ODE for the inverse function of quintic polynomial

[square173205]

It is well known that in general the quintic algebraic equation is not solved algebraically. It means that the inverse function of quintic polynomial is not expressed algebraically. However we also know some functions which are not elementally functions are expressed with nonlinear differential equations. Then is it possible to express the inverse function of quintic polynomial with any nonlinear differential equation? I tried to induce such nonlinear differential equation in the following site;

 

http://hecoaustralia.fortunecity.com/stbruno/stbruno.htm

[Country Boy]

??? The general quintic does NOT have an inverse function.

[Bignose]

Even if it did, what the heck is the point in making a 5th order, nonlinear ODE to solve something like this? It will have to be solved via computational method. That mess on your website is not going to have an analytic solution. Unless you know of one? It just looks awfully unlikely.

 

So, then, what is the point? You can use a computer to solve for a quintic polynomial pretty fast.

 

Again, like I said in my other post, if you are just posting these things so we click on your links, this seems awfully spam-ish to me. Do you have a larger point, square? Do you have something you want to discuss on the forum?

[square173205]

??? The general quintic does NOT have an inverse function.

Just not solvable with using radicals:cool:

[Country Boy]

Just not solvable with using radicals:cool:

 

No, in general, a quintic polynomial function does NOT have an inverse function. Just as a quadratic polynomial does NOT have an inverse function.

(Since the quintic is of odd degree it, unlike a quadratic, MIGHT have an inverse. That why I said "in general".)

[square173205]

No, in general, a quintic polynomial function does NOT have an inverse function. Just as a quadratic polynomial does NOT have an inverse function.

(Since the quintic is of odd degree it, unlike a quadratic, MIGHT have an inverse. That why I said "in general".)

 

You don't allow using imaginary number, do you?

[Country Boy]

You don't allow using imaginary number, do you?

Of course I do! I'm not talking about solving equations, I am talking about the inverse function! You, in your first post, referred to "the inverse function of quintic polynomial" You are aware, are you not, that f(x)= x^2, because it is not a one-to-one function, does NOT HAVE an inverse function.

[Bignose]

I'd still like to know how writing out a 5th order, highly non-linear ODE is better than having just the original polynomial. Both will still have to be solved by computer, if a solution exists. So, I don't see how one is better than the other...

[square173205]

Of course I do! I'm not talking about solving equations, I am talking about the inverse function! You, in your first post, referred to "the inverse function of quintic polynomial" You are aware, are you not, that f(x)= x^2, because it is not a one-to-one function, does NOT HAVE an inverse function.

 

Surely, but even in the case of quadratic polynomial there exists a pair of domain and region where the mapping is one-to-one.

[Country Boy]

Surely, but even in the case of quadratic polynomial there exists a pair of domain and region where the mapping is one-to-one.

 

Oh, yes. We can always restrict the domain so that the function becomes one-to-one, and so has an inverse function on that domain. I have no problem with that. I was only objecting to talking about the inverse function of a quintic polynomial.

[jamesvnguyen]

If I say that I can find the roots of the general form of quintic, hexic, septic, autic, nonic ... PRECISELY without guessing the initial values (involved exactly few iteration, of course); I'm wondering how useful it is? Don't think that I am crazy, just tell me if it is new, and worth to publicize. Thanks for your advice!

[D H]

If I say that I can find the roots of the general form of quintic, hexic, septic, autic, nonic ... PRECISELY without guessing the initial values (involved exactly few iteration, of course); I'm wondering how useful it is? Don't think that I am crazy, just tell me if it is new, and worth to publicize. Thanks for your advice!

No value whatsoever. That the roots of fifth order and higher polynomials cannot be expressed as radicals was independently proven by Abel and Galois at the beginning of the 19th century. Read up on Galois theory. The wiki article on this subject is a start.

[jamesvnguyen]

I understand why you said that. It is hard to believe. But how hard is it to prove it? If you give me a RANDOM set of coefficients of the polynomial (fifth order +), I will give you ALL the roots in nano second. How hard it is to know if they are the correct answers! All you need to do is plug them in to see if they yield to Zeros. Any way, If it is true what I just said, is it new and useful? I appriciate very much for your response!

(Yes, for the fifth order and lower, I can do it by hand.)

[D H]

From http://mathworld.wolfram.com/QuinticEquation.html

Unlike quadratic' date=' cubic, and quartic polynomials, the general quintic cannot be solved algebraically [b']in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions[/b], as rigorously demonstrated by Abel (Abel's impossibility theorem) and Galois. However, certain classes of quintic equations can be solved in this manner.

That does not mean that quintic equations cannot be solved using more advanced techniques, for example Bring radicals and the Jacobi theta function.

The general quintic can be solved in terms of Jacobi theta functions' date=' as was first done by Hermite in 1858. Kronecker subsequently obtained the same solution more simply, and Brioschi also derived the equation.[/quote']

 

 

That said, find the exact roots of

 

[math]x^5+2x+1[/math]

[jamesvnguyen]

the solutions of X^5 + 2X + 1 = 0 are:

 

-0.70187357 + -0.8796972 i = 1st Solution

-0.70187357 + 0.8796972 i = 2nd

0.94506809 + 0.8545175 i = 3rd

0.94506809 + -0.8545175 i = 4th

-0.48638904 = 5th

 

Only take me a few nano second to get them!

[D H]

I asked for the exact answers. Yours are not.

 

Hint: The decimal expansion is infinite. You must either represent the exact answer in terms of radicals (which cannot be done for this equation), in terms of an infinite series, or in terms of some special function such as the Jacobi theta function.

[jamesvnguyen]

I appreciate you to check me out. You can give me any set of coefficients of quintic, hexic, and septic; and I will give you the answers. (I stop at autic, since I don't have time to do it- Is it worth to continue?)

 

My goal is finding the roots of the polynomial of fifth order and higher. The accuracy is up to the floating point of the system allowed. Is it worth to continue? (It is not a big program or anything likes that. Everyone can have it in their calculator)

[D H]

You are not finding the exact solutions. You are finding approximations. It is relatively easy to approximate the zeros of a polynomial. There is nothing new here per se. There are a *lot* of approximation techniques out there. You may have something if you have found a new, fast algorithm for approximating the roots of a polynomial. I have my doubts that your approach will be new. Before you proceed further, I suggest you look through the literature (an incredible amount of literature) on approximation techniques.

[jamesvnguyen]

I'm not an expert about this. But this is very straight forward, no guessing, no looping, involves exact number of steps, small (can do by hand). I stop short to call it a formular (two big formular). Forget about it?

 

DH, I would like to learn from you.

I'm wondering, is there any technique can find the roots of the polynomial (for example, hexic) which does not have a real root (only complex roots)?

[D H]

Forget about it?

Not necessarily. As I said earlier, do a literature search. I, for one, am not going to do the literature search for you. It would be a waste of my time. On the other hand, it's your idea, so that might make it worth your time. A side benefit: you will learn something in the process.

[jamesvnguyen]

DH, here is my appreciation for your advice!

 

How to find square root of a number

Sqrt(b) ?

find (a) such that (a*a) is close to (b)

 

For example: if b=178, (a) can be 6, 7, 8, 9, 10, 11, 12, or 13……

 

Let V = a + b/a

Let Z = (V+4*b/V+16*b*V/(V*V+4*b)+64*b*V*(V*V+4*b)/(POWER((V*V+4*b),2)+16*b*V*V))

 

You will see Sqrt(b) = Z/32 + 8*b/Z [with either (a) that you chose]

 

Is it powerful?

 

DH, there is no body here but you and me. Why are you here, just curiosity?

I think that I have something that I would like to contribute. But I don't know how to start. I need to talk to some one who is an expert about this. And you seem like the one that I am looking for. Will you willing to discuss about this subject. If you do, here is my email: dog92707@yahoo.com