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From the Generating function for Legendre Polynomials:

$$\Phi(x,h)=(1-2xh+h^2)^{-1/2}\quad\text{for}\quad \mid{h}\,\mid\,\lt 1$$

My text states that:

For $x=1$ $$\Phi(1,h)=\color{red}{(1-2h+h^2)^{-1/2}=\frac{1}{1-h}}=1+h+h^2+\cdots$$

My question is about the justification of the equality marked $\color{red}{\mathrm{red}}$.

Since although $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((1-h)(1-h)\Big)^{-1/2}$$$$=\Big((1-h)^2\Big)^{-1/2}=\color{#180}{\frac{1}{1-h}}=1+h+h^2+\cdots\tag{1}$$

as required.

I could also write $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((h-1)(h-1)\Big)^{-1/2}$$$$=\Big((h-1)^2\Big)^{-1/2}=\color{blue}{\frac{1}{h-1}}=\frac{1}{h}\left(1-\frac{1}{h}\right)^{-1}\ne 1+h+h^2+\cdots\tag{2}$$

Why is it that $\Phi(1,h)$ is equal to $(1)$ but not equal to $(2)$? I think the answer lies somewhere with the fact that $\bbox[yellow]{\mid \,h\mid\,\lt 1}$ but I'm not quite sure how to justify it completely.

You may be wondering why I'm asking such a question, as one user pointed out that given my rep I should know this answer easily. The fact of the matter is I do not know the answer. I'm just looking for a simple English explanation as to why $\Phi(1,h)\ne \color{blue}{\dfrac{1}{h-1}}$. Although I usually prefer hints; I was given so many hints on this question but none of them helped. So if someone could literally spell out the answer to me I would be most grateful.

Thank you.

BLAZE
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  • When you take a square root, there are two possible signs. Which root do you (or your book) want? – noobermin Mar 23 '16 at 03:47
  • @noobermin Are you sure about that? Only the principle root is selected so it will be the positive root. So the book doesn't need to state which root (for this reason the book does not mention which root) – BLAZE Mar 23 '16 at 03:50
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    @BLAZE Uhm, doesn't that answer your question then? It's just chosen such that $\Phi$ is positive. I'm confused then what the issue is. – noobermin Mar 23 '16 at 03:57

2 Answers2

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Of course for real $x$, we have $\big(x^2\big)^{-1/2} = \,\mid x \,\mid^{-1}$ using the principal square root. In your case, it is assumed that $\mid \, h\mid \, \lt 1$, which tells you $$ \left|\frac{1}{1-h}\right| = \frac{1}{1-h} $$ but $$ \left|{\frac{1}{h-1}}\right| = -{\frac{1}{h-1}} $$

BLAZE
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GEdgar
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    If $|h|<1$, then $-1 < h < 1$, so $1-h > 1-1 > 0$, so $1/(1-h) > 0$. [Assuming $h$ is supposed to be real.] – GEdgar Mar 25 '16 at 12:56
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    $$|h|<1 \implies h < 1 \implies -1 < -h \implies 0 < 1-h \implies 0 < \frac{1}{1-h}$$ – Lee Mosher Mar 25 '16 at 12:58
  • @GEdgar Perfect answer; thanks very much. By the way, in your comment did you mean to write $1-h\gt h-h \gt 0$ instead of $1-h \gt 1-1 \gt 0$? – BLAZE Mar 29 '16 at 00:25
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    @BLAZE ... that would have been OK, but I meant $h < 1 \Longrightarrow -h > -1 \Longrightarrow 1-h > 1-1$. – GEdgar Mar 29 '16 at 00:49
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For $$\mid\, h \mid\lt 1$$ $$\implies -1\lt h \lt 1 $$ $$\implies 1\gt h \gt -1$$ $$\implies 1-h\gt h-h\gt 0$$ Since $$1-h\gt 0$$ $$\implies \frac{1}{1-h}\gt 0$$ $$\implies \left|\frac{1}{1-h}\right| = \bbox[#AFF]{\frac{1}{1-h}}$$

For $$\mid\, h \mid\lt 1$$ $$\implies -1\lt h \lt 1 $$ $$\implies h-1\lt 1-1 \lt 0$$

Since $$h-1\lt 0$$ $$\implies \frac{1}{h-1}\lt 0$$ $$\implies \left|{\frac{1}{h-1}}\right| = -{\frac{1}{h-1}} = \bbox[#AFF]{\frac{1}{1-h}}$$ as before.

So $$\Phi(1,h)=\underbrace{\Big((1-h)^2\Big)^{-1/2}=\left|\frac{1}{1-h}\right|}_{\color{red}{\Large\text{Using}\, \big(x^2\big)^{-1/2}\, = |x|^{-1}}}=\bbox[#AFA]{\frac{1}{1-h}}=1+h+h^2+\cdots$$

Also $$\Phi(1,h)=\underbrace{\Big((h-1)^2\Big)^{-1/2}=\left|\frac{1}{h-1}\right|}_{\color{red}{\Large\text{Using}\, \big(x^2\big)^{-1/2} =\, |x|^{-1}}}=\underbrace{-\frac{1}{h-1}=\bbox[#AFA]{\frac{1}{1-h}}}_{\large\color{#F80}{\text{By the argument shown above}}}=1+h+h^2+\cdots$$

So to summarize; writing $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((1-h)(1-h)\Big)^{-1/2}=\Big((1-h)^2\Big)^{-1/2}$$ or

$$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((h-1)(h-1)\Big)^{-1/2}=\Big((h-1)^2\Big)^{-1/2}$$ both result in the same answer of $$\bbox[yellow]{\frac{1}{1-h}}$$ as they should do.

A special thanks goes to @GEdgar and @A.S. for giving me the knowledge necessary to make this answer.

BLAZE
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