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

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- Jan 27th 2007, 12:55 PMdadonProof of approximation for function
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 - Jan 27th 2007, 02:01 PMCaptainBlack
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 - Jan 27th 2007, 02:16 PMThePerfectHacker
There is another thing you can do but it is similar to what Captain

**Blank**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.