How would you find the root of $\sqrt{26}$ by bisection method? A step by step solution would be greatly appreciated!
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3Here it's quite step by step so check it out Finding roots by binary search – Kaster Mar 01 '13 at 07:42
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Let $f(x)=x^2$. $$ f(5) = 25 < 26\\ f(6) = 36 > 26\\ f(5.5) = 30.25 > 26\\ f(5.2) = 27.04 > 26\\ f(5.1) = 26.01 > 26\\ f(5.05) = 25.5025 < 26\\ f(5.07) = 25.7049 <26\\ f(5.09) = 25.9081 < 26\\ f(5.095) = 25.959025 < 26\\ f(5.098) = 25.989604 < 26\\ f(5.099) = 25.999801 < 26\\ f(5.0995) = 26.00490025 > 26\\ \dots $$
Emanuele Paolini
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1note that $f(5.2)$ is not exactly bisection. A computer would use $f(5.25)$, $f(5.125)$ ... – John Dvorak Mar 01 '13 at 08:24
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But then the computer is working in binary (so it doesn't suffer working with more precision than is actually known) and computes with 2 bits of precision just as fast as 53 bits of precision anyways, so you can't really compare.... – Mar 01 '13 at 09:09