FYUG Even Semester Exam, 2025
STATISTICS: Mathematical and Numerical Analysis (STADSC-252)
Subject: Statistics
Paper Code: STADSC-252
Semester: 4th Semester (FYUG)
Full Marks: 70 | Pass Marks: 28
Time: 3 Hours
UNIT-I
Question 1 (a) [2 Marks]
Show that for a singleton set, supremum is equal to infimum.
Let S = {a} be a singleton set containing one real number a.
- Since 'a' is the only element, every element x in S satisfies x <= a and x >= a.
- By definition, the Supremum (Least Upper Bound) is the smallest value that is greater than or equal to all elements in the set. Here, sup(S) = a.
- Similarly, the Infimum (Greatest Lower Bound) is the largest value less than or equal to all elements. Here, inf(S) = a.
- Therefore, sup(S) = inf(S) = a.
Question 1 (b) [2 Marks]
Find the supremum and infimum of the set S = {1/n; n in N}.
The set is S = {1, 1/2, 1/3, 1/4, ...}.
- Supremum: The largest element in the set is 1 (when n=1). Thus, sup(S) = 1.
- Infimum: As n approaches infinity, 1/n approaches 0. Although 0 is not in the set, it is the greatest lower bound. Thus, inf(S) = 0.
Question 2 (a)(i) [5 Marks]
Define interior point and open set. Show that finite intersection of open sets is also an open set.
Interior Point: A point 'x' is an interior point of set S if there exists a neighborhood (x - epsilon, x + epsilon) contained entirely within S.
Open Set: A set is open if every point in the set is an interior point.
Proof: Let G1 and G2 be two open sets. Let G = G1 intersection G2.
- If G is empty, it is open by definition.
- If x belongs to G, then x belongs to G1 and x belongs to G2.
- Since G1 is open, there is an epsilon1 such that (x - epsilon1, x + epsilon1) is in G1.
- Since G2 is open, there is an epsilon2 such that (x - epsilon2, x + epsilon2) is in G2.
- Let epsilon = min(epsilon1, epsilon2). Then the neighborhood (x - epsilon, x + epsilon) is contained in both G1 and G2, and thus in G.
- Hence, G is an open set.
UNIT-II
Question 4 (a)(i) [5 Marks]
State and prove Leibnitz's theorem on alternating series.
Statement: An alternating series sum((-1)^(n-1) * u_n) converges if:
1. u_n is monotonically decreasing (u_n+1 <= u_n for all n).
2. limit of u_n as n approaches infinity is 0.
Proof: Consider the partial sums S_2n and S_2n+1.
- S_2n = (u1 - u2) + (u3 - u4) + ... + (u_2n-1 - u_2n). Since u_n is decreasing, each term in bracket is positive, so S_2n is an increasing sequence.
- Also, S_2n = u1 - (u2 - u3) - (u4 - u5) - ... - u_2n. This shows S_2n < u1.
- By Monotone Convergence Theorem, S_2n converges to a limit, say S.
- S_2n+1 = S_2n + u_2n+1. Since limit of u_n is 0, S_2n+1 also converges to S.
- Since both even and odd partial sums converge to the same limit, the series converges.
UNIT-III
Question 5 (a) [2 Marks]
State Lagrange's mean value theorem.
If a function f(x) is:
- Continuous in the closed interval [a, b].
- Differentiable in the open interval (a, b).
Then there exists at least one point 'c' in (a, b) such that:
f'(c) = [f(b) - f(a)] / (b - a)
Question 6 (a)(i) [5 Marks]
State and prove Rolle's theorem.
Statement: If f(x) is continuous in [a, b], differentiable in (a, b), and f(a) = f(b), then there exists at least one 'c' in (a, b) such that f'(c) = 0.
Proof:
- If f(x) is constant, then f'(x) = 0 for all x in (a, b).
- If f(x) is not constant, since it's continuous on [a, b], it attains its maximum M and minimum m.
- Since f(a) = f(b), at least one of M or m must be attained at an interior point 'c'.
- If f(c) is a maximum, then f(c+h) <= f(c).
- This implies the right-hand derivative is <= 0 and left-hand derivative is >= 0.
- Since f'(c) exists, both must equal 0. Thus, f'(c) = 0.
UNIT-IV
Question 7 (b) [2 Marks]
Evaluate (Delta^2 / E) e^x.
Using operators: Delta = E - 1.
- (Delta^2 / E) = (E - 1)^2 / E = (E^2 - 2E + 1) / E = E - 2 + E^(-1).
- Applying to e^x: E(e^x) - 2(e^x) + E^(-1)(e^x).
- Result: e^(x+h) - 2e^x + e^(x-h).
- Factor e^x: e^x [e^h - 2 + e^(-h)].
Question 8 (a)(i) [10 Marks] - Option A
State and prove Newton-Gregory forward interpolation formula.
f(a + nh) = f(a) + nC1 Delta f(a) + nC2 Delta^2 f(a) + ...
Proof: We use the shift operator E, where E f(x) = f(x+h).
- We know E = 1 + Delta.
- For any n, E^n = (1 + Delta)^n.
- Applying Binomial Theorem: E^n = 1 + nC1 Delta + nC2 Delta^2 + ... + nCn Delta^n.
- Applying this to f(a): E^n f(a) = f(a + nh) = [1 + nC1 Delta + nC2 Delta^2 + ...] f(a).
- Expanding gives the Newton-Gregory forward formula.
UNIT-V
Question 10 (b)(i) [6 Marks]
Derive general quadrature formula of numerical integration and hence obtain Newton's 1/3rd rule.
General Quadrature: Integral of f(x) dx from x0 to xn = nh [y0 + (n/2) Delta y0 + (n(2n-3)/12) Delta^2 y0 + ...].
Deriving Simpson's 1/3 Rule: Put n = 2 in the general formula and neglect 3rd and higher order differences.
Integral (x0 to x2) f(x)dx = (h/3) [y0 + 4y1 + y2]
Summing for all intervals (0 to 2, 2 to 4, ...):
I = (h/3) [ (y0 + yn) + 4(y1 + y3 + ...) + 2(y2 + y4 + ...) ]