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Suppose there is a cubic polynomial in x with roots A,B,C and another cubic polynomial (in t) with roots
1/(A-1), 1/(B-1), 1/(C-1)

is to be found.

My text book mentions two ways of doing this. One is to use vieta's relations, which is time consuming. Another way mentioned is to let t=1/(x-1), make x the subject, and then substitute and simplify. However, I cant really wrap my mind around this. Just because this is the relation between the roots doesnt mean that all values of x and t are in thay relation, does it?

Moreover, if I try this method with more generality i.e trying to find a polynomial with roots f(A),f(B),f(C) where f(u) is some invertible function like e^u, it appears to fail.

Could someone please elaborate on the proper method of tackling such questions.

Vulgar Mechanick
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    Friend, Vieta's relations, though time-consuming as it seems to you, is one of the easiest ways you can use to solve your problem. – Spectre Oct 02 '20 at 05:04
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    Another thing you can do (if you know $A$,$B$ and $C$) is find $\frac{1}{A - 1}, \frac{1}{B - 1}, \frac{!}{C - 1}$ , take the factors of the required polynomial as $(t - a)(t - b)(t - c) [a = \frac{1}{A - 1}, b = \frac{1}{B - 1}, etc.]$ and then multiply them. – Spectre Oct 02 '20 at 05:07

2 Answers2

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Suppose you had the cubic polynomial $P(x) = ax^3 + bx^2 + cx + d = 0$ for which you had three real or complex roots $x_1, x_2, x_3$, then you can say that $P(x_i) = ax_i^3 + bx_i^2 + cx_i + d = 0$ for $i = 1,2,3$.

Now say you wanted to construct a new polynomial equation that had roots $y_1, y_2, y_3$ where $y_i = \frac 1{x_i-1}$ (contingent on the assumption none of the roots of the original is $1$, of course), then as your reference said, you can change the subject to $x_i$, i.e. $x_i = 1 + \frac 1{y_i}$ and substitute that into the original equation. You know that $P(x_i) = 0$, so that implies that $P(1+\frac 1{y_i}) = 0$ also. By manipulation, you can show that this new equation is a cubic $Q(y_i) = 0$. By the explicit mapping that you've already derived, each of the three roots $y_i$ of this new cubic will correspond to each of the three roots $x_i$ of the original cubic.

Certainly, this method will work for any invertible transform as long as you are careful about domain and existence considerations, but you may not get a polynomial equation after transformation (like in your case with the exponential function).

Deepak
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  • How can you determine if a certain transformation will lead to a polynomial? Is the linear rational function just a special case, and the correct way of finding a POLYNOMIAL with transformed roots using vieta's relations? – Vulgar Mechanick Oct 02 '20 at 05:57
  • Any algebraic function (excluding transcendental functions) of a variable will definitely give you a polynomial after the transform. So even a transform like $\frac{\sqrt x + 5x}{x^6 -\frac 2{x^3}}$ is guaranteed to give you a polynomial after substitution and manipulation. This is because any such transform will always map an algebraic number to another algebraic number (i.e. it will not return a transcendental). However, unless the transform is invertible, you may not have an easy mapping of the roots, and you cannot be sure of the degree of the resulting polynomial until you do the algebra. – Deepak Oct 02 '20 at 06:12
  • @OVERWOOTCH If $P(x)=0$ is a polynomial equation and we make a transformation $x=r(y),$ where $r(y)$ is a rational function, then the resulting equation is also a polynomial equation. – Allawonder Oct 02 '20 at 06:17
  • Can you give a example in which invertible is not easily possible and we need to take help of vietas or something else to get the polynomial ? – Orion_Pax Apr 06 '22 at 03:34
  • @Deepak Sir ... – Orion_Pax Apr 06 '22 at 03:35
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Let $A$, $B$ and $C$ be roots of the polynomial $x^3+ax^2+bx+c.$

Thus, $$A+B+C=-a,$$ $$AB+AC+BC=b,$$ $$ABC=-c$$ and we obtain: $$\sum_{cyc}\frac{1}{A-1}=\frac{\sum\limits_{cyc}(A-1)(B-1)}{\prod\limits_{cyc}(A-1)}=\frac{b+2a+3}{-c-b-a-1},$$ which gives a coefficient before $x^2$ in new polynomial: $$\frac{2a+b+3}{a+b+c+1}.$$ Also, $$\sum_{cyc}\frac{1}{(A-1)(B-1)}=\frac{\sum\limits_{cyc}(A-1)}{\prod\limits_{cyc}(A-1)}=\frac{-a-3}{-a-b-c-1}=\frac{a+3}{a+b+c+1}$$ and $$\prod_{cyc}\frac{1}{A-1}=\frac{1}{-a-b-c-1}$$ and we got the answer: $$x^3+\frac{2a+b+3}{a+b+c+1}x^2+\frac{a+3}{a+b+c+1}x+\frac{1}{a+b+c+1}.$$ We used the Viete's theorem and this is a right method for solving your problem.

Also, you can see that by using a cyclic summation it turns out easy enough.

I think the second method takes more time.

Let $x=\frac{1}{y-1}$, where $y$ is a root if the polynomial $x^3+ax^2+bx+c.$

Thus, since $y=\frac{x+1}{x},$ we obtain: $$\left(\frac{x+1}{x}\right)^3+a\left(\frac{x+1}{x}\right)^2+b\left(\frac{x+1}{x}\right)+c,$$ which gives the same result, but with a bit of more computations.

Michael Rozenberg
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  • You didnt even read my question. I know how to do this with vieta's relations. – Vulgar Mechanick Oct 02 '20 at 05:53
  • @OVERWOOTCH I read. You are welcome to ask your questions. I used the Viete's theorem and this is a right method for solving your problem. Also, you can see that by using a cyclic summation it turns our easy. – Michael Rozenberg Oct 02 '20 at 05:58