About Fresnel's paper "Memoir on the diffraction of light" In Fresnell's paper "On the diffraction of light" he says that he express the difference RA+AF-RF of Fig.14 in series and he extract the equation (1).

Figure 14:

Let $\rm F$ be any point on the receiving screen outside the shadow. The difference of path traversed by the direct ray, and by the ray reflected at the edge of the opaque body, and meeting the direct ray at this point, is $\rm RA+AF-RF$. Let us represent $\rm FT$ by $x$ and and express in series the values of $\rm RF$, $\rm AR$, and $\rm AF$. Then, if we neglect all terms involving any power of $x$ or of $c$ higher than the second, since they are very small compared with distances $a$ and $b$, the terms which contain $c$ will disappear and we shall have for the difference of path traversed $$d=\frac{a}{2b(a+b)}x^2\tag{1}$$

I have tried to solve this with Maclaurin  binomial series $$\\(1+x)^m=1+mx+\frac{m(m-1)x^2}2  $$ but I do not conclude the same solution. Is there any ideas how is it solved?

For a more detailed description of the geometry, see this excerpt:


 A: First, we have that the triangles $RAB$ and $RTC$ are similar, so that
$$TC=RC\,\left(\frac{AB}{RB}\right)=(a+b)\,\left(\frac{c/2}{a}\right)=\frac{(a+b)c}{2a}\,.$$
This means
$$\begin{align}
RF&=\sqrt{FC^2+RC^2}=\sqrt{(FT+TC)^2+RC^2}=\sqrt{\left(x+\frac{(a+b)c}{2a}\right)^2+(a+b)^2}
\\
&=(a+b)\,\sqrt{1+\left(\frac{x}{a+b}+\frac{c}{2a}\right)^2}\approx(a+b)\left(1+\frac{1}{2}\,\left(\frac{x}{a+b}+\frac{c}{2a}\right)^2\right)
\\
&=a+b+\frac{x^2}{2(a+b)}+\frac{cx}{2a}+\frac{(a+b)c^2}{8a^2}\,.
\end{align}$$
Next, 
$$\begin{align}
RA&=\sqrt{RB^2+AB^2}=\sqrt{a^2+\left(\frac{c}{2}\right)^2}=a\,\sqrt{1+\left(\frac{c}{2a}\right)^2}
\\&\approx a\,\left(1+\frac{1}{2}\,\left(\frac{c}{2a}\right)^2\right)=a+\frac{c^2}{8a}
\end{align}$$
and
$$\begin{align}
AF&=\sqrt{AM^2+MF^2}=\sqrt{BC^2+(FC-MC)^2}=\sqrt{BC^2+(FC-AB)^2}
\\
&=\sqrt{b^2+\left(x+\frac{(a+b)c}{2a}-\frac{c}{2}\right)^2}=b\,\sqrt{1+\left(\frac{x}{b}+\frac{c}{2a}\right)^2}
\\
&\approx b\,\left(1+\frac{1}{2}\,\left(\frac{x}{b}+\frac{c}{2a}\right)^2\right)
=b+\frac{x^2}{2b}+\frac{cx}{2a}+\frac{bc^2}{8a^2}\,.
\end{align}$$
Finally, we get
$$\begin{align}d&=RA+AF-RF
\\
&\approx\small\left(a+\frac{c^2}{8a}\right)+\left(b+\frac{x^2}{2b}+\frac{cx}{2a}+\frac{bc^2}{8a^2}\right)-\left(a+b+\frac{x^2}{2(a+b)}+\frac{cx}{2a}+\frac{(a+b)c^2}{8a^2}\right)
\\
&=\frac{x^2}{2b}-\frac{x^2}{2(a+b)}=\frac{ax^2}{2b(a+b)}\,.
\end{align}$$
