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Let $$f_n(x)=\frac{\mathrm d^n}{\mathrm dx^n}((1-x^2)^n)$$

Any hints on how to show that it has $n$ distinct real roots?

Ali Caglayan
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Kal S.
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  • Arg, I am good at not reading tonight. I thought for some reason we were working with $\prod(k-x^2)$ or something. – Dan Uznanski Aug 22 '14 at 00:00
  • @DanUznanski Regarding your hint, I thought about using this result but I couldn't see how. Because I need to show that $f_{n-1}$ has n+1 distinct real roots.. – Kal S. Aug 22 '14 at 00:02
  • You'd actually have to show that there are n+1 distinct real roots, which is still not happening today. – Dan Uznanski Aug 22 '14 at 00:04
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    It'd be enough to show that if $x=a$ is a root of $f_n(x)$ then it can't be a root of $f_n'(x)$ (and vice versa). Though I don't see an obvious reason why that's true... – Semiclassical Aug 22 '14 at 00:08
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    These are multiples of http://en.wikipedia.org/wiki/Legendre_polynomials . In particular, they are orthogonal polynomial, hence each has full set of different (real) roots. – Quang Hoang Aug 22 '14 at 00:35

1 Answers1

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Show that there exist a relation (check for a similar relation with Legendre polynomials):

$$ f_{n+1}(x) = (2n+1)\, x\, f_n(x) - 2n^2 f_{n-1}(x) $$

Observe that $f_1(x) = -2x$ and $f_2(x) = 2(1-x^2)f_1(x)$ have different nonzero roots.

And now assume all nonzero roots of $f_n$ and $f_{n-1}$ are different, then, if $f_{n+1}(\alpha)=f_n(\alpha)=0$ for some $\alpha\neq 0$, then by the relation above $f_{n-1}(\alpha)=0$, a contradiction.

(Here's a proof of the relation above (or it's equivalent for Legendre polynomials, which differ only by a normalization factor from yours): http://www.phys.ufl.edu/~fry/6346/legendre.pdf)

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