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.


So first , let's start with the roots in Arithmetic Progression.

Consider a polynomial : α1x3+α2x2+α3x+α4\alpha_1x^3+\alpha_2x^2+\alpha_3x+\alpha_4.

Let its roots be β,γ,λ\beta , \gamma , \lambda such that they form an arithmetic progression.So we have a relation :

2γ=β+λ(1)\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:

σ1=β+γ+λ=α2α1(2)σ2=βγ+βλ+λγ=α3α1(3)σ3=βγλ=α4α1(4)\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)(1),(2) we have :

β+γ+λ=2γ+γ=3γ=α2α1γ=α23α1(5)\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)(1),(3),(5) we have :

βγ+βλ+λγ=α3α1γ(β+λ)+βλ=α3α1(6)2γ2+βλ=α3α12(α23α1)2+βλ=α3α1βλ=α3α12α229α12βλ=9α1α32α229α12(7)\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)(4) , (7) we have :

σ3βλ=γα4α19α1α32α229α12=α23α1(α4α1)×(9α129α1α32α22)=α23α19α12α49α1α32α2227α12α4=9α1α2α32α23\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 :

α4(2α23+27α12α4)=9i=14αi\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 β,γ,λ\beta , \gamma , \lambda form a Geometric progression.Thus we have the relation :

γ2=βλ(8)\gamma^2=\beta\lambda \dots (8)

Using (1)(1) we have :

β+λ=α2α1γβ+λ=(α2α1+γ)(9)\beta+\lambda = \dfrac{-\alpha_2}{\alpha_1} - \gamma \Rightarrow \beta+\lambda=-\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) \dots (9)

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

βγλ=α4α1γ3=α4α1γ=α4α13(10)\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)(6),(9),(10) we will have :

γ(β+λ)+βλ=α3α1γ(α2α1+γ)+γ2=α3α1α2γα1γ2+γ2=α3α1α2γα1=α3α1γ=α3α2α4α13=α3α2α4α1=α33α23α4α23=α1α33\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:

α1α4=(α2α3)3\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
4 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!

Nice read! Great work!

Next stop: Apply Newton's sum! 。◕‿◕。

Pi Han Goh - 4 years, 1 month ago

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

Archit Boobna - 4 years, 1 month ago

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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.

Nihar Mahajan - 4 years, 1 month ago

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

Prasun Biswas - 4 years, 1 month ago

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Thanks a lot :)

Nihar Mahajan - 4 years, 1 month ago

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

Nihar Mahajan - 4 years, 1 month ago

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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 a3 a ^3 , we may assume that the roots are 1,r,r2 1, r, r^2 , which means that the polynomial is A(x1)(xr)(xr2)=A[x3(1+r+r2)x2+(r+r2+r3)xr3] 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 α4α1=(α3α2)3 \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 nn roots are in geometric progression, what is the relationship between αn+1α1 \frac{\alpha_{n+1} } { \alpha_1 } and αnα2 \frac{ \alpha_{n } } { \alpha_{2 }} ?
How about αniα2+i \frac{ \alpha_{n -i } } { \alpha_{2 +i }} ?

Calvin Lin Staff - 4 years ago

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My thought: Its a good idea to assume the roots as 1,r,r21,r,r^2.But one can argue that we get the relationship α4α1=(α3α2)3\dfrac{\alpha_4}{\alpha_1} = \left(\dfrac{\alpha_3}{\alpha_2}\right)^3 only if one of the roots is 11.What about when it is not equal to 11? There is no guarantee that the relationship remains the same for all possible roots.

For a polynomial whose nn roots are in geometric progression, I think αn+1α1=(αnα2)n\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 αniα2+i\dfrac{\alpha_{n-i}}{\alpha_{2+i}}.

Nihar Mahajan - 4 years ago

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@Nihar Mahajan Right, as I said, some steps/explanation is still missing. One of the steps is explaining what "Dividing throughout by a3 a^3 " means. What this is, is that if the polynomial h(x) h(x) has roots a,ar,ar2 a , ar, ar^2 , then the polynomial h(ax) h(ax) will have roots of 1,r,r2 1, r, r^2 . Thus, we can apply the above to conclude about the coefficients of h(ax) h(ax) , from which that gives us information about h(x) h(x) . It is not immediately apparent that they have to satisfy the same equation, so we have to verify why h(x)h(x) still works.

Of course, another approach would be to check that

(xa)(xar)(xar2)=x3(a+ar+ar2)x2+(a2r+a2r2+a2r3)ar3. (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 ani/a2+ia_{n-i} / a_{2+i} .

Calvin Lin Staff - 4 years ago

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Gud... Thanks...

Abhay Agnihotri - 4 years, 1 month ago

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Welcome buddy!

Nihar Mahajan - 4 years, 1 month ago

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Excellent work!

Swapnil Das - 4 years ago

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gud work nihar

idris muhammed - 4 years ago

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@Nihar Mahajan @Pi Han Goh What about this?

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

3x=a2a13x=\dfrac{-a_2}{a_1}

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

Ankit Kumar Jain - 2 years, 2 months ago

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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.

Pi Han Goh - 2 years, 2 months ago

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

Ankit Kumar Jain - 2 years, 2 months ago

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@Ankit Kumar Jain I mean ...A different variable

Ankit Kumar Jain - 2 years, 2 months ago

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@Ankit Kumar Jain 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.

Pi Han Goh - 2 years, 2 months ago

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@Pi Han Goh I will try be through the articles that you have mentioned here.

Thanks a lot sir.!

Ankit Kumar Jain - 2 years, 2 months ago

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

Ankit Kumar Jain - 2 years, 2 months ago

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