Let's say you need to nd the square root of 4.1 You pull out your hand calculator for the Calculator app on your phone}, type in v4.7, and you have the answer. But how did the calculator know it? In fact, we take modern computational conveniences for granted. The hand calculator for the corresponding app]I is a fantastic product of engineering; and in this exercise, we're going to apply one of the techniques they use to calculate square roots: Taylor Series. Taylor series allow us to write any differentiable function find as an innite series of the form ; (I) = f (a) + J\" (a) (x a) + if\") (z w}? + m (a: aJ3+...+f""\"'l [I am... for values of x near x = a. Note that f in} {a} represents the n'th derivative of the function ffx} evaluated at x = a. Clearly, the details of a Taylor series expansion depend upon the function involved; for the square root funcon f {I} = , it can be shown that: - The Taylor series for f (I) = 3/5 will be an alternating series; that is, the signs of consecutive terms will alternate from positive to negative and back again, with the pattern repeating indenitely. - If we assume |$ all {i 1, then the absolute value of each term is smaller than the one before. Combined, these two facts tell us that the Taylor series for f {I} = ff converges; and if we tmncate the series after n terms, then the error in our approximation will be smaller than the absolute value of term n+1 in the series. . Determine the number of terms in the corresponding Taylor series expansion required to approximate the value of v4.7 to within 10 3, and state the resulting approximate value of V4.7 . . Use the absolute value of the first term you omitted to estimate the error in your approximation. Use this table to organize your work: nth term Evaluate nth of Taylor Cumulative Function function term Series sum of Approximation n and and n! (x - a)" of Error evaluated Taylor accurate to derivatives derivatives Taylor estimate at value Series f (?) (x) within 10^-5 Series f (?) (a) of terms interest 2 3 4 5