Binomial Series?

[huntsiekfind]

Use the binomial series to approximate ?6 to within .005 Hint is square root 6/2 = square root over the fraction 3/2, so square root 6 = 2 square root over fraction 3/2 = 2 square root over 1+1/2

Help would be greatly appreciated.

 

[BigGlenntheHeavy]

$\displaystyle \sqrt6 \ = \ 2.44948974278.... \ (TI-89).$

$\displaystyle \sqrt6 \ = \ 2\sqrt(1+\frac{1}{2}) \ = \ 2(1+\frac{1}{2})^{1/2}.$

$\displaystyle Hence, \ with \ only \ four \ iterations, \ we \ get:$

$\displaystyle 2[1+.25+\frac{[(.5)(-.5)(.25)]}{2}+\frac{[(.5)(-.5)(-1.5)(.125)]}{6}] \ = \ 2.453625$

$\displaystyle 2.453625-2.44948974278 \ = \ .00363525722, \ need \ about \ two \ more \ iterations.$

$\displaystyle Note: \ Binomial \ Series \ = \ (1+x)^{k} \ = \ 1+kx+\frac{k(k-1)x^{2}}{2!} \ + \ \frac{k(k-1)(k-2)x^{3}}{3!} \ + \ ...$

$\displaystyle Interval \ of \ Convergence \ is \ -1<x<1, \ and \ binomial \ series \ is \ valid \ for \ noninteger \ values \ of \ k.$

 

[Deleted member 4993]

I believe the error should be calculated from the upper bound of error in Taylor series.

 

[BigGlenntheHeavy]

You're right, Subhotosh Khan. I edited my above thread.