# Some generalized stuff by me (extension) - (2)

All of us have solved many problems on cubic polynomials who have their roots in Arithmetic Progression and Geometric Progression.So I have found a relationship between the coefficients of these polynomials which makes these roots special in this note.

Now by suggestion of Pi Han Goh , I have found a relationship between the coefficients of a cubic polynomial whose roots form an Harmonic Progression.

Consider a polynomial : $\alpha_1x^3+\alpha_2x^2+\alpha_3x+\alpha_4$.

Let its roots be $\beta , \gamma , \lambda$ such that they form a harmonic progression.So we have a relation :

$\Rightarrow \gamma = \dfrac{2\beta\lambda}{\beta + \lambda} \dots (1) \\ \Rightarrow \beta+\lambda = \dfrac{2\beta\lambda}{\gamma} \dots(2) \\ \Rightarrow \gamma(\beta+\lambda) = 2\beta\lambda \dots (3)$

Now our favorite polynomial tool cum friend that is , Vieta's formula will help us to have some more relations:

$\sigma_1=\beta +\gamma + \lambda = \dfrac{-\alpha_2}{\alpha_1} \dots (4) \\ \sigma_2=\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \dots (5) \\ \sigma_3= \beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \dots (6)$

Using $(3),(5)$ we have:

$\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow 2\beta\lambda + \beta\lambda= \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow 3\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\\Rightarrow \beta\lambda = \dfrac{\alpha_3}{3\alpha_1} \dots (7)$

Using $(6),(7)$ , we have:

$\beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \\\Rightarrow \dfrac{\alpha_3}{3\alpha_1} \gamma = \dfrac{-\alpha_4}{\alpha_1} \\ \Rightarrow \gamma = \dfrac{-\alpha_4(3\alpha_1)}{\alpha_1\alpha_3} \\ \Rightarrow \gamma = \dfrac{-3\alpha_4}{\alpha_3} \dots (8)$

Using $(2) , (4) , (7) , (8)$ , we have:

$\beta +\gamma + \lambda = \dfrac{-\alpha_2}{\alpha_1} \\\Rightarrow \dfrac{2\beta\lambda}{\gamma}+\gamma = \dfrac{-\alpha_2}{\alpha_1} \\\Rightarrow \dfrac{\left(\dfrac{2\alpha_3}{3\alpha_1}\right)}{ \dfrac{3\alpha_4}{\alpha_3}} - \dfrac{-3\alpha_4}{\alpha_3} = \dfrac{-\alpha_2}{\alpha_1} \\\Rightarrow \dfrac{-2\alpha_3^2}{9\alpha_1\alpha_4} - \dfrac{3\alpha_4}{\alpha_3} = \dfrac{-\alpha_2}{\alpha_1} \\\Rightarrow \dfrac{2\alpha_3^3+27\alpha_1\alpha_4^2}{9\alpha_1\alpha_3\alpha_4} = \dfrac{\alpha_2}{\alpha_1}$

This when cross multiplied we get the relation:

$\Large\boxed{ 2\alpha_1\alpha_3^3 + 27(\alpha_1\alpha_4)^2 = 9\displaystyle\prod_{i=1}^4 \alpha_i}$

My Observation: Since I have found this out for a cubic polynomial , for this reasons are $9,27$ are appearing in the relation? So for a $n^{th}$ degree polynomial , will $n^2 , n^3$ or more of $n^a$ types will appear in relation?

Note by Nihar Mahajan
6 years ago

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@Calvin Lin @Pi Han Goh @Prasun Biswas I did it as you wanted :) Cheers!

- 6 years ago

Did you notice that the final answer (in the box) is very similar to the one in arithmetic progression? Can you explain why?

- 6 years ago

If $a,b,c$ form an Arithmetic Progression , then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ form a Harmonic progression.May be this is responsible for the similarity.

If you observe what is changed you will see that the term $2\alpha_2^3\alpha_4$ got changed to the term $2\alpha_1\alpha_3^3$. So we see that $\alpha_2 \rightarrow \alpha_3 \ , \ \alpha_4 \rightarrow \alpha_1$.

- 6 years ago

- 6 years ago

Oh yes! So is my statement is valid?

- 6 years ago

I think it is because, Arithmetic Progression is nothing but the reciprocal of Harmonic Progression. But, I doubt whether this statement gives the complete explanation.

- 6 years ago

Well done Nihar! Here's a huge $\huge CHEERS!$

EDIT:Please go through Nihar's suggestion. :)

- 6 years ago

Its Nihar* not Bihar -_-

- 6 years ago

Lol

Ha ha! I am Sorry. . . My bad! the smart phone I'm using changes everything! And the small screen did the damage too!

- 6 years ago

Thanks. I will be equally satisfied if you posted just "Cheers". So Cheers!!!

- 6 years ago

Thanks! Modified it :P

- 6 years ago

Also I created a funny problem on you and me, while travelling I'll post that soon check that out!

P.S: keep watching, and stay tuned to brilliant.org :P

- 6 years ago

I hope you don't troll me in that problem :P :3 .Post it soon. $\ddot\smile$.

- 6 years ago

Sorry, due to signal problems I am not able to do much. Also I am sorry if I'd wasted you time, I'll surely post it within an hour or so and don't forget to check it out. . .

- 6 years ago

Yes! I have finally posted it! See this, hope you enjoy it!

- 6 years ago

You should keep in mind that, with increase in the degree of the polynomial considered, there will be increase in the number of coefficients of the terms of the polynomial too. If you really think that there might be some kind of pattern, try repeating the same procedure for a degree 4 polynomial and see if you get some pattern which agrees with this obtained result for degree $3$ polynomial.

If you think any such pattern exists (which I highly doubt since I think the relation will become more and more uncertain with increase in degree of the polynomial), try making a claim for the relation of coefficients of the degree $n$ polynomial and verify using induction on $n$.

As I mentioned, I doubt there's a simple general result for higher degree polynomials. But, then again, I might be wrong. ;)

- 6 years ago

Yes sir, I agree with you, I tried doing it for a bi quadratic polynomial, though I didn't finish it (my bad laziness!), I think the result may not be obtained. . .

But, then again, I might be wrong.

Well then,I might be wrong too! $\huge\ddot\smile$

- 6 years ago

Great work!

Thanks!!! Sorry I didn't notice your comment.

- 6 years ago

No worries bro . Do you have any plans on writing a paper on such results ? If so , I'm eager to read it !

Call the roots $\dfrac1{x-y} , \dfrac1x , \dfrac1{x+y}$ , first find product of roots taken two at a time , then product of all roots and then divide the two to obtain the value of $\dfrac1x$ . Satisfy the equation with this value as $\dfrac1x$ is a root of the equation. Clearing off fractions yields the desired result.

- 4 years, 2 months ago