How do you find a one-decimal place approximation for #root5 26#?
Use one step of Newton's method to find
To find an approximation for the
#a_(i+1) = a_i + (n - a_i^5)/(5 a_i^4)#
In our case let
If we want more accuracy, just apply the formula again to get
This method is a particular case of Newton's method for finding a zero of a continuous differentiable function
In our case