User:Michael Retriever/Roots

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Newton presented this manipulation of his binomial theorem, stating that it was a much shorter way to extract roots.


[edit] Extraction of roots

Given the binomial theorem where r is any real number


(x+y)^r=\sum_{k=0}^\infty{r \choose k}x^{(r-k)}y^k\,


{r \choose k} = \frac{1}{k!} \prod_{m=0}^{k-1}{(r - m)}\,


we can equally use it for any rational number. For this explanation we'll use r = 1/b


\sqrt[b]{x+y}=\,


=(x+y)^{1/b}=\sum_{k=0}^\infty{1/b \choose k}x^{(1/b-k)}y^k=\,


=x^{(1/b)}+{\frac{1/b}{1!}}x^{(1/b-1)}y+{\frac{(1/b)(1/b-1)}{2!}}x^{(1/b-2)}y^2+\dots\,


If we apply this to \sqrt[b]{1-x} we get


\sqrt[b]{1-x}=1+{\frac{1/b}{1!}}(-x)+{\frac{(1/b)(1/b-1)}{2!}}(-x)^2+{\frac{(1/b)(1/b-1)(1/b-2)}{3!}}(-x)^3+\dots\,


Now let's check the way to extract roots with an example. We know that


\sqrt{7}=\sqrt{9-2}=\sqrt{9\cdot\left(\frac{9-2}{9}\right)}=\,


=\sqrt{9\cdot\left(1-\frac{2}{9}\right)}=\sqrt{9}\cdot\sqrt{1-\frac{2}{9}}=3\cdot\sqrt{1-\frac{2}{9}}\,


Using the previous infinite expansion with \sqrt{1-(2/9)} we get that


\sqrt{1-(2/9)}=1+{\frac{1}{(1!)(2)}}\left(\frac{-2}{9}\right)+\frac{-1}{(2!)(2^2)}\left(\frac{2^2}{9^2}\right)+\frac{3}{(3!)(2^3)}\left(\frac{-2^3}{9^3}\right)+\frac{-3\cdot5}{(4!)(2^4)}\left(\frac{2^4}{9^4}\right)+\dots=\,


=1-\frac{1}{9}-\frac{1}{162}-\frac{1}{1458}-\frac{5}{52488}-\frac{7}{472392}-\frac{7}{2834352}-\dots


thus


\sqrt{7}=3\cdot\left(1-\frac{1}{9}-\frac{1}{162}-\frac{1}{1458}-\frac{5}{52488}-\frac{7}{472392}-\frac{7}{2834352}-\dots\right)


The true square root of 7 is 2.64575131.... If we make an approximation of the formula with the fractions shown, we get that the square root of 7 is 2.64575289..., which is correct up to the 5th digit after the dot. That is a fairly good approximation taking into account that we didn't need a calculator to do it.