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# Relation between roots and derivatives.

Statement

The arithmetic mean of roots of a polynomial in a single variable of degree $$n$$ 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 $$k^{th}$$ derivatives (where $$k \in \mathbb{N} : 1 \le k \le (n-1)$$ ).



Proof

Consider a general polynomial $$p(x)$$ of degree $$n$$, with the leading coefficient not equal to $$0$$. Let us say,

$p(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0$

Here, $$a_n \neq 0$$.

Using Vieta's formulas,

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

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

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

$p_1 (x) = p'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + (n-2) a_{n-2} x^{n-3} + \cdots + 2 a_2 x + a_1$

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

The arithmetic mean of roots of $$p_1 (x)$$ is thus: $$\dfrac{-(n-1) a_{n-1}}{n(n-1) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)$$

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

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

$p_2 (x) = p''(x) = n(n-1) a_n x^{n-2} + (n-1)(n-2) a_{n-1} x^{n-3} + (n-2)(n-3) a_{n-2} x^{n-4} + \cdots + 2 a_2$

The sum of roots of $$p_2 (x)$$ is: $$\dfrac{-(n-1)(n-2) a_{n-1}}{n(n-1) a_n} = \dfrac{-(n-2) a_{n-1}}{n a_n}$$

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

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

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

$p_k (x) = p^{k} (x) = n(n-1)(n-2) \cdots (n-k+1) a_n x^{n-k} + (n-1)(n-2)(n-3)\cdots (n-k) a_{n-1} x^{n-k-1} + (n-2)(n-3)(n-4)\cdots (n-k-1) a_{n-2} x^{n-k-2} + \cdots + k! a_k$

$\\ \text{OR} \\$

$p_k (x) = p^{k} (x) = \left(^n P_k\right) a_n x^{n-k} + \left(^{n-1} P_k\right) a_{n-1} x^{n-k-1} + \left(^{n-2} P_k\right) a_{n-2} x^{n-k-2} + \cdots + \left(^{n-(n-k)}P_{k}\right) a_k$

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

The sum of roots of $$p_k (x)$$ is: $$\dfrac{- \left( ^{n-1}P_k \right) a_{n-1}}{ \left( ^n P_k \right) a_n} = \dfrac{-(n-k) a_{n-1}}{n a_n}$$

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

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

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

$p_{n-1} (x) = p^{n-1} (x) = n! a_n x + (n-1)!$

Here,

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

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

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

However, if we differentiate $$p(x)$$ one more time, it will yield us a constant which is $$n! a_n$$. The following $$p^n (x)$$ 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 $$n$$ 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 $$k^{th}$$ derivatives, where $$k \in \mathbb{N}$$ and $$1 \le k \le (n-1)$$.



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

Problem

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



Solution

Instead of differentiating $$2$$ times, we can directly use our mean-value approach.

So, we have,

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

$$\implies \dfrac{-10}{1} \times \dfrac{1}{4} = \dfrac{\sigma}{2} \qquad \qquad \small \color{blue}{\text{where} \ \sigma \ \text{denotes the sum of roots of} \ f''(x)}$$

$$\implies \sigma = -5 \ \Box$$

We can check our answer as,

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

Where it is evident that,

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

Note by Tapas Mazumdar
4 weeks 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 · 3 weeks, 6 days ago

Thank you.

I'm surely gonna read your note right now. :) · 3 weeks, 6 days ago

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 $$x^2 + 2x + 1 = 0$$ is -1, but the mean of roots of $$2 = 0$$ 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. Staff · 4 weeks ago

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

Also, the conditions for my claims are:

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

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

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

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

2. Actually, the roots do not need to be real. And we do not need distinct $$n$$ 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 :) Staff · 3 weeks, 6 days ago

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. · 3 weeks, 6 days ago

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 $$a _ i$$ was real (and we only needed $$a_n \neq 0$$).

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 Staff · 3 weeks, 6 days ago

Thanks for your suggestions. I've formatted my note for it to look more presentable. · 3 weeks, 6 days ago

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 $$k = 1$$ and $$n-1$$. It would be helpful to write up the more general version. Staff · 3 weeks, 6 days ago

I got the proof. Making changes. :) · 3 weeks, 6 days ago

Can you provide me the proof for general $$k^{th}$$ derivative. It would be very helpful. · 3 weeks, 6 days ago

Actually, I'm not through with binomials as of now. And without it I think for a general $$k^{th}$$ derivative, it would be not such a great looking derivation. · 3 weeks, 6 days ago

Nice work! · 4 weeks ago