Math 2 | ek-sight-ing science

Math 2; Solving higher polynomials.

Solving higher order (n>2) polynomials with and without n-zero points by means of parabola.

If you are a little mathematical you already know what parabola are and also most probably what polynomials are. For the few who do not know, a little explaination:

A number "n" can be represented in a graph as a dot on some spot (Nx, Ny).

That same number can also represent a straight horizontal line by the formula Y= n.

A straight line that inclines or declines is represented as Y=A*X + B and in the previous example Y=n then A=1 and B = 0

Another type of curve is represented in what is known as a parabola Y = A*X*X + B* X + C = AX^2 + BX + C

and we can keep on doing this type of tricks the next step is Y = A*X*X*X + B*X*X + C*X + D which is AX^3 + BX^2 + CX + D

all these are pictured here below:

As you understand you can go higher and higher in X^n where n is bigger then 3. and some examples are pictured here above. These graphs are known as polynomials and have the general equation:

Y = A*X^n + B*X^(n-1) + ....... + C*X + D

Mathematically it is good to know what the formula is of such a type of graph so that is is easy to calculate with. And for up to X^3 all graphs can be solved analytical mathematically. That is to say that there is an exact way to determine/derive the formula from some basic information. for example the parabola is solved by means of the ABC formula:

X1,2 = (-B +/- square root( B^2 – 4*A*C)) / (2*A)

also for a formula with X^3 it can be solved exactly but which for,ula is too big to show it here.

But if one encounters a graph of the order of 4 or higher this cannot be solved analytically mathematical, and some fancy trick need to be applied in order to find a good solution. And the better this method the more accurate we will be able to find this formula.

In 2015 (thank God) I found a simple and good way of solving higher order (higher then 4) polynomial.

And for that I copied the paper that I wrote about it and will copy that here just as I wrote it. for the real deal you can look at:

https://www.academia.edu/28192153/Solving_higher_order_n_2_polynomials_with_and_without_n_zero_points_by_means_of_parabola

[PS -see for all pictures below this text because it is not possible in this text editor to get this inserted as in the original paper -]

Solving higher order (n>2) polynomials with and without n-zero points by means of parabola.

Edgar Korteweg.

Not affiliated to an institution, all in private, e.korteweg@yahoo.nl

ABSTRACT

Solving higher order polynomials (order n, with n zero points and with less than n zero points) which can be achieved by using 2nd order polynomials. Each zero point is approached by means of a parabola. Solving higher order polynomial without zero points need first to undergo a polynomial transformation in which a maximum 2nd order polynomial is subtracted to create (preferably n) artificial zero points. Then the parabola approach is applied and the whole is transformed back. This leads, in excel spreadsheets, to very high accuracies (10^-12 to 10^-15).

Keywords: polynomial solving with parabola.

INTRODUCTION.

In mathematics people deal on a regular basis with what is known as polynomials. In this way functions can be approached and problems solved as well in purely mathematical sense or also in the physical sense. The use of polynomials is widely spread out in all fields of science and there for very useful. But solving higher order polynomials is not an easy job to be done. Therefore a relatively easy way comes in handy on all levels, whether that is on high-school level or electronic or quantum science level or on whatever other science field level, a handy mathematical tool eases the way to further research and new fields. In the, as far as I know, new procedure that is shown here, such a tool and way is given. The whole concept of solving higher order polynomials is divided into two sections: Part I and Part II since both parts deal with a slightly different approach, since the problem was different in both cases although the Part I method could also be applied to the Part II problem in finding solutions.

THE THEORY, PART I

In order to solve any polynomial equation of any order higher then 2, provided that it has as many points that cross the X-line (n zero points) as its order is (n), we can use several 2nd order polynomials or parabola formulas AiX2+BiX+Ci [1] to solve the equation by means of the ABC formula [1] Xi1,2 = (-Bi +/- square root( Bi2 – 4*Ai*Ci)) / (2*Ai)

The axiom is than, that every higher order polynomial that has as much X-line crossings as its order is, has an equal amount of 2nd order polynomials to approach its x-line crossings (zero points). There may be a very slight remaining error. In this procedure an error is achieved in the atomic nuclear size order of magnitude, in using just excel sheet calculations. That means that in the unit of 1 the order of error magnitude lays around 10-12 to 10-15 which is very accurate considering that nothing more than a common excel program was used. For the most applications this is more than enough. I am confident that higher accuracies are possible with more mathematical specialized programs, but that was not the goal of this research.

The general solutions for any such polynomial is then:

F(X) = KXn +LXn-1 + MXn-2 + …PX+ Q =

(X-[ ( -B1+/-SQRT(B12-4*A1*C1)/(2*A1) ) ] )*

(X-[ ( -B2+/-SQRT(B22-4*A2*C2)/(2*A2) ) ] ) * …. *

(X-[ ( -Bn+/-SQRT(Bn2-4*An*Cn)/(2*An) ) ] ) + (slight) error (1)

Or

n

F(X)=KXn +LXn-1+…+PX+ Q = Π (Xi-[ (-Bi +/- SQRT(Bi2-4*Ai*Ci)/(2*Ai))] ) + err (2)

i=1

Provided that there are n X-line crossings as well. The accuracy obtained in this way, at least with excel spreadsheet type of calculations, can be at the most, at the level of hydrogen atom nuclear size. That is in the order of magnitude of 10-15 meter, if meters is the unit in use. Maybe with better programming this level of accuracy can be raised.

IN PRACTICE, PART I.

The general idea of this procedure is depictured in this next graph of a 6th order polynomial and its X-line crossings approaches: The blue line is the original line and the thick pinkish line it the approached line added with 2 to make a clear distinction between the 2 lines.

The original formula is: \

F(X) = 0.01X6 - 0.02X5 - 0.26X4 + 0.28X3 + 1.45X2 - 0.26X - 0.8 (3)

By approaching it with parabolas as pictured this method was able (in excel) to retrieve the next formula:

(X+3.97823229118505) * (X+2.160451734963370) * (X+0.73931607989855)*

(X-0.8227584931668120) * (X-3.068679302374810) * (X-4.986562310264870) (4)

Again for picturing convenience I added a constant of 2 to make a clear distinction in the picture between the original and the calculated lines.

Table 1. The parabola used for this approach (*)

Parameter Parabola 1 Parabola 2 Parabola 3 Parabola 4 Parabola 5 Parabola 6

A X2 23.582892 - 2.36145 0.10274489 1.0429138 - 3.141060 34.794760

B X 169.77934 -7.323459 -1.4158220 0.3760574 13.233580 - 317.7157

Constant 302.19095 -4.799778 -1.1028991 -1.015386 -11.03090 719.10958

(*) The values for the individual numerical parameters are rounded off to fit within the table and match the layout.

Figure. 1: The parabola-approach method in practice, to make distinction a constant of 2 was added.

This example was taken to show the proof of concept, but it works for any higher order polynomials higher than 2, provided that the polynomial has an equal X-line crossings or zero points, as the order of the polynomial is which is n.

Since the solution to a parabola has 2 solutions X1 and X2 it is found that there seems to be an alternating + and – sequence in the ABC formulas being used for the approaching parabolas.

As is known:

Xi1 = (-Bi + square root( Bi2 – 4*Ai*Ci)) / (2*Ai) (5)

then

Xi2 = (-Bi - square root( Bi2 – 4*Ai*Ci)) / (2*Ai) (6)

And visa versa:

It is found that the solutions in finding the higher order polynomial seems to demands a alternating + and – and thus using alternating use of the ABC formula.

So if:

X11 = (-B1 + square root( B12 – 4*A1*C1)) / (2*A1) (7)

then

X22 = (-B2 -- square root( B22 – 4*A2*C2)) / (2*A2) (8)

And:

X31 = (-B3 + square root( B32 – 4*A3*C3)) / (2*A3) (9)

And so on till the last Xi is found.

The same goes for if the X11 solution starts with:

X11 = (-B1 -- square root( B12 – 4*A1*C1)) / (2*A1) (10)

Then

X22 = (-B2 + square root( B22 – 4*A2*C2)) / (2*A2) (11)

And

X31 = (-B3 -- square root( B32 – 4*A3*C3)) / (2*A3) (12)

So in the usage of the ABC formula there appears to be an alternating sequence with respect to the + and – in this formula after the -Bi -term

Graphically this makes sense since also the polynomial curve looks kind of wavy and thus alternating, which is the primary cause of the alternation of the +’s and –‘s in the ABC formula. How this is exactly related and how to mold that into a formula, was not the scope of this research.

THE REALITY OF PART I.

In order to show how this works in real life, this procedure has been filmed and published as well on youtube [2], [3] and [4] ever since this procedure was found. It is found that it works very well in several different examples that were tested by means of this method. For that reason some youtube films have been made and published as mentioned. So far it is not known whether this type of method has been found and used in approaching higher order polynomials.

THE THEORY, PART II.

For polynomials without n zero points, a trick can be used to create (preferably n) artificial zero points (AZPs). For that the next axiom is postulated: for every high order polynomial (order n>2) a lower order polynomial (preferably n £ 2) can be constructed in such a way that the number of coinciding points of these 2 polynomials is also n. Due to this, Artificial Zero Points (AZPs) can be obtained by simply subtracting the low order polynomial (preferably maximum n=2) from the higher order polynomial. This procedure is called polynomial transformation.

The next step in solving the issue, the (artificial) zero points can be approached by the method as was described in The Theory, part I.

After this, the whole solution can be transformed back. So far the theory, in the example below it is shown how this procedure works in practice. Also here a very high accuracy was achieved in the same order as mentioned before (in the order of magnitude of 10-12 to 10-15) simply by Excel spreadsheet usage. Which can be used by any computer educated individual.

IN PRACTICE, PART II.

Through a lot of higher order polynomials, if not all, one can draw a lower order polynomial. This should be maximal preferably a 2nd order polynomial or parabola to keep finding solutions simpel. It can be drawn in such a way that the intersections of the higher polynomial will cross-sect the lower order polynomial so that, in that way, one can create artificial zero points, by simply reducing the higher order polynomial with the lower order polynomial (or parabola). The best or easiest way to do this, is so that the parabola has at least 1, but preferably 2 zero points itself.

Lets say there is a higher order polynomial of order n:

F(X)= AXn + BXn-1 + ... + DX2 + EX + F (13)

Then try to find a (preferably a maximum 2nd order polynomial) parabola;

G(X) = KX2 + LX + M (14)

In such a way that it cross-sects F(X) in order to create n artificial zero points as shown in figure 1 down here.

From the parabola (G(X) = KX2 + LX + M) the zero points are easily retrieved via the ABC formula, if needed.

Figure. 2: The polynomial transformation, graphically. - See below -

Now by subtracting the known parabola (figure 1 red line) from the higher order polynomial (figure 1 purple line) one can get a new polynomial (figure 1 - blue line)

H(X) = F(X)-G(X) (15)

H(X)= F(X)-G(X) = AXn + BXn-1 + ... + DX2 + EX + F - [KX2 + LX + M] = (16)

H(X) = AXn + BXn-1+ ... + (D-K)X2 + (E-L)X + (F-M) (17)

This newly constructed higher order polynomial (blue) does have (artificial) zero points which can be retrieved as discussed in the method of Part I, that was described in the first part of the paper, of course they can also be retrieved in other iterative ways and this is shown in figure 2.

Figure 3: Retrieving the (artificial) zero points by means of the method described in the 1st method as mentioned. - see below -

After this, the reconstruction can take place by adding to the original formula, the previously used parabola formula (if needed in its zero point form).

H(X) = AXn+..+ F - [KX2+LX+M] = (X - P)*(X-Q)* ...*(X-R) (18)

Suppose for the parabola we have the zero point form:

G(X) = KX2 + LX + M = (X-U)*(X-V) (19)

Than this can be used for constructing the final solution by reversing the procedure and add up the Parabola, and then we end up with the next solution:

F(X) = H(X) + G(X) = (X - P)*(X-Q)* ...*(X-R) + (X-U)*(X-V) (20)

Which is the final answer in terms of zero points.

As can be seen graphically, a whole series of different parabola’s or in some cases straight lines (AX+B) can be found in order to solve this problem. This means that there is no single answer to this issue, but rather a whole family for (7) of solutions.

Of course each solution is a unique combination, but non the less there is more than one solution. This is limited however to the boundaries of the original higher order polynomial to match with the lower order polynomial (line or parabola).

The number of solutions may be infinite due to that in between 1.00001 and 1.00002 there is always a number that fits in between, that also goes for this situation. But the range of parabolas and thus parameters (K,L,M) to be used is limited. Non the less no imaginary roots were required in this way.

THE REALITY OF PART II.

In order to also show how this final procedure works real life, also this procedure has been published as well on youtube [5], ever since this procedure was found. It is found that it works very well in several different examples that were tested by means of this method. For that reason some youtube films have been made and published as mentioned. So far it is not known to me that this type of method has been found and used in approaching higher order polynomials.

CONCLUSIONS AND DISCUSSION

Method of Part I:

For many higher order polynomials with equal zero-points (n) as the order is (n) this method proofs to be a valid way of retrieving the zero-point solutions (x-a)*(x-b)*… with very high accuracy in the order of magnitude of 10-12 to 10-15. This order of magnitude is, if compared to real life scales and taking meter as a unit on the size of hydrogen atom nuclear sizes. This however is achieved using nothing more than the commonly used Excel spreadsheet program.

The question arises why to use preferably 2nd order polynomials or parabola ? The answer is simple: with parabola there are 3 parameters that meet and coincide with higher order polynomials and at the same time keeping the solutions very simple and basic. This makes that these mathematical problems are easy to solve, using nothing more than high school math. The 3 parameters are the constant C, the linear component BX and the curvature of the high order polynomial expressed in AX2.

In this way the procedure is kept easy and low level, so everybody is able to do it in a simple and direct way, but very accurate non the less. One does not need to be a rocket scientists to perform this kind of math. High school education is enough, which makes this method elegantly simple, next to highly accurate, and readily to be used on all levels.

Of course one could also use 1st order approaches, but it turns out to be less accurate even though the math is even more simple. Naturally for not too accurate needs, this is acceptable. But for higher accuracy needs, the parabola method is far more accurate, even with a simple and easily available program as excel spreadsheets.

One could turn of course also to 3rd order polynomials, if one is willing to solve the 3rd order solutions, which are far more complex and more elaborate to work with.

The 2nd order polynomials-parabola’s is a method which lays exactly in the middle, for convenience, accuracy and simplicity and thus work speed.

However Excel is accurate for the most common use in this method, I am convinced that other more advanced programs can achieve a far more accurate level, if that is required. The limitations of excel is about in the order of magnitude of 10-15.

As mentioned there seem to be an alternating sequence in the “+” ‘s and the “-“ ‘s when the ABC formula is used. It was beyond the scope of this research to find the exact relationship and an appropriate formula for that. That is one of the next steps to be taken.

Method of Part II:

The next challenge was to find a solution to higher order polynomials without or with only a few zero-points.

With almost the same ease as in Part I the method in Part II could be applied.

Now it is not required to use the Part I method to retrieve the zero points, there are other methods to do this. But in the scope of this research of course the method of part I was used to do so. But again there are several other iterative methods.

For many (if not all) higher order polynomials with lesser zero points as the order is (n) this method proofs to be a valid way of retrieving a solution (x-a)*(x-b)*…. + (x-u)*(x-v) with very high accuracy. And what goes for the Part I method also goes for the Part II method; there may be computer programs that can handle even higher accuracy. But, for now, with excel it was possible to achieve a, for everyday use, very high accuracy.

The scope of finding an even more accurate method was not part of this research, that may be another objective in a further research. Based on the use of the excel spreadsheets it is estimated that the achieved accuracy of 10-15 is the best that can be retrieved from a program as excel. For even higher accuracies it is estimated that other programs may be required or developed.

Acknowledgement

There are a few persons I like to thank for this results and one Person in particular I like to thank for this knowledge and that is God-Jesus Christ-Holy Spirit Almighty. In my opinion He is the provider of all knowledge and He gave me this knowledge, to pass it on to humanity. This may be not a usual way of doing in the (predominantly secular) science community, but non the less I remain with this statement of gratitude to ward Him the keeper and giver of all knowledge. Since July 2004 I am a creationistic, Zionistic, messianic type of Christian and in my opinion God should get the honor of scientific findings on every level. Just as in the old days of people like Newton, Kepler, Galileo, Huygens and so on.

Finally I like to thank my father and (late) mother and my wife and daughter for their support in their daily support in our situation and these findings.

Biography of the author.

Edgar G.J.J. Korteweg

Education

Masters in Chemistry & Physics 2002-2004 (unfinished)

University of Utrecht, Netherlands

BSc in chemistry 1991-1995 BSc degree

University of Groningen, Netherlands

Physics 1991 (unfinished)

University of Groningen Netherlands

Medical chemistry 1988-1990 propeadeuse degree

University of Breda, Dr Struycken department Etten-Leur, Netherlands

Summary of experience