Final answer:
To evaluate the square root of 16.3 using the Taylor series, we expand the square root function around the point a=16, using up to the second derivative of the function to get an approximate value.
Step-by-step explanation:
The question requests an evaluation of square root of 16.3 using three terms of the Taylor series. The Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. To simplify, we can consider the square root function f(x) = sqrt(x) expanded around a where x=a is a point where we know f(a) and its derivatives. For this example, we may choose a=16 since that's a perfect square close to 16.3, and we know that sqrt(16) = 4.
To find the approximation using three terms (the function value, the first derivative, and the second derivative at the point a), we set up the expansion as follows:
f(x) = f(a) + f'(a)*(x-a)/1! + f''(a)*(x-a)^2/2!
For f(x) = sqrt(x), the first derivative f'(x) = 1/(2*sqrt(x)) and the second derivative f''(x) = -1/(4*x*sqrt(x)). Substituting a=16 and x=16.3 into the formula provides the approximation for sqrt(16.3).