solution of quadratic equations -- incorrect solution

On 07/03/2012 01:37 PM, Viktor T. Toth wrote:
> Krishna,
>
> The problem you are describing has to do with the sign ambiguity of the
> square root operator. Neither solutions are "right" or "wrong" per se; they
> are in fact equivalent up to the sign of sqrt. However, only one sign choice
> yields valid solutions of the original equation. Which sign is the right one
> depends on the numerical values of r1 and r2.
>
> Consider the following two cases:
>
> 	r1=1.2, r2=1.8,
>
> and
>
> 	r1=2.2, r2=-1.8.
>
> Note how in the first case, it's sola that is the "right" solution of e1a,
> whereas in the second case, it's solb:
>
> (%i1) display2d:false$
> (%i2) e1a: x^2+y^2=r1^2$
> (%i3) e2a: (x-1)^2+(y-1)^2=r2^2$
> (%i4) sola: solve([e1a,e1a-e2a],[x,y])$
> (%i5) radcan(ev([e1a,e2a],sola[1]));
> (%o5) [r1^2 = r1^2,r2^2 = r2^2]
> (%i6) radcan(ev([e1a,e2a],sola[2]));
> (%o6) [r1^2 = r1^2,r2^2 = r2^2]
> (%i7) e1b: x^2+y^2=(r1-err)^2$
> (%i8) e2b:(x-1)^2+(y-1)^2=(r2-err)^2$
> (%i9) solb: solve([e1b,e1b-e2b],[x,y])$
> (%i10) radcan(ev([e1b,e2b],solb[1]));
> (%o10) [r1^2-2*err*r1+err^2 = r1^2-2*err*r1+err^2,
>          r2^2-2*err*r2+err^2 = r2^2-2*err*r2+err^2]
> (%i11) radcan(ev([e1b,e2b],solb[2]));
> (%o11) [r1^2-2*err*r1+err^2 = r1^2-2*err*r1+err^2,
>          r2^2-2*err*r2+err^2 = r2^2-2*err*r2+err^2]
> (%i12) radcan(sola[1]-ev(solb[1],err=0));
> (%o12) [0 = 0,0 = 0]
> (%i13) radcan(sola[2]-ev(solb[2],err=0));
> (%o13) [0 = 0,0 = 0]
> (%i14) n1a:sola,r1=1.2,r2=1.8$
> (%i15) n1b:solb,r1=1.2,r2=1.8,err=0$
> (%i16) e1a,n1a,r1=1.2;
> (%o16) 1.44 = 1.44
> (%i17) e1a,n1b,r1=1.2;
> (%o17) 1.270589256539026 = 1.44
> (%i18) n2a:sola,r1=2.2,r2=-1.8$
> (%i19) n2b:solb,r1=2.2,r2=-1.8,err=0$
> (%i20) e1a,n2a,r1=2.2;
> (%o20) .2721120854381727 = 4.84
> (%i21) e1a,n2b,r1=2.2;
> (%o21) 4.84 = 4.84
>
> In reality, neither of these solutions is right or wrong, one just needs to
> take a precise look at the branch cut of the sqrt function and how it
> applies to the problem at hand. The numerical difference between the two
> solutions arises because of the order in which Maxima evaluates the square
> roots and takes the principal value.
>
> In many cases, calling radcan can help; it will not guarantee (of course)
> that the solution you find is the "right" one for a given set of values, but
> at least it improves consistency between expressions.
>
>
> Viktor
>
>
>
> -----Original Message-----
> From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu]
> On Behalf Of Krishna Myneni
> Sent: Tuesday, July 03, 2012 1:00 PM
> To: Viktor T. Toth
> Cc: maxima at math.utexas.edu
> Subject: Re: [Maxima] solution of quadratic equations -- incorrect solution
>
> On 07/03/2012 07:16 AM, Viktor T. Toth wrote:
>> If you check your y-solutions, you will find that the differences are due
> to
>> trivial algebra and the solutions are, in fact, identical when you set
>> err=0. You have expressions that are formally just like
> sqrt(A-B)*sqrt(C-D)
>> vs. sqrt(B-A)*sqrt(D-C), which may appear different at first but they
> really
>> aren't. You can actually verify this using Maxima itself, as well as
> verify
>> that the solutions that were obtained indeed solve your system of
> equations:
>> (%i1) display2d:false$
>> (%i2) e1a: x^2+y^2=r1^2$
>> (%i3) e2a: (x-1)^2+(y-1)^2=r2^2$
>> (%i4) sola: solve([e1a,e1a-e2a],[x,y])$
>> (%i5) radcan(ev([e1a,e2a],sola[1]));
>> (%o5) [r1^2 = r1^2,r2^2 = r2^2]
>> (%i6) radcan(ev([e1a,e2a],sola[2]));
>> (%o6) [r1^2 = r1^2,r2^2 = r2^2]
>> (%i7) e1b: x^2+y^2=(r1-err)^2$
>> (%i8) e2b:(x-1)^2+(y-1)^2=(r2-err)^2$
>> (%i9) solb: solve([e1b,e1b-e2b],[x,y])$
>> (%i10) radcan(ev([e1b,e2b],solb[1]));
>> (%o10) [r1^2-2*err*r1+err^2 = r1^2-2*err*r1+err^2,
>>           r2^2-2*err*r2+err^2 = r2^2-2*err*r2+err^2]
>> (%i11) radcan(ev([e1b,e2b],solb[2]));
>> (%o11) [r1^2-2*err*r1+err^2 = r1^2-2*err*r1+err^2,
>>           r2^2-2*err*r2+err^2 = r2^2-2*err*r2+err^2]
>> (%i12) radcan(sola[1]-ev(solb[1],err=0));
>> (%o12) [0 = 0,0 = 0]
>> (%i13) radcan(sola[2]-ev(solb[2],err=0));
>> (%o13) [0 = 0,0 = 0]
>>
>>
>> Viktor
>>
>>
>>
>>
>>
>>
>> -----Original Message-----
>> From: maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu]
>> On Behalf Of Krishna Myneni
>> Sent: Monday, July 02, 2012 7:39 PM
>> To: maxima at math.utexas.edu
>> Subject: solution of quadratic equations -- incorrect solution
>>
>> Previously, I posted that Maxima's solve() was unable to solve a system of
>> two quadratic equations directly. Several workarounds were suggested to
> find
>> the solution to the following system of equations:
>>
>> e1:   (x - a)^2 + (y - b)^2 = r1^2
>>
>> e2:  (x - c)^2 + (y - d)^2 = r2^2
>>
>> While the workarounds permit Maxima's solve() to obtain analytic solutions
>> to the above equations, we have found that Maxima gives an incorrect
>> solution to a slight variant of the above equations. Instead of the
> general
>> equations above, it is simpler to illustrate the problem with the
> following
>> eqns:
>>
>> e1:  x^2 + y^2 = (r1 - err)^2
>>
>> e2: (x - 1)^2 + (y - 1)^2 = (r2 - err)^2
>>
>> If we first find the analytic solutions to the set, [x^2 + y^2 = r1^2,
>> (x-1)^2 + (y-1)^2 = r2^2], and then find the solutions to the above
>> equations with "err", substituting err=0 into the latter gives solutions
>> which are not the same as the former, as shown below. Comparing the
>> algebraic expressions with err=0 show that the x-solutions are the same,
> but
>> the y-solutions have a sign error in product terms r1*r2.
>>
>>
>>
>> (%i1) e1: x^2 + y^2 = r1^2;
>>                                     2    2     2
>> (%o1)                            y  + x  = r1
>> (%i2) e2: (x-1)^2 + (y-1)^2 = r2^2;
>>                                     2          2     2
>> (%o2)                      (y - 1)  + (x - 1)  = r2
>>
>> (%i4) solve([e1,e1-e2],[x,y]);
>>
>>                          4        2        2     4       2          2     2
>>                 sqrt(- r2  + (2 r1  + 4) r2  - r1  + 4 r1  - 4) + r2  - r1
> -
>> 2
>> (%o4) [[x = -
>> ---------------------------------------------------------------,
>>                                                4
>>
>>                 2               2             2               2          2
>> 2
>> y = (sqrt(- r2  + 2 r1 r2 - r1  + 2) sqrt(r2  + 2 r1 r2 + r1  - 2) - r2  +
>> r1 + 2)/4],
>>
>> [x =
>>            4        2        2     4       2          2     2
>> sqrt(- r2  + (2 r1  + 4) r2  - r1  + 4 r1  - 4) - r2  + r1  + 2
>> ---------------------------------------------------------------,
>>                                  4
>>
>>                   2               2             2               2
> 2
>> y = - (sqrt(- r2  + 2 r1 r2 - r1  + 2) sqrt(r2  + 2 r1 r2 + r1  - 2) + r2
>>
>>                2
>>            - r1  - 2)/4]
>>
>> ]
>>
>> Generalizing these equations slightly, and finding the new solutions, we
>> have,
>>
>> (%i15) e1: x^2 + y^2 = (r1-err)^2;
>>                                 2    2             2
>> (%o15)                       y  + x  = (r1 - err)
>> (%i16) e2: (x-1)^2 + (y-1)^2 = (r2-err)^2;
>>                                 2          2             2
>> (%o16)                 (y - 1)  + (x - 1)  = (r2 - err)
>> (%i17) sol: solve([e1,e1-e2],[x,y]);
>>
>> (%o17) [[x = - (sqrt(- r2  + 4 err r2  + (2 r1  - 4 err r1 - 4 err  + 4)
> r2
>>                 2        2                    4           3             2
> 2
>>    + (- 4 err r1  + 8 err  r1 - 8 err) r2 - r1  + 4 err r1  + (4 - 4 err )
> r1
>>                      2          2                2
>>    - 8 err r1 + 8 err  - 4) + r2  - 2 err r2 - r1  + 2 err r1 - 2)/4,
>>
>>                 2                          2                   2
>> y = (sqrt(- r2  - 2 r1 r2 + 4 err r2 - r1  + 4 err r1 - 4 err  + 2)
>>           2               2          2                2
>>    sqrt(r2  - 2 r1 r2 + r1  - 2) - r2  + 2 err r2 + r1  - 2 err r1 + 2)/4],
>>    
>>                  4           3        2                   2        2
>> [x = (sqrt(- r2  + 4 err r2  + (2 r1  - 4 err r1 - 4 err  + 4) r2
>>                 2        2                    4           3             2
> 2
>>    + (- 4 err r1  + 8 err  r1 - 8 err) r2 - r1  + 4 err r1  + (4 - 4 err )
> r1
>>                      2          2                2
>>    - 8 err r1 + 8 err  - 4) - r2  + 2 err r2 + r1  - 2 err r1 + 2)/4,
>>
>>                   2                          2                   2
>> y = - (sqrt(- r2  - 2 r1 r2 + 4 err r2 - r1  + 4 err r1 - 4 err  + 2)
>>           2               2          2                2
>>    sqrt(r2  - 2 r1 r2 + r1  - 2) + r2  - 2 err r2 - r1  + 2 err r1 - 2)/4]]
>>
>>
>>
>> If we set err = 0, the equations to be solved are identical to the
> original
>> equations. However, setting err = 0 in the solutions does not give the
> same
>> results as before. The x-solutions are the same, but the y-solutions have
>> product terms r1*r2 which have a different sign from the solutions of the
>> original equations. The solutions to the problem are incorrect.
>>
>>
>> Krishna
>>
>>
>>
>>
> Victor,
>
> I reproduce your maxima output (Maxima 5.24), but the two solutions,
> sola and solb with err=0, are not the same. This can be seen
> algebraically and numerically.
>
> (%i16) ev(sola[1],[r1=1,r2=1]);
>
> (%o16) [x = 0,y = 1]
> (%i17) ev(solb[1],[r1=1,r2=1,err=0]);
>
> (%o17) [x = 0,y = 0]
>
>
> The x-solutions are the same, but the y-solutions are not. If you print
> the solutions, the comparison is not
>
> sqrt(A-B)*sqrt(C-D) vs. sqrt(B-A)*sqrt(D-C)
>
> but
>
> sqrt(A-B)*sqrt(C-D) vs. sqrt(A+B)*sqrt(C+D)
>
>
> (%i18) sola[1];
>
> (%o18) [x = -(sqrt(-r2^4+(2*r1^2+4)*r2^2-r1^4+4*r1^2-4)+r2^2-r1^2-2)/4,
>           y =
> (sqrt(-r2^2+2*r1*r2-r1^2+2)*sqrt(r2^2+2*r1*r2+r1^2-2)-r2^2+r1^2+2)
>             /4]
>
> (%i20) ev(solb[1],err=0);
>
> (%o20) [x = -(sqrt(-r2^4+(2*r1^2+4)*r2^2-r1^4+4*r1^2-4)+r2^2-r1^2-2)/4,
>           y =
> (sqrt(-r2^2-2*r1*r2-r1^2+2)*sqrt(r2^2-2*r1*r2+r1^2-2)-r2^2+r1^2+2)
>             /4]
>
>
>
>
> Regards,
> Krishna
>
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>
>
Victor,

Ok. I understand that there is an ambiguity here which depends on the 
sign of the product r1*r2, which is inside of the square root of the 
y-solutions. That is, in fact, the only difference between sola and solb 
with err=0. However, your assertion that the two solutions are 
equivalent up to the sign of the square root is not apparent to me. 
Nevertheless, it is true that the two solutions, sola and solb with 
err=0, are, in fact, different solutions. Each set gives two distinct 
x,y pairs.

Now, as to the question of the "correct" solution, thank you very much 
for pointing out that both sets of solutions may be correct under 
different sets of parameters, r1 and r2. This suggests that Maxima's 
solve() is returning an incomplete set of solutions for the two equations,

x^2 + y^2 = r1^2, (x-1)^2 + (y-1)^2 = r2^2

There is an arbitrary choice among the possible solutions, consisting of 
the union of sola and solb with err=0 (and perhaps other solutions if we 
want to admit complex r1 and r2 ?). The particular set of solutions 
returned for the above equation (sola) happen to work for r1 > 0 and r2 
 > 0.

In contrast, the set of solutions returned for the second set of 
equations containing (r1 - err)^2 and (r2 - err)^2, is a different 
choice of two solutions from the complete space of solutions. The 
solutions returned in this case are not valid solutions for r1 > 0, r2 > 
0, and err =0.

Is this the correct interpretation to take with regard to the set of 
solutions returned by solve, i.e. that they are an arbitrary selection 
of a subset of solutions from the complete set?

Krishna