\[1300\]
is approximately 36 - \[36^2=1296\]
, so the square root is just a little bit more.We can find a better estimate by writing
\[\sqrt{1300}=36+x\]
where \[x\]
is small.Then
\[1300=(36+x)^2=1296+72x+x^2\]
(1)Since
\[x\]
is small, \[x^2\]
is very small and we can ignore it, so\[1300 \simeq 1296+72x\]
.\[4 \simeq 72x\]
.\[x \simeq \frac{4}{72} =0.0555555...\]
.Then
\[\sqrt{1300} \simeq 36.055555...\]
.We can improve this estimate by taking
\[x^2\]
into account.
From (1) we have \[x^2+72x-4=0 \rightarrow x=\frac{4-x^2}{72}\]
.We can use this as an iteration expression
\[x_{n+1}=\frac{4-x_n^2}{72}\]
, and by putting \[x_1=\frac{1}{18}\]
on the right hand side we get \[x_2=\frac{4-(1/18)^2}{72} \simeq 0.055513\]
, so that \[\sqrt{1300} \simeq 37.055513\]
.This is correct to 8 significant figures.