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Undergraduate · Newton root iteration

Newton root iteration: A fractional starting estimate

Newton root iteration: investigate a fractional starting estimate with number whose root is sought a = 5; positive starting guess b = 1.5.

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Newton root iteration: A fractional starting estimate. Iteration: 0. Root estimate: 1.5. Residual x²−A: -2.75. Absolute root error: 0.736A positive-root iteration02.4203.57root estimateiteration n → · labeled axes rescale to this model
A≥1 and the initial guess is positive, so division by zero is avoided and the positive root is selected. The displayed iterates use the analytic Newton formula. This safe example does not establish convergence from arbitrary starts for arbitrary functions.

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Iteration
0
Root estimate
1.5
Residual x²−A
-2.75
Absolute root error
0.736

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Newton's method linearizes a nonlinear equation at the current estimate. For x²=A and a positive guess, the update averages x with A/x. The residual and the error in the root are related but not identical, so both are tracked rather than treating either one as the other. This investigation starts with Number whose root is sought A = 5; Positive starting guess b = 1.5. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.

A relationship to keep

xₙ₊₁=(xₙ+A/xₙ)/2; target √A

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  1. STEP 1

    Set up the mathematical model

    Specify a positive target A and a strictly positive starting estimate. The starting case is “A fractional starting estimate.”

  2. STEP 2

    Follow the changing quantity

    Apply seven Newton updates, plotting each estimate and the exact positive root for reference.

  3. STEP 3

    Explain and test the result

    Compare a guess below the root with one far above it. After the first update the arithmetic-geometric mean inequality places the estimate at or above the root.

Your turn to explain

Make a prediction. Test your reasoning.

Keep Number whose root is sought A = 5; Positive starting guess b = 1.5. Pause the timeline at 20%. Given iteration = 1, calculate root estimate, residual x²−a, absolute root error. Show the substitution into the displayed formula.

Use the values specified in the question. Reset restores the initial values for this investigation.

Compare your explanation

Starting at x₀=1.5, repeatedly use x←(x+5/x)/2 for 1 updates to obtain 2.416667. Substitution in the equation gives residual (2.416667)²−5=0.840278, while root error is |2.416667−√5|=0.180599. Results: Root estimate: 2.417; Residual x²−A: 0.84; Absolute root error: 0.181. Decimal values are rounded; retain the original parameters when checking.

Work through a full lesson

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Undergraduate · Calculus

Linearize a nonlinear root problem and inspect the update

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