tag:blogger.com,1999:blog-1364629727248557844.post2720963547231309547..comments2010-03-30T10:15:32.300+11:00Comments on The Mathematical Bricoleur: Incredibly simple ways to get rational approximations to square rootsGBhttp://www.blogger.com/profile/01100913350264083875noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-1364629727248557844.post-77798791886973249632009-04-07T13:23:00.000+10:002009-04-07T13:23:00.000+10:00If you're finding sqrt(7), you might start with 2 ...If you're finding sqrt(7), you might start with 2 if, say, you're starting off your thinking from a continued-fraction-approximation perspective, but it doesn't matter much to the algorithm, which tends to improve any reasonable approximation and will take 2:1 to 3:1 in the first few steps anyway.<BR/><BR/>on the rate of improvement:<BR/><BR/>If you compute the error after each addition step and take ratios of absolute values of successive errors, the ratio rapidly goes to that ratio. I haven't done the algebra to prove it converges at that rate.GBhttps://www.blogger.com/profile/01100913350264083875noreply@blogger.comtag:blogger.com,1999:blog-1364629727248557844.post-13341098329132488422009-04-02T08:23:00.000+11:002009-04-02T08:23:00.000+11:00I don't know how common the TI 89 [Texas Instr...I don't know how common the TI 89 [Texas Instruments] calculator is outside the US, but here's a program to do this method:<BR/><BR/>arch(r,a,b)<BR/>Prgm<BR/>1->n<BR/>While n<8<BR/> a+r*b->t<BR/> a+b->z<BR/> gcd(t,z)->g<BR/> t/g->a<BR/> z/g->b<BR/> Disp a,b<BR/> n+1->n<BR/>EndWhile<BR/>EndPrgm<BR/><BR/>Then from the home screen you need to type: arch(r,a,b), where r is the number you want the square root of, and a and b are your starting numbers. Of course a and b should be 3,1 if you're finding say, sqrt(7), since the result is closer to 3 than to 2.<BR/><BR/>Question: Can you explain the bit about the error improves by a factor approaching 2 + sqrt(3)?Peterhttps://www.blogger.com/profile/04875668909485036128noreply@blogger.com