How can I show $(1+4e^2)^{1/2}\approx 1+2e^2$ for $e<<1$?

Question

I tried using Taylor expansion, here's what I did: $$(1+4e^2)^{1/2}=1+\frac{4e^2}{(1+4e^2)^{1/2}}+...,$$ but I got nowhere from here. Are there any hints on how to solve it?

Answer 1

Writing $4e^2$ is more confusing than just writing $x$. The point is that you are expanding the function $(1+x)^{1/2}$ about $x=0$, and then plugging in $x=4\epsilon^2$. (I decline to use $e$ for this purpose because in standard notation $e$ is reserved for Euler's number, aka the base of the natural logarithm). Taylor expansion gives $1+\frac{x}{2}+o(x)$, since $\left. \frac{d}{dx}(1+x)^{1/2} \right |_{x=0}=1/2.$

Answer 2

For any $x>0$, $\sqrt{1+x}<1+\frac{x}{2}$ is trivial by squaring, but $$ 1+\frac{x}{2}-\sqrt{1+x} = \frac{\left(1+\frac{x}{2}\right)^2-(1+x)}{1+\frac{x}{2}+\sqrt{1+x}}< \frac{x^2}{8\sqrt{1+x}} $$ hence $$\boxed{\,1+\frac{x}{2}-\frac{x^2}{8\sqrt{1+x}}< \sqrt{1+x}<1+\frac{x}{2}\,}$$ Now, just replace $x$ by $4e^2$.