deep0199 Posted June 15, 2005 Share Posted June 15, 2005 What is the best way to manually find the square root of a positive rational number? I understand that it will be an approximation, what what is the way to get the closest possible answer rounded to 2-3 decimal places. Link to comment Share on other sites More sharing options...

fuhrerkeebs Posted June 15, 2005 Share Posted June 15, 2005 You could just memorize the roots of a bunch of prime numbers, and just factor the number, pull out all pairs of two and just multiply the roots of the remaining primes...or you could use taylor expansion. Link to comment Share on other sites More sharing options...

Dapthar Posted June 15, 2005 Share Posted June 15, 2005 What is the best way to manually find the square root of a positive rational number? It depends on what you mean by "best". If you want a simple' date=' and quick way to get a reasonably accurate estimate of a square root, try the first method on this page: http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html If you want a more accurate approach, you could use the Binomial Theorem, and the details of how to use such a method are also on the aforementioned page, under the heading [b']Square Roots Using Infinite Series[/b]. Link to comment Share on other sites More sharing options...

Ducky Havok Posted June 16, 2005 Share Posted June 16, 2005 If you only want it to a couple decimal places, I find this method useful. [math]\frac{1}{2\sqrt{x}}dx+ \sqrt{x}[/math] Here's an example: [math]\sqrt{24.8}[/math]. Pick the closest perfect square, in this case 25. So x=25, and dx=-.2. When you put the numbers in you get -.02+5, which is 4.98. When you put it in a calculator its really 4.97996, so it's pretty close. Link to comment Share on other sites More sharing options...

DQW Posted June 17, 2005 Share Posted June 17, 2005 Often, you could use a binomial approximation (as suggested previously) : choose the nearest perfect square to x; let it be n^{2}, where n is some convenient rational or known square root. Let x = n^{2} + d, where |d| << n^{2} Then [math]\sqrt{x} = \sqrt{n^2 + d} = n*\sqrt{1 + \frac{d}{n^2}} \approx n*(1+ \frac{d}{2n^2}) [/math] Link to comment Share on other sites More sharing options...

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