Multivariate analysis.

Erable implements the following functions:

• der with a list of variables at level 1 returns the gradient of the expression at level 2 with respect to these variables.
  Example:
  'X+2*Y' { X Y } der returns { 1 2 }= $(\frac{\partial f}{\partial x} ,\frac{\partial f}{\partial y})$.
• DIV returns the divergence of a list-vector at level 2 with respect to a list of variables at level 1.
  Example:
  { 'X+2*Y' 'X^2+3*Y^3'} { X Y } DIV returns '1+9*Y^2'= $\frac{\partial f}{\partial x} +\frac{\partial f}{\partial y}$.
• CURL returns the rotationnal (same arguments as DIV)
• LAPL returns the laplacian of a symbolic expression at level 2 with respect to a list of variables at level 1 (same arguments as der, LAPL is simply a shortcut for der DIV)
• HESS returns at level 1 the hessian of a symbolic expression with respect to a list of variables (same arguments as der). Level 2 is the gradient. This is useful to find local minima and local maxima of a function: you find first the solutions of gradient=0 (you may use the Gröbner basis program of ALG48 to simplify this system and use the SOLV instruction to find all solutions), then you compute with EXEC the hessian at these critical points, and you find the signature of the critical point using GAUSS (in the other directory: see section 14).
  Example:
  
  f(X,Y)= X⁴+XY+Y³
  
  
  { X Y } HESS returns at level 2:
  
  $$(4X^3+Y,X+3Y^2)= (\frac{\partial f}{\partial X},\frac{\partial f}{\partial Y} )$$
  
  
  and at level 1:
  
  $$\left(\begin{array}{cc} 12X^2 & 1\\ 1 & 6Y \end{array}\right)$$
  
  
  To find critical points, you have to solve level 2=(0,0):
  
  (4X³+Y,X+3Y²)=(0,0)
  
  
  hence X=-3Y². Swap level 2 and 1, type 'X=-3*Y^2' EXEC, then 1 GET to have the first coordinate:
  
  4(-3Y²)³+Y
  
  
  then Y SOLV. This equation has two real solutions: 0 and approximately 0.392026340842 giving two critical points. For (0,0), the hessian is:
  
  $$\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$
  
  
  hence (0,0) is not an extremum (signature (1,1)). For the second point ( Y=0.392026340842, X=-3Y²), the hessian is:
  
  $$\left(\begin{array}{rlc} 1.84421582297 & 1 \\ 1 & -2.76632373445 \end{array}\right)$$
  
  
  hence is not an extremum.