2^{(4-r)} \times 41^r[^2\\
&= ^4C_0.2^4\hspace0.5em\omega\hspace0.5em^4C_1.2^3.41\hspace0.5em\omega\hspace0.5em^4C_2.2^2.41^2\hspace0.5em\omega\hspace0.5em^4C_3.2.41^3\hspace0.5em\omega\hspace0.5em^4C_4.41^4\\
&= 16\omega\underline{13}12\omega\underline{403}44\omega\underline{5513}68\omega\underline{28257}61\\
&= 16\omega\underline{13}12\omega\underline{403}44\omega\underline{5796}25\omega61\\
&= 16\omega\underline{13}12\omega\underline{461}40\omega25\omega61\\
&= 16\omega\underline{17}73\omega40\omega25\omega61\\
&= 33\omega73\omega40\omega25\omega61\\
&= 3373402561
\end{align}
$$

Note, one important observation. Here you can immediately identify the last couple of digits of the final result by evaluating for the last term. Unlike in Binomial Theorem, where all the terms need to be added. So, there at least last few digits of all the terms need to be added to get the last few digits of the final result. This is an advantage for General Powering formula, but unfortunately we cannot say the for other digit groups which are at higher place order. The reason is simple, they may have to be added with carry-overs from the groups (terms) on the right.

A corollary of the General Formula of Transformation (for two digits)

With to Plus form

$$a\omega b = 10a+b$$

The above equation is true only when $b$ has one digit. The more general equation is:-

$$\boxed{a\omega b = a.10^d+b}\hspace1em\text{where d}=]b[\hspace1em\ldots\text{Corollary 1}$$

Plus to With form

$$a\omega b = 10a +b \Rightarrow a\omega b = 9a + a + b \Rightarrow \boxed{a+b = a\omega b -9a}\hspace1em\ldots\text{Corollary 2}$$

Negative digits

We all are aware of negative numbers but in digit equations we might end up with negative digits! I see no physical significance of that but let’s try to find out some mathematical meaning of this.

$$
\begin{align}
a\omega(-b) &= 10a + (-b)\\
&= 10a -b\\
&= 10a + b -2b\\
&= a\omega b -2b
\end{align}
$$

Altire,

$$
\begin{align}
a\omega(-b) &= -(-10a + b)\\
&= -(10a+b -20a)\\
&= -(a\omega b – 20a)\\
&= -(a\omega b) +20a
\end{align}
$$

$$
\therefore \boxed{a\omega(-b) = a\omega b-2b = -(a\omega b)+20a}
$$

Remember, $(-a)\omega b \neq -(a\omega b)$. The former means that only the digit $a$ is negative. The later means that the complete number is negative.

Similarly,

$$
\boxed{(-a)\omega b = a\omega b-20a = -(a\omega b)+2b}
$$

Also similarly evaluating, we get,

$$
\boxed{(-a)\omega(-b) = -(a\omega b)}\hspace1em\ldots\text{Corollary 3}
$$

So, we see that in a negative number all the digits too are negative. Not surprising, since in a positive number all digits are positive.

Proof of General Powering Formula

Since I already know this formula, so I just need to prove that it is correct. For this the best tool is Mathematical Induction. So, the formula to prove is

$$(a\omega b)^n = \omega\sum_{r=0}^n\hspace1em{]^nC_r \times a^{(n-r)} \times b^r[}^d$$

Case 1: n = 0

So,

$$
\begin{align}
L.H.S. &= (a\omega b)^0 = 1\\
\\
\\
R.H.S. &= \omega\sum_{r=0}^0\hspace1em ^0C_r \times a^{(0-r)} \times b^r\\
&= 1\times a^0 \times b^0\\
&= 1\tag{Proved}
\end{align}
$$

Case 2: n = 1

$$
\begin{align}
L.H.S.&= (a\omega b)^1 = a\omega b\\
\\
\\
R.H.S. &= \omega\sum_{r=0}^1\hspace1em ^1C_r \times a^{(1-r)} \times b^r\\
&= ^1C_0. a^1. b^0 \omega ^1C_1.a^0.b^1\\
&= a\omega b\tag{Proved}
\end{align}
$$

Case 3: For n

Suppose that the formula is true for all values of $n$, then by Mathematical Induction, the formula must be valid for $n+1$ too.

So,

$$
\begin{align}
&(a\omega b)^{(n+1)}\\
&= (a\omega b)^n \times (a\omega b)\\
&= (a\omega b)^n \times (a\times 10^d+b)\hspace1em\ldots\text{where d}=]b[\tag{By Corollary 1}\\
&= 10^da\times(a\omega b)^n + b\times (a\omega b)^n\\
&= 10^da\times(\omega\sum_{r=0}^n\hspace1em ^nC_r\times a^{(n-r)} \times b^r) +\\
&\hspace2em b\times(\omega\sum_{r’=0}^n\hspace1em ^nC_{r’}\times a^{(n-r’)} \times b^{r’}) \tag{Since formula is correct for n}\\
&= 10^d\times(a^{(n+1)} \hspace2pt\omega\hspace2pt \omega\sum_{r=1}^n\hspace1em ^nC_r\times a^{(n+1-r)} \times b^r) +\\
&\hspace2em(\omega\sum_{r’=0}^{(n-1)}\hspace1em ^nC_{r’}\times a^{(n-r’)} \times b^{(r’+1)} \hspace2pt\omega\hspace2pt b^{(n+1)})\tag{1}
\end{align}
$$

Since $]a[\hspace2pt=\hspace2pt]b[\hspace2pt=\hspace2pt d$, so every term of $\omega\sum$ too will have $d$ digits, as per General Powering Formula. This implies that term $b^{(n+1)}$ in equation (1) too has $d$ digits. So, if we can rewrite (1) as below:-

$$
\begin{align}
a^{(n+1)} \hspace2pt\omega\hspace2pt&\omega\sum_{r=1}^n\hspace1em ^nC_r\times a^{(n+1-r)} \times b^r\tag{2}\\
+\hspace2em\hspace2em &\omega\sum_{r’=0}^{(n-1)}\hspace1em ^nC_{r’}\times a^{(n-r’)} \times b^{(r’+1)} \hspace2pt\omega\hspace2pt b^{(n+1)}\tag{3}\\
a^{n+1} \hspace2pt\omega\hspace2pt & \overline{\omega\sum_{r=1}^n\hspace1em {^nC_r\times a^{n+1-r} \times b^r \brace + ^nC_{(r-1)}\times a^{n-(r-1)} \times b^{(r-1)+1}} \hspace2pt\omega\hspace2pt b^{n+1}}\\
\Rightarrow a^{n+1} \hspace2pt\omega\hspace2pt &\bigg( \omega\sum_{r=1}^n\hspace1em ^nC_r\times a^{n+1-r} \times b^r + ^nC_{r-1}\times a^{n+1-r} \times b^{r}\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\tag{4}
\end{align}
$$

Notice that in the above, I have removed the $10^d$ factor since all it did is to shift its term by $d$ digits to left, that is, by exact number of digits in the $b^{(n+1)}$ group. After removing the first digit group from equation (2) and last digit group from equation (3), $\omega\sum$ in both the equations have $n-1$ digit groups left. Each of these digit groups will get added term by term, i.e. group $1$ of (2) will get added to group $0$ of (3) and so on. So this implies $r’ = r – 1$.

Further simplifying,

$$
\begin{align}
(4)&= a^{n+1} \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em a^{n+1-r}.b^r(\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!})\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\\
&= a^{n+1} \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em a^{n+1-r}.b^r.\frac{(n+1)!}{r!(n+1-r)!}.(\frac{n+r-1}{n+1}+\frac{r}{n+1})\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\\
&= a^{n+1} \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em a^{n+1-r}.b^r.^{n+1}C_r.(1)\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\\
&= \bigg(C_0^{n+1}.a^{n+1-0}.b^0\bigg) \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em C_r^{n+1}.a^{n+1-r}.b^r\bigg) \hspace2pt\omega\hspace2pt \bigg(C_{n+1}^{n+1}.a{(n+1-(n+1))}.b{n+1}\bigg)\\
&= \omega\sum_{r=0}^{(n+1)}\hspace1em ^{(n+1)}C_r.a^{(n+1)-r}.b^r\\
\end{align}
$$

So, proved that $(a\omega b)^{(n+1)} = \omega\sum_{r=0}^{(n+1)}\hspace1em ^{(n+1)}C_r.a^{(n+1)-r}.b^r$ when $(a\omega b)^n = \omega\sum_{r=0}^n\hspace1em ^nC_r.a^{n-r}.b^r$.

So, all cases proved.

Application of Digit Math

Now time to see Digit Math in action. I will be using this to prove some empirical concepts.

Link to list of applications of Digit Math.

September 24, 2011June 9, 2018 by Nirupam Biswas

CInk version 2 finally released!

CInk, Coding, Featured, Javascript, My Work, Open Source, Regulars
Leave a comment

Finally CInk version 2 has been released.

Some key new things

New website with complete API documentation and guides on how you can use CInk JS code.
Finally released the full source code of CInk renderer and compiler. License – GPL v3.
CInk finally supports all the features of original CFDG, including Paths.
CInk has introduced support for texts which extends the capabilities of CFDG considerably. Check out the cool demo – Neon Letters. To learn more about it see – “Text transforms” section here.

Last but not the least, you can post comments on CInk website. The comments section is at the bottom of each page.

Goto Cink website – cink.applegrew.com.

September 23, 2011January 8, 2012 by Nirupam Biswas

Create your own Cyberoam client in Python!

This is a very old post. Reposting it since while migrating it from blogger.com to here it got lost. Found it recently. Not sure how relevant or correct is this now.

Cyberthon updated to version 2.5 by Vinit and Siddhartha Sahu:-
Get the latest code from https://github.com/siddharthasahu/Cyberthon-enhanced.

Cyberthon updated to version 1.2:-
Added a new feature in this version. Now Cyberthon won’t exit ever, once started, even when it can’t connect to Cyberoam server. If liked its old behavior then set never_quit to False

Code fixed now. Sorry, I didn’t notice before (for 2 months since posting it here) that the code here was not working. I think it is fixed now. The problem was that the tabs were being expaded here differently. So, I just ran expand cyberthon.py command to expand all tabs in it beforehand.

My college uses the software Cyberoam for controlling the bandwidth and timing allocated to us for surfing the internet. If you don’t know what is it then visit here. So, to access the internet we all need to install the client software and login via that. But I don’t like to install it because it feels like a college spy on my computer, you may find it irrational but I can’t help being a paranoid. I have always used the web login page of Cyberoam but it is not a very good solution, first because I will always need to keep the browser open. (The Cyberoam page refreshes at a specific time periodically. If it fails to do so then you get logged out.) Second problem is that on my Linux platform if I work from console mode then I can’t login because no browser can run without GUI support (I know of Lynx but I haven’t been able to use it for it.)

Recently I thought of investigating the working of the web login page of Cyberoam. After bringing up the login page of Cyberoam I changed the value of attribute method of form tag to GET from POST. So after filling-up the username and password when I hit enter key, there it was all the details that are sent from my browser to Cyberoam server in my browser address bar. I thought of simulating the POST action using a Python code. It required only a pinch of Python to do this. I tried the following Python code to simulate that and it worked, as it should. You too can try the following code and login to Cyberoam.

Code:

[code lang=”python” wraplines=”false”]
import urllib
urllib.urlopen("http://192.168.5.10:8090/corporate/servlet/CyberoamHTTPClient","mode=191&isAccessDenied=null&url=null&message=&username=YourUserName&password=YourPassword&saveinfo=saveinfo&login=Login")[/code]

Be sure to substitute YourUserName and YourPasswordfor your username and password respectively. The above will get you logged in but won’t keep you logged it because you need to re-post the login data at every (usually) 3 minutes. So, I got down to work again. Below is quick and dirty shell script to get around this problem. Note the above code has been distorted and compressed into single line and passed via piping to Python in script below.

Code:

[code lang=”bash” wraplines=”false”]
while [[ 1 == 1 ]]
do
printf "%s\n%s\n" "import urllib" "urllib.urlopen(\"http://192.168.5.10:8090/corporate/servlet/CyberoamHTTPClient\",\"mode=191&isAccessDenied=null&url=null&message=&username=YourUserName&password=YourPassword&saveinfo=saveinfo&login=Login\")" |python
sleep 3m
done[/code]

Again in the above code do replace YourUserName and YourPassword for your actual username and password respectively. Well now it was working great. 🙂 As usual I was again not satisfied because how will I know if I have logged in successfully or not? The only way for that was by parsing the returned page after logging-in for the message. I noticed that returned page’s source code contained…

[code lang=”html”][/code]

Please notice the part message. The value after that is the message I was seeking. So, all I needed is to parse this page for the value of attribute src of the tag frame. Also, note this very string also contains some more very useful values, viz. – loginstatus and liverequesttime. I noticed that value of loginstatus becomes true when logged in otherwise it remains false, and the value of liverequesttimegiave the time (in seconds) in which to re-post the login data.

The HTML parsing class in the Python script (which I wrote) below is based on the code from the great HTML parsing tutorial located here. Note the script uses the zenity command to display GUI dialog boxes. If you do not have this installed or simply don’t want to display GUI dialog boxes then start the script with -nogui argument. If you want the script to remain completely silent then use the -silent argument. I call this script Cyberthon (=Cyberoam+Python) 😉

Requirement(s):-
1) Python 2.5 (may work with previous versions, but not tested)
2) Zenity (needed to show the dialog boxes, use -nogui switch if you want the messages to appear on the console.)

Code:

[code lang=”python” wraplines=”false”]#!/usr/bin/python
#!/usr/bin/python

#Program Name: Cyberthon (Python Cyberoam Client)
#Coder AppleGrew
#License GPL
#Version 1.2
cyberroamIP = "192.168.5.10" #The IP of the Cyberoam site.
cyberroamPort = "8090" #Set to "" if not using.
username = "your_username" #Your username
passwordFile = "/home/you/.passwd" #Path file containing a single string, your password.
sleeptime = 0 #in minutes or set to 0, it will then parse this value from the cyberoam returned page dynamically.
never_quit = True #Once started cyberthon will never, even when the cyberoam server cannot be connected.

import sys

silent = False
nogui = False
for arg in sys.argv:
if "-silent" == arg:
silent = True
if "-nogui" == arg:
nogui = True

#Parsing and logging in too.
import sgmllib

class MyCyberroamParser(sgmllib.SGMLParser):
"A simple parser class."

def parse(self, s):
"Parse the given string ‘s’."
self.feed(s)
self.close()

def __init__(self, verbose=0):
"Initialise an object, passing ‘verbose’ to the superclass."

sgmllib.SGMLParser.__init__(self, verbose)
self.required_entities = [‘message’,’loginstatus’,’liverequesttime’]
self.frames_attr = []
self.in_required_entity = False
self.current_entity = ""
self.entity_values = {}

def do_frame(self, attributes):
for name, value in attributes:
if name == "src":
self.frames_attr.append(value)

def unknown_entityref(self,ref):
self.current_entity = ref
if ref in self.required_entities:
self.in_required_entity=True

def handle_data(self, data):
"Try to get the value of entity &message. Used in 2nd pass of parsing."

if self.in_required_entity:
self.entity_values[self.current_entity] = data[1:] #To remove the preceeding =
self.in_required_entity = False

def get_src(self,index=-1):
"Return the list of src targets."
if index == -1:
return self.frames_attr
else:
return self.frames_attr[index]

import urllib, sgmllib,time,commands,os

pf = open(passwordFile)
passwd = pf.readline()
pf.close()
if passwd[-1] == ‘\n’: #Removing terminating newline character.
passwd = passwd[:-1]

cyberroamAddress = cyberroamIP
if cyberroamPort != "":
cyberroamAddress = cyberroamAddress+":"+cyberroamPort

sec2sleep = 60*sleeptime
lastmsg = ""
msgChanged = True
lastMsgWasFailMsg = False
sec2sleepOnError = 6
while True:
try:
# Logging in and fetching the Cyberroam login page.
f = urllib.urlopen("http://"+cyberroamAddress+"/corporate/servlet/CyberoamHTTPClient","mode=191&isAccessDenied=null&url=null&message=&username="+username+"&password="+passwd+"&saveinfo=saveinfo&login=Login")
sec2sleepOnError = 6
except IOError, (errno, strerror):
if not silent:
print "Connection to Cyberoam server timed out. Error(%s): %s" % (errno, strerror)

if sec2sleepOnError > 30:
if not silent:
if nogui:
print "Quitting program."
else:
if never_quit:
if not lastMsgWasFailMsg:
os.popen(‘zenity –info –text="Failed to connect to server, but I am NOT quitting." –title="Cyberthon" >/dev/null’)
lastMsgWasFailMsg = True
else:
commands.getoutput(‘zenity –info –text="Could not connect to the server. Quitting program." –title="Cyberthon"’)
if not never_quit:
sys.exit(1)
else:
sec2sleepOnError = 6

if not silent:
print "Retrying in %s seconds" % sec2sleepOnError

time.sleep(sec2sleepOnError)
sec2sleepOnError = sec2sleepOnError*2
continue

s = f.read()

# Try and process the page.
# The class should have been defined first, remember.
myparser = MyCyberroamParser()
myparser.parse(s)

# Get the the src targets. It contains the status message. And then parse it again for entity &message.
qindex = myparser.get_src(1).index(‘?’)
srcstr = myparser.get_src(1)[:qindex+1]+’&’+myparser.get_src(1)[qindex+1:]

myparser.parse(srcstr)

message = myparser.entity_values[‘message’]
if lastmsg != message or lastMsgWasFailMsg:
lastmsg = message
msgChanged = True
lastMsgWasFailMsg = False

if (not silent) and msgChanged:
msgChanged = False

msg=”
i=0
while i < len(message): #Converting hex nos. to characters. t=message[i] if message[i]==’%’: no=int(message[i+1:i+3],16) t=chr(no) i=i+2 msg=msg+t i=i+1 message = "" for x in msg:#Changing all + to space. if x == ‘+’: x = " " message=message+x if nogui: print message else: os.popen(‘zenity –info –text="From Cyberoam: ‘+message+’" –title="Cyberthon" >/dev/null’)

if myparser.entity_values[‘loginstatus’].lower()!="true":
break;

if sleeptime==0:
sec2sleep = int(myparser.entity_values[‘liverequesttime’])
time.sleep(sec2sleep)
[/code]

Please copy the file very carefully. A single extra space here and there may throw Python haywire, making it throw up all sorts of nasty syntax errors. Paste the contents of this code in a file and name it – Cyberthon.py.

August 26, 2011June 9, 2018 by Nirupam Biswas

Migrate from Apache to Nginx: The new guide

`AG_BACKQUOTE_TURN_ON`

If you are here then most probably, like me you too want to migrate from Apache to Nginx. Well I have already migrated and I am loving it! You can get a quick recap of Apache vs Nginx comparison here. Via this blog post I will share and explain my Nginx conf, in the hope that this can prove helpful for you. I have assumed that you have Nginx and Php-FPM installed. You can read how to install Ngnix from here – http://goo.gl/tq6vT. I installed it from its source. You can install PHP-FPM as per the guide at – http://goo.gl/Fx9hJ. My repo had PHP-FPM so I was saved from this trouble.

The net is littered with loads of blog posts on this topic, but most of them are either out of date or make contradictory suggestions. I have scoured the net and cooked up my own Nginx config based on many helpful blog posts on them. My config has been progressively tweaked as per my needs. I now feel that it is good enough to be shared.

To see if my Nginx config suites your needs, you first need to understand my use case. I have four domains hosted on the same server – applegrew.com, cink.applegrew.com, fb.applegrew.com and this blog.

applegrew.com – This serves some static HTML, few XMLs and dynamic web pages (PHP).
cink.applegrew.com – This is my Chrome Experiment site and has no dynamic web pages, only static files like HTMLs, images, text files, etc.
fb.applegrew.com – I added this one recently to host my Facebook Apps. So, naturally this has dynamic web pages, coded in PHP. Since, FB has mandated that from 1 Oct, 2011, all FB Apps must be accessible via HTTPS, so, this domain is configured to be accessible via both HTTP and HTTPS. The Nginx config for this takes care of setting the PHP parameter `$_SERVER[‘HTTPS’]` when HTTPS is used.
blog.applegrew.com– Configuring Nginx for this blog was no easy task. This blog is powered by WordPress. If you too have a WordPress blog then you MUST install the following WP plugins for performance. The Nginx config that I have shared assumes that these plugins are installed and takes full advantage of it.

Must have WordPress plugins:-
- WP Super Cache– Excellent plugin which generates static HTML files for your blog. Nobody updates their blog every minute. It’s not a Twitter. So, why generate that same page again and again for every user who visits you blog? The solution is to cache the generate page. Later when any user visits your blog then that user will be served the cached page. This saves a ton of overhead. Particularly when you are using Nginx, since, we run PHP and Nginx processes separately. So we can configure Nginx to serve the generated file, if present, and completely bypass PHP. This plugin is smart enough to refresh the cache when you make a new post or update it.
  Tip: Install this plugin after you have finalized your sites design, else you will have to manually clean the cache to make the site changes available.
- WP Minify– This plugin strips out all the JS and CSS links from your blog and then combines them generate a unified CSS and JS. The result is cut down on the number of requests to your server for additional CSS and JS files. This plugin also minifies the combined CSS and JS files, which produces a much smaller file.
  Tip: If you install a new plugin after you install this one, and if that is not working, then try clearing the cache of this plugin. Since, it is possible that the new plugin will try to put some new CSS or JS which might get stripped out but not cached in the combined file.

Now its time for the configs.

nginx.conf

[code lang=”cink”]
user apache apache; #The uid and gid of the nginx process
worker_processes 4; #Number of worker processes that needs to be created.

error_log /var/log/error-n.log;

pid /usr/local/nginx/logs/nginx.pid;

events {
worker_connections 1000;
}

http {
include mime.types; #Includes a config file which is available with ngix’s default installation.
index index.html index.htm index.php index.shtml;

log_format main ‘$remote_addr – $remote_user [$time_local] "$request" ‘
‘$status $body_bytes_sent "$http_referer" ‘
‘"$http_user_agent" "$http_x_forwarded_for"’;

sendfile on;
keepalive_timeout 5;

gzip on;

# Sets the default type to text/html so that gzipped content is served
# as html, instead of raw uninterpreted data.
#default_type text/html;

server {#If someone tries to access the url http://applegrew.com/xxx then
#this will redirect him to http://www.applegrew.com/xxx.
server_name applegrew.com;
rewrite ^ http://www.applegrew.com$request_uri? permanent;
}

server {#The config for www.applegrew.com
server_name www.applegrew.com;

access_log /var/www/applegrew.com/access-n.log main; #Where access log will be written for this domain.
error_log /var/www/applegrew.com/error-n.log;

root /var/www/applegrew.com/html; #The document root for this domain.

location ~ /admin/ { deny all; } #Denies access to some www.applegrew.com/admin/ url
location ~ /private/ { deny all; }

include cacheCommon.conf; #This caches common static files. This config is given later in this post.
include drop.conf; #This config is given later in this post.
include php.conf; #Configures PHP access for this domain. This config is given later in this post.
include err.conf; #Some common custom error messages I show. This config is given later in this post.
}

server {#Config to serve HTTP traffic.
server_name fb.applegrew.com;

access_log /var/www/fb.applegrew.com/access.log main;
error_log /var/www/fb.applegrew.com/error.log;

root /var/www/fb.applegrew.com/html;

include cacheCommon.conf;
include php.conf;
include drop.conf;
include err.conf;
}

server {//Config to serve HTTPS traffic.
listen 443;
server_name fb.applegrew.com;

ssl on;
ssl_certificate /var/ssl/fb.applegrew.com.crt; #See http://goo.gl/mvHo7 to know how to create crt file.
ssl_certificate_key /var/ssl/fb_applegrew_com.key;

access_log /var/www/fb.applegrew.com/access.log main;
error_log /var/www/fb.applegrew.com/error.log;

root /var/www/fb.applegrew.com/html;

include cacheCommon.conf;
include phpssl.conf; #Notice the difference. This is not php.conf. This config will be provided later in this post.
include drop.conf;
include err.conf;
}

server {
server_name blog.applegrew.com;

access_log /var/www/blog.applegrew.com/access-n.log main;
error_log /var/www/blog.applegrew.com/error-n.log;

root /var/www/blog.applegrew.com/html;

#If tgz file mathcing the request already exists then that will be sent, skipping on the fly compression by nginx.
gzip_static on;

location / {
# does the requested file exist exactly as it is? if yes, serve it and stop here
if (-f $request_filename) { break; }

# sets some variables to help test for the existence of a cached copy of the request
set $supercache_file ”;
set $supercache_uri $request_uri;

# IF the request is a post, has a query attached, or a cookie
# then don’t serve the cache (ie: users logged in, or posting comments)
if ($request_method = POST) { set $supercache_uri ”; }
if ($query_string) { set $supercache_uri ”; }
if ($http_cookie ~* "comment_author_|wordpress|wp-postpass_" ) {
set $supercache_uri ”;
}

# if the supercache_uri variable hasn’t been blanked by this point, attempt
# to set the name of the destination to the possible cache file
if ($supercache_uri ~ ^(.+)$) {
set $supercache_file /wp-content/cache/supercache/$http_host/$1index.html;
}

# If a cache file of that name exists, serve it directly
if (-f $document_root$supercache_file) { rewrite ^ $supercache_file break; }

# Otherwise send the request back to index.php for further processing
if (!-e $request_filename) { rewrite . /index.php last; }

#try_files $uri $uri/ /index.php;
}
location ~ /wp-config\.php { deny all; }
location ~ /wp-content/bte-wb/.*\..* { deny all; }

include cacheCommon.conf;
include drop.conf;
include php.conf;
include err.conf;

#Let wordpress show its own error pages.
fastcgi_intercept_errors off;
}

server {
server_name cink.applegrew.com;

access_log /var/www/cink.applegrew.com/access-n.log main;
error_log /var/www/cink.applegrew.com/error-n.log;

root /var/www/cink.applegrew.com/html;

include cacheCommon.conf;
include drop.conf;
include err.conf;
}

server {#If none the above matched then maybe the url was accessed, (say) via the IP directly. We then show applegrew.com.
listen 80 default;
server_name _;
access_log /var/www/applegrew.com/access.log-n main;

server_name_in_redirect off;

rewrite ^ http://www.applegrew.com$request_uri? permanent;
include err.conf;
}
}
[/code]

cacheCommon.conf

[code lang=”cink”]
#Asks browsers to cache files with extension ico, css, gif, jpg, jpeg, png, txt and xml.
location ~* \.(?:ico|css|js|gif|jpe?g|png|txt|xml)$ {
# Some basic cache-control for static files to be sent to the browser
expires max;
add_header Pragma public;
add_header Cache-Control "public, must-revalidate, proxy-revalidate";
}
[/code]

drop.conf

[code lang=”cink”]
location = /favicon.ico { access_log off; log_not_found off; } #Don’t log this.
location ~ /\. { access_log off; log_not_found off; deny all; } #Block . (dot) files access
#Don’t log and deny access to files which end with ~, as these are usually backup files.
location ~ ~$ { access_log off; log_not_found off; deny all; }
[/code]

err.conf

[code lang=”cink”]
error_page 500 502 503 504 /50x.html;
error_page 403 404 /404.html; # Yes for 403 too we show 404 error, just to mislead.
location = /50x.html {
root /home/webadmin/err/;
}
location = /404.html {
root /home/webadmin/err/;
}
[/code]

php.conf

[code lang=”cink”]
location ~ \.php { #All requests that end with .php are directed to PHP process.
include phpparams.conf; #This file is described later in this post.
}
[/code]

phpssl.conf

[code lang=”cink”]
location ~ \.php {#This the same as php.conf but adds few ssl specific configs.
fastcgi_param HTTPS on; #This sets $_SERVER[‘HTTPS’] to ‘on’.
fastcgi_param SSL_PROTOCOL $ssl_protocol; #This sets the $_SERVER[‘SSL_PROTOCOL’].
fastcgi_param SSL_CIPHER $ssl_cipher; #This sets the $_SERVER[‘SSL_CIPHER’].
fastcgi_param SSL_SESSION_ID $ssl_session_id; #This sets the $_SERVER[‘SSL_SESSION_ID’].
fastcgi_param SSL_CLIENT_VERIFY $ssl_client_verify; #This sets the $_SERVER[‘SSL_CLIENT_VERIFY’].

include phpparams.conf;
}
[/code]
We need to set the `$_SESSION` ourselves since unlike mod_php (in Apache), Php-Fpm is not embedded in Nginx and it doesn’t have these information available to it unless we set it. The above config doesn’t set all the flags a script might expect, but the usual ones. If need to set some more then go to Nginx HttpSslModule’s Built-in variables section.

phpparams.conf

[code lang=”cink”]
#PHP FastCGI
fastcgi_param QUERY_STRING $query_string;
fastcgi_param REQUEST_METHOD $request_method;
fastcgi_param CONTENT_TYPE $content_type;
fastcgi_param CONTENT_LENGTH $content_length;

fastcgi_param SCRIPT_NAME $fastcgi_script_name;
fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
fastcgi_param REQUEST_URI $request_uri;
fastcgi_param DOCUMENT_URI $document_uri;
fastcgi_param DOCUMENT_ROOT $document_root;
fastcgi_param SERVER_PROTOCOL $server_protocol;

fastcgi_param GATEWAY_INTERFACE CGI/1.1;
fastcgi_param SERVER_SOFTWARE nginx;

fastcgi_param REMOTE_ADDR $remote_addr;
fastcgi_param REMOTE_PORT $remote_port;
fastcgi_param SERVER_ADDR $server_addr;
fastcgi_param SERVER_PORT $server_port;
fastcgi_param SERVER_NAME $server_name;

fastcgi_connect_timeout 60;
fastcgi_send_timeout 180;
fastcgi_read_timeout 180;
fastcgi_buffer_size 128k;
fastcgi_buffers 4 256k;
fastcgi_busy_buffers_size 256k;
fastcgi_temp_file_write_size 256k;

fastcgi_pass unix:/usr/local/nginx/logs/php5-fpm.sock; #I have configured both Php-Fpm and Nginx to communicate via file sockets.
[/code]

/etc/php-fpm.conf

[code]
include=/etc/php-fpm.d/*.conf
pid = /var/run/php-fpm/php-fpm.pid
error_log = /var/log/php-fpm/error.log
log_level = error
[/code]

/etc/php-fpm.d/www.conf

[code]
[www]
listen = /usr/local/nginx/logs/php5-fpm.sock
listen.allowed_clients = 127.0.0.1
user = apache
group = apache

pm = dynamic
pm.max_children = 6; #This can be increased on 512MB RAM. For 256MB you ca use 2.
pm.start_servers = 3; #This can be increased. For 256MB you can use 1.
pm.min_spare_servers = 3; #This can be increased. For 256MB you can use 1.
pm.max_spare_servers = 5; #This can be increased. For 256MB you can use 1.
pm.max_requests = 500
slowlog = /var/log/php-fpm/www-slow.log
php_admin_value[error_log] = /var/log/php-fpm/www-error.log #All PHP errors will go into this.
php_admin_flag[log_errors] = on
[/code]
Note: The settings above are indicative. You need to experiment with different settings on your system. I have a MySql DB too running on the same system. In my case the minimal settings for 256MB RAM too cause problem. The PHP process used to choke after 4-5 days of running. So, finally I was forced to increase server RAM to 512MB.

/etc/init.d/nginxd

I wrote this shell script to start stop nginx as a service on my CentOS server.
[code lang=”bash”]
#!/bin/bash
# chkconfig: 235 85 15
# description: The Nginx Server is an efficient and extensible \
# server implementing the current HTTP standards.

cmd=/usr/local/nginx/sbin/nginx #Change this to match your Nginx installation path.
start() {
pgrep ‘nginx$’ > /dev/null
if (( $? != 0 ))
then
echo ‘Staring nginx’
$cmd
RETVAL=$?
if (( $RETVAL == 0 ))
then
echo ‘Started successfully’
fi
else
echo ‘Nginx already running’
RETVAL=0
fi
}

RETVAL=0
case "$1" in
start)
start
;;
stop)
echo ‘Shutting down Nginx quickly’
$cmd -s stop
RETVAL=$?
;;
quit)
echo ‘Gracefully shutting down Nginx’
$cmd -s quit
RETVAL=$?
;;
restart)
echo ‘Stopping Nginx’
$cmd -s stop
start
;;
reload)
echo ‘Reloading cofig’
$cmd -s reload
RETVAL=$?
;;
reopen)
echo ‘Reopening log files’
$cmd -s reopen
RETVAL=$?
;;
help)
$cmd -?
RETVAL=$?
;;
test)
echo ‘Test config’
$cmd -t
RETVAL=$?
;;
*)
echo $"Usage: nginx {start|stop|quit|restart|reload|reopen|help|test}"
echo "stop – quick shutdown"
echo "quit – graceful shutdown"
echo "reload – close workers, load config, start new workers"
echo "reopen – reopen log files"
echo "test – only tests the config"
RETVAL=3
esac

exit $RETVAL
[/code]
You can install the above by copying the nginxd file to /etc/init.d then run
`sudo /sbin/chkconfig nginxd –add`
`sudo /sbin/chkconfig nginxd on`

You can the give commands to the script by
`sudo /sbin/service nginxd command here`

Well, I hope this post been helpful.

The problem

The proof

Case 1:

Case 2:

Check out other applications of Digit Math

The problem

The proof

Check out other applications of Digit Math

What is Digit Math?

Before we begin…

Symbols and Operators

\(\omega\) (With-operator)

\(]x[\) (Digit Group operator)

\(\omega\sum\) (Digits summation operator)

Precedence of With-operator

Properties of With-operator

Some Important Formulae

General formula of transformation

Bi-multiplication formula

Rules for using the bi-multiplication formula

Squaring formula

General Powering Formula

A corollary of the General Formula of Transformation (for two digits)

With to Plus form

Plus to With form

Negative digits

Proof of General Powering Formula

Case 1: n = 0

Case 2: n = 1

Case 3: For n

Application of Digit Math

Some key new things

Must have WordPress plugins:-

nginx.conf

cacheCommon.conf

drop.conf

err.conf

php.conf

phpssl.conf

phpparams.conf

/etc/php-fpm.conf

/etc/php-fpm.d/www.conf

/etc/init.d/nginxd