5. Using the Maclaurin series for sin x (see table on p. 740...

5. Using the Maclaurin series for sin x (see table on p. 740 in the text and posted on the Written Homework 17 page in Canvas), (a) (2 points) Find the power series representation for f(x) = sin(x^2). (b) (3 points) Find the power series representation for ?[0 to 0.2] sin(x^2) dx. (c) (1 points) Use the power series representation to approximate the definite integral in (b) so that the error in approximation is less than 1/10^4. (Hint: Use the Alternating Series Estimation Theorem.)

Transcript

00:01 Using the maclaurin series for sine of x, we're asked to find the power series representation for sine of x squared, and then to use that to find the power series representation for this integral, and then to use that to approximate the integral with an error less than 1 over 10 to the 4th.

00:16 So as a reminder, i've written down the maclaurin series for sine of x, and we'll use this to solve part a.

00:24 So sine of x squared.

00:28 Basically, wherever you see an x, just replace it with an x squared, and the result still holds.

00:36 So this is equal to the summation from n equals 0 to infinity.

00:42 I'll have negative 1 to the n, and then x squared.

00:48 That whole quantity to the 2n plus 1.

00:51 This is all over still 2n plus 1 factorial.

00:58 So i can simplify this a little bit.

01:05 I have negative 1 to the n, and then this will be x to the 4n plus 2.

01:14 All of that is over 2n plus 1 factorial.

01:21 So this is the power series for sine of x squared.

01:26 Now for the power series of this integral, let me write that below.

01:30 So the integral from 0 to 0 .2 of sine of x squared dx is equal to the integral from 0 to 0 .2 of this sum.

01:49 So let me just copy this real quick, put it here, and of course i can't forget dx.

02:01 And now because this is a power series, i can switch the order of integration and summation.

02:08 So this is equal to the sum from n equals 0 to infinity of the integral still from 0 to 0 .2 of negative 1 to the n, and then x to the 4n plus 2.

02:27 This is all over 2n plus 1 factorial dx.

02:36 All right, so remember that we're integrating with respect to x.

02:41 So this is the summation still from n equals 0 to infinity.

02:46 I'll have 4n plus 3 times 2n plus 1 factorial.

02:56 That's the denominator.

02:59 And in the numerator, i still have negative 1 to the n.

03:02 This is now x to the 4n plus 3, and this is evaluated from 0 to 0 .2, but i'm actually going to write 0 .2 as 1 fifth just so we can be exact.

03:16 And when i evaluate this expression at 0, i get 0.

03:19 So i only have it just evaluated at 1 fifth.

03:24 So still the summation from n is equal to 0 to infinity.

03:29 Let me write the denominator first...