# Engineering Mathematics Problem with Taylor's Series

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!

• Do you need to expand it in terms of $\frac lx$? Aug 21, 2015 at 11:47
• 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. Aug 21, 2015 at 11:51

There is no need to use Taylor series, just algebra and limits: $$H=\frac{M}{2l}\left(\frac{1}{(x-l)^2}-\frac{1}{(x+l)^2} \right) =\frac{M}{2l}\frac{(x+l)^2-(x-l)^2}{(x^2-l^2)^2} =\frac{M}{2l}\frac{4xl}{x^4(1-(l/x)^2)^2} \approx \frac{2M}{x^3}$$
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
$$H=\color{blue}{\frac{2M}{x^3}}+O\left(\frac{l^2}{x^5}\right). \tag4$$
• The binomial series have quadratic terms $+3u^2$ in both cases. They cancel in the next step, but as stated it is wrong. Dec 16, 2023 at 15:21