求解 Matlab 一个系数很小的二次方程

Question

I'm implementing a code in matlab to solve quadratic equations, using the resolvent formula:

Here´s the code:

clear all 
format short 
a=1; b=30000000.001; c=1/4; 
rdelta=sqrt(b^2-4*a*c); 
x1=(-b+rdelta)/(2*a);
x2=(-b-rdelta)/(2*a);
fprintf(' Roots of the polynomial %5.3f x^2 + %5.3f x+%5.3f \n',a,b,c)  
fprintf ('x1= %e\n',x1)
fprintf ('x2= %e\n\n',x2)
valor_real_x1= -8.3333e-009;
valor_real_x2= -2.6844e+007;

error_abs_x1 = abs (valor_real_x1-x1);
error_abs_x2 = abs (valor_real_x2-x2);

error_rel_x1 = abs (error_abs_x1/valor_real_x1);
error_rel_x2 = abs (error_abs_x2/valor_real_x2);

fprintf(' absolute_errorx1 = |real value - obtained value| = |%e - %e| = %e \n',valor_real_x1,x1,error_abs_x1)  
fprintf(' absolute_errorx2 = |real value - obtained value| = |%e - %e| = %e \n\n',valor_real_x2,x2,error_abs_x2) 

fprintf(' relative error_x1 = |absolut error / real value| = |%e / %e| = %e \n',error_abs_x1,valor_real_x1,error_rel_x1 ) 
 fprintf(' relative_error_x2 = |absolut error / real value|  = |%e / %e| = %e \n',error_abs_x2,valor_real_x2,error_rel_x2) 

The problem I have is that it gives me an exact solution, ie for values a = 1, b = 30000000,001 c = 1/4, the values of the roots are:

Roots of the polynomial 1.000 x^2 + 30000000.001 x+0.250 
 x1= -9.313226e-009
 x2= -3.000000e+007

Knowing that the exact value of the roots of the polynomial are:

x1= -8.3333e-009
x2= -2.6844e+007

Which gives me the following errors in the absolute and relative precision of the calculations:

 absolute_errorx1 = |real value - obtained value| = |-8.333300e-009 - -9.313226e-009| = 9.799257e-010 
 absolute_errorx2 = |real value - obtained value| = |-2.684400e+007 - -3.000000e+007| = 3.156000e+006 

 relative error_x1 = |absolut error / real value| = |9.799257e-010 / -8.333300e-009| = 1.175916e-001 
 relative_error_x2 = |absolut error / real value|  = |3.156000e+006 / -2.684400e+007| = 1.175682e-001

My question is: Is there an optimum method to obtain the roots of a quadratic equation?, ie I can make changes to my code to reduce the relative error between the expected solution and the resulting solution?

Answer 1

Using the quadratic formula directly in this cases results in a large loss of numerical precision from subtracting two values of very similar magnitude. This is because the expression

sqrt(b*b - 4*a*c)

is nearly the same as b. So you should use only one of these two roots, the one that does not involve subtracting two very close values, and for the other root you can use (for instance) the fact that the product of roots of a quadratic is c/a. I'll let you fill in the gaps.

Answer 2

Why does this sound like a homework problem from a first class in numerical analysis?

It has been a while since I was that young, but as I recall there is a trick. Anyway, you are wrong. The true roots of that polynomial are

solve('x^2 + 30000000.001*x + 0.25')
ans =
          -30000000.000999991666666666944442
 -0.0000000083333333330555578703796293981491

How well does roots do here?

p = [1 30000000.001 1/4];
format long g
roots(p)
ans =
             -30000000.001
     -8.33333333305556e-09

That actually seems pretty good. How does HPF do?

DefaultNumberOfDigits 64
a = hpf(1);
b = hpf('30000000.001');
c = hpf('0.25');

r1 = (-b + sqrt(b*b - 4*a*c))/(2*a)
r1 =
-0.000000008333333333055557870379629398149125529835186899898569329967

r2 = (-b - sqrt(b*b - 4*a*c))/(2*a)
r2 =
-30000000.000999991666666666944442129620370601850874470164813100101

Yep, HPF works nicely enough too.

So what happens when you use double precision numbers and the standard formula? Yeah, crapola arrives.

a = 1;
b = 30000000.001;
c = 0.25;

>> r1 = (-b + sqrt(b*b - 4*a*c))/(2*a)
r1 =
     -7.45058059692383e-09

>> r2 = (-b - sqrt(b*b - 4*a*c))/(2*a)
r2 =
             -30000000.001

Again, massive subtractive cancellation eats away at the result. (I seem to recall that was the problem you had in your last question.)

There is a trick you can use. See that the large solution was well estimated, just not the one near zero. So, what happens if you solved for the roots of fliplr(p) using the quadratic formula? How does this solve your problem? What transformation is implicitly done when you do that? (Sorry, but I won't do your homework. I think the above was enough of a hint anyway.)

Answer 3

i think your "real" values might be wrong (or maybe it's a precision thing... I dunno)

a*(valor_real_x1^2)+b*(valor_real_x1)+c

ans =

   9.9999e-07

a*(valor_real_x2^2)+b*(valor_real_x2)+c

ans =

  -8.4720e+13

Answer 4

A nice formula for this problem:

var q = sqrt(c*a)/b;
var f = .5 + .5 *sqrt(1-4*q*q);
var x1=-b*f/a;
var x2=-c/(f*b);