Taylor series centered at \(0\) – the most common power-series expansions.
A Maclaurin series is simply a Taylor series with center \(a=0\).
Tip: List the derivatives, evaluate them at \(0\), build the first several terms, then write the general series from the pattern.
A Maclaurin series is a special case of a Taylor series. The Taylor series can be centered at any point \(a\). When we choose the center to be \(a = 0\), the Taylor series becomes a Maclaurin series.
Taylor series (any center \(a\))
\[ f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!}(x-a)^n \]
Maclaurin series (center \(a = 0\))
\[ f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n \]
So every Maclaurin series is a Taylor series — just the one centered at zero. Not every Taylor series is a Maclaurin series; only those with center 0 get the special name.
We center at 0 because it makes the math simpler and more useful.
Simpler Calculations
When \(a = 0\), the formula loses the \((x-a)\) terms. The powers are just plain \(x^n\), which are easier to write, differentiate, integrate, and multiply.
Nicest Values at Zero
Many common functions (\(e^x\), \(\sin x\), \(\cos x\), \(\frac{1}{1-x}\)) have especially clean derivative values at \(x = 0\). This makes building the series much easier.
Standard Reference
Almost every textbook and computer program stores the Maclaurin series of the basic functions. Once you know the series at 0, you can shift it to any other center if needed.
Cleanest Form
Power series centered at 0 look like ordinary polynomials: \(c_0 + c_1 x + c_2 x^2 + \cdots\). This form is ideal for multiplying series or solving differential equations.
In short: Centering at 0 is a convenience that gives us the simplest, most widely used version of the Taylor series. That is why the Maclaurin series gets its own name and appears so often.
Yes — it is necessary (and very helpful) to memorize a few key Maclaurin series.
They appear constantly
On quizzes, tests, and the AP Calculus BC exam you will be expected to know the basic series cold. You rarely have time to re-derive them from scratch.
Most problems use substitution
The majority of Maclaurin problems say “use a known series…” rather than “take derivatives.” If you don’t already know the base series, you can’t solve these problems quickly.
They are the building blocks
Once you know the core series, you can generate dozens of others just by substituting, multiplying by \(x\), differentiating, or integrating.
They save time and reduce errors
Re-deriving \(e^x\) or \(\sin x\) every time is slow and invites algebra mistakes. Having them memorized lets you focus on the actual problem.
The ones you should know by heart
| Function | Maclaurin Series | Interval |
|---|---|---|
| \(e^x\) | \(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\) | all \(x\) |
| \(\sin x\) | \(\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}\) | all \(x\) |
| \(\cos x\) | \(\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}x^{2n}\) | all \(x\) |
| \(\dfrac{1}{1-x}\) | \(\displaystyle\sum_{n=0}^{\infty} x^n\) | \(\lvert x\rvert < 1\) |
| \(\ln(1+x)\) | \(\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}x^n\) | \(-1 < x \le 1\) |
Bottom line: You do not need to memorize every possible series. You do need to memorize the five core ones above. Everything else is built from them by substitution, differentiation, or integration.
Find the Maclaurin series for \(f(x) = e^x\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= e^x \\ f'(x) &= e^x \\ f”(x) &= e^x \\ f”'(x) &= e^x \\ &\vdots \\ f^{(n)}(x) &= e^x \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ f^{(n)}(0) = e^0 = 1 \quad \text{for every } n \]
Step 3: Build the first several non-zero terms
\[ \begin{align*} n=0: &\quad \dfrac{1}{0!}x^0 = 1 \\ n=1: &\quad \dfrac{1}{1!}x = x \\ n=2: &\quad \dfrac{1}{2!}x^2 = \dfrac{x^2}{2} \\ n=3: &\quad \dfrac{1}{6}x^3 \\ n=4: &\quad \dfrac{1}{24}x^4 \end{align*} \]
Beginning of the series: \[ e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \cdots \]
Step 4: Write the general series
\[ e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \]
\( e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \)
Find the Maclaurin series using derivatives evaluated at 0
Find the Maclaurin series for \(f(x) = \sin x\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \sin x \\ f'(x) &= \cos x \\ f”(x) &= -\sin x \\ f”'(x) &= -\cos x \\ f^{(4)}(x) &= \sin x \\ &\vdots \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ \begin{align*} f(0) &= 0 \\ f'(0) &= 1 \\ f”(0) &= 0 \\ f”'(0) &= -1 \\ f^{(4)}(0) &= 0 \\ f^{(5)}(0) &= 1 \end{align*} \]
All even-powered terms are zero.
Step 3: Build the first several non-zero terms
\[ \begin{align*} n=1: &\quad x \\ n=3: &\quad -\dfrac{x^3}{6} \\ n=5: &\quad \dfrac{x^5}{120} \\ n=7: &\quad -\dfrac{x^7}{5040} \end{align*} \]
\[ \sin x = x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \dfrac{x^7}{7!} + \cdots \]
Step 4: Write the general series
\[ \sin x = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \]
\( \sin x = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \)
Find the Maclaurin series for \(f(x) = \cos x\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \cos x \\ f'(x) &= -\sin x \\ f”(x) &= -\cos x \\ f”'(x) &= \sin x \\ f^{(4)}(x) &= \cos x \\ &\vdots \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ \begin{align*} f(0) &= 1 \\ f'(0) &= 0 \\ f”(0) &= -1 \\ f”'(0) &= 0 \\ f^{(4)}(0) &= 1 \end{align*} \]
All odd-powered terms are zero.
Step 3: Build the first several non-zero terms
\[ \begin{align*} n=0: &\quad 1 \\ n=2: &\quad -\dfrac{x^2}{2} \\ n=4: &\quad \dfrac{x^4}{24} \\ n=6: &\quad -\dfrac{x^6}{720} \end{align*} \]
\[ \cos x = 1 – \dfrac{x^2}{2!} + \dfrac{x^4}{4!} – \dfrac{x^6}{6!} + \cdots \]
Step 4: Write the general series
\[ \cos x = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}x^{2n} \]
\( \cos x = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}x^{2n} \)
Find the Maclaurin series for \(f(x) = \sin(2x)\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \sin(2x) \\ f'(x) &= 2\cos(2x) \\ f”(x) &= -4\sin(2x) \\ f”'(x) &= -8\cos(2x) \\ f^{(4)}(x) &= 16\sin(2x) \\ f^{(5)}(x) &= 32\cos(2x) \\ &\vdots \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ \begin{align*} f(0) &= 0 \\ f'(0) &= 2 \\ f”(0) &= 0 \\ f”'(0) &= -8 \\ f^{(4)}(0) &= 0 \\ f^{(5)}(0) &= 32 \end{align*} \]
All even-powered terms are zero.
Step 3: Build the first several non-zero terms
\[ \begin{align*} n=1: &\quad \dfrac{2}{1!}x = 2x \\ n=3: &\quad \dfrac{-8}{3!}x^3 = -\dfrac{4}{3}x^3 \\ n=5: &\quad \dfrac{32}{5!}x^5 = \dfrac{4}{15}x^5 \end{align*} \]
\[ \sin(2x) = 2x – \dfrac{4}{3}x^3 + \dfrac{4}{15}x^5 – \cdots \]
Step 4: Write the general series
The pattern of the coefficients is \(2^{2n+1}\) with alternating signs: \[ \sin(2x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n 2^{2n+1}}{(2n+1)!}x^{2n+1} \]
\( \sin(2x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n 2^{2n+1}}{(2n+1)!}x^{2n+1} \)
Find the Maclaurin series for \(f(x) = e^{3x}\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= e^{3x} \\ f'(x) &= 3e^{3x} \\ f”(x) &= 9e^{3x} \\ f”'(x) &= 27e^{3x} \\ f^{(4)}(x) &= 81e^{3x} \\ &\vdots \\ f^{(n)}(x) &= 3^n e^{3x} \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ f^{(n)}(0) = 3^n e^{0} = 3^n \]
Step 3: Build the first several non-zero terms
\[ \begin{align*} n=0: &\quad 1 \\ n=1: &\quad 3x \\ n=2: &\quad \dfrac{9}{2}x^2 \\ n=3: &\quad \dfrac{27}{6}x^3 = \dfrac{9}{2}x^3 \\ n=4: &\quad \dfrac{81}{24}x^4 = \dfrac{27}{8}x^4 \end{align*} \]
\[ e^{3x} = 1 + 3x + \dfrac{9}{2}x^2 + \dfrac{9}{2}x^3 + \dfrac{27}{8}x^4 + \cdots \]
Step 4: Write the general series
\[ e^{3x} = \sum_{n=0}^{\infty} \dfrac{3^n}{n!}x^n \]
\( e^{3x} = \sum_{n=0}^{\infty} \dfrac{3^n}{n!}x^n \)
Slightly more involved derivatives or pattern recognition
Find the Maclaurin series for \(f(x) = \ln(1+x)\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \ln(1+x) \\ f'(x) &= \dfrac{1}{1+x} \\ f”(x) &= -\dfrac{1}{(1+x)^2} \\ f”'(x) &= \dfrac{2}{(1+x)^3} \\ f^{(4)}(x) &= -\dfrac{6}{(1+x)^4} \\ f^{(5)}(x) &= \dfrac{24}{(1+x)^5} \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ \begin{align*} f(0) &= 0 \\ f'(0) &= 1 \\ f”(0) &= -1 \\ f”'(0) &= 2 \\ f^{(4)}(0) &= -6 \\ f^{(5)}(0) &= 24 \end{align*} \]
Step 3: Build the first several non-zero terms
\[ \begin{align*} n=1: &\quad x \\ n=2: &\quad -\dfrac{x^2}{2} \\ n=3: &\quad \dfrac{x^3}{3} \\ n=4: &\quad -\dfrac{x^4}{4} \\ n=5: &\quad \dfrac{x^5}{5} \end{align*} \]
\[ \ln(1+x) = x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \dfrac{x^4}{4} + \dfrac{x^5}{5} – \cdots \]
Step 4: Write the general series
\[ \ln(1+x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}x^n \]
\( \ln(1+x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}x^n \)
Find the Maclaurin series for \(f(x) = \dfrac{1}{1-x}\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= (1-x)^{-1} \\ f'(x) &= (1-x)^{-2} \\ f”(x) &= 2(1-x)^{-3} \\ f”'(x) &= 6(1-x)^{-4} \\ f^{(4)}(x) &= 24(1-x)^{-5} \\ &\vdots \\ f^{(n)}(x) &= n!(1-x)^{-(n+1)} \end{align*} \]
Step 2: Evaluate at \(a = 0\)
\[ f^{(n)}(0) = n! \]
Step 3: Build the first several terms
\[ \begin{align*} n=0: &\quad 1 \\ n=1: &\quad x \\ n=2: &\quad x^2 \\ n=3: &\quad x^3 \\ n=4: &\quad x^4 \end{align*} \]
\[ \dfrac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \cdots \]
Step 4: Write the general series
\[ \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]
\( \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n \quad (|x|<1) \)
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