Is there some equivalent to the quadratic eqaution for higher order polynomials? At best, I seem to recall some trick that works for some third order polynomials and even a “lesser” trick for fourth order polynomials? - Jinx
There’s a general solution known for polynomials of arbitrary degree, but it’s not something that’s useful for calculating roots. Realistically speaking, approximation methods are going to be favored over exact solutions for cubic and higher degree polynomials.
jinx
Here’s a method for solving cubic equations:
www.1728.com/cubic2.htm
and another for quartic equations:
www.1728.com/quartic2.htm
As you can see, they are very involved.
Jinx
And if you want to see the cubic formula written out in similar fashion as the quadratic formula, click here:
http://planetmath.org/encyclopedia/CubicFormula.html
And if that doesn’t look intimidating enough, look at the monstrosity known as the quartic formula:
http://planetmath.org/encyclopedia/QuarticFormula.html
Don’t forget to scroll all the way to the right to see the ENTIRE formula.
In a way, I guess these are what you wanted, but they aren’t very practical to use.
What the OP is referring to is commonly known as the quadratic formula, which is used for solving quadratic equations.
The quadratic formula is known and loved by millions of high school & college algebra students, but hardly anybody knows the corresponding formulas for third and fourth-degree equations. (I, for one, don’t!) Since those formulas are so much more complicated, there are usually easier ways of dealing with such equations. And, after centuries of searching, it was proved (by Abel) that there is no algebraic formula for the general solution of polynomial equations of degree 5 or more, like there is for degrees up through 4.
Sure, there are formulas for up to 4th degree polynomials for finding roots. But they’re so messy that it’s easier to use approximation methods. It’s a proven fact (thanks to Galois) that there are no such formulas for 5th degree or higher polynomials.
The advantage of the formulas for finding cubic and quartic roots is that they give you all the roots of the polynomial–including complex roots (if any), something that real approximation cannot do.
Let’s get picky here for a second. The issue is not whether there’s a general formula for quintics, but whether there’s one that only involves addition, subtraction, multiplication, division and square roots. There are very large classes of quintic equations for which there is such a formula, but there’s no such formula for the general quintic. However, if you’re willing to allow other operations, there is a general formula for the roots of a quintic equation, which you can see here. Again, it’s not much use for calculations–this is where Newton’s method and other numeric techniques come into play.
I seem to remember that there are numerical methods that can converge to the complex roots of a function. The one I saw was based on second-order polynomials, but I don’t remember its name. Maybe someone will come along with more information.
By definition. If a method can converge to a non-real number, then it’s not a real approximation method. But there are perfectly good complex approximation methods: If nothing else, one could use a variant of the method of bisection.
The real advantage to the exact equations is they enable you to find irrational roots exactly. There are methods guaranteed to find all of the rational roots (even complex ones) of any polynomial in a finite bounded time. If all but four of your roots are rational, you can then reduce the order of your equation and find the last irrational roots using the appropriate formula.
Oh, and ultrafilter, I think you meant roots of arbitrary order, not just square roots. Even something as simple as x[sup]5[/sup] - 2 = 0 can’t be solved with just +,-,*,/, and square roots. And while there are formulas using other operations for an arbitrary polynomial, that’s not very significant. At some point, you’re basically just defining some new operation to be the solution you’re looking for, and you’ve just transformed the problem into evaluating that operation, which might be just as hard.
The listed webpages say that too, but I’m not sure I buy that. If computers are good at anything, it’s multiplying lots of numbers together, which is all those formulae appear to be doing. Why bother with an approximation when you can get the real deal?
Floating point error. All computer calculations are approximations.