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Oct 16, 2022, 1:55:16 PM10/16/22

to

Hi all,

while helping my daughter with her school algebra problems I thought I might

get some benefit for myself and dabble in some CAS. (Back when I studied

math in the 90s I didn't even know such a thing as CAS existed, and I

always found algebra terribly tedious).

Topic of the day, circle equations. So why not try and figure out the

intersection points (if any) of two circles.

(%i1) e1: x^2 + y^2 + a*x + b*y + c = 0;

2 2

(%o1) y + b y + x + a x + c = 0

(%i2) e2: x^2 + y^2 + d*x + e*y + f = 0;

2 2

(%o2) y + e y + x + d x + f = 0

X and y in the two equations are the same because we're searching the point(s)

both have in common. So here goes:

(%i3) solve([e1, e2], [x, y]);

(%o3) []

That took several seconds and yielded nothing. So let's try a more homeworky

approach: subtract the equations to get rid of the square terms, solve for x

(or y) and substitute into one of the equations:

(%i5) solve(%, x);

(e - b) y + f - c

(%o5) [x = - -----------------]

d - a

Now I can already see that substituting this into %o1 or %o2 will bring up an

idiotically convoluted quadratic equation. Just the right thing for a CAS.

Nice. But now I'm stuck. I can't find anything in the docs on how to proceed

from here.

Questions:

1) Why didn't Maxima find any solution at %i3?

2) How does one substitute the expression for x at %o5 into %o1?

3) I'm using the Linux console. How can I make Maxima use readline, so that

arrow-up brings up an editable copy of an earlier command line?

Thanks!

while helping my daughter with her school algebra problems I thought I might

get some benefit for myself and dabble in some CAS. (Back when I studied

math in the 90s I didn't even know such a thing as CAS existed, and I

always found algebra terribly tedious).

Topic of the day, circle equations. So why not try and figure out the

intersection points (if any) of two circles.

(%i1) e1: x^2 + y^2 + a*x + b*y + c = 0;

2 2

(%o1) y + b y + x + a x + c = 0

(%i2) e2: x^2 + y^2 + d*x + e*y + f = 0;

2 2

(%o2) y + e y + x + d x + f = 0

X and y in the two equations are the same because we're searching the point(s)

both have in common. So here goes:

(%i3) solve([e1, e2], [x, y]);

(%o3) []

That took several seconds and yielded nothing. So let's try a more homeworky

approach: subtract the equations to get rid of the square terms, solve for x

(or y) and substitute into one of the equations:

(%i5) solve(%, x);

(e - b) y + f - c

(%o5) [x = - -----------------]

d - a

Now I can already see that substituting this into %o1 or %o2 will bring up an

idiotically convoluted quadratic equation. Just the right thing for a CAS.

Nice. But now I'm stuck. I can't find anything in the docs on how to proceed

from here.

Questions:

1) Why didn't Maxima find any solution at %i3?

2) How does one substitute the expression for x at %o5 into %o1?

3) I'm using the Linux console. How can I make Maxima use readline, so that

arrow-up brings up an editable copy of an earlier command line?

Thanks!

Oct 17, 2022, 4:36:54 AM10/17/22

to

Robert Latest wrote:

> [stuff]

My questions from before have largely answered themselves by playing around a

bit more with Maxima. Turns out that even the simpler general case of solving

the intersection(s) of a circle with a straight line yields unwieldy solutions.

When it comes to algebra, I try to teach my kids to "first do it only with

letters, then insert numbers at the end," but I had never tried it with circle

equations.

I still don't know why Maxima can't solve the general case of two (possibly)

intersecting circles, but I have a hunch that it would be a useless jumble of

symbols. Much like the surprisingly absurd roots of a cubic equation.

The remaining question is: Is there Readline support for Maxima on the Linux

console?

> [stuff]

My questions from before have largely answered themselves by playing around a

bit more with Maxima. Turns out that even the simpler general case of solving

the intersection(s) of a circle with a straight line yields unwieldy solutions.

When it comes to algebra, I try to teach my kids to "first do it only with

letters, then insert numbers at the end," but I had never tried it with circle

equations.

I still don't know why Maxima can't solve the general case of two (possibly)

intersecting circles, but I have a hunch that it would be a useless jumble of

symbols. Much like the surprisingly absurd roots of a cubic equation.

The remaining question is: Is there Readline support for Maxima on the Linux

console?

Oct 17, 2022, 10:56:50 AM10/17/22

to

>>>>> "RL" == Robert Latest <bobl...@yahoo.com> writes:

RL> The remaining question is: Is there Readline support for Maxima on the Linux

RL> console?

if running just 'maxima' fails to use readline (whether it does depends

on which lisp system was used to compile maxima), then there will be an

'rmaxima' command you can call instead.

rmaxima is similar to running 'rlwrap maxima'.

-JimC

--

James Cloos <cl...@jhcloos.com> OpenPGP: 0x997A9F17ED7DAEA6

RL> The remaining question is: Is there Readline support for Maxima on the Linux

RL> console?

if running just 'maxima' fails to use readline (whether it does depends

on which lisp system was used to compile maxima), then there will be an

'rmaxima' command you can call instead.

rmaxima is similar to running 'rlwrap maxima'.

-JimC

--

James Cloos <cl...@jhcloos.com> OpenPGP: 0x997A9F17ED7DAEA6

Oct 17, 2022, 12:02:09 PM10/17/22

to

Robert Latest schrieb:

long time and seems somehow impossible to fix), but feeding your input

into Derive 6.10:

e1 := x^2 + y^2 + a*x + b*y + c = 0

e2 := x^2 + y^2 + d*x + e*y + f = 0

SOLVE([e1, e2], [x, y])

[x = ((b - e)*SQRT(a^2*(e^2 - 4*f) + 2*a*d*(2*(c + f) - b*e)

+ b^2*(d^2 - 4*f) + 4*b*e*(c + f) - 4*(c^2 + c*(d^2 + e^2 - 2*f)

+ f^2))*SIGN(a - d) + a*(b*e - 2*c - e^2 + 2*f) - d*(b^2 - b*e

- 2*(c - f)))/(2*(a^2 - 2*a*d + b^2 - 2*b*e + d^2 + e^2))

AND y = - (SQRT(a^2*(e^2 - 4*f) + 2*a*d*(2*(c + f) - b*e)

+ b^2*(d^2 - 4*f) + 4*b*e*(c + f) - 4*(c^2 + c*(d^2 + e^2 - 2*f)

+ f^2))*ABS(a - d) + a^2*e - a*d*(b + e) + b*(2*c + d^2 - 2*f)

- 2*e*(c - f))/(2*(a^2 - 2*a*d + b^2 - 2*b*e + d^2 + e^2)),

x = - ((b - e)*SQRT(a^2*(e^2 - 4*f) + 2*a*d*(2*(c + f) - b*e)

+ b^2*(d^2 - 4*f) + 4*b*e*(c + f) - 4*(c^2 + c*(d^2 + e^2 - 2*f)

+ f^2))*SIGN(a - d) - a*(b*e - 2*c - e^2 + 2*f) + d*(b^2 - b*e

- 2*(c - f)))/(2*(a^2 - 2*a*d + b^2 - 2*b*e + d^2 + e^2))

AND y = (SQRT(a^2*(e^2 - 4*f) + 2*a*d*(2*(c + f) - b*e)

+ b^2*(d^2 - 4*f) + 4*b*e*(c + f) - 4*(c^2 + c*(d^2 + e^2 - 2*f)

+ f^2))*ABS(a - d) - a^2*e + a*d*(b + e) - b*(2*c + d^2 - 2*f)

+ 2*e*(c - f))/(2*(a^2 - 2*a*d + b^2 - 2*b*e + d^2 + e^2))]

yields an answer within seconds. Derive has been discontinued, however,

and was only available for Windows anyway. For your school algebra

problems on Linux, maybe you should try Giac?

Martin.

Oct 18, 2022, 3:44:07 PM10/18/22

to

Robert Latest <bobl...@yahoo.com> wrote:

> Hi all,

>

> while helping my daughter with her school algebra problems I thought I might

> get some benefit for myself and dabble in some CAS. (Back when I studied

> math in the 90s I didn't even know such a thing as CAS existed, and I

> always found algebra terribly tedious).

>

> Topic of the day, circle equations. So why not try and figure out the

> intersection points (if any) of two circles.

>

> (%i1) e1: x^2 + y^2 + a*x + b*y + c = 0;

> 2 2

> (%o1) y + b y + x + a x + c = 0

> (%i2) e2: x^2 + y^2 + d*x + e*y + f = 0;

> 2 2

> (%o2) y + e y + x + d x + f = 0

>

> X and y in the two equations are the same because we're searching the point(s)

> both have in common. So here goes:

>

> (%i3) solve([e1, e2], [x, y]);

> (%o3) []

>

Using FriCAS:
> Hi all,

>

> while helping my daughter with her school algebra problems I thought I might

> get some benefit for myself and dabble in some CAS. (Back when I studied

> math in the 90s I didn't even know such a thing as CAS existed, and I

> always found algebra terribly tedious).

>

> Topic of the day, circle equations. So why not try and figure out the

> intersection points (if any) of two circles.

>

> (%i1) e1: x^2 + y^2 + a*x + b*y + c = 0;

> 2 2

> (%o1) y + b y + x + a x + c = 0

> (%i2) e2: x^2 + y^2 + d*x + e*y + f = 0;

> 2 2

> (%o2) y + e y + x + d x + f = 0

>

> X and y in the two equations are the same because we're searching the point(s)

> both have in common. So here goes:

>

> (%i3) solve([e1, e2], [x, y]);

> (%o3) []

>

(37) -> eq1 := x^2 + y^2 + a*x + b*y + c

2 2

(37) y + b y + x + a x + c

Type: Polynomial(Integer)

(38) -> eq2 := x^2 + y^2 + d*x + e*y + f

2 2

(38) y + e y + x + d x + f

Type: Polynomial(Integer)

(39) -> res := solve([eq1, eq2], [x, y])

(39)

[

(- e + b)y - f + c

[x = ------------------,

d - a

2 2 2 2 2

(e - 2 b e + d - 2 a d + b + a )y

+

2 2 2

((2 e - 2 b)f + (- a d - 2 c + a )e + b d - a b d + 2 b c)y + f

+

2 2 2

(- a d - 2 c + a )f + c d - a c d + c

=

0

]

]

Type: List(List(Equation(Fraction(Polynomial(Integer)))))

This is not an explicit solution, but arguably is more useful: intead of

two equations of order 2 in two variables you get so called triangular

system, where last equation contains only single variable and other

equations reduce to equation in single variable when you plug in

solutions of later equation. In fact, here first equation gives

you x in terms of y.

If you want it more explicit we can solve quadratic equation:

(40) -> radicalSolve(res(1)(2), y)

(40)

[

y

=

(- d + a)

*

ROOT

2 2 2 2 2

- 4 f + (4 b e + 4 a d + 8 c - 4 b - 4 a )f + (- 4 c + a )e

+

2 2 2

(- 2 a b d + 4 b c)e + (- 4 c + b )d + 4 a c d - 4 c

+

2 2

(- 2 e + 2 b)f + (a d + 2 c - a )e - b d + a b d - 2 b c

/

2 2 2 2

2 e - 4 b e + 2 d - 4 a d + 2 b + 2 a

,

y

=

(d - a)

*

ROOT

2 2 2 2 2

- 4 f + (4 b e + 4 a d + 8 c - 4 b - 4 a )f + (- 4 c + a )e

+

2 2 2

(- 2 a b d + 4 b c)e + (- 4 c + b )d + 4 a c d - 4 c

+

2 2

(- 2 e + 2 b)f + (a d + 2 c - a )e - b d + a b d - 2 b c

/

2 2 2 2

2 e - 4 b e + 2 d - 4 a d + 2 b + 2 a

]

Type: List(Equation(Expression(Integer)))

This is solved using usual quadratic formula, but since coefficients

of the quadratic are complicated the result is large. One could

plug in this y into formula for x, but result would be equally large.

Note: when you look at "general solution" there are subtleties.

You can see this already from the first solution: there is

division by d - a. And indeed, depending on the other parameters

beside normal case of two solutions there may be one solution,

no solutions or infinitely many solutions. Similarely, quadratic

formula gives "generic solution" and ignores special cases.

Normally a CAS will give you "generic solution", that is solution

which is valid as long as there are no divisions and which may

miss some special cases. Some teachers insist of analysiong

all cases, that quickly gets tedious both for people and at

somewhat larger scale for computers. If you have numbers in

the problem, than it is possible to see during computation

if there are special cases and handle such cases apropriately,

this avoids purely theoretical difficulties. OTOH if you

really need general solution, then there is some extra work

to check usual CAS solution and write proper conditions.

--

Waldek Hebisch

Oct 20, 2022, 9:02:37 AM10/20/22

to

anti...@math.uni.wroc.pl wrote:

> This is solved using usual quadratic formula, but since coefficients

> of the quadratic are complicated the result is large. One could

> plug in this y into formula for x, but result would be equally large.

>

> Note: when you look at "general solution" there are subtleties.

> You can see this already from the first solution: there is

> division by d - a. And indeed, depending on the other parameters

> beside normal case of two solutions there may be one solution,

> no solutions or infinitely many solutions. Similarely, quadratic

> formula gives "generic solution" and ignores special cases.

I know all that. I didn't really *need* to find a general solution for the
> This is solved using usual quadratic formula, but since coefficients

> of the quadratic are complicated the result is large. One could

> plug in this y into formula for x, but result would be equally large.

>

> Note: when you look at "general solution" there are subtleties.

> You can see this already from the first solution: there is

> division by d - a. And indeed, depending on the other parameters

> beside normal case of two solutions there may be one solution,

> no solutions or infinitely many solutions. Similarely, quadratic

> formula gives "generic solution" and ignores special cases.

intersections of two circles. I just took my daughter's homework assignment as

inspiration to play around with a CAS (which I always wanted but didn't have

any reason to). I expected a screenful of messy solutions, but not an empty

result.

> Normally a CAS will give you "generic solution", that is solution

> which is valid as long as there are no divisions and which may

> miss some special cases. Some teachers insist of analysiong

> all cases, that quickly gets tedious both for people and at

> somewhat larger scale for computers.

> If you have numbers in

> the problem, than it is possible to see during computation

> if there are special cases and handle such cases apropriately,

> this avoids purely theoretical difficulties. OTOH if you

> really need general solution, then there is some extra work

> to check usual CAS solution and write proper conditions.

>

Oct 20, 2022, 4:36:04 PM10/20/22

to

Robert Latest <bobl...@yahoo.com> wrote:

> anti...@math.uni.wroc.pl wrote:

> > This is solved using usual quadratic formula, but since coefficients

> > of the quadratic are complicated the result is large. One could

> > plug in this y into formula for x, but result would be equally large.

> >

> > Note: when you look at "general solution" there are subtleties.

> > You can see this already from the first solution: there is

> > division by d - a. And indeed, depending on the other parameters

> > beside normal case of two solutions there may be one solution,

> > no solutions or infinitely many solutions. Similarely, quadratic

> > formula gives "generic solution" and ignores special cases.

>

> I know all that. I didn't really *need* to find a general solution for the

> intersections of two circles. I just took my daughter's homework assignment as

> inspiration to play around with a CAS (which I always wanted but didn't have

> any reason to). I expected a screenful of messy solutions, but not an empty

> result.

OK. So use a CAS with reasonable equation solver (and not Maxima
> anti...@math.uni.wroc.pl wrote:

> > This is solved using usual quadratic formula, but since coefficients

> > of the quadratic are complicated the result is large. One could

> > plug in this y into formula for x, but result would be equally large.

> >

> > Note: when you look at "general solution" there are subtleties.

> > You can see this already from the first solution: there is

> > division by d - a. And indeed, depending on the other parameters

> > beside normal case of two solutions there may be one solution,

> > no solutions or infinitely many solutions. Similarely, quadratic

> > formula gives "generic solution" and ignores special cases.

>

> I know all that. I didn't really *need* to find a general solution for the

> intersections of two circles. I just took my daughter's homework assignment as

> inspiration to play around with a CAS (which I always wanted but didn't have

> any reason to). I expected a screenful of messy solutions, but not an empty

> result.

which is (in)famous for failures in its solver).

--

Waldek Hebisch

Oct 21, 2022, 12:40:59 PM10/21/22

to

anti...@math.uni.wroc.pl wrote:

> OK. So use a CAS with reasonable equation solver (and not Maxima

> which is (in)famous for failures in its solver).

Thanks. Didn't know there were big differences.
> OK. So use a CAS with reasonable equation solver (and not Maxima

> which is (in)famous for failures in its solver).

Oct 29, 2022, 6:25:46 AM10/29/22

to

For instance

e1: x^2+y^2 + 1 =0

is not even a circle, yet obeys to your e1.

First of all, use wxmaxima. It provides menus, help and many other things

that simplify life to a beginner.

Then, perhaps you could find a more recent computer.

(%i14) solve([e1,e2],[x,y]);

Evaluation took 2.6020 seconds (1.6910 elapsed)

(%o14) []

Finally, you should use a correct approach, as solve is right returning an

empty set. For instance:

(%i10) eliminate([e1,e2],[x,y]);

(%o10) [((d-a)*sqrt(-4*f^2+(4*b*e+4*a*d+8*c-4*b^2-4*a^2)*f+

(a^2-4*c)*e^2+(4*b*c-2*a*b*d)*e+(b^2-4*c)*d^2+4*a*c*d-4*c^2)+(2*b-2*e)*f+

(a*d+2*c-a^2)*e-b*

d^2+a*b*d-2*b*c)/(2*e^2-4*b*e+2*d^2-4*a*d+2*b^2+2*a^2)]

(%i11) eliminate([e1,e2],[x]);

(%o11) [(e^2-2*b*e+d^2-2*a*d+b^2+a^2)*y^2+(2*e*f+b*(-2*f+d^2-a*d)-

a*d*e+a^2*e+c*(2*b-2*e))*y+f^2-a*d*f+a^2*f+c*(-2*f+d^2-a*d)+c^2]

(%i12) eliminate([e1,e2],[y]);

(%o12) [(e^2-2*b*e+d^2-2*a*d+b^2+a^2)*x^2+(2*d*f+a*(e^2-2*f)+b*(-d*e-a*e)

+b^2*d+c*(2*a-2*d))*x+f^2-b*e*f+b^2*f+c*(-2*f+e^2-b*e)+c^2]

Here are your quadratic equations. Maybe they have real solutions, maybe

not. Good luck exploring your 6 dimensional parameter space.

There are plenty of tutorials around for wxmaxima/maxima that could be

helpful.

HTH.

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