# Some Generalized stuff by me

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.

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

Note by Nihar Mahajan
5 years, 1 month ago

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I assume you're only solving for cubic polynomials.

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

Next stop: Apply Newton's sum! ｡◕‿◕｡

- 5 years, 1 month ago

Hey, I observed that even $\left\lceil { \pi }^{ \pi }\ln { \pi } \right\rceil$ is the meaning of life.

- 5 years, 1 month ago

Thanks a lot. I will for sure try it for Harmonic progression. And yeah , I will not make Newton's sums feel lonely. $\ddot\smile$.

- 5 years, 1 month ago

This is really cool. Good work, keep it up! :)

- 5 years, 1 month ago

Thanks a lot :)

- 5 years, 1 month ago

@Calvin Lin I will be happy if you read this. :)

- 5 years, 1 month ago

While the (tedious) algebra work has (hopefully) no mistakes, it ends up obscuring the "true beauty" of the equation. Having found the nice relation, you should think about alternative / better ways of proving the result. Often, the first solution (esp a brute force one) is not the best approach. There will be parts that can be discarded, or rewritten.

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

By dividing throughout by $a ^3$, we may assume that the roots are $1, r, r^2$, which means that the polynomial is $A(x-1)(x-r)(x-r^2) = A[x^3 - (1+r+r^2) x^2 + (r+r^2 + r^3 ) x - r^3]$. Hence, we can immediately conclude that $\frac{\alpha_4}{ \alpha_1} = \left( \frac{ \alpha_3}{\alpha_2} \right) ^3$.

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

For a polynomial whose $n$ roots are in geometric progression, what is the relationship between $\frac{\alpha_{n+1} } { \alpha_1 }$ and $\frac{ \alpha_{n } } { \alpha_{2 }}$?
How about $\frac{ \alpha_{n -i } } { \alpha_{2 +i }}$?

Staff - 5 years, 1 month ago

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 $\dfrac{\alpha_{n-i}}{\alpha_{2+i}}$.

- 5 years, 1 month ago

Right, as I said, some steps/explanation is still missing. One of the steps is explaining what "Dividing throughout by $a^3$" means. What this is, is that if the polynomial $h(x)$ has roots $a , ar, ar^2$, then the polynomial $h(ax)$ will have roots of $1, r, r^2$. Thus, we can apply the above to conclude about the coefficients of $h(ax)$, from which that gives us information about $h(x)$. It is not immediately apparent that they have to satisfy the same equation, so we have to verify why $h(x)$ still works.

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 $a_{n-i} / a_{2+i}$.

Staff - 5 years, 1 month ago

Gud... Thanks...

- 5 years, 1 month ago

Welcome buddy!

- 5 years, 1 month ago

Excellent work!

- 5 years, 1 month ago

gud work nihar

- 5 years ago

Let the roots be $x-y,x,x+y$ , by Vieta's relations we have

$3x=\dfrac{-a_2}{a_1}$

But since $x$ is a root of the equation , therefore substitute this value of $x$ in the equation and simplify by to get the desired result.

- 3 years, 3 months ago

No, you're abusing the math notations. You can't have "x" is a root the equation (with variable "x" again). That makes no sense.

You're on the right track, yes, if you let "p-q, p, p + q" be the 3 roots that follow an AP, then yes, you can prove all the results that Nihar has shown with less work.

- 3 years, 3 months ago

Oops...I should have used the same variable..Sorry!!

- 3 years, 3 months ago

I mean ...A different variable

- 3 years, 3 months ago

No worries! The next step from here is probably figure out how to identify whether a cubic equation has real and/or non-real roots.

Here are some relevant articles: Cardano's method and Cubic discriminant.

- 3 years, 3 months ago

I will try be through the articles that you have mentioned here.

Thanks a lot sir.!

- 3 years, 3 months ago

And even the GP case can be done easily with a similar analysis.

- 3 years, 3 months ago