February 20, 2010 10:06 AM Subscribe

Is it possible to solve for x in y=(x^t)/(x-1), for any value of t? Wolfram alpha and maxima failed, so I guess not, but I'm interested in what more general classes of unsolvable equations this falls in.

posted by jewzilla to Science & Nature (12 answers total) 2 users marked this as a favorite

posted by jewzilla to Science & Nature (12 answers total) 2 users marked this as a favorite

If t is not a positive integer, you have a different problem on your hands than what serathen says above. It looks to be unsolvable with your normal algebraic operations.

posted by King Bee at 10:28 AM on February 20, 2010

posted by King Bee at 10:28 AM on February 20, 2010

Fascinating question. Abel-Ruffini theorem

By the way, suppose t is not an integer. Not sure what formal solutions are available in this case.

posted by polymodus at 10:35 AM on February 20, 2010

By the way, suppose t is not an integer. Not sure what formal solutions are available in this case.

posted by polymodus at 10:35 AM on February 20, 2010

Oh, and the proof that there is no closed-form solution to finding roots of higher-degree polynomials comes from a branch of abstract algebra called Galois Theory. I found that problem fascinating in college, and took a semester of abstract algebra hoping to understand it, but I never got that far.

And of course, just because your equation has no closed form solution doesn't mean it can't be solved numerically for a particular*y* and *t* in an efficient way. I wouldn't be surprised if even 3rd degree polynomials are solved by general numerical methods in practice, rather than using the special-case cubic equation. It's pretty hairy.

posted by serathen at 10:38 AM on February 20, 2010

And of course, just because your equation has no closed form solution doesn't mean it can't be solved numerically for a particular

posted by serathen at 10:38 AM on February 20, 2010

King Bee: yeah, if *t* can be an arbitrary real number then the problem is way out of my league.

posted by serathen at 10:41 AM on February 20, 2010

posted by serathen at 10:41 AM on February 20, 2010

Mine too. I think the OP is better off numerically solving for x for particular y and t, as you suggest, instead of trying to solve the more general case.

posted by King Bee at 10:43 AM on February 20, 2010

posted by King Bee at 10:43 AM on February 20, 2010

Yes, you can solve the equation of course, but there are multiple solutions, the solutions have multiplicities, and solutions cannot necessarily be expressed using addition and multiplication and roots. If t=n/m is a rational with n>m>0, then, for each y, the equation y=(x^t)/(x-1) has exactly n solutions in the complex plane, counting multiplicity.

posted by jeffburdges at 11:17 AM on February 20, 2010

posted by jeffburdges at 11:17 AM on February 20, 2010

To clarify, Galois Theory tells us that not all polynomials of degree 5 or higher have a closed form solution. However, it certainly does not preclude the possibility of any polynomials of higher degree having closed form solutions.

I will leave finding a specific example of a lack of a closed form solution to a mathematician for whom Galois Theory is fresher in their mind. However, in at least one set of cases I can preclude a rational solution by Eisenstein's criterion: if y is prime and t>1.

posted by zxcv at 11:31 AM on February 20, 2010

I will leave finding a specific example of a lack of a closed form solution to a mathematician for whom Galois Theory is fresher in their mind. However, in at least one set of cases I can preclude a rational solution by Eisenstein's criterion: if y is prime and t>1.

posted by zxcv at 11:31 AM on February 20, 2010

So, what do you mean by solve? If *x* and *y* are both variables, then whether *t* is 2 or *n*, you have your closed form solution already: y = x^{t}/(x-1). You've got countably infinite integer solutions, except for *x=1*. (By solution, I mean a set of numbers (a,b) wherein the equation is satisfied for x=a, y=b .)

If you want your equation in terms of*y* rather than *x*, you're not going to be able to do that in a complete way, in closed form, except in the case where t = 1. In this case, you can easily solve for *x*: x = y/(y-1).

This isn't so for higher values of t because the function isn't one-to-one. That is, for higher values of*t*, you're not going to be able to uniquely determine *x* in terms of *y* because there are values of *x* (in fact, nearly all of them) for which there are two values of *y*. The function isn't invertible. The reason that it's not one-to-one is this: there's always going to be a vertical asymptote at x = 1. So consider values of x for x>1. As x approaches 1 from the right, all values of y are going to be covered because the limit goes to infinitey. However, the limit as (positive) x goes to infinity (heading right, on the x-axis) goes to infinity as well. I imagine that you could attempt to break up the graph into pieces to define it in terms of y rather than x...

The other side of the graph -- where x is less than 1 -- is sometimes one-to-one and sometimes not. It depends on t. The limit as x approaches 1 from the left goes to negative infinity. So we've got all of the negative reals covered already. Heading in the other direction along the x axis, the graph is going to behave approximately like x^(t-1). If t is even, all the y-values will be negative. If t is odd, all the y-values will be positive. In the later case, we do have a one-to-one situation. But in the former case, we're going to be hitting all the negative reals*again* -- not a one-to-one situation. Again, in the case where t is even, you could perhaps break up the graph into pieces in order to express the inverse, i.e., express it in terms of x rather than y.

If*y* is an integer (or even any real or irrational) rather than a variable, then nevermind; you are just trying to solve a potentially high degree polynomial.

posted by sentient at 6:57 PM on February 20, 2010

If you want your equation in terms of

This isn't so for higher values of t because the function isn't one-to-one. That is, for higher values of

The other side of the graph -- where x is less than 1 -- is sometimes one-to-one and sometimes not. It depends on t. The limit as x approaches 1 from the left goes to negative infinity. So we've got all of the negative reals covered already. Heading in the other direction along the x axis, the graph is going to behave approximately like x^(t-1). If t is even, all the y-values will be negative. If t is odd, all the y-values will be positive. In the later case, we do have a one-to-one situation. But in the former case, we're going to be hitting all the negative reals

If

posted by sentient at 6:57 PM on February 20, 2010

Just to clarify, there is no closed form solution for all polynomials of (say) degree 10, but x

And same goes for x

posted by flug at 8:42 AM on February 21, 2010

By the way it is entertaining to plug equations similar to x^{10} - yx + y into something Wolfram Alpha to see the roots. For instance, here is x^{10} - 50x + 50, which has 2 real roots.

posted by flug at 8:57 AM on February 21, 2010

posted by flug at 8:57 AM on February 21, 2010

If t is of the form 1/n, where n is an integer, it can pretty easily be transformed into an equivalent polynomial of degree n. So if n is a positive integer of range 0-5 you can definitely solve it; if n>5 then possibly though unlikely.

If t is of the form m/n, where both m & n are integers, it can easily be transformed into an equivalent polynomial of degree n*m so the same applies (if n*m is in the range 0-5 you can definitely find a closed form solution and if n*m is greater than 5 then a closed form solution is possible but unlikely).

And last if all, t is irrational you are pretty well sunk--no closed form solution possible.

posted by flug at 8:58 AM on February 21, 2010

This thread is closed to new comments.

^{t}- yx + y = 0. That is, we're trying to find the roots of a polynomial of degree t. For a fixed t < 5 it's possible to get this into a closed form, but not for degree 5 or higher. So it makes sense that it would not be possible in the general case either.posted by serathen at 10:27 AM on February 20, 2010