Question 2. Find a power series expansion for f
′
(x) and g(x) = R x
0
f(u)du given the
expansion for f(x) below
1. f(x) = (1 + x)
−1 =
P∞
n=0(−1)nx
n
2. f(x) = sin(x)
x =
P∞
k=0(−1)k x
2k
(2k+1)!

We are given two functions \( f(x) \), and we need to find the power series expansions for the derivative \( f'(x) \) and \( g(x) = \int_0^x f(u) \, du \), where \( f(u) \) is the original function in each case. Let's approach both problems step by step.

### Part 1: \( f(x) = (1 + x)^{-1} \)

The given power series expansion for \( f(x) \) is:
\[
f(x) = (1 + x)^{-1} = \sum_{n=0}^{\infty} (-1)^n x^n
\]
#### Step 1: Find \( f'(x) \)
We can differentiate term by term:

\[
f'(x) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} (-1)^n x^n \right)
= \sum_{n=0}^{\infty} (-1)^n \cdot n x^{n-1}
= \sum_{n=1}^{\infty} (-1)^n \cdot n x^{n-1}
\]
Thus, the power series expansion for \( f'(x) \) is:
\[
f'(x) = \sum_{n=1}^{\infty} (-1)^n \cdot n x^{n-1}
\]

#### Step 2: Find \( g(x) = \int_0^x f(u) \, du \)
Now, integrate the power series of \( f(x) \) term by term:
\[
g(x) = \int_0^x \left( \sum_{n=0}^{\infty} (-1)^n u^n \right) du
= \sum_{n=0}^{\infty} (-1)^n \int_0^x u^n \, du
= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}
\]
So the power series expansion for \( g(x) \) is:
\[
g(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}
\]

### Part 2: \( f(x) = \frac{\sin(x)}{x} \)

The given power series expansion for \( f(x) \) is:
\[
f(x) = \frac{\sin(x)}{x} = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!}
\]
#### Step 1: Find \( f'(x) \)
Differentiate term by term:
\[
f'(x) = \frac{d}{dx} \left( \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!} \right)
= \sum_{k=0}^{\infty} (-1)^k \frac{2k \cdot x^{2k-1}}{(2k+1)!}
\]
Thus, the power series expansion for \( f'(x) \) is:
\[
f'(x) = \sum_{k=1}^{\infty} (-1)^k \frac{2k \cdot x^{2k-1}}{(2k+1)!}
\]

#### Step 2: Find \( g(x) = \int_0^x f(u) \, du \)
Now, integrate term by term:
\[
g(x) = \int_0^x \left( \sum_{k=0}^{\infty} (-1)^k \frac{u^{2k}}{(2k+1)!} \right) du
= \sum_{k=0}^{\infty} (-1)^k \frac{1}{(2k+1)!} \int_0^x u^{2k} \, du
= \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1) \cdot (2k+1)!}
\]
So the power series expansion for \( g(x) \) is:
\[
g(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1) \cdot (2k+1)!}
\]

### Summary:
For \( f(x) = (1 + x)^{-1} \):
- \( f'(x) = \sum_{n=1}^{\infty} (-1)^n n x^{n-1} \)
- \( g(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} \)

For \( f(x) = \frac{\sin(x)}{x} \):
- \( f'(x) = \sum_{k=1}^{\infty} (-1)^k \frac{2k \cdot x^{2k-1}}{(2k+1)!} \)
- \( g(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1) \cdot (2k+1)!} \)

Would you like further details or clarifications on any part of this process?

Here are 5 related questions to extend your understanding:
1. How would you find the radius of convergence for the series expansion of \( f'(x) \)?
2. Can you find a closed-form expression for \( g(x) \) in the case of \( f(x) = (1 + x)^{-1} \)?
3. How does the series expansion of \( \frac{\sin(x)}{x} \) differ from that of \( \sin(x) \)?
4. What happens to the series expansion of \( g(x) \) if \( f(x) = \cos(x) \) instead of \( \sin(x)/x \)?
5. How would the series expansions change if \( f(x) = (1 - x)^{-1} \)?

**Tip:** When differentiating or integrating power series, ensure that the operations are valid within the interval of convergence.

Find a power series expansion for f'(x) and g(x) = ∫_0^x f(u) du given the expansion for f(x) below:
1. f(x) = (1 + x)^{-1} = Σ (−1)^n x^n
2. f(x) = sin(x)/x = Σ (−1)^k x^(2k)/(2k+1)!

Explore the power series expansions for the derivatives and integrals of the functions (1 + x)^{-1} and sin(x)/x. This detailed guide walks through the differentiation and integration of power series, providing clear formulas and step-by-step solutions. Ideal for students in Grades 11-12 or college calculus.

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