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Directed Graphs, or Digraphs, of Relation

Mathematics · FE Reference Handbook section

Mathematics
0 formulas
10 exam-style examples
~45 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • of V called edges (or arcs). For edge (a, b), the vertex a is called the initial vertex and vertex b is called the terminal vertex. An

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 4 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

  • f(x)=x2−5f(x) = x^{2} - 5
  • x0=4x_{0} = 4

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(4+5/4)=2.62500x_{1} = ½(4 + 5/4) = 2.62500

  3. Iteration 2

    x2=½(2.62500+5/2.62500)=2.26488x_{2} = ½(2.62500 + 5/2.62500) = 2.26488

  4. Exact value — √5 = 2.23607

  5. Error — |x₂ − √5| = 0.028813

Answer:
x2=2.26488witherror0.028813x_{2} = 2.26488 with error 0.028813

Why the other options are there

  • 1.2500 (single division, no averaging)
  • -7.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 2
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (2)

Use Newton's algorithm on f(x) = x² − 7 with a starting value x₀ = 4 to obtain two improved estimates of √7, then report the error after the second iteration.

Given

  • f(x)=x2−7f(x) = x^{2} - 7
  • x0=4x_{0} = 4

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(4+7/4)=2.87500x_{1} = ½(4 + 7/4) = 2.87500

  3. Iteration 2

    x2=½(2.87500+7/2.87500)=2.65489x_{2} = ½(2.87500 + 7/2.87500) = 2.65489

  4. Exact value — √7 = 2.64575

  5. Error — |x₂ − √7| = 0.009140

Answer:
x2=2.65489witherror0.009140x_{2} = 2.65489 with error 0.009140

Why the other options are there

  • 1.7500 (single division, no averaging)
  • -5.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 3
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (3)

Use Newton's algorithm on f(x) = x² − 9 with a starting value x₀ = 3 to obtain two improved estimates of √9, then report the error after the second iteration.

Given

  • f(x)=x2−9f(x) = x^{2} - 9
  • x0=3x_{0} = 3

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(3+9/3)=3.00000x_{1} = ½(3 + 9/3) = 3.00000

  3. Iteration 2

    x2=½(3.00000+9/3.00000)=3.00000x_{2} = ½(3.00000 + 9/3.00000) = 3.00000

  4. Exact value — √9 = 3.00000

  5. Error — |x₂ − √9| = 0.000000

Answer:
x2=3.00000witherror0.000000x_{2} = 3.00000 with error 0.000000

Why the other options are there

  • 3.0000 (single division, no averaging)
  • 3.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 4
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (4)

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 3 to obtain two improved estimates of √4, then report the error after the second iteration.

Given

  • f(x)=x2−4f(x) = x^{2} - 4
  • x0=3x_{0} = 3

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(3+4/3)=2.16667x_{1} = ½(3 + 4/3) = 2.16667

  3. Iteration 2

    x2=½(2.16667+4/2.16667)=2.00641x_{2} = ½(2.16667 + 4/2.16667) = 2.00641

  4. Exact value — √4 = 2.00000

  5. Error — |x₂ − √4| = 0.006410

Answer:
x2=2.00641witherror0.006410x_{2} = 2.00641 with error 0.006410

Why the other options are there

  • 1.3333 (single division, no averaging)
  • -2.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 5
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (5)

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 6 to obtain two improved estimates of √6, then report the error after the second iteration.

Given

  • f(x)=x2−6f(x) = x^{2} - 6
  • x0=6x_{0} = 6

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(6+6/6)=3.50000x_{1} = ½(6 + 6/6) = 3.50000

  3. Iteration 2

    x2=½(3.50000+6/3.50000)=2.60714x_{2} = ½(3.50000 + 6/3.50000) = 2.60714

  4. Exact value — √6 = 2.44949

  5. Error — |x₂ − √6| = 0.157653

Answer:
x2=2.60714witherror0.157653x_{2} = 2.60714 with error 0.157653

Why the other options are there

  • 1.0000 (single division, no averaging)
  • -24.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 6
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (6)

Use Newton's algorithm on f(x) = x² − 9 with a starting value x₀ = 6 to obtain two improved estimates of √9, then report the error after the second iteration.

Given

  • f(x)=x2−9f(x) = x^{2} - 9
  • x0=6x_{0} = 6

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(6+9/6)=3.75000x_{1} = ½(6 + 9/6) = 3.75000

  3. Iteration 2

    x2=½(3.75000+9/3.75000)=3.07500x_{2} = ½(3.75000 + 9/3.75000) = 3.07500

  4. Exact value — √9 = 3.00000

  5. Error — |x₂ − √9| = 0.075000

Answer:
x2=3.07500witherror0.075000x_{2} = 3.07500 with error 0.075000

Why the other options are there

  • 1.5000 (single division, no averaging)
  • -21.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 7
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (7)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 5 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

  • f(x)=x2−5f(x) = x^{2} - 5
  • x0=5x_{0} = 5

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(5+5/5)=3.00000x_{1} = ½(5 + 5/5) = 3.00000

  3. Iteration 2

    x2=½(3.00000+5/3.00000)=2.33333x_{2} = ½(3.00000 + 5/3.00000) = 2.33333

  4. Exact value — √5 = 2.23607

  5. Error — |x₂ − √5| = 0.097265

Answer:
x2=2.33333witherror0.097265x_{2} = 2.33333 with error 0.097265

Why the other options are there

  • 1.0000 (single division, no averaging)
  • -15.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 8
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (8)

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 3 to obtain two improved estimates of √6, then report the error after the second iteration.

Given

  • f(x)=x2−6f(x) = x^{2} - 6
  • x0=3x_{0} = 3

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(3+6/3)=2.50000x_{1} = ½(3 + 6/3) = 2.50000

  3. Iteration 2

    x2=½(2.50000+6/2.50000)=2.45000x_{2} = ½(2.50000 + 6/2.50000) = 2.45000

  4. Exact value — √6 = 2.44949

  5. Error — |x₂ − √6| = 0.000510

Answer:
x2=2.45000witherror0.000510x_{2} = 2.45000 with error 0.000510

Why the other options are there

  • 2.0000 (single division, no averaging)
  • 0.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 9
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (9)

Use Newton's algorithm on f(x) = x² − 3 with a starting value x₀ = 4 to obtain two improved estimates of √3, then report the error after the second iteration.

Given

  • f(x)=x2−3f(x) = x^{2} - 3
  • x0=4x_{0} = 4

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(4+3/4)=2.37500x_{1} = ½(4 + 3/4) = 2.37500

  3. Iteration 2

    x2=½(2.37500+3/2.37500)=1.81908x_{2} = ½(2.37500 + 3/2.37500) = 1.81908

  4. Exact value — √3 = 1.73205

  5. Error — |x₂ − √3| = 0.087028

Answer:
x2=1.81908witherror0.087028x_{2} = 1.81908 with error 0.087028

Why the other options are there

  • 0.7500 (single division, no averaging)
  • -9.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

Example 10
Newton's algorithm applied to a square root — Directed Graphs, or Digraphs, of Relation (10)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 3 to obtain two improved estimates of √5, then report the error after the second iteration.

Given

  • f(x)=x2−5f(x) = x^{2} - 5
  • x0=3x_{0} = 3

Find

x₁, x₂ and the absolute error |x₂ − √N|

Start with the thinking

  • Newton's algorithm is x_{k+1} = x_k − f(x_k)/f′(x_k); for f = x² − N it collapses to the averaging form.
  • Convergence is quadratic, so two iterations already give several correct digits.

Step-by-step solution

  1. Formula

    xk+1=xk−f(xk)/f′(xk)=½(xk+N/xk)x_{k+1} = x_k - f(x_k)/f'(x_k) = ½(x_k + N/x_k)

  2. Iteration 1

    x1=½(3+5/3)=2.33333x_{1} = ½(3 + 5/3) = 2.33333

  3. Iteration 2

    x2=½(2.33333+5/2.33333)=2.23810x_{2} = ½(2.33333 + 5/2.33333) = 2.23810

  4. Exact value — √5 = 2.23607

  5. Error — |x₂ − √5| = 0.002027

Answer:
x2=2.23810witherror0.002027x_{2} = 2.23810 with error 0.002027

Why the other options are there

  • 1.6667 (single division, no averaging)
  • -1.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Directed Graphs, or Digraphs, of Relation

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