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!