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CSIR NET Mathematical Science UNIT – 3 Flashcards
50 question-and-answer cards covering UNIT – 3 as it is examined in CSIR NET Mathematical Science. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the UNIT – 3 deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
Give the classification of the general second order linear PDE $A u_{xx}+B u_{xy}+C u_{yy}+\dots=0$.
Based on discriminant $B^2-4AC$: \n- $B^2-4AC>0$: hyperbolic (e.g. wave equation). \n- $B^2-4AC=0$: parabolic (e.g. heat equation). \n- $B^2-4AC<0$: elliptic (e.g. Laplace equation).
Classify the wave, heat, and Laplace equations and give their canonical forms.
Wave $u_{tt}=c^2u_{xx}$: hyperbolic. Heat $u_t=\alpha u_{xx}$: parabolic. Laplace $u_{xx}+u_{yy}=0$: elliptic. They are the prototypes of the three PDE classes.
How do you find the general solution of a higher order linear PDE with constant coefficients $F(D,D')z=0$ where $D=\partial_x,\,D'=\partial_y$?
If $F(D,D')$ factors into linear factors $(D-m_iD')$, each simple factor $(D-mD')$ contributes $\phi_i(y+mx)$. A factor repeated $k$ times contributes $\phi_1(y+mx)+x\phi_2(y+mx)+\dots+x^{k-1}\phi_k(y+mx)$, with arbitrary functions $\phi_i$.
For $F(D,D')z=0$, how are the roots $m$ found, and what is the complementary function for distinct roots $m_1,\dots,m_n$?
Put $D=m,\,D'=1$ in the homogeneous operator to get the auxiliary equation $F(m,1)=0$. For distinct roots: $$z = \phi_1(y+m_1x)+\phi_2(y+m_2x)+\cdots+\phi_n(y+m_n x).$$
Apply separation of variables to the 1-D heat equation $u_t=\alpha u_{xx}$. What ODEs and solution form arise?
Let $u=X(x)T(t)$. Then $\frac{T'}{\alpha T}=\frac{X''}{X}=-\lambda$. So $X''+\lambda X=0$ and $T'+\alpha\lambda T=0$. For $\lambda=k^2>0$: $T=e^{-\alpha k^2 t}$, giving solutions $u=e^{-\alpha k^2 t}(A\cos kx+B\sin kx)$.
Apply separation of variables to the 1-D wave equation $u_{tt}=c^2u_{xx}$. What ODEs result?
Let $u=X(x)T(t)$. Then $\frac{T''}{c^2T}=\frac{X''}{X}=-\lambda$, giving $X''+\lambda X=0$ and $T''+c^2\lambda T=0$. Both factors are oscillatory for $\lambda>0$, e.g. $u=(A\cos kx+B\sin kx)(C\cos ckt+D\sin ckt)$ with $\lambda=k^2$.
Apply separation of variables to Laplace's equation $u_{xx}+u_{yy}=0$ in Cartesian coordinates.
Let $u=X(x)Y(y)$. Then $\frac{X''}{X}=-\frac{Y''}{Y}=-\lambda$. So $X''+\lambda X=0$ (oscillatory) and $Y''-\lambda Y=0$ (exponential/hyperbolic) for $\lambda>0$, e.g. $u=(A\cos kx+B\sin kx)(Ce^{ky}+De^{-ky})$.
State the bisection method and its convergence behaviour for finding a root of $f(x)=0$.
If $f$ is continuous on $[a,b]$ with $f(a)f(b)<0$, repeatedly bisect: $c=\frac{a+b}{2}$, keep the subinterval where the sign change occurs. It always converges (linearly) with error halving each step: after $n$ steps error $\leq \frac{b-a}{2^{n}}$.
Write the Newton–Raphson iteration formula and state its order of convergence.
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$ It converges quadratically (order $2$) to a simple root, provided $f'\neq 0$ near the root and the initial guess is sufficiently close.
Define the fixed-point iteration method and state its convergence condition.
Rewrite $f(x)=0$ as $x=g(x)$ and iterate $x_{n+1}=g(x_n)$. It converges to a fixed point $\alpha$ (linearly) if $g$ is continuous and $|g'(x)|<1$ in a neighbourhood of $\alpha$; smaller $|g'(\alpha)|$ gives faster convergence.
Define the order (rate) of convergence $p$ of an iterative method with errors $e_n=x_n-\alpha$.
The method has order $p$ if $$\lim_{n\to\infty}\frac{|e_{n+1}|}{|e_n|^{p}} = C \neq 0$$ for some constant $C$ (the asymptotic error constant). $p=1$ is linear, $p=2$ quadratic, $p\approx1.618$ for the secant method.
Give the secant method formula and its order of convergence.
$$x_{n+1} = x_n - f(x_n)\,\frac{x_n - x_{n-1}}{f(x_n)-f(x_{n-1})}.$$ Order of convergence is the golden ratio $p=\frac{1+\sqrt{5}}{2}\approx 1.618$ (superlinear).
What is the order of convergence of Newton–Raphson at a root of multiplicity $m>1$, and how can quadratic convergence be restored?
At a multiple root, ordinary Newton–Raphson converges only linearly (rate $1-\tfrac{1}{m}$). Modified Newton $x_{n+1}=x_n - m\frac{f(x_n)}{f'(x_n)}$ restores quadratic convergence.
Describe the Gauss elimination method for solving $A\vec{x}=\vec{b}$.
Use elementary row operations (forward elimination) to reduce the augmented matrix $[A\,|\,b]$ to upper-triangular form, then solve by back substitution. Partial pivoting (swapping rows to put the largest pivot on the diagonal) is used to control round-off error. Operation count $\sim \frac{n^3}{3}$.
Write the Gauss–Seidel iteration formula for the system $A\vec{x}=\vec{b}$.
$$x_i^{(k+1)} = \frac{1}{a_{ii}}\left(b_i - \sum_{j<i} a_{ij}x_j^{(k+1)} - \sum_{j>i} a_{ij}x_j^{(k)}\right).$$ It uses the most recently updated values immediately within each sweep.
How does Gauss–Seidel differ from the Jacobi method, and what guarantees convergence?
Jacobi uses only old iterate values $x_j^{(k)}$ for all $j$; Gauss–Seidel uses updated values $x_j^{(k+1)}$ for $j<i$ within the same sweep, so it generally converges faster. Both converge if $A$ is strictly diagonally dominant (or symmetric positive definite for Gauss–Seidel).
Define the forward, backward, and central difference operators.
Forward: $\Delta f(x)=f(x+h)-f(x)$. Backward: $\nabla f(x)=f(x)-f(x-h)$. Central: $\delta f(x)=f\!\left(x+\tfrac{h}{2}\right)-f\!\left(x-\tfrac{h}{2}\right)$.
State the relations between the shift operator $E$ and the forward/backward difference operators.
$Ef(x)=f(x+h)$. Then $\Delta = E-1$, $\nabla = 1-E^{-1}$, and $E=e^{hD}$ where $D$ is the differentiation operator. Also $\delta = E^{1/2}-E^{-1/2}$ and the averaging operator $\mu=\tfrac{1}{2}(E^{1/2}+E^{-1/2})$.
Write the Newton's forward difference interpolation formula.
With $u=\frac{x-x_0}{h}$: $$f(x)=f_0 + u\Delta f_0 + \frac{u(u-1)}{2!}\Delta^2 f_0 + \frac{u(u-1)(u-2)}{3!}\Delta^3 f_0 + \cdots$$
State the Lagrange interpolation formula for nodes $x_0,\dots,x_n$.
$$P(x)=\sum_{i=0}^{n} f(x_i)\,L_i(x),\qquad L_i(x)=\prod_{\substack{j=0\\ j\neq i}}^{n}\frac{x-x_j}{x_i-x_j}.$$ The $L_i$ satisfy $L_i(x_j)=\delta_{ij}$.
How does Hermite interpolation differ from Lagrange interpolation, and what is the degree of the Hermite polynomial through $n+1$ nodes?
Hermite interpolation matches both function values $f(x_i)$ and derivative values $f'(x_i)$ at each node. With $n+1$ nodes it imposes $2(n+1)$ conditions, giving a unique polynomial of degree $\leq 2n+1$.
What is a cubic spline interpolant and what continuity/smoothness does it enforce?
A cubic spline is a piecewise cubic polynomial on each subinterval $[x_i,x_{i+1}]$ that interpolates the data and is $C^2$ globally: the function, first derivative, and second derivative are continuous across all interior nodes. A natural spline additionally sets $S''=0$ at the endpoints.
State the error/remainder term for polynomial interpolation at $n+1$ nodes.
$$f(x)-P_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^{n}(x-x_i),$$ for some $\xi$ in the smallest interval containing the nodes and $x$ (assuming $f\in C^{n+1}$).
Compare direct (Gauss elimination) and iterative (Gauss–Seidel) methods for linear systems.
Direct methods (Gauss elimination) give the exact solution in finitely many steps ($\sim n^3/3$ operations), best for small/dense systems. Iterative methods (Gauss–Seidel/Jacobi) refine an approximation each sweep, are memory-efficient for large sparse systems, but require a convergence condition (e.g. diagonal dominance).
What this deck covers
The UNIT – 3 deck follows the CSIR NET Mathematical Science UNIT – 3 syllabus — 6 chapters and 33 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 8.3 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 209 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
UNIT – 3 flashcards FAQ
How many UNIT – 3 flashcards are in this CSIR NET Mathematical Science deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these CSIR NET Mathematical Science flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the UNIT – 3 cards cover?
They follow the CSIR NET Mathematical Science UNIT – 3 syllabus — 6 chapters and 33 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.