## Palindrome Function

To get a number’s palindrome in a programming language like python is easy. There are ways to swap between integer and string and strings can be manipulated.

>>> n = 1234
>>> int(str(n)[::-1])
4321

But I wanted to create a mathematical function $p(n)$, which returns an integer’s palindrome. Thus $p(1234) = 4321$.

Firstly I needed a way of determining the number’s size. In base $10$ the length is calculated using the logarithm to said base.
$l(n) = \lfloor \log_{10}{n} \rfloor + 1$
$l(1234) = \lfloor \log_{10}{1234} \rfloor = \lfloor3.09 \rfloor + 1 = 4$

Secondly I need a way to isolate a specific digit. Using the floor function, this function returns the $i\text{-th}$ digit (starting on the right with $i=0$).
$d_i(n) = \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10$
$d_2(1234) = \lfloor \frac{1234}{10^2} \rfloor - \lfloor \frac{1234}{10^{2+1}} \rfloor \cdot 10 = \lfloor 12.34 \rfloor - \lfloor 1.23 \rfloor \cdot 10 = 12 - 1 \cdot 10 = 2$

Thirdly both of these functions can be used to split up the number into a sum.
$n = \sum\limits_{i=0}^{l(n)-1} \Big[ d_i(n) \cdot 10^{i} \Big] = \sum\limits_{i=0}^{\lfloor \log_{10}{n} \rfloor} \Big[ \big( \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{i} \Big]$

Fourthly I only need to swap the power of ten at the end to get my palindrome function.
$p(n) = \sum\limits_{i=0}^{l(n)-1} \Big[ d_i(n) \cdot 10^{l(n) - 1 - i} \Big] = \sum\limits_{i=0}^{\lfloor \log_{10}{n} \rfloor} \Big[ \big( \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{\lfloor \log_{10}{n} \rfloor - i} \Big]$

Thus the final function $p(n)$ is defined.
$p(n) = \sum\limits_{i=0}^{\lfloor \log_{10}{n} \rfloor} \Big[ \big( \lfloor \frac{n}{10^i} \rfloor - \lfloor \frac{n}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{\lfloor \log_{10}{n} \rfloor - i} \Big]$

To check if the formula is correct, I use $1234$ (as seen above).
$p(1234) = \sum\limits_{i=0}^{\lfloor \log_{10}{1234} \rfloor} \Big[ \big( \lfloor \frac{1234}{10^i} \rfloor - \lfloor \frac{1234}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{\lfloor \log_{10}{1234} \rfloor - i} \Big]$
$p(1234) = \sum\limits_{i=0}^{3} \Big[ \big( \lfloor \frac{1234}{10^i} \rfloor - \lfloor \frac{1234}{10^{i+1}} \rfloor \cdot 10 \big) \cdot 10^{3 - i} \Big]$
$p(1234) = d_0(1234) \cdot 10^3 + d_1(1234) \cdot 10^2 + d_2(1234) \cdot 10^1 + d_3(1234) \cdot 10^0$
$p(1234) = 4000 + 300 + 20 + 1 = 4321$

The series goes as follows: $\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots = \frac{\pi^2}{6}$
By rearranging the equation, you get the following: $\pi = \sqrt{6\cdot\big(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\big)}$
# Python 2.7.7 Code
# Jonathan Frech 14th of December, 2015