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.