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Let $Q_n(x) = (x^2-1)^n$ and $P_n(x) = Q_n^{(n)}(x)$.

Using Rolle's theorem, prove that $P_n$ has exactly $n$ roots.

Antonio Vargas
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user142205
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  • Can you state Rolle's Theorem? Can you check if your function sassiest it's hypotheses? If so, what conclusion about your function does it offer? – MPW Apr 10 '14 at 10:36
  • Rolle's theorem: Let $f$ be a continuous function on $[a,b]$ and n $f(a) = f(b)$. Then, there is $c \in [a,b]$ such that $f\prime(c) = 0$ – user142205 Apr 10 '14 at 10:39
  • $f$ has to be continuous on $[a,b]$ and differentiable on $]a,b[$, you forgot one of the hypotheses – Alessandro Codenotti Apr 10 '14 at 12:32

3 Answers3

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Notice $Q_n(x)$ has two roots at $-1$ and $1$, both with multiplicity $n$.

When one differentiate $Q_n(x)$, Rolle's theorem tells us $Q'_n(x)$ has a root $x_{11} \in (-1,1)$. $Q'_n(x)$ continue to have two roots at $-1$ and $1$ and the multiplicity there have dropped to $n-1$.

When one differentiate $Q'_n(x)$ again, Rolle's theorem tells us $Q''_n(x)$ has a root $x_{21} \in (0, x_{11})$ and another root $x_{22} \in (x_{11}, 1)$. $Q''_n(x)$ continue to have two roots at $0$ and $1$ but the multiplicity there become $n-2$.

In general, after we differentiate $Q_n(x)$ for $k < n$ times, $Q^{(k)}_n(x)$ will has at least $k$ roots in $(-1,1)$. If we call these roots as $x_{k1}, x_{k2}, \ldots, x_{kk}$ and the corresponding roots of $Q^{(k-1)}(x)$ in $(-1,1)$ as $x_{k-1,1}, x_{k-1,2},\ldots,x_{k-1,k-1}$, they will satisfy

$$0 < x_{k,1} < x_{k-1,1} < x_{k,2} < x_{k-1,2} < \cdots < x_{k-1,k-1} < x_{k,k} < 1$$

It is not hard to see $Q^{(k)}(x)$ has two extra roots at $-1$ and $1$, both with multiplicity $n-k$.

Finally, after the $n^{th}$ differentiation, we find $Q^{(n)}_n(x)$ has at least $n$ roots in $(-1,1)$. These roots will be sandwiched by a set of $n+1$ roots of $Q^{(n-1)}_n(x)$ in $[-1,1]$. Since $Q^{(n)}_n(x)$ is a polynomial of degree $n$, counting multiplicity, it has exactly $n$ roots . This means the $n$ roots of $Q^{(n)}_n(x)$ we obtained above exhaust all the roots of $Q^{(n)}_n(x)$ and they are all simple.

achille hui
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  • Why are you using the phrase "will have at least __ roots"? According to the fundamental theorom of algebra " every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots". Doesn't that contradict with the "at least" part? – noam Azulay Jan 12 '20 at 23:37
  • @noamAzulay Rolle's theorem is concerning about real roots of real differentiable functions. – achille hui Jan 13 '20 at 02:08
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You can see that $P_n$ has at least $n$ roots. Suppose then that it has $n+1$ roots. Use Rolle's Theorem and see that this is wrong. Then $P_n$ has at most $n$ roots.

Haha
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Hint : show that if $Q(X)$ is a polynomial that can be completely factorized as $\prod_{k}(X-a_k)$ where the $a_k$ are not necessarily distinct numbers, then this property also holds for $Q'$ (apply Rolle’s theorem to $\frac{Q'}{Q}$ on suitable intervals).

Ewan Delanoy
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