By William G. Hoover

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**Additional resources for Molecular Dynamics (Lecture Notes in Physics) **

**Sample text**

The function solve returns a list of solutions. Its first argument is an equation (or a list of equations), and the second one is an unknown variable (or a list of unknown variables). If the right-hand side of an equation is zero, then it may be omitted together with =. The second argument may be omitted, if the unknown variables are the only variables contained in the equations. Every solution in the list returned by solve is the equality with the unknown variable in the left-hand side, or the list of such equalities in the case of an equation system.

Operator f; x:=xx$ y:=yy$ let f(x)=y; clear x; f(x); f(xx); clear y; f(xx); clear f(xx); yy y x:=xx$ y:=yy$ f(x):=y; f(xx) := yy clear x; f(x); f(xx); clear y; f(xx); clear f(xx); yy yy The statements l e t x= ... and x: = ... both change the value of x (the first one, to the unevaluated right-hand side, while the second one, to its value). Therefore, naturally, any of these statements destroys the result of a previous one. At each moment, only one substitution for a variable x raised to an integer power can exist.

By the way, you can see that an assignment has a value, and may be used in an expression. % Taylor series for exp procedure expl(x,n); begin scalar s,u; s:=l; u:=l; for i:=l:n do « u:=u*x/i; s:=s+u » ; return s end$ on div,revpri; expl(x,10); clear expl; ! ^ 1 l 2 l 3 1 4 2 6 x9 + + x + 24 120 1 fi 720 1 « 7 5040 40320 x10 3628800 362880 3628800 °/0 method using small variables procedure exp2(x); begin scalar s,u,i; s:=l; u:=x; i:=l; repeat « s:=s+u; i:=i+l; u:=u*x/i » until u=0; return s end$ weight x=l$ wtlevel 10$ exp2(x); exp2(x+x~2); clear exp2; .