The Newton-Raphson method questions

Question 1

Write down the Newton-Raphson iteration formula for solving f(x) = 0.

Accepted Answer

x_n+1 = x_n − f(x_n)/f′(x_n). B1 for the formula, B1 for the tangent interpretation. Geometrically it slides down the tangent at the current point to where that tangent crosses the axis, then starts again from there. The minus sign is part of the formula, not an optional extra.

Question 2

The tangent to the curve y = f(x) at the point (x_n, f(x_n)) has equation y − f(x_n) = f′(x_n)(x − x_n). Show that this tangent meets the x-axis at x = x_n − f(x_n)/f′(x_n).

Accepted Answer

On the x-axis y = 0, so the equation of the tangent becomes −f(x_n) = f′(x_n)(x − x_n). M1 for setting y = 0 in the equation of the tangent. Dividing by f′(x_n) gives x − x_n = −f(x_n)/f′(x_n), so x = x_n − f(x_n)/f′(x_n), which is the printed result. A1 for the rearrangement. That crossing is what the method takes as the next estimate, so one Newton-Raphson step is one tangent drawn and followed to the axis, and the formula is a record of that construction rather than a rule to be memorised on its own. The division requires f′(x_n) to be non-zero, which is the case a horizontal tangent excludes.

Question 3

f(x) = x² − 5. Taking x₀ = 2, apply the Newton-Raphson method once, and give x₁ as an exact fraction.

Accepted Answer

f′(x) = 2x, so the iteration reads x_n+1 = x_n − (x_n² − 5)/(2x_n). With x₀ = 2: f(2) = −1 and f′(2) = 4, so x₁ = 2 − (−1)/4 = 9/4. M1 for the substitution into the formula, A1 for 9/4. Two minus signs meet in that step, so the estimate moves up towards √5 = 2.2360… rather than down.

Question 4

State what one step of the Newton-Raphson method does geometrically, and state what happens to that step when f′(x_n) is small but not zero.

Accepted Answer

The step takes the tangent to y = f(x) at the point (x_n, f(x_n)) and reads x_n+1 off where that tangent meets the x-axis. B1 for the tangent at the current point, B1 for the crossing of the x-axis. A small gradient makes a shallow tangent, which meets the axis a long way from x_n, so the next iterate can land far from the root asked for and later steps may settle on a different root instead. B1 for the long step and what it can lead to. Where f′(x_n) is exactly zero the tangent is horizontal and no next iterate exists at all.

Question 5

f(x) = x³ + x − 5. Using the Newton-Raphson method with x₀ = 1.5, find x₁ and x₂, giving x₂ to 5 decimal places.

Accepted Answer

f(1.5) = −0.125 and f′(1.5) = 3(1.5)² + 1 = 7.75, so x₁ = 1.5 + 0.125/7.75 = 1.516129…. Then f(x₁) = 0.00117…, f′(x₁) = 7.8959…, and x₂ = 1.51598 to 5 decimal places. M1 for f and f′ at 1.5, A1 for the first iterate, M1 for the second iteration, A1 for 1.51598. The root is 1.5159802…, so x₁ was right to 3 decimal places and x₂ is right to 7. Near a simple root, and once the iterates are sufficiently close under the usual smoothness conditions, Newton-Raphson normally has quadratic convergence. Keep unrounded intermediate values: rounding x₁ to 1.5161 before the second step corrupts the fifth decimal place of x₂.

Question 6

f(x) = ln x + x − 3. Show that f(x) = 0 has a root in the interval (2, 3), and use the Newton-Raphson method with x₀ = 2 to find x₁ and x₂, giving x₂ to 4 decimal places.

Accepted Answer

f(2) = ln 2 − 1 = −0.30685 < 0 and f(3) = ln 3 = 1.09861 > 0, and f is continuous for x > 0, so a root of f(x) = 0 lies in (2, 3). B1 for the two values with the sign change and the continuity. f′(x) = 1/x + 1, so f′(2) = 1.5 and x₁ = 2 + 0.30685/1.5 = 2.20457. Then f(x₁) = −0.004900 and f′(x₁) = 1.45360, so x₂ = 2.2079 to 4 decimal places. M1 for f and f′ at x₀ = 2, A1 for x₁, A1 for 2.2079. The answer lies inside (2, 3), which is where the sign change put the root, so the iteration has converged to the root the first part located. Locating the root before iterating is what supplies the starting value here, and testing the ends of the rounding interval settles the last figure: f(2.20785) = −0.000131 and f(2.20795) = 0.000014.

Question 7

f(x) = x³ − 2x + 2. Find the iterates x₁ and x₂ given by the Newton-Raphson method from x₀ = 0, and state what they show about the method from this starting value.

Accepted Answer

f′(x) = 3x² − 2. At x₀ = 0, f(0) = 2 and f′(0) = −2, so x₁ = 0 − 2/(−2) = 1. At x₁ = 1, f(1) = 1 and f′(1) = 1, so x₂ = 1 − 1/1 = 0. M1 for both steps. The iterates repeat 0, 1, 0, 1, … without end, so the sequence never approaches the only root of f(x) = 0, which is at x = −1.769 to 3 decimal places: the method does not converge from x₀ = 0. A1 for the repeating pair and that conclusion. Neither f′(0) nor f′(1) is zero, so both steps are defined and the arithmetic is correct throughout; a starting value can be unusable without any tangent being horizontal. f(−2) = −2 and f(−1) = 3, so a sign change locates the root in (−2, −1) and a start there reaches it.

Question 8

f(x) = e^x − 3x. Given that f(x) = 0 has a root between x = 1.5 and x = 2, use the Newton-Raphson method with x₀ = 1.5 to find x₁ and x₂, giving x₂ to 5 decimal places.

Accepted Answer

f′(x) = e^x − 3. At x₀ = 1.5: f(1.5) = 4.4817 − 4.5 = −0.018311 and f′(1.5) = 1.481689, so x₁ = 1.5 + 0.018311/1.481689 = 1.512358. Then f(x₁) = 0.00034… and f′(x₁) = 1.5374…, so x₂ = 1.51213 to 5 decimal places. M1 for f and f′ at 1.5, A1 for x₁, M1 for the second iteration, A1 for 1.51213. The value lies between 1.5 and 2, which is where the root was located, so the iteration has converged to the one asked for. Keep x₁ unrounded on the calculator: entering 1.5124 for the second step moves the fifth decimal place of x₂.

Question 9

f(x) = x³ − 6x² + 9x − 1. Explain why no Newton-Raphson iterate can be found from x₀ = 3, and find x₁ from x₀ = 3.2, giving your answer to 4 decimal places.

Accepted Answer

f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), so f′(3) = 0, while f(3) = 27 − 54 + 27 − 1 = −1 is not zero. The tangent at x = 3 is horizontal, so it never meets the x-axis and the formula divides by zero: no first iterate exists there. B1 for f′(3) = 0, B1 for f(3) not being zero and the tangent being horizontal. From x₀ = 3.2: f(3.2) = −0.872 and f′(3.2) = 1.32, so x₁ = 3.2 + 0.872/1.32 = 3.8606 to 4 decimal places. M1 for the iteration, A1 for 3.8606. The derivative is still small at 3.2, so the step is long: it travels 0.66 and passes the root at 3.5321, which later steps come back to.

Question 10

A parachutist falls from rest, and the distance fallen after t seconds is modelled by s = 50t + 250(e^−0.2t − 1) metres. Given that the time taken to fall 100 m lies between 5 and 6 seconds, use the Newton-Raphson method with t₀ = 5 to find that time to 3 significant figures.

Accepted Answer

Setting s = 100 gives 50t + 250(e^−0.2t − 1) = 100, and t cannot be made the subject because it appears both inside and outside the exponential. So take f(t) = 50t + 250(e^−0.2t − 1) − 100 and find a root. B1 for forming the equation. f′(t) = 50 − 50e^−0.2t, and at t₀ = 5: f(5) = −8.0301 and f′(5) = 31.6060, so t₁ = 5 + 8.0301/31.6060 = 5.25407, and a second step gives t₂ = 5.25048. M1 for f′, M1 for the iteration, A1 for 5.25 seconds. The value lies between 5 and 6, as the question said it would. Give the time in seconds to the 3 significant figures asked for: 5.25048 s claims a precision the model does not carry.

Question 11

f(x) = x³ − 2x − 5, and the root of f(x) = 0 is α = 2.094551 correct to 6 decimal places. Using the Newton-Raphson method with x₀ = 2, find x₁ and x₂, giving x₂ to 6 decimal places, and compare how close x₀, x₁ and x₂ each lie to α.

Accepted Answer

f′(x) = 3x² − 2. At x₀ = 2: f(2) = −1 and f′(2) = 10, so x₁ = 2 + 1/10 = 2.1 exactly. Then f(2.1) = 0.061 and f′(2.1) = 11.23, so x₂ = 2.1 − 0.061/11.23 = 2.094568 to 6 decimal places. M1 for f and f′ at 2, A1 for x₁ = 2.1, M1 for the second iteration, A1 for 2.094568. The distances from α are 0.0946, 0.0054 and 0.0000166, each one of the order of the square of the one before it, which is the quadratic convergence expected near a simple root from a suitable starting value. B1 for the comparison. Two steps have bought about four more correct decimal places than the starting value had.

Question 12

f(x) = x³ − 3x + 1. Explain why the Newton-Raphson method cannot be applied with x₀ = 1. Taking x₀ = 0.9 instead, find x₁ and comment on the result, given that f(x) = 0 has roots near −1.879, 0.347 and 1.532.

Accepted Answer

f′(x) = 3x² − 3, so f′(1) = 0 while f(1) = −1. The Newton-Raphson formula would divide by zero, so no next iterate is defined from x₀ = 1. The tangent there is horizontal. For x₀ = 0.9, f(0.9) = −0.971 and f′(0.9) = −0.57, giving x₁ = 0.9 − (−0.971)/(−0.57) = −0.8035… ≈ −0.804. The small derivative produces a large step. Continuing from this value converges to the root near 1.532, not the nearer root near 0.347. B1 for f′(1) = 0, B1 for the division being undefined, B1 for the horizontal tangent, M1 for the iteration from 0.9, A1 for the value of x₁, B1 for the comment on the root reached. A sign-change interval or sketch should be used to identify the intended root before iteration.

Question 13

f(x) = ln x, so that f(x) = 0 has the single root x = 1. Show that the Newton-Raphson method gives x_n+1 = x_n(1 − ln x_n), and explain why the method breaks down when it is started at x₀ = 3.

Accepted Answer

f′(x) = 1/x, so x_n+1 = x_n − ln x_n/(1/x_n) = x_n − x_n ln x_n = x_n(1 − ln x_n). M1 for the substitution into the formula, A1 for the printed form. From x₀ = 3: x₁ = 3(1 − ln 3) = 3(1 − 1.09861) = −0.29584. ln x is defined only for x > 0, so f(x₁) cannot be evaluated and no x₂ exists: the method breaks down at the second step rather than the first. B1 for the negative first iterate with that reason. The gradient at x = 3 is only 1/3, so the tangent there is shallow and meets the axis 3.29584 to the left of x₀, on the far side of the vertical asymptote of y = ln x at x = 0. From x₀ = 1.5 the same iteration gives 0.89180, 0.99392 and 0.99998, converging to the root at x = 1, so what fails here is the starting value and not the method.

Question 14

Show that applying the Newton-Raphson method to f(x) = x² − a, for a constant a > 0, gives the iteration x_n+1 = ½(x_n + a/x_n). Taking a = 10 and x₀ = 3, find x₁, x₂ and x₃, giving x₃ to 6 decimal places, and show that x_n+1 − √a = (x_n − √a)²/(2x_n).

Accepted Answer

With f(x) = x² − a and f′(x) = 2x, the formula gives x_n+1 = x_n − (x_n² − a)/(2x_n) = (x_n² + a)/(2x_n), which is ½(x_n + a/x_n). M1 for the substitution, A1 for the printed form. With a = 10 from x₀ = 3: x₁ = ½(3 + 10/3) = 3.166667, x₂ = 3.162281 and x₃ = 3.162278 to 6 decimal places, against √10 = 3.1622776…. M1 for a correct iterate, A1 for x₃. For the identity, subtract √a from the iteration and put the result over the common denominator: x_n+1 − √a = (x_n² + a − 2x_n√a)/(2x_n), and the numerator is the square (x_n − √a)², which is the printed result. M1 for the common denominator, A1 for recognising the square. The right-hand side is positive for every positive x_n, so every iterate after the first sits above √a, and the error at each step is the previous error squared and then divided by about 2√a.

Question 15

A spherical tank of radius 2 m holds liquid to a depth of h m, and the volume held is V = πh²(6 − h)/3 cubic metres for 0 ≤ h ≤ 4. Find the depth at which the tank holds 10 m³, to 3 significant figures, using the Newton-Raphson method from h₀ = 1.5, and explain why the equation has two further roots the model cannot use.

Accepted Answer

Setting V = 10 gives πh²(6 − h)/3 = 10, so take f(h) = πh²(6 − h)/3 − 10 and find a root. B1 for forming the equation. Differentiating, f′(h) = π(12h − 3h²)/3 = π(4h − h²). M1 for f′. At h₀ = 1.5: f(1.5) = 0.60288 and f′(1.5) = 11.7810, so h₁ = 1.5 − 0.60288/11.7810 = 1.44883, and a second step gives h₂ = 1.44846. M1 for the iteration, A1 for h₁, A1 for the depth 1.45 m. The cubic has two more roots, near h = −1.155 and h = 5.707, and neither describes the tank: a depth is not negative, and it cannot exceed the diameter of 4 m. B1 for rejecting both, with a reason. Reading the model before iterating is what fixes the starting value: 1.5 sits inside 0 ≤ h ≤ 4 and the other roots are outside it.