Solving a Simple nth Order Polynomial

[dimensionless]

Homework Statement

I'm trying to solve the following equation
$$C + x + x^{3} + x^{5} + x^{7} + x^{9} = 0$$

The Attempt at a Solution

I can factor out $$x+B$$, but the factoring process will be very tedious. A little too tedious if I'm barking up the wrong tree. Any suggestions?

 

[happyg1]

I don't think x+B (I think you mean x+C) factors out of that.

CC

 

[D H]

Multiply by $x^2-1$

 

[dimensionless]

I could have B = C

I multiplied by $$x^{2}-1$$ and I got
$$-C-x-Cx^{2}+x^{11} = 0$$

I good improvement, but still not a solution.

 

[christianjb]

I don't see how this can be done analytically. I would use a numerical method.

 

[d_leet]

I could have B = C

I multiplied by $$x^{2}-1$$ and I got
$$-C-x-Cx^{2}+x^{11} = 0$$

I good improvement, but still not a solution.

That can't be correct. If we call the original function p(x) then if we start with

p(x)=0

and multiply by x²-1 then you end up with

(x²-1)p(x)=0

It should be obvious that 1, and -1 are solutions of this equation, but you should notice that they are not solutions of the result you found after multiplying, thus you must have made a mistake somewhere.

 

[dimensionless]

I reworked the multiplication and got a very similar answer:
$$-C-x+Cx^{2}+x^{11} = 0$$

Maybe there is some 11th order 'quadratic' I can use to solve this.

 

[dimensionless]

I don't see how this can be done analytically. I would use a numerical method.

Yeah, I'm having trouble with it. I do expect, however, that an analytic solution can be found.

 

[matt grime]

I very much doubt your expectation is true. Analytic solutions (as in sensible functions of the coeffecients) probably cannot be found. They do not, in general, exist for any polynomial of degree 5 or more. (Though there are methods for quintics.)

There is, of course, Galois theory. If you could work out the Galois group you might have a chance.

 

[dimensionless]

I imagine that there is an algorithm that can be worked out by hand and not be tooo tedious. Mathematica seems to have a function that finds the roots of a polynomial exactly.

http://documents.wolfram.com/mathematica/functions/Roots

At any rate, I don't see any reason why this shouldn't be possible.

 

[HallsofIvy]

But I'm sure that function does not find roots in terms of an unknown parameter C! That's the problem. Roots and methods of soltution depend strongly on what C is. For one thing, any rational roots of this equation must be integers that evenly divide C.

 

[matt grime]

Mathematica seems to have a function that finds the roots of a polynomial exactly

Sorry, but that is nonsense. It may have functions to find roots to arbitrary precision, but that is not the same as finding the roots. And it won't even have that since a computer has a machine error to contend with.

At any rate, I don't see any reason why this shouldn't be possible.

apart from lots of people with maths phds telling you it can't be done? See, you've made me make proof be appeal to authority! Gah, what a rubbish answer. Let me expand on what I wrote before:

Google Galois Theory. I didn't put those words in at random.

It can be shown that there is no method that allows you to work out the roots of arbitrary polynomials by field operations and n'th roots of the coefficients for deg >4, and for deg 5 there exist methods by elliptic functions. BUT THAT IS IT. Any attempt at a solution by tactics akin to the quadratic formula is DOOMED to fail for arbitrary polys of degree >4.

 

[dimensionless]

It can be shown that there is no method that allows you to work out the roots of arbitrary polynomials by field operations and n'th roots of the coefficients for deg >4, and for deg 5 there exist methods by elliptic functions. BUT THAT IS IT. Any attempt at a solution by tactics akin to the quadratic formula is DOOMED to fail for arbitrary polys of degree >4.

This question is not aimed specifically at Matt, but I've done some more reading about this. Everything I seem to find says that polynomials of deg >4 are not solvable by using =, -, *, /, and ^, although Matt has very strongly implied that there are methods for deg 5 polynomials. I still wonder if there may be solutions by other means. What about the use of transforms and operator functions? I'm not totally convinced that it is possible to proove that a solution does not exist by means of transforms, operator functions, and other methods.