As far as I remember the speed of light in vacuum is determined by the formula c= \frac{1}{\sqrt{\epsilon_{0}\cdot \mu_{0}}}. The actual numerical value follows from that.
BTW the speed of light in any environment is given by c = \frac{1}{\sqrt{\epsilon \cdot \mu}} where ε is the dielectric...
Thhis is the trouble with "objectified C". If you want to do this in C, you would put this code in a separate module and declare the pointer as "static" (which means that only code in that module can access it).
As I have noted in an Insight, just create the Fourier serien from the function \pi ^{2}\cdot x - x^{3} . This ends up in 12\cdot \sum_{n=1}^{\infty}\frac{\sin(n(\pi - x))}{n^{3}} . Evaluate both expressions at π/2. and you have the desired proof.
As I read the OP, he stated "the bare pads are used to wire bond (using aluminium wire) to the DUT". I replied to that part, not the derived problem of removing the solder.
Turn the problem around - if you solder wire-wrap pins to the problematic areas, you can use wire-wrap techniques to connect - or is there something that I do not understand in the problem description?
Some number systems are better than others in expressing mathemathical conjectures. The Peano axioms are meaningless in roman numerals - leaving aside the fact that 0 does not exist, the notion of a "successor" is not trivial (what is the successor of VIII?)
Sorry, no computer language has any sort of logic built-in. What it has, is a way to describe how to do things. The "logic" must reside in the brain of the implementer. How to transform that logic into working code may depend on the tool chosen - you can write a huge application in assembly...
Obviously, if \mathbf{A}\mathbf{x}=\mathbf{y} then \mathbf{x}=\mathbf{A}^{-1}\mathbf{y} as long as \mathbf{A}^{-1} exists. Usually, this just means that det(\mathbf{A})\neq 0 .
x+iy\equiv Re^{i\theta} where R=\sqrt{x^{2}+y^{2}} and \theta =\arctan(\frac{y}{x}) . This is the easiest representation for complex multiplication (you multiply the argumets and add the angles). Complex addition is easiest in the cartesian notation.