confrac_samples.html

The Brouncker algorithm for the continued fractions of a root of a quadratic.

> with(linalg):

>

Basic calculations and functions.

To start, we create a matrix from

> A:=matrix(2,4,[1,0,1,0,0,1,0,-7]);

>

The function "fun" produces a polynomial in x,y from a 2 x 2 matrix. We need its values to calculate steps.

>

> fun:=m-> unapply(m[1,3]*x^2+2*m[1,4]*x*y+m[2,4]*y^2,x,y);

> fun(A)(x,y);fun(A)(3,1);

The function ch(m,t,n) changes matrix m as follows:

If n=1 then it add t times row 1 to row 2.

If n=2 it adds t times row 2 to row 1.

Then it makes a symmetric matrix operation on the last two columns.

> ch:=proc(m,t,n) local tmp:
tmp:=copy(m); if n=1 then tmp:=addrow(tmp,1,2,t); tmp:=addcol(tmp,3,4,t);
elif n=2 then tmp:=addrow(tmp,2,1,t);tmp:=addcol(tmp,4,3,t);fi;op(tmp);end:

>

Case of

We start work on .

We shall record the t-values as we proceed .

> A:=matrix(2,4,[1,0,1,0,0,1,0,-7]);

We check the function at (1,1);

If it is negative, then we try values at (t,1) and stop just before the values change sign.

> fun(A)(1,1); fun(A)(2,1);fun(A)(3,1);

So set t=2 and n=1(because the value at (1,1) was less than 1.)

> tvals:=[2];

> A1:=ch(A,2,1);

Repeat with new matrix. Here, the value at (1,1) is bigger than 1 , so we try (1,t), until sign changes.

> fun(A1)(1,1);fun(A1)(1,2);

So we set t=1 and n=2 (because the value at (1,1) was bigger than 1.)

> tvals:=[op(tvals),1];

> A2:=ch(A1,1,2);

Repeat! Again, we get t=1, n=1.

> fun(A2)(1,1);fun(A2)(2,1);

> tvals:=[op(tvals),1];

> A3:=ch(A2,1,1);

> fun(A3)(1,1);

This is a special case. When the function evalutes to 1, then set t=1 and reverse the order of row operations (i.e. change n from 1 to 2 or 2 to 1.

> tvals:=[op(tvals),1];