William:
As I mentioned in my first reply, I didn't spend much time trying to
understand your math. I took another quick look at it this morning and I
must confess I don't understand what you are trying to do. The maxima
statements that were listed at the top of your original post:
-------------------------------------------------------
(%i1) x^2+d*y*4+d^2*(-4)$
(%i2) e1: rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
(%i3) en1: 10$
(%i4) en2: 5$
(%i5) %phi[1] : first(e2)$
(%i6) %phi[1] : first(e2)$
(%i7) %phi[2] : second(e2)$
(%i8) [(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))]$
[(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))]$
((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1)));
(%i9) %,numer;
-------------------------------------------------------
I changed slightly to get the results you mentioned in that post
-------------------------------------------------------
x^2+d*y*4+d^2*(-4);
e1:rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
/*Added this line from 'f1()' below*/
e2:subst([x=en1, d=en2],[first(diff(e1,x)), diff(e1,x)]);
en1: 10;
en2: 15;
/*Using %phi conflicts with a constant name maybe?? but didn't change it*/
%phi[1] : first(e2);
/*%phi[1] : first(e2); Removed this line*/
%phi[2] : second(e2);
[(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))];
[(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))];
((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1)));
%,numer;
-------------------------------------------------------
but after further inspection I notice that a couple of lines are not
effective
-------------------------------------------------------
x^2+d*y*4+d^2*(-4);
e1:rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
/*Added this line from 'f1()' in your post*/
e2:subst([x=en1, d=en2],[first(diff(e1,x)), diff(e1,x)]);
en1: 10;
en2: 15;
/*Using %phi conflicts with a constant name maybe?? but didn't change it*/
%phi[1] : first(e2);
/*%phi[1] : first(e2); Removed this line*/
%phi[2] : second(e2);
/*The following two lines aren't doing anything that affects the value
of the last line
[(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))];
[(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))]; */
/*This line is actually the value you are outputting*/
((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1)));
%,numer;
-------------------------------------------------------
I simply modified the function 'f1(ang)' that you provided
-------------------------------------------------------
x^2+d*y*4+d^2*(-4)$
e1: rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
en1: 10$ /* set x = 10 */
f1(ang) := block([n, en2:5],
e2: subst([x=en1, d=en2],[first(diff(e1,x)), diff(e1,x)]),
%phi[1] : first(e2),
%phi[2] : second(e2),
ang : [(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))],
[(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))],
((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1))),
for i:1 thru n while ang # 0.178092938 do en2 : en2+i,
return (en2)); /*Should return en2:55. I do not
understand what's going on here. */
-------------------------------------------------------
to look like this so that it would output the value of the statement
where I thought 'ang' should be assigned.
In the process, I changed the initial value of en2 to 1, made the
parameter 'ang' to be 'nIn', and initialized the value of 'n' to 'nIn'
minus 1 when 'f1(nIn)' is invoked.
-------------------------------------------------------
x^2+d*y*4+d^2*(-4)$
e1: rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
en1: 10$ /* set x = 10 */
/*f1(ang) := block([n, en2:5], Use a different parameter name */
f1(nIn) := block([ang, n:nIn-1, en2:1], /*add initialization for n,
and modify en2 value*/
/*for i:1 thru n while ang # 0.178092938 do ( */
for i:0 thru n do (
e2:subst([x=en1, d=en2],[first(diff(e1,x)), diff(e1,x)]),
%phi[1] : first(e2),
%phi[2] : second(e2),
[(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))],
[(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))],
ang :
((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1))),
en2 : en2 + 1 /*Need variable increment here that can be less
than 1.0*/
),
return( float(ang) )
);
-------------------------------------------------------
I added a few more explanatory notes this morning to illustrate:
-------------------------------------------------------
x^2+d*y*4+d^2*(-4)$
e1: rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
en1: 10$ /* set x = 10 */
/*f1(ang) := block([n, en2:5], Use a different parameter name */
/*nIn is just the number of iterations*/
f1(nIn) := block([ang, n:nIn-1, en2:1], /*add initialization for n,
and modify en2 val*/
/*for i:1 thru n while ang # 0.178092938 do ( */
for i:0 thru n do (
e2:subst([x=en1, d=en2],[first(diff(e1,x)), diff(e1,x)]),
%phi[1] : first(e2),
%phi[2] : second(e2),
/* The following two lines don't seem to be doing anything
[(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))],
[(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))], */
/*If this is the value you want, then 'ang' has to be assigned here*/
ang :
((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1))),
/*en2 is just acting as a counter that is incremented at each
iteration*/
en2 : en2 + 1 /*Need variable increment here that can be less
than 1.0*/
),
return( float(ang) )
);-------------------------------------------------------
If indeed the value of 'ang' as it's shown in the modified 'f1(nIn)' is
what you want, then to get more accuracy would require the value of
'en2' to increment at a rate less than 1.0 as you get closer to the
value you desire.
I made no accommodations to supply a target value or to provide the
means to converge to one. That would require checking the value of 'ang'
after each assignment and taking steps to change the increment rate of
'en2' and also to to accommodate overshooting the target value as you
tried to converge to the target.
When I invoked 'f1(56)' I just tried a few values to see if it was
producing the same output as the statements that were given external to
those in 'f1(nIn)'.
I hope this helps,
Paul Bowyer
On 09/06/2010 09:01 PM, William Porter wrote:
> Thank you Paul and Robert for your input. It is greatly appreciated.
>
> I am too inexperienced with Maxima to carry it much farther than what
> Paul has outlined. I have tried to use the find_root (<some
> expression>, en2, 10, 80); at the end of Paul's outline. Unfortunately
> I am not sure what to insert in for <some expression>. Guessing at
> what expression to enter, I have tried f1, ang, nIn, etc. but just
> guessing with no results.
>
> On the last line f1(56); how did you know to insert the value 56? Is
> there an expression that would generate the value?
>
> Also on line en2 : en2 + 1), /*Need variable increment here that can
> be less than 1.0*/... what would be an example?
>
> x^2+d*y*4+d^2*(-4)$
> e1: rhs(first(solve(x^2+d*y*4+d^2*(-4),y)));
> en1: 10$ /* set x = 10 */
> f1(nIn) := block([ang, n:nIn-1, en2:1],
> for i:0 thru n do (
> e2:subst([x=en1, d=en2],[first(diff(e1,x)), diff(e1,x)]),
> %phi[1] : first(e2),
> %phi[2] : second(e2),
> [(%phi[1]*d*(-2)),(d+((%phi[1])^(2)*d*(-1)))],
> [(%phi[2]*d*(-2)),(d+((%phi[2])^(2)*d*(-1)))],
> ang :
> ((%pi*1/2)+(atan((((%phi[1]+(%phi[2]*(-1))))^((-1))*(1+(%phi[2]*%phi[1]))))*(-1))),
> en2 : en2 + 1), /*Need variable increment here
> that can be less than 1.0*/
> return( float(ang) ) );
>
> f1(56); /* How do you know to enter the 56 value? Can the
> expressions generate this value? */
>
> William Porter
>
> wporter at omegapar.com
>
>
>
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