Approximating square roots using binomial expansion. | square root expansion
Personally,Iwouldnthavedoneitthatway.SohereishowIwouldvedoneit:Method1:$$sqrt2=sqrt{1+1}=1+frac12-frac18+dotsapprox1+frac12-frac18=frac{11}8=1.375$$whichismuchclearertome,sinceitavoidshavingtotakedecimalsraisedtopowersandgivesyousomethingyoucaneasilydobyhand.$$1.3752=1.890625$$Obviouslyitapproachesthecorrectvalueasyoutakemoreterms.Method2:Thisiscalledfixed-pointiteration/Newtonsmethod,anditbasicallygoeslikethis:$$x=sqrt2impliesx2=2$$$$2x2=2+x2$$Dividebothsidesby$2x$andweget$$x=frac{2+x2}{2x}...
Personally, I wouldnt have done it that way. So here is how I wouldve done it:
Method 1:$$sqrt2=sqrt{1+1}=1+frac12-frac18+dotsapprox1+frac12-frac18=frac{11}8=1.375$$
which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.
$$1.3752=1.890625$$
Obviously it approaches the correct value as you take more terms.
Method 2:This is called fixed-point iteration/Newtons method, and it basically goes like this:
$$x=sqrt2implies x2=2$$
$$2x2=2+x2$$
Divide both sides by $2x$ and we get
$$x=frac{2+x2}{2x}$$
Now, interestingly, Im going to call the $x$s on the left $x_{n+1}$ and the $x$s on the right $x_n$, so
$$x_{n+1}=frac{2+(x_n)2}{2x_n}$$
and with $x_0approxsqrt2$, we will then have $x=lim_{n oinfty}x_n$. For example, with $x_0=1$,
$x_0=1$
$x_1=frac{2+12}{2(1)}=frac32=1.5$
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