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Generalizing polynomial formulas?

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I heard it's for some reason mathematically impossible to have a formula for a polynomial beyond a quartic for some reason, but its a consistant pattern as the degree goes up so I don't see why it wouldn't be possible. Basically I'm looking for a formula that generalizes polynomials and I hear that's impossible for some reason.

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I'm not sure exactly what you are getting at, but you can generalize polynomials:

 

[math]f(x)=\sum_{j=0}^{n-1} a_j \, x^{p_j}[/math]

 

where [math]a_j[/math] and [math]p_j[/math] are any real numbers and [math]n[/math] is the count of terms, or more accurately defined as:

 

[math]f(x)=\sum_{j=0}^{n} a_j \, x^{n-j}[/math]

 

where [math]a_j[/math] are real numbers, [math]n+1[/math] is the count of terms, and [math]n[/math] is the highest power.

 

You can even have quintic polynomials and beyond:

 

Quintic Polynomials

 

Polynomials

 

Unless of course you are referring to a generalized equation for finding the roots of any polynomial.

Edited by Daedalus

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Unless of course you are referring to a generalized equation for finding the roots of any polynomial.

 

That's actually exactly what I mean. I formula for any polynomial that =0, which I would imagine would be very complex since it's so general. I don't see why there wouldn't be equations for like an octic polynomial, and then if we can get that far, where there isn't an equation to generate those formulas. I mean we have a formula for 5 different types of polynomials as far as I know, surely mathematicians have noticed the patterns in those by now.

Edited by questionposter

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This link may be an interesting starting point for you.

 

So in other words, its possible to have a formula for any degree of a polynomial equation, but you actually can't have a consistant pattern with the ways the formulas for them are written, you'd have to figure them out all of them by hand?

Edited by questionposter

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No, that's not what it says:

 

The theorem says that not all solutions of higher-degree equations can be obtained by starting with the equation's coefficients and rational constants, and repeatedly forming sums, differences, products, quotients, and radicals (n-th roots, for some integer n) of previously obtained numbers. This clearly excludes the possibility of having any formula that expresses the solutions of an arbitrary equation of degree 5 or higher in terms of its coefficients, using only those operations, or even of having different formulas for different roots or for different classes of polynomials, in such a way as to cover all cases.

 

You can't even have a general formula for just degree 5 polynomials, say.

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If you really enjoy number crunching and unwieldy formulae - there is a method for determining if a quintic is solvable with radicals, and another which (3 page equation) which allows the solvable quintics to be solved

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No, that's not what it says:

 

 

 

You can't even have a general formula for just degree 5 polynomials, say.

 

Well obviously a quintic is going to cross the x axis at some point, so there has to be 0s. There has to be something going one which causes the process of y being equal to an x value, and the way that happens is you use y to solve for x. If x is equal to y, then in some way or another, y has to be equal to x, and since y can be 0, x has to equal something at that point. But that's number theory; its indisputable that if ...x...=...y..., then ...y...=...x..., but otherwise I'm not quite sure why having something like a 5th root means I'm not actually seeing real 0s on my calculator when I see the curve cross the x axis.

 

 

http://en.wikipedia....Ruffini_theorem

 

"It does not assert that higher-degree polynomial equations are unsolvable."

 

There's no way that after thousands of years that there wouldn't be a way to figure out the 0s of a mere 5th or even 6th degree equation, even the quadratic formula was developed during the ancient Greece era.

Edited by questionposter

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Well obviously a quintic is going to cross the x axis at some point, so there has to be 0s. There has to be something going one which causes the process of y being equal to an x value, and the way that happens is you use y to solve for x. If x is equal to y, then in some way or another, y has to be equal to x, and since y can be 0, x has to equal something at that point. But that's number theory; its indisputable that if ...x...=...y..., then ...y...=...x..., but otherwise I'm not quite sure why having something like a 5th root means I'm not actually seeing real 0s on my calculator when I see the curve cross the x axis.

 

QP you are conflating the idea that a quintic must have at least one real root with the existence of an method to find it. All quintics must cross the x axis and will have a real value of x for which y is 0 - that does not mean that there is a method for find that x value in radical terms. The equation y = x^5-x+1 must cross the x axis - ie it has a real root - but that root is not expressible in terms of radicals or quotients of integers; ie what ever value you put will be an approximation.

 

There's no way that after thousands of years that there wouldn't be a way to figure out the 0s of a mere 5th or even 6th degree equation, even the quadratic formula was developed during the ancient Greece era.

No we haven't found a method - but we have shown that a general method is an impossibility - that is Abel-Ruffini which you have mentioned, which prove that some quintics have non radical solutions.

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QP you are conflating the idea that a quintic must have at least one real root with the existence of an method to find it. All quintics must cross the x axis and will have a real value of x for which y is 0 - that does not mean that there is a method for find that x value in radical terms. The equation y = x^5-x+1 must cross the x axis - ie it has a real root - but that root is not expressible in terms of radicals or quotients of integers; ie what ever value you put will be an approximation.

 

 

No we haven't found a method - but we have shown that a general method is an impossibility - that is Abel-Ruffini which you have mentioned, which prove that some quintics have non radical solutions.

 

But it says right in that wikipedia article that the Abel Ruffini theorem does not state that higher degree polynomials are unsolvable, it just states you can't find all the roots using radical forms.

 

How does a graphing calculator know where the 0s are if we don't? We programmed all our knowledge into a graphing calculator so there must be something within our knowledge that can tell us the answer, unless calculators are artificial intelligence...

Edited by questionposter

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You can use the Newton-Raphson method or the bisection method and root-finding methods by hand and get the location to any accuracy you desire, no problem. It's just a bit tiresome, and computers are a bit faster at tiresome additions, multiplications, and divisions.

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It estimates them using an inductive method where each guess is an improvement on the last - (Newtoniam Method or Laguerre or one of others); it does not arrive at the correct answer because there isn't one. The methods for solving polynomials of a lower order will only give radical solutions - and quintics have non-radical solutions so a general formula to solve all quintics cannot exist

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It estimates them using an inductive method where each guess is an improvement on the last - (Newtoniam Method or Laguerre or one of others); it does not arrive at the correct answer because there isn't one. The methods for solving polynomials of a lower order will only give radical solutions - and quintics have non-radical solutions so a general formula to solve all quintics cannot exist

 

Well I tried doing something simple myself, and I found that I can in fact find the 0s of an 8th degree polynomial.

 

x^8+256 is an 8th degree polynomial with no real roots, but just by looking at it I can tell that you can get the roots by doing this

 

[math](x^4+16i)(x^4-16i)[/math]

 

So I'm thinking now that there can be a formula which factors a large degree polynomial into smaller factors and uses the lower degree polynomial formulas on it, like

 

roots of (x^8+constant) = quint(x_to_the_power_of_(1/2)n)+sqrt(constant)i)*quint(x_to_the_power_of_(1/2)n)+sqrt(constant)i)

 

 

quint=quintic formula

 

It would be pretty tedious work, but you can probably just program the process into your calculator.

 

Something tells me there's some way to find the roots if I happen to add another part to the polynomial like a 3x^7.

 

Also the math thing is broken, it won't let me use complicated exponents.

Edited by questionposter

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Well obviously a quintic is going to cross the x axis at some point, so there has to be 0s.

 

Any polynomial of degree n with complex coefficients has exactly n complex roots, counting multiplicity. This is the Fundamental Theorem of Algebra, which is usually proved in a class on complex analysis.

 

There exists no "solution by radicals" (i.e. no solution involving just roots addition and multiplication) for general polynomials of degree 5 or higher. This is usually proved in an algebra course on the Galois theory of fields.

 

So, roughly speaking, the roots always exist, but you can't find them exactly.

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Any polynomial of degree n with complex coefficients has exactly n complex roots, counting multiplicity. This is the Fundamental Theorem of Algebra, which is usually proved in a class on complex analysis.

 

There exists no "solution by radicals" (i.e. no solution involving just roots addition and multiplication) for general polynomials of degree 5 or higher. This is usually proved in an algebra course on the Galois theory of fields.

 

So, roughly speaking, the roots always exist, but you can't find them exactly.

 

Ok, we can't find the roots using conventional methods, but why should innovation be limited from looking towards other methods like the one I mentioned before? Maybe an eighth degree polynomial isn't solvable, but two 4th degree polynomials are.

Edited by questionposter

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QP - if you look back to the first answers from Capn R and myself we did say there was no general formula. Various forms of higher order polynomials can be solved with radicals (per your example); but it has been shown that it is not possible to find a general formula -

 

ie there is no general formula in which you can plug in a, b, c, d, e, and f for the equation ax^5+bx^4+cx^3+dx^2+ex +f =0 and guarantee that it will spit out the roots of the equation.

 

People have worked for centuries to make formulae that solve specific forms of higher order polynomials by reducing to two solvable parts etc - (I am not pulling your leg here) solving quintics and quartics was a spectator sport in renaissance italy - and many formula exist for various forms. As per my links above, for those quintics that do have radical solutions there is a formula - but (labouring a point) not all quintics have radical solutions. There is no possible formula or method that will give you anything other than an approximation of the roots for y = x^5-x+1

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QP - if you look back to the first answers from Capn R and myself we did say there was no general formula. Various forms of higher order polynomials can be solved with radicals (per your example); but it has been shown that it is not possible to find a general formula -

 

ie there is no general formula in which you can plug in a, b, c, d, e, and f for the equation ax^5+bx^4+cx^3+dx^2+ex +f =0 and guarantee that it will spit out the roots of the equation.

 

People have worked for centuries to make formulae that solve specific forms of higher order polynomials by reducing to two solvable parts etc - (I am not pulling your leg here) solving quintics and quartics was a spectator sport in renaissance italy - and many formula exist for various forms. As per my links above, for those quintics that do have radical solutions there is a formula - but (labouring a point) not all quintics have radical solutions. There is no possible formula or method that will give you anything other than an approximation of the roots for y = x^5-x+1

 

How about breaking down x^5-x+1 into (x^3...)(x^2...), how would I do that? I would make it solvable with the other formulas.

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How about breaking down x^5-x+1 into (x^3...)(x^2...), how would I do that? I would make it solvable with the other formulas.

 

You can't! That has been proved. Abel-Ruffini proved that some quintics (and other higher orders) can never be broken down to radicals. Personally I think it shows the study of mathematics in a brilliant light that in some circumstances we can show how to do things, but in others we can prove that things are impossible to do (not just difficult, nor just undiscovered, but impossible)

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You can't! That has been proved. Abel-Ruffini proved that some quintics (and other higher orders) can never be broken down to radicals. Personally I think it shows the study of mathematics in a brilliant light that in some circumstances we can show how to do things, but in others we can prove that things are impossible to do (not just difficult, nor just undiscovered, but impossible)

 

Well it can't be impossible because I've seen a graphing calculator that's been programmed to find the exact roots of higher degree polynomials, even the imaginary ones, which essentially proves that within our own knowledge is the ability to find the roots of higher degree polynomials. And then if I even type in a septic equation and also y=0 and go to 2nd --> trace --> intersect, it shows the number at exactly x=0. Regardless of whatever knowledge we think we currently have, septic equations regardless of how they are written, such as with x^7-x+57, will definitely have at least 1 x intercept. Doesn't matter if you theorize it's not possible because by the logic of graphing itself, its shown otherwise. Since we programmed only whatever knowledge we have into a calculator, and the calculator can know x intercepts, then there has to be something we know that can give us the actual x intercepts.

 

It's just like how people say x/0 doesn't have an answer even though it does have an answer but it's just now what you'd expect. That's probably also true for generalizing polynomial formulas. Since y=0, and 0= the equation, something has to be able to be done to 0 in order to get x since both sides of the equation have the same value. In order for the equation to equal 0, x has to be at least 1 definite number no matter what, even if it's imaginary.

 

Just like if 1+x=3, then 3-1=x.

 

1+x could not = 3 unless x was a number.

 

Somehow, all polynomials have roots, whether we know how to find them or not, but I think that we do have the capability to find them based on the facts that x has to equal something in order for y to = 0, and that calculators can find them.

Edited by questionposter

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One can numerically solve for roots, but the answers will not be exact -- the calculator gives x=0 because it is able to compute an answer sufficiently close to zero.

 

There's a number of numerical algorithms for estimating roots, one of which is used by your calculator:

 

http://en.wikipedia.org/wiki/Root-finding_algorithm

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One can numerically solve for roots, but the answers will not be exact -- the calculator gives x=0 because it is able to compute an answer sufficiently close to zero.

 

There's a number of numerical algorithms for estimating roots, one of which is used by your calculator:

 

http://en.wikipedia....nding_algorithm

 

Even though it seems we do have limits to our current mathematical knowledge, isn't it true that in y=x^7+x+27 that x has to be equal to some number(s) in order for the equation to = 0 or even any number? And since that's true, doesn't there have to be a specific process that's causing that statement to be true? It's just a matter of finding it I think.

Edited by questionposter

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Even though it seems we do have limits to our current mathematical knowledge, isn't it true that in y=x^7+x+27 that x has to be equal to some number(s) in order for the equation to = 0 or even any number?

Yes.

 

And since that's true, doesn't there have to be a specific process that's causing that statement to be true? It's just a matter of finding it I think.

Why should this necessarily be true? There are a number of problems in math where solutions are easy to verify but incredibly difficult to find. The Abel-Ruffini theorem proves that one cannot create an algebraic root-finding equation for, say, seventh-order polynomials that is general to any seventh-order polynomial, like the quadratic equation is general to all quadratics. Certainly there are some seventh-order polynomials which can be factored, but there will always exist some polynomials which don't fit in your root-finding equation.

 

A version of the proof is presented here if you're interested, although I can't vouch for its accuracy.

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Yes.

 

 

Why should this necessarily be true? There are a number of problems in math where solutions are easy to verify but incredibly difficult to find. The Abel-Ruffini theorem proves that one cannot create an algebraic root-finding equation for, say, seventh-order polynomials that is general to any seventh-order polynomial, like the quadratic equation is general to all quadratics. Certainly there are some seventh-order polynomials which can be factored, but there will always exist some polynomials which don't fit in your root-finding equation.

 

A version of the proof is presented here if you're interested, although I can't vouch for its accuracy.

 

Well what if I don't use a single equation, but a process?

 

Say I have x^7+x^6-x+10=0

 

And then in the calculator I find the intersection between

 

x^7+x^6 = x-10?

 

Or perhaps we me even need to use higher dimensional mathematics.

Edited by questionposter

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That's exactly equivalent. I suspect your calculator will use a numerical algorithm to solve that as well.

 

You should look into how common numerical algorithms work.

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That's exactly equivalent. I suspect your calculator will use a numerical algorithm to solve that as well.

 

You should look into how common numerical algorithms work.

 

But it's not like this is quantum mechanics where we can't even know a specific location, these are specific and definite numbers in a specific process yielding specific results. I just don't see how y=x couldn't always be the same as x=y even after looking at the abel-ruffini theorem. One way or another x has to = something in order for y to = 0, no matter what. Maybe we can't do it using the conventional methods as in starting out with the coefficients and changing how they are summed, but if x does = y and we can't find x even though it still has to be true that y=x, then there's probably just some way to change the look of expressions and equations that we haven't thought of yet. I've read about the Abel Ruffini theorem and it seems to state in multiple sources that the main problem is that we can't find the roots based on how we currently write them and try to solve them, which is still just how a value "looks", and as we have already found many times, changing the "look" is the key to balancing equations.

Edited by questionposter

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