Calculus II: Series Cheat Sheet

The Geometric Series Key

The foundation for representing functions as power series is the formula for the sum of an infinite geometric series.

$$ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} $$

This formula is valid only when the series converges, which happens when the absolute value of the common ratio r is less than 1.

$$ |r| < 1 $$

GOAL: Algebraically manipulate a given function `f(x)` until it matches the form `a / (1-r)`. Then, you can directly convert it into a power series.

Example: Basic Conversion

Let's represent `f(x) = 1 / (1 + x)` as a power series.

Step 1: Rewrite the function to match the `a / (1-r)` form. The key is to turn the `+` into a `-`.

$$ f(x) = \frac{1}{1+x} = \frac{1}{1 - (-x)} $$

Step 2: Identify `a` and `r`.

By comparing, we see that a = 1 and r = -x.

Step 3: Write the power series.

$$ \sum_{n=0}^{\infty} (1)(-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n $$

Step 4: Find the Interval and Radius of Convergence (IoC & RoC).

Use the condition `|r| < 1`:

$$ |-x| < 1 \implies |x| < 1 $$

This means the IoC is (-1, 1) and the RoC is R = 1.

Centering at c != 0

To center a series at a value `c`, you need to create an `(x-c)` term.

Example: `f(x) = 1/x` centered at `c = 1`.

Step 1: Introduce `x-1` into the denominator.

$$ f(x) = \frac{1}{x} = \frac{1}{1 + (x-1)} = \frac{1}{1 - (-(x-1))} $$

Step 2: Identify `a` and `r`.

a = 1 and r = -(x-1).

Step 3: Write the series.

$$ \sum_{n=0}^{\infty} (-(x-1))^n = \sum_{n=0}^{\infty} (-1)^n (x-1)^n $$

Step 4: Find IoC.

$$ |-(x-1)| < 1 \implies |x-1| < 1 \implies -1 < x-1 < 1 $$

The IoC is (0, 2) and RoC is R = 1.

Dealing with Constants

If the `a` position isn't 1, factor it out from the denominator.

Example: `f(x) = 1 / (3 - x)` centered at `c = 0`.

Step 1: Factor out the `3` from the denominator to make it a `1`.

$$ f(x) = \frac{1}{3(1 - x/3)} = \frac{1/3}{1 - x/3} $$

Step 2: Identify `a` and `r`.

a = 1/3 and r = x/3.

Step 3: Write and simplify the series.

$$ \sum_{n=0}^{\infty} \frac{1}{3} \left(\frac{x}{3}\right)^n = \sum_{n=0}^{\infty} \frac{x^n}{3 \cdot 3^n} = \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}} $$

Step 4: Find IoC.

$$ |x/3| < 1 \implies |x| < 3 $$

The IoC is (-3, 3) and RoC is R = 3.

Operations on Series

You can perform operations like multiplication and addition on series.

Multiplication

Example: `f(x) = x^3 / (x+2)`

First, find the series for `1/(x+2)`, which is `\sum (-1)^n x^n / 2^(n+1)`. Then, multiply the result by `x^3`.

$$ x^3 \cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^{n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+3}}{2^{n+1}} $$

The IoC is determined by the original series, which for `1/(x+2)` is `(-2, 2)`.

Addition (via Partial Fractions)

Example: `f(x) = 3 / (x^2 + x - 2)`

Step 1: Decompose the fraction.

$$ \frac{3}{(x+2)(x-1)} = \frac{-1}{x+2} + \frac{1}{x-1} $$

Step 2: Find the power series for each part.

For `\frac{-1}{x+2}`: $$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1} x^n}{2^{n+1}} \quad \text{IoC: } (-2, 2) $$

For `\frac{1}{x-1}`: $$ \sum_{n=0}^{\infty} -x^n \quad \text{IoC: } (-1, 1) $$

Step 3: Combine the series and find the final IoC.

$$ \sum_{n=0}^{\infty} \left( \frac{(-1)^{n+1}}{2^{n+1}} - 1 \right) x^n $$

The final IoC is the intersection of the individual intervals. The intersection of `(-2, 2)` and `(-1, 1)` is (-1, 1).