Problem 3: Taylor series.

3 . a)

> taylor(x*cos(x) - sin(x),x,12);

The easiest way to get the general pattern is to recall that one of the top-10 Taylor series isthat of the cosine, with terms

(-1)^k x^(2k) /(2k)!

so the terms in the series for x cos(x) are

(-1)^k x^(2k+1) /(2k)!

Since the terms in the sine series are

(-1)^k x^(2k+1) /(2k+1)!,

the coefficient of x^(2k+1) in the Taylor series of our function are

> simplify((-1)^k *(1/(2*k)! - 1/(2*k+1)! ));

The mysterious denominator of this is the same as (2k+1) ! .

b) The radius of convergence is infinity . Reason: we can use the ratio test, to find that the ratio of consecutive terms in the Taylor series is

> simplify(- x ( (2*(k+1)/ ((2*(k+1))!)) / (2*k/ ((2*k)!))));

>

As k -> infinity, this -> 0 regardless of x. hence the series converges for every x.

c) The coefficient of x^11 (happens when k=5) in the series is f^(11) (0) / 11!, but from the above we see that it is also equal to . Therefore the answer is:

> - 11! /3991680 ;