Solving the quintic equation

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We start with the quintic equation in Bring-Jerrard form f(x)=x5+x+t, tεR.

We will assume that x=x(t) is a suitably smooth function of t, that x(0)=0, and that 5x4+1≠0. If we then differentiate x as a function of t, via f(x) = 0, we get x'(t)=-1/(5x4+1). Clearly x(0) = 0 implies that x'(0)=-1.

Further differentiation of x(t), with evaluation at t = 0, shows that x”(0) = 0 and x”'(0) = 0. These values will provide the initial conditions for the solution of a 4th order LDE satisfied by x(t).

The next stage is to write/expand the expression -1/(5x4+1)as a power series in x and use the fact that for powers of x which are 5 or higher these can be expressed as linear combinations of x to the power 0, 1, 2, 3, or 4. This follows from the fact that x5= -x-t, x6= -x2-tx etc. So, for example x9 = x5-tx4 = -x-t-tx4 = -tx4-x-t.

The discriminant ∆2 appears in the following calculations. ∆2 =Π1≤i<j≤5 (xi – xj)2, where xi and xj are pairs of distinct roots of f(x)=0. For the quintic x5+x+t = 0, ∆2 = 44+55t4 .

2 can be shown to be equal to (-1)10Π1≤i≤5f ‘(xi).

This expression in turn can be shown to be equal to 55Π1≤i≤4 f (yi) where f ‘(yi)=0.

It is easily shown that f(yi)=(4/5)(yi+5t/4), and so ∆2 = 55(4/5)4Π1≤i≤4 (yi+5t/4) or,

2= 55(-4/5)4Π1≤i≤4 (-5t/4-yi) .

Now f ‘(x) = 5Π1≤i≤4 (x-yi) where f ‘(yi)=0.

Hence Δ2 = 55(-4/5)4(1/5)f ‘(-5t/4) or, when simplified, Δ2 = 44 + 55t4.

The details of the calculation are tedious to write out but here is the resulting expression …

Δ2x ‘ = -5(43)x4+5242tx3-534t2x2+54t3x-44.

Proof (short version)

Assume that x≠o and 5x4+1≠0.

Start with x ‘(t) = -1/(5x4+1).

Then, Δ2x ‘ = -(55t4 + 44)/(5x4+1)+44-44.



=-44-5x((4x)4-(5t)4)/(4x+5t) Note: 5x4+1≠0 implies that 4x+5t≠0.


If we differentiate both sides of this equation, then after some rearrangement we end up with an expression where the left-hand side is a linear combination of x” and x’, while the right hand side has no term in x where the power of x is greater than 3.

This is … Δ2x” + 10(54)t3x’ = 5(42)x3 – 3(4)52tx2 + 6(53)t2x.

Repeating this process again yields …

Δ2x”’ + 30(54)t3x” + 135(53)t2x’ = -15(42)x2 + 51(52)tx.

And finally, Δ2xiv + 50(54)t3x”’ + 117(54)t2x” + 51(54)tx’ – 1155x = 0.

This equation can be solved using standard power series methods, with the initial conditions above, i.e. x(0)=0, x'(0)=-1, x”(0)=0, x”'(0)=0, to give …

x =∑0≤n<∞ (-1)n+1 (5n)!/(n!(4n+1)!)t4n+1 ,|t|<4/(5 4√5).

The initial condition(s) for the other roots come from the non-zero complex roots of x(0)4 = -1,

i.e. x(0)=(1+i)/√2, … , -(1+i)/√2.

The existence and uniqueness theorems for ODE+LDE’s ensure that this solution to the above LDE is also a solution to

the quintic equation x5 + x + t = 0, tεR.

A method of solving polynomial equations in this style was first discovered in the mid 19th century by Cockle and Harley.

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