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not suitable format for function definitions, to display symbolic results (CAS) - compsystems - 01-20-2016 05:20 PM
Hello sorry for my bad English
I wish to make a program, which shows the step by step, yet not 2D expressions (math book) viewer on terminal, I am not satisfied with the current impression 1D (linear)
First consider the following definition
vector5(n) := seq (z^k, k, 1, n) [ENTER]
It is placed on history as:
(n)->seq( z ^ k, k, 1, n)
if you want to modify the above definition k => (k+1) , the expression (n)->seq( z^k, k, 1, n) is useless because it does not include the name of the function to overwrite function
I think it should include the name of the function to edit, otherwise you have to go to the original definition (ENTRY ANS). The idea is to edit in two ways
vector5(n) := seq (z^k, k, 1, n)
or
vector5(n)->seq( z^(k+1), k, 1, n)
... by showing a step (TERMINAL VIEW), creating implicit vector vector5 shown as
vec5 = (n)->seq( z^(k+1), k, 1, n) , it's confusing as a teaching tool
Now when I show the symbolic dot product, the expression is much more confusing
vec5(n) · vec6(n) = dot( ( n->seq(z^k,k,1,n) ) (n) , ( n->seq((1/(a+z)^k),k,1,n) )(n) )
Why is placing a postfix (n)?
To solve this problem, I suggest that the function RIGHT, get the right side of a definition, so I can capture the expression and not the definition
right( n->seq(z^k,k,1,n) ) => seq(z^k,k,1,n) =)
PHP Code:
#cas
vectorExamples()
BEGIN
//implicit definition of a vector
// explicit *
//vector1 := seq(3*k^3-5*k^2+17*k,k,-14,2,3);
// implicit *
//vector2:= seq(3k^3-5k^2+17k,k,-14,2,3);
//print("vec1 = " + vector1); wait; print("vec2 = " + vector2); wait; print("");
vector3 := seq(k^2-3*k,k,1,5);
vector4 := seq(2*k^3+5*k,k,1,5);
print("vec3 = " + vector3); wait; // print("vec3 = " + right( vector3) );
print("vec4 = " + vector4); wait; // print("vec4 = " + right( vector4) );
//scalar product
scalar1 := dot(vector3,vector4);
print("vec3 · vec4 = " + scalar1); wait; // print("vec3 · vec4 = " + right( scalar1) );
print("");
vector5(n) := seq(z^k,k,1,n);
vector6(n) := seq(1/(a+z)^k,k,1,n);
print("vec5 = " + vector5); wait;
print("vec6 = " + vector6); wait;
print("");
scalar2(n):= dot(vector5(n),vector6(n));
print(scalar2);
print("");
print("scalar2(3): " + scalar2(3));
print("");
print("scalar2(3)|(a=8,z=9)): " + approx(scalar2(3)|(a=8,z=9)));
wait;
scalar3(n,a_,z_):= scalar2(n)|(a=a_,z=z_);
print(scalar3);
print(scalar3(3,8,9));
//vectorial product
END;
#end
// terminal view
vector1: [-9450,-4785,-1992,-585,-78,15]
vector2: [-9450,-4785,-1992,-585,-78,15]
vec3 = [-2,-2,0,4,10]
vec4 = [7,26,69,148,275]
vec3 · vec4 = 3276
vector5: (n)->seq (z^k, k, 1, n)
vector6: (n)->seq((1/(z+a)^k),k,1,n)
scalar2:
(n)-> dot( ( n->seq(z^k,k,1,n) )(n), ( n->seq((1/(a+z)^k),k,1,n) )(n) )
vector5(3) = [z,z^2,z^3]
vector6(3) = [1/(a+z),1/(a+z)^2,1/(a+z)^3]
vector5(3) · vector6(3) = (z/(a+z))+(z^2/(a+z)^2)+z^3/(a+z)^3
ans | (a=8,z=9)
4707/4913
0.958070425402