Girolamo Cardano and the Solution to the Cubic

Girolamo Cardano was born September 24, 1501 in Italy. Throughout his youth, Cardano (or more commonly referred to as Cardan) studied mathematics under the guidance of his father. Cardan studied medicine at Pavia University and later at the University of Padua. During this time Cardan gained a reputation as a difficult man, and developed a gambling problem. The College of Physicians in Milan would not admit him due to the fact that he was an illegitimate child; despite this, Cardan became a successful, renowned physician. At the request of fellow colleagues, the College of Physicians admitted Cardan in 1539. During this time, Cardan published two mathematical books and became acquainted with many mathematicians, including Niccolo Tartaglia. Tartaglia had won a mathematics contest in which he utilized the solution to the cubic. Cardan urged Tartaglia to show him the method. After making a solemn promise not to publish his solution, Tartaglia agreed. Interestingly, Tartaglia's solution to the cubic was not his original work. The solution to the cubic has been traced back to Scipione del Ferro, a mathematics professor at Bologna around the beginning of the sixteenth century. Ferro didn't publicize his research, but he taught a student how to solve the cubic before he died in 1526.

After discovering in 1543 that Tartaglia's discovery was not really his own, Cardan published the solutions to the cubic and quartic equations after six years of study. In 1545, Cardan's most famous work, Ars Magna revealed these and other solutions. The solutions to these equations were the first major breakthroughs in mathematics since the time of the Greeks.

Solving the Cubic

Cardan's solution to the cubic is demonstrated in the following steps.

The cubic equation is of the form

Cardan changed this equation to one with no quadratic term.

Using the substitution x = y - a/3

We get

x^3 + ax^2 + bx + c = 0

= ( y - a/3)^3 + a(y-a/3)^2 + b(y-a/3) + c

Aside

( y-a/3)^2 = (y-a/3) (y-a/3)

= y^2 - ay/3 - ay/3 + a^2/9

= y^2 - 2ay/3 + a^2/9

( y-a/3)^3 = ( y-a/3) (y-a/3) (y-a/3)

= (y-a/3)^2 (y-a/3)

= y^2 - 2ay/3 + a^2/9 (y-a/3)

= y^3 - 2ay^2/3 + a^2y/9 - ay^2/3 + 2a^2y/9 - a^3/27

= y^3 - 3ay^2/3 + 3a^2/9 - a^3/27

= y^3 - ay^2 +a^2y/3 - a^3/27

Now back to the cubic. Substitute the results from the Aside section into the original cubic.

y^3 - ay^2 +a^2y/3 - a^3/27 + a ( y^2 - 2ay/3 ) + a^2/2)+ b ( y - a/3 ) + c = 0

Multiply through to get

= y^3 - ay^2 + a^2y/3 - a^3/27 + ay^2 - 2a^2y/3 + a^3/9 + by - ba/3 +c

Simplify and you are left with

= y^3 - a^2y/3 - 2a^3/27 + by - ba/3 + c

The coefficients in front of the y terms and the coefficients that remain by themselves can be

renamed as the new coefficients p and q

This equation has now become

y^3 = py + q

Now let y = u + v

So now we have y^3 = ( u^3 + v^3 ) + 3uv (u + v)

which is equal to 3uvy + ( u^3 + v^3) (remember y = u+v)

This equation needs to equal the right hand side which is py+q

In order for this to hold true, p = 3uv and q = u^3+v^3

If you eliminate the v using the equality p = 3uv, you get v = p/3u

Substitute this to get u^3 + (p/3u)^3 = q

> z:=(u^3+(p/(3*u))^3-q=0);

> solve(z,u);

The above shows the roots of the equation. So, in general by using u^3 + ( p/3u)^3 = v,

the roots of this are q/2 + or - [ ( q/2 )^2 - ( p/3 )^3 ]^(1/2) . So these are our values of u^3. Because of symmetry, we

get these same values for v^3. Keeping in mind that u^3 + v^3 = q , let u^3 = q/2 +[( q/2 )^2 - ( p/3 )^3 ]^(1/2)

and let v^3 = q/2 - [ ( q/2 )^2 - (p/3)^3 ]^(1/2) .

Once all of these equations are substituted where appropriate, the following results are obtained.

y = u + v = [ ( q/2 ) + [ ( q/2 )^2 - ( p/3 )^3 ]^(1/2) ] ^(1/3) + [ (q/2) - [ ( q/2 )^2 - ( p/3 )^3 ]^( 1/2 ) ] ^ ( 1/3 )

which can be written this way and be refferred to as cardan.

> cardan := (p,q) -> (sqrt((p/3)^3+(q/2)^2)+q/2)^(1/3)-
(sqrt((p/3)^3+(q/2)^2)-q/2)^(1/3);

Complex Number Portion

Cardan went through many different cases of cubic equations in his book Ars magna, there are 13 cases to be exact. Tartaglia only gave him the solution to the one of the form x^3 + px = q . He then had to expand on Tartaglias information to solve the other cases. When he came to the form x^3 = px + p, he made the substitution x = u + v as opposed to x = u - v. This is illustrated in a later example.

> cardan := (p,q) -> (sqrt((p/3)^3+(q/2)^2)+q/2)^(1/3)-
(sqrt((p/3)^3+(q/2)^2)-q/2)^(1/3);

This was Cardan's solution to the cubic x^3 + px = q.

Cardan noticed a problem with this formula that arose when (q/2)^2 < (p/3)^3. This situatiion leads to dealing with square roots of negative numbers.

Working with the equation x^3 = 15x + 4 a substitution is made making x = u + v and the product of u and v being one third the coefficient of the x term, or 1/3(15) = 5. This leaves:

x^3 = 15x + 4 using substitution

(u + v)^3 = 15(u + v) + 4 expanding the equation leaves

u^3 + 3u^2v + 3uv^2 + v^3 = 15u + 15v + 4 using the fact that uv = 5 results in

u^3 + 15u + 15v + v^3 = 15u + 15v + 4 which simplifies to

u^3 + v^3 = 4 with uv = 5 it is known that v = 5/u

u^3 + (5/u)^3 = 4 which simplifies to

u^3 + 15/u^3 = 4 multipling through by u^3 leaves

u^6 + 15 = 4u^3 which is a quadratic in u^3, letting u^3 = p

p^2 + 15 = 4p solving this quadratic results in

> z:=(x^2-4*x+15=0);

> solve(z,x);

This is where I ran into the sqrt(11) instead of 121.

This shows that u^3 = 2 + sqrt (-11) and v^3 = 2 - sqrt (-11). Substituting back into the u^3 +v^3 = 4 equation results in: x = (2 + sqrt (-11))^1/3 + (2 - sqrt (-11))^1/3

> solx :=cardan(-15,4);

> solve(x^3=15*x+4,x);

I still haven't solved the problem that we ran into today so this portion of our project is pretty rough and will be better once I understand why my substition isn't working out. This is just to give you an idea of where we are.

Cardan had established that you couldn't square and number and get a negative number but was unable to understand his answer. He called these expressions "sophistic" and referred to his result "as subtile as it is useless." He later wrote Tartaglia on the subject:

I have sent to enquire after the solution to various problems for which you have given me

no answer, one of which concerns the cube equal to an unknown plus a number. I have

certainly grasped this rule, but when the cube of one-third of the coefficient of the unknown

is greater in valus than the square of one-half of the number, then, it appears, I cannot make

it fit into the equation.

This shows that Cardan knew the conditions that resulted in square rots of negative numbers but his appeal to Tartaglia fell on bitter ears because Tartaglia still felt betrayed so he tried to confuse Cardan writing:

... and thus I say in reply that you have not mastered the true way of solving problems of this

kind, and indeed I would say that your methods are totally false.

Exploring the effects of coefficients of cubic equations on the roots

QUESTION: In the cubic equation x^3 + x^2 + x + a, where a is a constant, how do the roots change as a varies over the real numbers? When a represents a coefficient of either the x, x^2, or x^3 term, how do the roots change?

Investigation:

Given the standard cubic equation

> c:=x^3+x^2+x+a;

we can examine the effects of changing the constant term, a, on the roots of the equation.

> showroots := proc(a)
local eqn,x, i,sol,rts;

> eqn := x^3+x^2+x+a ;

> sol :=solve(eqn,x);

> rts := seq([Re(sol[i]),Im(sol[i])],i=1..3);

> plot({rts},style=point,symbol=circle);
end:

reserved word `end` unexpected

To show the roots when a = 2, that is, our equation is x^3 + x^2 + x + 2:

> showroots(2);

Now, let's vary a between -200 and 200. (We've chosen -200 and 200 because they are large numbers that can be indicative of the way a affects the equation as a varies over the real numbers.) We have an animation with 41 frames, as follows:

> with(plots):

> plots[display]([seq(showroots(-4999+10000/40*i),i=0..40)],insequence=true);

Notice that as a changes from very small number to a large number, the roots are real, until a certain point. As a gets larger, two of the roots have complex portions. As a gets very large, all three roots approach zero.

Now let's look at the effect of changing the coefficient of the x term in our equation, which we will call "g" (we set the constant term of the equation equal to zero):

> g:=x^3+x^2+a*x;

> showrootsB := proc(b)
local eqn,x, i,sol,rts;

> eqn := x^3+x^2+b*x;

> sol :=solve(eqn,x);

> rts := seq([Re(sol[i]),Im(sol[i])],i=1..3);

> plot({rts},style=point,symbol=circle);
end:

> with(plots):

> plots[display]([seq(showrootsB(-199+400/40*i),i=0..40)],insequence=true);

The three roots of the equation are real up to a certain point, at which two roots have complex portions. As b increases, the real part of the two roots stays between 0 and -2, while the complex portion of the roots increases indefinitely.

Now let's change the coefficient of the x^2 term:

> h:=x^3+a*x^2+x+20;

> showrootsC := proc(c)
local eqn,x, i,sol,rts;

> eqn := x^3+c*x^2+x ;

> sol :=solve(eqn,x);

> rts := seq([Re(sol[i]),Im(sol[i])],i=1..3);

> plot({rts},style=point,symbol=circle);
end:

> with(plots):

> plots[display]([seq(showrootsC(-199+400/40*i),i=0..40)],insequence=true);

The roots are all real, except for a brief interval when a is near zero.

Lastly, a look at the way the coefficient of the x^3 term of the cubic equation changes the roots:

> m := d*x^3 + x^2 + x;

> showrootsD := proc(d)
local eqn,x, i,sol,rts;

> eqn := d*x^3+x^2+x ;

> sol :=solve(eqn,x);

> rts := seq([Re(sol[i]),Im(sol[i])],i=1..3);

> plot({rts},style=point,symbol=circle);
end:

> with(plots):

> plots[display]([seq(showrootsD(-199+400/40*i),i=0..40)], insequence=true);

Again, the roots are real up to a certain point, at which the real and complex portions of the root change dramatically from zero to a small real number and a large complex number.

Question: At which a,b,c, and d do the roots become complex? What is the value of the real root at this point?

BIBLIOGRAPHY

Boyer,

Burton, David M. The History of Mathematics . McGraw-Hill, pp. 297-302.

Stillwell, John. Mathematics and Its History . New York: Springer-Verlag, 1989, pp.54-55, 59-62.