Series and Convergence, Linear Transformations, Nonlinear First-Order ODEs, and Correlation–Regression

This chapter is written for the GATE Agricultural Engineering (AG) paper alone. Its Section 1, Engineering Mathematics, names five things that the shared mathematics chapters do not teach, and they are gathered here: sequences and infinite series with the tests for convergence; the Taylor and Maclaurin series of a single-variable function; linear and orthogonal transformations; the nonlinear first-order equations beyond the separable, homogeneous and linear forms — exact equations and the Bernoulli equation; and correlation and regression analysis. Everything else in AG’s mathematics — matrices, eigenvalues and Cayley–Hamilton, partial derivatives and Euler’s theorem, vector calculus, higher-order ODEs, Laplace transforms, the three PDEs, distributions and the numerical methods — is in the shared chapters listed on your paper’s syllabus page. Each section ends in the form a GATE question takes: a test to apply, a coefficient to read, a line to identify.

1. Sequences, infinite series and the tests for convergence

A sequence aₙ converges if aₙ approaches a finite limit as n → ∞. An infinite series Σaₙ converges if its sequence of partial sums Sₙ = a₁ + a₂ + … + aₙ converges. The first thing to check is always the nth-term test: if aₙ does not tend to zero, the series diverges. The converse is false — the harmonic series Σ1/n has terms tending to zero and still diverges, because its partial sums grow like ln n.

Two benchmark series decide most comparisons. The geometric series Σrⁿ converges (to a/(1 − r) from first term a) exactly when |r| < 1. The p-series Σ1/nᵖ converges exactly when p > 1: Σ1/n² converges, Σ1/n and Σ1/√n diverge. For series of positive terms, the comparison test says a series smaller term-by-term than a convergent one converges, and one larger than a divergent one diverges; the limit comparison test says that if aₙ/bₙ tends to a finite non-zero limit, the two series converge or diverge together.

The tests, and when each is the right one
Test	Statement	Use it when
Ratio (D’Alembert)	L = lim \|aₙ₊₁/aₙ\|: converges if L < 1, diverges if L > 1, no decision if L = 1	factorials and powers such as n!, aⁿ, nⁿ appear
Root (Cauchy)	L = lim \|aₙ\|^(1/n): same verdicts as the ratio test	the whole term is raised to the power n
Integral	Σf(n) and ∫f(x)dx from 1 to ∞ converge together, f positive and decreasing	the term is a function you can integrate, such as 1/(n ln n)
Leibniz (alternating)	Σ(−1)ⁿ⁺¹bₙ converges if bₙ decreases to zero	the signs alternate

A series is absolutely convergent if Σ|aₙ| converges, and absolute convergence implies convergence. A series that converges while Σ|aₙ| diverges is conditionally convergent: the alternating harmonic series 1 − 1/2 + 1/3 − … converges (to ln 2) although Σ1/n diverges.

⚠️ L = 1 is not "converges"

The ratio and root tests give L = 1 for every p-series, convergent (p = 2) or divergent (p = 1). An option that reads "the ratio test shows Σ1/n² converges" is wrong for the reason, even though the series does converge — the p-series test or the integral test decides it.

2. Taylor and Maclaurin series

If f has derivatives of all orders at x = a, its Taylor series about a is f(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … = Σ f⁽ⁿ⁾(a)(x − a)ⁿ/n!. The Maclaurin series is the special case a = 0. The coefficient of (x − a)ⁿ is therefore f⁽ⁿ⁾(a)/n!, and reading a coefficient off a known series is usually faster than differentiating n times.

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/(1 − x)	1 + x + x² + x³ + …	\|x\| < 1
(1 + x)ᵐ	1 + mx + m(m − 1)x²/2! + …	\|x\| < 1 (m not a whole number)

The interval where a power series Σcₙxⁿ converges is found with the ratio test applied to the terms: the radius of convergence is R = lim |cₙ/cₙ₊₁| when that limit exists. For Σxⁿ/n the ratio of successive coefficients tends to 1, so R = 1; the end-points are checked separately (at x = −1 it is the alternating harmonic series and converges, at x = 1 the harmonic series and diverges). Substitution builds new series: e^(2x) is the eˣ series with x replaced by 2x, so its x³ coefficient is 2³/3! = 4/3.

ℹ️ Where the Fourier series sits

AG’s Calculus heading names "Fourier, Taylor and MacLaurin series" together. The Fourier series — coefficients by integration, even and odd functions, the value at a jump — is taught in the first section of the shared Transforms chapter; read that section and stop, since AG names no Fourier transform.

3. Linear and orthogonal transformations

A linear transformation y = Ax maps a vector x to a vector y and satisfies A(αu + βv) = αAu + βAv. Applying B after A is the single transformation BA, so composition is matrix multiplication and the order matters. The transformation can be undone exactly when A is non-singular (det A ≠ 0), and then x = A⁻¹y; a singular A collapses some non-zero vector to zero and cannot be inverted. The determinant is also the factor by which the transformation scales areas (in two dimensions) or volumes (in three).

A transformation is orthogonal when AᵀA = AAᵀ = I. Then A⁻¹ = Aᵀ, the columns (and rows) are mutually perpendicular unit vectors, lengths and angles are preserved (|Ax| = |x|), det A = ±1 (+1 for a pure rotation, −1 when a reflection is included), and every eigenvalue has modulus 1 — so a real eigenvalue can only be +1 or −1, and a rotation by θ in the plane has eigenvalues e^(±iθ). The plane rotation R = [[cos θ, −sin θ], [sin θ, cos θ]] is the standard example: at θ = 90° it sends (1, 0) to (0, 1).

🧠 Check orthogonality without multiplying out

Look at the columns: each must have length 1 and each pair must have a zero dot product. [[0.6, −0.8], [0.8, 0.6]] passes both (0.36 + 0.64 = 1; 0.6 × −0.8 + 0.8 × 0.6 = 0), so its inverse is simply its transpose.

4. Nonlinear first-order ODEs: exact equations and the Bernoulli equation

The shared first-order chapter covers the variable-separable, homogeneous (y = vx) and linear (integrating factor e^(∫P dx)) forms. AG’s wording is "linear and non-linear first order ODE", and two further forms complete it. An equation M(x, y)dx + N(x, y)dy = 0 is exact when ∂M/∂y = ∂N/∂x; there is then a function u(x, y) with ∂u/∂x = M and ∂u/∂y = N, and the solution is u(x, y) = c. To build u, integrate M with respect to x (holding y constant) and add those terms of N that do not contain x, integrated with respect to y.

Worked: (2xy + 3)dx + (x² + 4y)dy = 0. Here ∂M/∂y = 2x = ∂N/∂x, so it is exact. ∫M dx = x²y + 3x; the term of N free of x is 4y, whose integral is 2y². The solution is x²y + 3x + 2y² = c. An inexact equation can sometimes be made exact by an integrating factor: if (∂M/∂y − ∂N/∂x)/N is a function of x alone, e^(∫ that dx) is one.

The Bernoulli equation dy/dx + P(x)y = Q(x)yⁿ (n ≠ 0, 1) is nonlinear because of yⁿ, but dividing by yⁿ and substituting v = y^(1−n) makes it linear: dv/dx + (1 − n)P v = (1 − n)Q, solved with the integrating factor. For dy/dx + y = xy³, n = 3 and the substitution is v = y⁻²; the equation becomes dv/dx − 2v = −2x.

⚠️ The exponent in the Bernoulli substitution is 1 − n, not n − 1 or n

For n = 3 the substitution is v = y⁻², and v = y² or v = y³ both leave the equation nonlinear. The sign is the whole point: v = y^(1−n) is what cancels the yⁿ on the right.

5. Correlation and regression analysis

For paired data (xᵢ, yᵢ) the covariance is cov(x, y) = Σ(xᵢ − x̄)(yᵢ − ȳ)/n and Karl Pearson’s correlation coefficient is r = cov(x, y)/(σₓσᵧ). It is dimensionless, lies between −1 and +1, is unchanged by shifting or rescaling either variable by a positive factor, and measures only linear association: r = 0 means no linear relationship, not no relationship (y = x² on a symmetric range of x has r = 0).

The least-squares regression line of y on x minimises the sum of squared vertical deviations and is y − ȳ = bᵧₓ(x − x̄), with bᵧₓ = r σᵧ/σₓ = cov(x, y)/σₓ². The line of x on y is x − x̄ = bₓᵧ(y − ȳ), with bₓᵧ = r σₓ/σᵧ. Three facts follow and are examined constantly: both lines pass through (x̄, ȳ), so the means are found by solving the two line equations simultaneously; r² = bᵧₓ · bₓᵧ, and r takes the common sign of the two coefficients; and since r² ≤ 1, the product of the coefficients can never exceed 1 — which is how you tell which given line is which.

For ranked data, Spearman’s rank correlation is ρ = 1 − 6Σdᵢ²/(n(n² − 1)), where dᵢ is the difference between the two ranks of item i. With n = 5 and Σd² = 8, ρ = 1 − 48/120 = 0.6.

🧠 Identifying the two lines

Given x + 2y = 5 and 2x + 3y = 8: taking the first as y on x gives bᵧₓ = −1/2, and the second as x on y gives bₓᵧ = −3/2; the product 3/4 ≤ 1 is allowed. The other assignment gives bₓᵧ = −2 and bᵧₓ = −2/3, product 4/3 > 1, impossible. So r = −√0.75 = −0.866, and solving the pair gives the means x̄ = 1, ȳ = 2.

Key takeaways

If aₙ does not tend to zero the series diverges; if it does, you still need a test. Σ1/nᵖ converges only for p > 1, and Σrⁿ only for |r| < 1.
The ratio and root tests are silent when the limit is 1 — every p-series lands there.
The coefficient of (x − a)ⁿ in a Taylor series is f⁽ⁿ⁾(a)/n!; read it off a known series by substitution rather than differentiating.
An orthogonal matrix has Aᵀ as its inverse, determinant ±1, preserves length, and has eigenvalues of modulus 1.
Exact means ∂M/∂y = ∂N/∂x; Bernoulli becomes linear with v = y^(1−n).
Both regression lines pass through the means, r² = bᵧₓbₓᵧ ≤ 1, and r carries the sign of the coefficients.

Practice questions (12)

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.

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/n² is a p-series with p = 2 > 1, so it converges (its sum is π²/6). Σ1/n and Σ1/√n are p-series with p = 1 and p = 1/2, both ≤ 1, so they diverge. Σn/(n + 1) fails the nth-term test outright, since its terms tend to 1, not 0.
For the series Σ n!/nⁿ, the limit L = lim |aₙ₊₁/aₙ| used in the ratio test equals:
1. e, so the series diverges
2. 1/e, so the series converges
3. 1, so the test is inconclusive
4. 0, so the series converges
Show answer
Answer: B — 1/e, so the series converges
aₙ₊₁/aₙ = [(n + 1)!/(n + 1)ⁿ⁺¹]·[nⁿ/n!] = (n + 1)nⁿ/(n + 1)ⁿ⁺¹ = (n/(n + 1))ⁿ = 1/(1 + 1/n)ⁿ, which tends to 1/e ≈ 0.368 < 1, so the series converges. Cancelling the factorials but forgetting the (n + 1) in the power gives 1, the tempting "inconclusive"; inverting the ratio gives e.
What is the coefficient of x³ in the Maclaurin series of e^(2x)? Give the answer to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 1.33
Replace x by 2x in eˣ = Σxⁿ/n!: the x³ term is (2x)³/3! = 8x³/6, so the coefficient is 4/3 = 1.33. Forgetting to cube the 2 gives 2/6 = 0.33, and dropping the factorial gives 8.
A is a real orthogonal matrix. Which of the following statements are necessarily true?
1. AᵀA = I
2. det A = +1 or −1
3. A⁻¹ = Aᵀ
4. All eigenvalues of A are real
Show answer
Answer: A — AᵀA = I; B — det A = +1 or −1; C — A⁻¹ = Aᵀ
Orthogonality is AᵀA = I, which gives A⁻¹ = Aᵀ and (det A)² = 1, so det A = ±1. The eigenvalues have modulus 1 but need not be real: the 90° plane rotation [[0, −1], [1, 0]] is orthogonal with eigenvalues ±i. Ticking the fourth option confuses orthogonal with symmetric matrices, whose eigenvalues are always real.
The transformation with matrix [[cos θ, −sin θ], [sin θ, cos θ]] at θ = 90° maps the vector (1, 0) to:
1. (0, 1)
2. (0, −1)
3. (−1, 0)
4. (1, 1)
Show answer
Answer: A — (0, 1)
At θ = 90° the matrix is [[0, −1], [1, 0]], and multiplying by the column (1, 0) picks out the first column, (0, 1): an anticlockwise quarter-turn. (0, −1) is the clockwise turn, which the transpose would give.
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 + 3xy + 2y² = c
Show answer
Answer: A — x²y + 3x + 2y² = c
∂M/∂y = 2x = ∂N/∂x, so the equation is exact. ∫(2xy + 3)dx = x²y + 3x; the only term of N without x is 4y, and ∫4y dy = 2y². So x²y + 3x + 2y² = c. Writing 4y² forgets to integrate, and adding x² from N again double-counts the x²y term already obtained.
The substitution that reduces dy/dx + y = xy³ to a linear equation is:
1. v = y³
2. v = y²
3. v = y⁻²
4. v = y/x
Show answer
Answer: C — v = y⁻²
This is a Bernoulli equation with n = 3, and v = y^(1−n) = y⁻² gives dv/dx = −2y⁻³ dy/dx; dividing the equation by y³ and substituting yields dv/dx − 2v = −2x, which is linear. v = y/x is the substitution for a homogeneous equation, and positive powers of y leave the y³ term in place.
The two regression coefficients of a bivariate data set are bᵧₓ = 0.8 and bₓᵧ = 0.45. What is the correlation coefficient r?
Numerical answer — type the value.
Show answer
Answer: 0.6
r² = bᵧₓ · bₓᵧ = 0.8 × 0.45 = 0.36, so |r| = 0.6, and r is positive because both coefficients are positive. Averaging the two coefficients (0.625) is the common slip; the correlation is their geometric mean, not their arithmetic mean.
The two regression lines of a data set are x + 2y = 5 and 2x + 3y = 8. What is the correlation coefficient r? Give the answer to three decimal places.
Numerical answer — type the value.
Show answer
Answer: -0.866
Try x + 2y = 5 as y on x: y = -x/2 + 5/2, so bᵧₓ = -0.5; then 2x + 3y = 8 as x on y: x = -1.5y + 4, so bₓᵧ = -1.5. The product 0.75 is at most 1, so this assignment is valid (the reverse gives 4/3, impossible). r = -√0.75 = -0.866, negative because both coefficients are. Reporting +0.866 ignores the sign rule.
Two judges rank 5 entries. The sum of the squares of the rank differences is 8. What is Spearman’s rank correlation coefficient?
Numerical answer — type the value.
Show answer
Answer: 0.6
ρ = 1 − 6Σd²/(n(n² − 1)) = 1 − 6 × 8/(5 × 24) = 1 − 48/120 = 0.6. Using n² instead of n² − 1 in the denominator gives 1 − 48/125 = 0.616, and forgetting the 6 gives 0.933.
Which of the following series converge?
1. Σ 1/n²
2. Σ (1/2)ⁿ
3. Σ 1/n
4. Σ (−1)ⁿ/√n
Show answer
Answer: A — Σ 1/n²; B — Σ (1/2)ⁿ; D — Σ (−1)ⁿ/√n
Σ1/n² is a p-series with p = 2; Σ(1/2)ⁿ is geometric with r = 1/2; Σ(−1)ⁿ/√n alternates with 1/√n decreasing to zero, so it converges by Leibniz’s test — conditionally, since Σ1/√n diverges. Σ1/n is the harmonic series and diverges. Leaving out the fourth option treats conditional convergence as divergence.
The Maclaurin series ln(1 + x) = x − x²/2 + x³/3 − … represents the function for:
1. all real x
2. −1 < x ≤ 1
3. −1 ≤ x < 1
4. |x| < 2
Show answer
Answer: B — −1 < x ≤ 1
The coefficients have ratio n/(n + 1) → 1, so the radius of convergence is 1. At x = 1 the series is 1 − 1/2 + 1/3 − …, which converges (to ln 2); at x = −1 it is −(1 + 1/2 + 1/3 + …), which diverges — as it must, since ln 0 is undefined. So the interval is −1 < x ≤ 1; the interval with the end-points swapped, −1 ≤ x < 1, belongs to the series of ln(1 − x).