$= \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$

This is a finite polynomial that approximates $f$ near $x = a$.

Taylor Series (Infinite!)

If we let $n \to \infty$, we get the Taylor series:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$

$= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$

💡 Key Idea: Taylor series is the "best" power series representation of $f$ centered at $a$!

Maclaurin Series (Special Case)

When $a = 0$, we get a Maclaurin series:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$

$= f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

Most common Taylor series are Maclaurin series!

Finding Taylor Series - The Recipe

Step 1: Find derivatives $f'(x), f''(x), f'''(x), \ldots$

Step 2: Evaluate at center: $f(a), f'(a), f''(a), \ldots$

Step 3: Look for pattern in derivatives

Step 4: Write general term $\frac{f^{(n)}(a)}{n!}(x-a)^n$

Step 5: Sum from $n = 0$ to $\infty$

Example 1: Maclaurin Series for $e^x$

Step 1: Find derivatives

$f(x) = e^x, \quad f'(x) = e^x, \quad f''(x) = e^x, \quad \ldots$

All derivatives are $e^x$!

Step 2: Evaluate at $a = 0$

$f(0) = e^0 = 1$ $f'(0) = 1$ $f''(0) = 1$ $f^{(n)}(0) = 1$ for all $n$

Step 3: Write Taylor series

$e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

Converges for all $x$! (radius $R = \infty$)

Example 2: Maclaurin Series for $\sin x$

Step 1: Find derivatives

$f(x) = \sin x$ $f'(x) = \cos x$ $f''(x) = -\sin x$ $f'''(x) = -\cos x$ $f^{(4)}(x) = \sin x$ (pattern repeats!)

Step 2: Evaluate at $x = 0$

$f(0) = \sin 0 = 0$ $f'(0) = \cos 0 = 1$ $f''(0) = -\sin 0 = 0$ $f'''(0) = -\cos 0 = -1$ $f^{(4)}(0) = \sin 0 = 0$ $f^{(5)}(0) = \cos 0 = 1$

Pattern: $0, 1, 0, -1, 0, 1, 0, -1, \ldots$

Step 3: Only odd powers survive!

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

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

Converges for all $x$!

Example 3: Maclaurin Series for $\cos x$

Pattern in derivatives at $x = 0$:

$f(0) = 1, \quad f'(0) = 0, \quad f''(0) = -1, \quad f'''(0) = 0, \quad f^{(4)}(0) = 1$

Pattern: $1, 0, -1, 0, 1, 0, -1, 0, \ldots$

Only even powers!

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

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

Converges for all $x$!

Example 4: Taylor Series for $\ln x$ centered at $a = 1$

Step 1: Find derivatives

$f(x) = \ln x$ $f'(x) = \frac{1}{x} = x^{-1}$ $f''(x) = -x^{-2}$ $f'''(x) = 2x^{-3}$ $f^{(4)}(x) = -6x^{-4} = -3!x^{-4}$ $f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}$ for $n \geq 1$

Step 2: Evaluate at $x = 1$

$f(1) = \ln 1 = 0$ $f'(1) = 1$ $f''(1) = -1$ $f'''(1) = 2$ $f^{(n)}(1) = (-1)^{n-1}(n-1)!$ for $n \geq 1$

Step 3: Write Taylor series

$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n$

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

$= (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$

Converges for $0 < x \leq 2$ (interval: $(0, 2]$)

Relationship Between $\sin x$ and $\cos x$

Notice: $\frac{d}{dx}[\sin x] = \cos x$

Differentiate the series for $\sin x$:

$\frac{d}{dx}\left[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right]$

$= 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \cdots$

$= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \cos x$ ✓

Series confirm calculus relationships!

Using Known Series (Smart Way!)

Instead of computing all derivatives, use:

• Substitution
• Differentiation/Integration
• Algebraic manipulation

Example 5: Find Series for $e^{-x^2}$

Start with: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

Substitute $x \to -x^2$:

$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

$= 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots$

Much easier than computing derivatives!

Example 6: Find Series for $x \sin x$

Start with: $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

Multiply by $x$:

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

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

$= x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} + \cdots$

Example 7: Find Series for $\int e^{-x^2} dx$

From Example 5: $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

Integrate term by term:

$\int e^{-x^2} dx = C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)}$

$= C + x - \frac{x^3}{1! \cdot 3} + \frac{x^5}{2! \cdot 5} - \frac{x^7}{3! \cdot 7} + \cdots$

Note: This integral has no elementary antiderivative, but we can express it as a series!

Taylor Series Centered at $a \neq 0$

Example: Find Taylor series for $e^x$ centered at $a = 2$.

All derivatives of $e^x$ are $e^x$, so:

$f^{(n)}(2) = e^2$ for all $n$

Taylor series:

$e^x = \sum_{n=0}^{\infty} \frac{e^2}{n!}(x-2)^n$

$= e^2 \left[1 + (x-2) + \frac{(x-2)^2}{2!} + \frac{(x-2)^3}{3!} + \cdots\right]$

$= e^2 \sum_{n=0}^{\infty} \frac{(x-2)^n}{n!}$

When Does Taylor Series Equal the Function?

The Taylor series equals $f(x)$ when:

$\lim_{n \to \infty} R_n(x) = 0$

where $R_n(x)$ is the remainder (error) after $n$ terms.

For most common functions (like $e^x, \sin x, \cos x, \ln(1+x)$), this happens on their interval of convergence.

⚠️ Common Mistakes

Mistake 1: Wrong Factorial in General Term

For $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$:

WRONG: General term is $\frac{(-1)^n x^n}{n!}$

RIGHT: General term is $\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ (only odd powers!)

Mistake 2: Wrong Starting Index

For $\sin x$, first nonzero term is $x$ (when $n = 0$).

WRONG: Sum starts at $n = 1$

RIGHT: $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ (starts at $n = 0$)

Mistake 3: Forgetting Center for Taylor Series

For $\ln x$ centered at $a = 1$:

WRONG: $\ln x = \sum \frac{(-1)^{n-1} x^n}{n}$

RIGHT: $\ln x = \sum \frac{(-1)^{n-1} (x-1)^n}{n}$ (must use $(x-1)^n$!)

Mistake 4: Computing All Derivatives Unnecessarily

To find series for $e^{3x}$:

WRONG: Compute $f'(x), f''(x), f'''(x), \ldots$

RIGHT: Use $e^x = \sum \frac{x^n}{n!}$, substitute $x \to 3x$:

$e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} = \sum_{n=0}^{\infty} \frac{3^n x^n}{n!}$

📝 Practice Strategy

1. Memorize basic series: $e^x, \sin x, \cos x, \frac{1}{1-x}, \ln(1+x)$
2. Use substitution: Replace $x$ with $f(x)$ in known series
3. Multiply/divide by powers: For $x^k f(x)$, just multiply series by $x^k$
4. Differentiate/integrate: Term by term when needed
5. For derivatives at $a$: Look for patterns to avoid computing all
6. Check first few terms: Make sure they match Taylor polynomial
7. Odd/even functions: $\sin x$ has only odd powers, $\cos x$ only even
8. Write general term: Essential for summation notation

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

Step 2: Find $e^{2x}$

Substitute $x \to 2x$ in series for $e^x$:

$e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}$

$= 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \cdots$

$= 1 + 2x + 2x^2 + \frac{4x^3}{3} + \cdots$

Step 3: Multiply series

$e^{2x} \cos x = \left(1 + 2x + 2x^2 + \frac{4x^3}{3} + \cdots\right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right)$

Multiply term by term (collect like powers):

Constant term: $1 \cdot 1 = 1$

$x$ term: $2x \cdot 1 = 2x$

$x^2$ term: $2x^2 \cdot 1 + 1 \cdot (-\frac{x^2}{2}) = 2x^2 - \frac{x^2}{2} = \frac{3x^2}{2}$

$x^3$ term: $\frac{4x^3}{3} \cdot 1 + 2x \cdot (-\frac{x^2}{2}) = \frac{4x^3}{3} - x^3 = \frac{x^3}{3}$

$x^4$ term: (from multiple products) $1 \cdot \frac{x^4}{24} + 2x^2 \cdot (-\frac{x^2}{2}) + (\text{higher order terms})$ $= \frac{x^4}{24} - x^4 + \cdots$

(This gets tedious!)

First few terms:

$e^{2x} \cos x = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{3} + \cdots$

Answer: $e^{2x}\cos x = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{3} + \cdots$

(Usually only first 3-4 terms are requested)

Let $u = x - 1$, so $x = 1 + u$:

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

Step 2: Use geometric series

We know: $\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n$ for $|r| < 1$

Replace $r$ with $-u$:

$\frac{1}{1+u} = \frac{1}{1-(-u)} = \sum_{n=0}^{\infty} (-u)^n = \sum_{n=0}^{\infty} (-1)^n u^n$

Step 3: Substitute back $u = x-1$

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

$= 1 - (x-1) + (x-1)^2 - (x-1)^3 + \cdots$

Step 4: Find interval of convergence

Need $|u| < 1$: $|x-1| < 1$ $-1 < x - 1 < 1$ $0 < x < 2$

Answer: $\frac{1}{x} = \sum_{n=0}^{\infty} (-1)^n(x-1)^n$ for $0 < x < 2$