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The Generating function for Legendre Polynomials is:

$$\Phi(x,h)=(1-2xh+h^2)^{-1/2}\quad\text{for}\quad \mid{h}\,\mid\,\lt 1\tag{1}$$ and Legendres' Differential equation is $$\begin{align*} (1-x^2)y^{\prime\prime}-2xy^{\prime}+l(l+1)y=0\tag{2} \end{align*}$$ The question in my text asks me to:

Verify the identity $$\color{blue}{\left(1-x^2\right)\frac{\partial^2 \Phi}{\partial x^2}-2x\frac{\partial\Phi}{\partial x}+h\frac{\partial^2}{\partial h^2}\left(h\Phi \right)=0}\tag{A}$$ by straightforward differentiation of $(1)$ and some algebra.

End of question.


My attempt:

I started by taking partial derivatives of $(1)$ first with respect to $x$ then with respect to $h$: $$\color{#180}{\frac{\partial \Phi}{\partial x}}=-\frac12 \left(1-2xh +h^2 \right)^{-3/2}\cdot(-2h)=\color{red}{h\Phi^3}$$

$$\color{#180}{\frac{\partial^2 \Phi}{\partial x^2}}=-\frac32 h\left(1-2xh+h^2\right)^{-5/2}\cdot (-2h)=\color{red}{3h^2\Phi^5}$$

$$\color{#180}{\frac{\partial \Phi}{\partial h}}=-\frac12 \left(1-2xh +h^2 \right)^{-3/2}\cdot(2h-2x)=\color{red}{(x-h)\Phi^3}$$

$$\begin{align}\color{#180}{\frac{\partial^2 \Phi}{\partial h^2}} & = -\frac32 \left(1-2xh+h^2\right)^{-5/2}\cdot (2h-2x)\cdot(x-h)-\left(1-2xh +h^2 \right)^{-3/2}\\ & =3\left(1-2xh+h^2\right)^{-5/2}\cdot(x-h)^2-\left(1-2xh +h^2 \right)^{-3/2}\\&=3\Phi^5(x-h)^2-\Phi^3\\&=\color{red}{\Phi^3\Big(3\Phi^2(x-h)^2-1\Big)}\end{align}$$

I can see the similarity between $(\mathrm{A})$ and $(2)$; but I am confused how to obtain $(\mathrm{A})$.

Could someone please help me reach equation $(\mathrm{A})$?


EDIT:

Now that the question has been answered a special thanks goes to @Semiclassical for the helpful comment and @enthdegree for explaining to me that $$\begin{array}{rcl}\frac{\partial^2}{\partial h^2} h \Phi &=& \frac{\partial}{\partial h} ( \frac{\partial}{\partial h} h \Phi) \\ &=& \frac{\partial}{\partial h} \left( \Phi + h \frac{\partial}{\partial h} \Phi \right) \\ &=& \frac{\partial}{\partial h} \Phi + \frac{\partial}{\partial h} \Phi + h\frac{\partial^2}{\partial h^2} \Phi \end{array}$$ I would also like to thank @JJacquelin and especially @Markus Scheuer for giving such a well constructed argument. All the answers given were great; but since I could only accept one answer I think Markus gave the most intuitive answer so the answer credit goes to him. In any case I have up-voted all your answers :)


Much to my shame I should have known better that substitution of the $\color{#180}{\mathrm{green}}$ terms into $(\mathrm{A})$ was all that was required to verify the identity $(\mathrm{A})$. Instead of verifying it seems I was trying to derive $(\mathrm{A})$ by noting the striking similarity between $${\left(1-x^2\right)\frac{\partial^2 \Phi}{\partial x^2}-2x\frac{\partial\Phi}{\partial x}+h\frac{\partial^2}{\partial h^2}\left(h\Phi \right)=0}\tag{A}$$ and $$\begin{align*} (1-x^2)y^{\prime\prime}-2xy^{\prime}+l(l+1)y=0\tag{2} \end{align*}$$

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    $\begingroup$ I haven't worked the algebra by hand, but I'd suggest, upon plugging in the green terms, dividing through by $\Phi^5$: what you end up with upon simplifying will involve only $\Phi^-2=1-2xh+h^2$ which should be tractable by hand. $\endgroup$ – Semiclassical Mar 28 '16 at 5:10
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We already know \begin{align*} \frac{\partial\Phi}{\partial x}&=h\Phi^3\qquad&\qquad\frac{\partial\Phi}{\partial h}&=(x-h)\Phi^3\tag{1}\\ \frac{\partial^2\Phi}{\partial x^2}&=3h^2\Phi^5\qquad&\qquad\frac{\partial^2\Phi}{\partial h^2}&=3(x-h)^2\Phi^5-\Phi^3\tag{2}\\ &&\qquad\frac{\partial^2}{\partial x^2}\left(h\Phi\right)&=2\frac{\partial \Phi}{\partial h}+h\frac{\partial^2\Phi}{\partial h^2}\tag{3} \end{align*}

We obtain \begin{align*} (1-x^2)&\frac{\partial^2\Phi}{\partial x^2}-2x\frac{\partial \Phi}{\partial x}+h\frac{\partial ^2}{\partial h^2}\left(h\Phi\right)\\ &=(1-x^2)\frac{\partial^2\Phi}{\partial x^2}-2x\frac{\partial \Phi}{\partial x} +2h\frac{\partial\Phi}{\partial h}+h^2\frac{\partial^2\Phi}{\partial h^2}\tag{4}\\ &=(1-x)^23h^2\Phi^5-2xh\Phi^3+2h(x-h)\Phi^3+3h^2(x-h)^2\Phi^5-h^2\Phi^3\tag{5}\\ &=3h^2(1-x^2+x^2-2hx+h^2)\Phi^5+(-2xh+2xh-2h^2-h^2)\Phi^3\tag{6}\\ &=3h^2\Phi^3-3h^2\Phi^3\\ &=0 \end{align*}

Comment:

  • In (4) we use (3)

  • In (5) we use (1) and (2)

  • In (6) we collect according to powers of $\Phi$

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  • $\begingroup$ Excellent answer; It's very elegant the way you write proofs, compliments :) $\endgroup$ – BLAZE Mar 28 '16 at 23:51
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    $\begingroup$ @BLAZE: Many thanks for your nice comment. Good to see, that answers are sometimes really appreciated! :-) $\endgroup$ – Markus Scheuer Mar 29 '16 at 7:39
  • $\begingroup$ Hi, I have another question (not related to this question); since you answer questions so well I was wondering if you would take a look at it, thanks. $\endgroup$ – BLAZE Apr 9 '16 at 11:13
  • $\begingroup$ @BLAZE: Good question! I do hope my answer helps to clarify things. $\endgroup$ – Markus Scheuer Apr 9 '16 at 15:35
  • $\begingroup$ Yes it did, thank you so much. I have now added an answer of my own and was wondering if you could check it for plausibility? The reason I ask is because I think that my answer may not be rigorous enough as I have only considered the set of functions $\sin(nx)$, where in the link you posted a proof with a general $f(x)$ which I struggle to follow. Is it sufficient proof to just consider the set of functions $\sin(nx)$? Thanks again for your answer which has now been accepted :-) $\endgroup$ – BLAZE Apr 11 '16 at 22:43
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You've done everything right. All that is left to do is plug these formulae into $(1)$ to see that the identity holds. We see that: $$\begin{array}{rl} & \left(1-x^2\right)\frac{\partial^2 \Phi}{\partial x^2}-2x\frac{\partial\Phi}{\partial x}+h\frac{\partial^2}{\partial h^2}\left(h\Phi \right) \\ = & (1-x^2) \cdot (3h^2 \phi^5)-2x \cdot (h\phi^3) + h\cdot \left(2\frac{\partial\Phi}{\partial h}+h\frac{\partial^2\Phi}{\partial h^2}\right) \\ =&(1-x^2) \cdot (3h^2 \phi^5)-2x \cdot (h\phi^3) + h\cdot \left\lbrace2(x-h)\phi^3+h\cdot \phi^3 \cdot [3\phi^2\cdot (x-h)^2-1]\right\rbrace \\ = & -3 h^2 \Phi^3+3 h^2 \Phi^5+3 h^4 \Phi^5-6 h^3 \Phi^5 x \end{array}$$

At this point we can realize that this is $0$ if the following is (by factoring out $3h^2\Phi^3$): $$\begin{array}{rl} & -1+ \Phi^2+ h^2 \Phi^2-2 h \Phi^2 x \end{array}$$

which expanded out equals 0.

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  • $\begingroup$ Thanks for your answer. Since when does $\dfrac{\partial^2}{\partial h^2}\left(h\Phi \right)=\left(2\dfrac{\partial\Phi}{\partial h}+h\dfrac{\partial^2\Phi}{\partial h^2}\right)$? Can you elaborate please? $\endgroup$ – BLAZE Mar 28 '16 at 5:29
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    $\begingroup$ Using the product rule twice: \begin{array}{rcl}\frac{\partial^2}{\partial h^2} h \Phi &=& \frac{\partial}{\partial h} ( \frac{\partial}{\partial h} h \Phi) \\ &=& \frac{\partial}{\partial h} \left( \Phi + h \frac{\partial}{\partial h} \Phi \right) \\ &=& \frac{\partial}{\partial h} \Phi + \frac{\partial}{\partial h} \Phi + h\frac{\partial^2}{\partial h^2} \Phi \end{array} $\endgroup$ – enthdegree Mar 28 '16 at 5:32
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The simplest way is to start from the generating function definition :

enter image description here

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  • $\begingroup$ Thank you for your answer, I appreciate seeing a different approach. $\endgroup$ – BLAZE Mar 29 '16 at 0:14

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