Where does this approximation come from?

[fereopk]

$$\frac{\sqrt{1-a}}{\sqrt{1-b}}\approx \left ( 1-\frac{1}{2}a\right )\left ( 1+\frac{1}{2}b\right )$$

I know that the binomial approximation is first used,

$$\frac{\sqrt{1-a}}{\sqrt{1-b}}\approx \frac{1-\frac{1}{2}a}{1-\frac{1}{2}b}$$

But how does one approximate:

$$\frac{1-\frac{1}{2}a}{1-\frac{1}{2}b}\approx \left ( 1-\frac{1}{2}a\right )\left ( 1+\frac{1}{2}b\right )$$?

 

[slider142]

Do you know the series for $\frac{1}{1-x}$ ?

 

[fereopk]

Do you know the series for $\frac{1}{1-x}$ ?

No, unfortunately. Is there a name for this approximation?

 

[Curious3141]

$$\frac{\sqrt{1-a}}{\sqrt{1-b}}\approx \left ( 1-\frac{1}{2}a\right )\left ( 1+\frac{1}{2}b\right )$$

I know that the binomial approximation is first used,

$$\frac{\sqrt{1-a}}{\sqrt{1-b}}\approx \frac{1-\frac{1}{2}a}{1-\frac{1}{2}b}$$

But how does one approximate:

$$\frac{1-\frac{1}{2}a}{1-\frac{1}{2}b}\approx \left ( 1-\frac{1}{2}a\right )\left ( 1+\frac{1}{2}b\right )$$?

The expression can be rearranged to: $\displaystyle {(1-a)}^{\frac{1}{2}}{(1-b)}^{-\frac{1}{2}}$. Now apply the binomial approximation to each term.

 

[fereopk]

The expression can be rearranged to: $\displaystyle {(1-a)}^{\frac{1}{2}}{(1-b)}^{-\frac{1}{2}}$. Now apply the binomial approximation to each term.

Ahh, I see. Thanks!