Hey MathsForum
If $\displaystyle x$ is small, show that $\displaystyle \sqrt{{\left\{\frac{1+x}{1-x}\right\}}} \approx 1 + x + \frac{x^2}{2}$
Can anyone help?
Thanks
Method 1 - consider the truncated Taylor expansion of
$\displaystyle \sqrt{{\left\{\frac{1+x}{1-x}\right\}}}$
about $\displaystyle 0$
Method 2 - Multiply together the first few terms (and the remainders) of the Taylor series of $\displaystyle \sqrt{1+x}$ and $\displaystyle 1/\sqrt{1-x}$ and discard the terms of order $\displaystyle x^3$ and higher.
We have:
$\displaystyle (1+x)^{1/2}=1+(1/2)x+(1/2)(-1/2)x^2/2+O(x^3)$
and:
$\displaystyle (1-x)^{-1/2}=1+(-1/2)(-x)+(-1/2)(-3/2)(-x)^2/2+O(x^3)$
so:
$\displaystyle \sqrt{{\left\{\frac{1+x}{1-x}\right\}}}=1+x+(1/4)x^2-(1/8)x^2+(3/8)x^2+O(x^3)$$\displaystyle \approx 1+x+x^2/2$
RonL
There is another thing you can do but it is similar to what CaptainBlank said. You can parabolize, remeber in Calculus I you did linearization, that is the best line. Here you do the best parabola, which turns out to be the same coefficients as in the Taylor series.