How to Approximate √(1 - x) / √x when x is Small: A Taylor Series Approach

In this article, we will delve into the mathematical technique of using Taylor series to approximate the expression √(1 - x) / √x when x is a small value. This technique is particularly useful in understanding series approximations and providing a practical example through the given value x 1/8. We will perform algebraic manipulations and series expansions to derive our approximation.

Taylor Series Expansion

The Taylor series expansion is a method to represent a function as an infinite sum of terms. For a function ( f(x) ) around a point ( x a ), the Taylor series is given by:

( f(x) f(a) f'(a)(x-a) frac{f''(a)}{2!}(x-a)^2 frac{f'''(a)}{3!}(x-a)^3 cdots )

In this case, we are looking for the expansion of √(1 - x) / √x around ( x 0 ).

Step-by-Step Analysis

First, we start by rewriting the given expression:

[frac{1-x}{1x} frac{2}{1x} - 1]

We know that the series expansion of ( frac{2}{1x} ) around ( x 0 ) is:

[frac{2}{1x} 2 cdot (1 - x x^2 - x^3 cdots)]

Substituting this into the expression:

[frac{1-x}{1x} 2(1 - x x^2 - x^3 cdots) - 1]

Expanding this, we get:

[frac{1-x}{1x} 1 - 2x 2x^2 - 2x^3 cdots]

Next, we take the square root of both sides:

[sqrt{frac{1-x}{1x}} sqrt{1 - 2x 2x^2 - 2x^3 cdots}]

Using the Taylor series expansion for the square root function ( sqrt{1 t} approx 1 frac{1}{2}t - frac{1}{8}t^2 cdots ) when ( t ) is small, we set ( t -2x 2x^2 - 2x^3 cdots ):

[sqrt{frac{1-x}{1x}} approx 1 frac{1}{2}(-2x 2x^2 - 2x^3 cdots) - frac{1}{8}(-2x 2x^2 - 2x^3 cdots)^2 cdots]

Simplifying the expression, we discard higher-order terms and obtain:

[sqrt{frac{1-x}{1x}} approx 1 - x frac{x^2}{2}]

Applying the Value x 1/8

Now, let's substitute ( x 1/8 ) into the approximation:

[sqrt{frac{1 - frac{1}{8}}{1 frac{1}{8}}} 1 - frac{1}{8} frac{(frac{1}{8})^2}{2}]

Calculating each term:

[1 - frac{1}{8} frac{frac{1}{64}}{2} 1 - frac{1}{8} frac{1}{128}]

Combining the fractions:

[1 - frac{1}{8} frac{1}{128} frac{112}{128} - frac{16}{128} frac{1}{128} frac{112 - 16 1}{128} frac{97}{128}]

Hence, the approximate value of ( sqrt{frac{7}{9}} ) is given by:

[sqrt{frac{7}{9}} approx frac{113}{128}]

And therefore:

[sqrt{7} approx frac{339}{128}]

This approximation is quite accurate, given the small value of x.

Conclusion

By applying the Taylor series expansion for the given function and substituting the value ( x 1/8 ), we have shown that ( sqrt{7} approx frac{339}{128} ). This method provides a practical approach to approximate square roots for small values of x.