A Type I error in a hypothesis test is:

rejecting the null hypothesis when it is true

accepting the null hypothesis when it is false

rejecting the alternative hypothesis when it is false

using a one-tailed test where a two-tailed test was needed

Which of the following statements about iterative methods for Ax = b are correct? (Select all that apply.)

Strict diagonal dominance of A guarantees convergence of the Jacobi and Gauss–Seidel methods

Gauss–Seidel uses updated values within the same iteration

Both methods need every diagonal element aᵢᵢ to be non-zero

Strict diagonal dominance of A guarantees convergence of the Jacobi and Gauss–Seidel methods

Gauss–Seidel uses updated values within the same iteration

Both methods need every diagonal element aᵢᵢ to be non-zero

Gauss–Seidel converges for every non-singular matrix

Engineering Mathematics — what only the MN paper adds: series, Taylor’s theorem, nonlinear first-order equations, hypothesis tests, regression and interpolation

Section 1 of the GATE Mining Engineering (MN) paper is mostly the engineering list the shared mathematics chapters already teach, but six of its sub-items are taught in none of them, and this chapter is written for this paper alone to cover them. From the Calculus heading: sequences and series and Taylor’s theorem. From Differential Equations: the non-linear first-order forms — exact equations and the Bernoulli equation — which the borrowed first-order chapter does not reach. From Probability and Statistics: sampling theory and hypothesis testing, and Pearson correlation and linear regression. From Numerical Methods: interpolation, with the iterative solution of linear equations beside it. Read the shared chapters first — linear algebra, limits and derivatives, partial derivatives, the Fourier-series section of Transforms, vector calculus, differential equations, probability and numerical methods — and come here for the rest. The examples lean on mining data where they can: a borehole assay table is where a regression line or an interpolated grade is actually needed.

1. Sequences, series and the tests for convergence

A sequence aₙ converges to L if its terms get and stay arbitrarily close to L; a series Σaₙ converges if its sequence of partial sums Sₙ = a₁ + … + aₙ does. The first question to ask of any series is the nth-term test: if aₙ does not tend to 0, the series diverges. The converse is false — the harmonic series Σ1/n has aₙ → 0 and still diverges — so aₙ → 0 is necessary, never sufficient.

The standard tests, and when each is the one to reach for
Test	Statement	Use it when
Geometric series	Σarⁿ converges to a/(1 − r) iff \|r\| < 1	the ratio of successive terms is constant
p-series	Σ1/nᵖ converges iff p > 1	the term is a power of n
Comparison / limit comparison	if aₙ/bₙ → a finite non-zero limit, Σaₙ and Σbₙ behave alike	the term is a rational function of n
Ratio (d’Alembert)	L = lim \|aₙ₊₁/aₙ\|: L < 1 converges, L > 1 diverges, L = 1 no decision	factorials or exponentials appear
Root (Cauchy)	L = lim \|aₙ\|^1/n, same verdicts as the ratio test	the whole term is raised to the nth power
Integral	for positive decreasing f(n) = aₙ, Σaₙ and ∫₁^∞ f dx converge together	ln n appears, e.g. Σ1/(n ln n)
Leibniz (alternating)	Σ(−1)ⁿbₙ converges if bₙ decreases to 0	the signs alternate

A series is absolutely convergent if Σ|aₙ| converges, and conditionally convergent if Σaₙ converges but Σ|aₙ| does not; the alternating harmonic series 1 − ½ + ⅓ − … (sum ln 2) is the standard example of the second. A power series Σcₙxⁿ converges for |x| < R, the radius of convergence, with R = lim |cₙ/cₙ₊₁| from the ratio test; the end points x = ±R must be tested separately.

⚠️ L = 1 decides nothing

Both Σ1/n (divergent) and Σ1/n² (convergent) give a ratio-test limit of exactly 1. When the ratio or root test returns 1, change test — a p-series comparison or the integral test almost always settles it.

2. Taylor’s theorem

If f has n + 1 derivatives near a, then f(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n! + Rₙ, and the Lagrange form of the remainder is Rₙ = f⁽ⁿ⁺¹⁾(ξ)(x − a)ⁿ⁺¹/(n + 1)! for some ξ between a and x. With a = 0 it is the Maclaurin series. The theorem is the mean value theorem carried to higher order: n = 0 gives f(x) = f(a) + f′(ξ)(x − a) exactly.

Maclaurin series worth knowing by heart
Function	Series	Valid for
eˣ	1 + x + x²/2! + x³/3! + …	all x
sin x	x − x³/3! + x⁵/5! − …	all x
cos x	1 − x²/2! + x⁴/4! − …	all x
ln(1 + x)	x − x²/2 + x³/3 − …	−1 < x ≤ 1
(1 + x)ⁿ	1 + nx + n(n − 1)x²/2! + …	\|x\| < 1 (any real n)
1/(1 − x)	1 + x + x² + x³ + …	\|x\| < 1

🧠 Bounding the error

For an alternating series whose terms decrease, the error of stopping is smaller than the first term left out: sin 0.1 ≈ 0.1 − 0.1³/6 is in error by less than 0.1⁵/120 ≈ 8 × 10⁻⁸. For eˣ use the Lagrange bound with the largest value of eᶿ on the interval.

3. Non-linear first-order equations: exact and Bernoulli

Exact equations. M(x, y) dx + N(x, y) dy = 0 is exact when ∂M/∂y = ∂N/∂x; then there is a function F with ∂F/∂x = M and ∂F/∂y = N, and the solution is F(x, y) = c. Find F by integrating M with respect to x, then add whatever terms in y alone are needed to make ∂F/∂y equal N. If the test fails, an integrating factor may repair it: if (M_y − N_x)/N is a function of x alone, μ = exp∫[(M_y − N_x)/N] dx; if (N_x − M_y)/M is a function of y alone, μ = exp∫[(N_x − M_y)/M] dy.

The Bernoulli equation dy/dx + P(x)y = Q(x)yⁿ (n ≠ 0, 1) is non-linear, but the substitution v = y¹⁻ⁿ turns it into the linear equation dv/dx + (1 − n)P v = (1 − n)Q, which the integrating factor exp∫(1 − n)P dx solves. Solve for v, then return to y = v^1/(1−n).

⚠️ Forgetting the (1 − n)

Differentiating v = y¹⁻ⁿ gives dv/dx = (1 − n)y⁻ⁿ dy/dx, so both P and Q are multiplied by (1 − n) in the linear equation. For n = 2 that factor is −1, and dropping it flips the sign of every term on the right.

4. Sampling theory and hypothesis testing

The mean x̄ of a random sample of size n from a population with mean μ and standard deviation σ is itself a random variable, with mean μ and standard error σ/√n. By the central limit theorem x̄ is approximately normal for large n (n ≥ 30 is the usual working rule) whatever the population’s shape. A confidence interval for μ is x̄ ± z·σ/√n, with z = 1.645, 1.96 and 2.576 for 90%, 95% and 99% confidence; when σ is unknown and n is small, the sample standard deviation s replaces σ and Student’s t with n − 1 degrees of freedom replaces z.

Null hypothesis H₀ — the claim tested, e.g. μ = μ₀; the alternative H₁ is μ ≠ μ₀ (two-tailed) or μ > μ₀, μ < μ₀ (one-tailed).
Test statistic z = (x̄ − μ₀)/(σ/√n), or t = (x̄ − μ₀)/(s/√n) with n − 1 degrees of freedom when σ is estimated from a small sample.
Significance level α — the probability of rejecting H₀ when it is true (a Type I error). At α = 0.05 two-tailed, reject if |z| > 1.96.
Type II error β — accepting H₀ when it is false; power = 1 − β. For a fixed α, a larger sample lowers β.
Other tests on the same logic: the χ² test for goodness of fit and independence, Σ(O − E)²/E; the F test for the ratio of two variances.

🎯 Why √n and not n

Averaging n independent values divides the variance by n, so the standard deviation of the mean falls only as √n: to halve the standard error of a grade estimate, four times as many samples are needed.

5. Pearson correlation and linear regression

For paired data (xᵢ, yᵢ), write Sxy = Σ(x − x̄)(y − ȳ), Sxx = Σ(x − x̄)² and Syy = Σ(y − ȳ)². Pearson’s correlation coefficient is r = Sxy/√(Sxx·Syy), with −1 ≤ r ≤ 1; it measures linear association only, so r = 0 does not mean the variables are unrelated. The least-squares regression line of y on x is y − ȳ = b_yx(x − x̄) with b_yx = Sxy/Sxx = r·σy/σx; the line of x on y has b_xy = Sxy/Syy = r·σx/σy.

Both regression lines pass through the point of means (x̄, ȳ).
r² = b_yx · b_xy, and r takes the common sign of the two coefficients, which always agree in sign.
r² is the coefficient of determination: the fraction of the variance of y that the line explains.
r is unchanged by a change of origin or scale of either variable (a positive scale factor); the regression coefficients are not unchanged by scale.

⚠️ Two lines, not one

The line of x on y is not the line of y on x solved for x; the two coincide only when |r| = 1. To predict y from a given x, use b_yx.

6. Interpolation, and iterative solution of linear equations

Interpolation fits the polynomial through tabulated points and reads values between them. For equally spaced x with step h, forward differences are Δy₀ = y₁ − y₀, Δ²y₀ = Δy₁ − Δy₀, and so on, and Newton’s forward formula with u = (x − x₀)/h is y = y₀ + uΔy₀ + u(u − 1)Δ²y₀/2! + u(u − 1)(u − 2)Δ³y₀/3! + …; the backward formula uses differences at the end of the table. For unequal spacing, Lagrange’s formula writes y = Σ yᵢ Lᵢ(x), where Lᵢ(x) = Π_j≠i(x − xⱼ)/(xᵢ − xⱼ). Through n + 1 points the interpolating polynomial of degree ≤ n is unique, so every method gives the same answer.

The direct solution of Ax = b by Gauss elimination and LU decomposition is in the Linear Algebra chapter. Iterative methods start from a guess and improve it. Jacobi computes every new component from the previous iterate: xᵢ⁽ᵏ⁺¹⁾ = (bᵢ − Σ_j≠i aᵢⱼxⱼ⁽ᵏ⁾)/aᵢᵢ. Gauss–Seidel uses each new component as soon as it is found, so it usually converges about twice as fast. Both converge for any starting guess when A is strictly diagonally dominant (|aᵢᵢ| > Σ_j≠i|aᵢⱼ| in every row); reordering the equations to put the largest coefficients on the diagonal is often the first step.

ℹ️ Where the rest of Numerical Methods is

The trapezoidal and Simpson’s rules and the single-step (Euler, Runge–Kutta) and multi-step (Adams–Bashforth) methods for differential equations are in the shared Numerical Methods chapter, and Gauss elimination and LU in the Linear Algebra chapter.

Key takeaways

A series can converge only if its terms tend to zero, but terms tending to zero prove nothing; a ratio-test limit of 1 means change test.
Taylor’s theorem with the Lagrange remainder f⁽ⁿ⁺¹⁾(ξ)(x − a)ⁿ⁺¹/(n + 1)! both builds the approximation and bounds its error.
Test M_y = N_x before anything else for an exact equation; for Bernoulli substitute v = y¹⁻ⁿ and multiply P and Q by (1 − n).
The standard error of a mean is σ/√n; a test statistic is the distance from the hypothesised value measured in standard errors.
Both regression lines pass through (x̄, ȳ) and r² = b_yx·b_xy; interpolation through n + 1 points gives one polynomial whichever formula builds it.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The sum of the infinite series 1 + 2/3 + (2/3)² + (2/3)³ + … is ____.
Numerical answer — type the value.
Show answer
Answer: 3
It is geometric with first term a = 1 and ratio r = 2/3, and |r| < 1, so the sum is a/(1 − r) = 1/(1/3) = 3. Summing only a few terms (1 + 0.667 + 0.444 ≈ 2.1) or using 1/(1 + r) = 0.6 are the usual slips.
Which of the following series converges?
1. Σ 1/n
2. Σ 1/√n
3. Σ 1/n²
4. Σ n/(n + 1)
Show answer
Answer: C — Σ 1/n²
Σ1/nᵖ converges only for p > 1, so Σ1/n² (p = 2) converges while Σ1/n (p = 1) and Σ1/√n (p = ½) diverge. Σn/(n + 1) fails the nth-term test outright, since its terms tend to 1, not 0.
Which of the following series converge? (Select all that apply.)
1. Σ (−1)ⁿ⁺¹/n
2. Σ n!/nⁿ
3. Σ 2ⁿ/n²
4. Σ 1/(n ln n), n ≥ 2
Show answer
Answer: A — Σ (−1)ⁿ⁺¹/n; B — Σ n!/nⁿ
The alternating harmonic series converges (conditionally) by the Leibniz test. For n!/nⁿ the ratio is (n/(n + 1))ⁿ → 1/e < 1, so it converges. 2ⁿ/n² grows without bound and fails the nth-term test, and Σ1/(n ln n) diverges by the integral test, since ∫dx/(x ln x) = ln(ln x) → ∞.
The radius of convergence of the power series Σ xⁿ/(n·2ⁿ), n = 1 to ∞, is ____.
Numerical answer — type the value.
Show answer
Answer: 2
With cₙ = 1/(n·2ⁿ), R = lim |cₙ/cₙ₊₁| = lim 2(n + 1)/n = 2. The factor n affects only the end points (the series diverges at x = 2 and converges at x = −2), not the radius; answering 1/2 inverts the ratio.
Using the Maclaurin polynomial of eˣ up to and including the x³ term, the approximate value of e^0.5, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.646
1 + 0.5 + 0.5²/2 + 0.5³/6 = 1 + 0.5 + 0.125 + 0.020833 = 1.645833 ≈ 1.646. The true value is 1.648721, and the next term, 0.5⁴/24 ≈ 0.0026, accounts for almost all of the difference. Dividing by 3 instead of 3! = 6 gives 1.667.
The coefficient of x³ in the Maclaurin series of ln(1 + x) is:
1. 1/3
2. −1/3
3. 1/6
4. 1/2
Show answer
Answer: A — 1/3
ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + …, so the x³ coefficient is +1/3: the nth coefficient is (−1)ⁿ⁺¹/n, with no factorial. The value 1/6 = 1/3! belongs to the sine and exponential series, and −1/3 has the sign of an even power.
The general solution of the exact equation (2xy + 3) dx + (x² + 4y) dy = 0 is:
1. x²y + 3x + 2y² = c
2. x²y + 3x + 4y² = c
3. 2x²y + 3x + 2y² = c
4. x²y + 2y² = c
Show answer
Answer: A — x²y + 3x + 2y² = c
M_y = 2x = N_x, so it is exact. ∫M dx = x²y + 3x; its y-derivative is x², and N = x² + 4y needs an extra 4y, whose integral is 2y². So F = x²y + 3x + 2y² = c. Writing 4y² forgets to integrate 4y, and dropping 3x loses the part of M with no y in it.
The solution of the Bernoulli equation dy/dx + y = y² with y(0) = 0.5 is evaluated at x = ln 3. The value of y, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
n = 2, so v = 1/y gives dv/dx − v = −1, whose solution is v = 1 + Ceˣ. y(0) = 0.5 means v(0) = 2, so C = 1 and y = 1/(1 + eˣ). At x = ln 3, y = 1/(1 + 3) = 0.25. Check: y′ + y = 1/(1 + eˣ)² = y². Omitting the (1 − n) = −1 factor gives dv/dx + v = 1 and the wrong curve.
Assay values of a coal seam have a known standard deviation of 12 units. A sample of 36 cores gives a mean of 104 against a claimed population mean of 100. The z statistic for testing the claim is ____.
Numerical answer — type the value.
Show answer
Answer: 2
The standard error is σ/√n = 12/6 = 2, so z = (104 − 100)/2 = 2. Dividing by σ itself gives 0.33, and dividing by σ/n gives 12; since |z| = 2 > 1.96, the claim would be rejected at the 5% level (two-tailed).
For a population with σ = 10, a sample of n = 100 is drawn. The half-width of the 95% confidence interval for the mean (z = 1.96), to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.96
Half-width = z·σ/√n = 1.96 × 10/10 = 1.96. The full interval is x̄ ± 1.96, which is 3.92 wide; quoting 3.92 answers a different question, and using σ/n = 0.1 gives 0.196.
A Type I error in a hypothesis test is:
1. rejecting the null hypothesis when it is true
2. accepting the null hypothesis when it is false
3. rejecting the alternative hypothesis when it is false
4. using a one-tailed test where a two-tailed test was needed
Show answer
Answer: A — rejecting the null hypothesis when it is true
A Type I error rejects a true H₀, and its probability is the significance level α. Accepting a false H₀ is the Type II error, whose probability is β; rejecting a false alternative is simply a correct decision, and choosing the wrong tail is a design mistake, not an error type.
Five boreholes give depth x (in units of 10 m) = 1, 2, 3, 4, 5 and grade y (%) = 2, 4, 5, 4, 5. Pearson’s correlation coefficient r, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.77
x̄ = 3 and ȳ = 4. Deviations: x − x̄ = −2, −1, 0, 1, 2 and y − ȳ = −2, 0, 1, 0, 1. Sxy = 4 + 0 + 0 + 0 + 2 = 6, Sxx = 10, Syy = 6. r = 6/√60 = 0.7746 ≈ 0.77. The slope Sxy/Sxx = 0.6 is the regression coefficient, not r.
For the borehole data x = 1, 2, 3, 4, 5 and y = 2, 4, 5, 4, 5, the slope of the least-squares regression line of y on x, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.6
b_yx = Sxy/Sxx = 6/10 = 0.6, so the line is y = 4 + 0.6(x − 3) = 2.2 + 0.6x, and it passes through the means (3, 4). Using Syy in the denominator gives b_xy = 1.0, the slope of the other line.
The two regression coefficients of a data set are b_yx = 0.8 and b_xy = 0.45. The correlation coefficient r, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.6
r² = b_yx·b_xy = 0.8 × 0.45 = 0.36, so r = +0.6, positive because both coefficients are positive. Averaging the two coefficients gives 0.625, and forgetting the square root gives 0.36.
A function takes the values f(0) = 1, f(1) = 3 and f(2) = 7. Using Newton’s forward interpolation formula, the value of f(1.5), to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 4.75
Δy₀ = 2, Δy₁ = 4, Δ²y₀ = 2, and u = 1.5. y = 1 + 1.5 × 2 + 1.5 × 0.5 × 2/2 = 1 + 3 + 0.75 = 4.75. The data fit x² + x + 1 exactly, which gives the same 4.75. Stopping at the linear term gives 4.0, and joining f(1) to f(2) by a straight line gives 5.0.
The system 4x + y = 9, x + 3y = 7 is solved by the Gauss–Seidel method starting from x = 0, y = 0. The value of y after the first iteration, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.58
x₁ = (9 − 0)/4 = 2.25, and Gauss–Seidel uses it at once: y₁ = (7 − 2.25)/3 = 1.5833 ≈ 1.58. Jacobi would still use x = 0 and give y₁ = 7/3 = 2.33. The exact solution is x = 20/11 ≈ 1.818, y = 19/11 ≈ 1.727, and the matrix is diagonally dominant, so the iteration converges to it.
Which of the following statements about iterative methods for Ax = b are correct? (Select all that apply.)
1. Strict diagonal dominance of A guarantees convergence of the Jacobi and Gauss–Seidel methods
2. Gauss–Seidel uses updated values within the same iteration
3. Both methods need every diagonal element aᵢᵢ to be non-zero
4. Gauss–Seidel converges for every non-singular matrix
Show answer
Answer: A — Strict diagonal dominance of A guarantees convergence of the Jacobi and Gauss–Seidel methods; B — Gauss–Seidel uses updated values within the same iteration; C — Both methods need every diagonal element aᵢᵢ to be non-zero
Diagonal dominance is a sufficient condition for both methods; Gauss–Seidel differs from Jacobi precisely by using each new value immediately; and both divide by aᵢᵢ, so a zero diagonal must first be removed by reordering. Non-singularity alone does not make either method converge — many invertible matrices make the iterates diverge.