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Calculus II: Series Cheat Sheet

Calculus II: Series Cheat Sheet

















Calculus II: Functions as Power Series





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…