a. Find the macluarin polynomial of sin (x)

b. Find the maclaurin polynomial of sin (x^2)

c. Substitue x^2 for x in the solution to problem a. How does it compare to the solution of problem b?


Ok, a is easy enough. C is the easy way, but in b, I take it they mean to find the maclaurin series the traditional way, by taking the 1st, 2nd, 3rd.... derivatives, evaluating it at zero, and inserting it in the maclaurin polynomial.

Do you have any idea what a nightmare it is to find the maclaurin series of sin(x^2)?   f(0) is 0, f'(0 is 0, f''(0) is 2, everything else is zero all the way up to the 6th derivative. Try finding the 6th derivative of sin(x^2) it takes FOREVER! I did it twice and I've now got a screaming headache! Both times the sixth derivative evaluated at zero came to -24.

so the second derivative evaluated at zero is 2, and the 6th derivative evaluated at zero is -24. All others below 6 are zero.

So these would be inserted in the second and 6th term in the maclaurin polynomial respectively.

2(x^2)/2!      (-24X^6)/6!

So the first term in the macluarin series would be x^2, the next term would be -X^6/30.

The first term x^2 is correct, but the last term is supposed to be x^6/3!       and obviously its not.

TWICE I did this nightmarish problem and twice I got the same result. Any idea's?