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Question : Find an approximate value of $\sqrt[3]{25}$ using Bisection Method.

Since it doesnt state the accuracy in the question,how many iterations am I going to do to get that approximate value? Also since it doesnt state the interval,how am I even going to apply the bisection method on this one?

user3543012
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2 Answers2

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the accuracy is usually 3 decimal points or so.

you need to choose the function and the interval (they didn't give you function as well). the natural function to such a problem is $f(x)=x^3-25$, and since $2<25^{1/3}<3$, you can take the interval $[2,3]$

tzoorp
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Let $f$ the function defined by: $\forall x\in\mathbb{R},\,f(x)=x^3-25$. We know that $f( \sqrt[3]{25})=0$ and the function has only one root since it's strictly increasing and continuous.

$f(0)$ for example is negative and $f(3)$ is positive so you can consider the interval $[0,3]$. You can take any other numbers.

The number of iterations is as you wish. As much it's bigger, the value you get is closer to $\sqrt[3]{25}$.

Scientifica
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  • Then to find f(p1) and so on,should I just do for example (p1)^3 ? – user3543012 Jan 03 '15 at 12:37
  • I'm sorry I didn't understand what you said. If you're talking about the next steps then $\dfrac{0+3}{2}=\dfrac{3}{2}$ so you calculate $f\left(\dfrac{3}{2}\right)=\left(\dfrac{3}{2}\right)^3-25$ if it's negative then consider the interval $\left[\dfrac{3}{2},3\right]$ if not then consider the interval $\left[0,\dfrac{3}{2}\right]$ and so on. – Scientifica Jan 03 '15 at 15:06
  • Lets suppose I take the interval [2;3].I calculate (2+3)/2 which is 2.5. We have 2.5^3 which is 15.6.Now the interval is [2.5;3].Here we have 2.75^ 3 which is 20.7.I keep doing this until I get the value of 25.Is this alright? – user3543012 Jan 03 '15 at 16:15
  • The way you're going is right, but you'll never get the value of 25. You'll just keep getting more and more close to it. $\sqrt[3]{25}=2,...$ and you reached $2,75$. In fact $\sqrt[3]{25}=2,92...$ so you still have to iterate. When you'll be satisfied with the result, then stop, like if you want to know the 2 numbers after , then once you get $2,92.$ you stop. If you want three then you must iterate more. If you write a program and execute it showing the value you get in each iteration you'll notice that as more iterating more numbers after , are correct. That's approximating $\sqrt[3]{25}$. – Scientifica Jan 03 '15 at 17:49
  • I understand ,thank you very much for your time and elaborated answer :) – user3543012 Jan 03 '15 at 18:31
  • It's a pleasure. Welcome to stackexchange :D – Scientifica Jan 03 '15 at 19:54