9514 1404 393
Answer:
see attached
Step-by-step explanation:
For some integer N whose root we want to estimate, let s, r, f be defined as follows:
N = s² +r . . . . . . the integer N is the sum of the nearest smaller square and a remainder
√N = s +f . . . . . the square root of N is the root of that square plus a fraction
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If we square the second equation, we get ...
N = (s +f)² = s² +2fs +f²
Then the fraction f can be found to be representable as a continued fraction ...
N -s² = 2fs +f²
f(2s +f) = N -s² . . . . . factor out f
The first equation tells us N -s² = r, so this can become the iterative formula ...
f = r/(2s +f)
If we add the integer part of the root, we get ...
N = s +f
N ≈ s +r/(2s +f) . . . . . . where f is an approximation of the fractional part of the root.
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We can get a "linear approximation" using ...
f = r/(2s +1)
If we use this to improve the approximation, we get
[tex]\boxed{\sqrt{N}\approx s+\cfrac{r}{2s+\cfrac{r}{2s+1}}}[/tex]
For the numbers on the list in this problem, this approximation gives a result accurate to thousandths or better.
The attached spreadsheet does this calculation and rounds to hundredths. The value in the "frac" column is the linear approximation fraction shown above. Here is an example of the calculation for √27:
[tex]\sqrt{27}\approx 5+\cfrac{2}{2\cdot5+\cfrac{2}{2\cdot5+1}}=5+\cfrac{2}{10+\cfrac{2}{11}}=5+\cfrac{2\cdot11}{112}=5\dfrac{11}{56}\\\\\approx5.19643\quad\text{approximation with continued fraction}\\\\\sqrt{27}=5.19615\quad\text{calculator result for root}[/tex]
