Where does this approximation come from?

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Discussion Overview

The discussion revolves around the approximation of the expression \(\frac{\sqrt{1-a}}{\sqrt{1-b}}\) and how it can be expressed in terms of a product involving binomial approximations. Participants explore the steps involved in deriving the approximation and seek clarification on the underlying series used.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant presents the approximation \(\frac{\sqrt{1-a}}{\sqrt{1-b}} \approx \left ( 1-\frac{1}{2}a\right )\left ( 1+\frac{1}{2}b\right )\) and questions how to derive this from the binomial approximation.
  • Another participant inquires about the series for \(\frac{1}{1-x}\), suggesting it may be relevant to the approximation.
  • A later reply reiterates the initial approximation and suggests rearranging the expression to \({(1-a)}^{\frac{1}{2}}{(1-b)}^{-\frac{1}{2}}\) before applying the binomial approximation to each term.
  • One participant expresses understanding after receiving clarification on the rearrangement and application of the binomial approximation.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the specific series or name for the approximation, and the discussion remains exploratory with multiple viewpoints on the derivation process.

Contextual Notes

Some assumptions about the validity of the binomial approximation in this context are not explicitly stated, and the discussion does not resolve the conditions under which these approximations hold.

fereopk
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\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 )?
 
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Do you know the series for \frac{1}{1-x} ?
 
slider142 said:
Do you know the series for \frac{1}{1-x} ?

No, unfortunately. Is there a name for this approximation?
 
fereopk said:
\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.
 
Curious3141 said:
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!
 

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