$c^3 \ll l^3$ prove that $\sqrt{l\over{l+c}}+\sqrt{l \over{l-c}}=2+{3c^2\over 4l^2} $

Question

If $c^3$ is negligible compared to $l^3$, how may I prove that

$$\sqrt{l\over{l+c}}+\sqrt{l \over{l-c}}=2+{3c^2\over 4l^2}?$$

This might be a problem involving binomial series.

Answer 1

$$\sqrt{\frac{l}{l+c}} + \sqrt{\frac{l}{l - c}} = \left(\frac{l+c}{l}\right)^{-1/2} + \left(\frac{l-c}{l}\right)^{-1/2} = (1 + c/l)^{-1/2} + (1 - c/l)^{-1/2}$$

Answer 2

The $\frac{3}{4}$ looks wrong to me. Maybe I missed something. Anyway, here is a solution We have $$ \sqrt{\frac{l}{l+c}} = \sqrt{1 - \frac{c}{l+c}} \approx 1 - \frac{c}{2(l+c)}. $$ Simlarly $$ \sqrt{\frac{l}{l-c}} = \sqrt{1 + \frac{c}{l-c}} \approx 1 + \frac{c}{2(l-c)}. $$ Summing the two terms gives us $$ \sqrt{\frac{l}{l+c}} + \sqrt{\frac{l}{l-c}} \approx 1 - \frac{c}{2(l+c)} + 1 + \frac{c}{2(l-c)} = 2 + \frac{c^2}{(l^2-c^2)} \approx 2 + \frac{c^2}{l^2} $$

Answer 3

Good catch in identifying the usefulness of binomial series here. Consider:$$\sqrt{\frac{l}{l+c}}=\left(1+\frac{c}l\right)^{-1/2}=\sum_{k=0}^\infty(-1)^k\binom{1/2}k\left(\frac{c}l\right)^k$$and similarly:$$\sqrt{\frac{l}{l-c}}=\left(1-\frac{c}l\right)^{-1/2}=\sum_{k=0}^\infty\binom{1/2}k\left(\frac{c}l\right)^k$$... so it is clear their sum produces:$$\sqrt\frac{l}{l+c}+\sqrt\frac{l}{l-c}=2\sum_{k=0}^\infty\binom{1/2}{2k}\frac{c^{2k}}{l^{2k}}=2+\frac{3c^2}{4l^2}+O\left(\frac{c^4}{l^4}\right)$$For $c\ll l$ we have $|c/l|\ll1$ hence the remaining terms quickly grow insignificant, giving us:$$\sqrt\frac{l}{l+c}+\sqrt\frac{l}{l-c}\sim 2+\frac{3c^2}{4l^2}$$