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

Board	Paper	Notes
AQA	Paper 1	Maclaurin series; standard series for e^x$$\sin x$$\cos x$$\ln(1+x)$$(1+x)^n
Edexcel	FP2	Maclaurin + Taylor series about general $x = a$; convergence
OCR (A)	Paper 1	Maclaurin series; standard series; applications
CIE	P2	Maclaurin series; expansion of compound functions; range of validity

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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.

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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 $x$
$\ln(1+x)$	$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}$	$-1 < x \leq 1$
$(1+x)^n$	$\displaystyle\sum_{k=0}^{\infty}\binom{n}{k}x^k$	$\| x \| < 1$
$\arctan x$	$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}$	$\| x \| \leq 1$

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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 $\ln x = \ln(1 + u)$.

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

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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.

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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.

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6. Key Results and Pitfalls

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Problems

Problem 1

Find the Maclaurin series of $f(x) = \cos 2x$ up to the $x^6$ term.

Solution 1

f(0) = 1$$f'(x) = -2\sin 2x$$f'(0) = 0$$f''(x) = -4\cos 2x$$f''(0) = -4 f'''(x) = 8\sin 2x$$f'''(0) = 0$$f^{(4)}(x) = 16\cos 2x$$f^{(4)}(0) = 16$$f^{(5)}(0) = 0 $f^{(6)}(0) = -64$.

Alternatively, substitute $2x$ into $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots$:

$$\boxed{\cos 2x = 1 - 2x^2 + \frac{2x^4}{3} - \frac{4x^6}{45} + \cdots}$$

If you get this wrong, revise: Standard Series, Section 2.

Problem 2

Find the Maclaurin series of $\ln(1 - x^2)$ up to the $x^6$ term, stating the range of validity.

Solution 2

Substitute $-x^2$ into $\ln(1+u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots$:

$$\ln(1 - x^2) = -x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \cdots$$

Range: $|x^2| < 1 \implies |x| < 1$. Also converges at $x = \pm 1$ (alternating harmonic at $x = 1$ when substituted, but here $-x^2 - x^4/2 - x^6/3 - \cdots$ at $x = 1$ is $-\sum 1/n$ Diverges).

Valid for $-1 < x < 1$.

If you get this wrong, revise: Standard Series, Section 2.

Problem 3

Use the Maclaurin series for $e^x$ to show that $\displaystyle\sum_{n=0}^{\infty}\frac{2^n}{n!} = e^2$.

Solution 3

$e^x = \displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$.

Setting $x = 2$: $e^2 = \displaystyle\sum_{n=0}^{\infty}\dfrac{2^n}{n!}$. $\blacksquare$

If you get this wrong, revise: Exponential Function, Section 2.1.

Problem 4

Find the Taylor series of $\sin x$ about $x = \dfrac{\pi}{3}$ up to the $(x - \pi/3)^3$ Term.

Solution 4

$f(x) = \sin x$. $f(\pi/3) = \dfrac{\sqrt{3}}{2}$. $f'(x) = \cos x$ $f'(\pi/3) = \dfrac{1}{2}$. $f''(x) = -\sin x$, $f''(\pi/3) = -\dfrac{\sqrt{3}}{2}$. $f'''(x) = -\cos x$, $f'''(\pi/3) = -\dfrac{1}{2}$.

$$\sin x = \frac{\sqrt{3}}{2} + \frac{1}{2}\left(x-\frac{\pi}{3}\right) - \frac{\sqrt{3}}{4}\left(x-\frac{\pi}{3}\right)^2 - \frac{1}{12}\left(x-\frac{\pi}{3}\right)^3 + \cdots$$

If you get this wrong, revise: Taylor Series About $x = a$, Section 3.

Problem 5

Use series to evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^3}$.

Solution 5

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

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

$\dfrac{\sin x - x}{x^3} = -\dfrac{1}{6} + \dfrac{x^2}{120} - \cdots \to \boxed{-\dfrac{1}{6}}$ as $x \to 0$.

If you get this wrong, revise: Evaluating Limits, Section 5.2.

Problem 6

Find the Maclaurin series of $(1 + 2x)^{-1/2}$ up to the $x^3$ term and state the range of validity.

Solution 6

Using the binomial series with $n = -1/2$:

$(1+u)^{-1/2} = 1 + \left(-\dfrac{1}{2}\right)u + \dfrac{(-1/2)(-3/2)}{2}u^2 + \dfrac{(-1/2)(-3/2)(-5/2)}{6}u^3 + \cdots$

$= 1 - \dfrac{u}{2} + \dfrac{3u^2}{8} - \dfrac{5u^3}{16} + \cdots$

Substituting $u = 2x$:

$$\boxed{(1+2x)^{-1/2} = 1 - x + \frac{3x^2}{2} - \frac{5x^3}{2} + \cdots}$$

Valid for $|2x| < 1 \implies |x| < \dfrac{1}{2}$.

If you get this wrong, revise: Binomial Series, Section 2.5.

Problem 7

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{(2x)^n}{n^2 + 1}$.

Solution 7

$a_n = \dfrac{2^n}{n^2+1}$.

$\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{2^{n+1}(n^2+1)}{2^n(n^2+2n+2)} = \dfrac{2(n^2+1)}{n^2+2n+2} \to 2$ as $n \to \infty$.

$R = \dfrac{1}{2}$. The series converges for $\boxed{|x| < \dfrac{1}{2}}$.

If you get this wrong, revise: The Ratio Test, Section 4.2.

Problem 8

Find the Maclaurin series of $e^x \sin x$ up to the $x^5$ term.

Solution 8

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

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

Multiply and collect terms:

$x$: $1 \cdot x = x$

$x^2$: $x \cdot x = x^2$

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

$x^4$: $\dfrac{x^3}{6} \cdot x + x \cdot \left(-\dfrac{x^3}{6}\right) = 0$

$x^5$: $\dfrac{x^4}{24} \cdot x + \dfrac{x^2}{2}\left(-\dfrac{x^3}{6}\right) + 1 \cdot \dfrac{x^5}{120} = \dfrac{x^5}{24} - \dfrac{x^5}{12} + \dfrac{x^5}{120} = \dfrac{5 - 10 + 1}{120}x^5 = -\dfrac{x^5}{30}$

$$\boxed{e^x \sin x = x + x^2 + \frac{x^3}{3} - \frac{x^5}{30} + \cdots}$$

If you get this wrong, revise: Compound Functions

• Section 5.3.

Problem 9

Show that $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!} = \sin 1$.

Solution 9

$\sin x = \displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$.

Setting $x = 1$: $\sin 1 = \displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)!}$. $\blacksquare$

If you get this wrong, revise: Sine Function, Section 2.2.

Problem 10

Use series to evaluate $\displaystyle\lim_{x\to 0}\frac{e^{-x^2} - 1}{x^2}$.

Solution 10

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

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

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

If you get this wrong, revise: Evaluating Limits, Section 5.2.

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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.

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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.

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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.

Solution

(a) $\ln\!\left(\dfrac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)$

$\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \dfrac{x^5}{5} - \cdots$

$\ln(1-x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - \dfrac{x^4}{4} - \dfrac{x^5}{5} - \cdots$

$$\ln\!\left(\frac{1+x}{1-x}\right) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots$$

(b) Valid for $-1 < x < 1$ (intersection of validity of $\ln(1+x)$ and $\ln(1-x)$).

Question 12

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

Solution

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

$$\frac{1 - \cos x - x^2/2}{x^4} = \frac{1 - (1 - x^2/2 + x^4/24 - \cdots) - x^2/2}{x^4} = \frac{-x^4/24 + \cdots}{x^4} \to \boxed{-\frac{1}{24}}$$

Question 13

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

Solution

Base case ($n = 0$): $\dfrac{d^0}{dx^0}(xe^x) = xe^x = (x+0)e^x$. True.

Inductive step. Assume $\dfrac{d^k}{dx^k}(xe^x) = (x+k)e^x$.

$$\frac{d^{k+1}}{dx^{k+1}}(xe^x) = \frac{d}{dx}[(x+k)e^x] = e^x + (x+k)e^x = (x+k+1)e^x \quad \blacksquare$$

Question 14

Find the Maclaurin series of $\sqrt{\cos x}$ up to the $x^4$ term.

Solution

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

$\sqrt{\cos x} = (1 - x^2/2 + x^4/24 - \cdots)^{1/2}$

Using $(1+u)^{1/2} = 1 + u/2 - u^2/8 + \cdots$ with $u = -x^2/2 + x^4/24$:

$u^2 = x^4/4 + \cdots$

$$\sqrt{\cos x} = 1 + \frac{1}{2}\!\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) - \frac{1}{8}\!\left(\frac{x^4}{4}\right) + \cdots$$$$= 1 - \frac{x^2}{4} + \frac{x^4}{48} - \frac{x^4}{32} + \cdots = 1 - \frac{x^2}{4} - \frac{x^4}{96} + \cdots$$

Question 15

Use the Maclaurin series for $\arctan x$ to show that $\displaystyle 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}$.

Solution

$\arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dfrac{x^7}{7} + \cdots$ for $|x| \leq 1$.

Setting $x = 1$:

$$\arctan 1 = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}$$

Therefore $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} = \frac{\pi}{4}$. $\blacksquare$

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10. Advanced Worked Examples

Example 10.1: Maclaurin series of $e^x\sin x$

Problem. Find the Maclaurin series of $e^x\sin x$ up to the term in $x^5$.

Solution. $e^x = 1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\dfrac{x^5}{120}+\cdots$

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

Multiplying and collecting terms up to $x^5$:

$e^x\sin x = \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\!\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)$

$= x - \frac{x^3}{6} + \frac{x^5}{120} + x^2 - \frac{x^4}{6} + \frac{x^5}{120} + \frac{x^3}{2} - \frac{x^5}{12} + \frac{x^4}{6} - \frac{x^5}{72} + \frac{x^5}{24} - \frac{x^6}{144} + \cdots$

$= x + x^2 + \dfrac{x^3}{3} - \boxed{\dfrac{x^5}{30} + O(x^6)}$

Wait, let me be more careful:

$x^5$ coefficient: $\dfrac{1}{120} + \dfrac{1}{120} - \dfrac{1}{12} - \dfrac{1}{72} + \dfrac{1}{24}$

$= \dfrac{6 + 6 - 30 - 5 + 15}{720} = \dfrac{-8}{720} = -\dfrac{1}{90}$.

So $e^x\sin x = x + x^2 + \dfrac{x^3}{3} - \dfrac{x^5}{30} + \cdots$

Let me recompute the $x^4$ and $x^5$ terms carefully:

$x^4$: from $(-\frac{1}{6}x^4)(1) + (x^4/24)(x/x)...$ Actually:

• $1 \times (-\frac{x^3}{6})$ contributes nothing to $x^4$
• $x \times (-\frac{x^3}{6}) = -\frac{x^4}{6}$ ✓
• $\frac{x^2}{2} \times x = \frac{x^3}{2}$ (not $x^4$)

$x^4$ terms: $x \cdot (-\frac{x^3}{6}) = -\frac{x^4}{6}$ and $\frac{x^4}{24} \cdot x$… No, that’s $x^5$.

Actually $x^4$ comes from: $x \cdot (-\frac{x^3}{6}) = -\frac{x^4}{6}$.

So coefficient of $x^4$ is $-\dfrac{1}{6}$.

$$\boxed{e^x\sin x = x + x^2 + \frac{x^3}{3} - \frac{x^4}{6} - \frac{x^5}{30} + O(x^6)}$$

Example 10.2: Taylor series about a non-zero point

Problem. Find the Taylor series of $\ln x$ about $x = 1$ up to the term in $(x-1)^4$.

Solution. $f(x) = \ln x$, $f(1) = 0$.

f'(x) = \dfrac{1}{x}$$f'(1) = 1. f''(x) = -\dfrac{1}{x^2}$$f''(1) = -1. f'''(x) = \dfrac{2}{x^3}$$f'''(1) = 2. f^{(4)}(x) = -\dfrac{6}{x^4}$$f^{(4)}(1) = -6.

$$\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$$

Example 10.3: Using series to evaluate a limit

Problem. Find $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$.

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

$$\frac{e^x - 1 - x - x^2/2}{x^3} = \frac{x^3/6 + x^4/24 + \cdots}{x^3} = \frac{1}{6} + \frac{x}{24} + \cdots$$ $$\boxed{\lim_{x \to 0} = \frac{1}{6}}$$

Example 10.4: Convergence of a Maclaurin series

Problem. Find the radius of convergence of the Maclaurin series of $\ln(1+x)$.

Solution. $\ln(1+x) = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}$.

Ratio test: $\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{n}{n+1}|x| \to |x|$.

Converges when $|x| < 1$. At $x = 1$: alternating harmonic series (converges). At $x = -1$: $-\sum \dfrac{1}{n}$ (diverges).

Radius of convergence: $\boxed{1}$ Interval: $(-1, 1]$.

Example 10.5: Series solution to estimate a definite integral

Problem. Use a Maclaurin series to estimate $\displaystyle\int_0^{0.5} e^{-x^2}\,dx$ to 4 Decimal places.

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

$$\int_0^{0.5} e^{-x^2}\,dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216} - \cdots\right]_0^{0.5}$$

$= 0.5 - \dfrac{0.125}{3} + \dfrac{0.03125}{10} - \dfrac{0.0078125}{42} + \dfrac{0.001953125}{216}$

$= 0.5 - 0.04167 + 0.003125 - 0.000186 + 0.000009 = \boxed{0.4613}$ (4 d.p.)

Example 10.6: Differentiating a known series

Problem. By differentiating the Maclaurin series for $\dfrac{1}{1-x}$ Find the Maclaurin series For $\dfrac{1}{(1-x)^2}$.

Solution. $\dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$

Differentiating: $\dfrac{1}{(1-x)^2} = \sum_{n=1}^{\infty} nx^{n-1} = \sum_{n=0}^{\infty} (n+1)x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots$

Example 10.7: Multiplying two Taylor series

Problem. Find the Maclaurin series of $\cos x \cdot e^{-x}$ up to $x^4$.

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

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

Multiplying: $(1 - \frac{x^2}{2} + \frac{x^4}{24})(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24})$

$= 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^2}{2} + \frac{x^3}{2} - \frac{x^4}{4} + \frac{x^4}{24} + \cdots$

$= 1 - x + 0 - \frac{x^3}{3} + \left(\frac{1}{24}-\frac{1}{4}+\frac{1}{24}\right)x^4 + \cdots$

Wait: $\frac{1}{24}-\frac{6}{24}+\frac{1}{24} = -\frac{4}{24} = -\frac{1}{6}$.

$$\boxed{\cos x \cdot e^{-x} = 1 - x - \frac{x^3}{3} - \frac{x^4}{6} + O(x^5)}$$

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11. Common Pitfalls

Pitfall	Correct Approach
Confusing the Maclaurin series (about $x=0$) with a general Taylor series	Maclaurin: $a=0$; Taylor: $f(x) = \sum \dfrac{f^{(n)}(a)}{n!}(x-a)^n$
Forgetting the factorial in the denominator	Each term has $\dfrac{f^{(n)}(0)}{n!}x^n$ Not $\dfrac{f^{(n)}(0)}{n}x^n$
Using a series outside its radius of convergence	Always check: e.g., $\ln(1+x)$ converges only for $-1 < x \leq 1$
Incorrectly multiplying series	Collect like powers carefully; use a table if needed

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12. Additional Exam-Style Questions

Question 8

Find the Maclaurin series of $(1+x)^{-1/2}$ and determine its radius of convergence.

Solution

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

Using the general binomial: $\displaystyle\sum_{n=0}^{\infty} \binom{-1/2}{n} x^n$.

Radius of convergence: $|x| < 1$ (from the binomial series convergence condition).

Question 9

Prove that $\displaystyle\int_0^1 \frac{\ln(1+x)}{x}\,dx = \frac{\pi^2}{12}$.

Solution

$\ln(1+x) = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}$ for $|x| < 1$.

$\dfrac{\ln(1+x)}{x} = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^{n-1}}{n}$.

$\displaystyle\int_0^1 \frac{\ln(1+x)}{x}\,dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = \eta(2) = \frac{\pi^2}{12}$.

This is the Dirichlet eta function evaluated at 2. $\blacksquare$

Question 10

Use the Maclaurin series for $\sin x$ to find $\sin 0.1$ correct to 8 decimal places.

Solution

$\sin 0.1 = 0.1 - \dfrac{0.001}{6} + \dfrac{0.00001}{120} - \cdots = 0.1 - 0.00016667 + 0.00000008 - \cdots$

$\boxed{\sin 0.1 \approx 0.09983342}$

The $x^7$ term contributes approximately $10^{-12}$ Which is negligible.

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13. Advanced Topics

13.1 Lagrange form of the remainder

The error in truncating a Taylor series after $n$ terms is:

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

For some $c$ between $a$ and $x$. This gives a bound on the truncation error.

13.2 Power series solutions of ODEs

The Maclaurin series method can solve ODEs that cannot be solved by standard methods. Substitute $y = \sum a_n x^n$ into the ODE and equate coefficients.

13.3 Standard Maclaurin series (reference)

Function	Series	Radius
$e^x$	$\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$	$\infty$
$\sin x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$	$\infty$
$\cos x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$	$\infty$
$\ln(1+x)$	$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}$	$1$
$(1+x)^\alpha$	$\displaystyle\sum_{n=0}^{\infty} \binom{\alpha}{n}x^n$	$1$
$\arctan x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$	$1$
$\dfrac{1}{1-x}$	$\displaystyle\sum_{n=0}^{\infty} x^n$	$1$

13.4 Using series to prove identities

Many trigonometric identities can be derived from series. For example, $e^{ix} = \cos x + i\sin x$ Implies all the standard addition formulae.

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14. Further Exam-Style Questions

Question 11

Find the Maclaurin series of $\dfrac{\sin x}{x}$ and determine $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$.

Solution

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

$\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots$

$\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$ (the constant term).

Question 12

Prove that the Maclaurin series of $\cos x$ converges to $\cos x$ for all real $x$.

Solution

The $n$-th derivative of $\cos x$ is one of $\pm\cos x$ or $\pm\sin x$ So $|f^{(n)}(c)| \leq 1$ for all $c$ and $n$.

By the Lagrange remainder: $|R_n(x)| \leq \dfrac{|x|^{n+1}}{(n+1)!}$.

For any fixed $x$: $\displaystyle\lim_{n \to \infty} \frac{|x|^{n+1}}{(n+1)!} = 0$ (factorial grows faster than exponential).

Therefore $R_n(x) \to 0$ and the series converges to $\cos x$. $\blacksquare$

Question 13

Use series to evaluate $\displaystyle\lim_{x \to 0} \frac{\tan x - x}{x^3}$.

Solution

$\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots$

$\dfrac{\tan x - x}{x^3} = \dfrac{x^3/3 + 2x^5/15 + \cdots}{x^3} = \dfrac{1}{3} + \dfrac{2x^2}{15} + \cdots$

$\boxed{\displaystyle\lim_{x \to 0} \frac{\tan x - x}{x^3} = \frac{1}{3}}$

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15. Advanced Topics

15.1 Power series solutions of differential equations

For an ODE $y'' + p(x)y' + q(x)y = 0$ Assume $y = \sum_{n=0}^{\infty} a_n x^n$.

Substitute into the ODE and equate coefficients of each power of $x$ to zero. This gives a Recurrence relation for $a_n$.

15.2 The exponential generating function

$E(x) = \sum_{n=0}^{\infty} \dfrac{a_n x^n}{n!}$.

This is useful in combinatorics and probability (e.g., the exponential generating function of the Bernoulli numbers).

15.3 Binomial series, convergence at the endpoints

$(1+x)^\alpha$ converges at $x = 1$ when $\alpha > -1$ and at $x = -1$ when $\alpha > 0$.

Example: $(1+x)^{1/2}$ converges at $x = 1$ (giving $\sqrt{2}$) but diverges at $x = -1$.

15.4 Series acceleration

Techniques like Euler acceleration or Shanks transformation can speed up the convergence of slowly Converging alternating series.

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16. Further Exam-Style Questions

Question 14

Find the Maclaurin series of $\ln(1-x^2)$ up to $x^6$ and state the radius of convergence.

Solution

$\ln(1-u) = -\displaystyle\sum_{n=1}^{\infty} \frac{u^n}{n}$ for $|u| < 1$. With $u = x^2$:

$\ln(1-x^2) = -\displaystyle\sum_{n=1}^{\infty} \frac{x^{2n}}{n} = -x^2 - \dfrac{x^4}{2} - \dfrac{x^6}{3} - \cdots$

Radius of convergence: $|x^2| < 1 \implies |x| < 1$. At $x = 1$: $-\sum \dfrac{1}{n}$ diverges.

$\boxed{\ln(1-x^2) = -x^2 - \dfrac{x^4}{2} - \dfrac{x^6}{3} + O(x^8)}$ Radius $= 1$.

Question 15

Prove that $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = 2$.

Solution

$S = \displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = \sum_{n=1}^{\infty} n x^n\Big|_{x=1/2}$.

We know $\displaystyle\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$ for $|x| < 1$.

So $\displaystyle\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}$.

At $x = 1/2$: $S = \dfrac{1/2}{(1/2)^2} = \dfrac{1/2}{1/4} = 2$. $\blacksquare$

flowchart TD
    A[08 Maclaurin And Taylor Seriesx] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to maclaurin and taylor series, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• quadratic equations and the discriminant
• simultaneous equations
• polynomial division and the factor theorem
• partial fractions
• binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

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Cross-References

• Further Calculus provides the differentiation and integration foundations upon which the Maclaurin and Taylor series are built.
• Differential Equations uses power series solutions to solve ODEs that cannot be solved by standard methods.
• Complex Numbers connects exponential and trigonometric series through Euler’s formula, unifying the standard Maclaurin series.
• Further Algebra supplies the binomial expansion and partial fraction techniques used in deriving and manipulating series expansions.