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Claim

Let $P_n: \Bbb R \mapsto \Bbb R $, $n \in \Bbb N\cup\{0\}$

$P_n(x):={1\over{2^n{n!}}}{d^n\over dx^n}[(x^2-1)^n]$, then $P_n$ has n distinct roots in ]-1, 1[

Proof

$1$. Base

When $n=1$, $P_1(x) = {1 \over 2}{d \over dx}[x^2-1]=x$ and $P_1(x)$ has $1$ distinct root. Thus,

$Claim$ holds for the case of $n=1$.

$2$. Now assume the $Claim$ holds for the case of n=k. Then,

$P_k(x):={1\over{2^k{k!}}}{d^k\over dx^k}[(x^2-1)^k]$, then $P_k$ has k distinct roots in ]-1, 1[ .

Since $P_k(1)=P_k(-1)=0$, there $\exists c \in \Bbb N\cup\{0\}$ s.t. ${d \over dx}P_k(c)=0$ in $]-1,1[$ (by Rolle's Theorem).

Let $c$ be the root of $P_k(x)$. Then

$P_k(c)={1\over{2^k{k!}}}{d^k\over dc^k}[(c^2-1)^k]=0$

Question From abvoe proof, I had stucked in the middle of showing that $c$ is not one of the $n$ distinct roots of $P_k(x)$

any adivce or hint to proceed?

Question' Or, following the comment, I had found

$P_{k+1} = {1\over{2^k{k!}}}{d^k\over dx^k}[(x^2-1)^kx]$ which is slightly different from $P_k(x):={1\over{2^k{k!}}}{d^k\over dx^k}[(x^2-1)^k]$.

How could I use the fact that $P_k(x)$ has n distinct root and conclude $P_{k+1}(x)$ has one more root other than that of $P_k(x)$?

Beverlie
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    To use induction you need some kind of relation between $P_k$ and $P_{k+1}$, possibly (likely?) also involving the derivatives. Have you found one? Recurrence formulas for orthogonal polynomials $Q_n$ (if this is a sequence of one) often lead to simple proofs for something that could be described as interlacing property of zeros (= the zeros of $Q_k$ are between those of $Q_{k+1}$). May be that can be done here as well? – Jyrki Lahtonen May 14 '17 at 04:45
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    Probably worth noting that these are the so-called Legendre polynomials, generated according to Rogrigues' formula. These indeed are a sequence of orthogonal polynomials as @JyrkiLahtonen suggested (specifically with respect to the $L^2$-norm on $[-1,1]$). – Semiclassical May 14 '17 at 05:11
  • Hint: For $0 \le k \le n$, let $S_k$ be the statement "$\frac{d^k}{dx^k} (1-x^2)^{\color{red}{n}}$ has $k$ simple roots in $(-1,1)$ and two roots of multiplicity $n-k$ at $\pm 1$". Repeat applying Rolle's theorem to show $S_0 \implies S_1 \implies \cdots \implies S_n$. – achille hui May 14 '17 at 05:30
  • @achille hui That statement is exactly what I want to show. But the problem I have is can not generally prove your statement because of the complicay of too many terms are coming out – Beverlie May 14 '17 at 06:00
  • @jackerysmith you need three things. 1) Rolle's theorem, 2) if $f(x)$ has a root of multiplicity $m$ at $a$, then $f'(x)$ has a root of multiplicity at least $m-1$ at $a$. 3), the sum of multiplicity of roots of a polynomial is bounded from above by its degree. Statement $S_0$ is trivial, try to do $S_0 \implies S_1$ using above 3 facts. Once you understand how to do that, the argument for the rest is similar. – achille hui May 14 '17 at 06:09
  • @achillehui now I am following you guidance, but for instance in case of k=2, I can easily show that ${d \over dx^2}(x^2-1)^n$ has two roots of multiplicity n-2 and also at least one more roots by Rolle's theorem, however i) how can I show that this root from Rolle's is simple and how to find ii)one more root? – Beverlie May 18 '17 at 00:23
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    $\frac{d}{dx}(x^2-1)^n$ has three roots, one root with multiplicity $n-1$ at $x_{10} = -1$, one root with multiplicity $1$ at $x_{11} \in (-1,1)$, the last one with multiplicity $n-1$ at $x_{12} = 1$. Apply Rolle's theorem to $[x_{10},x_{11}]$ and $[x_{11},x_{12}]$. You get two roots for $\frac{d^2}{dx^2}(x^2-1)^n$ at $x_{21}$ and $x_{22}$ which satisfy $x_{10} < x_{21} < x_{11} < x_{22} < x_{12}$. – achille hui May 18 '17 at 01:27
  • @achillehui duoxie – Beverlie May 18 '17 at 01:31
  • @achillehui would you help me with another legendary example - https://math.stackexchange.com/questions/2285812/for-legendary-equation-p-nx-1-x2p-nx-2xp-nx-nn1p-n want to know above linked equation holds any special meaning – Beverlie May 18 '17 at 01:37

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