# 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
4 years, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

- 4 years, 4 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?

- 4 years, 4 months 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 $\alpha_2 \rightarrow \alpha_3 \ , \ \alpha_4 \rightarrow \alpha_1$.

- 4 years, 4 months ago

- 4 years, 4 months ago

Oh yes! So is my statement is valid?

- 4 years, 4 months 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.

- 4 years, 4 months ago

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

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

- 4 years, 4 months ago

Its Nihar* not Bihar -_-

- 4 years, 4 months ago

Lol

- 4 years, 4 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!

- 4 years, 4 months ago

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

- 4 years, 4 months ago

Thanks! Modified it :P

- 4 years, 4 months 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

- 4 years, 4 months ago

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

- 4 years, 4 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. . .

- 4 years, 4 months ago

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

- 4 years, 4 months 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. ;)

- 4 years, 4 months 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! $\huge\ddot\smile$

- 4 years, 4 months ago

Great work!

- 4 years, 4 months ago

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

- 4 years, 4 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 !

- 4 years, 4 months ago

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.

- 2 years, 6 months ago