Roots of equations

Page 1

Roots of equations texts by Patrizio Gravano

Index 0. Introduction 1. Quadratic equations 2. Algebraic equations in x of degree n > 2

Autumn 2014

A new little note on equations in x of degree n ≼ 2, studied in R. Possible location of positive roots. by Patrizio Gravano

0. Introduction In this paper, I would like to consider the possible location of positive roots of algebraic equations of degree n ≼ 2, studied in R. Firstly, I would like to consider the case n = 2 (quadratic equations). Secondly, I would like to discuss the general case n > 2.

1. Quadratic equations To solve a quadratic equation as ađ?‘Ľ 2 + bx + c = 0 you can use the following formula x=

−đ?‘?Âąâˆšđ?‘? 2 −4đ?‘Žđ?‘? 2đ?‘Ž

.

It is a very important formula. However, using elementary algebraic theory you can say if exists real number x, called unknown, for example đ?‘Ľđ?‘œ that ađ?‘Ľ0 2 + bđ?‘Ľ0 + c = 0 is true. In general, you can say when x exists and when it is an real number! In fact, in this speech I would observe that if you have the following equation (x ∓ a) ( x ∓b) = 0 ⇒ x = Âą a and x = Âą đ?‘?. You can observe that in general AB = 0 ⇒ A = 0 or B = 0 or A = B = 0. More, you can use this symbol ⇔ (necessary and sufficient condition). In general, you have j algebraic products of sums such as (x ∓đ?‘&#x;đ?‘ ) for any r real number and for every integer s ≤ j you can obviusly extend this procedure at the case j > 2.

When you consider the case that you have j =2 products of sums you can note that it is not necessary use formulas as x =

−đ?‘?Âąâˆšđ?‘? 2 −4đ?‘Žđ?‘? 2đ?‘Ž

.

Now, I would like to explain a particular case to demonstrate no real solutions in a specific case. In fact, in this one I admit a = b = c ≠0. We can write a ∈ ( -∞, 0) âˆŞ ( 0 , -∞), or a ∈ R {0}. In this case there is no solution in R. R denote the set which contains all real numbers (continuous real set). If you admit a = b = c ≠0 you can write đ?‘Ľ 2 + x + 1 = 0. x = 0 is not a solution. In fact for x = 0 we have 1 = 0. I consider x ≠0. I suppose x > 0 and observe đ?‘Ľ 2 + x = - 1. When x > 0 this relation is never true. In fact, we obtain that a positive real (as đ?‘Ľ 2 + x) equal to a negative integer, as - 1. Now we can consider the subcase following. For x < 0 we can write đ?‘Ľ 2 = - 1 - x = - 1 + |x|, formally correct (but only formally!) when x < 0. But when we consider ( -∞, 0) we can admit (-∞ , -1] âˆŞ (-1 , 0). We have to consider two different subcases. The first one is when x ∈ (-∞, -1]. Using elementary algebraic rules you can have the following formal equality: |x | - đ?‘Ľ 2 = 1. We can observe no real solution in this one when x ∈ (-∞ , 0). In fact in this case the real number - |x | - đ?‘Ľ 2 = 1. What I have just written is right when we consider the second possible case, when x ∈ ( 1, 0). However, we can unify these two different subcases. In fact we can write đ?‘Ľ 2 = |x||x|. We have |x| - |x||x| = 1, and then |x| (1 - |x|) = 1. This equality is not true. In fact when |x| > 1, the number |x| (1 - |x|) < 0, then, a fortiori, |x| (1 - |x|) < 1. But you can note that from |x| (1 - |x|) = 1 you have (1/s) + s where s = |x| ∈ đ?‘… + with đ?‘… + to formalize the set of the positive real numbers. But (1/s) + s = 1+s /s > 1, for each positive real s, as definite.

Then we can observe that it is impossible to have (1+s/s) = 1 when we put s = |x| I would like to say how I can write đ?‘Ľ 2 + x + 1 = 0 from ađ?‘Ľ 2 + bx + c = 0 when you suppose a = b = c ≠0. You can take ađ?‘Ľ 2 + bx + c = 0 and divide by a. You must remember that 0/a = 0. We obtain đ?‘Ľ 2 + x + 1 = 0 immediatly. Finally, I would to remember that đ?‘Ľ 2 = - 1 - x = - 1 + |x|, formally admissible (but only formally!) when x < 0. But there are not some x real numbers that verify this equality. Next days I will send a new speech to demonstrate there are no real solutions to quadratic equation when you consider (a, b, c) ≠(0, 0, 0) in case a>0, b>0 and c > 0 and in different case too, when you consider a < 0, b < 0 and c < 0, as Cartesius said. It is easier to solve two different cases, when we have b = 0 or c = 0. But in general we must consider a ≠0. But the general case is the following: ađ?‘Ľ 2 + bx + c = 0 → đ?‘Ľ 2 + Îśx + Ďƒc = 0 with Îś = b/a and Ďƒ = c/a. I will demonstrate the following statement: There are no positive real numbers, such as đ?‘Ľ0 > 0, that đ?‘Ľ0 2 +Îś đ?‘Ľ0 + Ďƒ c = 0 is true when Îś > 0 and Ďƒ > 0. I continue to discuss about inexistence of the real number đ?‘Ľ0 > 0 as solution of the second order algebraic equations. I will demonstrate the following statement: There are not positive real numbers, such as đ?‘Ľ0 > 0, that đ?‘Ľ0 2 +Îś đ?‘Ľ0 + Ďƒ c = 0 is true when Îś > 0 and Ďƒ > 0. Now I wish consider the general case, when a≠b ≠c ≠0. In this case we can note there are no real solutions to quadratic equation when you consider (a, b, c) ≠(0, 0, 0) in case a>0, b>0 and c > 0 and in different case too, when you consider a < 0, b < 0 and c < 0, as Cartesius said. From a≠b ≠c ≠0 we have Îś â‰ Ďƒ. đ?‘Ľ0 ∈ (0, + ∞) is not a solution of đ?‘Ľ0 2 +Îś đ?‘Ľ0 + Ďƒ = 0 because from đ?‘Ľ0 2 +Îś đ?‘Ľ0 + Ďƒ = 0 we can write đ?‘Ľ0 2 = - Îś đ?‘Ľ0 - Ďƒ. But this equality is not true because đ?‘Ľ0 2 > 0 but - Îś đ?‘Ľ0 - Ďƒ < 0 when đ?‘Ľ0 ∈ (0, + ∞) and Îś > 0 and Ďƒ > 0 too. đ?‘Ľ0 = 0 is not a solution. It is very easy to demonstrate it by substitution.

Now we can consider the case đ?‘Ľ0 ∈ ( - ∞, 0). In fact, you can consider the following algebraic steps. đ?‘Ľ0 2 = - Îś đ?‘Ľ0 - Ďƒ → đ?‘Ľ0 2 + Îś đ?‘Ľ0 = - Ďƒ →

(đ?‘Ľ0 )<0

đ?‘Ľ0 2 - Îś |đ?‘Ľ0 |= - Ďƒ → |đ?‘Ľ0 ||đ?‘Ľ0 |- |đ?‘Ľ0 | Îś =

- Ďƒ → |đ?‘Ľ0 |(|đ?‘Ľ0 |- Îś )= - Ďƒ. Former equality is not true when we consider the case |đ?‘Ľ0 |> Îś. In fact in this case |đ?‘Ľ0 |(|đ?‘Ľ0 |Îś ) > 0 but â€“Ďƒ < 0 when Ďƒ > 0. When we consider |đ?‘Ľ0 |= Îś = b/a we have |đ?‘Ľđ?‘œ |0 = - Ďƒ. Hence 0 = - Ďƒ (impossible when Ďƒ ≠0). This statement is not possible. Then đ?‘Ľ0 = - Îś = - (b/a) is not a solution. I remember I have just written đ?‘Ľ0 2 - Îś |đ?‘Ľ0 |= - Ďƒ. I obtained it from đ?‘Ľ0 2 = - Îś đ?‘Ľ0 - Ďƒ when đ?‘Ľ0 < 0. For Îś < |đ?‘Ľ0 | we note đ?‘Ľ0 2 - Îś |đ?‘Ľ0 |= - Ďƒ is a consistent equality when Ďƒ > 0. From this formal consistency we can write the following equality Ďƒ = Îś| đ?‘Ľ0 |- (đ?‘Ľ0 )2 . From these one we have c/a = (b/a)r - đ?‘&#x; 2 when r = |đ?‘Ľ0 |. From last equality we obtain c = br đ?‘&#x; 2 and then ađ?‘&#x; 2 - br + c = 0. For every (a, b, c) ≠(0, 0, 0) we can solve ađ?‘&#x; 2 - br +c = 0 respect to unknown r. However, we have obtain this quadratic equation considering the case r = |đ?‘Ľ0 | when đ?‘Ľ0 < 0. This polynomial f( r) = đ?‘Žđ?‘&#x; 2 - br + c is formally different (Abel’s identity polynomial criterion) from polynomial f(x) = đ?‘Ľ 2 + (b/a)x +(c/a). In fact if we consider the first one for x < 0 we have r = |x| ⇒ x = - r or identically r = - x. Then from f(r) =ađ?‘&#x; 2 - br + c, consistent when r > 0, we have f(-x) = ađ?‘Ľ 2 + bx + c. In general f(∓x) = f(Âą r) and f(x) = f(-r). Now I would like to consider ađ?‘&#x; 2 - br + c = 0 noting that đ?‘&#x; 2 (b/a) r + (c/a) = 0 → đ?‘&#x; 2 = (b/a) r - (c/a). I remember I have made it with a > 0, b > 0 , c > 0. I have admit the condition r > 0. Considering former equality, I observe đ?‘&#x; 2 > 0, always. We consider the case r > 0. r = b/a is not a possible solution, in fact đ?‘&#x; 2 = đ?‘&#x; 2 - (c/a) is true only and only if you have (c/a) = 0. It is impossible. r < o is not a solution. In fact for any r < o we have a positive number such as đ?‘&#x; 2 equal to a negative real number as (b/a) r - (c/a). We have a negative number because the sum of two negative numbers is a negative number (closure). From đ?‘&#x; 2 = (b/a) r - (c/a) we can say r > b/a can be a solution of this equation. But from (b/a)r - (c/a) > 0 we can also write (b/a)r > c/a → r > c/b.

If you consider these conditions you can admit r is a solution for r > max ( c/b ; b/a). Why x = - r we have just demonstrate there are no positive real solutions for quadratic equation such as ađ?‘Ľ 2 + bx + c = 0 when we admit a > 0, b > 0 and c > 0 or we admit a < 0, b < 0, c < 0. When (a, b, c) ≠(0, 0, 0) the number đ?‘Ľ0 = 0 is not a solution for this equation. We remember in this case f(x) = f(-x) and for every (a, b, c) we have lim đ?‘“(đ?‘Ľ) = Âą ∞ in đ?‘Ľâ†’ Âąâˆž

according to a > 0 or a < 0. In this paragraph about quadratic equations and its possible real roots, I have just demonstrated the following statement: There are no positive real numbers, such as đ?‘Ľ0 > 0, that đ?‘Ľ0 2 +Îś đ?‘Ľ0 + Ďƒ c = 0 is true when Îś > 0 and Ďƒ > 0. It is very easy to demonstrate it in more simple way. In fact from ađ?‘Ľ0 2 + bđ?‘Ľ0 + c = 0 we obtain ađ?‘Ľ0 2 + bđ?‘Ľ0 = - c. However - c < 0 when c > 0. Moreover, when a > 0 and b > 0 for đ?‘Ľ0 < 0 last equality is not true. Similarly, we can consider the subcase c < 0. For c < 0 we have - c > 0. However, when a < 0 and b < 0 for every đ?‘Ľ0 > 0 the number ađ?‘Ľ0 2 + bđ?‘Ľ0 is a negative number but - c is, for c < 0, a positive number. Then we have just obtained an equality impossible. In fact is not possible a positive number equal to a negative number.

2. Algebraic equations in x of degree n > 2 Now, I would like to generalize this method considering a polynomial function such as P(x) = ∑đ?‘›0 đ?‘Žđ?‘› đ?‘Ľ đ?‘› in x of degree n. We can write P(x) = ∑đ?‘›0 đ?‘Žđ?‘› đ?‘Ľ đ?‘› = 0. From this one we can obtain the following: ∑đ?‘›đ?œ™âˆˆđ?‘‹ đ?‘Žđ?œ™ đ?‘Ľ đ?œ™ đ?œ— = - (∑đ?‘›âˆ’1 đ?œ—∈đ?‘Œ đ?‘Žđ?œ— đ?‘Ľ ) when n is an even positive integer. In this case we have X = (0 , 2, 4, 6, ‌.

, 2u=n) and Y = (1, 3, 5, ‌. , 2u - 1). But when you consider the case n = 2u + 1 for every positive integer u we have the đ?œ— following relation: ∑đ?‘›đ?œ™âˆˆđ?‘‡ đ?‘Žđ?œ™ đ?‘Ľ đ?œ™ = - (∑đ?‘›âˆ’1 đ?œ—∈đ?‘? đ?‘Žđ?œ— đ?‘Ľ ) . In this case we have T = (1, 3, 5, ‌. , n)

and Z = (0, 2, 4, ‌.. , n - 1).

When you consider the first one you can observe that if đ?‘Žđ?œ™ > 0 and đ?‘Žđ?œ— > 0 last equality can be

right

if

x

<

0.

For

x

>

0

this

one

cannot

be

true

because

in

general a positive number such as ∑nĎ•âˆˆT aĎ• x Ď• cannot be equal to a negative number such as Ď‘ - (∑n−1 Ď‘âˆˆZ a Ď‘ x ). In a similar way we can demonstrate that x > 0 cannot be a solution (root)

when you consider the special case for n = 2u + 1 with u an integer positive number. If you consider these cases you can admit it exists one or more numbers, e.g. Ď• or Ď‘ we have đ?‘Žđ?œ— = 0 or đ?‘Žđ?œ™ = 0. In fact if you consider polynomial function such as P(x) = đ?‘Žđ?‘› đ?‘Ľ đ?‘› + đ?‘Žđ?‘›âˆ’1 đ?‘Ľ đ?‘›âˆ’1 + đ?‘Žđ?‘›âˆ’2 đ?‘Ľ đ?‘›âˆ’2 + ‌‌ + đ?‘Ž0 đ?‘Ľ 0 . We consider P(x) = 0 with đ?‘Žđ?‘› ≠0 to have an algebraic equation in x of degree n. From P(x) = đ?‘Žđ?‘› đ?‘Ľ đ?‘› + đ?‘Žđ?‘›âˆ’1 đ?‘Ľ đ?‘›âˆ’1 + đ?‘Žđ?‘›âˆ’2 đ?‘Ľ đ?‘›âˆ’2 + ‌‌ +đ?‘Ž0 đ?‘Ľ 0 we can write (dividing by đ?‘Žđ?‘› ≠0 ) P(x) = đ?‘Ľ đ?‘› + đ?œŒđ?‘›âˆ’1 đ?‘Ľ đ?‘›âˆ’1 + đ?œŒđ?‘›âˆ’2 đ?‘Ľ đ?‘›âˆ’2 + ‌.. + đ?œŒ0 = 0, when đ?œŒđ?‘– = đ?‘Žđ?‘– / đ?‘Žđ?‘–−1 for i positive integer and i ≤ n-1. Thus đ??ź0 = { n-1, n-2, ‌‌. , 0 }. I can consider the case đ?œŒđ?‘– < 0 for one and only one i < n. I consider the case đ?œŒ0 ≠0. Hence đ?‘Ľ đ?‘› + đ?œŒđ?‘›âˆ’1 đ?‘Ľ đ?‘›âˆ’1 + đ?œŒđ?‘›âˆ’2 đ?‘Ľ đ?‘›âˆ’2 +đ?œŒĂŹ+1 đ?‘Ľ ĂŹ+1 + đ?œŒĂŹâˆ’ 1 đ?‘Ľ ĂŹâˆ’1 +‌.. + đ?œŒ0 = |đ?‘Žđ?‘– |đ?‘Ľ đ?‘– for 0 < i < n. I observe đ?œŒđ?‘›âˆ’1 đ?‘Ľ đ?‘›âˆ’1 + đ?œŒđ?‘›âˆ’2 đ?‘Ľ đ?‘›âˆ’2 + ‌.. + đ?œŒ0 > 0 when we consider x > 0. But from it I can write đ?‘Ľ đ?‘› < |đ?œŒđ?‘– |đ?‘Ľ đ?‘– → đ?‘Ľ đ?‘›âˆ’đ?‘– < |đ?œŒđ?‘– | → x < đ?‘›âˆ’đ?‘–√|đ?œŒđ?‘– |. Thus if đ?‘Ľ0 > 0 is a postitive real number to have P(đ?‘Ľ0 ) = 0, called a zero of P(x), I have just obtained đ?‘Ľ0 ∈ (0 ,

đ?‘›âˆ’đ?‘–

√|đ?œŒđ?‘– |). When the solution exists this one is a number conteined in

former open interval. But we can generalize. In fact we can consider the case there are two or more đ?œŒđ?‘˜ < 0. We can consider the case đ?œŒđ?‘– < 0 and đ?œŒđ?‘— < 0. For the same reason generally we have đ?‘Ľ đ?‘› < |đ?œŒđ?‘– |đ?‘Ľ đ?‘– + |đ?œŒđ?‘— |đ?‘Ľ đ?‘— . A fortiori đ?‘Ľ đ?‘› < min ( |đ?œŒđ?‘– |đ?‘Ľ đ?‘– , |đ?œŒđ?‘— |đ?‘Ľ đ?‘— ). Obviously, we can study the case when there are more than two real parameters đ?œŒđ?‘˜ < 0.

publication by Pascal McLee

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