Series and Taylor Expansion, Area and Volume, Nonlinear First-Order ODEs, and Numerical Approximation
1. Infinite series and the tests for convergence
An infinite series Σaₙ converges when its partial sums Sₙ = a₁ + a₂ + … + aₙ approach a finite limit. Always apply the nth-term test first: if aₙ does not tend to zero, the series diverges and nothing more needs checking. The converse fails — the harmonic series Σ1/n has terms that tend to zero and still diverges, because its partial sums grow like ln n. Two benchmark families settle most comparisons: the geometric series Σarⁿ converges, to a/(1 − r), exactly when |r| < 1, and the p-series Σ1/nᵖ converges exactly when p > 1.
| Test | Statement | Reach for it when |
|---|---|---|
| Comparison and limit comparison | For positive terms: smaller than a convergent series converges; larger than a divergent one diverges. If aₙ/bₙ → a finite non-zero limit, both behave alike | the term looks like a p-series for large n, e.g. (n + 1)/(n³ + 2) |
| Ratio (D’Alembert) | L = lim |aₙ₊₁/aₙ|: converges if L < 1, diverges if L > 1, undecided if L = 1 | factorials or powers such as n!, 2ⁿ, 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 ∫₁^∞ f(x)dx converge or diverge together, f positive and decreasing | the term integrates easily, e.g. 1/(n ln n) |
| Leibniz (alternating) | Σ(−1)ⁿ⁺¹bₙ converges if bₙ decreases steadily to zero | the signs alternate |
2. Taylor and Maclaurin series of a function of one variable
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! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n! + …. About a = 0 it is called the Maclaurin series. The coefficient of (x − a)ⁿ is always f⁽ⁿ⁾(a)/n!, which is how a GATE question asking for one particular coefficient is answered without writing the whole series. Truncating after the nth term leaves the remainder Rₙ = f⁽ⁿ⁺¹⁾(ξ)(x − a)ⁿ⁺¹/(n + 1)! for some ξ between a and x — the truncation error that numerical methods then inherit.
| 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 when n is not a positive integer |
| 1/(1 − x) | 1 + x + x² + x³ + … | |x| < 1 |
3. The definite integral applied to area and volume
The area between two curves y = f(x) above and y = g(x) below, from x = a to x = b, is A = ∫ₐᵇ [f(x) − g(x)] dx; find the crossing points first, because they are usually the limits and because the upper and lower curves swap there. For y = x and y = x² the curves meet at 0 and 1, x lies above x² between them, and A = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6. A definite integral counts area below the x-axis as negative, so an area that crosses the axis must be split at the crossing and the parts added as absolute values.
A solid of revolution is swept when a region turns about an axis. The disc method about the x-axis gives V = π∫ₐᵇ [f(x)]² dx; if the region lies between two curves it becomes the washer method, V = π∫ₐᵇ ([f(x)]² − [g(x)]²) dx. The shell method about the y-axis gives V = 2π∫ₐᵇ x f(x) dx. More generally, a solid whose cross-section at x has area A(x) has volume ∫ A(x) dx — which is how the volume of a reservoir is computed from the areas enclosed by its contours, and how a landfill of sloping sides is sized.
4. Nonlinear first-order equations: exact and Bernoulli
The shared first-order chapter teaches the separable, homogeneous and linear forms. Two nonlinear forms remain. An equation 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, adding an unknown h(y), and fixing h by matching ∂F/∂y to N. For (2xy + 3)dx + (x² + 4y)dy = 0, ∂M/∂y = 2x = ∂N/∂x; integrating M gives x²y + 3x + h(y), and matching gives h′(y) = 4y, so the solution is x²y + 3x + 2y² = C. When the test fails, an integrating factor can sometimes restore exactness: if (M_y − N_x)/N depends on x alone, e^(∫ that dx) is one.
The Bernoulli equation dy/dx + P(x)y = Q(x)yⁿ, n ≠ 0, 1, becomes linear under the substitution v = y^(1 − n): dividing by yⁿ and using dv/dx = (1 − n)y^(−n) dy/dx gives dv/dx + (1 − n)P v = (1 − n)Q, which the linear-equation method then solves. Environmental science meets two Bernoulli equations constantly. Second-order decay, dC/dt = −kC², separates directly to 1/C = 1/C₀ + kt. Logistic growth, dN/dt = rN(1 − N/K), is Bernoulli with n = 2 and solves to N = K/(1 + [(K − N₀)/N₀]e^(−rt)) — the curve behind logistic population forecasting and microbial growth that saturates.
5. Approximation, errors, accuracy and precision
If x is the true value and x* an approximation, the absolute error is |x − x*|, the relative error |x − x*|/|x|, and the percentage error 100 times that. Round-off error comes from carrying a finite number of digits; truncation error from stopping an infinite process — a Taylor series cut after n terms, a derivative replaced by a difference, an integral by a sum. Truncation error shrinks as the step h shrinks, while round-off error grows as h shrinks (because nearly equal numbers are subtracted), so every numerical method has an optimum step rather than "the smaller the better". Errors propagate: in a sum or difference the absolute errors add; in a product or quotient the relative errors add, so a relative error of 1% in each of two measured factors gives up to 2% in their product.
| Term | What it asks | Spoilt by |
|---|---|---|
| Accuracy | How close is the result to the true value? | systematic error (bias) — an uncalibrated DO meter reading 0.5 mg/L high every time |
| Precision | How close are repeated results to one another? | random error — scatter between replicate titrations |
| Significant figures | How many digits carry information? | reporting more digits than the method resolves |
6. Interpolation, least-squares curve fitting and numerical differentiation
Interpolation passes a polynomial exactly through tabulated points. For equally spaced x with step h, Newton’s forward-difference formula is y(x₀ + ph) = y₀ + pΔy₀ + p(p − 1)Δ²y₀/2! + p(p − 1)(p − 2)Δ³y₀/3! + …, where Δy₀ = y₁ − y₀ and Δ²y₀ = Δy₁ − Δy₀; the backward form, built on ∇ at the end of the table, is used near the last entry. From x = 0, 1, 2 with y = 1, 3, 7, Δy₀ = 2 and Δ²y₀ = 2, so y(0.5) = 1 + 0.5 × 2 + 0.5(−0.5)/2 × 2 = 1.75. For unequal spacing use Lagrange’s form, y(x) = Σ yᵢ Lᵢ(x) with Lᵢ(x) = Π_(j≠i) (x − xⱼ)/(xᵢ − xⱼ); n + 1 points always determine a unique polynomial of degree at most n, so the two methods give the same polynomial.
Curve fitting does not pass through the data; it finds the curve that minimises the sum of squared vertical deviations. For a straight line y = a + bx through n points the normal equations Σy = na + bΣx and Σxy = aΣx + bΣx² give b = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²) and a = ȳ − bx̄, so the fitted line always passes through (x̄, ȳ). Curves that are linear after a transformation are fitted the same way: y = ae^(bx) becomes ln y = ln a + bx, and y = axᵇ becomes log y = log a + b log x — which is how a first-order rate constant is read from the slope of ln C against t.
| Formula | Expression | Truncation error |
|---|---|---|
| Forward difference | f′(x) ≈ [f(x + h) − f(x)]/h | O(h) |
| Backward difference | f′(x) ≈ [f(x) − f(x − h)]/h | O(h) |
| Central difference | f′(x) ≈ [f(x + h) − f(x − h)]/(2h) | O(h²) |
| Second derivative, central | f″(x) ≈ [f(x + h) − 2f(x) + f(x − h)]/h² | O(h²) |
Key takeaways
- Check the nth term first; if it does not tend to zero the series diverges. For powers of n compare with a p-series (convergent only for p > 1); use the ratio test for factorials and exponentials.
- The coefficient of (x − a)ⁿ in a Taylor series is f⁽ⁿ⁾(a)/n!, and stopping the series leaves a truncation error of the order of the first term dropped.
- Area is ∫(upper − lower)dx split at crossings; volume of revolution is π∫r²dx by discs or 2π∫x f(x)dx by shells — square the radius.
- M dx + N dy = 0 is exact when M_y = N_x; a Bernoulli equation becomes linear with v = y^(1 − n). Second-order decay gives 1/C = 1/C₀ + kt and a half-life of 1/(kC₀).
- Interpolation passes through the points (Newton for equal steps, Lagrange for unequal); least squares minimises squared deviations and its line passes through (x̄, ȳ); the central difference is O(h²), the one-sided differences O(h).
Practice questions (16)
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?
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Answer: C — Σ 1/n^1.1
A p-series Σ1/nᵖ converges exactly when p > 1, and 1.1 > 1, so Σ1/n^1.1 converges, however slowly. Σ1/√n (p = 0.5) and the harmonic series (p = 1) diverge, and n/(n + 1) tends to 1 rather than 0, so the last series fails the nth-term test outright.Which of the following series converge?
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Answer: A — Σ (−1)ⁿ⁺¹/n; C — Σ n/2ⁿ
(a) converges by Leibniz’s test (1/n decreases to zero), though only conditionally. (c) converges by the ratio test: aₙ₊₁/aₙ = (n + 1)/(2n) → 1/2 < 1. (b) fails the nth-term test, since (1 + 1/n)ⁿ → e, not 0. (d) diverges by the integral test: ∫dx/(x ln x) = ln(ln x), which grows without bound.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.
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Answer: 1.33
The coefficient is f‴(0)/3!. Each derivative of e^(2x) brings a factor 2, so f‴(0) = 8 and the coefficient is 8/6 = 1.33. Equivalently, substitute 2x into eˣ: (2x)³/3! = 8x³/6. Forgetting the factorial gives 8.Using the Maclaurin series of eˣ up to and including the x² term, what is the approximate value of e^0.1? Give the answer to three decimal places.
Numerical answer — type the value.
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Answer: 1.105
e^0.1 ≈ 1 + 0.1 + 0.1²/2 = 1 + 0.1 + 0.005 = 1.105. The true value is 1.10517, so the truncation error is about 0.00017, close to the first omitted term 0.1³/6 = 0.000167.What is the area enclosed between the curves y = x and y = x²? Give the answer to three decimal places.
Numerical answer — type the value.
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Answer: 0.167
The curves meet where x = x², at x = 0 and x = 1, and x ≥ x² between them. A = ∫₀¹ (x − x²)dx = 1/2 − 1/3 = 1/6 = 0.167. Integrating x² − x instead gives −1/6: the sign says the curves were taken in the wrong order.The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. What is the volume of the solid generated? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 25.13
By discs, V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π × 16/2 = 8π = 25.13. Forgetting to square the radius gives π∫√x dx = 16π/3 = 16.76, which is the classic error.The differential equation (2xy + 3)dx + (x² + 4y)dy = 0 is:
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Answer: B — exact, with solution x²y + 3x + 2y² = C
M = 2xy + 3 and N = x² + 4y, so ∂M/∂y = 2x = ∂N/∂x and the equation is exact. ∫M dx = x²y + 3x + h(y); ∂/∂y gives x² + h′(y) = x² + 4y, so h = 2y². Option C forgets that ∫4y dy = 2y².The equation dy/dx + y = xy³ is reduced to a linear equation by the substitution:
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Answer: B — v = y⁻²
It is a Bernoulli equation with n = 3, and the substitution is v = y^(1 − n) = y⁻². Then dv/dx = −2y⁻³ dy/dx, and dividing the equation by y³ turns it into dv/dx − 2v = −2x, which is linear. v = y/x is the substitution for a homogeneous equation, a different form.A pollutant decays by second-order kinetics, dC/dt = −kC², with k = 0.02 L/(mg·d). If its initial concentration is 10 mg/L, what is its concentration after 5 days, in mg/L?
Numerical answer — type the value.
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Answer: 5
Separating, 1/C = 1/C₀ + kt = 1/10 + 0.02 × 5 = 0.1 + 0.1 = 0.2, so C = 5 mg/L. Treating it as first order, C = 10e^(−0.1) = 9.05, uses the wrong law; and here 5 days happens to be exactly the half-life 1/(kC₀) = 1/(0.02 × 10).A standard solution of true concentration 2.50 mg/L is measured as 2.45 mg/L. What is the percentage error of the measurement?
Numerical answer — type the value.
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Answer: 2
Percentage error = |true − measured|/true × 100 = 0.05/2.50 × 100 = 2 per cent. Dividing by the measured value instead gives 2.04, which is a relative error taken against the wrong reference.Replicate titrations of one sample agree closely with one another but all lie well above the certified value. The results are:
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Answer: C — precise but inaccurate
Close agreement among replicates is precision; closeness to the true value is accuracy. A consistent offset is a systematic error, which lowers accuracy without spreading the results, so they are precise but inaccurate. Option D describes large scatter, which is the opposite of what is observed.A quantity is tabulated as y = 1, 3 and 7 at x = 0, 1 and 2. Using Newton’s forward-difference interpolation, what is y at x = 0.5? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 1.75
Δy₀ = 3 − 1 = 2, Δy₁ = 7 − 3 = 4, Δ²y₀ = 2. With p = 0.5: y = 1 + 0.5 × 2 + 0.5 × (0.5 − 1)/2 × 2 = 1 + 1 − 0.25 = 1.75. Check: the interpolating polynomial is x² + x + 1, and at 0.5 it gives 0.25 + 0.5 + 1 = 1.75. Linear interpolation between the first two points gives 2, ignoring the curvature.The function values f(1) = 2, f(2) = 5 and f(4) = 17 are known. Using Lagrange interpolation through these three points, what is f(3)?
Numerical answer — type the value.
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Answer: 10
L₀(3) = (3 − 2)(3 − 4)/[(1 − 2)(1 − 4)] = −1/3, L₁(3) = (3 − 1)(3 − 4)/[(2 − 1)(2 − 4)] = 1, L₂(3) = (3 − 1)(3 − 2)/[(4 − 1)(4 − 2)] = 1/3. f(3) = 2(−1/3) + 5(1) + 17(1/3) = 10. The data are x² + 1, and the quadratic through three of its points is x² + 1 itself, so 3² + 1 = 10 confirms it.A straight line y = a + bx is fitted by least squares to the points (1, 2), (2, 3), (3, 5) and (4, 6). What is the slope b?
Numerical answer — type the value.
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Answer: 1.4
n = 4, Σx = 10, Σy = 16, Σxy = 2 + 6 + 15 + 24 = 47, Σx² = 30. b = (4 × 47 − 10 × 16)/(4 × 30 − 10²) = (188 − 160)/(120 − 100) = 28/20 = 1.4, and a = (16 − 1.4 × 10)/4 = 0.5. Joining only the end points gives (6 − 2)/3 = 1.33, which is not the least-squares slope.Estimate the derivative of f(x) = x³ at x = 2 by the central-difference formula with h = 0.1. Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 12.01
f′(2) ≈ [f(2.1) − f(1.9)]/(2 × 0.1) = (9.261 − 6.859)/0.2 = 2.402/0.2 = 12.01. The exact value is 3 × 2² = 12, and the error 0.01 equals h²f‴/6 = 0.01 × 6/6, as the O(h²) error term predicts.When the step h in a forward-difference estimate of a derivative is halved, the truncation error approximately:
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Answer: A — halves, while round-off error tends to grow
The forward difference has truncation error hf″/2 + …, which is O(h), so halving h halves it. At the same time f(x + h) and f(x) get closer, their difference loses significant digits, and the round-off error, of order ε/h, grows. Option B describes the central difference, which is O(h²).