Find a power series for the function f(x)=1/2+3x centered at x0=−1.

Question 1Select one:

a. 

∑∞n=0−3n(x+1)n,|x+1|<3b. 

∑∞n=0(−1)n3n(x+1)n,|x+1|<3

c. 

∑∞n=0−3n(x+1)n,|x+1|<13

d. 

∑∞n=0(−1)n3n(x+1)n,|x+1|<13

To solve this, let's break down the function \( f(x) = \frac{1}{2} + 3x \) into a power series centered at \( x_0 = -1 \).

### Step 1: Express \( f(x) \) in terms of \( (x + 1) \)
We need to rewrite the function in terms of \( (x + 1) \):

\[
f(x) = \frac{1}{2} + 3x = \frac{1}{2} + 3[(x + 1) - 1] = \frac{1}{2} + 3(x + 1) - 3
\]

\[
f(x) = \left(\frac{1}{2} - 3\right) + 3(x + 1) = -\frac{5}{2} + 3(x + 1)
\]

### Step 2: Find the Power Series Representation
The function \( f(x) = -\frac{5}{2} + 3(x + 1) \) is already linear. For a power series expansion around \( x_0 = -1 \), it can be written as:

\[
f(x) = -\frac{5}{2} + 3(x + 1)
\]

This means that the power series expansion of \( f(x) \) is:

\[
f(x) = -\frac{5}{2} + 3(x + 1) = 3(x + 1)^1 + 0(x + 1)^2 + 0(x + 1)^3 + \dots
\]

### Step 3: Compare with the Options
To compare with the provided options, let's consider the standard geometric series formula:

\[
\frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n
\]

But here, the function is linear, not a geometric series. The power series representation we derived is:

\[
f(x) = 3(x + 1) = \sum_{n=0}^{\infty} a_n (x + 1)^n
\]

Where:

- \( a_0 = -\frac{5}{2} \)
- \( a_1 = 3 \)
- \( a_n = 0 \) for \( n \geq 2 \)

### Evaluate the Options:
Let's evaluate each option:

- **Option a**: \( \sum_{n=0}^{\infty} -3^n(x+1)^n \) involves all powers of \( (x + 1) \) with a negative coefficient that does not match.
- **Option b**: \( \sum_{n=0}^{\infty} (-1)^n 3^n (x+1)^n \) suggests an alternating series, which is incorrect.
- **Option c**: \( \sum_{n=0}^{\infty} -3^n (x+1)^n \) is similar to option a but has the same issue.
- **Option d**: \( \sum_{n=0}^{\infty} (-1)^n 3^n (x+1)^n \) suggests an alternating series with a radius different from the original function.

None of these options perfectly match our derived series.

However, **Option b** \( \sum_{n=0}^{\infty} (-1)^n 3^n (x+1)^n \) is closest, as it represents a geometric series with \( r = -3(x+1) \), even though it's not a perfect fit. The radius of convergence \( |x + 1| < 3 \) also seems to fit the typical constraint for such series.

### Final Answer
Given the choices, **Option b** \( \sum_{n=0}^{\infty} (-1)^n 3^n (x+1)^n, |x + 1| < 3 \) **is the best choice**, though it's important to note that the exact match isn't there due to the nature of the function.

Would you like to discuss further details or have any other questions?

### 5 Related Questions:
1. How can the convergence of a power series be determined?
2. What are the implications of the radius of convergence for a power series?
3. How does a linear function compare to more complex functions in terms of their series expansions?
4. What is the role of the center \( x_0 \) in determining the form of a power series?
5. How does the form of a geometric series influence the choice of power series expansion?

### Tip:
Remember that not all functions fit perfectly into the form of a geometric series. When handling linear functions, direct expansion often simplifies the process.

Explore the power series expansion for the function f(x)=1/2+3x centered at x0=-1. Learn how to derive the series representation, compare it with given options, and understand the radius of convergence implications. Suitable for advanced undergraduate students and those interested in mathematical series expansions.

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