Once we know the Maclaurin series for basic functions like $e^x$, $\sin x$, and $\frac{1}{1-x}$, we can create new series by substitution, differentiation, and integration. These manipulations are powerful tools for finding series representations of more complicated functions.
🎯 In this section you will learn
- How to use substitution to find series for composite functions
- How to differentiate and integrate power series term by term
- How the radius of convergence changes with these operations
- How to find series for functions like $\frac{1}{1+x^2}$, $\ln(1+x)$, and $\arctan x$
📌 Key Properties of Power Series
$$ \sum_{n=0}^\infty c_n (x-a)^n \quad \text{has radius of convergence } R $$
- Substitution: Replace $x$ with $u(x)$ (e.g., $x^2$, $2x$, $-x$). The new radius adjusts accordingly.
- Differentiation: $\frac{d}{dx} \sum c_n (x-a)^n = \sum n c_n (x-a)^{n-1}$ (same $R$).
- Integration: $\int \sum c_n (x-a)^n dx = \sum \frac{c_n}{n+1} (x-a)^{n+1} + C$ (same $R$, endpoints may change).
💡 Important
Differentiation and integration of power series can be performed term‑by‑term within the interval of convergence. The radius of convergence stays the same, but convergence at endpoints may change.
📌 Substitution
Find the Maclaurin series for $\frac{1}{1+x}$.
①
Known series
$\frac{1}{1-u} = \sum_{n=0}^\infty u^n$, $|u| < 1$
②
Substitute $u = -x$
$\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty (-1)^n x^n$
③
Interval
$|-x| < 1$ → $|x| < 1$, converges for $|x| < 1$
Find the Maclaurin series for $e^{x^2}$.
- $e^u = \sum_{n=0}^\infty \frac{u^n}{n!}$
- Let $u = x^2$: $e^{x^2} = \sum_{n=0}^\infty \frac{(x^2)^n}{n!} = \sum_{n=0}^\infty \frac{x^{2n}}{n!}$
Find the Maclaurin series for $\sin(3x)$.
- $\sin u = \sum_{n=0}^\infty \frac{(-1)^n u^{2n+1}}{(2n+1)!}$
- Let $u = 3x$: $\sin(3x) = \sum_{n=0}^\infty \frac{(-1)^n (3x)^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n 3^{2n+1} x^{2n+1}}{(2n+1)!}$
📌 Differentiation of Power Series
Find the series for $\frac{1}{(1-x)^2}$ by differentiating $\frac{1}{1-x}$.
①
Known series
$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$
②
Differentiate term by term
$\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2} = \frac{d}{dx} \sum_{n=0}^\infty x^n = \sum_{n=1}^\infty n x^{n-1} = \sum_{n=0}^\infty (n+1) x^n$
The Maclaurin series for $\ln(1+x)$ can be found by integrating $\frac{1}{1+x}$.
📌 Integration of Power Series
Find the Maclaurin series for $\ln(1+x)$.
①
Known series
$\frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n$
②
Integrate term by term
$\ln(1+x) = \int \frac{1}{1+x} dx = \int \sum_{n=0}^\infty (-1)^n x^n dx = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} + C$
③
Determine $C$
At $x=0$, $\ln(1)=0$, so $C=0$. Thus $\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n}$
Find the Maclaurin series for $\arctan x$.
- $\frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}$ (substitute $u = -x^2$ into $1/(1-u)$)
- Integrate: $\arctan x = \int \frac{1}{1+x^2} dx = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} + C$
- At $x=0$, $\arctan 0 = 0$, so $C=0$
- $\arctan x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$, $|x| \le 1$
⚠️ Important Notes
- Substitution: When substituting $u = cx^k$, the new radius is $R^{1/k} / |c|$ (if $R$ is the original radius).
- Differentiation/Integration: The radius of convergence stays the same, but convergence at endpoints may change.
- Always check endpoints separately after integration — the interval may expand or contract at the endpoints.
🔍 Key Takeaways
- Substitution: Replace $x$ with a function of $x$ in a known series.
- Differentiation: Differentiate term by term (same $R$).
- Integration: Integrate term by term (same $R$, endpoints may change).
- These techniques allow us to find series for many functions from a few basic ones.
← Back to Module Page