Problem 47 - College Algebra and Calculus: An Applied Approach

Question

Write the partial fraction decomposition for the rational expression. Check your result algebraically. Then assign a value to the constant \(a\) and use a graphing utility to check the result graphically. $$ \frac{1}{a^{2}-x^{2}} $$

Step-by-Step Solution

Answer

The partial fraction decomposition of \( \frac{1}{a^{2}-x^{2}} \) is \( \frac{1}{2a(a+x)} + \frac{1}{2a(a-x)} \)

Step 1: Expression Breakdown

Rewrite the given expression \( \frac{1}{a^{2}-x^{2}} \) in the difference of squares form: \( \frac{1}{{(a+x)(a-x)}} \)

Step 2: Apply Partial Fraction Decomposition

Apply the decomposition to \(\frac{1}{{(a+x)(a-x)}}\), such that it can be written as: \( \frac{A}{{a+x}} + \frac{B}{{a-x}} \). Here, A and B are constants that we need to find.

Step 3: Common Denominator

Next, find the common denominator for \( \frac{A}{{a+x}} + \frac{B}{{a-x}} \). This gives us \( \frac{A(a-x) + B(a+x)}{{(a+x)(a-x)}} \)

Step 4: Equate Expressions for Comparison

For the partial fractions to be identical to the original fraction, equate \( \frac{A(a-x) + B(a+x)}{{(a+x)(a-x)}} \) to \( \frac{1}{{a^{2}-x^{2}}} \), which simplifies to: \( A(a-x) + B(a+x) = 1 \)

Step 5: Solve for A and B

Using the equation obtained in Step 4, we solve for A and B. One way to do this is by substitution of two suitable values of \( x \). Choosing \( x = a \) gives \( A=1/2a \) and choosing \( x = -a \) gives \( B = 1/2a \)

Step 6: Substitute A and B Back

Substitute A and B back into \( \frac{A}{{a+x}} + \frac{B}{{a-x}} \) to yield the final solution: \( \frac{1}{2a(a+x)} + \frac{1}{2a(a-x)} \)

Step 7: Checking Algebraically

Multiply the solution by \( a^{2}-x^{2} \) and if it equals 1, then it is correct algebraically.

Step 8: Checking Graphically

Assign a value to \( a \) and plot both the original function and the decomposed function. If they are identical, then the decomposed function is correct graphically.

Key Concepts

Rational Expressions

Rational expressions are mathematical expressions where the numerator and the denominator are polynomials. Understanding them is crucial because most algebraic manipulations involve fractions, and this is no different in the world of algebra and calculus. In our given problem, we have a rational expression \( \frac{1}{a^{2}-x^{2}} \). Here, the numerator is 1, a simple constant, while the denominator is \( a^2 - x^2 \), a polynomial expression. It represents how one quantity changes in relation to another.

• Numerator: Can be any polynomial, even a constant.
• Denominator: Also a polynomial, but it needs to be non-zero.

Handling these expressions often involves simplifying or rewriting them, finding common denominators, or performing partial fraction decomposition. This makes subsequent calculations, like integrations, much more manageable.

Difference of Squares

The difference of squares is a special case in algebra where a polynomial looks like \( a^2 - b^2 \). This can always be factored into \( (a + b)(a - b) \), a useful identity when simplifying expressions or setting up partial fraction decomposition.In our exercise, the expression \( a^2 - x^2 \) fits this pattern neatly because:

• \( a^2 - x^2 = (a + x)(a - x) \)

This factorization is essential. It allows us to break down the original rational expression into simpler parts, setting the stage for partial fraction decomposition. Recognizing and applying this pattern not only simplifies expressions but also eases computational tasks that rely on simpler forms.

Constants in Fractions

In partial fraction decomposition, we rewrite a rational expression as a sum of fractions. Each fraction has constants in the numerator that we have to figure out.For the expression \( \frac{1}{a^{2}-x^{2}} \), its decomposition form is \( \frac{A}{a+x} + \frac{B}{a-x} \). We need to determine the constants \( A \) and \( B \). This involves:

• Solving simultaneous equations derived by equating the original expression to the decomposed form.
• Selecting strategic values of \( x \), like \( x = a \) and \( x = -a \), to simplify calculations.
• Determining \( A \) and \( B \) as \( \frac{1}{2a} \), achieved by solving \( A(a-x) + B(a+x) = 1 \).

Mastering this step is crucial for understanding integration techniques involving rational expressions. It provides insight into the importance of algebraic manipulation in finding these constants.

Graphing Utility

A graphing utility is a tool, often found in calculators and software, that visually represents mathematical functions. This can help check the accuracy of algebraic manipulations by ensuring the graphical output matches expectations.To verify the partial fraction decomposition of \( \frac{1}{a^{2} - x^{2}} \), we can:

• Assign a specific value to \( a \).
• Graph the original function \( \frac{1}{a^2 - x^2} \) and the decomposed function \( \frac{1}{2a(a+x)} + \frac{1}{2a(a-x)} \).
• Compare the graphs to see if they overlap.

If both graphs look identical, it confirms the decomposition is correct. This visual confirmation adds another layer of understanding beyond algebraic manipulation, showcasing the power of modern technology in learning and verifying mathematical concepts.