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At some general time k∆t

Mathematics · FE Reference Handbook section

Mathematics
3 formulas
10 exam-style examples
~51 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • which can be used with starting condition xo to solve recursively for x(∆t), x(2∆t), …, x(n∆t).
  • The method can be extended to nth order differential equations by recasting them as n first-order equations.
  • which can be expressed as the recursive equation

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 — At some general time k∆t

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 5 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=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+4/5)=2.90000x_{1} = ½(5 + 4/5) = 2.90000

  3. Iteration 2

    x2=½(2.90000+4/2.90000)=2.13966x_{2} = ½(2.90000 + 4/2.90000) = 2.13966

  4. Exact value — √4 = 2.00000

  5. Error — |x₂ − √4| = 0.139655

Answer:
x2=2.13966witherror0.139655x_{2} = 2.13966 with error 0.139655

Why the other options are there

  • 0.8000 (single division, no averaging)
  • -16.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

Example 2
Newton's algorithm applied to a square root — At some general time k∆t (2)

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 → At some general time k∆t

Example 3
Newton's algorithm applied to a square root — At some general time k∆t (3)

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 → At some general time k∆t

Example 4
Newton's algorithm applied to a square root — At some general time k∆t (4)

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 → At some general time k∆t

Example 5
Newton's algorithm applied to a square root — At some general time k∆t (5)

Use Newton's algorithm on f(x) = x² − 5 with a starting value x₀ = 2 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=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+5/2)=2.25000x_{1} = ½(2 + 5/2) = 2.25000

  3. Iteration 2

    x2=½(2.25000+5/2.25000)=2.23611x_{2} = ½(2.25000 + 5/2.25000) = 2.23611

  4. Exact value — √5 = 2.23607

  5. Error — |x₂ − √5| = 0.000043

Answer:
x2=2.23611witherror0.000043x_{2} = 2.23611 with error 0.000043

Why the other options are there

  • 2.5000 (single division, no averaging)
  • 3.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

Example 6
Newton's algorithm applied to a square root — At some general time k∆t (6)

Use Newton's algorithm on f(x) = x² − 3 with a starting value x₀ = 3 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=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+3/3)=2.00000x_{1} = ½(3 + 3/3) = 2.00000

  3. Iteration 2

    x2=½(2.00000+3/2.00000)=1.75000x_{2} = ½(2.00000 + 3/2.00000) = 1.75000

  4. Exact value — √3 = 1.73205

  5. Error — |x₂ − √3| = 0.017949

Answer:
x2=1.75000witherror0.017949x_{2} = 1.75000 with error 0.017949

Why the other options are there

  • 1.0000 (single division, no averaging)
  • -3.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

Example 7
Newton's algorithm applied to a square root — At some general time k∆t (7)

Use Newton's algorithm on f(x) = x² − 6 with a starting value x₀ = 5 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=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+6/5)=3.10000x_{1} = ½(5 + 6/5) = 3.10000

  3. Iteration 2

    x2=½(3.10000+6/3.10000)=2.51774x_{2} = ½(3.10000 + 6/3.10000) = 2.51774

  4. Exact value — √6 = 2.44949

  5. Error — |x₂ − √6| = 0.068252

Answer:
x2=2.51774witherror0.068252x_{2} = 2.51774 with error 0.068252

Why the other options are there

  • 1.2000 (single division, no averaging)
  • -14.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

Example 8
Newton's algorithm applied to a square root — At some general time k∆t (8)

Use Newton's algorithm on f(x) = x² − 7 with a starting value x₀ = 6 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=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+7/6)=3.58333x_{1} = ½(6 + 7/6) = 3.58333

  3. Iteration 2

    x2=½(3.58333+7/3.58333)=2.76841x_{2} = ½(3.58333 + 7/3.58333) = 2.76841

  4. Exact value — √7 = 2.64575

  5. Error — |x₂ − √7| = 0.122660

Answer:
x2=2.76841witherror0.122660x_{2} = 2.76841 with error 0.122660

Why the other options are there

  • 1.1667 (single division, no averaging)
  • -23.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

Example 9
Newton's algorithm applied to a square root — At some general time k∆t (9)

Use Newton's algorithm on f(x) = x² − 4 with a starting value x₀ = 6 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=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+4/6)=3.33333x_{1} = ½(6 + 4/6) = 3.33333

  3. Iteration 2

    x2=½(3.33333+4/3.33333)=2.26667x_{2} = ½(3.33333 + 4/3.33333) = 2.26667

  4. Exact value — √4 = 2.00000

  5. Error — |x₂ − √4| = 0.266667

Answer:
x2=2.26667witherror0.266667x_{2} = 2.26667 with error 0.266667

Why the other options are there

  • 0.6667 (single division, no averaging)
  • -26.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

Example 10
Newton's algorithm applied to a square root — At some general time k∆t (10)

Use Newton's algorithm on f(x) = x² − 8 with a starting value x₀ = 2 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=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+8/2)=3.00000x_{1} = ½(2 + 8/2) = 3.00000

  3. Iteration 2

    x2=½(3.00000+8/3.00000)=2.83333x_{2} = ½(3.00000 + 8/3.00000) = 2.83333

  4. Exact value — √8 = 2.82843

  5. Error — |x₂ − √8| = 0.004906

Answer:
x2=2.83333witherror0.004906x_{2} = 2.83333 with error 0.004906

Why the other options are there

  • 4.0000 (single division, no averaging)
  • 6.0000 (forgot to divide by f′)

Reference: FE Reference Handbook — Mathematics → At some general time k∆t

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