Interval: $(-\infty, \infty)$

Radius: $R = \infty$

2. Sine Function

$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

Interval: $(-\infty, \infty)$

Only odd powers!

3. Cosine Function

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

Interval: $(-\infty, \infty)$

Only even powers!

4. Geometric Series

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

Interval: $(-1, 1)$

Most versatile for substitutions!

5. Natural Logarithm

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

Interval: $(-1, 1]$ (includes 1!)

6. Binomial Series (General)

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

Interval: $(-1, 1)$ when $p$ is not a positive integer

When $p$ is a positive integer, this is a finite polynomial (binomial theorem).

Application 1: Approximating Functions

Example: Approximate $\sin(0.1)$ using the first three terms.

$\sin x \approx x - \frac{x^3}{6}$

$\sin(0.1) \approx 0.1 - \frac{(0.1)^3}{6} = 0.1 - \frac{0.001}{6}$

$\approx 0.1 - 0.000167 = 0.099833$

Actual value: $\sin(0.1) \approx 0.0998334...$

Very close! ✓

Application 2: Evaluating Limits

Example: Find $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Substitute series for $e^x$:

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

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

$\frac{e^x - 1 - x}{x^2} = \frac{\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots}{x^2}$

$= \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} + \cdots$

As $x \to 0$:

$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{1}{2}$

No L'Hôpital's needed!

Application 3: Finding Derivatives at a Point

Example: Find $f^{(10)}(0)$ if $f(x) = x^2 e^{3x}$.

Step 1: Find Maclaurin series for $f(x)$

$e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots$

$x^2 e^{3x} = x^2\left(1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \cdots\right)$

$= x^2 + 3x^3 + \frac{9x^4}{2} + \frac{9x^5}{2} + \cdots$

Step 2: General term has form $\frac{f^{(n)}(0)}{n!}x^n$

The coefficient of $x^{10}$ is $\frac{f^{(10)}(0)}{10!}$.

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

For $x^{10}$ term: $n + 2 = 10$, so $n = 8$.

Coefficient: $\frac{3^8}{8!}$

$\frac{f^{(10)}(0)}{10!} = \frac{3^8}{8!}$

$f^{(10)}(0) = \frac{3^8 \cdot 10!}{8!} = 3^8 \cdot 10 \cdot 9 = 3^8 \cdot 90$

$= 6561 \cdot 90 = 590,490$

Application 4: Integrating Non-Elementary Functions

Example: Express $\int_0^1 \frac{\sin x}{x} dx$ as an infinite series.

Step 1: Write series for $\sin x$

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

Step 2: Divide by $x$

$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots$

Step 3: Integrate from 0 to 1

$\int_0^1 \frac{\sin x}{x} dx = \int_0^1 \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots\right) dx$

$= \left[x - \frac{x^3}{18} + \frac{x^5}{600} - \frac{x^7}{35280} + \cdots\right]_0^1$

$= 1 - \frac{1}{18} + \frac{1}{600} - \frac{1}{35280} + \cdots$

$\approx 0.9461$ (using first 4 terms)

Substitution Techniques

Example 5: Find series for $\frac{1}{1+x^2}$

Start with: $\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n$

Substitute $u = -x^2$:

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

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

$= 1 - x^2 + x^4 - x^6 + x^8 - \cdots$

Interval: $|x| < 1$

Example 6: Find series for $\arctan x$

From Example 5: $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots$

We know: $\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$

Integrate both sides:

$\arctan x = \int \left(1 - x^2 + x^4 - x^6 + \cdots\right) dx$

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

At $x = 0$: $\arctan 0 = 0$, so $C = 0$.

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

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

Interval: $[-1, 1]$

Fun Application: Approximating $\pi$

From $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$

At $x = 1$: $\arctan 1 = \frac{\pi}{4}$

$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots$

$\pi = 4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots\right)$

(This converges very slowly, but it's a cool formula!)

Differentiation Example

Example 7: Find series for $\frac{1}{(1-x)^2}$

Method 1: Differentiate $\frac{1}{1-x}$

$\frac{d}{dx}\left[\frac{1}{1-x}\right] = \frac{1}{(1-x)^2}$

$\frac{d}{dx}\left[\sum_{n=0}^{\infty} x^n\right] = \sum_{n=1}^{\infty} n x^{n-1}$

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

Or reindex: $\sum_{n=0}^{\infty} (n+1)x^n$

Multiplying Series

Example 8: Find first 3 terms of $e^x \sin x$

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

$\sin x = x - \frac{x^3}{6} + \cdots$

Multiply:

$(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots)(x - \frac{x^3}{6} + \cdots)$

Constant term: None (since $\sin x$ starts with $x$)

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

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

$x^3$ term: $\frac{x^2}{2} \cdot x + 1 \cdot (-\frac{x^3}{6}) + x \cdot 0 = \frac{x^3}{2} - \frac{x^3}{6} = \frac{x^3}{3}$

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

⚠️ Common Mistakes

Mistake 1: Wrong Interval After Substitution

For $\frac{1}{1+x^2}$ from $\frac{1}{1-u}$ with $u = -x^2$:

WRONG: Interval is still $(-1, 1)$

RIGHT: Need $|u| < 1$, so $|-x^2| < 1$, which means $|x| < 1$ (still $(-1,1)$!)

But if you substitute $u = 2x$, interval changes!

Mistake 2: Forgetting Integration Constant

When integrating series: $\int \sum c_n x^n dx = C + \sum \frac{c_n x^{n+1}}{n+1}$

Must find $C$ using initial condition!

Mistake 3: Using Series Outside Interval of Convergence

$\ln(1+x) = x - \frac{x^2}{2} + \cdots$ only for $-1 < x \leq 1$

WRONG: $\ln 3 = 2 - \frac{4}{2} + \frac{8}{3} - \cdots$ (using $x = 2$)

$x = 2$ is outside the interval!

Mistake 4: Wrong Power After Substitution

For $e^{x^2}$ from $e^x = \sum \frac{x^n}{n!}$:

WRONG: $e^{x^2} = \sum \frac{x^{2n}}{n!}$

RIGHT: $e^{x^2} = \sum \frac{(x^2)^n}{n!} = \sum \frac{x^{2n}}{n!}$ ✓

(In this case same answer, but be careful with the logic!)

📝 Practice Strategy

1. Memorize the Big 5: $e^x, \sin x, \cos x, \frac{1}{1-x}, \ln(1+x)$
2. Know intervals: Most are $(-\infty, \infty)$ except geometric and log
3. Substitution: Replace $x$ with anything, adjust interval
4. Term by term operations: Valid within radius of convergence
5. For limits: Expand both numerator and denominator
6. For integrals: Integrate series term by term
7. For derivatives at 0: Find coefficient of $x^n$ in series
8. Check first 3-4 terms: Usually sufficient for problems

$= x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots$

Step 2: Simplify numerator

$\sin x - x + \frac{x^3}{6} = \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) - x + \frac{x^3}{6}$

$= \frac{x^5}{120} - \frac{x^7}{5040} + \cdots$

Step 3: Divide by $x^5$

$\frac{\sin x - x + \frac{x^3}{6}}{x^5} = \frac{\frac{x^5}{120} - \frac{x^7}{5040} + \cdots}{x^5}$

$= \frac{1}{120} - \frac{x^2}{5040} + \cdots$

Step 4: Take limit

$\lim_{x \to 0} \left(\frac{1}{120} - \frac{x^2}{5040} + \cdots\right) = \frac{1}{120}$

Answer: $\frac{1}{120}$

Substitute $x \to -x^2$:

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

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

Step 2: Integrate term by term

$\int_0^{0.5} e^{-x^2} dx = \int_0^{0.5} \left(1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \frac{x^8}{24} - \cdots\right) dx$

$= \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216} - \cdots\right]_0^{0.5}$

Step 3: Evaluate at $x = 0.5$

$= 0.5 - \frac{(0.5)^3}{3} + \frac{(0.5)^5}{10} - \frac{(0.5)^7}{42} + \frac{(0.5)^9}{216} - \cdots$

$= 0.5 - \frac{0.125}{3} + \frac{0.03125}{10} - \frac{0.0078125}{42} + \frac{0.001953125}{216} - \cdots$

$= 0.5 - 0.041667 + 0.003125 - 0.000186 + 0.000009 - \cdots$

Step 4: Sum terms

$\approx 0.5 - 0.041667 + 0.003125 - 0.000186$

$\approx 0.461272$

To three decimal places: 0.461

Answer: $\int_0^{0.5} e^{-x^2} dx \approx 0.461$