Simple Mathematics

Let us consider now some simple mathematics.
We all want to find some easy way to remember formulae. So what we usually do is to "logically link" those hard formulae to something simpler, that our mind can easily remember.

This is not the case, when you are talking to an engineer.

In fact, consider the simple equality:

$\Large 1+1=2$

Every engineer should learn, since its early age, to refactor it in a more professional fashion.
In fact, taking into account calculus and trigonometry, we all know that $1 = \ln(e)$ and that $1 = \sin^2(\vartheta) + \cos^2(\vartheta)$. Further, remember that

$\large 2 = \sum\limits_{n=0}^\infty \left( \frac{1}{2} \right)^n$

So the equality can be rewritten as:

$\large \ln(e) + \sin^2(\vartheta) + \cos^2(\vartheta) = \sum\limits_{n=0}^\infty \left( \frac{1}{2} \right)^n$

Let us now consider hyperbolic functions. Remember that $1 = \cosh(\vartheta) * \sqrt{1 - \tanh^2(\vartheta)}$.
Again, from calculus we know that:

$\large e = \lim_{z\to0}\left(1+\frac{1}{z}\right)$

So, our equality can be rewritten as:

$\small \ln\left(\lim\limits_{z\to0} \left(1+\frac{1}{z}\right)\right) + \sin^2(\vartheta) + \cos^2(\vartheta) = \sum\limits_{n=0}^\infty \frac{\cosh(\vartheta) * \sqrt{1 - \tanh^2(\vartheta)}}{2^n}$

Consider now that $0! = 1$. Further, remember that the inverse matrix of a transposed matrix is equal to the transposed matrix of the inverse. Then, in a one-dimensional geometric space, we have that:

$(\bar{x}^T)^{-1} - (\bar{x}^{-1})^T$

Then, combining we get:

$\large \left[ (\bar{x}^T)^{-1} - (\bar{x}^{-1})^T \right]! = 1$

Thus, inserting this into the previous equality, we get:

$\ln\left(\lim\limits_{z\to0}\left( ((\bar{x}^T)^{-1} - (\bar{x}^{-1})^T) ! +\frac{1}{z}\right)\right) + \sin^2(\vartheta) + \cos^2(\vartheta) = $
$ = \sum\limits_{n=0}^\infty \frac{\cosh(\vartheta) * \sqrt{1 - \tanh^2(\vartheta)}}{2^n}$

Which is a de-facto simpler way for an engineer to remember some simple mathematics such as $1+1=2$.