Is there any especial operator on the function $f(x)$ to represent the following matrix:
\begin{bmatrix} f & \frac{\mathrm{d}}{\mathrm{d} x}f & \frac{\mathrm{d^2}}{\mathrm{d} x^2}f & ... \\ f^2 & \frac{\mathrm{d}}{\mathrm{d} x} f^2 & \frac{\mathrm{d^2}}{\mathrm{d} x^2} f^2 & \cdots \\ f^3 & \frac{\mathrm{d} }{\mathrm{d} x}f^3 & \frac{\mathrm{d^2}}{\mathrm{d} x^2} f^3 & \cdots\\ \vdots & \vdots &\vdots &\ddots \end{bmatrix}
If there is not such operator, how could we express this by means of available operators?

  • $\begingroup$ What is an "available" operator? (Also, depending on whether your concept of "operator" requires it to be linear, what you've got there may not be one at all). $\endgroup$ – Henning Makholm Aug 24 '13 at 13:26
  • $\begingroup$ @HenningMakholm By available operator I mean something like Laplacian, Gradient, etc, and others which I am not familiar with. $\endgroup$ – Amir Kazemi Aug 24 '13 at 13:30
  • $\begingroup$ "Something like" is not really an definition that one can use to decide whether you consider a given operator "available" or not. $\endgroup$ – Henning Makholm Aug 24 '13 at 14:04
  • $\begingroup$ @HenningMakholm You are right. What exactly I mean is that you can use a combination of operators you know. No restriction to use any established operator in calculus, etc. $\endgroup$ – Amir Kazemi Aug 24 '13 at 14:08
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    $\begingroup$ Assume it's an $(n\times n)$-matrix. Then its determinant is the Wronskian (see here: en.wikipedia.org/wiki/Wronskian) of the $n$-tuple $(f,f^2,\ldots, f^n)$. $\endgroup$ – Christian Blatter Aug 24 '13 at 14:21

Often in such cases, and trivially applied by physicists, I just use the index form for $i$ row and $j$ column. In that notation the operator can be made distinctive in its appearance

$${\bf{A_{i,j}}} =\frac{d^{(j-1)}}{dx^{(j-1)}} f^{(i)}$$

That helps to gain at least a concise operator notation and in a simple way.


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