AP Calculus BC: New Series from Known Series

Mastering Taylor series is crucial for solving complex problems in engineering, physics, and advanced mathematics, but recomputing derivatives for every new function is tedious. Instead, you can build an entire library of series by strategically manipulating a few memorized, foundational ones. This approach transforms series from a rote calculation into a powerful and flexible problem-solving tool, saving immense time on exams and in applications.

The Foundational Series: Your Building Blocks

Before you can generate new series, you must have a solid grasp of the core, memorized series. These are typically derived from the Maclaurin series (a Taylor series centered at $x=0$) for key functions. The most essential are:

• Exponential: $e^x={\sum }_{n=0}^{\infty }\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ for all $x$.
• Sine: $\sin x={\sum }_{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ for all $x$.
• Cosine: $\cos x={\sum }_{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ for all $x$.
• Geometric: $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n=1+x+x^2+x^3+\cdots$ for $|x|<1$.

Think of these as your primary tools. Every other technique is about adapting these tools to fit new tasks without starting from scratch.

Substitution: The Most Direct Method

Substitution is the simplest and most common technique. You replace the $x$ in a known series with a more complicated expression, provided it fits within the series' interval of convergence. This is analogous to evaluating a function: if you know $f(x)$, you can find $f(u)$ where $u$ is some expression.

Example: Find the Maclaurin series for $e^{-3x^2}$. You know the series for $e^u$. Simply let $u=-3x^2$ and substitute it into the series for $e^u$: $$e^{-3x^2}=\sum _{n=0}^{\infty }\frac{(-3x^2{)}^n}{n!}=\sum _{n=0}^{\infty }\frac{(-3{)}^n{x}^{2n}}{n!}=1-3x^2+\frac{9x^4}{2!}-\frac{27x^6}{3!}+\cdots$$

The convergence remains for all $x$ because the original $e^x$ series converges for any real number input. However, you must be cautious with the geometric series. If you want the series for $\frac{1}{1+4x}$, you substitute $u=-4x$: $$\frac{1}{1+4x}=\frac{1}{1-(-4x)}=\sum _{n=0}^{\infty }(-4x{)}^n=\sum _{n=0}^{\infty }(-4{)}^n{x}^n$$ This new series converges when $|-4x|<1$, or $|x|<\frac{1}{4}$.

Multiplication and Division of Series

You can find the series for a product of two functions by multiplying their corresponding Taylor series, much like multiplying polynomials. You collect all terms that contribute to a given power of $x$. Division is more complex and often involves algebraic long division of the series or relating the function to a geometric series.

Example (Multiplication): Find the first four non-zero terms for $e^x\sin x$. Multiply the two infinite series: $$(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots )\cdot (x-\frac{x^3}{6}+\cdots )$$ To get the terms up to $x^4$:

• Constant term: None.
• $x$ term: $1\cdot x=x$.
• $x^2$ term: $1\cdot 0+x\cdot x=x^2$.
• $x^3$ term: $(1\cdot (-\frac{x^3}{6}))+(x\cdot 0)+(\frac{x^2}{2}\cdot x)=-\frac{1}{6}{x}^3+\frac{1}{2}{x}^3=\frac{1}{3}{x}^3$.
• $x^4$ term: $(\frac{x^2}{2}\cdot (-\frac{x^3}{6}))+(\frac{x^3}{6}\cdot x)=-\frac{1}{12}{x}^5+\frac{1}{6}{x}^4$. We only take the $x^4$ term: $\frac{1}{6}{x}^4$.

Thus, $e^x\sin x=x+x^2+\frac{1}{3}{x}^3+\frac{1}{6}{x}^4+\cdots$.

Differentiation and Integration of Series

Within their interval of convergence (excluding endpoints, which require separate check), a power series can be differentiated or integrated term-by-term. This is a powerful theorem that lets you move between related functions.

• Differentiation: The series for $f^{\prime }(x)$ is found by differentiating each term of the series for $f(x)$.
• Integration: The series for $\int f(x)dx$ is found by integrating each term of the series for $f(x)$, remembering the "+ C".

Example: Find the Maclaurin series for $\frac{1}{(1-x{)}^2}$. Notice that $\frac{1}{(1-x{)}^2}$ is the derivative of $\frac{1}{1-x}$. Start with the geometric series and differentiate both the function and its series representation term-by-term: $$\frac{d}{dx}\left[\frac{1}{1-x}\right]=\frac{1}{(1-x{)}^2}$$ $$\frac{d}{dx}\left[\sum _{n=0}^{\infty }{x}^n\right]=\sum _{n=1}^{\infty }n{x}^{n-1}\text{(Note the index shift)}$$ Therefore, $\frac{1}{(1-x{)}^2}={\sum }_{n=1}^{\infty }n{x}^{n-1}={\sum }_{n=0}^{\infty }(n+1)x^n=1+2x+3x^2+4x^3+\cdots$, and it converges for $|x|<1$.

Composition of Functions

Composition involves substituting one entire series into another. This is useful for functions like $e^{\sin x}$ or $\cos (x^3)$. The process is straightforward but requires careful algebra to collect like terms. You substitute the entire inner series (e.g., for $\sin x$) in for the variable in the outer series (e.g., for $e^u$), then expand and simplify.

Example: Find the first three non-zero terms of the Maclaurin series for $e^{\sin x}$.

1. Inner series: $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$
2. Outer series: $e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$
3. Substitute $u=(x-\frac{x^3}{6}+\cdots )$:

$$e^{\sin x}=1+(x-\frac{x^3}{6}+\cdots )+\frac{1}{2}(x-\frac{x^3}{6}+\cdots {)}^2+\frac{1}{6}(x-\cdots {)}^3+\cdots$$ Now expand, keeping terms up to $x^4$:

• The $1$ is $1$.
• The linear term is $x$.
• The squared term: $\frac{1}{2}(x^2-2\cdot x\cdot \frac{x^3}{6}+\cdots )=\frac{x^2}{2}-\frac{x^4}{6}+\cdots$. We keep the $x^2/2$ and the $x^4/6$ for now.
• The cubed term: $\frac{1}{6}(x^3+\cdots )=\frac{x^3}{6}+\cdots$.

1. Combine terms: $1+x+\frac{x^2}{2}+(-\frac{x^3}{6}+\frac{x^3}{6})+(-\frac{x^4}{6})+\cdots =1+x+\frac{x^2}{2}-\frac{x^4}{6}+\cdots$

Common Pitfalls

1. Ignoring the Radius of Convergence: When you manipulate a series, the interval of convergence may change. For substitution, the new series converges when the substituted expression is within the original interval. For differentiation, the radius of convergence stays the same, but convergence at the endpoints may be lost. For integration, it stays the same, but convergence may be gained at an endpoint. Always state or consider the new interval.
2. Incorrect Multiplication or Composition: When multiplying or composing series, it's easy to miss terms that contribute to a specific power. Use a systematic method: write terms in ascending order and align them by power. For composition, substitute the entire inner series and expand carefully, truncating only after collecting all terms of the desired degree.
3. Index Errors during Calculus Operations: Differentiating ${\sum }_{n=0}^{\infty }{x}^n$ gives ${\sum }_{n=1}^{\infty }n{x}^{n-1}$. The index shifts because the constant term ($n=0$) becomes zero. You can re-index back to $n=0$ if needed, but you must adjust the exponent and coefficient correctly: ${\sum }_{n=1}^{\infty }n{x}^{n-1}={\sum }_{n=0}^{\infty }(n+1)x^n$.
4. Overcomplicating the Problem: Always look for the simplest known series to manipulate. For $\frac{x}{2+3x}$, rewrite it as $\frac{x}{2}\cdot \frac{1}{1-(-\frac{3x}{2})}$ to use the geometric series. For $\arctan (x^2)$, recognize its derivative is $\frac{2x}{1+x^4}$ and use a substitution into the geometric series, then integrate.

Summary

• Work from a core set of memorized series (for $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1-x}$) to generate infinite others through algebraic and calculus operations.
• Substitution is your first and most frequent tool: replace $x$ with any expression that fits within the convergence interval.
• Differentiate or integrate series term-by-term to find series for derivatives and antiderivatives; the radius of convergence is preserved.
• Multiply or compose series by treating them like elongated polynomials, carefully collecting all terms that contribute to each power of $x$.
• Always track the interval of convergence after any manipulation, as it is as important as the series itself for determining where your representation is valid.