To demonstrate that the $n$-th derivative of the polynomial $(x^2 - 1)^n$ has all real roots that are simple (i.e., no repeated roots), we can approach it by analyzing the nature of the roots of successive derivatives of this polynomial.

Let's proceed step-by-step.

1. Define the Polynomial

Let: $f(x) = (x^2 - 1)^n$ where $f(x)$ is a polynomial of degree $2n$ with roots at $x = \pm 1$, each of multiplicity $n$.

2. Properties of Successive Derivatives and Real Zeros

By the properties of differentiable functions and the Rolle's Theorem, we know that:

• The $n$-th derivative of $f(x)$, denoted $f^{(n)}(x)$, will have degree $2n - n = n$.
• Due to the specific structure of $(x^2 - 1)$ and the multiple roots at $x = \pm 1$, each successive derivative reduces the multiplicity of the roots.

3. Applying Rolle's Theorem for Real Roots and Simplicity

• Induction Step: For the initial function $f(x) = (x^2 - 1)^n$, we know it has $2n$ roots, all at $x = \pm 1$.
• Each differentiation step reduces the multiplicity of the roots at $x = \pm 1$.
• By the time we reach the $n$-th derivative, $f^{(n)}(x)$, the function will have $n$ distinct roots, all real, and located in the interval $(-1, 1)$.

4. Applying Sturm's Theorem for Simplicity of Roots

Sturm's Theorem states that for real-rooted polynomials of successive derivatives, the roots will alternate between the intervals. The $n$-th derivative $f^{(n)}(x)$ of $f(x) = (x^2 - 1)^n$ has exactly $n$ distinct roots in $(-1, 1)$, meaning they are simple (not repeated).

Conclusion

Thus, the $n$-th derivative of $f(x) = (x^2 - 1)^n$ has all real roots, which are simple and lie in $(-1, 1)$.

Would you like more details or have any questions? Here are some related questions:

1. How does Rolle's theorem specifically help in proving the real roots of derivatives?
2. Could we use another method to show that these roots are real?
3. What if the initial polynomial had complex roots? How would this affect derivatives?
4. What does it mean for roots to be simple in mathematical terms?
5. How does the structure of $(x^2 - 1)$ ensure all roots lie in $(-1, 1)$?

Tip: Differentiation of polynomials with symmetrical forms often reveals patterns in root behavior, especially with powers of binomials like $(x^2 - 1)^n$.