Taylor’s theorem with remainders

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2023, 2024)
• Priority tier: T3
• Marks (count): 10 (2)
• Average solve time: ~6 min
• Difficulty mix: easy 1, medium 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Taylor’s theorem appears in Section A (compulsory) and is always 10 marks — fully algorithmic and perfectly reproducible. Both question types are mechanical: match Taylor coefficients to determine parameters, or expand a standard function about a point and sum to a numerical answer. Neither type requires any creative work beyond knowing the $f^{(n)}(a)/n!$ coefficient formula and the standard series for $\ln x$, $\sin x$, $\cos x$. Two clean examples cover the full historical range.

Minimum Theory

Taylor’s theorem. If $f$ is $(n+1)$ times differentiable on an interval containing $a$, then for $x$ near $a$:

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+R_n(x)$

The three standard remainder forms are: Lagrange $R_n=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$; Cauchy $R_n=\dfrac{f^{(n+1)}(c)}{n!}(x-c)^n(x-a)$; Peano $R_n=o\bigl((x-a)^n\bigr)$ as $x\to a$.

Standard Maclaurin series ($a=0$):

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,\qquad\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$

Taylor series for $\ln x$ about $a=1$. Since $f^{(n)}(1)=(-1)^{n-1}(n-1)!$ for $n\ge 1$ and $f(1)=0$:

$\ln x=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots,\quad 0<x\le 2.$

Alternating-series truncation error. For a convergent alternating series with decreasing terms, the truncation error after $n$ terms is bounded in absolute value by the first omitted term.

Question Archetypes

taylor-coefficient-matching (1 question(s); 2023)

Recognition Cues

• $\displaystyle\lim_{x\to 0}\frac{\text{(expression with parameters)}\cdot\sin x,\cos x\text{ terms}}{x^k}=L$
• The limit is $0/0$ (and possibly $0/0^2$ or higher); the parameters must simultaneously kill all singular terms and hit the target value.
• The denominator is $x^3$: expand numerator through $x^3$ terms.

Solution Template

1. Expand $\sin x$ and $\cos x$ as Maclaurin series to the required order (at least $x^3$).
2. Substitute into the numerator. Collect powers of $x$.
3. Divide by $x^k$. Identify which coefficients must vanish for the limit to be finite.
4. Write the linear system. (Zero condition for each negative-power coefficient; value condition for the $x^0$ coefficient.)
5. Solve and state $p$, $q$.

Worked Example

2023 Paper 1, 2023-P1-Q1c (10 marks)

Find the values of $p$ and $q$ for which $\displaystyle\lim_{x\to 0}\dfrac{x(1+p\cos x)-q\sin x}{x^3}=1$.

Step 1 — Expand. Using $\cos x=1-\dfrac{x^2}{2}+O(x^4)$ and $\sin x=x-\dfrac{x^3}{6}+O(x^5)$:

$x(1+p\cos x)=(1+p)x-\frac{p}{2}x^3+O(x^5),$

$q\sin x=qx-\frac{q}{6}x^3+O(x^5).$

Step 2 — Numerator.

$\text{Numerator}=(1+p-q)x+\left(-\frac{p}{2}+\frac{q}{6}\right)x^3+O(x^5).$

Step 3 — Divide by $x^3$.

$\frac{\text{Num}}{x^3}=\frac{1+p-q}{x^2}+\left(-\frac{p}{2}+\frac{q}{6}\right)+O(x^2).$

Step 4 — Conditions. For the limit to be finite, kill the $1/x^2$ term:

$1+p-q=0\qquad\Rightarrow\qquad q=1+p.\qquad(1)$

For the limit to equal $1$:

$-\frac{p}{2}+\frac{q}{6}=1.\qquad(2)$

Step 5 — Solve. Substitute (1) into (2): $-p/2+(1+p)/6=1$, giving $(1-2p)/6=1$, so $p=-5/2$, $q=-3/2$.

$\boxed{\;p=-\tfrac{5}{2},\quad q=-\tfrac{3}{2}.\;}$

Common Traps

• The $1/x^2$ term (coefficient of $x$ in the numerator) must vanish for the limit to exist at all — if you skip this condition and only equate the $x^0$ coefficient, you get an inconsistent answer.
• Expand $\cos x$ to $x^2$ (and keep the $x\cdot x^2=x^3$ product); similarly expand $\sin x$ to $x^3$. Stopping too early loses the $x^3$ terms.
• Verify: substitute $p=-5/2$, $q=-3/2$ back: coefficient of $x$ is $1-5/2-(-3/2)=1-5/2+3/2=0$ ✓; coefficient of $x^3$ divided by $x^3$ is $-(-5/2)/2+(-3/2)/6=5/4-1/4=1$ ✓.

taylor-expansion-evaluation (1 question(s); 2024)

Recognition Cues

• “Expand $f(x)$ in powers of $(x-a)$ by Taylor’s theorem; hence find $f(a+h)$ to $k$ decimal places.”
• The function is $\ln x$, $e^x$, $\sin x$, or similar; the expansion point $a$ has a known value ($\ln 1=0$, $e^0=1$, etc.).
• The numerical evaluation step requires summing enough terms of an alternating series.

Solution Template

1. Compute $f^{(n)}(a)$ for $n=0,1,2,\ldots$ Find the pattern.
2. Write the Taylor series $f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(a)}{n!}(x-a)^n$.
3. Set $x=a+h$. Write out the first few terms numerically.
4. Sum until the next term is below the accuracy threshold (alternating-series bound: error $\le|$next term$|$).
5. Round to the required number of decimal places.

Worked Example

2024 Paper 1, 2024-P1-Q1d (10 marks)

Expand $\ln x$ in powers of $(x-1)$ by Taylor’s theorem and hence find $\ln(1.1)$ correct to four decimal places.

Step 1 — Derivatives at $a=1$. $f(x)=\ln x$; $f^{(n)}(x)=(-1)^{n-1}(n-1)!/x^n$ for $n\ge 1$, so $f^{(n)}(1)=(-1)^{n-1}(n-1)!$ and $f(1)=0$.

Step 2 — Taylor series.

$\ln x=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n$

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

valid for $0<x\le 2$.

Step 3 — Evaluate at $x=1.1$ (so $(x-1)=0.1$).

Step 4 — Truncation. The series is alternating with decreasing terms; the 6th term $\approx 1.7\times10^{-7}<5\times10^{-5}$ (the 4th-decimal threshold). Sum of terms 1–5:

$0.1000000-0.0050000+0.0003333-0.0000250+0.0000020=0.0953103.$

$\boxed{\;\ln(1.1)\approx 0.0953.\;}$

Common Traps

• Keep enough decimal places during summation — intermediate rounding to 4 places propagates errors. Carry at least 6 places until the final rounding.
• The series converges for $0<x\le 2$; state this validity range.
• The alternating-series truncation bound: include enough terms so that the first omitted term $<5\times10^{-5}$ for 4-decimal accuracy. Here term 6 is $\approx1.7\times10^{-7}$; terms 1–5 suffice, but terms 1–4 give $0.09530$ which rounds to $0.0953$ correctly anyway — verify to be sure.

Marks-Aware Writing

Both questions are 10 marks. For coefficient matching (2023): Step 3 (setting up the $1/x^2$ condition) and Step 4 (the linear system) carry most marks. An answer that correctly identifies both conditions but makes an arithmetic error in solving earns 7 marks. For expansion-evaluation (2024): writing the general Taylor formula with the $f^{(n)}(1)=(-1)^{n-1}(n-1)!$ pattern earns 4 marks; the numeric summation earns 4 marks; citing the alternating-series bound earns 2 marks. An answer that writes out the series correctly but makes a summation error earns 7 marks.

Practice Set

(No additional practice items in the corpus beyond the two worked examples.)