How to prove that $\frac{d^n}{dx^n}(x^2-1)^n=0$ has $n$ real roots?

Question

How do I prove that $$\frac{d^n}{dx^n}(x^2-1)^n=0$$ has $n$ real roots?

Answer 1

Firstly, $\partial(\dfrac{\text{d}^n}{\text{d}x^n}[(x^2-1)^n])=n$, so $\dfrac{\text{d}^n}{\text{d}x^n}[(x^2-1)^n]$ has at most $n$ roots.
Applying Leibniz Identity, $$\forall k<n,k\in\mathbb{N},\quad \dfrac{\text{d}^k}{\text{d}x^k}[(x^2-1)^n]=\sum\limits_{i=0}^k\binom{k}{i}\dfrac{\text{d}^i(x-1)^n}{\text{d}x^i}\dfrac{\text{d}^{k-1}(x+1)^n}{\text{d}x^{k-i}}$$ Thus, $\pm1$ are roots of $\dfrac{\text{d}^k}{\text{d}x^k}[(x^2-1)^n]$ when $k<n$.
Applying Rolle's Theorem, $\dfrac{\text{d}}{\text{d}x}[(x^2-1)^n]$ has roots $\xi_{11}\in(-1,1)$ and $\pm1$.
Applying Rolle's Theorem again, $\dfrac{\text{d}^2}{\text{d}x^2}[(x^2-1)^n]$ has roots $\xi_{21}\in(-1,\xi_{11})$, $\xi_{22}\in(\xi_{11},1)$ and $\pm1$.
Similarly, $\dfrac{\text{d}^{n-1}}{\text{d}x^{n-1}}[(x^2-1)^n]$ has $n+1$ roots $$-1=\xi_{n-1,0}<\xi_{n-1,1}<\dots<\xi_{n-1,n-1}<\xi_{n-1,n}=1$$ At last, Applying Rolle's Theorem again, we get the $n$ roots of $\dfrac{\text{d}^n}{\text{d}x^n}[(x^2-1)^n]$:
$$\xi_{n,1}\in(-1,\xi_{n-1,1}),\dots,\xi_{n,n}\in(\xi_{n-1,n-1},1)$$ Q.E.D.

Answer 2

Hint: Try Induction on $n$.

Also on the above remark, between any two real roots of $f$, there will be a real root of $f'$.

Answer 3

For those who are interested, there is a more general result one can use:

For any polynomial $f(z)$ with complex coefficients, the roots
of $f'(z)$ is contained within the convex hull of the roots of $f(z)$.

Since $p(x) = (x^2-1)^n$ has only 2 distinct roots $\pm 1$, all roots of $p'(x)$ belongs to the line segment joining $\pm 1$. Namely, the interval $[-1,1]$ on the real axis. Apply this theorem $n$ times, we find all roots of $\frac{d^n}{dx^n} p(x)$ are not only real but belong to $[-1,1]$.

The proof of the general case is pretty short, let I give it here.

Let $f(z) = C \prod_i (z - \alpha_i)^{m_i}$ be a complex polynomial with roots $\alpha_i$, each with multiplicity $m_i$.
Let $w$ be a root of $f'(z)$. If $w$ is one of $\alpha_i$, then we are done. If not, we have: $$0 = f'(w)/f(w) = \sum_i \frac{m_i}{w - \alpha_i} = \sum_i \frac{m_i (\bar{w} - \bar{\alpha_i})}{|w - \alpha_i|^2}$$ Taking complex conjugate and rearrange terms, we get $w = \frac{\sum_i \rho_i \alpha_i}{\sum_i \rho_i}$ where $\rho_i = \frac{m_i}{|w - \alpha_i|^2}$ are all real and positive.

Answer 4

Let $P(x)$ be a polynomial of degree $n$. Suppose $P(x)$ has a zero of degree $m$ at $x=a$. This means that $$ P(x)=(x-a)^mQ(x) $$ where $Q(x)$ is a polynomial of degree $n-m$ and $Q(a)\ne0$. Thus, $$ \begin{align} P'(x) &=(x-a)^{m-1}\left[mQ(x)+(x-a)Q'(x)\right]\\ &=(x-a)^{m-1}R(x) \end{align} $$ where $R(x)$ is degree $n-m$ and $R(a)=mQ(a)\ne0$.

That is, each zero of $P(x)$ is reduced by one degree. However, Rolle's Theorem insures that between two adjacent zeros of $P(x)$, $P'(x)$ has a zero.

Let $k$ be the number of zeros of $P(x)$ not counting multiplicities. Differentiation will reduce the degree of each of the $k$ zeros by one, and will add $k-1$ zeros between the $k$ zeros by Rolle's Theorem. Thus, the total number of zeros of $P'(x)$, counting multiplicities, is at least the number of zeros of $P(x)$, counting multiplicities, minus one. New zeros might be introduced to $P'(x)$.

The number of zeros not counting multiplicities of $P'(x)$ is at least $k-1$. In addition, each multiple zero of $P(x)$ persists. Thus, $P'(x)$ has at least $k+m-1$ roots not counting multiplicities where $P(x)$ has $k$ zeros, not counting multiplicities including $m$ multiple zeros.

Counting multiplicities, $(x^2-1)^n$ has $2$ zeros of degree $n$. Therefore, these two roots persist until the $n^{\text{th}}$ derivtive. After $n-1$ derivatives, there should be at least $n+1$ roots with the multiple roots gone. After the $n^{\text{th}}$ derivative, there should be at least $n$ roots, not counting multiplicity. As a degree $n$ polynomial, it has at most $n$ roots. Therefore, it must have exactly $n$ roots.