Calculus II: Power Series and Coordinate Systems

To convert from polar coordinates ( $r, \theta$ ) to Cartesian coordinates ( $x, y$ ), the fundamental identities are used:
- $x = r \cos(\theta)$
- $y = r \sin(\theta)$
Trigonometric signs depend on the quadrant in which the angle $\theta$ resides:
- In the third quadrant, $\cos(\theta)$ is negative and $\sin(\theta)$ is negative.
- In the fourth quadrant, $\cos(\theta)$ is positive and $\sin(\theta)$ is negative.
Example conversion discussed: converting $r=3$ in a specific quadrant (e.g., fourth quadrant where cosine is positive).

The Ratio Test is the primary tool for finding the radius of convergence ( $R$ ) and the interval of convergence ( $I$ ) for power series.
General Procedure:
- Apply the limit of the absolute value of the ratio of successive terms: $L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$ .
- Ensure the series is in the form of a fraction to simplify calculations (though the speaker notes that "fraction bars look terrible").
Case Study Analysis:
- For the series centered at $x=4$ : The ratio test simplifies to |x - 4| < 1.
- Center ( $c$ ): $4$ .
- Radius ( $R$ ): $1$ .
- Initial Interval: $(3, 5)$ , because $4-1=3$ and $4+1=5$ .
Testing Endpoints:
- At $x = 3$ : The term becomes $(-1)^n$ . We must determine if this alternating series converges.
- The Alternating Series Test requires:
1. $\lim_{n \to \infty} b_n = 0$ (where $b_n$ is the absolute value of the term).
2. The sequence $b_n$ must be decreasing (i.e., absolute value of the $n + 1$ term is less than the $n$ term).
- Example check: \frac{n+1}{(n+1)^3+1} < \frac{n}{n^3+1}. Cross-multiplication and expanding terms ( $n + n^4 + 1 \dots$ ) help verify this inequality.
Limit Comparison Test:
- If unsure about convergence at an endpoint, compare it to a known series, such as a P-series ( $1/n^2$ ), identifying that since the degree of the denominator is higher, the series likely converges.

To represent a rational function as a power series, use the geometric series formula: $\frac{a}{1 - r} = \sum_{n=0}^{\infty} a r^n$ , which converges when |r| < 1.
Partial Fraction Decomposition:
- For a function like $\frac{3x^2 + x}{\dots}$ , first decompose into simpler fractions.
- Example result from transcript: $\frac{1}{x - 1} - \frac{1}{x + 2}$ .
Algebraic Manipulations to Match Geometric Form:
- For $\frac{1}{x-1}$ : To get the form $1 - r$ , rewrite as $-\frac{1}{1 - x}$ . This becomes $-\sum x^n$ .
- For $\frac{1}{x+2}$ : Factor out a $2$ to change the leading constant to $1$ : $\frac{1}{2(1 + x/2)}$ . Then rewrite as $\frac{1}{2(1 - (-x/2))}$ .
- This gives the series: $\frac{1}{2} \sum_{n=0}^{\infty} (-\frac{x}{2})^n$ .
- Note on centering: If the centers are different, the combined series becomes problematic. It is best to maintain a consistent center for both parts of the partial fraction.

The objective is to find the power series for $f(x) = \text{arctan}(\frac{x}{3})$ .
Step 1: Differentiation:
- Using the chain rule: $\frac{d}{dx} \text{arctan}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}$ .
- For $u = \frac{x}{3}$ , the derivative is $\frac{1}{1 + (x/3)^2} \cdot \frac{1}{3} = \frac{1}{3(1 + x^2/9)} = \frac{1}{3 + x^2/3}$ . More simply, this simplifies to the form $\frac{1/3}{1 - (-x^2/9)}$ .
Step 2: Geometric Expansion:
- Expand the derivative into a series: $\frac{1}{3} [1 - \frac{x^2}{9} + \frac{x^4}{9^2} - \frac{x^6}{9^3} + \dots]$ .
Step 3: Integration:
- Integrate the series term-by-term to recover the arc-tangent series.
- General Form: $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1) \cdot 3^{2n+1}}$ .
- The denominator includes $(2n+1)$ from integration and powers of $9$ (which are $3^{2n}$ ) multiplied by the initial $1/3$ , resulting in $3^{2n+1}$ .

The Maclaurin series for $f(x) = (1 + x)^{1/2}$ involves finding successive derivatives and evaluating them at $x=0$ .
Finding Derivatives:
- $f'(x) = \frac{1}{2}(1+x)^{-1/2}$
- $f''(x) = (\frac{1}{2})(-\frac{1}{2})(1+x)^{-3/2}$
- $f'''(x) = (\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})(1+x)^{-5/2}$
Coefficients at $x=0$ :
- $f(0) = 1$
- $f'(0) = 1/2$
- $f''(0) = -1/4$
- $f'''(0) = 3/8$
Constructing the Series:
- The general term for the numerator after the first few terms is the product of odd integers: $1 \times 3 \times 5 \times \dots \times (2n - 3)$ .
Peeling Off Terms:
- Since the first one or two terms often do not fit the general alternating pattern found in the rest of the derivatives, they must be "peeled off" (written separately) from the summation notation.
Index Shifting:
- If a series is initially indexed such that terms are redundant or cumbersome (e.g., using $n+2$ everywhere), shift the index to simplify.
- To subtract $2$ from the variable $n$ within the formula, you must add $2$ to the starting index value (e.g., moving from $n=0$ to $n=2$ ).