Lagrange’s interpolation

At a Glance

• Frequency: 5 sub-parts across 5 of 13 years (2015, 2016, 2017, 2020, 2025)
• Priority tier: T2
• Marks (count): 15 (3), 20 (2)
• Average solve time: ~14 min
• Difficulty mix: easy 1, medium 2, hard 2
• Section: B | Dominant type: computation

Why This Chapter Matters

Lagrange interpolation appears in 5 of the last 13 years and always in Section B with high mark values (15–20 marks). The questions span five distinct patterns — from a straightforward computation to a confluent (Hermite-type) formula derived by a limiting argument — so this is one of the most varied numerical-analysis chapters in the corpus. The two 20-mark questions (2015, 2016) are the clearest tests of the core formula and its error bound; the three 15-mark questions (2017, 2020, 2025) require deeper structural understanding. The divided-difference route to Newton’s form (2025) and the confluent-limit derivation (2020) are the two patterns most likely to catch candidates who only practised the basic formula.

Minimum Theory

Lagrange interpolating polynomial. Given $n+1$ distinct nodes $x_0,x_1,\ldots,x_n$ and values $f_i=f(x_i)$, the unique polynomial $P_n$ of degree $\le n$ satisfying $P_n(x_i)=f_i$ is $P_n(x)=\sum_{i=0}^n f_i\,L_i(x),\qquad L_i(x)=\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}.$ The basis polynomials $L_i$ satisfy $L_i(x_j)=\delta_{ij}$ and $\sum_i L_i(x)\equiv1$ (partition of unity, useful check).

Interpolation error. If $f$ has $(n+1)$ continuous derivatives, the error at $x$ is $f(x)-P_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\,\omega(x),\qquad \omega(x)=\prod_{i=0}^n(x-x_i),$ for some $\xi$ in the smallest interval containing $\{x_0,\ldots,x_n,x\}$. Hence $|f(x)-P_n(x)|\le\frac{M_{n+1}}{(n+1)!}\,|\omega(x)|,\qquad M_{n+1}=\max|f^{(n+1)}|$ over that interval.

Newton’s divided-difference form. An equivalent representation (especially efficient for unequally spaced nodes): $P_n(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+\cdots$ where the divided differences are built recursively: $f[x_k]=f_k,\qquad f[x_i,\ldots,x_k]=\frac{f[x_{i+1},\ldots,x_k]-f[x_i,\ldots,x_{k-1}]}{x_k-x_i}.$

Confluent (osculatory) interpolation. When two nodes $x_0$ and $x_0+\epsilon$ coalesce ($\epsilon\to0$), the Lagrange polynomial on $\{x_0,x_0+\epsilon,x_1\}$ converges to a formula matching $f(x_0)$, $f'(x_0)$, and $f(x_1)$ — the simplest Hermite interpolant. The error term picks up a double root at $x_0$: $E(x)=\tfrac{1}{6}(x-x_0)^2(x-x_1)f'''(\xi)$.

Question Archetypes

Five patterns cover every Lagrange interpolation question in the corpus.

lagrange-interpolation (1 question(s); 2015)

Recognition Cues

• A table of $(x_i, f_i)$ values; “find the Lagrange interpolating polynomial” or “find $f$ at a new point.”
• Four nodes → cubic polynomial.

Solution Template

1. Write down the four Lagrange basis polynomials $L_i(x)$ using the node values.
2. Evaluate each $L_i$ at the target point.
3. Multiply by $f_i$ and sum.
4. Optional: guess a low-degree polynomial fitting the data — if $f$ values are integers, often the data lies on an exact polynomial of manageable degree.

Worked Example(s)

2015 Paper 2, 2015-P2-Q6c (20 marks)

Find the Lagrange interpolating polynomial fitting the data below. Find $f(1.5)$.

Lagrange basis at $x=1.5$ (nodes $x_0=-1,x_1=2,x_2=3,x_3=4$):

$L_0(x)=\frac{(x-2)(x-3)(x-4)}{(-3)(-4)(-5)}=\frac{(x-2)(x-3)(x-4)}{-60},$ $L_1(x)=\frac{(x+1)(x-3)(x-4)}{(3)(-1)(-2)}=\frac{(x+1)(x-3)(x-4)}{6},$ $L_2(x)=\frac{(x+1)(x-2)(x-4)}{(4)(1)(-1)}=\frac{(x+1)(x-2)(x-4)}{-4},$ $L_3(x)=\frac{(x+1)(x-2)(x-3)}{(5)(2)(1)}=\frac{(x+1)(x-2)(x-3)}{10}.$

At $x=1.5$: $L_0=0.03125$, $L_1=1.5625$, $L_2=-0.78125$, $L_3=0.1875$. (Sum $=1$ ✓.)

$P(1.5)=(-1)(0.03125)+(11)(1.5625)+(31)(-0.78125)+(69)(0.1875)$ $=-0.03125+17.1875-24.21875+12.9375=\boxed{5.875.}$

Shortcut: check whether the data fits a low-degree exact polynomial. Test $f(x)=x^3+x+1$: $f(-1)=-1$ ✓, $f(2)=11$ ✓, $f(3)=31$ ✓, $f(4)=69$ ✓. So $P(x)=x^3+x+1$ and $P(1.5)=3.375+1.5+1=5.875$ ✓ — much faster once the guess works.

Common Traps

• Sign tracking in denominators. $(-3)(-4)(-5)=-60$, not $+60$.
• Each $L_i(x_j)=\delta_{ij}$ — verify at the nodes as a quick sanity check.
• Check $\sum L_i(x)=1$ at the evaluation point as a running total.

lagrange-with-error (1 question(s); 2016)

Recognition Cues

• “Estimate $f(a)$ using a Lagrange polynomial of degree $n$ over given nodes.”
• “Compute the error bound over $[l,r]$” and “the actual error $E(a)$.”
• Three deliverables: the estimate, a derivative-based bound, and the actual error.

Solution Template

1. Compute node values $f(x_i)$ numerically.
2. Evaluate basis polynomials $L_i$ at the target point; check $\sum L_i=1$.
3. Compute $P_n = \sum f(x_i)L_i$.
4. Derive $f^{(n+1)}(x)$ and find (or bound) $M_{n+1}=\max|f^{(n+1)}|$. Look for amplitude form $A\cdot g(x)$ with known $\max|g|$.
5. Evaluate $|\omega(x)|$ at the target point and over the interval.
6. State the bound and compute $E=|f(\text{target})-P_n(\text{target})|$.

Worked Example(s)

2016 Paper 2, 2016-P2-Q6c (20 marks)

$f(x)=e^{2x}\cos3x$ on $[0,1]$. Estimate $f(0.5)$ using degree-3 Lagrange interpolation over nodes $0,\ 0.3,\ 0.6,\ 1$. Give the error bound over $[0,1]$ and the actual error $E(0.5)$.

Node values ($x$ in radians throughout):

Basis values at $x=0.5$ (check $\sum L_i=1$ ✓):

$P_3(0.5)=(1)(-0.055556)+(1.132643)(0.396825)+(-0.754338)(0.694444)+(-7.315129)(-0.035714)$ $\approx\boxed{0.1313.}$

Error bound. Differentiate $f=e^{2x}\cos3x$ four times: $f^{(4)}(x)=e^{2x}(120\sin3x-119\cos3x).$ Recognise this as $e^{2x}\cdot A\sin(3x+\delta)$ with amplitude $A=\sqrt{120^2+119^2}=\sqrt{28561}=169$. So $|f^{(4)}(x)|\le169\,e^{2x}\le169e^2\approx1248.75\quad\text{on }[0,1].$ At $x=0.5$: $|\omega(0.5)|=(0.5)(0.2)(0.1)(0.5)=0.005=1/200$.

$|E(0.5)|\le\frac{169e^2}{4!}\cdot\frac{1}{200}=\frac{169e^2}{4800}\approx\boxed{0.260.}$

Actual error. $f(0.5)=e^1\cos1.5\approx2.71828\times0.07074\approx0.1923.$ $E(0.5)=0.1923-0.1313\approx\boxed{0.061,}$ well inside the bound $0.260$ ✓.

Common Traps

• Keep all trig arguments in radians. Using degree mode wrecks every node value.
• $\sqrt{120^2+119^2}=169$ exactly — this clean amplitude is the key to the error bound. Bounding $|120\sin3x|+|119\cos3x|\le239$ is valid but over-conservative.
• Two different “error bounds” arise: pointwise at $x=0.5$ uses $|W(0.5)|=1/200$; uniform over $[0,1]$ uses $\max_{[0,1]}|W|\approx0.0176$. The question asks for the bound over $[0,1]$ so report $M_4=169e^2$ and the resulting estimate.

lagrange-form-proof (1 question(s); 2017)

Recognition Cues

• “Show that the Lagrange interpolation formula for nodes $u_{-1},u_0,u_1,u_2$ can be written in the form $u_x=yu_0+xu_1+\tfrac{y(y^2-1)}{3!}\Delta^2u_{-1}+\tfrac{x(x^2-1)}{3!}\Delta^2u_0$.”
• The variables $x$ and $y=1-x$ are fractional positions between the nodes.

Solution Template

1. Write the four Lagrange basis polynomials explicitly for nodes $-1,0,1,2$.
2. Express $\Delta^2u_{-1}=u_1-2u_0+u_{-1}$ and $\Delta^2u_0=u_2-2u_1+u_0$.
3. Expand the right-hand side of the claimed formula, substituting $y=1-x$.
4. Match the coefficient of each $u_i$ on both sides against $L_i(x)$.

Worked Example(s)

2017 Paper 2, 2017-P2-Q6b (15 marks)

For equidistant values $u_{-1},u_0,u_1,u_2$, show the Lagrange interpolant satisfies $u_x=yu_0+xu_1+\frac{y(y^2-1)}{3!}\Delta^2u_{-1}+\frac{x(x^2-1)}{3!}\Delta^2u_0$ with $x+y=1$.

Four equidistant nodes at $-1,0,1,2$; interpolate at argument $x$ (measured in spacing units from $u_0$); write $y=1-x$.

Lagrange basis polynomials: $L_{-1}=-\frac{x(x-1)(x-2)}{6},\quad L_0=\frac{(x+1)(x-1)(x-2)}{2},$ $L_1=-\frac{(x+1)x(x-2)}{2},\quad L_2=\frac{(x+1)x(x-1)}{6}.$

Coefficient of $u_{-1}$ in the claimed formula: $\frac{y(y^2-1)}{6}=\frac{(1-x)(1-x^2)}{6}=\frac{(1-x)(-x)(2-x)}{6}=-\frac{x(1-x)(x-2)}{6}=L_{-1}.\;\checkmark$

Coefficient of $u_2$: $\frac{x(x^2-1)}{6}=\frac{x(x-1)(x+1)}{6}=L_2.\;\checkmark$

Coefficient of $u_0$: collect contributions from $y$, the $-2$ in $\Delta^2u_{-1}$, and the $+1$ in $\Delta^2u_0$: $y-\frac{2y(y^2-1)}{6}+\frac{x(x^2-1)}{6}\;\xrightarrow[y=1-x]{}\;\frac{(x+1)(x-1)(x-2)}{2}=L_0.\;\checkmark$

Coefficient of $u_1$: similarly equals $L_1$. ✓

Since all four coefficients match, the formula is the Lagrange polynomial. $\blacksquare$

Common Traps

• The bridge identity is $y^2-1=(y-1)(y+1)=(-x)(2-x)$, which converts $\tfrac{y(y^2-1)}{6}$ into $L_{-1}$. Miss this and the verification stalls.
• The denominator is $3!=6$ despite the differences being second differences; don’t reduce it to $2!=2$.
• Verify each coefficient separately — expanding everything at once leads to algebra errors.

confluent-interpolation (1 question(s); 2020)

Recognition Cues

• “Write the Lagrange polynomial on nodes $x_0$, $x_0+\epsilon$, $x_1$; take $\epsilon\to0$ to establish the formula…” with a coefficient involving $f'(x_0)$.
• The result is a Hermite (osculatory) interpolant matching $f$ at $x_0,x_1$ and $f'$ at $x_0$.
• The error term has $(x-x_0)^2$ rather than $(x-x_0)$.

Solution Template

1. Write the three-node Lagrange basis polynomials on $x_0,\,x_0+\epsilon,\,x_1$.
2. Taylor-expand $f(x_0+\epsilon)=f(x_0)+\epsilon f'(x_0)+O(\epsilon^2)$.
3. Group terms: the $1/\epsilon$ singularities from $\ell_0$ and $\ell_1$ cancel (they both multiply $f(x_0)$); the $\epsilon f'(x_0)$ piece of $\ell_1$ survives as a finite limit.
4. Take $\epsilon\to0$ in each coefficient and in the node factor of the error term $(x-x_0)(x-x_0-\epsilon)(x-x_1)\to(x-x_0)^2(x-x_1)$.

Worked Example(s)

2020 Paper 2, 2020-P2-Q8b (15 marks)

Three-point Lagrange on $x_0$, $x_0+\epsilon$, $x_1$; take $\epsilon\to0$ to derive the stated formula with error $E(x)=\tfrac16(x-x_0)^2(x-x_1)f'''(\xi)$.

Step 1 (three-node Lagrange). Basis: $\ell_0=\frac{(x-x_0-\epsilon)(x-x_1)}{(-\epsilon)(x_0-x_1)},\quad \ell_1=\frac{(x-x_0)(x-x_1)}{\epsilon(x_0+\epsilon-x_1)},\quad \ell_2=\frac{(x-x_0)(x-x_0-\epsilon)}{(x_1-x_0)(x_1-x_0-\epsilon)}.$

Step 2 (Taylor expand). $f(x_0+\epsilon)=f(x_0)+\epsilon f'(x_0)+O(\epsilon^2)$.

Step 3 (group and limit). The $f(x_0)$ contributions from $\ell_0$ and $\ell_1$ are $\ell_0 f(x_0)+\ell_1 f(x_0)$; the $1/\epsilon$ pieces cancel. The surviving derivative term gives coefficient $\tfrac{(x-x_0)(x-x_1)}{x_0-x_1}$ for $f'(x_0)$. The $\ell_2$ coefficient converges to $\tfrac{(x-x_0)^2}{(x_1-x_0)^2}$.

Taking $\epsilon\to0$: $f(x)=\frac{(x_1-x)(x+x_1-2x_0)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)+E(x).$

Step 4 (error). The three-node error is $\tfrac{f'''(\xi)}{6}(x-x_0)(x-x_0-\epsilon)(x-x_1)$; as $\epsilon\to0$ the node product becomes $(x-x_0)^2(x-x_1)$: $\boxed{\;E(x)=\frac{1}{6}(x-x_0)^2(x-x_1)\,f'''(\xi).\;}$

Verification: substituting $f(t)=t^2$ (so $f'''=0$) should give exact reproduction — it does when the $f'(x_0)$ coefficient carries the sign $\tfrac{(x-x_0)(x-x_1)}{x_0-x_1}$ (negative of what the printed paper states). The corrected sign has been verified.

Common Traps

• The $1/\epsilon$ singularities cancel only because both multiply $f(x_0)$. Keep terms grouped by which node value they multiply.
• The error has $(x-x_0)^2$ (double root at the merged node), not $(x-x_0)$.
• Printed-paper note (2020): the middle $f'(x_0)$ coefficient in the question has the wrong sign; the correct coefficient is $\tfrac{(x-x_0)(x-x_1)}{x_0-x_1}$, verified by exactness on $f=t^2$.

divided-difference-interpolation (1 question(s); 2025)

Recognition Cues

• Unequally spaced nodes; “find the unique polynomial of degree $\le n$ fitting the data.”
• “Obtain the bound on the truncation error.”
• Newton’s divided-difference form is the cleanest approach for unequal spacing.

Solution Template

1. Build the divided-difference table: first-order $f[x_i,x_{i+1}]=(f_{i+1}-f_i)/(x_{i+1}-x_i)$, then second-order $f[x_0,x_1,x_2]=(f[x_1,x_2]-f[x_0,x_1])/(x_2-x_0)$, etc.
2. Write Newton’s form: $P_n(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+\cdots$
3. Expand and verify at all nodes.
4. Error bound: $|E(x)|\le\tfrac{M_{n+1}}{(n+1)!}|\omega(x)|$. Find $\max|\omega|$ on the interval by setting $\omega'=0$.

Worked Example(s)

2025 Paper 2, 2025-P2-Q7b (15 marks)

Find the unique polynomial of degree $\le2$ fitting $x:0,1,3$; $f:1,3,55$. Obtain the truncation-error bound.

Divided-difference table:

Newton’s form: $P_2(x)=1+2(x-0)+8(x-0)(x-1)=1+2x+8x(x-1)=\boxed{8x^2-6x+1.}$

Check: $P_2(0)=1$ ✓, $P_2(1)=3$ ✓, $P_2(3)=72-18+1=55$ ✓.

Error bound. For 3-node interpolation: $E(x)=\frac{f'''(\xi)}{6}\,\omega(x),\qquad \omega(x)=x(x-1)(x-3).$ Find $\max|\omega|$ on $[0,3]$: set $\omega'(x)=3x^2-8x+3=0$, giving $x=(4\pm\sqrt7)/3$. At the interior critical points: $\max_{[0,3]}|\omega|=\left|\omega\!\left(\tfrac{4+\sqrt7}{3}\right)\right|\approx2.1126.$

$|E(x)|\le\frac{M_3}{6}\cdot2.1126\approx0.3521\,M_3,\qquad M_3=\max_{[0,3]}|f'''|.$

Common Traps

• Divided-difference arithmetic: $f[x_1,x_2]$ uses the gap $x_2-x_0$ in the denominator for the second-order term, not $x_2-x_1$.
• The $1/(n+1)!$ factor in the error is $1/3!=1/6$, not $1/2!$.
• Finding $\max|\omega|$ requires solving $\omega'=0$; the maximum is at an interior point of $[0,3]$, not an endpoint.

Marks-Aware Writing

20-mark questions (2015, 2016): Expect to write out all four basis polynomials explicitly, evaluate each at the target point, and assemble the sum. For the error question (2016): derive $f^{(4)}$ symbolically, bound it by recognising the amplitude form $A\cdot e^{2x}$, then give both the pointwise and interval bounds.

15-mark questions (2017, 2020, 2025): Structure matters more than computation. For the form-proof (2017): verify each coefficient in sequence — that is the whole proof. For the confluent limit (2020): a clear description of the $\epsilon\to0$ mechanism (singularities cancel because both multiply $f(x_0)$) is worth as many marks as the final formula. For divided differences (2025): write the difference table explicitly; the Newton polynomial and error bound follow mechanically.

Practice Set