Quintic root

Here’s a curious result I ran across the other day. Suppose you have a quintic equation of the form

z5z − 1 = 0.

(It’s possible to reduce a general quintic equation to this form, known as Bring-Jerrard normal form.) There is no elementary formula for the roots of this equation, but the following infinite series does give a root as a function of the leading coefficient z:

\sum_{n=0}^\infty {5n \choose n} \frac{z^n}{4n+1}

One reason this is interesting is that the series above has a special form that makes it a hypergeometric function of z. You can read more about it here.

I could imagine situations where having such an expression for a root is useful, though I doubt the series would be much use if you just wanted to find the roots of a fifth degree polynomial numerically. Direct application of something like Newton’s method would be much simpler.

10 thoughts on “Quintic root

  1. Typo in your quintic equation (which, as written, would has a root of -1 when z=0 in contradiction to the series which is positive for positive z). The equation should be z x^5 – x + 1 = 0.


  2. Dan, when z = 0, the equation becomes -x + 1 = 0, so x = 1. The series also evaluates to 1 when z = 0, so I think it’s OK.

  3. An explicit *symbolic* expansion of Newton’s method will produce a series. And likely it will be equally simple if not the same.

  4. SteveBrooklineMA

    I may be confused, by it seems you would need |z|<1 at least for convergence.

  5. This summation looks divergent to me in all cases.

    The function (5n choose n) / (4n + 1) diverges at a factorial rate. Even the (z^n) exponential when z << 1 can't overcome that asymptotically.

    Here's a wolfram|alpha that graphs the function with z = 0.1 (admittedly, not the sum, but it helps to see that the series is divergent): http://www.wolframalpha.com/input/?i=%285n+choose+n%29*%280.1%5En%2F%284n%2B1%29%29+from+0+to+20

    And even when z = 0, z^n = 1, but the rest of the function still diverges. Maybe I'm missing something?

  6. Joe, (5n choose n) <= sum over all k of (5n choose k) = 2^(5n) so when z<1/32 the series converges.

    If you replace 0.1 with 0.03 in your Wolfram|Alpha plot, you'll see that the divergence goes away.

  7. Just to follow up on series convergence… at first glance it does look like convergence is an issue. Playing with Stirling’s approximation you can show that asymptotically (5n-choose-n) is
    sqrt(5/(8*pi*n))* (3125/256)^n. So one would expect convergence as long as |z| < (256/3125) == 0.08192.

  8. Plugged this into SymPy to get the explicit form as a hypergeometric function:

    >>> print(summation(binomial(5*n, n)*z**n/(4*n + 1), (n, 0, oo)))
    Piecewise((hyper((1/5, 2/5, 3/5, 4/5), (1/2, 3/4, 5/4), 3125*z/256), 3125*Abs(z)/256 <= 1), (Sum(z**n*binomial(5*n, n)/(4*n + 1), (n, 0, oo)), True))

    The convergence condition is the same as given by others here (although note that convergence conditions given by SymPy are not guaranteed to be sharp).

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