Maclaurin series and approximation

Derive the coefficients before using them

Start with the requested highest power. For a polynomial through $x^n$, find the derivatives at zero through order $n$, then divide each by its factorial. A derivative value is not yet a series coefficient.

Keep three questions separate: can the coefficients be formed, where does the infinite series equal the function, and how accurate is the chosen finite truncation?

A power series has real constant coefficients and ascending powers:

$a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots=\sum_{r=0}^{\infty}a_rx^r.$

Assuming it represents f near zero, differentiate term by term within its interval of convergence. The complete source sequence is:

$f^{\prime}(x)=a_1+2a_2x+3a_3x^2+\cdots+na_nx^{n-1}+\cdots,$

$f^{\prime\prime}(x)=2a_2+3(2)a_3x+4(3)a_4x^2+\cdots+n(n-1)a_nx^{n-2}+\cdots,$

$f^{\prime\prime\prime}(x)=3(2)a_3+4(3)(2)a_4x+5(4)(3)a_5x^2+\cdots+n(n-1)(n-2)a_nx^{n-3}+\cdots.$

After n differentiations the first surviving term is $n!a_nx^0=n!a_n$. Substitution at zero gives the source coefficient table:

Evaluation: $f(0)=a_0$

Coefficient: $a_0=f(0)$

Evaluation: $f^{\prime}(0)=a_1$

Coefficient: $a_1=f^{\prime}(0)$

Evaluation: $f^{\prime\prime}(0)=2!a_2$

Coefficient: $a_2=f^{\prime\prime}(0)/2!$

Evaluation: $f^{\prime\prime\prime}(0)=3!a_3$

Coefficient: $a_3=f^{\prime\prime\prime}(0)/3!$

Evaluation: $f^{(n)}(0)=n!a_n$

Coefficient: $a_n=f^{(n)}(0)/n!$

Putting these coefficients back gives the Maclaurin formula shown above and identified by the source as being in MF27. Maclaurin series are Taylor series centred at zero.

Independent coefficient check. If $f^{\prime\prime}(0)=6$, the coefficient of $x^2$ is $6/2!=3$. Divide by the factorial even when a derivative value looks simple.

Higher derivatives are repeated operations

The full source notation table is:

Order: First

Equivalent derivative notation: $f^{\prime}(x)=f^{(1)}(x)=dy/dx=d(y)/dx$

Order: Second

Equivalent derivative notation: $f^{\prime\prime}(x)=f^{(2)}(x)=d^2y/dx^2=d(dy/dx)/dx$

Order: Third

Equivalent derivative notation: $f^{\prime\prime\prime}(x)=f^{(3)}(x)=d^3y/dx^3=d(d^2y/dx^2)/dx$

Order: Fourth

Equivalent derivative notation: $f^{(4)}(x)=d^4y/dx^4=d(d^3y/dx^3)/dx$

Order: nth

Equivalent derivative notation: $f^{(n)}(x)=d^ny/dx^n=d(d^{n-1}y/dx^{n-1})/dx$: differentiate n times

In general, $d^ny/dx^n$ is not $(dy/dx)^n$ for $n>1$. For example, independently taking $y=x^3$ gives $y^{\prime\prime}=6x$, while $(y^{\prime})^2=9x^4$.

The source also warns that the first-derivative reciprocal rule $dx/dy=1/(dy/dx)$ does not extend by simply taking reciprocals of higher derivatives. The first rule assumes a differentiable local inverse and a nonzero finite derivative.

For $y=x^2$ on $x>0$, independently $d^2x/dy^2=-1/(4x^3)$, whereas $1/(d^2y/dx^2)=1/2$.

Existence of coefficients does not by itself prove equality

The source lists three necessary conditions: derivatives of every order exist, their values at zero are defined, and $f(0)$ is defined. These let us form coefficients, but are not sufficient by themselves to prove that the series equals f near zero.

That equality needs additional justification, such as a Taylor remainder tending to zero. This qualification is separate from the retained source wording.

For example, define $g(0)=0$ and $g(x)=e^{-1/x^2}$ when $x\ne0$. Every derivative at zero is 0, so its Maclaurin series is identically 0, yet $g(x)>0$ away from zero. This counterexample explains why the coefficient derivation starts by assuming a valid power-series representation.

Derive the logarithm expansion

Follow the same three moves for each standard function: differentiate, evaluate at zero, then divide by the factorial. Read the validity condition after deriving the coefficients; it is part of the result.

The source lists derivations of $\ln(1+x)$, $e^x$, $\sin x$, $\cos x$, $\tan x$ and $(1+x)^n$. The logarithm, exponential, sine and cosine derivations are incorporated below; the remaining tangent and binomial derivations follow.

The Student copy prints the derivatives but leaves their values at zero and the expansion blank. The Teacher completes every entry:

Function or derivative: $f(x)=\ln(1+x)$

Value at zero: $f(0)=\ln1=0$

Function or derivative: $f^{\prime}(x)=(1+x)^{-1}$

Value at zero: $f^{\prime}(0)=1$

Function or derivative: $f^{\prime\prime}(x)=(-1)(1+x)^{-2}$

Value at zero: $f^{\prime\prime}(0)=-1$

Function or derivative: $f^{\prime\prime\prime}(x)=(-1)(-2)(1+x)^{-3}$

Value at zero: $f^{\prime\prime\prime}(0)=(-1)(-2)=2$

Function or derivative: $f^{(4)}(x)=(-1)(-2)(-3)(1+x)^{-4}$

Value at zero: $f^{(4)}(0)=(-1)(-2)(-3)=-6$

Substitution gives the Teacher’s full displayed chain:

$\ln(1+x)=0+1x+\frac{-1}{2!}x^2+\frac{(-1)(-2)}{3!}x^3+\frac{(-1)(-2)(-3)}{4!}x^4+\cdots$

$=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots.$

Both copies state the representation interval $-1<x\le1$. Independently, at $x=1$ the alternating harmonic series converges to $\ln2$; at $x=-1$ the harmonic terms diverge and $\ln(1+x)$ is undefined. For $|x|>1$ the series terms do not tend to zero. Retain the endpoint distinction rather than replacing the printed interval by $|x|<1$.

Compare logarithm partial sums graphically

The source uses $Y_1=\ln(1+X)$ and successively adds the following curves. Both copies print all five input-screen/display/observation rows:

Source row: 1

Input or selection: $Y_2=X$ (red) with $Y_1$ (blue)

Printed observation: The line does not resemble the logarithm across the shown window.

Source row: 2

Input or selection: Add $Y_3=X-X^2/2$ (black)

Printed observation: It starts to resemble the logarithm.

Source row: 3

Input or selection: Add $Y_4=X-X^2/2+X^3/3$ (magenta)

Printed observation: It resembles the logarithm more closely.

Source row: 4

Input or selection: Add $Y_5=X-X^2/2+X^3/3-X^4/4$ (green)

Printed observation: It resembles the logarithm still more closely.

Source row: 5

Input or selection: Hide the other polynomial graphs, leaving $Y_1$ and $Y_5$

Printed observation: Compare the fourth-degree partial sum directly with the logarithm.

The reconstruction below uses the same inputs and successive selections. Its graph window is illustrative: the source does not specify the original calculator settings. Near zero, even the first line is a local approximation because the logarithm’s derivative there is 1.

The source remark calls the approximation valid only for $-1<x\le1$; this is the infinite series’ representation interval.

Keep two checks separate. A graph comparison does not prove convergence. Being inside the convergence interval also does not guarantee that a finite truncation meets a chosen error tolerance. These distinctions qualify the printed observation.

Examples 1–2 retain the full Teacher factorisation, logarithm law, substitution through $x^6$, transformed inequality and closed-endpoint answer.

Derive the exponential series

Both copies print the complete derivative table and derivation:

Order: 0

Derivative: $f(x)=e^x$

Value at zero: $f(0)=1$

Order: 1

Derivative: $f^{\prime}(x)=e^x$

Value at zero: $f^{\prime}(0)=1$

Order: 2

Derivative: $f^{\prime\prime}(x)=e^x$

Value at zero: $f^{\prime\prime}(0)=1$

Order: 3

Derivative: $f^{\prime\prime\prime}(x)=e^x$

Value at zero: $f^{\prime\prime\prime}(0)=1$

Order: 4

Derivative: $f^{(4)}(x)=e^x$

Value at zero: $f^{(4)}(0)=1$

Every later derivative also equals $e^x$, with value 1 at zero. Substitution into the coefficient formula gives

$e^x=1+1x+\frac1{2!}x^2+\frac1{3!}x^3+\frac1{4!}x^4+\cdots+\frac1{n!}x^n+\cdots$

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

The representation is valid for every $x\in\mathbb R$. This does not mean that a low-degree polynomial is accurate at every real input.

Derive the sine series

Both copies show the repeating derivative cycle:

Order: 0

Derivative: $\sin x$

Value at zero: 0

Order: 1

Derivative: $\cos x$

Value at zero: 1

Order: 2

Derivative: $-\sin x$

Value at zero: 0

Order: 3

Derivative: $-\cos x$

Value at zero: −1

Order: 4

Derivative: $\sin x$

Value at zero: 0

Order: 5

Derivative: $\cos x$

Value at zero: 1

Thus

$\sin x=0+1x+\frac0{2!}x^2+\frac{-1}{3!}x^3+\frac0{4!}x^4+\frac1{5!}x^5+\cdots$

$=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,\qquad x\in\mathbb R.$

For x in radians sufficiently small that cubic and higher terms can be neglected, the source gives $\sin x\approx x$. Radians are essential; all-real convergence of the infinite series is a different statement from small-angle accuracy of this linear truncation.

Derive the cosine series

The complete source cycle is:

Order: 0

Derivative: $\cos x$

Value at zero: 1

Order: 1

Derivative: $-\sin x$

Value at zero: 0

Order: 2

Derivative: $-\cos x$

Value at zero: −1

Order: 3

Derivative: $\sin x$

Value at zero: 0

Order: 4

Derivative: $\cos x$

Value at zero: 1

Order: 5

Derivative: $-\sin x$

Value at zero: 0

Hence

$\cos x=1+0x+\frac{-1}{2!}x^2+\frac0{3!}x^3+\frac1{4!}x^4+\frac0{5!}x^5+\cdots$

$=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots,\qquad x\in\mathbb R.$

When x is in radians and cubic and higher powers may be neglected, the source gives $\cos x\approx1-x^2/2$. Independently, the cubic coefficient is zero, so the first omitted nonzero term is $x^4/24$, compared with the sine approximation’s first omitted term $-x^3/6$.

Compare cosine partial sums graphically

The source writes the general term as $(-1)^r x^{2r}/(2r)!$, with the cosine series valid for all real x. Both copies give this five-row calculator comparison:

Source row: 1

Input or selection: $Y_1=\cos X$ (blue), $Y_2=1$ (red)

Printed observation: The constant function does not look like cosine across the shown window.

Source row: 2

Input or selection: Add $Y_3=1-X^2/2!$ (red); $Y_2$ becomes magenta

Printed observation: The quadratic starts to look like cosine.

Source row: 3

Input or selection: Add $Y_4=1-X^2/2!+X^4/4!$ (black)

Printed observation: It looks more like cosine.

Source row: 4

Input or selection: Add $Y_5=1-X^2/2!+X^4/4!-X^6/6!$ (green)

Printed observation: It looks even more like cosine.

Source row: 5

Input or selection: Hide $Y_2,Y_3,Y_4$, leaving cosine and $Y_5$

Printed observation: Compare the final partial sum directly with cosine.

The reconstruction uses an explicitly illustrative common window. Near zero even the constant 1 approximates cosine: its error begins at order $x^2$. Adding successive even powers improves local agreement, but a plot is not a proof of convergence or a guarantee of accuracy at every displayed input. The original screens and their changing colours are retained in the source artwork archive.

The source asks: Can you think of an alternative method to obtain the series expansion of $\cos x$? Its full answer differentiates the sine series:

$$\cos x=\frac{d}{dx}(\sin x)=\frac{d}{dx}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots.$$

Termwise differentiation is justified within the power series’ radius of convergence, which is infinite for sine. Thus this produces cosine’s all-real expansion, rather than a new finite-polynomial identity.

Derive the tangent series

Both copies mark this expansion not in MF27 and print every step through the fifth derivative. Writing derivatives as polynomials in $\tan x$ makes evaluation at zero immediate. Start with $f(x)=\tan x$, so $f(0)=0$.

First derivative:

$$f^{\prime}(x)=\sec^2x=1+\tan^2x,\qquad f^{\prime}(0)=1.$$

Second derivative:

$$f^{\prime\prime}(x)=2\tan x\sec^2x$$ $$=2\tan x(1+\tan^2x)=2\tan x+2\tan^3x,\qquad f^{\prime\prime}(0)=0.$$

Third derivative:

$$f^{\prime\prime\prime}(x)=2\sec^2x+6\tan^2x\sec^2x$$ $$=(2+6\tan^2x)(1+\tan^2x)$$ $$=2+8\tan^2x+6\tan^4x,\qquad f^{\prime\prime\prime}(0)=2.$$

Fourth derivative:

$$f^{(4)}(x)=16\tan x\sec^2x+24\tan^3x\sec^2x$$ $$=(16\tan x+24\tan^3x)(1+\tan^2x)$$ $$=16\tan x+40\tan^3x+24\tan^5x,\qquad f^{(4)}(0)=0.$$

Fifth derivative:

$$f^{(5)}(x)=16\sec^2x+120\tan^2x\sec^2x+120\tan^4x\sec^2x$$ $$=(16+120\tan^2x+120\tan^4x)(1+\tan^2x)$$ $$=16+136\tan^2x+240\tan^4x+120\tan^6x,\qquad f^{(5)}(0)=16.$$

Substitute these values into the Maclaurin formula:

$$\tan x=0+1x+\frac0{2!}x^2+\frac2{3!}x^3+\frac0{4!}x^4+\frac{16}{5!}x^5+\cdots$$ $$=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\cdots=x+\frac{x^3}{3}+\frac{2x^5}{15}+\cdots.$$

Use radians. When x is small enough to neglect cubic and higher powers, the source gives $\tan x\approx x$. Independently, its first omitted nonzero term is $x^3/3$.

The tangent series is a local expansion: unlike sine and cosine, its representation interval is $|x|<\pi/2$, bounded by the nearest poles. The derivative calculation by itself does not prove that interval. Do not apply the all-real statement from the preceding cosine page to tangent.

The source also offers a QR-linked exploration of Maclaurin expansions and validity. This is the destination decoded from the original QR; its current external content has not been reviewed. The original artwork is preserved with the source images.

Know which formulae may be quoted

The source summary lists the binomial, exponential, sine, cosine and logarithm expansions as available in MF27. They may be quoted without proof unless the question asks for their derivation. Tangent is not included in that list. The general terms and representation intervals are:

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^r}{r!}+\cdots,\qquad x\in\mathbb R.$$ $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+\frac{(-1)^r x^{2r+1}}{(2r+1)!}+\cdots,$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+\frac{(-1)^r x^{2r}}{(2r)!}+\cdots.$$

The sine and cosine expansions apply for all real x in radians. The source repeats the small-angle forms $\sin x\approx x$, $\cos x\approx1-x^2/2!$ and $\tan x\approx x$, assuming radians and that cubic and higher powers can be neglected.

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+\frac{(-1)^{r+1}x^r}{r}+\cdots,\qquad -1<x\le1.$$

The binomial expansion is derived next. Its printed $|x|<1$ condition must be distinguished from the unrestricted finite polynomial when the exponent is a positive integer.

Derive the binomial expansion

Let n be a constant and $f(x)=(1+x)^n$. Both source copies print the following complete derivative sequence and values at zero:

$$f(x)=(1+x)^n,\qquad f(0)=1.$$ $$f^{\prime}(x)=n(1+x)^{n-1},\qquad f^{\prime}(0)=n.$$ $$f^{\prime\prime}(x)=n(n-1)(1+x)^{n-2},\qquad f^{\prime\prime}(0)=n(n-1).$$ $$f^{\prime\prime\prime}(x)=n(n-1)(n-2)(1+x)^{n-3},$$ $$f^{\prime\prime\prime}(0)=n(n-1)(n-2).$$ $$f^{(4)}(x)=n(n-1)(n-2)(n-3)(1+x)^{n-4},$$ $$f^{(4)}(0)=n(n-1)(n-2)(n-3).$$

In general, for a positive integer r,

$$f^{(r)}(x)=n(n-1)\cdots(n-r+1)(1+x)^{n-r},$$ $$f^{(r)}(0)=n(n-1)\cdots(n-r+1).$$

Substitution into the Maclaurin formula gives the source’s MF27 expansion:

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$ $$\hspace{1em}+\frac{n(n-1)\cdots(n-r+1)}{r!}x^r+\cdots.$$

The source states the representation range $-1<x<1$ (also written $|x|<1$ in its summary). Independently, this is a sufficient interval for general real n; endpoint behaviour depends on n and cannot be decided by copying the logarithm’s endpoint rule. The positive-integer case below terminates and has no such restriction.

Positive integer powers give a finite binomial theorem

For a positive integer n, the source gives

$$(a+b)^n=a^n+\binom n1 a^{n-1}b+\binom n2 a^{n-2}b^2+\cdots$$ $$\hspace{1em}+\binom nr a^{n-r}b^r+\cdots+\binom n{n-1}ab^{n-1}+b^n.$$

For integers r with $0\le r\le n$, the binomial coefficient is

$$\binom nr={}^{n}C_r=\frac{n!}{(n-r)!\,r!}.$$

The source identifies this formula as being in MF27. The resulting expansion is finite, with n+1 terms before numerical specialisations or cancellation. The polynomial identity holds for all real a and b; it is not confined by an infinite series’ convergence interval.

The printed factoring remark is

$$(a+b)^n=a^n\left(1+\frac ba\right)^n.$$

This lets an expansion in the form $(1+x)^n$ generate one for a general binomial by setting $x=b/a$. Independently, this rewriting requires $a\ne0$. If $a=0$ for positive integer n, use the original finite identity directly. For a nonterminating binomial expansion, transform the sufficient condition to $|b/a|<1$ as well; the factoring step alone does not remove it.

The source asks: What is the position of the term with binomial coefficient $\binom nr$? The Student leaves the answer blank; the Teacher gives the (r+1)th term. Counting starts with $a^n$ at r=0 as the first term, so $T_{r+1}=\binom nr a^{n-r}b^r$. Do not confuse the exponent index r with the term’s position.

Choose the expansion form and the required powers

1. Decide whether the exponent makes the expansion finite.
2. For the standard infinite form, factor the bracket so its constant term is 1.
3. Name the substituted argument and transform its validity condition.
4. List the powers needed in the final answer before expanding.

The source compares finite binomial expansion with the general binomial Maclaurin form in three rows:

Feature: Basic form

Finite binomial expansion: $(a+b)^n$

General binomial Maclaurin form: $(1+x)^n$; the bracket’s constant term must be 1.

Feature: Power n

Finite binomial expansion: Positive integer $n\in\mathbb Z^+$

General binomial Maclaurin form: n is a constant.

Feature: Number of terms

Finite binomial expansion: Finite, n+1 terms

General binomial Maclaurin form: The source says infinitely many terms.

The last entry describes a nonterminating binomial series; independently, a nonnegative integer exponent makes the Maclaurin expansion terminate too. These are overlapping cases, not mutually exclusive types of function. The requirement to begin with 1 concerns the bracket used for the standard binomial formula, not every Maclaurin series: sine, for example, has constant term 0.

Example 3 applies the two source callouts: factor $(a+bx)^n$ to $a^n(1+bx/a)^n$ when its bracket does not start with 1, then transform the standard argument condition to $|bx/a|<1$. The rewriting requires a nonzero a and suitable domain for fractional real powers.

It retains all the source’s cube-root coefficients and both inverse-expression methods. Its cube-root part repeats the expansion in Discussion question 6(i); that existing problem’s further approximation asks remain intact.

For $(1-1/x)^{-1}$, directly substituting $-1/x$ gives descending powers and a representation away from zero. Rewriting as $-x(1-x)^{-1}$ produces the requested ascending powers for $|x|<1$, while preserving the original $x\ne0$ condition.

Self-Practice 1 then contrasts a finite fourth power, an inverse-square expression and a negative fractional power. Its printed fourth-degree answer for part (a) is retained separately from the requested cubic truncation.

Combine series only where all required expansions apply

Write a power checklist before multiplying. To keep the quadratic term, collect constant-times-quadratic, linear-times-linear, and quadratic-times-constant contributions. Then intersect the validity intervals: every expansion used must apply at the same input.

Example 4 decomposes a rational function into two partial fractions before expanding. Both copies give the partial fractions; the Teacher supplies the remaining expansion and every interval step. Their closing remark stresses AND: intersect $-1<x<1$ with $-4<x<4$, giving $-1<x<1$. Four terms include the constant and run through the cubic term. The rational function’s domain is a separate question from its series interval.

Example 5, attributed to 2011/CJC/JC1/MYE/5(modified), multiplies two normalized binomial series. To obtain the quadratic coefficient, include both individual quadratic terms and the product of the linear terms. Intersect the two sufficient intervals before attempting any numerical substitution. The source’s second interval line prints the wrong exponent sign; the lesson records the typo explicitly and uses the reciprocal cube-root factor from the actual derivation.

Independent validity check. For $\ln(1-2x)$, put $u=-2x$ into $-1<u\le1$. Reversing the inequalities when dividing by $-2$ gives $-1/2\le x<1/2$. For the nonterminating binomial expansion of $(1+2x)^{-1/2}$, the sufficient condition $|2x|<1$ gives $-1/2<x<1/2$. Their common interval excludes both endpoints. Check endpoints for the particular series rather than copying a condition from another formula.

Check the substitution before judging numerical accuracy

In Example 5, first check the interval. Brian’s x=3 falls outside $-1<x<1$, so reject it. The alternatives $8/11$ and $-1/3$ are inside that interval, but their quadratic estimates have substantially different errors. The Teacher’s closer-to-zero explanation is retained with an independent error comparison.

The printed first decimal, 1.4958, is retained as source evidence; the rational estimate $181/121$ correctly rounds to 1.4959. The second estimate $31/18$ rounds to 1.7222.

Rounding an estimate to four decimal places is different from proving it is accurate to four decimal places.

The source’s general closer-to-zero remark is a useful local guide for a fixed expansion and truncation. It is not a universal error-ordering rule for every function, sign or truncation. Convergence admits use of the infinite series; a finite approximation still needs an accuracy assessment.

Obtain coefficients from a differential relation

1. Record the given value at zero.
2. Differentiate the relation before substituting zero; retain every product-rule term.
3. Evaluate each new derivative at zero using values already found.
4. Divide by the matching factorial and stop at the requested power.

As an independent check, compare coefficients directly in the original relation when possible. Differentiating a truncated series lowers the known order by one.

To form a series through $x^r$ by this direct method, obtain derivatives through order r. If the equation already supplies the first derivative, fewer additional differentiations are needed. Known series or a recurrence may also provide the required coefficients.

Example 6 starts from $y^{\prime}=2xy-1$ and y(0)=1. The Teacher completes every intermediate product-rule line and zero value, then supplies the quartic approximation and 0.90938 at x=0.1. Factorials convert derivative values into coefficients. Rounding that polynomial value is distinct from bounding the omitted remainder.

Example 7 uses $y=\tan2x$ and integrates the shared callout into its final deduction: relate the requested $\sec^2(2x)$ to the derivative of the tangent series. Keep the chain-rule factor 2, then divide by it. Termwise differentiation applies within the power series’ convergence radius, rather than being an unrestricted operation on an arbitrary approximation.

Verify a series by a genuinely different route

Example 8 is marked Self-Reading and attributed to 2022/ACJC/I/7(modified). Both copies already print the full implicit-derivative solution and its MF27 sine/logarithm verification. Comparing their cubic coefficients checks the product-rule calculation independently. All four asks and marks 2, 3, 3, 2 are retained.

For its final quotient, read the denominator as $\sqrt{1-x}$, giving exponent $-1/2$ in the reciprocal binomial factor. Truncate both factors only after identifying the requested quadratic order, and retain the linear-times-linear contribution; it cancels the numerator’s non-logarithmic quadratic term. The final answer is $\ln2+(1+\tfrac12\ln2)x+(\tfrac38\ln2)x^2$.

Choose a strategy for an extended expression

The source gives three common strategies and explicitly says the list is not exhaustive.

Expression: $\cos x/x^2$

Strategy: Rewrite as a power of x times the standard cosine expansion.

Expression: $e^{1-2x}$

Strategy: Apply the law of indices, then replace the exponential argument by −2x.

Expression: $(1-x)^{-2}$

Strategy: Recognize the derivative of the geometric series.

The complete source illustrations are:

$$\frac{\cos x}{x^2}=\frac1{x^2}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)=\frac1{x^2}-\frac12+\frac{x^2}{24}-\cdots.$$

This representation uses x≠0 and has a negative power: it is not a Maclaurin series of a function defined at zero.

$$e^{1-2x}=e\,e^{-2x}=e\left[1+(-2x)+\frac{(-2x)^2}{2!}+\cdots\right]=e(1-2x+2x^2-\cdots).$$ $$\frac1{(1-x)^2}=\frac d{dx}\left(\frac1{1-x}\right)=\frac d{dx}(1+x+x^2+\cdots)=1+2x+\cdots.$$

For the geometric series, termwise differentiation is used inside |x|<1. Self-Practice 2 applies the coefficient method and standard-series composition; its part (b) target is printed in the ask, rather than supplied as a separate worked answer.

Use exact angle identities before small-angle truncations

Example 9 and Self-Practice 3 concern an angle with a fixed offset. Small x does not make x+π/6 or π/4+x/3 a small angle. Apply the sine, cosine or tangent addition formula first, then approximate only the small increment in radians. The source’s special-angle values and both cosine addition identities are retained in the relevant working.

The closing remark distinguishes exact equality from approximation: use = for an exact transformation and ≈ when replacing an expression by a truncation. The square-area proof in Example 11 likewise uses exact tangent subtraction and exact triangle areas before its final approximation.

The complete printed geometry solution appears on Student PDF 26 and Teacher PDF 25; Student PDF 25 is an extra blank page. Their shared proof and callouts are integrated into the example, and the source square/angle diagram is reconstructed with a text alternative.

Track order after division and cancellation

Work backwards from the requested final power. If division removes a factor $x^m$, the numerator must be known $m$ powers further. Cancellation may expose terms that looked unnecessary earlier.

For example, obtaining a quotient through $x^2$ after cancelling $x^2$ can require fourth-degree terms in both numerator and denominator. Keep $x\ne0$ during that cancellation, then consider a limit separately if asked.

Example 10 rewrites a reciprocal denominator as a negative power and retains only needed terms. Part (a) has a denominator nonzero at zero and gives sinθ/cos2θ≈θ through quadratic order. Part (b) has a denominator vanishing quadratically.

Its printed answer 2 is retained as a leading approximation; to retain every quadratic term of the final quotient, fourth-degree terms must first be kept in the numerator and denominator. The independently derived refinement is $2+\tfrac56\theta^2+O(\theta^4)$ for small nonzero θ.

Self-Practice 4 keeps the positive square root and the √2 prefactor when normalizing $\sqrt{1+\cos\theta}$. Example 11 has an outer factor θ, so its reciprocal bracket is needed only through the linear term to obtain a²θ(1−θ).

The complete source review and closing practice map are retained with this chapter; the historical partial-review notes describe earlier ingestion stages.

The closing reading suggestion names The Manga Guide to Calculus by Hiroyuki Kojima; its cover also credits Shin Togami and Becom Co., Ltd. The source suggests borrowing it from the CJC Library, which is a printed historical suggestion rather than a check of current availability.

Its illustrated extract connects the local linear approximation $f(x)\approx f(a)+f^{\prime}(a)(x-a)$ near $a$ with higher-degree approximations. Maclaurin uses $a=0$. It also models compound interest: principal $M$ at annual rate $x$ grows to $M(1+x)^n$ after $n$ years.

For a positive integer $n$, the binomial expansion is a finite polynomial, with $\binom nr=n!/[r!(n-r)!]$; its first coefficients are $1,n,n(n-1)/2,n(n-1)(n-2)/6$. This finite case has no infinite-series convergence restriction. Original cover/comic artwork remains in the source evidence archive.

Review each of the eight skills at the end of lecture and again at the end of tutorial, then use the printed practice map below.