One can expand the function a differentiable function, f(x), around xo as a Taylor power series as follows: Or in short 1 (x) = f(xo) + '(x)(x xo) + f'(x)(x xo) + ... = N f(x) = f(x0) + [=1/1 if(i) (x)(x xo)" - n=1 Given f(x) ex and x = 0, show that the Taylor series expansion of e* around xo N ex = 1+ n=1 1 n! xn +0(xN+1) = 0 is given by a) Write a code to evaluate the series expansion of e* for any x up to any order: N = 1, 2, 3, 4, ... 10 (sequentially increasing the order up to N = 10) b) Using this code, evaluate the values of the series expansion for values of x = 0.1, 1.0 and 10.0 for each N. c) Using the numpy provided exp(x) as the 'exact' value of e* calculate the 'true' error due to truncation of series at specific N value, plot the 'true' error for each x and N. d) Comment on the 'true' error as a function of N and the value of x.