Relation between roots and derivatives.

Statement

The arithmetic mean of roots of a polynomial in a single variable of degree nn with real and/or complex coefficients whose leading coefficient is not zero will always be the same as the arithmetic mean of roots of its kthk^{th} derivatives (where kN:1k(n1)k \in \mathbb{N} : 1 \le k \le (n-1) ).

Proof

Consider a general polynomial p(x)p(x) of degree nn, with the leading coefficient not equal to 00. Let us say,

p(x)=anxn+an1xn1+an2xn2++a2x2+a1x+a0p(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0

Here, an0a_n \neq 0.

Using Vieta's formulas,

Its sum of roots is given by: an1an\dfrac{-a_{n-1}}{a_n}

The arithmetic mean of its roots is thus: an1nan ()\dfrac{-a_{n-1}}{n a_n} \ (\bigstar)


Let us differentiate this polynomial w.r.t xx once. Let us denote the new polynomial as p1(x)p_1 (x). This polynomial has exactly n1n-1 roots.

p1(x)=p(x)=nanxn1+(n1)an1xn2+(n2)an2xn3++2a2x+a1p_1 (x) = p'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + (n-2) a_{n-2} x^{n-3} + \cdots + 2 a_2 x + a_1

The sum of roots of p1(x)p_1 (x) is: (n1)an1nan\dfrac{-(n-1) a_{n-1}}{n a_n}

The arithmetic mean of roots of p1(x)p_1 (x) is thus: (n1)an1n(n1)an=an1nan ()\dfrac{-(n-1) a_{n-1}}{n(n-1) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)

Which is exactly the same as the arithmetic mean of roots of p(x)p(x).


Let us differentiate it one more time w.r.t xx. Call this polynomial p2(x)p_2 (x). This polynomial has exactly n2n-2 roots.

p2(x)=p(x)=n(n1)anxn2+(n1)(n2)an1xn3+(n2)(n3)an2xn4++2a2p_2 (x) = p''(x) = n(n-1) a_n x^{n-2} + (n-1)(n-2) a_{n-1} x^{n-3} + (n-2)(n-3) a_{n-2} x^{n-4} + \cdots + 2 a_2

The sum of roots of p2(x)p_2 (x) is: (n1)(n2)an1n(n1)an=(n2)an1nan\dfrac{-(n-1)(n-2) a_{n-1}}{n(n-1) a_n} = \dfrac{-(n-2) a_{n-1}}{n a_n}

The arithmetic mean of roots of p2(x)p_2 (x) is thus: (n2)an1n(n2)an=an1nan ()\dfrac{-(n-2) a_{n-1}}{n(n-2) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)

Which is exactly the same as the arithmetic mean of roots of p(x)p(x).


Let us prove this for a general kthk^{th} derivative, where (kN:1k(n1))(k \in \mathbb{N} : 1 \le k \le (n-1)). Call this polynomial pk(x)p_k (x). This polynomial has exactly nkn-k roots.

pk(x)=pk(x)=n(n1)(n2)(nk+1)anxnk+(n1)(n2)(n3)(nk)an1xnk1+(n2)(n3)(n4)(nk1)an2xnk2++k!akp_k (x) = p^{k} (x) = n(n-1)(n-2) \cdots (n-k+1) a_n x^{n-k} + (n-1)(n-2)(n-3)\cdots (n-k) a_{n-1} x^{n-k-1} + (n-2)(n-3)(n-4)\cdots (n-k-1) a_{n-2} x^{n-k-2} + \cdots + k! a_k

OR \\ \text{OR} \\

pk(x)=pk(x)=(nPk)anxnk+(n1Pk)an1xnk1+(n2Pk)an2xnk2++(n(nk)Pk)akp_k (x) = p^{k} (x) = \left(^n P_k\right) a_n x^{n-k} + \left(^{n-1} P_k\right) a_{n-1} x^{n-k-1} + \left(^{n-2} P_k\right) a_{n-2} x^{n-k-2} + \cdots + \left(^{n-(n-k)}P_{k}\right) a_k

Here, nPr^n P_r denotes the permutation function which is equal to n!(nr)!\dfrac{n!}{(n-r)!}.

The sum of roots of pk(x)p_k (x) is: (n1Pk)an1(nPk)an=(nk)an1nan\dfrac{- \left( ^{n-1}P_k \right) a_{n-1}}{ \left( ^n P_k \right) a_n} = \dfrac{-(n-k) a_{n-1}}{n a_n}

The arithmetic mean of roots of pk(x)p_k (x) is thus: (nk)an1n(nk)an=an1nan ()\dfrac{-(n-k) a_{n-1}}{n(n-k) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)

Which is exactly the same as the arithmetic mean of roots of p(x)p(x).


Continuing differentiating further w.r.t. xx, for n1n-1 times yields us a linear polynomial. Let us denote this polynomial as pn1(x)p_{n-1} (x). This polynomial has exactly 11 root.

pn1(x)=pn1(x)=n!anx+(n1)!p_{n-1} (x) = p^{n-1} (x) = n! a_n x + (n-1)!

Here,

Sum of roots of pn1(x)p_{n-1} (x) == Arithmetic Mean of roots of pn1(x)p_{n-1} (x) == The only root of pn1(x)p_{n-1} (x).

Hence, the value of all three is equal to: (n1)!an1n!an=an1nan ()\dfrac{-(n-1)! a_{n-1}}{n! a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)

Which is exactly the same as the arithmetic mean of roots of p(x)p(x).


However, if we differentiate p(x)p(x) one more time, it will yield us a constant which is n!ann! a_n. The following pn(x)p^n (x) will not be a polynomial which has any roots and thus this process fails upon further differentiation.

Therefore, we come to the conclusion that the arithmetic mean of roots of a polynomial in a single variable of degree nn with real and/or complex coefficients and the leading coefficient not equal to zero will always be the same as the arithmetic mean of roots of its kthk^{th} derivatives, where kNk \in \mathbb{N} and 1k(n1)1 \le k \le (n-1).

Let us look at an example to be familiar with one of its useful applications.

Problem

Given that the polynomial f(x)=x4+10x3+35x2+50x+24f(x) = x^4 + 10 x^3 +35 x^2 + 50 x +24. Find the sum of the roots of its 2nd2nd derivative polynomial.

Solution

Instead of differentiating 22 times, we can directly use our mean-value approach.

So, we have,

Mean of roots of f(x)=Mean of roots of f(x)\text{Mean of roots of} \ f(x) = \text{Mean of roots of} \ f''(x)

    101×14=σ2where σ denotes the sum of roots of f(x)\implies \dfrac{-10}{1} \times \dfrac{1}{4} = \dfrac{\sigma}{2} \qquad \qquad \small \color{#3D99F6}{\text{where} \ \sigma \ \text{denotes the sum of roots of} \ f''(x)}

    σ=5 \implies \sigma = -5 \ \Box


We can check our answer as,

f(x)=x4+10x3+35x2+50x+24f(x)=4x3+30x2+70x+50f(x)=12x2+60x+70f(x) = x^4 + 10 x^3 +35 x^2 + 50 x +24 \\ f'(x) = 4 x^3 + 30 x^2 + 70x + 50 \\ f''(x) = 12 x^2 + 60x + 70

Where it is evident that,

Sum of roots of f(x) (σ)=6012=5 \text{Sum of roots of} \ f''(x) \ (\sigma) = \dfrac{-60}{12} = -5 \ \Box

Note by Tapas Mazumdar
3 years ago

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I really loved this proof :) ... keep up the good work. This Also reminded me of my own note . Check it out

Sabhrant . - 3 years ago

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Thank you.

I'm surely gonna read your note right now. :)

Tapas Mazumdar - 3 years ago

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I am sure that tha note is 100%correct.....

md mainu - 2 years, 11 months ago

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Nice work!

Sahil Silare - 3 years ago

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Appreciated your complements.

Tapas Mazumdar - 3 years ago

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(Edit: The concern raised in this comment has been dealt with)

Are you sure that "Mean of roots of f(x) = mean of roots of f''(x)"? That doesn't make sense to me.

E.g. the mean of roots of x2+2x+1=0 x^2 + 2x + 1 = 0 is -1, but the mean of roots of 2=0 2 = 0 is 0 (no roots).


In particular, can you figure out the condition that would make your claim true? Look through your proof, and see what implicit assumptions you have made about the conditions.

Calvin Lin Staff - 3 years ago

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I have mentioned about the condition where the derivative becomes a constant. And also showed that why it wouldn't work in that case.

Also see that 2=02=0 has 00 as the sum of roots and arithmetic mean as 00\dfrac 00, which is undefined. This is based on our false assumption that 2=02=0.

Also, the conditions for my claims are:

  • Roots must be real and there must exist exactly nn roots.
  • Only derivatives till n1n-1 times are to be considered.

In the case of the polynomial x2+2x+1x^2 + 2x + 1, we cannot say this because it doesn't contain exactly 22 real and distinct roots. Both roots are same.

However, let's take for example the polynomial x24x+1x^2 - 4x + 1, the AM of its roots is 22. We can only differentiate this polynomial once from our previous condition, so upon differentiating, we obtain 2x42x-4 whose AM of roots is also 22. We have already differentiated it 21=12-1=1 times and hence, our claim will not satisfy further for all cases.

Tapas Mazumdar - 3 years ago

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Great analysis of your work!

  1. The reason that we only allow for "derivatives till k=n1 k = n-1 times" is because we have a nknk \frac{ n - k} { n-k} term in the expression which needs to cancel out. Hence, it will not work for n=k n = k .

  2. Actually, the roots do not need to be real. And we do not need distinct nn roots, we just need to count with multiplicity. If you look through your proof, none of it requires root roots or distinct roots!

Can you clean up the note further to make this a much clearer read so that someone can easily understand what you are saying? It would be interesting to make a problem related to it too :)

Calvin Lin Staff - 3 years ago

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@Calvin Lin I got your point. Each complex root will have its conjugate and that will cancel the imaginary part of the other one. Furthermore, the coefficients of the polynomial are all real numbers and hence upon further differentiation will remain real and due to Vieta's formulas, the sum of roots will always be real too.

Can you suggest me some tips on how to make this note representable. This is my 2nd note, so I'm basically not used to creating notes.

Tapas Mazumdar - 3 years ago

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@Tapas Mazumdar Actually, we do not even need to use the theorem about complex conjugates. The claim will hold true even for polynomials of complex coefficients! We didn't need to use that fact that ai a _ i was real (and we only needed an0 a_n \neq 0 ).

Here are some suggestions to improve the note:
1. What is it that you want to explain? How can we make this more obvious? Providing clearer sections will allow the reader to understand where you are going.
2. Avoid mixing things up. Make it clear what the objective of each paragraph is. 3. Great that you're using Latex formatting. If you need help with more complex scenarios, let me know

Calvin Lin Staff - 3 years ago

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@Calvin Lin Thanks for your suggestions. I've formatted my note for it to look more presentable.

Tapas Mazumdar - 3 years ago

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@Tapas Mazumdar Great! This is much better! I can now clearly see the statement that you want to prove.

The proof currently only deals with the special case of k=1 k = 1 and n1 n-1 . It would be helpful to write up the more general version.

Calvin Lin Staff - 3 years ago

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@Calvin Lin Actually, I'm not through with binomials as of now. And without it I think for a general kthk^{th} derivative, it would be not such a great looking derivation.

Tapas Mazumdar - 3 years ago

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@Calvin Lin Can you provide me the proof for general kthk^{th} derivative. It would be very helpful.

Tapas Mazumdar - 3 years ago

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@Calvin Lin I got the proof. Making changes. :)

Tapas Mazumdar - 3 years ago

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