Approximate Square Roots

We can find quite an accurate estimate of a square root using simple algebra. The square root of  
\[1300\]
  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.

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