Why can sqrt(1+x) be approximated by 1+x/2 for small x?

[de1irious]

For small x, it seems sqrt(1+x) can be approximated by 1+x/2. Why exactly is this? Is there a theorem that I can refer to? Some kind of infinite series where the x^4 power term dies out?

Thanks!

 

[MaWM]

Sure. It's exactly what you said. It's just the binomial theorem and the binomial expansion. Remember that for small x, x^4 is much smaller than x^2 and can be neglected if an approximation is desired.

 

[Gib Z]

Or you could think of it even more easily. Draw a rough sketch of the graph. And then draw the graph of 1 + x/2. I think you will notice something :)

 

[de1irious]

nvm this is easy. thanks for the help

 

[ObsessiveMathsFreak]

You can also get the taylor series of the function about 0, but as has already been mentioned, this would be given by the inverse power binomial series anyway.

 

[Kummer]

Let $$f(x) = \sqrt{1+x}$$. Let $$-r<x<r$$ ($$0<r<1$$). The Taylor Series of $$f(x)$$ centered at 0 is:
$$1 + \frac{x}{2} - \frac{x^2}{4}+\frac{3x^3}{8}-...$$.
The Lagrange Remainder as you posted is,
$$R_1(x) = \frac{f''(y)}{2!}x^2$$ for some $$y$$ between $$0$$ and $$x\not =0$$.
Now,
$$|R_1(x)| = \left| \frac{f''(y)}{2!}x^2\right| \leq \left|-\frac{1}{4} \cdot \frac{(1+y)^{-3/2}}{2!}\right|r^2\leq \frac{r^2}{8}$$

The term $$\frac{r^2}{8}$$ determines the accuracy. Say we want your approximation to work for 2 decimal points then we require that $$\frac{r^2}{8} \leq .009$$ thus $$r\leq .26$$. Which means if you pick a fourth, that is, $$r=.25$$ then on the interval $$(-.25,.25)$$ your approximation must be accurate to at least two decimal places.