1
$\begingroup$

I'm attempting to understand Interval bisection. I'm given a simple question in my textbook, and I can do the process easily, I just don't know when to stop. The question is "Use Interval bisection to find the positive root of $x^2 - 7 = 0$, correct to one decimal place" (basically find the square root of 7 to 1 dp)

This is the solution I'm given: enter image description here

How is it known that it is 2.6?

The last line shows that the root is between 2.640625(from (a+b)/2) and 2.65625(from b).
2.640625 rounds to 2.6 but
2.65625 rounds to 2.7

Surely I would have to keep going until both the upper and lower limit of the interval round to 2.6?

If it's just simple truncation why didn't the solution stop on the second last line?

(this is just a simple question, so it is as if you can't just do root 7 on a calculator)

  • 0
    Please post the exact wording of the question. Currently you've posted the part of the question that you already understand verbatim but only paraphrased the part of the question that you're asking about. That's a bad idea, since one likely cause of the problem is that you misunderstood the question, and we can't help you with that unless you quote the question. The phrase "to 1 decimal place" in the answer could refer either to $x^2-7=0$ or to $x=2.6$; in the former case, the solution would be correct, since the error in the equation rounds to $0.0$ for the first time in that iteration.2012-05-07
  • 0
    By the way, if this is an exact copy of the answer, there's another error in it; $\frac{f(a+b)}2$ should be $f\left(\frac{a+b}2\right)$; that's what the column actually contains.2012-05-07
  • 0
    @Joriki, sorry I managed to leave off the "correct to one dp", but now the question is exaclty what I've quoted.2012-05-07
  • 0
    Hmm, actually, you know the root is in $[2.640625, 2.65625]$; so indeed the root is $2.6$, correct to the first decimal place (although the root may be closer to 2.7 than to 2.6, for all we know at this point).2012-05-07
  • 0
    @David, I see how you got that, but it seems more common sense than mathematical. As if that was the way to get the answer then it should have stopped on the third line from the bottom when it is know to in [2.625, 2.6875]2012-05-07
  • 0
    It seems the solution is off no matter how you interpret it. If they mean "find the digit in the tenth's place of the decimal expansion of $\sqrt 7$", then, as you said, they've done too much work. If they mean "find $\sqrt 7$ with error at most $\pm0.05$", the answer"x=2.6" is not justified by the process, and another iteration is needed to justify it. Perhaps Joriki is right...2012-05-07

3 Answers 3

1

I would share your concern that this is not complete. As David Mitra says, one further iteration is needed.

In fact since at the third line you know $$2.625^2-7 \lt 0 \lt 2.75^2-7,$$ and so you know the rounded answer is $2.6$ or $2.7$ and could save some effort by checking $$2.65^2-7=+0.0225,$$ so $2.6$ is the rounded solution.

1

Judging from your comment under hardmath's answer (which it would have been a good idea to include in the original question), you're right that the solution is incorrect. If the task is to determine the result of rounding $\sqrt7$ to one decimal place, then, as you say, you'd need to keep going until both bounds of the interval yield the same result upon rounding.

0

It seems likely that the original problem was to repeat the bisection until the length of the enclosing interval was less than half of $0.1$. That would be one interpretation of finding the root within "one decimal place".

Note that the lengths of the enclosing intervals decrease regularly. You started with an interval of length $1$, and divided the length by $2$ with each iteration. Therefore at the sixth line of the table (fifth bisection), the interval $[a,b]$ has been reduced to length $2^{-5} \lt 0.05$, and this is the earliest step at which that criterion is met.

  • 0
    So I should keep going until I get to an interval width of 0.05 (if the question is 1dp, 0.005 for 2dp ,etc), and then what? Truncate or round to the first dp? What if the first dp is different on the upper and lower "bounds"2012-05-07
  • 0
    Also in an example given in the textbook it says "Since both 3.3125 and 3.34375 [$a$ and $b$ in the final row] when corrected to 1dp are 3.3: the answer is 3.3" (it's a different question hence the different numbers) So it seems to want me to keep going until the interval "bounds" round to the same number.2012-05-07
  • 0
    Clearly if you are trying to answer problems posed in a textbook, that's the best authority on the meaning of terms used there.2012-05-08