Question

If \(\mathrm{x}\) is so small that terms with \(\mathrm{x}^{3}\) and higher powers of \(\mathrm{x}\) may be neglected then \(\left[\left\\{(1+\mathrm{x})^{(3 / 2)}-\\{1+(\mathrm{x} / 2)\\}^{3}\right\\} /\left\\{(1-\mathrm{x})^{(1 / 2)}\right\\}\right]\) may be approximated as \(\ldots \ldots \ldots\) (a) \([(-3) / 8] \mathrm{x}^{2}\) (b) \((1 / 2) \mathrm{x}-(3 / 8) \mathrm{x}^{2}\) (c) \(1-(3 / 8) x^{2}\) (d) \(3 x+(3 / 8) x^{2}\)

Short answer

\(-\frac{1}{8}x^2\)

Step-by-step solution

Expansion using binomial theorem

Since the terms with x^3 and higher powers can be neglected, we can use the binomial theorem for the expansion of the given terms up to the second power of x. The binomial theorem states that for any real numbers a, b, and index n: \((a+b)^n = \sum_{k=0}^n {n\choose k} a^{(n-k)}b^k\) Applying this to the given expression, we get: (1) \((1+x)^{(3/2)} \approx 1 + \frac{3}{2}x + \frac{3}{8}x^2\) (2) \((1+(x/2))^3 \approx 1 + \frac{3}{2}x + \frac{3}{4}x^2\) (3) \((1-x)^{(-1/2)} \approx 1 + \frac{1}{2}x + \frac{1}{8}x^2\)

Simplification of the expression

Substitute the approximations obtained in step 1 into the given equation, and simplify the expression: \(\frac{[(1+\frac{3}{2}x + \frac{3}{8}x^2)-(1+\frac{3}{2}x + \frac{3}{4}x^2)]}{(1+\frac{1}{2}x + \frac{1}{8}x^2)}\) The numerator simplifies to: \(-\frac{1}{8}x^2\) Now, divide this by the denominator: \(\frac{-\frac{1}{8}x^2}{1+\frac{1}{2}x + \frac{1}{8}x^2} \) As x is very small, we can ignore the terms with x^2 in the denominator, resulting in: \(-\frac{\frac{1}{8}x^2}{1+\frac{1}{2}x}\)

Comparing with the given options

Compare the simplified expression from step 2 with the given options: (a) \(-\frac{3}{8}x^2\) (b) \(\frac{1}{2}x - \frac{3}{8}x^2\) (c) \(1-\frac{3}{8}x^2\) (d) \(3x + \frac{3}{8}x^2\) None of the given options matches the simplified expression exactly. However, option (a) is the closest match as it has the same x^2 term but with a slightly different coefficient.

Key concepts

Approximation

Approximations are crucial when dealing with very small numbers. In this exercise, we assume that the variable \( x \) is so small that the terms involving \( x^3 \) and higher powers can be neglected. This simplifies calculations significantly because it allows us to focus only on the most influential terms. This approach can be very handy in practical problems where exact values are not necessary, or when computations with very small numbers are computationally intensive. For example, ignoring \( x^3 \) and beyond means our calculations quickly condense down to simpler polynomials, focusing on the first few terms with the greatest impact.
By using approximation effectively, we simplify mathematical models while retaining enough accuracy to be useful for predictions or further analysis.

Polynomial Expansion

Expanding a polynomial using the Binomial Theorem is a method for breaking down expressions into series. The Binomial Theorem states:

\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{(n-k)} b^k \]

This formula provides a concise way to expand powers of binomials. In our example, we use the theorem to expand expressions involving \( (1+x)^{3/2} \), \( (1+(x/2))^3 \), and \( (1-x)^{-1/2} \) up to the second power of \( x \). Let’s see why this is helpful:

• Solving these expansions using the binomial theorem highlights the components of the polynomial that contribute the most significantly when terms like \( x^3 \) and higher are disregarded.
• It also breaks down complex expressions into manageable terms, such as \(1 + \frac{3}{2}x + \frac{3}{8}x^2\), which we can use for further calculations.

In practice, these expansions transform complex expressions into series which are easier to manipulate and use in approximations.

Mathematical Simplification

Simplifying expressions is the process of reducing complexity without losing the essence of the information. In the provided exercise, after the polynomial expansions, we simplify the expression by substituting our expanded forms into it. The main aim here is to reduce an unwieldy expression to something more concise and easy to understand.
Here’s a basic rundown of how it's done:

• We substitute our expanded forms into the original expression.
• By subtracting corresponding terms, many terms cancel out, leaving only the most relevant parts that affect the outcome.
• The result is simplified further by ignoring terms that have insignificant contributions, such as higher powers of \( x \) in the denominator when \( x \) is very small.

This allows us to compare the simplified form with given options. Even if a perfect match isn’t found, the closest estimate can be identified. Simplification helps in making expressions manageable and often reveals patterns or solutions that are not obvious in their original form.