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Difference Equations

Mathematics · FE Reference Handbook section

Mathematics
1 formulas
10 exam-style examples
~47 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • Any system whose input v(t) and output y(t) are defined only at the equally spaced intervals

Core formulas for this FE topic

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

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 — Difference Equations

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 2 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=2x_{0} = 2

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(2+6/2)=2.50000x_{1} = ½(2 + 6/2) = 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

  • 3.0000 (single division, no averaging)
  • 4.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 2
Newton's algorithm applied to a square root — Difference Equations (2)

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 → Difference Equations

Example 3
Newton's algorithm applied to a square root — Difference Equations (3)

Use Newton's algorithm on f(x) = x² − 7 with a starting value x₀ = 3 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=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+7/3)=2.66667x_{1} = ½(3 + 7/3) = 2.66667

  3. Iteration 2

    x2=½(2.66667+7/2.66667)=2.64583x_{2} = ½(2.66667 + 7/2.66667) = 2.64583

  4. Exact value — √7 = 2.64575

  5. Error — |x₂ − √7| = 0.000082

Answer:
x2=2.64583witherror0.000082x_{2} = 2.64583 with error 0.000082

Why the other options are there

  • 2.3333 (single division, no averaging)
  • 1.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 4
Newton's algorithm applied to a square root — Difference Equations (4)

Use Newton's algorithm on f(x) = x² − 9 with a starting value x₀ = 4 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=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+9/4)=3.12500x_{1} = ½(4 + 9/4) = 3.12500

  3. Iteration 2

    x2=½(3.12500+9/3.12500)=3.00250x_{2} = ½(3.12500 + 9/3.12500) = 3.00250

  4. Exact value — √9 = 3.00000

  5. Error — |x₂ − √9| = 0.002500

Answer:
x2=3.00250witherror0.002500x_{2} = 3.00250 with error 0.002500

Why the other options are there

  • 2.2500 (single division, no averaging)
  • -3.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 5
Newton's algorithm applied to a square root — Difference Equations (5)

Use Newton's algorithm on f(x) = x² − 9 with a starting value x₀ = 5 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=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+9/5)=3.40000x_{1} = ½(5 + 9/5) = 3.40000

  3. Iteration 2

    x2=½(3.40000+9/3.40000)=3.02353x_{2} = ½(3.40000 + 9/3.40000) = 3.02353

  4. Exact value — √9 = 3.00000

  5. Error — |x₂ − √9| = 0.023529

Answer:
x2=3.02353witherror0.023529x_{2} = 3.02353 with error 0.023529

Why the other options are there

  • 1.8000 (single division, no averaging)
  • -11.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 6
Newton's algorithm applied to a square root — Difference Equations (6)

Use Newton's algorithm on f(x) = x² − 3 with a starting value x₀ = 5 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=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+3/5)=2.80000x_{1} = ½(5 + 3/5) = 2.80000

  3. Iteration 2

    x2=½(2.80000+3/2.80000)=1.93571x_{2} = ½(2.80000 + 3/2.80000) = 1.93571

  4. Exact value — √3 = 1.73205

  5. Error — |x₂ − √3| = 0.203663

Answer:
x2=1.93571witherror0.203663x_{2} = 1.93571 with error 0.203663

Why the other options are there

  • 0.6000 (single division, no averaging)
  • -17.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 7
Newton's algorithm applied to a square root — Difference Equations (7)

Use Newton's algorithm on f(x) = x² − 3 with a starting value x₀ = 2 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=2x_{0} = 2

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(2+3/2)=1.75000x_{1} = ½(2 + 3/2) = 1.75000

  3. Iteration 2

    x2=½(1.75000+3/1.75000)=1.73214x_{2} = ½(1.75000 + 3/1.75000) = 1.73214

  4. Exact value — √3 = 1.73205

  5. Error — |x₂ − √3| = 0.000092

Answer:
x2=1.73214witherror0.000092x_{2} = 1.73214 with error 0.000092

Why the other options are there

  • 1.5000 (single division, no averaging)
  • 1.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 8
Newton's algorithm applied to a square root — Difference Equations (8)

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 4 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=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+4/4)=2.50000x_{1} = ½(4 + 4/4) = 2.50000

  3. Iteration 2

    x2=½(2.50000+4/2.50000)=2.05000x_{2} = ½(2.50000 + 4/2.50000) = 2.05000

  4. Exact value — √4 = 2.00000

  5. Error — |x₂ − √4| = 0.050000

Answer:
x2=2.05000witherror0.050000x_{2} = 2.05000 with error 0.050000

Why the other options are there

  • 1.0000 (single division, no averaging)
  • -8.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 9
Newton's algorithm applied to a square root — Difference Equations (9)

Use Newton's algorithm on f(x) = x² − 8 with a starting value x₀ = 3 to obtain two improved estimates of √8, then report the error after the second iteration.

Given

  • f(x)=x2−8f(x) = x^{2} - 8
  • 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+8/3)=2.83333x_{1} = ½(3 + 8/3) = 2.83333

  3. Iteration 2

    x2=½(2.83333+8/2.83333)=2.82843x_{2} = ½(2.83333 + 8/2.83333) = 2.82843

  4. Exact value — √8 = 2.82843

  5. Error — |x₂ − √8| = 0.000004

Answer:
x2=2.82843witherror0.000004x_{2} = 2.82843 with error 0.000004

Why the other options are there

  • 2.6667 (single division, no averaging)
  • 2.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → Difference Equations

Example 10
Newton's algorithm applied to a square root — Difference Equations (10)

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 → Difference Equations

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