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Consider the function $$f_n(x)=(x^2-1)^n$$..........(20)

Differentiating this equation we get the second order differential equation, $$(1-x^2)f_n''+2(n-1)xf_n'+2nf_n=0$$..................(22)

We wish to differentiate this n times by use of Leibniz's formula, $$\frac{d^n}{dx^n}A(x)B(x)=\sum^n_{k=0}\frac{n!}{k!(n-k)!}\frac{d^kA}{dx^k}\frac{d^{n-k}B}{dx^{n-k}}$$......................(23)

Applying this to (22) we easily get $$(1-x^2)f_n^{(n+2)}-2xf_n^{(n+1)}+n(n+1)f_n^{(n)}=0$$......................................(24)

which is exactly Lergendre's differential equation (1-49). This equation is therefore satisfied by the polynomials $$y=\frac{d^n}{dx^n}(x^2-1)^n$$.....................(25)

The Legendre polynomials $P_n(x)$ are normalized by the requirement $P_n(1)=1$. Using $$y=2^nn!$$...............(26) for x=1,

We get $$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$...........................(27)

How does the author relate (24) with (25)? Where did he get equation (25)? And how do you do this normalization? Is this normalization the same as that in physics whereby we ensure that the probability does not exceed 1? This is the source.

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  • $\begingroup$ Have you plugged $(25)$ into $(24)$ to see what happens? $\endgroup$ Oct 9, 2016 at 6:10
  • $\begingroup$ I will get the Legendre DE. But where did the author come up with (25)? $\endgroup$
    – newbie125
    Oct 9, 2016 at 6:38
  • $\begingroup$ In many problems in analysis, we proceed by trial and error. $\endgroup$ Oct 9, 2016 at 8:09
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    $\begingroup$ The only term that survives, in the expansion of $\frac{d^n}{dx^n}(x^2-1)^n$ by the Cauchy formula you mentioned above, is when $k=n$, in this case we have $\binom n k=1$, $\frac{d^k}{dx^k}(x-1)^n=n!$ and $\frac{d^{n-k}}{dx^{n-k}}(x+1)^n=(1+1)^n=2^n.$ Do you see it? $\endgroup$ Oct 9, 2016 at 10:20
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    $\begingroup$ It is not equal to $0$. You have $\frac{d^1}{dx^1}(x-1)^1=1$, $\frac{d^2}{dx^2}(x-1)^2=2 \times 1$ (you differentiate twice), $\frac{d^3}{dx^3}(x-1)^3=3\times 2 \times 1$, and so on. $\endgroup$ Oct 9, 2016 at 13:20

1 Answer 1

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The author noticed that $f_n^{(n)}$ is a solution of Legendre's equation, according to (24), as when you take Legendre's equation $(1-x^2)y{''} -2xy{'} + n(n+1)y + 0$ and substitute $y=f_n^{(n)}$, you get (24). Normalizing is not at all like in quantum physics; it is just a simple choice to restrict the values of the functions between -1 and +1. Now apply rule (23) to $$f_n^{(n)} = ((x^2-1)^n)^{(n)} = ((x-1)^n \cdot (x+1)^n)^{(n)}$$, leading to \begin{equation} \begin{split} f_n^{(n)} (x) &= [(x+1)^n\cdot(x-1)^n]^{(n)} \\ &= \sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)} \\ &= \sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k \\ &= n!\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k \end{split} \end{equation}

Plugging in $x=1$, we get $$f_n^{(n)} (1) = n!\binom{n}{0}^2 \cdot 2^n= 2^nn!$$, so in order to get 1 we need to divide $f_n$ by $2^nn!$

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