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

Answer 1

Isaiah Allen

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

For sin x, the Maclaurin series 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,

The Maclaurin series is an even series, and f(x) = x sin x is an even function.

power series that is distinct, where f''(0)/(2!) = 1, f(0) = 0, and

Consequently, the series does not contain a constant term, 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)!)+...#

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Answer 2

Emily Ammerman

To find the Maclaurin series for $f(x) = x\sin(x)$ , follow these steps:

Recall the Maclaurin series for $\sin(x)$ :
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$
This series expansion is derived from the Taylor series of $\sin(x)$ centered at 0.
Multiply each term of the $\sin(x)$ series by $x$ :
$x\sin(x) = x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} + \dots$

The resulting series is the Maclaurin series for $f(x) = x\sin(x)$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.