4.08a, 4.08b - Maclaurin series | OCR A-Level Further Mathematics A

Replace a function by a power series built from its derivatives at zero. In 4.08a you will calculate coefficients and general terms; in 4.08b you will adapt standard series, combine them, and distinguish an infinite series' interval of validity from a finite approximation's accuracy.

Matching a function at zero

Can the cubic polynomial $1+x+x^2/2+x^3/6$ stand in for $e^x$ , and how far from zero would you trust it?

At $x = 0.2$ , it gives $1.221333\ldots$ , while $e^{0.2} = 1.221402\ldots$ . The agreement is striking because the polynomial uses only the value and first three derivatives of $e^x$ at one point. At $x = 2$ , however, it gives $19/3$ while $e^2 = 7.38906\ldots$ . The same polynomial is much less convincing there.

You already know

Differentiate powers, exponentials and trigonometric functions using A-level Maths rules; angles are in radians. Also $\sinh x = (e^x-e^{-x})/2$ , $\cosh x = (e^x+e^{-x})/2$ , with $\sinh0 = 0$ and $\cosh0 = 1$ .

Suppose a power series begins $a_0+a_1x+a_2x^2+\cdots$ . At zero its value is $a_0$ , its first derivative is $a_1$ , and its second derivative is $2a_2$ . More generally its derivative of order $r$ at zero is $r!a_r$ . Matching these to the function's derivatives gives $a_r = f^{(r)}(0)/r!$ .

Maclaurin series

The Maclaurin series of a function is the power series about $x = 0$ whose coefficient of $x^r$ is $f^{(r)}(0)/r!$ . Here $f^{(r)}$ is the derivative of order $r$ , $f^{(0)} = f$ , and $0! = 1$ . When the series represents the function, their values agree throughout its interval of validity.

For a polynomial the process can stop exactly. For instance, $f(x) = 2-3x+4x^2$ has $f(0) = 2$ , $f'(0) = -3$ and $f''(0) = 8$ ; dividing the second derivative by $2!$ recovers the coefficient $4$ , not $8$ .

Maclaurin expansion

Where the Maclaurin series represents $f$ ,

\begin{aligned} f(x) &= f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots \\ &= \sum_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}x^r \end{aligned}

The derivatives used must exist at zero. The general term is $\frac{f^{(r)}(0)}{r!}x^r$ ; a finite truncation is generally only an approximation.

In the formula booklet.

Having derivatives at zero lets you form coefficients, but it does not by itself justify an equality for every input. Use the standard intervals or the validity information supplied in a question. Proofs of those intervals are not needed here.

Key idea

A derivative value is not a series coefficient: divide the derivative of order $r$ at zero by $r!$ .

Locate that missing step in the following calculation.

Derivatives and the general term

Make a short derivative table at zero, then put those values into the Maclaurin formula with the factorials still visible.

For $e^{kx}$ , each differentiation multiplies by $k$ , so $f^{(r)}(0) = k^r$ . The exponential series in the booklet is valid for every real input; replacing that input by $kx$ shows that the resulting series is valid for every real $x$ when $k$ is real.

Worked example 1: finding a general term

Question. Find the Maclaurin series for $f(x) = e^{2x}$ , including its general term and the terms up to and including $x^3$ .

Plan. Differentiate repeatedly, identify the pattern at zero, and divide the derivative of order $r$ by $r!$ .

\begin{aligned} f^{(r)}(x) &= 2^re^{2x} && \text{factor }2\text{ per derivative} \\ f^{(r)}(0) &= 2^r && e^0 = 1 \\ e^{2x} &= \sum_{r=0}^{\infty}\frac{2^rx^r}{r!} && \text{Maclaurin formula} \\ &= 1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+\cdots && \text{first four terms} \end{aligned}

Why this step? The derivative supplies $2^r$ , but the coefficient is $2^r/r!$ ; the factorial remains essential even when the derivative pattern is simple.

Answer. The general term is $2^rx^r/r!$ , for integers $r \ge 0$ . The expansion begins $1+2x+2x^2+\frac43x^3+\cdots$ and is valid for all real $x$ .

Check. At $x = 0$ the series gives $1$ ; differentiating its displayed terms once and twice gives initial derivative values $2$ and $4$ , matching $e^{2x}$ .

Change only the sign of the inner scale. The derivative pattern now alternates, so every odd-power term changes sign while every even-power term stays the same.

Worked example 2: an alternating general term

Question. Find the Maclaurin series for $g(x) = e^{-2x}$ , including its general term and the terms up to and including $x^3$ .

Plan. Replace the factor $2$ per differentiation by $-2$ , then use the same factorial denominators.

\begin{aligned} g^{(r)}(x) &= (-2)^re^{-2x} && \text{chain rule repeatedly} \\ g^{(r)}(0) &= (-2)^r && e^0 = 1 \\ e^{-2x} &= \sum_{r=0}^{\infty}\frac{(-2)^rx^r}{r!} && \text{general term} \\ &= 1-2x+2x^2-\frac43x^3+\cdots && \text{simplify factorials} \end{aligned}

Why this step? Keep $(-2)^r$ in brackets: writing $-2^r$ would make every non-constant coefficient negative instead of alternating.

Answer. The general term is $(-2)^rx^r/r!$ , for integers $r \ge 0$ , and the series is valid for all real $x$ .

Check. Replacing $x$ by $-x$ in the first example changes exactly the odd-power terms, giving the second expansion.

A derivative formula given elsewhere in a question can save repeated differentiation. If it gives $f^{(8)}(0) = 1024$ , for example, the coefficient of $x^8$ is $1024/8! = 8/315$ . The coefficient is the number; the term includes $x^8$ .

Exam tip

For a “determine” or “show that” coefficient question, display $f^{(r)}(0)/r!$ before simplifying. A derivative value on its own is not the requested coefficient.

Apply the same pattern after changing the inner scale again.

Adapting the standard series

Use a booklet series directly when it matches the function; differentiating a complicated composite function repeatedly usually creates unnecessary work.

The five standard series below come from the same derivative rule. For $e^x$ , every derivative at zero is $1$ . For sine and cosine, the derivative cycle and the values at zero remove alternate powers: sine keeps odd powers and cosine keeps even powers. The logarithm and binomial coefficients follow their own repeated-derivative patterns.

The five standard series

For real $x$ ,

\begin{aligned} e^x &= \sum_{r=0}^{\infty}\frac{x^r}{r!} && \text{all }x \\ \sin x &= \sum_{r=0}^{\infty}\frac{(-1)^rx^{2r+1}}{(2r+1)!} && \text{all }x \\ \cos x &= \sum_{r=0}^{\infty}\frac{(-1)^rx^{2r}}{(2r)!} && \text{all }x \\ \ln(1+x) &= \sum_{r=1}^{\infty}\frac{(-1)^{r+1}x^r}{r} && -1 < x \le 1 \\ (1+x)^n &= 1+nx+\frac{n(n-1)}{2!}x^2+\cdots \\ &\quad+\frac{n(n-1)\cdots(n-r+1)}{r!}x^r+\cdots && |x| < 1 \end{aligned}

In the binomial series, $n$ is real. If $n$ is a non-negative integer, it terminates as a polynomial and is valid for all real $x$ ; $n = 0$ gives the constant polynomial $1$ . For other exponents, $|x| < 1$ is the standard guaranteed interval; any endpoint claim needs separate justification or supplied information.

In the formula booklet.

For direct use, their first terms are $\sin x = x-x^3/6+x^5/120-\cdots$ , $\cos x = 1-x^2/2+x^4/24-\cdots$ and $\ln(1+x) = x-x^2/2+x^3/3-\cdots$ . Replace the entire input everywhere: $\sin(2x) = 2x-(2x)^3/6+\cdots$ , not $2x-x^3/6+\cdots$ . These are substitutions into the booklet results, rather than additional results to memorise.

Worked example 3: a fractional power

Question. Find the expansion of $\sqrt{1+2x}$ up to and including $x^3$ , and state its standard open interval of validity.

Plan. Use the binomial series with exponent $1/2$ and replace its input by $2x$ .

\begin{aligned} (1+2x)^{1/2} &= 1+\frac12(2x)+\frac{(1/2)(-1/2)}{2!}(2x)^2 \\ &\quad+\frac{(1/2)(-1/2)(-3/2)}{3!}(2x)^3+\cdots && \text{binomial coefficients} \\ &= 1+x-\frac12x^2+\frac12x^3+\cdots && \text{collect powers} \\ |2x| < 1 &\iff -\frac12 < x < \frac12 && \text{transform the condition} \end{aligned}

Why this step? The input is $2x$ , so its square and cube contribute factors $4$ and $8$ ; the condition also applies to $2x$ , not to $x$ alone.

Answer. $\sqrt{1+2x} = 1+x-\frac12x^2+\frac12x^3+\cdots$ , for the standard open interval $-\frac12 < x < \frac12$ .

Check. Squaring the displayed polynomial gives $1+2x$ through the cubic term: the quadratic and cubic terms cancel, as they should.

Change only the exponent to $-1/2$ . The input and its validity condition stay the same, but the signs and sizes of the coefficients change.

Worked example 4: a reciprocal fractional power

Question. Find the expansion of $(1+2x)^{-1/2}$ up to and including $x^3$ , and state its standard open interval of validity.

Plan. Use the same binomial substitution with exponent $-1/2$ .

\begin{aligned} (1+2x)^{-1/2} &= 1-\frac12(2x)+\frac{(-1/2)(-3/2)}{2!}(2x)^2 \\ &\quad+\frac{(-1/2)(-3/2)(-5/2)}{3!}(2x)^3+\cdots && \text{new exponent} \\ &= 1-x+\frac32x^2-\frac52x^3+\cdots && \text{simplify} \\ |2x| < 1 &\iff -\frac12 < x < \frac12 && \text{same input condition} \end{aligned}

Why this step? The exponent changes every factor in each binomial coefficient; it does not simply negate the previous expansion.

Answer. $(1+2x)^{-1/2} = 1-x+\frac32x^2-\frac52x^3+\cdots$ on the stated open interval.

Check. Multiplying this polynomial by the previous example's polynomial gives $1$ through the cubic term, consistent with the two functions being reciprocals.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

4.08a, 4.08b - Maclaurin series

Matching a function at zero

Derivatives and the general term

Adapting the standard series

Continue this lesson.

Learn faster with Nova