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Newton root iteration · Number whose root is sought A=15; Positive starting guess b=7.5
Perform exactly n=4 updates. Find x₄, its equation residual x₄²−A, and its absolute error relative to √A. Givens: Number whose root is sought A=15; Positive starting guess b=7.5.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Seek the positive root of x²=15 using Newton's method with starting estimate x₀=7.5. Perform exactly n=4 updates. Find x₄, its equation residual x₄²−A, and its absolute error relative to √A.
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A starting point
Repeat x←(x+A/x)/2 four times without rounding intermediate iterates. Residual and root error measure different things.
Work through the reasoning
Step 1
Identify the model and target
Seek the positive root of x²=15 using Newton's method with starting estimate x₀=7.5. Perform exactly n=4 updates. The governing relation is xₙ₊₁=(xₙ+A/xₙ)/2; target √A. Repeat x←(x+A/x)/2 four times without rounding intermediate iterates. Residual and root error measure different things.
Step 2
Substitute and calculate
Starting at x₀=7.5, repeatedly use x←(x+15/x)/2 for 4 updates to obtain 3.872983. Substitution in the equation gives residual (3.872983)²−15=0.000001, while root error is |3.872983−√15|=0.
Step 3
Check the mathematical meaning
Positivity prevents division by zero. After the first update AM–GM puts the estimate at or above √15. The identity |x²−A|=|x−√A|·|x+√A| explains the different error readouts. Iteration: 4; Root estimate: 3.873; Residual x²−A: 0; Absolute root error: 0. Decimal values are rounded, so use unrounded intermediate values.
The answer
Starting at x₀=7.5, repeatedly use x←(x+15/x)/2 for 4 updates to obtain 3.872983. Substitution in the equation gives residual (3.872983)²−15=0.000001, while root error is |3.872983−√15|=0. Positivity prevents division by zero. After the first update AM–GM puts the estimate at or above √15. The identity |x²−A|=|x−√A|·|x+√A| explains the different error readouts. Animation check: Iteration: 4; Root estimate: 3.873; Residual x²−A: 0; Absolute root error: 0. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
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. Seek the positive root of x²=15 using Newton's method with starting estimate x₀=7.5. Perform exactly n=4 updates. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
xₙ₊₁=(xₙ+A/xₙ)/2; target √A
03 · Reflect and transfer
Explain what changes and why.
Why can a small residual and a small root error have different numerical sizes even for this safely convergent example?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.