Ps5 - Standard Maclaurin series | OCR A-Level Further Mathematics B (MEI)

Use the five standard expansions in the formula booklet rather than differentiating afresh. You will substitute and simplify exponential, logarithmic, trigonometric and binomial series, track their validity, and calculate a signed approximation error.

Exponential and logarithmic substitutions

Is $1-0.02-0.0002$ above or below $\sqrt{0.96}$ ?

The expression looks rather unlike a square root. A binomial series will explain where it comes from, and a signed error will tell us which side it lies on. First, practise the shared skill behind all five standard expansions: substitute into the whole term, including every power.

You already know

A Maclaurin coefficient is $f^{(r)}(0)/r!$ . A finite truncation approximates the function; the infinite equality has a validity set. You have already derived logarithm coefficients and seen that its upper endpoint is included.

The booklet lets you begin with known series. Use a temporary input $u$ so the formula's condition remains attached to the expression being substituted.

Exponential and logarithm series

For every real $u$ ,

e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots

For $-1<u\le1$ ,

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

The logarithm has no constant term; the exponential does.

In the formula booklet.

“First three non-zero terms” counts a non-zero constant as one term. It does not mean “through the cubic term” for every function.

Worked example 1: one input, two series

Question. Write the first three non-zero terms for $e^{2x}$ and $\ln(1+2x)$ , and state where each infinite expansion is valid.

Plan. Set $u=2x$ , simplify its powers and translate the logarithm inequality.

\begin{aligned} e^{2x}&=1+2x+\frac{(2x)^2}{2!}+\cdots =1+2x+2x^2+\cdots && \text{exponential terms} \\ \ln(1+2x)&=2x-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}+\cdots && \text{logarithm terms} \\ &=2x-2x^2+\frac83x^3+\cdots && \text{simplify powers} \\ -1<2x\le1&\iff-\frac12<x\le\frac12 && \text{logarithm validity} \end{aligned}

Why this step? The square of $2x$ is $4x^2$ , so the quadratic coefficient is not merely doubled.

Answer. $e^{2x}$ : $1+2x+2x^2$ , valid for all real $x$ . Logarithm: $2x-2x^2+\frac83x^3$ , valid for $-\frac12<x\le\frac12$ .

Check. Both constant values agree with the functions at zero: $e^0=1$ and $\ln1=0$ .

Reverse only the input's sign. The logarithm's included endpoint moves from the right to the left.

Worked example 2: a negative substituted input

Question. Give the first three non-zero terms and validity sets for $e^{-2x}$ and $\ln(1-2x)$ .

Plan. Substitute $u=-2x$ in both series and reverse the inequalities when dividing by a negative number.

\begin{aligned} e^{-2x}&=1-2x+\frac{(-2x)^2}{2!}+\cdots =1-2x+2x^2+\cdots && \text{even power positive} \\ \ln(1-2x)&=-2x-\frac{(-2x)^2}{2}+\frac{(-2x)^3}{3}+\cdots && \text{retain signs} \\ &=-2x-2x^2-\frac83x^3+\cdots && \text{all three negative} \\ -1<-2x\le1&\iff-\frac12\le x<\frac12 && \text{reverse inequality signs} \end{aligned}

Why this step? The logarithm's alternating signs combine with the input's powers; they do not remain alternating in the simplified result.

Answer. The displayed three-term expressions; exponential validity is all real $x$ , while logarithm validity is $-\frac12\le x<\frac12$ .

Check. At $x=-1/2$ , the logarithm input is $1$ , giving the included expansion of $\ln2$ . At $x=1/2$ , the function is undefined.

These direct substitutions are useful when no derivative-based method is requested. If a question says “hence” after supplying derivatives, use that earlier work instead.

Watch out

Wrong: $\ln(1-2x)=-2x+2x^2-\cdots$ . Right: its quadratic term is $-(-2x)^2/2=-2x^2$ . Substitute first, simplify signs second.

Change the coefficient and retain the exact logarithm endpoint information.

Sine and cosine in radians

Sine has only odd powers and cosine has only even powers, so the number of non-zero terms differs from the polynomial degree.

Their Maclaurin series use radians. This is built into the familiar derivatives $d(\sin x)/dx=\cos x$ and $d(\cos x)/dx=-\sin x$ .

Trigonometric series

For every real input $u$ , measured in radians,

\begin{aligned} \sin u&=u-\frac{u^3}{3!}+\frac{u^5}{5!}-\cdots \\ \cos u&=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots \end{aligned}

The general terms are $(-1)^r u^{2r+1}/(2r+1)!$ for sine and $(-1)^r u^{2r}/(2r)!$ for cosine, with $r\ge0$ .

In the formula booklet.

Both expansions are valid for all real inputs. A short truncation is usually most useful near zero, where the omitted powers are small.

Worked example 3: scaling the angle

Question. Find the first three non-zero terms of $\sin2x$ and $\cos2x$ .

Plan. Substitute $u=2x$ into each standard series and simplify the factorial coefficients.

\begin{aligned} \sin2x&=2x-\frac{8x^3}{6}+\frac{32x^5}{120}+\cdots && \text{odd powers} \\ &=2x-\frac43x^3+\frac4{15}x^5+\cdots && \text{simplify} \\ \cos2x&=1-\frac{4x^2}{2}+\frac{16x^4}{24}+\cdots && \text{even powers} \\ &=1-2x^2+\frac23x^4+\cdots && \text{simplify} \end{aligned}

Why this step? The input multiplier is raised to the same power as $x$ ; the cubic term gets a factor $2^3$ .

Answer. The displayed expressions, valid for all real $x$ .

Check. At zero they give $0$ and $1$ , matching sine and cosine.

Change only the requested sine expression by dividing it by $x$ . Every power decreases by one, so a constant term appears.

Worked example 4: a quotient near zero

Question. Find the first three non-zero terms for $\sin(2x)/x$ , and hence find its limit as $x\to0$ .

Plan. Divide every term of the sine expansion by $x$ , then let the remaining positive powers vanish.

\begin{aligned} \frac{\sin2x}{x} &=\frac{2x-\frac43x^3+\frac4{15}x^5-\cdots}{x} && x\ne0 \\ &=2-\frac43x^2+\frac4{15}x^4-\cdots && \text{divide each term} \\ \lim_{x\to0}\frac{\sin2x}{x}&=2 && \text{positive powers vanish} \end{aligned}

Why this step? The original quotient is undefined at $x=0$ , but its nearby values can still have a limit.

Answer. $2-\frac43x^2+\frac4{15}x^4$ , with limit $2$ . The infinite expansion is valid for every non-zero real $x$ .

Check. With $x=0.01$ , $\sin(0.02)/0.01=1.999866669\ldots$ , close to $2$ .

This quotient is a useful example of the difference between a value and a limit. Extending its value to $2$ at zero would remove the discontinuity, but that is an extra definition.

Watch out

Wrong: dividing the sine series by $x$ leaves its powers unchanged. Right: $x^3/x=x^2$ and $x^5/x=x^4$ . Divide every term.

Now change the angle multiplier.

Binomial powers and their validity

Fractional and negative powers can also be expanded: the binomial coefficients continue the same falling-product pattern you know for integer powers.

Binomial series

For real $p$ ,

(1+u)^p=1+pu+\frac{p(p-1)}{2!}u^2+ \frac{p(p-1)(p-2)}{3!}u^3+\cdots

If $p$ is a non-negative integer, this terminates and is valid for all real $u$ . Otherwise, $|u|<1$ guarantees validity. Endpoint convergence depends on $p$ ; do not automatically include or exclude either endpoint.

The formula and the non-terminating series' open-interval condition are provided.

In the formula booklet.

For example, $p=2$ gives the finite polynomial $1+2u+u^2$ . With $p=-1/2$ , none of the falling factors becomes zero, so the series continues indefinitely. These are different boundary cases of the same coefficient rule.

Worked example 5: a reciprocal square root

Question. Expand $(1+2x)^{-1/2}$ through the term in $x^3$ , and state the interval guaranteed by the booklet.

Plan. Set $p=-1/2$ and $u=2x$ , calculate the three coefficients and impose $|2x|<1$ .

\begin{aligned} (1+2x)^{-1/2} &=1-\frac12(2x) +\frac{(-\frac12)(-\frac32)}2(2x)^2 +\frac{(-\frac12)(-\frac32)(-\frac52)}6(2x)^3+\cdots && \text{binomial coefficients} \\ &=1-x+\frac32x^2-\frac52x^3+\cdots && \text{simplify} \\ |2x|<1&\iff-\frac12<x<\frac12 && \text{guaranteed interval} \end{aligned}

Why this step? The coefficient includes both the falling factors from $p$ and the power of the inner multiplier $2$ .

Answer. $1-x+\frac32x^2-\frac52x^3$ , guaranteed for $-\frac12<x<\frac12$ .

Check. The derivative at zero is $-1$ , matching the linear coefficient.

Change only the exponent from $-1/2$ to $1/2$ . This gives the square-root expansion needed for the opening puzzle.

Worked example 6: a square root

Question. Expand $(1+2x)^{1/2}$ through $x^3$ , stating the interval guaranteed by the booklet.

Plan. Replace the exponent in the falling factors while keeping the same input and interval calculation.

\begin{aligned} (1+2x)^{1/2} &=1+\frac12(2x) +\frac{(\frac12)(-\frac12)}2(2x)^2 +\frac{(\frac12)(-\frac12)(-\frac32)}6(2x)^3+\cdots && \text{new exponent} \\ &=1+x-\frac12x^2+\frac12x^3+\cdots && \text{simplify products} \\ |2x|<1&\iff-\frac12<x<\frac12 && \text{same guaranteed interval} \end{aligned}

Why this step? After the first factor, $p-1$ and $p-2$ are negative. Their signs make the quadratic term negative and the cubic term positive.

Answer. $1+x-\frac12x^2+\frac12x^3$ , guaranteed for $-\frac12<x<\frac12$ .

Check. Squaring the cubic polynomial gives $1+2x$ through the cubic term; the quadratic and cubic contributions cancel.

The open interval is a guarantee, not a universal verdict about its endpoints. For the reciprocal square-root series the full validity set is $-\frac12<x\le\frac12$ ; for the square-root series it is $-\frac12\le x\le\frac12$ . These endpoint facts are not stated in the booklet. For a general exponent, use separate endpoint information if the question requires it; no proof of convergence is needed here.

Watch out

Wrong: apply $|x|<1$ after substituting $u=2x$ . Right: apply $|2x|<1$ . Also distinguish a guaranteed open interval from a separately established full endpoint-inclusive set.

Keep the reciprocal square root but change the inner coefficient and its sign.

Approximations and signed error

Choose a small substituted input, truncate at the requested term, and compare the resulting value with the function using the stated error convention.

For a square root, writing the target as $(1+u)^{1/2}$ makes the useful input visible. The quadratic truncation is $1+u/2-u^2/8$ , obtained from the binomial formula above.

Error in an approximation

\text{error}=\text{approximate value}-\text{exact value}

A positive error means an overestimate; a negative error means an underestimate. Retain unrounded values when calculating it.

Not in the formula booklet: learn it.

A calculator value of the function supplies the comparison to as many figures as needed; it does not turn a truncated series into an exact equality.

Worked example 7: a small positive input

Question. Use $1+u/2-u^2/8$ to approximate $\sqrt{1.04}$ . Find the signed error to two significant figures.

Plan. Set $u=0.04$ , evaluate the polynomial and subtract the function value.

\begin{aligned} \sqrt{1.04}&\approx1+\frac{0.04}2-\frac{0.04^2}{8} && \text{small valid input} \\ &=1+0.02-0.0002=1.0198 && \text{quadratic truncation} \\ E&=1.0198-\sqrt{1.04} && \text{approximate minus exact} \\ &=-0.000003902718\ldots && \text{unrounded comparison} \end{aligned}

Why this step? The sign of the subtraction tells you whether the approximation lies above or below the true value.

Answer. Approximation $1.0198$ ; error $-3.9\times10^{-6}$ , an underestimate.

Check. The next term is $u^3/16>0$ , consistent with the omitted first correction raising the value.

Change only the input's sign to $-0.04$ . The quadratic term keeps its sign, but the direction of the error changes.

Worked example 8: the opening square root

Question. Use the same truncation to approximate $\sqrt{0.96}$ and find the signed error to two significant figures.

Plan. Write $0.96=1-0.04$ , substitute $u=-0.04$ , then subtract the exact function value.

\begin{aligned} \sqrt{0.96}&\approx1-\frac{0.04}2-\frac{(-0.04)^2}{8} && u=-0.04 \\ &=1-0.02-0.0002=0.9798 && \text{same quadratic sign} \\ E&=0.9798-\sqrt{0.96} && \text{signed convention} \\ &=0.000004102886\ldots && \text{positive difference} \end{aligned}

Why this step? A negative input does not make its square negative. The next omitted cubic term is negative, unlike in the previous example.

Answer. $0.9798$ , with error $4.1\times10^{-6}$ ; the opening expression is an overestimate.

Check. $\sqrt{0.96}=0.979795897\ldots<0.9798$ .

A negative error is useful information, not a mistake to erase by taking an absolute value. Only take the absolute error if the question asks for it.

Watch out

Wrong: error means the positive difference regardless of order. Right: here it means approximate minus exact. Reversing that subtraction reverses the interpretation.

Change the negative input's magnitude and retain enough figures to decide the signed error.

Quick Recap

Substitute the entire input into each power, then simplify.
Count non-zero terms carefully; constants count, zero coefficients do not.
Use radians for sine and cosine and apply each series' actual validity condition.
For binomial series, distinguish terminating powers, guaranteed open intervals and separate endpoint facts.
Signed error is approximate minus exact.

The opening expression is $0.9798$ , slightly above $\sqrt{0.96}$ .

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Ps5 - Standard Maclaurin series

Exponential and logarithmic substitutions

Sine and cosine in radians

Binomial powers and their validity

Approximations and signed error

Quick Recap

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