Average of roots of a polynomial

[apple2357]

Can't seem to find anything online about this, so will be grateful if someone knows anything about it?

If i take a quadratic that has two roots, if i find the average of the two roots, lets call it k, the line x= k is the line of symmetry of the quadratic. This can be seen graphically and through algebra.
I did the same thing for a cubic and the average gives the x coordinate of the point of inflexion ( this surprised me a little), I can't see this graphically but algebraically it falls out.
Is there some general result about quartics or higher-order polynomials and average value of roots ( where they exist in the real sense anyway!)

Thanks

 

[JeffM]

Let's review your quadratic and cubic calculations.

Quadratic.

[MATH]g(x) = a(x - p)(x - q) \implies g'(x) = a\{2x - (p + q)\} \implies g'(x) = 0 \text { iff } \dfrac{p + q}{2} = 0.[/MATH]
The arithmetic mean of the roots of a polynomial of degree 2 is the abcissa of the root of the polynomial's first derivative.

Cubic.

[MATH]g(x) = a(x - p)(x^2 - bx + c) \implies g(x) = 0 \text { if } x = p, \ x = \dfrac{b + \sqrt{b^2 - 4c}}{2}, \text { or } x = \dfrac{b - \sqrt{b^2 - 4c}}{2}.[/MATH]
The arithmetic mean of those roots is (b + p)/3.

[MATH]\text {But } g(x) = a\{x^3 - bx^2 + cx - px^2 + bpx + q = a\{x^3 - (b + p)x^2 + (c + bp)x + q\} \implies[/MATH]
[MATH]g'(x) = a\{3x^2 - 2(b + p)x + r\} \implies g''(x) = a\{6x - 2(b + p)\} \implies g''(x) = 0 \text { iff } x = \dfrac{2(b + p)}{6} = \dfrac{b + p}{3}.[/MATH]
The arithmetic mean of the roots of a polynomial of degree 3 is the abcissa of the root of the polynomial's second derivative.

From the fundamental theorem of algebra, we can express any quartic in the form

[MATH]g(x) = a(x^2 - bx + c)(x^2 - dx + e), \ \text { where } a \ne 0.[/MATH].

[MATH]\therefore g(x) = 0 \text { if } x = \dfrac{b \pm \sqrt{b^2 - 4c}}{2} \text { or } \dfrac{b \pm \sqrt{d^2 - 4e}}{2}.[/MATH]
If we average the first two roots we get [MATH]b.[/MATH]
If we average the second two roots we get [MATH]d[/MATH].

Because complex roots come in conjugate pairs, the averages will always be real.

So, obviously, the average of all four roots is [MATH]\frac{b + d}{4}[/MATH]............... edited

[MATH]\text{But } g(x) = a\{x^4 - (b + d)x^3 + px^2 + rx + q\} \implies g'(x) = a\{4x^3 - 3(b + d)x^2 + px + r\} \implies[/MATH]
[MATH]g''(x) = a\{12x^2 - 6(b + d)x + p\} \implies g'''(x) = 24x - 6(b + d) \implies g'''(x) = 0 \text { iff } x = \dfrac{6(b + d)}{24} = \dfrac{b + d}{4}.[/MATH]
I think we have something to be proved by induction.

The arithmetic mean of the roots of a polynomial of degree n > 1 is the abcissa of the root of the polynomial's (n - 1)^th derivative.

 

[JeffM]

There is a typo in the quartic section. It should say that the average of the roots is (b + d)/4. It is the sum of the roots that is b + d. ...... typo fixed in #2

 

[apple2357]

Thanks, that's the kind of conclusion i was wondering about.

I am left wondering whilst in the quadratic and cubic you can 'see' an interpretation of the average of the roots, does it make sense to think like this with quartics etc ? What i mean is, the average of the roots for a quadratic relates to the line of symmetry, for a cubic - the point of inflection, but what about a quartic?

 

[JeffM]

Thanks, that's the kind of conclusion i was wondering about.

I am left wondering whilst in the quadratic and cubic you can 'see' an interpretation of the average of the roots, does it make sense to think like this with quartics etc ? What i mean is, the average of the roots for a quadratic relates to the line of symmetry, for a cubic - the point of inflection, but what about a quartic?

The answer to that question lies in whether the root of the (n - 1)th derivative of a polynomial of degree n has some general significance about the polynomial. That is, given the roots, we know what the average of the roots locates, but what I at least do not know is the general significance of what has been located. So I know what is true, but I do not know whether it means anything relevant.

Below is a sketch of a proof.

[MATH]\text {Given } n \in \mathbb Z, \ n > 1,\ a \ne 0, \text { and } f(x) = ax^n + bx^{(n-1)} + \sum_{j=0}^{n-2} c_jx^{\{(n - 2) - j\}}.[/MATH]
By induction, show that the (n - 1)th derivative of f(x) is

[MATH]ax * n! + b(n - 1)!.[/MATH]
By the fundamental theorem of algebra, we know

[MATH]f(x) = a * \left ( \prod_{j=1}^n x - r_j \right ), \text { where } f(r_j) = 0.[/MATH]
Now prove, again by induction, that

[MATH]b = -\ a * \left ( \sum_{j=1}^n r_j \right )[/MATH].

From there it is trivial to show that

[MATH]n! * ax + b(n - 1)! = 0 \iff x = \dfrac{1}{n} * \sum_{j=1}^n r_j.[/MATH]

 

[apple2357]

Ok, still thinking about this from almost exactly a year ago... I am now thinking about a cubic polynomial with one real root and two complex roots.
It appears the result of averaging the roots still gives the x coordinate of the point of inflection.
And, can someone help me make sense of the Argand diagram, if i plot the roots ( one real and two complex) on the Argand diagram for a cubic with one real root, does it give any insight?

Take a cubic like f(x)= 2x^3+4x^2+x+2,




What does this argand diagram tell us about how the roots relate to the point of inflection?

 

[JeffM]

I think we are simply reiterating the same conclusion.

When we average these complex roots we get zero. So the average in this case is necessarily -2 * (1/3).

More generally, a cubic with roots at p, q, and r

[MATH]f(x) = a\{x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr\} \implies \\ f’(x) = a\{3x^2 - 2(p + q + r)x + pq + pr + qr\} \implies \\ f’’(x) = a\{6x - 2(p + q + r)\} \implies \\ f’’(x) = 0 \iff x = \dfrac{p + q + r}{3}.[/MATH]Even if q and r are complex roots, we still get a real result because q = s + ti and r = s - ti so the imaginary terms drop out.

However, there is no necessary line of symmetry in polynomials of degree > 2, at least not in the sense of a vertical line. A cubic has such a line of symmetry only if it has one real root. And that line is a vertical through the point of inflection. Thus, in that special case, the mean of the real and complex roots finds the axis of symmetry.

I don’t know a general condition for a polynomial to have an axis of symmetry, but possibly there is one. If so, then you can see whether the mean of the roots determine the location of that axis. Possibly you will have to consider polynomials of odd and even degree as separate cases.

 

[apple2357]

Can you explain what you mean when you talk about a line of symmetry in a cubic?
Can you give me an example?