# Showing this approximation holds

We see from the formula $(1+t)^2=1+2t+t^2$ that for small $t$, we have the approximate equality $$(1+t)^2\approx 1+2t$$ hence for small $u$, we have $$\sqrt{1+u} \approx 1+\frac u2$$

I know that I could get that second approximation using the Taylor series, but this book implies that it follows directly from the first approximation. I don't see how. Could something explain how we can see that $$(1+t)^2\approx 1+2t \implies \sqrt{1+u} \approx 1+\frac u2$$

• It is imperative to quantify what $\approx$ means before performing any rigorous proof. – Zhanxiong Nov 22 '15 at 4:18
• See binomial series. – Lucian Nov 22 '15 at 4:53

Let $u = 2t$ (so: $t = \frac{u}{2}$), substitute into the first approximation, square root both sides, and then flip the sides.