StatementThe 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 \(k \in \mathbb{N} : 1 \le k \le (n-1)\) ).

\(\)

ProofConsider 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.*

ProblemGiven 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.

\(\)

SolutionInstead 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,

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

## Comments

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TopNewestI really loved this proof :) ... keep up the good work. This Also reminded me of my own note . Check it out – Sambhrant Sachan · 5 months, 4 weeks ago

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I'm surely gonna read your note right now. :) – Tapas Mazumdar · 5 months, 4 weeks ago

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I am sure that tha note is 100%correct..... – Md Mainu · 4 months, 1 week ago

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Nice work! – Sahil Silare · 6 months ago

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– Tapas Mazumdar · 6 months ago

Appreciated your complements.Log in to reply

(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 \( x^2 + 2x + 1 = 0 \) 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. – Calvin Lin Staff · 6 months ago

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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:

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. – Tapas Mazumdar · 6 months ago

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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 \( n = k \).

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 :) – Calvin Lin Staff · 6 months ago

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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 · 6 months ago

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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 · 6 months ago

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– Tapas Mazumdar · 6 months ago

Thanks for your suggestions. I've formatted my note for it to look more presentable.Log in to reply

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. – Calvin Lin Staff · 6 months ago

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– Tapas Mazumdar · 6 months ago

I got the proof. Making changes. :)Log in to reply

– Tapas Mazumdar · 6 months ago

Can you provide me the proof for general \(k^{th}\) derivative. It would be very helpful.Log in to reply

– Tapas Mazumdar · 6 months 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.Log in to reply