(Taylor Series/Polynomials, Exponential Function). (i) Find the representation of e3.49 in the FPA 6 by generating a suitable number of terms of the sequence ( Zn ) recursively defined by Z0=1 and Zn=Zn13.49n! where n1 is a natural number. As we did during the lectures, first try to estimate the number of steps to obtain the representation in question, and then carry out the calculations till the condition Zk=Zk1 will become true at some step k. The number Zk will then be the required representation e3.49 in the FPA6. (ii) Show your work by entering the terms Zn, suitable 6-digit floating-point numbers, you have calculated in (i) in the input fields below; if a particular input field is not necessary, please enter an asterisk in it: Z0= Z1=Z2= Z3= Z4= Z5= Z6= Z7= Z8= Z9= Z10= Z11= Z12= Z13= Z14= Z15= Z16= Z17= Z18= Z19= (iii) Consequently, the required representation of e3.49 in the FPA6 is (Taylor Series/Polynomials, Exponential Function). (i) Find the representation of e3.49 in the FPA 6 by generating a suitable number of terms of the sequence ( Zn ) recursively defined by Z0=1 and Zn=Zn13.49n! where n1 is a natural number. As we did during the lectures, first try to estimate the number of steps to obtain the representation in question, and then carry out the calculations till the condition Zk=Zk1 will become true at some step k. The number Zk will then be the required representation e3.49 in the FPA6. (ii) Show your work by entering the terms Zn, suitable 6-digit floating-point numbers, you have calculated in (i) in the input fields below; if a particular input field is not necessary, please enter an asterisk in it: Z0= Z1=Z2= Z3= Z4= Z5= Z6= Z7= Z8= Z9= Z10= Z11= Z12= Z13= Z14= Z15= Z16= Z17= Z18= Z19= (iii) Consequently, the required representation of e3.49 in the FPA6 is