Solution of cubic and higher degree equations has always engaged attention of mathematicians from ancient times. Different approaches have been adopted to solve cubic equation, some transformed this equation to depressed form, some eliminated all the coefficients of this equation except cubic and constant term. By and large, all solutions seem tedious and requires constant concentration in their derivation. With the purpose of finding a solution which is easily doable, use of combined two continued fractions have been adopted and this has given inspiring results.

A fraction as we generally consider, is a part of whole quantity. Extending it, it can also be said, a fraction is a part plus complete quantity also and can be expressed as complete quantity plus fraction but certainly, it is never a whole in itself. A fraction thus can be written as ‘p’ divided by ‘q’ or p/q where p and q are whole number and q does not equal 1.

Let us take a fraction p/q = 71/137 and 71 being smaller can not be divided by 137 but 137 can be divided by 71. One can write it as 1 divided by1/(137/71) or (1)/1/(71/137). It is based on the principle that reciprocal of a reciprocal is same quantity.

Also (1)/1/(71/137) can be written as 1/137/71.

137 on being divided by 71 equals 1+ 66/71, therefore,

71/137 = 1/(1+ 66/71).

Again 66 can not be divided by 71 as 66 is smaller than 71. Again taking reciprocal of reciprocal of 66/71 as was done above, it can be written as 1/(71/66) and the procedure can be repeated.

Fraction 71/66 equals 1+ 5/66, thus

Fraction 71/137 = 1/{1+ 1/(1+ 5/66)}

5/66 can be written as 1/(66/5) or 1/(13+ 1/5), therefore,

71/137 = 0 + (1)/[1+1/{1+ 1/(13+ 1/5)}] or the fractions

0 + 1/[1+ 1/{1+ 1/(13+ 1/5)}] is called continued fractions of 71/137.

0 is added to this continued fraction as 71 could not be divided by 137 or it can be said that it is divided 0 time.

It is evident from above that procedure adopted above is same as finding the greatest common divisor (g c d) of 71 and 137. That is why great mathematician Euler described continued fractions as determining g c d.

By adopting the same procedure, fraction

15/11 can be written as 1+ 1/{2+ 1/(1+ 1/3)}.

Continued fraction can be written in generalised form as

pn/qn = a0 + b0/[a1 + b1/{a2 + b2/{a3+……an + bn/1)}}]………(1)

where coefficients a0, a1, a3………. are called partial quotients.

When b0, b1,b2,b3 all are equal to 1, continued fraction can be written as

pn/qn = a0 + 1/[a1 + 1/{a2 + 1/{a3+…………an-1 +1/an)}}]……(2).

Such a fraction is also abbreviated as (a1; a2, a3, a4, ……an).

Continued fractions which terminate at a point as we have seen, 71/137 terminated at 1/5, 15/11 at 1/3 and a0…….., b0 at bn/1, are called close continued fractions. But there are continued fractions which do not terminate but continue infinitely. These are called infinitely continued fractions or periodic continued fractions. Some examples are given below.

y = 1/[2+1/{2+ 1/(2+ 1 /2………… up to Infinity……………….(3),

y = 2+ 1/[2+ 1/{2+ 1/(2+ 1………up to Infinity………………….(4).

Partial fractions (3) and (4) are infinitely continued but on examination, it is observed that fraction (3) repeats all its terms even after first quotient whereas equation (4) repeats all its terms after two quotients, therefore equation (3) has period one and equation (4) has period 2. Period is number of quotients or terms after which the fraction continues all its terms.

It is easily discernible from equation (3) that it can be written as

y = 1/(2+ y) as it is infinitely continued after first quotient.

Or y^2 + 2y -1 = 0 which is a quadratic in y and has two roots as

y = -1+ 2^1/2 or -1- 2^1/2.

Since fraction (3) only contains positive signs, its sum must be positive, therefore fraction (3) equals -1+ 2^1/2.

Similarly fraction (4) has value

y = 2+ 1/y

y^2- 2y – 1 = 0.

This is also a quadratic in y and has value equal to

(1+ 2^1/2) ignoring the negative value.

We can thus write fractions (3) as

(2^1/2- 1) = 1/[2+1/{2+ 1/(2+ 1 /2………………. up to infinity or 2^1/2 =1+ 1/[2+1/{2+ 1/(2+ 1 /2………………. up to infinity……(5)

and equation (4) as

(1+ 2^1/2) = 2+ 1/[2+ 1/{2+ 1/(2+ 1………up to Infinity.

Mathematician Lagrange on the basis of this nature of infinitely continued fraction stated that all algebraic irrationalities can be expressed only and only in quadratic equations. Conversely, only and only quadratic equations express the algebraic irrationalities of infinitely continued fractions.

This property of infinitely continued fraction is utilised in solving quadratic equations and also finding square roots of quantities.

Let us say square root of quantity 2 is to be calculated. First step is to find that value which approximates 2. If I assume approximate value of square root of 2 as 1, on squaring, it becomes equal to 1 but if I assume its value as 2, on squaring, it becomes 4, therefore, 1 is in better proximity of square root of 2 than 2. Let x is that quantity when added to 1 gives the exact value of square root of 2. That means

1+ x = 2^1/2.

Or x^2 + 2x + 1 = 2

Or x^2 + 2x = 1

Or x = 1/(2 + x)

Here we find x in the denominator term x+2 but x = 1/(2 + x) and on substituting this value in the denominator, we get

x = 1/{2 + 1/(2+ x)}.

On successively substituting value of x = 1/(2 + x), we get never ending fractions as we substitute the value of x, another x further appears in subsequent denominator. Therefore x can be written as

x = 1/{2 + 1/(2+ 1/2+…………. up to infinity.

Therefore 2^1/2 = 1+ x = 1 + 1/{2 + 1/(2+ 1/2+…………. up to infinity.

Let us generalise the formula for determining square root of a quantity n.

Let m be any integer which approximates square root of n and x is an added quantity that exacts it to square root of n.

Then (x + m) = n^1/2

Or x^2 + 2.m.x + m^2 = n

Or x = (n- m^2)/(2m + x).

Substituting the value of x successively, we get

x = (n- m^2)/[2m + (n- m^2)/{2m+(n- m^2)/(2m+……………….up to infinity.

To determine convergence of a continued fraction, let us inspect fraction (5), sum of first row

p0/q0= 1,

sum of first plus second row, p1/q1 = 1+1/2 = 3/2.

Similarly, p2/q2= 7/5, p3/q3 =17/12 , p4/q4 =41/29 , p5/q5 = 99/70……….so on.

Therefore, p0/q0- p1/q1= 1- 3/2 = – ½,

p1/q1 – P2/q2 = 3/2- 7/5 = 1/10,

p2/q2- p3/q3 = 7/5- 17/12 = -1/84,

p3/q3- p4/q4=17/12- 41/29 = 1/348,

p4/q4- p5/q5 = 41/29 – 99/70 = -1/2030,

…………………………………………………………..

…………………………………………………………..

So on.

From above, it is clear that as sum of more quotients is taken up, difference between consecutive sum decreases and that means fraction is convergent. Pn/qn is also called nth convergent. It also reveals a pattern where difference alternates from negative to positive like that of swing span of a physical balance. This will be explained in detail in example section.

By mathematical induction, we can make a generalised statement from fraction (2) that

po /qo = a0/1, p1/q1=(a1a0+1)/a1, p2/q2 = (a2a1a0 +a2 +a0)/(a2a1 + 1) p3/q3 = (a3a2a1a0 +a3a2 +a3a0 +a1a0 +1)/ (a3a2a1 + a3 + a1)

………………………………………………………………….so on.

Convergence of this fraction can also be checked the way we did for fraction (5).

**Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis—where y = 0). Here the function is f(x) = (x3 + 3×2 − 6x − 8)/4.**

**Determining Roots Of a Cubic Equation**

With this background, we proceed to determine Roots of a cubic equation which can in general be written as

x^3 + px^2 + qx + r =0 ………………………(6),

where p, q and are coefficients of x^2, x and constant term.

Values of x that satisfy this equation are called its roots and this being a cubic equation has three roots as number of roots equals degree of equation. If one of its roots is determined, the equation can be reduced to a quadratic and its remaining two roots can easily be found out.

**Approximation Of A Root**

A root can be approximated, if x is assigned some value and that value is put in the cubic equation, it will either have positive value or negative value or even zero value. If the equation, on substitution, has zero value, then we are lucky enough as it would be its roots and the matter to find a root ends. But in the rare cases, such root is accidentally determined. Therefore, we take that case, say, where the value of equation is not zero but is positive or negative. To proceed further to find a root, a theorem is put forth here under.

Theorem: There is a value nearby its root that makes an equation on threshold positive or negative and increment or decrement by one in that value, reverses the sign of the equation then its root lies between that value and the value increased or decreased by one.

Let f(x) = a0x^n +a1x^n-1 +………………= 0, is an equation,

then there will be a value α, that will make the function

a0 α^n + a1 α^n-1 +……… at threshold positive or negative

and on putting the value as α -1 or α + 1 in equation, either the value of a0 (α+1)^n + a1 (α+1)^n-1 +……… will be opposite in sign

or

the value a0 (α-1)^n + a1 (α-1)^n-1 +………will be opposite in sign to that of a0 α^n + a1 α^n-1 +………

By applying this theorem and putting a random value of x and then noting the sign of equation and then decreasing or increasing its value by one so that a stage reaches when its sign is reversed, approximate value of its root lying between α and α- 1 or α and α+ 1 can be found out. This will bring us closer to the value of the root of equation.

**Transformation Of Cubic Equation**

Coming to equation (6), let the value of its root approximated is m. Then there will be a quantity y which will always be less than 1, such that

y+ m will be its exact root. On putting this value in equation (6), it takes the form

(y+ m)^3 + p.(y+ m)^2 + q.(y+ m) + r = 0.

On simplification, a cubic equation in y is formed as

y^3 +y^2(3m +p) + y.( 3m^2 + 2pm + q) + m^3+ p. m^2 + q.m + r = 0.

This transformed equation can be rewritten as

y^3 + a.y^2 + b.y+ c =0…………………………………………………(7),

where a = 3m +p,

b = 3m^2 + 2pm + q,

c = m^3+ p. m^2 + q.m + r.

Only discernible difference of this equation from equation (6), is that the cubic equation in y has a root which is less than 1 and root of this equation is related to equation (6) by

x = y + m.

**Determination Of Root By Combination Of Two Fractions**

To ascertain value of y, we can write the equation (7) as

y^3 + a.y^2 = -(b.y+ c),

or y = -(b.y + c)/ y.(1+ a.y).

*Expanding In Partial Fraction*

Above equation has numerator of degree one and denominator of degree two and, therefore, can be decomposed to partial fractions as

y =-(b.y + c)/ y.(1+ a.y) = -[ A/y + B/ (1+ a. y)]

Or by + c = A(1+ a.y) + B.y.

On comparing coefficient of y and constant term,

b = A.a + B and c = A.

On simplification,

A = c and B = b – a.c.

Therefore, y = – [ c/y + (b- ac)/(1+ a.y)

= (a.c- b)/(1+ a.y) – c/y…………………………….(8).

Equation (8) has two fractions (a.c- b)/(1+ a.y) and – c/y which have y in in the denominators indicating that these will continue infinitely. On successively putting the value of y as given by equation (8) in equation (8) itself, it will form never ending fractions as

y = (a.c- b)/[1+ a. {(a.c- b)/(1+ a.y) – c/y}] – c/{ (a.c- b)/(1+ a.y) – c/y}.

To show that this fraction is continuing infinitely, I put sign ……. In place of y.

y = (a.c- b)/[1+ a. {(a.c- b)/(1+ a…..) – c/…….}] – c/{ (a.c- b)/(1+ a……)} – c/……..}……………………………..,..(9).

y, therefore, can be calculated from convergent pn/qn as we discussed earlier where n will be number of times we want to consider partial quotients of fractions depending upon accuracy of result desired.

Alternatively, the equation (7) can also be rearranged so as to make it

y(y- s).(y- t) = – c + u.y

by factorising LHS of equation y^3 + a.y^2 = -(b.y+ c).

y therefore equals (- c + u.y)/(y- s).(y- t).

This can be decomposed to partial fraction as

y = A/(s- y) + B/(s- y)

On successively putting the value of y as A/(s- y) + B/(s- t) in itself, it will form never ending fractions as

= A/[s- {A/(s- y) + B/(s- y)}]

+ B/[t- {A/(s- y) + B/(s- y)}].

Since y is being successively substituted, dots are marked in place of y showing its infinite continuation. Then

y = = A/[s- {A/(s- …..)+ B/(s- …..)}]

+ B/[t- {A/(s- …….) + B/(s- ……..)}]…………………………………(9/1)

This method will be preferred as it can indicate approximate value of y by putting y as zero in right hand side and will be used in all calculations.

**Determination Of A Root Without Approximation**

Notwithstanding, the method where a root was approximated, a cubic equation can also be solved straight forward.

Let the equation be

x^3 + px^2 + qx + r =0.

It can also be written as x^3 + ax^2+ bx^2 + abx + q’x + r = 0

where p = a+ b and q = ab + q’.

This equation can also be factorised as

x(x+ a).(x+ b) = – (q’x + r).

This can then be decomposed to partial fraction as

x = A/(a+ x) + B/(b+ x),

where A and B can be determined by comparing coefficients of x and constant term. This is also an infinite continued fractions as x figures in both left hand side and right hand side. On successive substitution of value of x as A/(a+ x) + B/(b+ x) in its own equation, we get continued fraction as

= A/[a+ {A/(a+ x) + B/(b+ x)}]

+ B/[b+ A/(a+ x) + B/(b+ x)}]

= A/[a+ {A/(a+ …..) + B/(b+ …..)}]

+ B/[b+ {A/(a+ …..) + B/(b+ ….)}]…………………………..(9/2)

However when we compare fraction (9/2) with (9/1) and (9), we find that y (x in case of fraction 9/2), is always less than 1 but it may be more than 1 in case of equation 9/2. Y, in case of fractions 9 and 9/1 can be ignored but in fraction 9/2, it may result in crude approximation particularly in earlier convergents like p0/q0 and p1/q1. With this fact in mind, method of approximation of root is preferred and recommended.

**Theory And Concept**

If in a mathematical equation, its right hand side contains a part or whole of the quantity in its left hand side then the equation regenerates left hand side in right hand side and the regenerated quantity again contains left hand side, and regeneration again occurs and the cycle continues infinitely. The equation thus forms a close loop that continues infinitely.

For example, equation x = 1/(1+ x) has x in left hand side and 1/(1+ x) in right hand side. The output 1/(1+x) contains the input x. That means x in denominator of 1/(1+ x) can be replaced by 1/(1+ x) and the equation takes the form

x = 1/{1+ 1/(1+ x)}.

Again output or right hand side RHS has x and this can again be substituted by 1/(1+ x) and this will again contain x in RHS. Such a quantity is infinitely continued.

If x = 1+ x^2,

then x in RHS can be substituted by 1+ x^2 and

x = 1+ (1+ x)^2 = 1 +1 + 2x + x^2.

On successive substitution, a power series is generated.

However, when x is of the form A/(B + x), it is then a fraction with numerator A and denominator 1/(B+ x). Since it also contains x in RHS, it is also an infinitely continued fraction similar to the one as mentioned at (1) with the only difference that the present fraction continues infinitely and has

a0 = a1 = ……. b0= b1 = …… = 1.

But an equation of the form,

x= A/(B + x)

can be simplified as

x^2 + Bx – A = 0,

and this forms a quadratic equation in x. This deduces us to make a statement that infinite continued fraction is always expressible in quadratic equation. This quadratic equation has roots

= ½. [ – B + (B^2 + 4 A)^1/2] or

1/2.[- B – (B^2 + 4 A)^1/2].

If the quantity (B^2 + 4 A) is a perfect square, the roots will not have any square root term. Otherwise these will always have a square root term.

Such square root quantities are irrational as these can never be calculated exactly and their decimal portion continues infinitely. Such quantities give birth to fractions that continue endlessly. Being born form quadratic equations, these are called quadratic irrationalities or algebraic irrationalities.

With these concepts in mind, Great Mathematician Lagrange gave the theorem,

“Every periodic continued fraction represents a quadratic irrational number and every quadratic irrational number is represented by a periodic continued fraction.”

Period of a function has already been defined earlier by me. Every infinite continued fraction is periodic, that means quadratic equation is essential requirement for infinite continued fractions. But in this paper, cubic equations which are a degree higher than quadratic are being dealt. Does it not then contradict Lagrange theorem?

This is the moot question which will be convincingly settled here. A cubic equation in general is expressed as

y^3 + ly^2 +m y + n =0,

and its approximate root can always be estimated as it has already been explained in detail. Let the root approximated is y’ then on substituting

y = x + y’, a cubic equation can be found out. Let this equation be

x^3 + px^2 + qx + r =0.

It can also be written on rearrangement as

x = -r/(x^2+ px+ q)………………………………………………………(T)

This equation as it is and without any operation and in its present form, is not expandable to continued fraction, is admittedly true. Obviously, it contains a quadratic equation in denominator and that will result in a cubic equation whereas Lagrange Theorem requires formation of quadratic equation. A quadratic equation will only form when denominator of equation (T) contains x in first degree. Therefore, it is essential to transform it so that it may contain x in first degree in its denominator.

That is possible if denominator -r/(x^2+ px+ q) is decomposed into partial fractions as that will give it the form A/(a+ x) + B/(b+ x) where a+ x and b+ x correspond to first degree in x.

By equating -r/(x^2+ px+ q) = A/(a+ x) + B/ (b+ x) and comparing coefficient of x and constant term, A and B can be determined.

Then x can be written as

x = A/(a+ x) + B/(b+ x).

In calculus, while integrating a quantity of the form -r/(x^2+ px+ q) where x is variable, it is split up in partial fractions as A/(a+ x) + B/(b+ x) and each part is integrated separately without affecting the result. Same procedure is applied here.

It is explicit that the value of x is related to both partial fractions A/(a+ x) and B/(b+ x). It is also clear, x when approximated in the form p0/q0 will result as sum of A/a and B/b and this will also be true when x is small. As each A/(a+ x) and B/(b+ x) has a term x, these are also interconnected. Approximation of x as p/q, improves as number of quotients considered are increased, since both interconnected fractions are continued infinitely and also solutions given in examples corroborate the convergence. Variation in x on account of improvement in approximation will cause changes in A/(a+ x) and B/(b+ x). As improvement in approximation is affected by both A/(a+ x) and B/(b+ x), changes in A/(a+ x) will cause changes in B/(b+ x) and vice versa. Conclusion is that both A/(a+ x) and B/(b+ x) although separate are integrated with each other so that their resultant effect is same as that of -r/(x^2+ px+ q).

In other words, this can also be stated that A/(a+ x) and B/(b+ x) are interrelated two continued fractions and approximation of x on account of A/(a+ x) also affects B/(b+ x) and vice versa. This is due to the fact that expansion of A/(a+ x) into infinite continued fraction contains term B/(b+ x) and likewise expansion of B/(b+ x) into continued fraction contains term A/(a+ x).

Apart from this, except for their interrelation, A/(a+ x) is an infinite continued fraction with periodicity one and similarly B/(b+ x) is also an infinite continued fraction with periodicity one. These two fractions behaves like normal periodic fractions. However, it is their interrelation, that brings the effect of cubic equation.

Had there been a single periodic continued fraction formed from cubic equation, it would have violated Lagrange Theorem. On the other hand, formation of two periodic fractions has not only solved cubic equation but has kept principle employed in the theorem intact. It is a special case where a cubic equation is solvable and also meets the requirement of the theorem.

It is thus essential, for solving cubic equation by continued fractions, there must be two interrelated periodic continued fractions.

While solving cubic equation, two periodic interrelated continued convergent fractions were generated from A/(a+ x) and B/(b+ x) and how these two Infinite continued fractions are convergent have been dealt while solving cubic equations in example section.

This concept is also applicable to Quartic ( degree four) equations but for solving such equations, three interrelated continued fractions will be needed. These three partial fractions A/(a+x) + B/(b+x) + C/(c+ x) can always be expanded and x can be solved from equation

x = A/(a+x) + B/(b+x) + C/(c+ x)

as it was solved in case of cubic equation.

How terms A/(a+ x) and B/(b+ x) progress and what their algorithm is shown in figure.

{Figure 1 showing progression of terms of A/(a+ x) and B/(b+ x) in algorithm of x = A/(a+ x) and B/(b+ x }

Red line corresponds to term A/(a+ x) and black to B/(b+ x). Point A that corresponds to p0/q0 is first approximation point where x approximation is A/a + B/b. At this point, x equals A/(a+ x) + B/(b + x) and approximation is made by neglecting x, otherwise also x is small due to the fact transformation has been done after estimating the root. As the terms of fraction increase , points B of both branches (p1/q1) are reached, these contain four terms and at points C of four branches ( approximation p2/q2), these contains 8 terms. In this way, number of terms increases in geometric progression. However the figure is drawn up to point E, thereafter, dotted lines indicate that it is continuing up to infinity. The figure illustrates how the terms A/(a+ x), and B/(b+ x) are generated in geometrical progression.

**Examples And Applications**

Keeping in mind, the theory described above, cubic equation f(x) = x^3- 6x^2+ 10x -9 = 0 is solved here.

Applying theorem as stated above, at x = 4,

f(4) = 64 – 96 + 40 – 9 = – 1

and at x = 4 + 1 or 5,

f(5) = 125 – 150 + 50 – 9 = 41.

Obviously, when x increases from 4 to 5, equation changes sign from negative to positive. That means one of the root of the equation lies between 4 and 5. Let the root be 4 + y where y will be less than 1. 4+ y then must satisfy the equation,

(y + 4)^3 – 6. (y + 4)^2 + 10 (y + 4) -9 = 0.

On simplification,

y^3 + 6y^2 + 10 y -1 =0……………………………..(10).

Or y = 1/( y^2 + 6y + 10).

Since one of the root lies between 4 and 5, that means value of y is always less than 1. If we assume y to be very small, then rough estimate of y is,

y = 1/(0+ 0 + 10) =1/10 or .1 and x will be equal to 4.1.

But for finding exact value of y, continued fractions will be resorted to. Equation

y^3+ 6y^2 + 10y -1 = 0,

can be written as

y(y^2+ 6y + 8)+ 2y -1 = 0 on splitting 10 y into 8y + 2y so as to facilitate factors of y^2+ 6y + 8 as (y + 4).(y+ 2).

The equation therefore can be written as

y(y + 4).(y+ 2) = 1- 2y,

y = (1-2y)/(y+ 4).(y+ 2).

This can be decomposed in to partial fraction as

y =(1-2y)/(y+ 4).(y+ 2) = A/(y+ 2) + B/(y+ 4)

= (5/2)/(2+ y) – (9/2)/(4+ y)

Both these fractions contain y in the denominator and therefore, can be expanded as infinite continued fractions on successive substituting the value of y in the above equation.

y = (5/2)/[2+ {(5/2)/(2+ y)} – (9/2)/(4+ y)}] – (9/2)/[(4+ {(5/2)/(2+ y) – (9/2)/(4+ y)}]………………………………………………(11)

y = convergent p0/qo = .125

= convergent p1/q1 = .0855

= convergent p2/q2 = .094.

The pattern of above values indicates that the difference

p0/q0 – p1/q1 = .0395,

p1/q1- p2/q2 = -.0085,

is alternating from negative to positive as p/q convergent increases. Swing for first correction is .0395 and for second, correction is -.0085, it is analogous to swing of span of physical balance which also experiences shortening amplitude swing from its central position when an object of same mass as that the standard weight is kept on the span.

Convergence of y is also illustrated in figure 2 which is not drawn according to scale. Proving of convergence from this example establishes that the combined effect of two interrelated infinite continued fractions completely and correctly theorise the solution of cubic equations. It is illustrated in figure 2 that as approximation p/q of x pertains to large convergent, x attains the exact value from 0.125 to .0855 to .094 to……………….exact value.

Also second part Of figure 2, shows how swing reduces to zero from .0395 to -0.0085 to …………….0.

**Figure 2 shows improvement in approximation and swing with increase in number of convergents**

It is also worthwhile to check deviation of calculated value of root from its actual value. At y = .094, function y^3+ 6y^2 + 10y -1 has value of .00131 and at .0945, it has value of .0006 and assuming, x as .0945 satisfying the equation, value of x will calculate as 4 + y = 4.095. But y from continued fractions calculates as 4 + .094. Therefore at p2/q2, calculations by continued fraction has an error of .122 percent. As the number of terms for calculations will increase, value of root improves. That further proves that method of determining roots of a cubic equation as theorised, does not suffer from any infirmities.

**Calculating Cube Root Of A Quantity by Continued Fractions**

Applying the theory as described above for finding a root of cubic equation, let us calculate cube root of 9. It is explicit that 9^1/3 equals approximately 2. Three will not be good approximation as 3^3 makes 27 which differs by 18 from 9 whereas 2^3 makes 8 and that differs by 1 from 9. Let x which will be less than 1, be the addendum that makes

x + 2 = 9^1/3.

On cubing both sides, we get

x^3 + 6x^2 + 12 x + 8 = 9.

Or x^3 + 6x^2 + 12 x = 1.

Or x = 1/( x^2 + 6x + 12).

The fact that at x small and less than 1, terms containing x can be ignored. That gives approximate value of x as 1/12 but we are interested in determining the actual value of x, therefore on rearranging, the equation can be written as

x^3 + 6x^2 + 8 x = 1- 4x.

Or x(x+ 4).(x+ 2) = 1- 4x.

Or x = (1- 4x)/(x+ 4).(x+ 2).

On expanding it to partial fraction,

x= (1- 4x)/(x+ 4).(x+ 2) = A/(x+ 4) + B/(x+ 2).

Or x = (9/2)/(2+x) – (17/2)/(4 +x).

In the right hand side RHS, both the partial fractions have x, therefore, on successive substitution of

x in RHS as (9/2)/(2+x) – (17/2)/(4 +x),

we get

x = (9/2)/[{2+{(9/2)/(2+………..)}– {(17/2)/(4 +……….. )}]

– (17/2)/[4 +{ (9/2)/(2+………….) – (17/2)/(4 +……)}]……………………(12),

Where sign dots indicates that fraction is continuing infinitely.

Since x is small and less than 1 as explained earlier, first approximation

x = p0/q0 = 1/8 = 0.125,

second approximation,

x = p1/q1 = 0.0570410,

third approximation,

x = p2/q2 = 0.07357277.

It is explicit from above, as p/q convergent increases employing more terms, difference p0/q0 – p1/q1, p1/q1- p2/q2 alternates as convergent improves. That corroborates, continued infinite fraction is convergent and is correctly defined by infinite continued fraction (12).

It is also worthwhile to check deviation of calculated value of cube root 9 from its actual value. At approximated value 2 + .07357277 = 2.07357277, its cube that is (2.07357277)^3 equals 8.9157492 and therefore, has an error of .95 percent but as convergent p/q will increase, it reaches its exact and correct value. Here also pattern of values of p/q reveals that convergence is analogous to swings of span of a physical balance. After a few swings, the span settles at its mid position as illustrated in figure 2.

Examples how to determine a root of quartic equation has not been given, however quartic equations can also be solved following the method described in section theory and concept.

End

References:

1) NON-PERIODIC CONTINUED FRACTIONS FOR QUADRATIC IRRATIONALITIESMICHAEL O. OYENGO at http://www.math.illinois.edu/~mchlyng2/Non-Periodic%20continued%20fractions.pdf

2) The Topsy-Turvy World of Continued Fractions by MICHAEL O. OYENGO

3) General Method for Extracting Roots using (Folded) Continued Fractions by Manny Sardina

4) Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis—where y = 0). Here the function is f(x) = (x3 + 3×2 − 6x − 8)/4. Courtesy N Mori at https://en.m.wikipedia.org/wiki/Cubic_function

About Author:

Author is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.

Excellent. You’ve taken a very complex topic and made it consumable. It took me awhile to read, I had to keep looking things up, but you’ve done a masterful job!

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Thanks, sir for taking pains in going through the article. I get inspiration from you. Regards.

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