Suppose we have a set of functions with their coefficients. The coefficients can be decomposed into two parts, one of them is unknown, and one is known to be a power of a fixed number; the other part is always linear, but changing to each of the numbers raised to a power. Now, the coefficients multiply a function which is always the same. Now imagine that you have a set of polynomial equations, where you now that for each equation there corresponds a function. Let us say that these functions are comparable to discrete differences, and that the distinction for each polynomial is that this function gets a higher and higher order of the difference. So, the point is, in principle one could construct the Wronskian, right?
The question now is, if the Wronskian is zero, can we assume that we have and eigensystem?
yes. The source LaTex:
"
Perhaps a hint on proving that the system of (Eq. \ref{EigensystemEq} ) is indeed an eigensystem, could be to evaluate it simbolically as if it were a Wronskian, because if the "Wronskian" \textit{does not} vanishes, then linear independence would be proven, thus showing that the system in (Eq. \ref{EigensystemEq} ) is akin to a complete set of eigenvectors. The motivation for it to be considered as a candidate for evaluating the Wronskian comes from the fact that each recursion becomes every time of larger order, just as in the Wronskian there is a set of derivatives from the zeroth order (the primitive function) up to order $n-1$ (Eq. \ref{WronskianEq}).
\begin{equation}\label{WronskianEq}
W(f_1,\cdots,f_n)(x)= \left| \begin{array}{cccc} f_1(x) & f_2(x) &\cdots &f_n(x)\\
f'_1(x) & f'_2(x) &\cdots &f'_n(x)\\
\vdots & \vdots &\ddots &\vdots \\
f^{(n-1)}_1(x) & f^{(n-1)}_2(x) &\cdots &f^{(n-1)}_n(x)\\
\end{array} \right|
\end{equation}
where each $f_i(x) = v_i$, that is, each of the functions which are solutions of the aforementioned recursive functions.
"
- Badges
- Science topic
- Similar topics
- Engineering
- Control Theory