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This is a problem from Engineering Mathematics book by K. A. Stroud, 7th edition, Exercise 18, Chapter 12 Further problems. It has been given in a physics manner, but it just requires manipulation of Taylor series to get the result, which is what I can’t figure out. It doesn’t require any physics knowledge really to find the answer, that’s why I posted it on M.S.E. It states:

The field strength of a magnet $(H)$ at a point on the axis, distance $x$ from its center, is given by: $$H=\frac{M}{2l}\left(\frac{1}{(x-l)^2}-\frac{1}{(x+l)^2}\right),$$ where $2l =$ length of magnet and $M =$ moment. Show that if $l$ is very small compared with $x$, then $ H \approx \frac{2M}{x^3} $.

As far as I’m concerned, $H$ is a fraction of $x$ there (but I’m not sure, maybe it’s $H(x,l)$?), so this is $H(x)$. And so i have to find the Taylor series representation of $H(x+l)$. What I get is this: $$H(x+l) = \frac{M}{2l}\left(\frac{1}{(x-l)^2}-\frac{1}{(x+l)^2}\right) + M\left(\frac{1}{(x+l)^3}-\frac{1}{(x-l)^3}\right) + \frac{3Ml}{2}\left(\frac{1}{(x-l)^4}-\frac{1}{(x+l)^4}\right)$$ (since it says $l$ is small, I took only terms until the $x^2$ only).

I really don’t know how to prove what is needed in this. I would be very grateful for any help. Thanks in advance!

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    $\begingroup$ Do you need to expand it in terms of $\frac lx$? $\endgroup$ Aug 21, 2015 at 11:47
  • $\begingroup$ hmm no.. Just to apply the taylor's series expansion (i think): $$ f(x+h) = f(x) + hf'(x) + \frac{h^2}{2!}f''(x) + \frac{h^3}{3!}f'''(x) + ... $$ until the $h^{2}$ term. In general I'm unsure of how to prove what is need here, but i think i need Taylor's series. $\endgroup$
    – KeyC0de
    Aug 21, 2015 at 11:51

2 Answers 2

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There is no need to use Taylor series, just algebra and limits: $$H=\frac{M}{2l}\left(\frac{1}{(x-l)^2}-\frac{1}{(x+l)^2} \right) =\frac{M}{2l}\frac{(x+l)^2-(x-l)^2}{(x^2-l^2)^2} =\frac{M}{2l}\frac{4xl}{x^4(1-(l/x)^2)^2} \approx \frac{2M}{x^3} $$

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By the Taylor series expansion near $0$, say for $|u|<1$, we have $$ \frac{1}{(1-u)^2}=1+2u+O(u^3) \tag1 $$$$ \frac{1}{(1+u)^2}=1-2u+O(u^3) \tag2 $$ giving $$ \frac{1}{(1-u)^2}-\frac{1}{(1+u)^2}=4u+O(u^3) \tag3 $$ then, setting $u:=\dfrac{l}x$, we get from $(3)$, $$ \frac{M}{2l}\left\{\frac{1}{(x-l)^2}-\frac{1}{(x+l)^2}\right\}=\frac{M}{2lx^2}\left\{\frac{1}{(1-l/x)^2}-\frac{1}{(1+l/x)^2}\right\}=\frac{M}{2lx^2}\times \left(4\frac{l}x+O(l/x)^3\right) $$ or

$$ H=\color{blue}{\frac{2M}{x^3}}+O\left(\frac{l^2}{x^5}\right). \tag4$$

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  • $\begingroup$ The binomial series have quadratic terms $+3u^2$ in both cases. They cancel in the next step, but as stated it is wrong. $\endgroup$ Dec 16, 2023 at 15:21

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