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?

## Comments

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TopNewest@Calvin Lin @Pi Han Goh @Prasun Biswas I did it as you wanted :) Cheers! – Nihar Mahajan · 2 years, 3 months ago

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– Pi Han Goh · 2 years, 3 months ago

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

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 \(\alpha_2 \rightarrow \alpha_3 \ , \ \alpha_4 \rightarrow \alpha_1\). – Nihar Mahajan · 2 years, 3 months ago

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@Pi Han Goh Please reply whether this is correct or not. – Nihar Mahajan · 2 years, 3 months ago

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– Sravanth Chebrolu · 2 years, 3 months ago

Oh yes! So is my statement is valid?Log in to reply

of Harmonic Progression. But, I doubt whether this statement gives the complete explanation. – Sravanth Chebrolu · 2 years, 3 months agoreciprocalLog in to reply

P.S:keep watching, and stay tuned to brilliant.org :P – Sravanth Chebrolu · 2 years, 3 months agoLog in to reply

– Nihar Mahajan · 2 years, 3 months ago

I hope you don't troll me in that problem :P :3 .Post it soon. \(\ddot\smile\).Log in to reply

this, hope you enjoy it! – Sravanth Chebrolu · 2 years, 3 months ago

Yes! I have finally posted it! SeeLog in to reply

– Sravanth Chebrolu · 2 years, 3 months 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. . .Log in to reply

EDIT:Please go through Nihar's suggestion. :) – Sravanth Chebrolu · 2 years, 3 months agoLog in to reply

– Nihar Mahajan · 2 years, 3 months ago

Its Nihar* not Bihar -_-Log in to reply

– Azhaghu Roopesh M · 2 years, 3 months ago

LolLog in to reply

– Sravanth Chebrolu · 2 years, 3 months ago

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

– Nihar Mahajan · 2 years, 3 months ago

Thanks. I will be equally satisfied if you posted just "Cheers". So Cheers!!!Log in to reply

– Sravanth Chebrolu · 2 years, 3 months ago

Thanks! Modified it :PLog in to reply

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. ;) – Prasun Biswas · 2 years, 3 months ago

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Well then,I might be wrong too! \(\huge\ddot\smile\) – Sravanth Chebrolu · 2 years, 3 months ago

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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. – Ankit Kumar Jain · 5 months, 2 weeks ago

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Great work! – Azhaghu Roopesh M · 2 years, 3 months ago

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– Nihar Mahajan · 2 years, 3 months ago

Thanks!!! Sorry I didn't notice your comment.Log in to reply

– Azhaghu Roopesh M · 2 years, 3 months ago

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