OpenStudy (anonymous):

6 years ago
OpenStudy (anonymous):

There are two ways you can do this. One: take the fourth derivative of the function, and then compute f''''(0)/4! Two: recognize a Maclaurin series takes the form (a0 + a1*x + a2*x^2/2 + a3*x^3/6 + a4*x^4/24 + ...). Then set that equal to the function: 1/(1-sin(x^2)) = (a0 + a1*x + a2*x^2/2 + a3*x^3/6 + a4*x^4/24 + ...) 1 = (1-sin(x^2)) * (a0 + a1*x + a2*x^2/2 + a3*x^3/6 + a4*x^4/24 + ...) Then replace sin(x^2) with the first four coefficients of the MacLauren series for sin(u), plugging in x^2 for u. From here you should be able to solve for each power of x one at a time. That is, you look at the constant term 1 on the left, and set it equal to the (1-s0)*a0 on the right, solving for a0. Then solve for a1, and so on, down to a4. This solution seems more in depth but may actually be less computationally heavy.

6 years ago
OpenStudy (anonymous):

Hey thanks for the answer! but this was actually a multiple choice question with choices of: -1/2, 0, 1/2, 1, and -1. would there be an easier way to compute the answer without having to solve up to the 4th derivative? Otherwise it seems like it would take too much time to solve this one multiple choice question.

6 years ago
OpenStudy (anonymous):

Sorry but I cannot think of an easier way.

6 years ago