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Numerical Methods / LEVEL 1 · DIFFICULTY 4/5

Numerical Methods

Read the prerequisite, study a worked example and justify each step in these selected university topics.

2 stages · 16 practice problems · two 4-question assessment forms

STAGE UNUM 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Bisection, Newton & Error

Useful preparation: Derivatives, Chain Rule & Implicit Change

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Finding a root numerically means producing an approximation together with evidence about its accuracy. Bisection buys certainty by repeatedly shrinking an interval known to contain a root. Newton uses local slope information to move much faster near a suitable root, but that speed comes without a general guarantee from an arbitrary starting value.

Root and residual
A root r satisfies f (r)=0\displaystyle f\, \left(r\right) = 0. The residual at an approximation x is f(x); it measures failure to satisfy the equation, not directly the distance from a root.
Bisection bracket
A bracket [a,b] has a continuous function with opposite nonzero signs at the endpoints. The intermediate value theorem guarantees at least one root inside.
Newton update
Newton replaces the curve near x by its tangent line and uses that line’s zero as the next approximation. The derivative at x must be nonzero.

Conditions before calculation

  • Bisection requires continuity throughout the bracket, not just evaluations with opposite signs. If an endpoint already has value zero, the root is known without iteration.
  • Local quadratic convergence of Newton is guaranteed near a simple root under suitable smoothness, such as a twice continuously differentiable function with nonzero derivative at the root and a sufficiently close starting point.
  • A small residual implies a small position error only with additional conditioning information. For example, a derivative magnitude bounded below on the interval connecting the approximation and root gives an error bound.

Why the method works

1. Keep the invariant

After evaluating the midpoint, retain a half whose endpoints have opposite signs. The root remains bracketed; the interval width halves each time.

2. State the error convention

After k halvings the bracket width is its initial width divided by 2 to the k. Reporting the midpoint of that final bracket guarantees half that width as an error bound.

∣mk−r∣≤b0−a02k+1|m_k-r|\le\frac{b_0-a_0}{2^{k+1}}

3. Derive and monitor Newton

Setting the tangent approximation to zero yields the update. Monitor residual, step size, and domain validity; safeguards can return to a known bracket.

0≈f(xn)+f′(xn)(xn+1−xn),xn+1=xn−f(xn)f′(xn)0\approx f(x_n)+f'(x_n)(x_{n+1}-x_n),\quad x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

Worked examples

EXAMPLE 1

Bracket the positive root of x squared-5, beginning with [2,3], using three bisection halvings.

  1. Endpoint values are -1 and 4. The midpoint 2.5 gives 1.25, so keep [2,2.5].

  2. The midpoint 2.25 gives 0.0625, so keep [2,2.25].

  3. The midpoint 2.125 gives -0.484375, so the final bracket is [2.125,2.25]. Its midpoint is 2.1875.

    ∣2.1875−5∣≤0.1252=0.0625|2.1875-\sqrt5|\le\frac{0.125}{2}=0.0625

Conclusion: The final midpoint has a certified error at most 0.0625, even without knowing the exact root numerically.

EXAMPLE 2

Take two Newton steps toward the positive root of x squared-10 from x 0=3\displaystyle x\, 0 = 3.

  1. Differentiate f; the update averages x with 10x\displaystyle \frac{10}{x}.

    f′(x)=2x,xn+1=12(xn+10xn)f'(x)=2x,\qquad x_{n+1}=\frac12\left(x_n+\frac{10}{x_n}\right)

  2. The first step is 196\displaystyle \frac{19}{6}, using a nonzero derivative at 3.

    x1=3−−16=196x_1=3-\frac{-1}{6}=\frac{19}{6}

  3. The next step simplifies exactly, allowing a residual check.

    x2=721228≈3.162280702,f(x2)=151984x_2=\frac{721}{228}\approx3.162280702,\quad f(x_2)=\frac1{51984}

Conclusion: The second approximation is close to the positive root. Its residual is small; the slope near the root is safely away from zero.

Pitfalls & counterexamples

Tempting claim: Opposite signs always guarantee a root between two points.

Correction: Continuity on the entire interval is essential.

Test the claim: The function 1x\displaystyle \frac{1}{x} has opposite signs at -1 and 1 but is undefined at zero and has no root.

Check your understanding

Does Newton necessarily converge from every starting point?

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Compare your reasoning

No.

For f (x)=x\displaystyle f\, \left(x\right) = x cubed-2 x+2\displaystyle 2\, x + 2, the starting value 0 alternates between 0 and 1 instead of converging.

Connect this idea

Conditioning describes sensitivity of the mathematical answer; convergence describes the algorithm. Floating-point rounding, poor scaling, and derivative errors can affect both practical stopping decisions and reported digits.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    A bisection bracket initially has width 16. What width remains after 3 bisection steps?

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  2. Problem 2

    Use one Newton step for f (x)=x2−4\displaystyle f\, \left(x\right) = x^{2} - 4, starting at x0=4\displaystyle x_{0} = 4. Find x1\displaystyle x_{1}.

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  3. Problem 3

    A computed value 21 for an exact value 20. Find the relative error as a fraction.

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  4. Problem 4

    For f (x)=x2+2\displaystyle f\, \left(x\right) = x^{2} + 2, can the ordinary Newton formula take a step from x0=0\displaystyle x_{0} = 0? Enter 1 for yes, 0 for no.

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  5. Problem 5New context

    A root-finding bracket initially has width 24. What width remains after 3 bisection steps?

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  6. Problem 6New context

    A square-root algorithm takes one Newton step for f (x)=x2−9\displaystyle f\, \left(x\right) = x^{2} - 9, starting at x0=6\displaystyle x_{0} = 6. Find x1\displaystyle x_{1}.

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  7. Problem 7New context

    A numerical calculation gives 31 for an exact value 30. Find the relative error as a fraction.

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  8. Problem 8New context

    A root solver uses f (x)=x2+3\displaystyle f\, \left(x\right) = x^{2} + 3, can the ordinary Newton formula take a step from x0=0\displaystyle x_{0} = 0? Enter 1 for yes, 0 for no.

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Open stage UNUM 1.1 in the student workspace →

STAGE UNUM 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Interpolation & Numerical Integration

Useful preparation: Bisection, Newton & Error

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Numerical integration replaces a difficult curve by simpler pieces whose areas can be calculated. Linear interpolation joins neighboring data points with a straight line; integrating these lines produces trapezoids. Simpson’s rule integrates quadratic interpolants on pairs of equal-width panels. The approximation quality depends on the function between sample points, not merely the precision of the given samples.

Interpolation
An interpolant matches the sampled function values exactly at specified nodes. For two distinct nodes, there is one line passing through both data points.
Composite trapezoidal rule
For equal spacing h, add the areas under linear interpolants. Interior node values receive full weight h because two neighboring half-weights meet there.
Composite Simpson rule
For an even number n of equal-width panels, use weights 1,4,2,4,...,2,4,1 multiplied by h3\displaystyle \frac{h}{3}. The rule is exact for every polynomial of degree at most three.

Conditions before calculation

  • Interpolation nodes must be distinct. Standard composite Simpson weights require equally spaced nodes and an even number of panels; they do not apply unchanged to arbitrary data spacing.
  • The usual trapezoidal error bound assumes a continuous second derivative; Simpson’s usual error bound assumes a continuous fourth derivative on the full interval.
  • A formula based only on samples cannot certify accuracy for an unrestricted function between nodes. Discontinuities, singularities, and rapid oscillations require additional analysis or adapted methods.

Why the method works

1. Build the local model

Between a and b, the interpolating line weights the two endpoint values by distance. Its area is width times average endpoint height.

L(x)=f(a)+x−ab−a(f(b)−f(a))L(x)=f(a)+\frac{x-a}{b-a}(f(b)-f(a))

2. Assemble consistent weights

Let xj=a+jh and h=b−an\displaystyle h = \frac{b - a}{n}. For the trapezoidal sum, count each shared interior endpoint only through the two half-contributions.

Tn=h[f(a)+f(b)2+∑j=1n−1f(xj)]T_n=h\left[\frac{f(a)+f(b)}2+\sum_{j=1}^{n-1}f(x_j)\right]

3. Bound what remains

If M2 bounds the absolute second derivative and M4 bounds the absolute fourth derivative, these inequalities quantify truncation error, excluding measurement and rounding errors.

∣I−Tn∣≤(b−a)h2M212,∣I−Sn∣≤(b−a)h4M4180|I-T_n|\le\frac{(b-a)h^2M_2}{12},\qquad |I-S_n|\le\frac{(b-a)h^4M_4}{180}

Worked examples

EXAMPLE 1

Use two equal panels to integrate x squared from 0 to 2 by trapezoidal and Simpson rules.

  1. The nodes are 0,1,2, step size is 1, and sampled values are 0,1,4.

  2. Compute both weighted sums.

    T2=12(0)+1+12(4)=3,S2=13(0+4⋅1+4)=83T_2=\frac12(0)+1+\frac12(4)=3,\qquad S_2=\frac13(0+4\cdot1+4)=\frac83

  3. Compare with the antiderivative and apply the trapezoidal error bound using M 2=2\displaystyle M\, 2 = 2.

    I=[x33]02=83,∣I−T2∣=13=2⋅12⋅212I=\left[\frac{x^3}{3}\right]_0^2=\frac83,\qquad |I-T_2|=\frac13=\frac{2\cdot1^2\cdot2}{12}

Conclusion: Simpson is exact for this quadratic; the trapezoidal estimate is an overestimate with error 13\displaystyle \frac{1}{3}.

EXAMPLE 2

Measurements are f (1)=4\displaystyle f\, \left(1\right) = 4 and f (3)=10\displaystyle f\, \left(3\right) = 10. Linearly interpolate at x=2\displaystyle x = 2 and integrate the line from 1 to 3.

  1. The slope is the change in value divided by the nonzero node separation.

    m=10−43−1=3m=\frac{10-4}{3-1}=3

  2. The line through the first node gives the interpolated value.

    L(x)=4+3(x−1),L(2)=7L(x)=4+3(x-1),\qquad L(2)=7

  3. Its integral is the trapezoid area.

    ∫13L(x) dx=(3−1)4+102=14\int_1^3L(x)\,dx=(3-1)\frac{4+10}{2}=14

Conclusion: The interpolant has value 7 at the midpoint and area 14. These are exact for the fitted line, not necessarily for the unknown underlying function.

Pitfalls & counterexamples

Tempting claim: A rule exact at the sample points must have the exact integral.

Correction: Matching finitely many points leaves the behavior between them undetermined.

Test the claim: On [0,2], zero and x squared times (x−1)\displaystyle \left(x - 1\right) squared times (x−2)\displaystyle \left(x - 2\right) squared agree at 0,1,2, but the latter has positive integral.

Check your understanding

Can standard composite Simpson weights be used with five equal panels?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

The standard construction groups panels in pairs, so the panel count must be even.

Connect this idea

Interpolation also supports data reconstruction and differential-equation solvers. Comparing refined grids can suggest an error trend, but a proven bound still requires assumptions about the function.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Interpolate (0,2) and (4,10). Using linear interpolation, find y at x=1\displaystyle x = 1.

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  2. Problem 2

    Use one trapezoid for f (x)=x2\displaystyle f\, \left(x\right) = x^{2} on [0,2]. What is the approximation?

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  3. Problem 3

    Use Simpson’s rule for f (x)=2 x2\displaystyle f\, \left(x\right) = 2\, x^{2} on [0,2], with midpoint 1. Find the result.

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  4. Problem 4

    A linear interpolant on [0,2] has endpoint values 1 and 7. What is its integral over that interval?

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  5. Problem 5New context

    A measurement table has points (0,3) and (4,11). Using linear interpolation, find y at x=1\displaystyle x = 1.

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  6. Problem 6New context

    A numerical area estimate uses one trapezoid for f (x)=x2\displaystyle f\, \left(x\right) = x^{2} on [0,3]. What is the approximation?

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  7. Problem 7New context

    A numerical accumulation uses Simpson’s rule for f (x)=3 x2\displaystyle f\, \left(x\right) = 3\, x^{2} on [0,2], with midpoint 1. Find the result.

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  8. Problem 8New context

    A calibrated linear interpolant on [0,2] has endpoint values 2 and 8. What is its integral over that interval?

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