Estimate the cube root of 270

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Question from Tom, a student:

Question 1.

(a) By plotting suitable graphs, estimate to one decimal place the cube root of 270.
(b) With reference to your answer to part (a), use 8 iterations of a Bisection Search to refine your estimate. Use the nearest whole numbers either side of your estimate from part (a) as starting values.
(c) Using either of your starting values from part (b) as first guess, use the Newton-Raphson method to find the true value of the root (to 6 decimal places). Repeat using the other starting value from part (b) and compare the two results.

Hi Tom,

I would plot f(x) = x³ - 270. Suppose x* is the point that satisfies f(x*) = 0. When I did this I found that x* is some point between x = 6.4 and x = 6.5. f(6.4) < 0 and f(6.5) > 0. In fact the graph shows me that x* is closer to 6.4 then 6.5.

Now start your bisection search.

Iteration 1:
Try the point midway between 6.4 and 6.5, namely 6.45. Is f(6.45) < 0? If so then x* is between 6.45 and 6.50. Is f(6.45) > 0? If so then x* is between 6.40 and 6.45.

Iteration 2:
If x* is between 6.40 and 6.45 check the sign of f(6.425). The sign will tell you whether x* is between 6.400 and 6.425 or between 6.425 and 6.450.
If x* is between 6.45 and 6.50 check the sign of f(6.475). The sign will tell you whether x* is between 6.450 and 6.475 or between 6.475 and 6.500.

Continue for 8 iterations.

For part (c) use 6.4 or 6.5 as the starting value and use the Newton-Raphson method to approximate x*.

Harley

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