Question

How do you find the Maclaurin Series for `f(x)= x sinx`?

Answer 1

`x sin x = x^2-x^4/(3!)+x^6/(5!)-...+(-1)^(n-1) x^(2n)/((2n-1)!)+...`

Explanation:

Maclaurin series for sin x is

`x -x^3/(3!)+x^5/(5!)-...+ (-1)^(n-1) x^(2n-1)/((2n-1)!)+...`. So,

`x sin x = x^2-x^4/(3!)+x^6/(5!)-...+(-1)^(n-1) x^(2n)/((2n-1)!)+...`

sin x is an odd function. So, x sin x is even #. And so,

f(x) = x sin x is an even function and the Maclaurin series is an even

power series that is unique. Note that f(0) = 0 and f''(0)/(2!) = 1, and

so, there is no constant term in the series and the coefficient of

`x^2` is 1. Thus, the Maclaurin series for

`x sin x = x^2-x^4/(3!)+x^6/(5!)-...+(-1)^(n-1) x^(2n)/((2n-1)!)+...`