🇮🇳 GATE Mathematics · subject
GATE Mathematics Interpolation Syllabus
Every chapter and topic of Interpolation examined in GATE Mathematics — 2 chapters, 2 topics and 2 sub-topics, plus 51 flashcards written against it.
Interpolation syllabus — full chapter and topic list
Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Interpolation in GATE Mathematics, not a summary of it.
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Lagrange and Newton forms of interpolating polynomial
1 topic- Interpolating Polynomial
- Lagrange Form
- Newton Form
- Interpolating Polynomial
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Error in polynomial interpolation of a function
1 topic- Error Analysis
Interpolation flashcards for GATE Mathematics
23 of 51 cards from the Interpolation deck — real questions with worked answers.
What is interpolation in numerical analysis?
Interpolation is the process of constructing a function (typically a polynomial) that passes exactly through a given set of known data points $(x_i, y_i)$, so it can be used to estimate values at intermediate points within the data range.
State the existence-and-uniqueness theorem for the interpolating polynomial.
Given $n+1$ distinct nodes $x_0, x_1, \dots, x_n$ with values $y_0, \dots, y_n$, there exists a unique polynomial $P(x)$ of degree at most $n$ such that $P(x_i) = y_i$ for all $i = 0, 1, \dots, n$.
Why must the interpolation nodes $x_0, x_1, \dots, x_n$ be distinct?
If two nodes coincide ($x_i = x_j$ with $y_i \neq y_j$) the data is contradictory, and the Vandermonde system becomes singular, so a unique interpolating polynomial fails to exist.
What is the maximum degree of the polynomial interpolating $n+1$ distinct data points?
At most $n$ (degree $\leq n$). With $n+1$ points the unique interpolant has degree at most $n$.
What linear system arises when finding interpolating polynomial coefficients directly via the monomial basis, and what matrix governs it?
Writing $P(x)=\sum_{j=0}^{n} a_j x^{j}$ and imposing $P(x_i)=y_i$ gives the linear system $V\mathbf{a}=\mathbf{y}$, governed by the Vandermonde matrix $V_{ij}=x_i^{\,j}$.
What is the determinant of the Vandermonde matrix for nodes $x_0, \dots, x_n$?
$$\det(V)=\prod_{0 \leq i < j \leq n}(x_j - x_i),$$ which is nonzero precisely when all nodes are distinct, guaranteeing a unique interpolant.
Why is solving the Vandermonde system directly to find the interpolating polynomial usually avoided?
The Vandermonde matrix is typically ill-conditioned (especially for many or clustered nodes), making the direct monomial-coefficient approach numerically unstable; Lagrange or Newton forms are preferred.
Define the Lagrange basis polynomial $\ell_i(x)$ for nodes $x_0, \dots, x_n$.
$$\ell_i(x)=\prod_{\substack{j=0 \\ j \neq i}}^{n} \frac{x - x_j}{x_i - x_j}.$$ It is the degree-$n$ polynomial satisfying $\ell_i(x_k)=\delta_{ik}$.
Write the Lagrange form of the interpolating polynomial.
$$P(x)=\sum_{i=0}^{n} y_i\, \ell_i(x),\quad \text{where } \ell_i(x)=\prod_{\substack{j=0 \\ j\neq i}}^{n}\frac{x-x_j}{x_i-x_j}.$$
What value does the Lagrange basis polynomial $\ell_i(x)$ take at the nodes?
$\ell_i(x_k)=\delta_{ik}$, i.e. $\ell_i(x_i)=1$ and $\ell_i(x_k)=0$ for $k \neq i$ (the cardinality / Kronecker-delta property).
What identity do the Lagrange basis polynomials satisfy when summed (partition of unity)?
$$\sum_{i=0}^{n} \ell_i(x)=1 \quad \text{for all } x,$$ since the constant function $1$ is interpolated exactly by its own Lagrange representation.
Write the linear (two-point) Lagrange interpolation formula through $(x_0,y_0)$ and $(x_1,y_1)$.
$$P(x)=y_0\frac{x-x_1}{x_0-x_1}+y_1\frac{x-x_0}{x_1-x_0}.$$
State a key disadvantage of the Lagrange interpolation form.
Adding a new data point requires recomputing all basis polynomials $\ell_i(x)$ from scratch (no incremental update), and direct evaluation is less efficient than Newton's form. Newton's form fixes this.
Define the divided difference $f[x_i, x_{i+1}]$ of first order.
$$f[x_i, x_{i+1}]=\frac{f(x_{i+1})-f(x_i)}{x_{i+1}-x_i}=\frac{f[x_{i+1}]-f[x_i]}{x_{i+1}-x_i}.$$
Give the recursive formula for the $k$-th order divided difference.
$$f[x_i, \dots, x_{i+k}]=\frac{f[x_{i+1},\dots,x_{i+k}]-f[x_i,\dots,x_{i+k-1}]}{x_{i+k}-x_i}.$$
Write Newton's divided-difference form of the interpolating polynomial.
$$P(x)=f[x_0]+\sum_{k=1}^{n} f[x_0,\dots,x_k]\prod_{j=0}^{k-1}(x-x_j).$$
In Newton's divided-difference form, what are the coefficients of the polynomial?
The leading divided differences $f[x_0], f[x_0,x_1], f[x_0,x_1,x_2], \dots, f[x_0,\dots,x_n]$ — the top diagonal of the divided-difference table.
What is the main advantage of Newton's form over Lagrange's form?
Newton's form is incremental: adding a new data point only requires computing one additional divided difference and appending one term, without recomputing existing coefficients.
Are the Lagrange and Newton interpolating polynomials (for the same data) the same polynomial?
Yes. By the uniqueness theorem, both forms represent the identical degree-$\leq n$ polynomial; they differ only in how it is expressed and computed.
What is the symmetry property of divided differences?
Divided differences are symmetric in their arguments: $f[x_0, x_1, \dots, x_k]$ is invariant under any permutation of the nodes $x_0, \dots, x_k$.
How is the $n$-th divided difference related to the $n$-th derivative of $f$?
There exists $\xi$ in the interval spanning the nodes with $$f[x_0,\dots,x_n]=\frac{f^{(n)}(\xi)}{n!}.$$
Define the forward difference operator $\Delta$.
$\Delta f(x_i)=f(x_{i+1})-f(x_i)=f_{i+1}-f_i$. Higher orders: $\Delta^{k}f_i=\Delta^{k-1}f_{i+1}-\Delta^{k-1}f_i$.
State the Newton's forward difference interpolation formula for equally spaced nodes with step $h$ and $s=\frac{x-x_0}{h}$.
$$P(x)=f_0+\sum_{k=1}^{n}\binom{s}{k}\Delta^{k}f_0,\quad \binom{s}{k}=\frac{s(s-1)\cdots(s-k+1)}{k!}.$$
Planning Interpolation for GATE Mathematics
Interpolation is about 2% of the GATE Mathematics syllabus by topic count — 2 of 110 topics, spread over 2 chapters. At roughly 45 minutes per topic plus 12 minutes per sub-topic, a first pass runs to about 2 hours.
Work top-down: read the chapter, then tick topics off individually rather than marking the whole chapter done. Sub-topics are where silent gaps hide.
Interpolation (GATE Mathematics) FAQ
What is in the GATE Mathematics Interpolation syllabus?
Interpolation is split into 2 chapters — Lagrange and Newton forms of interpolating polynomial and Error in polynomial interpolation of a function, containing 2 topics and 2 sub-topics in total.
How many chapters are there in Interpolation for GATE Mathematics?
2 chapters. Interpolation accounts for about 2% of the topics in the whole GATE Mathematics syllabus (2 of 110).
How long should I spend on Interpolation for GATE Mathematics?
Budget around 2 hours for a first pass through Interpolation — about 45 minutes per topic plus 12 minutes per sub-topic across its 2 topics. Add revision cycles on top.
Are there flashcards for GATE Mathematics Interpolation?
Yes — a 51-card Interpolation deck. Sample cards are printed on this page, and the full deck is free in the Examius app with spaced repetition scheduling.