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

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

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

### Step 1: Factor the Denominator
First, let's factor the denominator \( x^2 - 2x - 15 \):

\[
x^2 - 2x - 15 = (x - 5)(x + 3).
\]

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

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

### Step 2: Partial Fraction Decomposition
Next, we apply partial fraction decomposition to express \( f(x) \) in a simpler form:

\[
\frac{8}{(x - 5)(x + 3)} = \frac{A}{x - 5} + \frac{B}{x + 3}.
\]

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

\[
8 = A(x + 3) + B(x - 5).
\]

Expanding and equating coefficients, we get:

1. For \( x \)-terms: \( A + B = 0 \),
2. For constant terms: \( 3A - 5B = 8 \).

Solving these equations, we find:

- From \( A + B = 0 \): \( B = -A \),
- Substitute \( B = -A \) into \( 3A - 5B = 8 \):

\[
3A - 5(-A) = 8 \Rightarrow 3A + 5A = 8 \Rightarrow 8A = 8 \Rightarrow A = 1.
\]

Thus, \( B = -1 \).

So we have:

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

### Step 3: Represent Each Term as a Power Series
Now we express each term in the form of a power series. For \( |x| < 5 \), we can write:

\[
\frac{1}{x - 5} = -\frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}} = -\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{5}\right)^n = -\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}.
\]

Similarly, for \( |x| < 3 \):

\[
\frac{1}{x + 3} = \frac{1}{3} \cdot \frac{1}{1 - \left(-\frac{x}{3}\right)} = \frac{1}{3} \sum_{n=0}^{\infty} \left(-\frac{x}{3}\right)^n = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{3^{n+1}}.
\]

### Step 4: Define \( g(x, n) \)
From the above, we see that:

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

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

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

Therefore,

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

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

---

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 \( f(x) \)?
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 \( x \) for accurate function representation.

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

Explore the process of finding a function \( g(x, n) \) such that \( f(x) = \sum_{n=0}^{\infty} g(x, n) \), where \( f(x) = \frac{8}{x^2 - 2x - 15} \). This solution involves partial fraction decomposition and power series expansion, suitable for Calculus II students.

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