The question that follows is needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L$$ are the associated Legendre functions on $[−1, 1]$.

The question in my textbook asks me to

Show that $$\begin{align}\require{enclose}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L&=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\end{align}$$ Where $L\,\text&\,m\, \text{are constants}$ and $0\leq m\leq L$.

Hint: Write $$(x^2-1)^L=(x-1)^L(x+1)^L$$ and find the derivatives by Leibniz' rule.

So this is what I have tried:

$$\begin{align}\require{enclose}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L&=\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}(x^2-1)^L \\&=\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}(x-1)^L(x+1)^L\quad\quad\longleftarrow\bbox[#F8A]{\text{Using the Hint}} \\&=(x-1)^L\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}(x+1)^L +(L-m)\frac{\mathrm{d}}{\mathrm{d}x}\left(x-1\right)^L\frac{\mathrm{d}^{L-(m+1)}}{\mathrm{d}x^{L-(m+1)}}(x+1)^L \\&\phantom{Abcde}+\frac{(L-m)(L-[m+1])}{2}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\left(x-1\right)^L\frac{\mathrm{d}^{L-(m+2)}}{\mathrm{d}x^{L-(m+2)}}(x+1)^L+\ldots\,.\end{align}$$ But this could go on forever and I have no idea how to evaluate (or simplify) terms like $$\frac{\mathrm{d}^{L-(m+2)}}{\mathrm{d}x^{L-(m+2)}}(x+1)^L\,.$$

Is there any chance someone could please give me some hints or advice on how to show that $$\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L=\color{#180}{\fbox{$\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L$}}$$ by starting at one side of the equation and showing that it is equal to the other side?

Best Regards.


We show the following is valid for $0\leq m\leq L$ \begin{align*} \frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L =\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L \end{align*}

We apply the Leibniz Product Rule \begin{align*} (f\cdot g)^{(n)}=\sum_{j=0}^n\binom{n}{j}f^{(j)}g^{(n-j)} \end{align*} to $(x^2-1)^L=(x+1)^L(x-1)^L$ according to the hint.

We obtain \begin{align*} &\frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L\\ &\quad=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L\tag{1}\\ &\quad=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{2}\\ &\quad=\frac{(L-m)!}{(L+m)!} \sum_{j=0}^{L-m}\frac{(L+m)!}{j!(L-m-j)!}\cdot\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{3}\\ &\quad=\frac{(L-m)!}{(L+m)!} \sum_{j=m}^{L}\frac{(L+m)!}{(j-m)!(L-j)!}\cdot\frac{L!}{(L+m-j)!}(x+1)^{L+m-j}\frac{L!}{j!}(x-1)^{j}\tag{4}\\ &\quad=\frac{(L-m)!}{(L+m)!}(x^2-1)^m \sum_{j=m}^{L}\binom{L+m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(j-m)!}(x-1)^{j-m}\tag{5}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=m}^{L}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{6}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=0}^{L+m}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{7}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L\tag{8}\\ \end{align*} and the claim is finished.


  • In (1) we apply the Leibniz product rule.

  • In (2) we do the differentiation which is not too hard since we have powers of linear factors only \begin{align} \frac{d^j}{dx^j}(x+1)^L &= L(L-1)(L-2)\cdots(L-j+1)(x+1)^{L-j}\\ &=\frac{L!}{(L-j)!}(x+1)^{L-j}\\ \frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L &= \frac{L!}{(L-(L-m-j))!}(x-1)^{L-(L-m-j)}\\ &=\frac{L!}{(m+j)!}(x-1)^{m+j} \end{align}

  • In (3) we factor out $(L-m)!$ from $\binom{L-m}{j}$ and expand the expression with $\frac{(L+m)!}{(L+m)!}$.

  • In (4) we shift the index $j$ to start from $m$ and substitute in the expression $j\rightarrow j-m$ accordingly.

  • In (5) we factor out $(x+1)^m(x-1)^m=(x^2-1)^m$ and rearrange the factorials.

  • In (6) we write the expression using derivatives.

  • In (7) we extend the range of $j$ to $0\leq j \leq L+m$ without changing anything, since we are just adding zeros.

More detailed: Since $(x+1)^L$ and $(x-1)^L$ are polynomials in $x$ of degree $L$ we get \begin{align*} \frac{d^j}{dx^j}(x+1)^L&=0\qquad\qquad L<j\leq L+m\\ \frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L&=0\qquad\qquad 0\leq j<m\\ \end{align*}

  • In (8) we use Leibniz' rule again.
  • $\begingroup$ @BLAZE: I've updated it in a way which seems to be more convenient for you (mathematically it's the same). To your other question: It's as simple as the answer of siddharth-venu. The answer of AndrewDHwang brings into play two different aspects, which is from my point of view not helpful. He moans about "two column style" proofs. My original answer wasn't of this type. He gives as example a contradictory expression ($-1=1$) and doesn't always perform equivalence relation (squaring) so that information get's lost and can't be reverted. This example addresses different things. $\endgroup$
    – epi163sqrt
    Apr 29 '16 at 14:54
  • $\begingroup$ @BLAZE: I've removed my now outdated comments. Regarding your question about upvote: From my point of view the top answer is clearly that from siddharth-venu. Regarding your question: Why my answer get's no more upvotes? I don't know. In fact I have some answers like this one which was much more time-consuming and of which I think provides helpful insight, but it never saw an upvote. ... Maybe in the future somebody will find it useful and valuable for an upvote. Regards, $\endgroup$
    – epi163sqrt
    Apr 29 '16 at 15:02
  • $\begingroup$ Hi again, sorry it took a while to respond. I have now read your proof and the only part I don't understand is (7). You mention that $\sum_{j=m}^{j=L}\rightarrow\sum_{j=0}^{j=L+m}$ and nothing else changes since we are just adding zeros. But suppose I have $\sum\limits{_{x=2}^{x=5}}x^2$: this is not the same sum as $\sum\limits{_{x=0}^{x=5+2}}x^2$. Clearly I'm missing the point; Could you please elaborate on this in your answer? Many thanks. - Removed my outdated comments as well :-) $\endgroup$
    – BLAZE
    May 2 '16 at 13:13
  • $\begingroup$ @BLAZE: Hint added. The point is we have polynomials of degree $L$ and at least one of the polynomials is differentiated more than $L$ times, so that always a factor $0$ occurs. So we add only zeros. :-) $\endgroup$
    – epi163sqrt
    May 2 '16 at 13:30
  • $\begingroup$ That makes sense but only if $L\ge 0$ & $m\ge 0$ which I am guessing they are, right? Can you please access your list of edits and post as an answer edit # 9. It was your latest version for proving the formula by transforming both sides. It will be useful to see both of your proofs to this question for comparison. It will also serve as a useful example of why the answer to this question is yes. I would do this myself, but it wouldn't be fair. Many thanks, as ever! :) $\endgroup$
    – BLAZE
    May 2 '16 at 16:01

Here's a hint: evaluating $$\frac{\mathrm{d}^{L-(m+2)}}{\mathrm{d}x^{L-(m+2)}}(x+1)^L$$ is easier than you think.

You can figure out what happens when you differentiate $f(t)=t^L$ repeatedly, right? You get $f'(t)=L t^{L-1}$, $f''(t)=L(L-1) t^{L-2}$, $f'''(t)=L(L-1)(L-2)t^{L-3}$, and so on. And $L(L-1)(L-2)$ can be written as $L!/(L-3)!$ if you prefer that, and so on. Now just write down what the pattern would be if you did it $L$ times, then replace $t$ by $x+1$ and $L$ by $L-(m+2)$.

  • $\begingroup$ Thank you so much for giving such a helpful hint; regards. $\endgroup$
    – BLAZE
    Apr 26 '16 at 8:53

Note: Please find here a former variation of my current answer. OP was asking to republish it to make the connection with some of OPs comments and his related question plausible.

Both answers are more or less the same. Originally I've decided to apportion it into two somewhat smaller pieces to enhance readability. The current answer was only written since OP was explicitely asking for a one sided approach.

Here we transform the left hand side to get a convenient representation. Then we transform the right hand side to obtain the same representation and equality is shown.

The following is valid for $0\leq m\leq L$ \begin{align*} \frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L =\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L \end{align*}

We apply the Leibniz Product rule \begin{align*} (f\cdot g)^{(n)}=\sum_{j=0}^n\binom{n}{j}f^{(j)}g^{(n-j)} \end{align*} according to the hint to $(x^2-1)^L=(x+1)^L(x-1)^L$.

We obtain from the LHS \begin{align*} \frac{d^{L-m}}{dx^{L-m}}&\left(x^2-1\right)^L\\ &=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L\tag{1}\\ &=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{2}\\ &=(L-m)!\sum_{j=0}^{L-m}\binom{L}{j}\binom{L}{m+j}(x+1)^{L-j}(x-1)^{m+j}\tag{3} \end{align*}


  • In (1) we apply the Leibniz product rule

  • In (2) we do the differentiation which is not too hard since we have powers of linear factors only \begin{align*} \frac{d^j}{dx^j}(x+1)^L&=L(L-1)(L-2)\cdots(L-j+1)(x+1)^{L-j}\\ &=\frac{L!}{(L-j)!}(x+1)^{L-j} \end{align*}

  • In (3) we factor out $(L-m)!$ from $\binom{L-m}{j}$ and rearrange the factorials to get nice binomial coefficients.

And now the RHS \begin{align*} &\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=0}^{L+m}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{1}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=m}^L\binom{L+m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(j-m)}!(x-1)^{j-m}\tag{5}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=0}^{L-m}\binom{L+m}{j+m}\frac{L!}{(L-j-m)!}(x+1)^{L-j-m}\frac{L!}{j!}(x-1)^j\tag{6}\\ &\quad=\frac{(L-m)!}{(L+m)!} \sum_{j=0}^{L-m}\binom{L+m}{j+m}\frac{L!}{(L-j-m)!}(x+1)^{L-j}\frac{L!}{j!}(x-1)^{j+m}\tag{7}\\ &\quad=(L-m)!\sum_{j=0}^{L-m}\binom{L}{j}\binom{L}{m+j}(x+1)^{L-j}(x-1)^{m+j}\tag{8} \end{align*} and the claim follows.


  • In (5) we do it similarly to (2). Note, that we also shrink the range of the index variable to $m\leq j\leq L$, since for all other values of $j$ the differentiation give zero.

  • In (6) we shift the index $j$ to start from $0$ and substitute in the expression $j\rightarrow j+m$

  • In (7) we apportion $(x^2-1)^m=(x+1)^m(x-1)^m$

  • In (8) we cancel $(L+m)!$ and reorder the factorials similarly to (3)


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