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.

So first , let's start with the roots in Arithmetic 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 an arithmetic progression.So we have a relation :

\[\Rightarrow 2\gamma = \beta + \lambda \dots (1)\]

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

\[\sigma_1=\beta +\gamma + \lambda = \dfrac{-\alpha_2}{\alpha_1} \dots (2) \\ \sigma_2=\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \dots (3) \\ \sigma_3= \beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \dots (4)\]

Using \((1),(2)\) we have :

\[\beta +\gamma + \lambda = 2\gamma+ \gamma = 3\gamma= \dfrac{-\alpha_2}{\alpha_1} \Rightarrow \gamma = \dfrac{-\alpha_2}{3\alpha_1} \dots (5)\]

Using \((1),(3),(5)\) we have :

\[\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \dots(6) \\ \Rightarrow 2\gamma^2 + \beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow 2\left( \dfrac{-\alpha_2}{3\alpha_1}\right)^2 + \beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \beta\lambda = \dfrac{\alpha_3}{\alpha_1} - \dfrac{2\alpha_2^2}{9\alpha_1^2} \\ \Rightarrow \beta\lambda=\dfrac{9\alpha_1\alpha_3-2\alpha_2^2}{9\alpha_1^2} \dots (7)\]

Using \((4) , (7)\) we have :

\[\dfrac{\sigma_3}{\beta\lambda} = \gamma \\ \Rightarrow \dfrac{\dfrac{-\alpha_4}{\alpha_1}}{\dfrac{9\alpha_1\alpha_3-2\alpha_2^2}{9\alpha_1^2}} = \dfrac{-\alpha_2}{3\alpha_1} \\ \Rightarrow \left(\dfrac{-\alpha_4}{\alpha_1}\right) \times \left(\dfrac{9\alpha_1^2}{9\alpha_1\alpha_3-2\alpha_2^2}\right)=\dfrac{-\alpha_2}{3\alpha_1} \\ \Rightarrow \dfrac{-9\alpha_1^2\alpha_4}{9\alpha_1\alpha_3-2\alpha_2^2} \\ \Rightarrow 27\alpha_1^2\alpha_4=9\alpha_1\alpha_2\alpha_3 - 2\alpha_2^3 \]

Thus , we can write the relation between coefficients as :

\[\Large\boxed{\alpha_4(2\alpha_2^3+27\alpha_1^2\alpha_4) = 9\displaystyle\prod_{i=1}^4 \alpha_i}\]

Now let the roots \(\beta , \gamma , \lambda\) form a Geometric progression.Thus we have the relation :

\[\gamma^2=\beta\lambda \dots (8)\]

Using \((1)\) we have :

\[\beta+\lambda = \dfrac{-\alpha_2}{\alpha_1} - \gamma \Rightarrow \beta+\lambda=-\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) \dots (9)\]

Using \((4),(8)\) we have :

\[\beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \Rightarrow \gamma^3= \dfrac{-\alpha_4}{\alpha_1} \Rightarrow \gamma= \sqrt[3]{ \dfrac{-\alpha_4}{\alpha_1}} \dots (10)\]

Using \((6),(9),(10)\) we will have :

\[\gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow -\gamma\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) + \gamma^2 = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \dfrac{-\alpha_2\gamma}{\alpha_1} - \gamma^2+\gamma^2 = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \dfrac{-\alpha_2\gamma}{\alpha_1} = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma = \dfrac{-\alpha_3}{\alpha_2} \\ \Rightarrow \sqrt[3]{ \dfrac{-\alpha_4}{\alpha_1}} = \dfrac{-\alpha_3}{\alpha_2} \\ \Rightarrow \dfrac{-\alpha_4}{\alpha_1} = \dfrac{-\alpha_3^3}{\alpha_2^3} \\ \alpha_4\alpha_2^3 = \alpha_1\alpha_3^3 \]

This when rearranged properly we get the following relationship between the coefficients:

\[\Large\boxed{\dfrac{\alpha_1}{\alpha_4}=\left(\dfrac{\alpha_2}{\alpha_3}\right)^3}\]

See the extention of this note by clicking here

## Comments

Sort by:

TopNewestI assume you're only solving for cubic polynomials.

Now try for Harmonic progression as well! And post some challenges!

Nice read! Great work!

Next stop: Apply Newton's sum! ｡◕‿◕｡ – Pi Han Goh · 1 year, 10 months ago

Log in to reply

– Archit Boobna · 1 year, 10 months ago

Hey, I observed that even \[\left\lceil { \pi }^{ \pi }\ln { \pi } \right\rceil \] is the meaning of life.Log in to reply

– Nihar Mahajan · 1 year, 10 months ago

Thanks a lot. I will for sure try it for Harmonic progression. And yeah , I will not make Newton's sums feel lonely. \(\ddot\smile\).Log in to reply

This is really cool. Good work, keep it up! :) – Prasun Biswas · 1 year, 10 months ago

Log in to reply

– Nihar Mahajan · 1 year, 10 months ago

Thanks a lot :)Log in to reply

@Calvin Lin I will be happy if you read this. :) – Nihar Mahajan · 1 year, 10 months ago

Log in to reply

For example, let's take the Geometric progression case. Here is a nice simple solution:

Now, I have still hidden the beauty of the solution. If you truly understand it, the following problem has a "one line" solution.

– Calvin Lin Staff · 1 year, 9 months agoLog in to reply

My thought:Its a good idea to assume the roots as \(1,r,r^2\).But one can argue that we get the relationship \(\dfrac{\alpha_4}{\alpha_1} = \left(\dfrac{\alpha_3}{\alpha_2}\right)^3\) only if one of the roots is \(1\).What about when it is not equal to \(1\)? There is no guarantee that the relationship remains the same for all possible roots.For a polynomial whose \(n\) roots are in geometric progression, I think \(\dfrac{\alpha_{n+1}}{\alpha_1} = \left(\dfrac{\alpha_n}{\alpha_2}\right)^n\). Am I right?

Also I didn't understand what do you want me to find about \(\dfrac{\alpha_{n-i}}{\alpha_{2+i}}\). – Nihar Mahajan · 1 year, 9 months ago

Log in to reply

Of course, another approach would be to check that

\[ (x-a)(x-ar)(x-ar^2) = x^3 - (a +ar+ar^2) x^2 + (a^2r + a^2r^2 + a^2r^3) - ar^3. \]

from which the identity follows. However, this is still "tedious magical algebra", instead of getting to the "actual reason why this relationship should exist" which would allow us to easily generalize to \(a_{n-i} / a_{2+i} \). – Calvin Lin Staff · 1 year, 9 months ago

Log in to reply

gud work nihar – Idris Muhammed · 1 year, 9 months ago

Log in to reply

Excellent work! – Swapnil Das · 1 year, 10 months ago

Log in to reply

Gud... Thanks... – Abhay Agnihotri · 1 year, 10 months ago

Log in to reply

– Nihar Mahajan · 1 year, 10 months ago

Welcome buddy!Log in to reply