Newton’s method practice question 16
A student uses two steps of Newton’s method to approximate , as shown. Find the first line that is wrong.
Check each line against the recipe for a Newton step: the value and the slope are both taken at the current estimate.
As a check on the result, is about 1.77, well short of 2.
For each step, note the point at which the slope in the denominator was computed. Is it the current estimate?
Theory UBC Math 100 §10.1Key results
Corrected working
Strategy
Take each step with the value and the slope at the current estimate.
Working
f(1) = −1 and f′(1) = 3.
Value and slope at the new estimate.
(10/27)(3/16) = 30/432 = 5/72. For comparison, the cube root of 2 is 1.2599 to four places.
Answer
The answer is x two equals 91 over 72, about 1.264.
Check
(91/72)³ ≈ 2.019, close to 2, while the student’s 98/81 cubes to only about 1.77.
Where this goes wrong
Keeping the first slope for every step. That is a different and slower method; Newton’s method redraws the tangent line at each new estimate.
The idea behind it
The Newton’s method lesson, a figure to move, in a few steps
Leans on
The power rule, from Differentiation rules
Try a similar question
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