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If, by reverse square, you mean square root, there is no simple answer.It is the number x, such that x^2 = 13.


You could use a formula based on logarithms:

Using natural logs, x = exp[0.5*ln(13)}

or, using base 10, x = 10^{0.5*log10(13)}


Alternatively, there is an iterative formula, based on the Newton-Raphson method:

Start with a reasonable guess, x(0).

Then for each n, calculate x(n+1) = x(n) - [x(n)^2 - 13]/2*x(n)

which simplifies to x(n+1) = x(n)/2 + 13/[2*x(n)].

With a reasonable starting value x(3) should be accurate to 3 decimal places.


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