Maclaurin and Taylor Series

Intuition

Any smooth function can be rebuilt from its derivatives at a single point: Taylor series say that if you know all the derivatives of a function at one point, you can reconstruct the entire function as an infinite polynomial. It’s like having a complete description of a curve encoded in a single number and its rates of change. The Maclaurin series is the special case where we expand about $x = 0$.

Why it matters: Taylor series let you approximate complicated functions ($e^x$, $\sin x$, $\cos x$) as simple polynomials. This is how calculators compute $\sin(0.5)$ — they don’t use geometry, they use the Maclaurin series. They’re also essential for solving differential equations and understanding convergence.

The key insight: More terms in the series give better approximations near the expansion point, but worse approximations far away. The radius of convergence tells you how far the approximation is valid. For $e^x$, the series converges everywhere; for $\ln(1+x)$, it only converges for $-1 < x \leq 1$.

Adjust the parameters in the graph above to explore the relationships between variables.

Board Coverage

1. Maclaurin Series

1.1 Definition

Definition. The Maclaurin series of a function $f$ that is infinitely differentiable at $x = 0$ is

$\boxed{f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \cdots = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\,x^n}$

This is a special case of the Taylor series about $x = a = 0$.

1.2 Derivation from the Taylor series

Assume $f$ can be written as a power series $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$.

Differentiating repeatedly and evaluating at $x = 0$:

• $f(0) = a_0$
• $f'(0) = a_1$
• $f''(0) = 2a_2$
• $f'''(0) = 6a_3 = 3!\,a_3$

$f^{(n)}(0) = n!\,a_n$Giving $a_n = \dfrac{f^{(n)}(0)}{n!}$.

Substituting back gives the Maclaurin series.

2. Standard Maclaurin Series

2.1 Exponential function

Proof of the Maclaurin series for $e^x$

Let $f(x) = e^x$. Then $f^{(n)}(x) = e^x$ for all $n$So $f^{(n)}(0) = 1$ for all $n$.

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

$\boxed{e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}} \quad \mathrm{for all } x \in \mathbb{R}$

$\square$

Radius of convergence: $\infty$ (converges for all real $x$).

2.2 Sine function

Proof of the Maclaurin series for $\sin x$

Let $f(x) = \sin x$. The derivatives cycle: \sin x$$\cos x$$-\sin x$$-\cos x$$\sin x…

At $x = 0$: $0, 1, 0, -1, 0, 1, 0, -1, \ldots$

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

$\boxed{\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}} \quad \mathrm{for all } x \in \mathbb{R}$

$\square$

Radius of convergence: $\infty$.

2.3 Cosine function

Let $f(x) = \cos x$. The derivatives cycle: \cos x$$-\sin x$$-\cos x$$\sin x$$\cos x…

At $x = 0$: $1, 0, -1, 0, 1, 0, -1, 0, \ldots$

$\boxed{\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}} \quad \mathrm{for all } x \in \mathbb{R}$

2.4 Natural logarithm

Let $f(x) = \ln(1+x)$ for $|x| < 1$.

f'(x) = \dfrac{1}{1+x}$$f''(x) = -\dfrac{1}{(1+x)^2}$$f'''(x) = \dfrac{2}{(1+x)^3}…

$f^{(n)}(x) = \dfrac{(-1)^{n-1}(n-1)!}{(1+x)^n}$So $f^{(n)}(0) = (-1)^{n-1}(n-1)!$.

$\boxed{\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}} \quad \mathrm{for } -1 < x \leq 1$

Radius of convergence: 1. The series also converges at $x = 1$ (alternating harmonic series).

2.5 Binomial series

For $|x| < 1$ and any real $n$:

$\boxed{(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots = \sum_{k=0}^{\infty}\binom{n}{k}x^k}$

Where $\dbinom{n}{k} = \dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$.

Radius of convergence: 1.

2.6 Arctangent

$\boxed{\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}} \quad \mathrm{for } |x| \leq 1$

Radius of convergence: 1. Converges at $x = \pm 1$ by the alternating series test.

Derivation. Since $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots$ for $|x| < 1$Integrate term by term.

2.7 Summary table

| Function | Maclaurin Series | Valid for | | ----------- | ----------------------------------------------------------------- | --------------- | --- | ------- | | $e^x$ | $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ | all $x$ | | $\sin x$ | $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ | all $x$ | | $\cos x$ | $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}$ | all $x$ | | $\ln(1+x)$ | $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}$ | $-1 < x \leq 1$ | | $(1+x)^n$ | $\displaystyle\sum_{k=0}^{\infty}\binom{n}{k}x^k$ | $| x | < 1$ | | $\arctan x$ | $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}$ | $| x | \leq 1$ |

3. Taylor Series About $x = a$

3.1 Definition

Definition. The Taylor series of $f$ about $x = a$ is

$\boxed{f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2!}f''(a) + \frac{(x-a)^3}{3!}f'''(a) + \cdots = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n}$

The Maclaurin series is the special case $a = 0$.

3.2 Taylor’s theorem with remainder

Theorem. If $f$ is $(n+1)$-times differentiable on an interval containing $a$ and $x$Then

$f(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k + R_n$

Where the remainder in Lagrange form is

$\boxed{R_n = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}}$

For some $c$ between $a$ and $x$.

Intuition. The remainder term tells us the error when truncating the series after $n$ terms. If $|R_n| \to 0$ as $n \to \infty$Then the Taylor series converges to $f(x)$.

3.3 Worked example

Find the Taylor series of $e^x$ about $x = 1$ up to the $x^3$ term.

f(x) = e^x$$f'(x) = e^x$$f''(x) = e^x$$f'''(x) = e^x.

At $x = 1$: $f(1) = f'(1) = f''(1) = f'''(1) = e$.

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

3.4 Converting between Maclaurin and Taylor series

Example. Find the Taylor series of $\ln x$ about $x = 1$.

Let $u = x - 1$So $\ln x = \ln(1 + u)$.

$\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$

$\boxed{\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots} \quad \mathrm{for } 0 < x \leq 2$

4. Interval and Radius of Convergence

4.1 Definitions

Definition. The radius of convergence $R$ of a power series $\displaystyle\sum_{n=0}^{\infty}a_n x^n$ is the value such that the series converges for $|x| < R$ And diverges for $|x| > R$.

The interval of convergence is the set of all $x$ for which the series converges. The endpoints $x = \pm R$ must be tested separately.

4.2 The ratio test

Theorem (Ratio test). For the series $\displaystyle\sum_{n=0}^{\infty}a_n x^n$Let

$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$

If $L$ exists, the radius of convergence is $R = \dfrac{1}{L}$.

Equivalently, if $L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}x^{n+1}}{a_n x^n}\right| = |x|\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$:

• If $L < 1$: the series converges absolutely.
• If $L > 1$: the series diverges.
• If $L = 1$: the test is inconclusive.

4.3 Worked examples

Example 1. Find the radius of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}$.

a_n = \dfrac{(-1)^{n-1}}{n}$$a_{n+1} = \dfrac{(-1)^n}{n+1}.

$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{n}{n+1} \to 1$ as $n \to \infty$.

$R = 1$. The series converges for $|x| < 1$.

At $x = 1$: alternating harmonic series, converges. At $x = -1$: $-\sum 1/n$Diverges.

Interval of convergence: $-1 < x \leq 1$.

Example 2. Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$.

$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{1/(n+1)!}{1/n!} = \dfrac{1}{n+1} \to 0$.

$R = \infty$. Converges for all $x$.

4.4 Convergence of the geometric series

Proof of convergence of the geometric series

For $|x| < 1$The geometric series is $\displaystyle\sum_{n=0}^{\infty}x^n$.

The partial sum is $S_N = 1 + x + x^2 + \cdots + x^N$.

$S_N = \dfrac{1 - x^{N+1}}{1 - x}$ (standard formula).

Since $|x| < 1$: $x^{N+1} \to 0$ as $N \to \infty$So $S_N \to \dfrac{1}{1-x}$.

$\boxed{\sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \quad \mathrm{for } |x| < 1}$

$\square$

Tip: tip $\dfrac{1}{1+x}$, $\ln(1+x)$And $\arctan x$ all derive from it.

5. Applications

5.1 Approximating values

Example. Approximate $e^{0.1}$ using the first four terms of the Maclaurin series.

$e^{0.1} \approx 1 + 0.1 + \dfrac{0.01}{2} + \dfrac{0.001}{6} = 1 + 0.1 + 0.005 + 0.000167 \approx 1.10517$.

The actual value is $\approx 1.10517$So the error is negligible with just 4 terms.

Example. Approximate $\sin(0.1)$ and bound the error.

$\sin(0.1) \approx 0.1 - \dfrac{0.001}{6} = 0.1 - 0.000167 = 0.099833$.

The next term is $\dfrac{(0.1)^5}{120} \approx 8.3 \times 10^{-8}$So the error is at most this.

5.2 Evaluating limits using series

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

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

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

$\dfrac{e^x - 1 - x}{x^2} = \dfrac{1}{2} + \dfrac{x}{6} + \cdots \to \dfrac{1}{2}$ as $x \to 0$.

5.3 Series expansions of compound functions

Example. Find the Maclaurin series of $e^{x^2}$ up to the $x^6$ term.

Substitute $x^2$ into the series for $e^x$:

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

Example. Find the Maclaurin series of $\cos(x^2)$ up to the $x^8$ term.

$\cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots$

Example. Find $\dfrac{e^x}{1-x}$ up to the $x^3$ term.

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

$\dfrac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

Multiply (Cauchy product):

$\dfrac{e^x}{1-x} = \left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right)\left(1 + x + x^2 + x^3\right) + O(x^4)$

$= 1 + (1+1)x + \left(1+1+\dfrac{1}{2}\right)x^2 + \left(1+1+\dfrac{1}{2}+\dfrac{1}{6}\right)x^3$

$= 1 + 2x + \dfrac{5}{2}x^2 + \dfrac{8}{3}x^3 + \cdots$

5.4 L’Hôpital’s rule and series

Intuition. L’Hôpital’s rule is a consequence of Taylor’s theorem. If $f(a) = g(a) = 0$Then Near $a$:

$\frac{f(x)}{g(x)} \approx \frac{f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2}{g'(a)(x-a) + \frac{g''(a)}{2}(x-a)^2} = \frac{f'(a) + \frac{f''(a)}{2}(x-a)}{g'(a) + \frac{g''(a)}{2}(x-a)}$

Taking $x \to a$ gives $\dfrac{f'(a)}{g'(a)}$Which is L’Hôpital’s rule. The series approach often Gives more information (higher-order terms) than applying L’Hôpital’s rule repeatedly.

6. Key Results and Pitfalls

Problems

7. Advanced Worked Examples

Example 7.1: Maclaurin series of a composite function

Problem. Find the Maclaurin series of $f(x) = \dfrac{1}{\sqrt{1 + x^2}}$ up to the $x^6$ term.

Solution. Use the binomial series with $n = -1/2$:

$(1 + u)^{-1/2} = 1 - \dfrac{u}{2} + \dfrac{3u^2}{8} - \dfrac{5u^3}{16} + \dfrac{35u^4}{128} - \cdots$

Substituting $u = x^2$:

$\frac{1}{\sqrt{1+x^2}} = 1 - \frac{x^2}{2} + \frac{3x^4}{8} - \frac{5x^6}{16} + \cdots$

Valid for $|x^2| < 1$I.e., $|x| < 1$.

Example 7.2: Taylor series and error bounds

Problem. Use the Taylor series of $\cos x$ about $x = 0$ to approximate $\cos(0.2)$ and bound The error.

Solution. $\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^6}{720} + \cdots$

Using three terms: $\cos(0.2) \approx 1 - 0.02 + \dfrac{0.0016}{24} = 1 - 0.02 + 0.0000667 = 0.980067$.

Error bound: the next term is $\dfrac{(0.2)^6}{720} = \dfrac{6.4 \times 10^{-7}}{720} \approx 8.9 \times 10^{-10}$.

So $|\text{error}| < 10^{-9}$.

Example 7.3: Product of two Maclaurin series

Problem. Find the Maclaurin series of $e^x \cos x$ up to the $x^4$ term.

Solution. $e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \cdots$

$\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$

Multiplying and collecting terms:

$1$: $1 \times 1 = 1$ $x$: $1 \cdot x = x$ $x^2$: $1 \cdot (-x^2/2) + x \cdot x = -x^2/2 + x^2 = x^2/2$ $x^3$: $1 \cdot 0 + x \cdot (-x^2/2) + (x^2/2) \cdot x = -x^3/2 + x^3/2 = 0$ $x^4$: $1 \cdot (x^4/24) + x \cdot 0 + (x^2/2)(-x^2/2) + (x^3/6) \cdot x = x^4/24 - x^4/4 + x^4/6 = (1 - 6 + 4)x^4/24 = -x^4/24$

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

Example 7.4: Series expansion of an inverse function

Problem. Find the Maclaurin series of $\sec x$ up to the $x^4$ term.

Solution. Write $\sec x = \dfrac{1}{\cos x} = (1 - x^2/2 + x^4/24 - \cdots)^{-1}$.

Using $(1 - u)^{-1} = 1 + u + u^2 + \cdots$ with $u = x^2/2 - x^4/24 + \cdots$:

$\sec x = 1 + \left(\frac{x^2}{2} - \frac{x^4}{24}\right) + \left(\frac{x^2}{2}\right)^2 + \cdots$

$= 1 + \frac{x^2}{2} + \left(-\frac{x^4}{24} + \frac{x^4}{4}\right) + \cdots = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$

Example 7.5: Using series to find a limit

Problem. Evaluate $\displaystyle\lim_{x\to 0}\frac{x - \sin x}{x^3}$.

Solution. $\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$

$\frac{x - \sin x}{x^3} = \frac{x - x + x^3/6 - x^5/120 + \cdots}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \cdots \to \boxed{\frac{1}{6}}$

Example 7.6: Maclaurin series by differentiation

Problem. Find the Maclaurin series of $f(x) = (1 + x)^3 e^x$ up to the $x^3$ term.

Solution. $f(x) = (1 + 3x + 3x^2 + x^3)e^x$.

$e^x = 1 + x + x^2/2 + x^3/6 + \cdots$

$f(x) = (1 + 3x + 3x^2 + x^3)(1 + x + x^2/2 + x^3/6)$

$1$: $1 \times 1 = 1$ $x$: $1 + 3 = 4$ $x^2$: $1/2 + 3 + 3 = 13/2$ $x^3$: $1/6 + 3/2 + 3 + 1 = 1/6 + 9/6 + 18/6 + 6/6 = 34/6 = 17/3$

$f(x) = 1 + 4x + \frac{13x^2}{2} + \frac{17x^3}{3} + \cdots$

Example 7.7: Convergence of alternating series

Problem. How many terms of $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ are needed to Approximate $\ln 2$ with an error less than $0.001$?

Solution. This is the alternating harmonic series, with $S = \ln 2$. The error after $N$ terms Is bounded by the absolute value of the $(N+1)$-th term:

$|\text{error}| \leq \frac{1}{N+1} < 0.001 \implies N + 1 > 1000 \implies N \geq 1000$

So at least 1000 terms are needed.

Example 7.8: Taylor series and approximation of definite integrals

Problem. Use the Maclaurin series of $e^{-x^2}$ to approximate $\displaystyle\int_0^{1/2} e^{-x^2}\,dx$ to 5 decimal places.

Solution. $e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \dfrac{x^8}{24} - \cdots$

$\int_0^{1/2}e^{-x^2}\,dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216}\right]_0^{1/2}$

$= \frac{1}{2} - \frac{1}{24} + \frac{1}{320} - \frac{1}{5376} + \frac{1}{110592}$

$= 0.5 - 0.041667 + 0.003125 - 0.000186 + 0.000009 = 0.461281$

The error is bounded by the next term: $\dfrac{1}{11 \times 2^{11}} = \dfrac{1}{22528} \approx 0.000044$.

So the integral $\approx 0.4613$ to 4 decimal places.

8. Connections to Other Topics

8.1 Maclaurin series and differential equations

The Maclaurin series provides a method for solving differential equations by substituting a power Series ansatz. See Differential Equations.

8.2 Taylor series and complex numbers

Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$ is the link between exponential series and Trigonometric series. See Complex Numbers.

8.3 Binomial series and further algebra

The binomial expansion is essential for partial fraction decomposition and generating functions. See Further Algebra.

9. Additional Exam-Style Questions

Question 11

(a) Find the Maclaurin series of $\ln\!\left(\dfrac{1 + x}{1 - x}\right)$ up to the $x^5$ term.

(b) State the range of validity.

Question 12

Use series to find $\displaystyle\lim_{x\to 0}\frac{1 - \cos x - \frac{x^2}{2}}{x^4}$.

Question 13

Prove by induction that $\displaystyle\frac{d^n}{dx^n}(xe^x) = (x + n)e^x$ for all $n \geq 0$.