Numerical Methods

Board Coverage

Board	Paper	Notes
AQA	Paper 2	Sign change, iteration, Newton-Raphson, Simpson’s rule
Edexcel	P2	Similar
OCR (A)	Paper 2	Includes fixed-point iteration and convergence
CIE (9709)	P2, P3	Numerical solutions of equations, integration in P2/P3

:::info The formula booklet gives the Newton-Raphson formula and the trapezium/Simpson’s rule. You Must know when each method is applicable and its limitations. :::

1. Locating Roots: Sign Change

1.1 Sign change theorem

Theorem. If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then there Exists at least one root $\alpha \in (a,b)$ such that $f(\alpha) = 0$ .

1.2 Limitations

The sign change theorem tells us a root exists but says nothing about:

How many roots are in the interval (there could be 1, 3, 5, …).
The exact location of the root.

:::caution A sign change is sufficient but not necessary for a root. If $f(x) = x^2$ Then $f(-1) = f(1) = 1$ (no sign change), but there is a root at $x = 0$ . Additionally, a sign change Could arise from a discontinuity rather than a root: $f(x) = 1/x$ has $f(-1) = -1$ and $f(1) = 1$ But no root. :::

Intuition. The sign change theorem is the Intermediate Value Theorem applied to the special case Of crossing zero. If you walk from a point below sea level to one above sea level, you must cross Sea level at some point — provided the ground is continuous (no teleporting).

2. Fixed-Point Iteration

2.1 Method

To solve $f(x) = 0$ Rewrite as $x = g(x)$ and iterate:

$x_{n+1} = g(x_n)$

Starting from an initial guess $x_0$ . If the sequence converges to $\alpha$ Then $f(\alpha) = 0$ .

2.2 Convergence condition

Theorem. If $g$ is continuously differentiable near a fixed point $\alpha$ and $|g'(\alpha)| \lt 1$ Then the iteration $x_{n+1} = g(x_n)$ converges to $\alpha$ for all starting Values sufficiently close to $\alpha$ .

If $|g'(\alpha)| \gt 1$ The iteration diverges.

Proof (linear convergence). Near $\alpha$ By Taylor’s theorem:

$g(x_n) = g(\alpha) + g'(\alpha)(x_n - \alpha) + O((x_n-\alpha)^2)$

Since $g(\alpha) = \alpha$ :

$x_{n+1} - \alpha = g'(\alpha)(x_n - \alpha) + O((x_n-\alpha)^2)$

For $x_n$ close to $\alpha$ :

$|x_{n+1} - \alpha| \approx |g'(\alpha)| \cdot |x_n - \alpha|$

If $|g'(\alpha)| \lt 1$ Then $|x_{n+1} - \alpha| \lt |x_n - \alpha|$ : the error shrinks, so the Iteration converges. If $|g'(\alpha)| \gt 1$ The error grows and the iteration diverges. $\blacksquare$

2.3 Rearrangement choices

Different rearrangements of $f(x) = 0$ give different $g(x)$ And some converge while others Diverge.

Example. Solve $x^3 + x - 1 = 0$ .

$g(x) = 1 - x^3$ : $g'(x) = -3x^2$ . Near the root $\alpha \approx 0.68$ : $|g'(\alpha)| \approx 1.39 \gt 1$ . Diverges.
$g(x) = \sqrt[3]{1-x}$ : $g'(x) = \dfrac{-1}{3(1-x)^{2/3}}$ . Near $\alpha$ : $|g'(\alpha)| \approx 0.72 \lt 1$ . Converges.

:::tip In exams, if a question asks you to show that a particular rearrangement converges, compute $g'(x)$ at the root and show $|g'(\alpha)| \lt 1$ . If asked why a rearrangement fails, show $|g'(\alpha)| \gt 1$ . :::

2.4 Geometric interpretation

The fixed-point iteration $x_{n+1} = g(x_n)$ can be visualised using the cobweb diagram. Plot $y = g(x)$ and $y = x$ . Starting from $x_0$ on the $x$ -axis, go vertically to $y = g(x_0) = x_1$ Then horizontally to $y = x$ Then vertically to $y = g(x_1) = x_2$ And so on.

If $0 \lt g'(\alpha) \lt 1$ : the cobweb spirals inward (monotone convergence).
If $-1 \lt g'(\alpha) \lt 0$ : the cobweb zigzags inward (oscillatory convergence).
If $|g'(\alpha)| \gt 1$ : the cobweb spirals or zigzags outward (divergence).

The closer $|g'(\alpha)|$ is to zero, the faster the convergence. When $g'(\alpha) = 0$ The Iteration achieves quadratic convergence (similar to Newton-Raphson), since the leading error term In the Taylor expansion vanishes.

3. Newton-Raphson Method

3.1 Derivation from the tangent line

To solve $f(x) = 0$ Start from $x_0$ and draw the tangent to $y = f(x)$ at $x_0$ . The tangent line Is:

$y - f(x_n) = f'(x_n)(x - x_n)$

Setting $y = 0$ (where the tangent crosses the $x$ -axis):

$0 - f(x_n) = f'(x_n)(x_{n+1} - x_n)$

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

This is the Newton-Raphson formula.

3.2 Proof of local quadratic convergence

Theorem. If $f$ is twice continuously differentiable, $f(\alpha) = 0$ , $f'(\alpha) \neq 0$ And $x_0$ is sufficiently close to $\alpha$ Then Newton-Raphson converges quadratically: $|x_{n+1} - \alpha| \leq C|x_n - \alpha|^2$ .

Proof (sketch). By Taylor’s theorem about $x_n$ :

$0 = f(\alpha) = f(x_n) + f'(x_n)(\alpha - x_n) + \frac◆LB◆f''(\xi)◆RB◆◆LB◆2◆RB◆(\alpha - x_n)^2$

For some $\xi$ between $\alpha$ and $x_n$ . Rearranging:

$\frac{f(x_n)}{f'(x_n)} = (\alpha - x_n) + \frac◆LB◆f''(\xi)◆RB◆◆LB◆2f'(x_n)◆RB◆(\alpha - x_n)^2$

$x_n - \frac{f(x_n)}{f'(x_n)} = \alpha - \frac◆LB◆f''(\xi)◆RB◆◆LB◆2f'(x_n)◆RB◆(\alpha - x_n)^2$

$x_{n+1} - \alpha = -\frac◆LB◆f''(\xi)◆RB◆◆LB◆2f'(x_n)◆RB◆(x_n - \alpha)^2$

Taking absolute values: $|x_{n+1} - \alpha| = \dfrac◆LB◆|f''(\xi)|◆RB◆◆LB◆2|f'(x_n)|◆RB◆|x_n - \alpha|^2 \leq C|x_n - \alpha|^2$ . $\blacksquare$

Intuition. Quadratic convergence means the number of correct decimal places roughly doubles With each iteration. If you have 2 correct digits, the next iteration gives about 4, then 8, Then 16. This is dramatically faster than the linear convergence of fixed-point iteration (which Adds roughly a fixed number of correct digits per step).

3.3 Failures

Newton-Raphson fails when:

$f'(x_n) = 0$ (the tangent is horizontal — division by zero).
$f'(x_n)$ is close to zero (the next iterate jumps far away).
The starting point is not close enough to the root.

:::caution Warning A different starting point. :::

3.4 Horizontal tangent failure

When $f'(x_n) = 0$ at some iterate, the Newton-Raphson formula requires division by zero and the Method breaks down entirely. Even when $f'(x_n)$ is merely small, the step $x_{n+1} = x_n - f(x_n)/f'(x_n)$ becomes very large, sending the iterate far from the root.

Example. Let $f(x) = x^3 - 3x + 3$ . Then $f'(x) = 3x^2 - 3$ So $f'(1) = 0$ . Starting at $x_0 = 1$ :

The formula gives $x_1 = 1 - f(1)/f'(1) = 1 - 1/0$ Which is undefined.

Even starting at $x_0 = 0.9$ : $f(0.9) = 0.729 - 2.7 + 3 = 1.029$ , $f'(0.9) = 2.43 - 3 = -0.57$ . $x_1 = 0.9 - 1.029/(-0.57) = 0.9 + 1.805 = 2.705$ .

The root is near $\alpha \approx -2.10$ So the iterate has been sent in the wrong direction. The Next iterate $x_2$ will be pulled back, but convergence is erratic compared to a well-chosen Starting point.

:::caution Warning Tangent is not close to horizontal near your starting point. :::

3.5 Slow convergence near inflection points

The quadratic convergence proof in Section 3.2 requires $f'(\alpha) \neq 0$ . When the root coincides With an inflection point, so that $f'(\alpha) = 0$ Convergence degrades from quadratic to linear.

Theorem. If $f(\alpha) = 0$$f'(\alpha) = 0$$f''(\alpha) \neq 0$ And $x_0$ is sufficiently Close to $\alpha$ Then Newton-Raphson converges linearly with rate $1/2$ :

$|x_{n+1} - \alpha| \approx \frac{1}{2}|x_n - \alpha|$

Proof sketch. Expanding by Taylor’s theorem to third order about $\alpha$ :

$f(x_n) = \frac◆LB◆f''(\alpha)◆RB◆◆LB◆2◆RB◆(x_n - \alpha)^2 + \frac◆LB◆f'''(\alpha)◆RB◆◆LB◆6◆RB◆(x_n - \alpha)^3 + O((x_n - \alpha)^4)$

$f'(x_n) = f''(\alpha)(x_n - \alpha) + \frac◆LB◆f'''(\alpha)◆RB◆◆LB◆2◆RB◆(x_n - \alpha)^2 + O((x_n - \alpha)^3)$

Therefore:

$x_{n+1} - \alpha = x_n - \alpha - \frac{f(x_n)}{f'(x_n)}$

$= (x_n - \alpha) - \frac◆LB◆\frac{f''(\alpha)}{2}(x_n - \alpha)^2 + O((x_n - \alpha)^3)◆RB◆◆LB◆f''(\alpha)(x_n - \alpha) + O((x_n - \alpha)^2)◆RB◆$

$= (x_n - \alpha) - \frac◆LB◆\frac{f''(\alpha)}{2}(x_n - \alpha) + O((x_n - \alpha)^2)◆RB◆◆LB◆f''(\alpha) + O(x_n - \alpha)◆RB◆$

$= (x_n - \alpha) - \frac{1}{2}(x_n - \alpha)\left(1 + O(x_n - \alpha)\right)$

$= \frac{1}{2}(x_n - \alpha) + O((x_n - \alpha)^2)$

So $|x_{n+1} - \alpha| \to \frac{1}{2}|x_n - \alpha|$ as $x_n \to \alpha$ : the error is halved each Step (linear convergence with rate $1/2$ ), not squared. $\blacksquare$

Example. $f(x) = (x-1)^3$ has a root at $x = 1$ where $f'(1) = 0$ . The Newton-Raphson iteration Becomes:

$x_{n+1} = x_n - \frac{(x_n - 1)^3}{3(x_n - 1)^2} = x_n - \frac{x_n - 1}{3} = \frac{2x_n + 1}{3}$

Starting at $x_0 = 4$ : $x_1 = 3$$x_2 = 7/3 \approx 2.333$$x_3 = 17/9 \approx 1.889$ $x_4 = 37/27 \approx 1.370$$x_5 = 75/81 \approx 1.210$ …

The error is multiplied by $2/3$ each step (linear, not quadratic). Compare: standard Newton-Raphson With $f'(\alpha) \neq 0$ would give roughly 1, then 2, then 4, then 8 correct digits. Here each step Only adds a fixed fraction of a digit.

4. The Trapezium Rule

4.1 Formula

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + 2y_1 + \cdots + 2y_{n-1} + y_n\right]$

Where $h = (b-a)/n$ and $y_i = f(a+ih)$ .

4.2 Error analysis

Theorem. The error in the composite trapezium rule is:

$E_T = -\frac{(b-a)^3}{12n^2}\,f''(\eta)$

For some $\eta \in (a, b)$ Provided $f$ is twice continuously differentiable on $[a, b]$ .

Derivation. Consider a single strip $[x_i, x_{i+1}]$ of width $h$ . The trapezium rule Approximates $\int_{x_i}^{x_{i+1}} f(x)\,dx$ by the area of a trapezium:

$\int_{x_i}^{x_{i+1}} f(x)\,dx \approx \frac{h}{2}\left[f(x_i) + f(x_{i+1})\right]$

To find the error, expand $f$ about the midpoint $m_i = x_i + h/2$ . Let $\delta = h/2$ :

$f(x_i) = f(m_i - \delta) = f(m_i) - \delta\, f'(m_i) + \frac◆LB◆\delta^2◆RB◆◆LB◆2◆RB◆f''(m_i) - \frac◆LB◆\delta^3◆RB◆◆LB◆6◆RB◆f'''(\zeta_1)$

$f(x_{i+1}) = f(m_i + \delta) = f(m_i) + \delta\, f'(m_i) + \frac◆LB◆\delta^2◆RB◆◆LB◆2◆RB◆f''(m_i) + \frac◆LB◆\delta^3◆RB◆◆LB◆6◆RB◆f'''(\zeta_2)$

Adding these:

$f(x_i) + f(x_{i+1}) = 2f(m_i) + \delta^2 f''(m_i) + O(h^3)$

The trapezium approximation for this strip is:

$\frac{h}{2}\left[f(x_i) + f(x_{i+1})\right] = h\,f(m_i) + \frac◆LB◆h\,\delta^2◆RB◆◆LB◆2◆RB◆f''(m_i) + O(h^4) = h\,f(m_i) + \frac{h^3}{8}f''(m_i) + O(h^4)$

The exact integral over this strip is (by Taylor expansion of the integral):

$\int_{x_i}^{x_{i+1}} f(x)\,dx = h\,f(m_i) + \frac{h^3}{24}f''(m_i) + O(h^5)$

Therefore the error on a single strip is:

$E_i = \frac{h}{2}\left[f(x_i) + f(x_{i+1})\right] - \int_{x_i}^{x_{i+1}} f(x)\,dx = \left(\frac{1}{8} - \frac{1}{24}\right)h^3 f''(m_i) + O(h^4) = \frac{h^3}{12}f''(m_i) + O(h^4)$

Summing over all $n$ strips and applying the Intermediate Value Theorem to $f''$ (which is Continuous), there exists $\eta \in (a, b)$ such that:

$E_T = \sum_{i=0}^{n-1} E_i = \frac{h^3}{12}\sum_{i=0}^{n-1} f''(m_i) + O(n \cdot h^4) = \frac{h^3}{12}\cdot\frac◆LB◆n\, f''(\eta)◆RB◆◆LB◆1◆RB◆ + O(h^4 \cdot n)$

Since $\sum_{i=0}^{n-1} f''(m_i) \cdot h \approx \int_a^b f''(x)\,dx$ and $nh = b - a$ :

$E_T = -\frac{(b-a)^3}{12n^2}\,f''(\eta)$

(The negative sign arises from the exact derivation via the Euler-Maclaurin formula; the key point Is the $h^2$ scaling.) $\blacksquare$

Key consequence. The error is proportional to $h^2 = (b-a)^2/n^2$ . Doubling the number of strips ( $n \to 2n$ ) reduces the error by a factor of 4, since $h \to h/2$ and $h^2 \to h^2/4$ .

4.3 Error bound

From the error formula, since $|f''(\eta)| \leq M$ for all $\eta \in [a,b]$ :

$|E_T| \leq \frac{(b-a)^3}{12n^2}\,M$

This gives a guaranteed upper bound on the absolute error. If we require the error to satisfy $|E_T| \lt \varepsilon$ We need:

$n \gt \sqrt◆LB◆\frac{(b-a)^3\, M}{12\,\varepsilon}◆RB◆$

Example. Approximate $\displaystyle\int_0^1 e^{-x^2}\,dx$ with the trapezium rule. Here $f(x) = e^{-x^2}$ So $f'(x) = -2x\,e^{-x^2}$ and $f''(x) = (4x^2 - 2)e^{-x^2}$ . On $[0,1]$ : $|f''(x)| \leq 2$ (achieved at $x = 0$ Where $f''(0) = -2$ ).

For error $\lt 10^{-4}$ :

$n \gt \sqrt◆LB◆\frac{1^3 \times 2}{12 \times 10^{-4}}◆RB◆ = \sqrt◆LB◆\frac{2}{0.0012}◆RB◆ = \sqrt◆LB◆1666.\bar{6}◆RB◆ \approx 40.8$

So $n = 42$ strips suffice (rounding up to the nearest even number, which is convenient if one later Wishes to compare with Simpson’s rule).

5. Simpson’s Rule

5.1 Formula

For an even number $n$ of strips:

$\int_a^b f(x)\,dx \approx \frac{h}{3}\left[y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + \cdots + 4y_{n-1} + y_n\right]$

The coefficients follow the pattern: $1, 4, 2, 4, 2, \ldots, 4, 1$ .

5.2 Derivation

Simpson’s rule approximates $f$ by quadratic arcs over pairs of strips. Over $[x_{2k}, x_{2k+2}]$ A Unique quadratic passes through $(x_{2k}, y_{2k})$$(x_{2k+1}, y_{2k+1})$$(x_{2k+2}, y_{2k+2})$ . Integrating this quadratic gives the area $\dfrac{h}{3}(y_{2k} + 4y_{2k+1} + y_{2k+2})$ .

5.3 Error bound

$|E| \leq \frac{(b-a)^5}{180n^4}M$

Where $|f^{(4)}(x)| \leq M$ on $[a,b]$ .

Intuition. Simpson’s rule has error $O(1/n^4)$ compared to $O(1/n^2)$ for the trapezium rule. Doubling the number of strips reduces the error by a factor of 16 for Simpson’s rule vs. 4 for the Trapezium rule. This is because quadratic approximations match the curvature of the function much Better than linear ones.

:::tip Simpson’s rule requires an even number of strips. The trapezium rule works with any Number. Simpson’s rule is exact for cubics (since the error depends on $f^{(4)}$ ). :::

6. Comparison of Methods

Method	Convergence	Requirement	Speed
Sign change	N/A	Continuous $f$	Locates only
Bisection	Linear	Sign change	Slow
Fixed-point	Linear	$\\|g'\\| \lt 1$	Moderate
Newton-Raphson	Quadratic	$f'(\alpha)\neq 0$	Fast
Trapezium rule	$O(1/n^2)$	Any $f$	Integration
Simpson’s rule	$O(1/n^4)$	Even $n$ strips	Integration

7. Comparison of Root-Finding Methods

7.1 When to use each method

Sign change (bisection): Use when you need a guaranteed, robust result and have a sign change. Always converges but slowly (halving the interval each step). No derivative needed. Excellent for Initial bracketing before switching to a faster method.

Fixed-point iteration: Use when the equation is rearranged into $x = g(x)$ and $|g'(\alpha)| \lt 1$ can be verified. Simple to implement but may converge slowly or diverge Entirely depending on the rearrangement. Different rearrangements of the same equation can give Radically different convergence behaviour.

Newton-Raphson: Use when $f'(x)$ is easy to compute and a good starting estimate is available. Extremely fast (quadratic convergence) when it works, but can fail catastrophically if $f'(x_n)$ is Small, if the starting point is poor, or if the root is at an inflection point.

7.2 Practical strategy

In practice, numerical software often combines methods:

Bracket the root using sign change to find an interval $[a, b]$ containing the root.
Refine using Newton-Raphson or fixed-point iteration starting from the midpoint or a favourable endpoint.
Verify the result by checking $f(x_n)$ is sufficiently close to zero.

:::info Newton-Raphson is the method of choice when derivatives are available and the function is Well-behaved. Fixed-point iteration is useful when the problem gives a contraction Mapping. Bisection is the reliable fallback when nothing else is guaranteed to work. :::

7.3 Cost comparison

Method	Cost per step	Convergence rate	Total cost to reach tolerance $\varepsilon$
Bisection	1 evaluation	$1/2$ (linear)	$O(\log_2(1/\varepsilon))$ steps
Fixed-point	1 evaluation	$\\|g'(\alpha)\\|$ (linear)	$O(\log_{1/\\|g'\\|}(1/\varepsilon))$ steps
Newton-Raphson	2 evaluations	Quadratic	$O(\log_2 \log_2(1/\varepsilon))$ steps

Newton-Raphson requires both $f$ and $f'$ per step (two evaluations), but its quadratic convergence Means it needs far fewer steps in total. For high accuracy requirements, Newton-Raphson is Overwhelmingly more efficient despite the higher per-step cost.

Problem Set

Problem 1

Show that $x^3 - 2x - 5 = 0$ has a root in the interval $[2, 3]$.

Solution 1

$f(x) = x^3 - 2x - 5$. $f(2) = 8-4-5 = -1 \lt 0$$f(3) = 27-6-5 = 16 \gt 0$.

Since $f$ is continuous and changes sign on $[2,3]$ By the sign change theorem there is a root in $(2,3)$ .

If you get this wrong, revise: Sign Change Theorem — Section 1.1.

Problem 2

Use fixed-point iteration $x_{n+1} = \sqrt[3]{2x_n + 5}$ with $x_0 = 2$ to find $x_3$. Give your answer to 4 decimal places.

Solution 2

$x_0 = 2.0000$ $x_1 = \sqrt[3]{9} = 2.0801$ $x_2 = \sqrt[3]{2(2.0801)+5} = \sqrt[3]{9.1602} = 2.0924$ $x_3 = \sqrt[3]{2(2.0924)+5} = \sqrt[3]{9.1848} = 2.0943$

If you get this wrong, revise: Fixed-Point Iteration — Section 2.

Problem 3

Show that the iteration $x_{n+1} = \dfrac{x_n^3 + 5}{2}$ for solving $x^3 - 2x - 5 = 0$ will not converge near the root.

Solution 3

$g(x) = \dfrac{x^3+5}{2}$$g'(x) = \dfrac{3x^2}{2}$.

Near the root $\alpha \approx 2.09$ : $g'(\alpha) = \dfrac{3(2.09)^2}{2} \approx \dfrac◆LB◆3 \times 4.37◆RB◆◆LB◆2◆RB◆ \approx 6.55 \gt 1$ .

Since $|g'(\alpha)| \gt 1$ The iteration diverges near $\alpha$ .

If you get this wrong, revise: Convergence Condition — Section 2.2.

Problem 4

Use Newton-Raphson with $x_0 = 2$ to find $x_2$ for $f(x) = x^3 - 2x - 5 = 0$. Give your answer to 4 decimal places.

Solution 4

$f'(x) = 3x^2 - 2$. $x_{n+1} = x_n - \dfrac{x_n^3 - 2x_n - 5}{3x_n^2 - 2}$.

$x_0 = 2$ : $f(2) = -1$$f'(2) = 10$ . $x_1 = 2 - (-1/10) = 2.1000$ .

$x_1 = 2.1$ : $f(2.1) = 9.261-4.2-5 = 0.061$$f'(2.1) = 13.23-2 = 11.23$ . $x_2 = 2.1 - 0.061/11.23 = 2.0946$ .

If you get this wrong, revise: Newton-Raphson Method — Section 3.

Problem 5

Use Simpson's rule with 4 strips to approximate $\displaystyle\int_0^2 e^{-x^2}\,dx$ to 4 decimal places.

Solution 5

$h = 0.5$. Values: $y_0 = 1$$y_1 = e^{-0.25} \approx 0.7788$$y_2 = e^{-1} \approx 0.3679$$y_3 = e^{-2.25} \approx 0.1054$$y_4 = e^{-4} \approx 0.0183$.

$\int_0^2 e^{-x^2}\,dx \approx \frac{0.5}{3}[1 + 4(0.7788) + 2(0.3679) + 4(0.1054) + 0.0183]$ $= \frac{0.5}{3}[1 + 3.1152 + 0.7358 + 0.4216 + 0.0183] = \frac{0.5}{3}(5.2909) \approx 0.8818$

If you get this wrong, revise: Simpson’s Rule — Section 5.

Problem 6

Show that the equation $e^x = 3x$ has exactly two real roots, and locate them.

Solution 6

Let $f(x) = e^x - 3x$. $f'(x) = e^x - 3$.

$f'(x) = 0 \implies x = \ln 3 \approx 1.099$ .

$f(\ln 3) = 3 - 3\ln 3 \approx 3 - 3.296 = -0.296 \lt 0$ .

Since $f''(x) = e^x \gt 0$ This is a global minimum. The minimum value is negative, and $f(x) \to \infty$ as $x \to \pm\infty$ So $f(x) = 0$ has exactly two roots.

$f(0) = 1 \gt 0$$f(1) = e-3 \lt 0$ : root in $(0,1)$ . $f(1) \lt 0$$f(2) = e^2-6 \gt 0$ : Root in $(1,2)$ .

If you get this wrong, revise: Sign Change Theorem — Section 1.

Problem 7

The Newton-Raphson formula fails when applied to $f(x) = x^{1/3}$ starting from $x_0 = 1$. Explain why.

Solution 7

$f(x) = x^{1/3}$$f'(x) = \dfrac{1}{3}x^{-2/3}$.

$x_{n+1} = x_n - \dfrac◆LB◆x_n^{1/3}◆RB◆◆LB◆\frac{1}{3}x_n^{-2/3}◆RB◆ = x_n - 3x_n = -2x_n$ .

So $x_1 = -2$$x_2 = 4$$x_3 = -8$ … The iterates oscillate and diverge.

The problem is that $f'(0) = \infty$ — the tangent at the root $x=0$ is vertical, so the Newton-Raphson step sends the iterate to $-\infty$ .

If you get this wrong, revise: Failures — Section 3.3.

Problem 8

Use the trapezium rule with 6 strips to approximate $\displaystyle\int_1^4 \ln x\,dx$.

Solution 8

$h = 0.5$. Values: $\ln 1 = 0$$\ln 1.5 \approx 0.4055$$\ln 2 \approx 0.6931$$\ln 2.5 \approx 0.9163$$\ln 3 \approx 1.0986$$\ln 3.5 \approx 1.2528$$\ln 4 \approx 1.3863$.

$\mathrm{Approx} = \frac{0.5}{2}[0 + 2(0.4055+0.6931+0.9163+1.0986+1.2528) + 1.3863]$ $= 0.25[0 + 2(4.3663) + 1.3863] = 0.25[8.7326 + 1.3863] = 0.25 \times 10.1189 \approx 2.5297$

(Exact: $[x\ln x - x]_1^4 = 4\ln 4 - 4 + 1 = 8\ln 2 - 3 \approx 2.5452$ .)

If you get this wrong, revise: The Trapezium Rule — Section 4.

Problem 9

For the equation $x^3 + x - 3 = 0$Show that $x_{n+1} = \sqrt[3]{3 - x_n}$ converges near the root.

Solution 9

$f(1) = -1 \lt 0$$f(1.2) = 1.728 + 1.2 - 3 = -0.072 \lt 0$$f(1.3) = 2.197 + 1.3 - 3 = 0.497 \gt 0$.

Root near $\alpha \approx 1.21$ .

$g(x) = (3-x)^{1/3}$$g'(x) = -\dfrac{1}{3}(3-x)^{-2/3}$ .

$|g'(\alpha)| = \dfrac{1}{3}(3-1.21)^{-2/3} = \dfrac{1}{3}(1.79)^{-2/3} \approx \dfrac◆LB◆1◆RB◆◆LB◆3 \times 1.489◆RB◆ \approx 0.224 \lt 1$ .

Converges since $|g'(\alpha)| \lt 1$ .

If you get this wrong, revise: Convergence Condition — Section 2.2.

Problem 10

Explain why the sign change theorem does not guarantee a root of $f(x) = \dfrac{1}{x-2}$ in the interval $[1, 3]$.

Solution 10

$f(1) = -1 \lt 0$ and $f(3) = 1 \gt 0$So there is a sign change. However, $f$ is **not continuous** on $[1,3]$ — it has a vertical asymptote at $x = 2$. The sign change theorem requires continuity, so it does not apply here. There is no root of $1/(x-2) = 0$.

If you get this wrong, revise: Limitations — Section 1.2.

Problem 11

Consider $f(x) = (x-2)^3$. The root is $\alpha = 2$. Apply Newton-Raphson starting from $x_0 = 5$. Show that the convergence is linear, find the convergence rate, and explain why quadratic Convergence is lost.

Solution 11

$f(x) = (x-2)^3$$f'(x) = 3(x-2)^2$.

$x_{n+1} = x_n - \dfrac{(x_n-2)^3}{3(x_n-2)^2} = x_n - \dfrac{x_n - 2}{3} = \dfrac{2x_n + 4}{3}$

Starting from $x_0 = 5$ : $x_1 = 14/3 \approx 4.667$$x_2 = 40/9 \approx 4.444$ $x_3 = 116/27 \approx 4.296$$x_4 = 344/81 \approx 4.198$ .

Error: $e_0 = 3$$e_1 = 8/3 \approx 2.667$$e_2 = 16/9 \approx 1.778$ $e_3 = 32/27 \approx 1.185$$e_4 = 64/81 \approx 0.790$ .

Ratio: $e_{n+1}/e_n = 2/3$ for all $n$ . This is linear convergence with rate $2/3$ .

Quadratic convergence is lost because $f'(\alpha) = f'(2) = 0$ . The root coincides with a stationary Point (inflection point), violating the condition $f'(\alpha) \neq 0$ in the quadratic convergence Theorem. As shown in Section 3.5, when $f'(\alpha) = 0$ and $f''(\alpha) \neq 0$ Newton-Raphson Degrades to linear convergence with rate $1/2$ . Here $f''(x) = 6(x-2)$ So $f''(2) = 0$ as well (triple root), giving rate $2/3$ instead of $1/2$ .

If you get this wrong, revise: Slow Convergence Near Inflection Points — Section 3.5.

Problem 12

(a) Use the trapezium rule with $n = 4$ strips to approximate $\displaystyle\int_0^2 \frac{1}{1+x^2}\,dx$.

(b) Given that $f''(x) = \dfrac{6x^2 - 2}{(1+x^2)^3}$ Find an upper bound $M$ for $|f''(x)|$ on $[0,2]$ and hence bound the error in your approximation.

Solution 12

(a) $h = 0.5$.

$y_0 = f(0) = 1$$y_1 = f(0.5) = 1/1.25 = 0.8$$y_2 = f(1) = 0.5$ $y_3 = f(1.5) = 1/3.25 \approx 0.3077$$y_4 = f(2) = 0.2$ .

$\mathrm{Approx} = \frac{0.5}{2}[1 + 2(0.8 + 0.5 + 0.3077) + 0.2] = 0.25[1 + 2(1.6077) + 0.2] = 0.25 \times 4.4154 \approx 1.1039$

(Exact value: $\arctan 2 \approx 1.1071$ .)

(b) $f''(x) = \dfrac{6x^2 - 2}{(1+x^2)^3}$ . On $[0, 2]$ The numerator $6x^2 - 2$ is maximised at $x = 2$ where it equals $6(4) - 2 = 22$ . The denominator $(1+x^2)^3$ is minimised at $x = 0$ where It equals 1. We need to maximise $|f''(x)|$ .

Checking critical points: $f'''(x) = 0$ gives potential extrema of $f''$ . Alternatively, evaluate at Endpoints and critical points. $f''(0) = -2$$f''(1) = 4/8 = 0.5$$f''(2) = 22/125 = 0.176$ .

So $M = 2$ (taking $|f''(x)| \leq 2$ ).

Error bound: $|E_T| \leq \dfrac◆LB◆(2-0)^3◆RB◆◆LB◆12 \times 4^2◆RB◆ \times 2 = \dfrac{8}{192} \times 2 = \dfrac{1}{12} \approx 0.0833$ .

The actual error is $|1.1071 - 1.1039| = 0.0032$ Well within the bound.

If you get this wrong, revise: Error Analysis — Section 4.2 and Error Bound — Section 4.3.

Problem 13

The equation $e^{-x} = x$ has a single real root $\alpha$.

(a) Show that $\alpha \in (0.5, 0.7)$ .

(b) Consider the rearrangement $x_{n+1} = e^{-x_n}$ . Show that this iteration converges near $\alpha$ .

Solution 13

(a) $f(x) = e^{-x} - x$. $f(0.5) = e^{-0.5} - 0.5 \approx 0.6065 - 0.5 = 0.1065 \gt 0$. $f(0.7) = e^{-0.7} - 0.7 \approx 0.4966 - 0.7 = -0.2034 \lt 0$. By the sign change theorem, $\alpha \in (0.5, 0.7)$.

(b) $g_1(x) = e^{-x}$$g_1'(x) = -e^{-x}$ . At $\alpha \approx 0.567$ : $|g_1'(\alpha)| = e^{-\alpha} = \alpha \approx 0.567 \lt 1$ . Converges.

(c) $g_2(x) = -\ln x$$g_2'(x) = -1/x$ . At $\alpha \approx 0.567$ : $|g_2'(\alpha)| = 1/\alpha \approx 1.763 \gt 1$ . Diverges.

Both rearrangements solve the same equation, but only $x_{n+1} = e^{-x_n}$ converges near the root.

If you get this wrong, revise: Rearrangement Choices — Section 2.3.

Problem 14

Newton-Raphson is applied to $f(x) = x^3 - 3x + 2$ (which has a double root at $x = 1$ and a single Root at $x = -2$). Starting from $x_0 = 0.5$Compute $x_1$ and $x_2$And comment on the rate of Convergence towards $x = 1$.

Solution 14

$f(x) = x^3 - 3x + 2 = (x-1)^2(x+2)$$f'(x) = 3x^2 - 3$.

$x_0 = 0.5$ : $f(0.5) = 0.125 - 1.5 + 2 = 0.625$$f'(0.5) = 0.75 - 3 = -2.25$ . $x_1 = 0.5 - 0.625/(-2.25) = 0.5 + 0.2778 = 0.7778$ .

$x_1 = 0.7778$ : $f(0.7778) = 0.4703 - 2.3334 + 2 = 0.1369$$f'(0.7778) = 1.8151 - 3 = -1.1849$ . $x_2 = 0.7778 - 0.1369/(-1.1849) = 0.7778 + 0.1155 = 0.8933$ .

The iterates are approaching $x = 1$ but slowly. Errors: $e_0 = 0.5$$e_1 \approx 0.2222$ $e_2 \approx 0.1067$ . The ratio $e_2/e_1 \approx 0.48$ And $e_1/e_0 \approx 0.44$ . This is Approximately linear convergence.

At the double root $x = 1$ : $f(1) = 0$$f'(1) = 0$ . Since $f'(\alpha) = 0$ Quadratic convergence Is lost (as in Section 3.5). For a double root, Newton-Raphson converges linearly with rate Approximately $1/2$ .

If you get this wrong, revise: Slow Convergence Near Inflection Points — Section 3.5.

Problem 15

(a) Use the trapezium rule with $n = 2$ strips to approximate $\displaystyle\int_0^1 \sqrt{x}\,dx$.

(b) The exact value is $2/3$ . Compute the actual error.

Solution 15

(a) $h = 0.5$. $y_0 = \sqrt{0} = 0$$y_1 = \sqrt{0.5} \approx 0.7071$$y_2 = \sqrt{1} = 1$.

$\mathrm{Approx} = \frac{0.5}{2}[0 + 2(0.7071) + 1] = 0.25 \times 2.4142 = 0.6036$

(b) Exact: $2/3 \approx 0.6667$ . Actual error: $|0.6667 - 0.6036| = 0.0631$ .

On $(0, 1]$ : $|f''(x)| = \frac{1}{4}x^{-3/2}$ Which is unbounded as $x \to 0^+$ . The error bound Requires $f''$ to be bounded on $[a,b]$ But $f''(x) \to \infty$ as $x \to 0$ .

If we instead apply the bound on $[\varepsilon, 1]$ for small $\varepsilon \gt 0$ : $M = \frac{1}{4}\varepsilon^{-3/2}$ Which blows up as $\varepsilon \to 0$ . This illustrates a Limitation of the error bound: it requires $f''$ to be bounded, which fails when $f$ has a vertical Tangent at an endpoint.

If you get this wrong, revise: Error Bound — Section 4.3.

Problem 16

For the equation $\cos x = x$Let $g(x) = \cos x$.

(a) Verify that a fixed point $\alpha$ exists in $(0, \pi/2)$ .

(b) Show that $|g'(\alpha)| \lt 1$ And hence that the iteration $x_{n+1} = \cos x_n$ converges.

Solution 16

(a) $g(0) = \cos 0 = 1 \gt 0$ and $g(\pi/2) = \cos(\pi/2) = 0 \lt \pi/2$. Since $g$ is continuous and $g(x) - x$ changes sign on $(0, \pi/2)$ (check: $g(0) - 0 = 1 \gt 0$ $g(\pi/2) - \pi/2 = -\pi/2 \lt 0$), a fixed point exists.

(b) $g'(x) = -\sin x$ . At the fixed point $\alpha \approx 0.7391$ : $|g'(\alpha)| = \sin(0.7391) \approx 0.6736 \lt 1$ . Converges.

If you get this wrong, revise: Convergence Condition — Section 2.2.

Problem 17

A student uses Newton-Raphson to solve $f(x) = \tan x - 1 = 0$ starting from $x_0 = 1.3$.

(a) Compute $x_1$ .

(b) The root is $\alpha = \pi/4 \approx 0.7854$ . Explain why starting at $x_0 = 1.3$ may not be Ideal, and identify a value of $x_0$ between $0$ and $\pi/2$ where Newton-Raphson would fail Entirely.

Solution 17

(a) $f(x) = \tan x - 1$$f'(x) = \sec^2 x = 1/\cos^2 x$.

$x_0 = 1.3$ : $f(1.3) = \tan(1.3) - 1 \approx 3.6021 - 1 = 2.6021$ . $f'(1.3) = \sec^2(1.3) = 1/\cos^2(1.3) \approx 1/0.0754 \approx 13.26$ . $x_1 = 1.3 - 2.6021/13.26 \approx 1.3 - 0.1962 = 1.1038$ .

This is still far from $\alpha = 0.7854$ . The function is very steep here (large $f'$ ), so the step Is small but the iterate is far from the root.

(b) Newton-Raphson fails when $f'(x_0) = 0$ I.e., $\sec^2 x_0 = 0$ Which never happens since $\sec^2 x \geq 1$ for all $x$ . However, as $x_0 \to \pi/2^-$$\cos x_0 \to 0$ and $f'(x_0) \to \infty$ While $f(x_0) = \tan x_0 - 1 \to \infty$ . The ratio $f(x_0)/f'(x_0) = (\tan x_0 - 1)\cos^2 x_0$ tends to a finite limit, but the function becomes Extremely steep near $\pi/2$ Making the iteration numerically unstable. Starting very close to $\pi/2$ sends the iterate far away.

A better starting point is $x_0 = 1.0$ : $f(1) = \tan 1 - 1 \approx 0.5574$ $f'(1) = \sec^2 1 \approx 3.426$ . $x_1 = 1 - 0.5574/3.426 \approx 0.8373$ Already close to $\alpha = 0.7854$ .

If you get this wrong, revise: Horizontal Tangent Failure — Section 3.4.

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