I did a google search and the first few hits I clicked on weren't helpful and that led me to prematurely abandon my efforts. Upon further investigation I've now found all you have to do is type in "knoppix lang=uk" at the boot prompt stage. Thanks.
In the preface of Volume 1: Two-Spinor Calculus and Relativistic Fields Roger Penrose, Wolfgang Rindler Penrose/Rindler say
Given the diagrammatic notation was, it seems, intended for private calculations it may be difficult answering the question if anybody else uses it.
Yes, that is how you obtain the Fourier coefficients. I was thinking
\begin{align*}
c_0 = \frac{1}{2 \pi} \int_{-\pi}^\pi \dfrac{r - \cos x}{r^2 - 2 r \cos x + 1} dx
\end{align*}
so you can't know the value of the integral you are after (##\int_{-\pi}^\pi \dfrac{1}{r^2 - 2 r \cos x + 1} dx##)...
Try the function
\begin{align*}
v(x) = A + u(x) e^{ix}
\end{align*}
where ##A## is an appropriately chosen constant. You only need the real part of the function.
I think you would apply the Intermediate Value Theorem in this context. It states:
If ##f## is a continuous function whose domain contains the interval ##[c, b]##, then it takes on any given value between ##f(c)## and ##f(b)## at some point within the interval.
If you happen to be a cartoon that has been murdered, there is the remarkably similar looking inspector John Gadget. Though it would be his niece Penny and his dog brain who would actually solve the murder. Although, to be honest, I don't recall there being a multitude of murders in the programme.
Then there is inspector Jacques Clouseau. Though being a bumbling and inept police detective, he would end up successfully solving the case by sheer luck.
Talking of John Thaw, there is the brilliant Detective Chief Inspector Morse.
Inspector Morse was a detective with the Oxford City Police, known for his intellectual acumen, love of classical music, taste for the finer things, and his fondness for real ale. Morse often solved complex murder...
Yep. It is possible to diagonalize a real symmetric matrix by a real orthogonal similarity transformation. If ##U^T M_S U = D##, then ##M_S = UDU^T = (UD^{1/2}U^T) (U D^{1/2}U^T) = M_S^{1/2} M_S^{1/2}##.