Math Problem Statement

Consider the function \( f(x) = \frac{8}{x^2 - 2x - 15} \). Find a function \( g(x, n) \) such that \( f(x) = \sum_{n=0}^{\infty} g(x, n) \).

Solution

To find a function g(x,n)g(x, n) such that

f(x)=n=0g(x,n),f(x) = \sum_{n=0}^{\infty} g(x, n),

we can start by examining the function f(x)=8x22x15f(x) = \frac{8}{x^2 - 2x - 15}.

Step 1: Factor the Denominator

First, let's factor the denominator x22x15x^2 - 2x - 15:

x22x15=(x5)(x+3).x^2 - 2x - 15 = (x - 5)(x + 3).

Thus, we can rewrite f(x)f(x) as:

f(x)=8(x5)(x+3).f(x) = \frac{8}{(x - 5)(x + 3)}.

Step 2: Partial Fraction Decomposition

Next, we apply partial fraction decomposition to express f(x)f(x) in a simpler form:

8(x5)(x+3)=Ax5+Bx+3.\frac{8}{(x - 5)(x + 3)} = \frac{A}{x - 5} + \frac{B}{x + 3}.

Multiplying both sides by (x5)(x+3)(x - 5)(x + 3), we get:

8=A(x+3)+B(x5).8 = A(x + 3) + B(x - 5).

Expanding and equating coefficients, we get:

  1. For xx-terms: A+B=0A + B = 0,
  2. For constant terms: 3A5B=83A - 5B = 8.

Solving these equations, we find:

  • From A+B=0A + B = 0: B=AB = -A,
  • Substitute B=AB = -A into 3A5B=83A - 5B = 8:

3A5(A)=83A+5A=88A=8A=1.3A - 5(-A) = 8 \Rightarrow 3A + 5A = 8 \Rightarrow 8A = 8 \Rightarrow A = 1.

Thus, B=1B = -1.

So we have:

f(x)=1x51x+3.f(x) = \frac{1}{x - 5} - \frac{1}{x + 3}.

Step 3: Represent Each Term as a Power Series

Now we express each term in the form of a power series. For x<5|x| < 5, we can write:

1x5=1511x5=15n=0(x5)n=n=0xn5n+1.\frac{1}{x - 5} = -\frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}} = -\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{5}\right)^n = -\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}.

Similarly, for x<3|x| < 3:

1x+3=1311(x3)=13n=0(x3)n=n=0(1)nxn3n+1.\frac{1}{x + 3} = \frac{1}{3} \cdot \frac{1}{1 - \left(-\frac{x}{3}\right)} = \frac{1}{3} \sum_{n=0}^{\infty} \left(-\frac{x}{3}\right)^n = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{3^{n+1}}.

Step 4: Define g(x,n)g(x, n)

From the above, we see that:

f(x)=n=0xn5n+1+n=0(1)nxn3n+1.f(x) = -\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}} + \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{3^{n+1}}.

So we can define g(x,n)g(x, n) as:

g(x,n)=xn5n+1+(1)nxn3n+1.g(x, n) = -\frac{x^n}{5^{n+1}} + \frac{(-1)^n x^n}{3^{n+1}}.

Therefore,

f(x)=n=0g(x,n),f(x) = \sum_{n=0}^{\infty} g(x, n),

where g(x,n)=xn5n+1+(1)nxn3n+1g(x, n) = -\frac{x^n}{5^{n+1}} + \frac{(-1)^n x^n}{3^{n+1}}.


Would you like a further breakdown of these steps or details on a specific part of the solution?

Related Questions

  1. How do you apply partial fraction decomposition to other rational functions?
  2. What are the conditions for representing a function as a power series?
  3. How can we determine the radius of convergence for each series in f(x)f(x)?
  4. How would this approach change if the denominator had complex roots?
  5. Can we generalize this method to other forms of rational functions?

Tip

When working with power series, ensure the series converges within the desired range of xx for accurate function representation.

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Math Problem Analysis

Mathematical Concepts

Algebra
Partial Fraction Decomposition
Power Series Expansion

Formulas

Partial fraction decomposition formula
Power series expansion: \( \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \)

Theorems

Power Series Expansion Theorem

Suitable Grade Level

Undergraduate (Calculus II level)