Solving cubic equations

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Assume that , where a, b, and c are Real and that the roots of f(x) = 0 are smooth (i.e. differentiable) functions of at least ‘c’ and are distinct. Denote the root functions by x1, x2 and x3.

If x is any one of these roots, it is easy to show that ∂x/∂c=-1/f ‘(x).

Also, ∆2=(x1-x2)2(x2-x3)2(x1-x3)2 can be shown to be independent of the roots x1, x2 and x3 and that it is just a function of a, b, and c. can also be shown to satisfy the equation .

Now .

The right-hand side of this equation simplifies to , i.e. it is independent of c. Hence the left-hand side can be differentiated with respect to c to give 0. If we do this, and use the fact that , we get the final LDE, (which is true for any root x)

. It is straightforward to show that .

If we now apply this to the reduced cubic equation f(x)=x3+ax+b=0, then it follows that

.

The LDE above then boils down to .

Solution of the above LDE.

Case 1 where , a<0 and .

The initial conditions for the LDE come from the reduced cubic with b=0,

i.e.so .

If b=0 and x=0, then . If b=0 and then

. These initial conditions are enough to allow the LDE to be solved using standard methods to give …

,

, and

.

Case 2 where and a>0.

If b=0, then from the reduced cubic we get one real value for x, i.e. x=0, so

when b=0.

Again solving the LDE using standard methods gives the Cardano solution …

.

This is also a valid solution if a<=0. The complex valued solutions can also be obtained in a similar way, i.e. when b = 0 take and .

—————————————-

If we further reduce the cubic to , where t is Real, then we can try to obtain a series solution to the LDE about t=0.

The LDE now becomes .

When t = 0, take x(0) = 0 for a Real solution and so x'(0) = -1.

Again using standard methods for solving LDE’s for series solutions we get …

for .

This must have the same value as the Cardano formula when a=1 and b=t, i.e.

Hence

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