Can 1/(1 + x) be simplified to 1-x for a very small value of x?

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The discussion centers on the simplification of the expression 1/(1 + x) to 1 - x for very small values of x, specifically when x is approximately 1.0091532×10-12. It is established that using the Taylor series expansion for 1/(1 + x) allows for the approximation 1 - x, as higher-order terms become negligible for such small x values. The equation can be rearranged to express x in terms of the fraction, confirming that for practical calculations, 1 - x suffices due to the convergence of the series for |x| < 1.

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If you have a fraction, for example,
\frac{1}{{1.0091532\times10^{-12}} + 1}

Is there a simple way to convert it to a more easily calculated form, specifically, 1-x (where x is a very small number)
 
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1/(1+a) = 1 - a/(1 + a), which for tiny a is about 1 - a. More precisely,
1/(1+a) = 1 - a/(1 + a) = 1 - a + a^2/(1 + a)
which is about 1 - a + a^2 for tiny a.
 
Matuku said:
If you have a fraction, for example,
\frac{1}{{1.0091532\times10^{-12}} + 1}

Is there a simple way to convert it to a more easily calculated form, specifically, 1-x (where x is a very small number)

Just arrange an equation based on that.
1 - x = \frac{1}{{1.0091532\times10^{-12}} + 1}

Determine an expression for the value of x, and then rewrite the left-hand expression.
 
Expanding on what CRGreathouse says, the Taylor series (around 0) for 1/(1 + x) is 1 - x + x2 - x3 + ... (and it converges if |x| < 1). You can approximate it by truncating the Taylor series, and since you have x ~ 10-12, for practical purposes 1 - x should be enough (any higher-order terms will be smaller than the precision you give anyway).
 

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