A Taylor series represents a function as an infinite sum of terms calculated from its derivatives at a single point. If the center is $a=0$, it's called a Maclaurin series. These series allow us to approximate complicated functions using polynomials, which are much easier to work with.
🎯 In this section you will learn
- The formula for Taylor series: $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$
- Maclaurin series (center at $0$)
- How to find Taylor series by computing derivatives
- The relationship between Taylor series and power series
📌 Taylor Series Formula
$$ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots $$
Taylor series of $f$ centered at $x=a$
💡 Maclaurin Series
When $a=0$, the series becomes $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$. This is the Maclaurin series.
Find the Maclaurin series for $f(x) = e^x$.
①
Find derivatives at $0$
$f^{(n)}(x) = e^x$, so $f^{(n)}(0) = 1$ for all $n$.
②
Apply the formula
$e^x = \sum_{n=0}^\infty \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$
③
Radius of convergence
$R = \infty$ (converges for all $x$)
Find the Maclaurin series for $f(x) = \sin x$.
①
Find derivatives at $0$
$f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=-1$, $f^{(4)}(0)=0$, pattern repeats every 4.
②
Apply the formula
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$
Find the Maclaurin series for $f(x) = \cos x$.
- Derivatives at $0$: $f(0)=1$, $f'(0)=0$, $f''(0)=-1$, $f'''(0)=0$, $f^{(4)}(0)=1$, etc.
- $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$
Find the Taylor series for $f(x) = \ln x$ centered at $a=1$.
①
Find derivatives at $x=1$
$f(1)=0$, $f'(x)=1/x$ → $f'(1)=1$
$f''(x)=-1/x^2$ → $f''(1)=-1$
$f'''(x)=2/x^3$ → $f'''(1)=2$
$f^{(4)}(x)=-6/x^4$ → $f^{(4)}(1)=-6$
②
Apply the formula
$\ln x = 0 + 1\cdot(x-1) - \frac{1}{2}(x-1)^2 + \frac{2}{6}(x-1)^3 - \frac{6}{24}(x-1)^4 + \cdots$
$= (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$
③
General form
$\ln x = \sum_{n=1}^\infty \frac{(-1)^{n-1} (x-1)^n}{n}$, for $0 < x \le 2$
⚠️ Important Notes
- Not all functions equal their Taylor series: The Taylor series may converge to something else if the function is not analytic.
- Computing derivatives: For simple functions like $e^x$, $\sin x$, $\cos x$, the pattern repeats.
- Memorize Maclaurin series for $e^x$, $\sin x$, $\cos x$, $\frac{1}{1-x}$, $\ln(1+x)$, $\arctan x$ — they appear frequently on the AP exam.
🔍 Key Takeaways
- Taylor series: $f(x) = \sum \frac{f^{(n)}(a)}{n!}(x-a)^n$.
- Maclaurin series: Taylor series with $a=0$.
- Finding a Taylor series requires computing derivatives at the center.
- Memorize the Maclaurin series for common functions.
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