Problem 17 - Calculus

Question

expand the quantity about 0 in terms of the variable given. Give four nonzero terms. $$\frac{1}{(a+r)^{2}} \text { in terms of } \frac{r}{a}$$

Step-by-Step Solution

Answer

The first four nonzero terms are \( \frac{1}{a^2} - \frac{2r}{a^3} + \frac{3r^2}{a^4} - \frac{4r^3}{a^5} \).

Step 1: Understanding the Expression

We want to expand \( \frac{1}{(a+r)^2} \) in terms of \( \frac{r}{a} \). To do this, we'll perform a series expansion about \( r=0 \).

Step 2: Variable Substitution

First, rewrite \( r \) in terms of \( \frac{r}{a} \). Let \( x = \frac{r}{a} \) so that \( r = ax \). The expression becomes \( \frac{1}{(a + ax)^2} = \frac{1}{a^2(1 + x)^2} \).

Step 3: Expanding Using Binomial Series

Use the binomial series expansion \( (1+x)^{-n} = 1 - nx + \frac{n(n+1)}{2}x^2 - \frac{n(n+1)(n+2)}{6}x^3 + \ldots \).For our expression, \( n = 2 \), so:\[ \frac{1}{(1+x)^2} = (1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \ldots \]

Step 4: Applying the Expansion

Substitute the expansion from Step 3 back into the expression from Step 2:\[ \frac{1}{a^2} \left( 1 - 2x + 3x^2 - 4x^3 + \ldots \right) = \frac{1}{a^2} - \frac{2x}{a^2} + \frac{3x^2}{a^2} - \frac{4x^3}{a^2} + \ldots \]

Step 5: Simplifying the Expansion

Substitute \( x = \frac{r}{a} \) back into the expansion:\[ \frac{1}{(a+r)^2} \approx \frac{1}{a^2} - \frac{2r}{a^3} + \frac{3r^2}{a^4} - \frac{4r^3}{a^5} \]

Step 6: Extracting Four Nonzero Terms

The first four nonzero terms of the expansion are:1. \( \frac{1}{a^2} \)2. \( -\frac{2r}{a^3} \)3. \( \frac{3r^2}{a^4} \)4. \( -\frac{4r^3}{a^5} \)

Key Concepts

Taylor Series

The Taylor Series is a fundamental concept in calculus that allows us to approximate functions. For a function that is infinitely differentiable at a point, the Taylor series expansion provides a polynomial approximation of the function around that point. This approximation becomes more accurate as more terms are included. Taylor series are expressed as an infinite sum of terms calculated from the values of the function's derivatives at a single point. For a function \( f(x) \), centered at \( a \), the Taylor series is given by:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\]In our problem, the expansion of \( \frac{1}{(a+r)^2} \) involved finding a series around \( r = 0 \), equivalent to the Taylor series expansion. By considering more terms, we achieve a closer approximation of the original function. This expansion gives us a tool to understand complex expressions by breaking them into more manageable polynomial parts.

Binomial Theorem

The Binomial Theorem provides a powerful way to expand expressions involving powers of sums. It is expressed using the formula:\[(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots\]For our exercise, we needed to expand \( \frac{1}{(1+x)^2} \) using the binomial series, which is a specific case of this theorem for expressing \((1+x)^{-n}\). This is useful for negative or fractional exponents. In the solution, \( n = -2 \), so the expansion formula provided us the series: - \((1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \ldots\)These terms are key when deriving polynomial approximations for expressions with negative exponents, helping simplify complex algebraic fractions. Understanding how we arrived at these coefficients via the Binomial Theorem is crucial for comprehending how the series relates back to the original expression.

Power Series

Power Series are infinite series that look like polynomials and represent functions as sums of powers of variables. A typical power series centered at 0 has the form:\[ a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots\]In our problem, after substituting \( x = \frac{r}{a} \), we expanded \( \frac{1}{a^2(1+x)^2} \) as a power series:- \( \frac{1}{a^2} - \frac{2}{a^3}r + \frac{3}{a^4}r^2 - \frac{4}{a^5}r^3 + \ldots \)This series showcases how variable substitution can aid in representing complicated functions in a more digestible form. Power series are exceptionally useful in calculus for function approximations, particularly when leveraging convergence properties. For practical problems, one often considers only the first few terms of the series, sufficient for a good approximation of the function's behavior near a point. The understanding of power series is essential for deeper analyses in mathematical modeling, physics, and engineering.