Section 9.8. First let s get some practice with determining the interval of convergence of power series.

Transcription

1 First let s get some practice with determining the interval of convergence of power series.

2 First let s get some practice with determining the interval of convergence of power series. Example (1) Determine the radius and interval of convergence of each power series. ( 1) n+1 (x + 5) n 1 n2 n n=1 (3x) n 2 (2n)! n=1 Try this on your own and check with the solution on the course website.

3 The last topic to cover in this section is the process of differentiating and integrating power series.

4 The last topic to cover in this section is the process of differentiating and integrating power series. Theorem If the function given by f (x) = a n (x c) n = a 0 +a 1 (x c)+a 2 (x c) 2 +a 3 (x c) 3 + has a radius of convergence of R > 0, then on the interval (c R, c + R), f is differentiable (and therefore it is continuous) and integrable. The radius of convergence of the series obtained by differentiation or integration of a power series is the same as the original power series. However, the interval of convergence may be different.

5 Given f (x) = a n (x c) n = a 0 +a 1 (x c)+a 2 (x c) 2 +a 3 (x c) 3 + then we have f (x) = a 1 + 2a 2 (x c) + 3a 3 (x c) 2 + f (x) = na n (x c) n 1 n=1

6 Given f (x) = a n (x c) n = a 0 +a 1 (x c)+a 2 (x c) 2 +a 3 (x c) 3 + then we have f (x) dx = C + a 0 (x c) + a 1(x c) 2 2 a n (x c) n+1 f (x) dx = C + n a 2(x c) 3 3 +

7 Let s go back to the power series we worked on at the end on x n Monday s class,. Recall that it is centered at x = 0, has a n2 n=1 radius of convergence R = 1, and its interval of convergence is [ 1, 1]. x n So, let f (x) = n 2. Before clicking next, find f (x) and its n=1 interval of convergence. Note it may help to write out the terms of your series like so, f (x) = n=1 x n n 2 = x + x x x and differentiate it term by term. Then write it in summation form again.

8 f (x) = x + x x x x x f (x) = 1 + 2x x x x x f (x) = 1 + x 2 + x x x x 6 6

9 f (x) = x + x x x x x f (x) = 1 + 2x x x x x f (x) = 1 + x 2 + x x x x 6 6 Next, we want to write this in summation form, like a n x n.

10 f (x) = x + x x x x x f (x) = 1 + 2x x x x x f (x) = 1 + x 2 + x x x x 6 6 Next, we want to write this in summation form, like a n x n. One tip for this is that you want to think of the power of x as being n when trying to write your formula. In particular, for this example n = 0, 1 n = 1, x 2 n = 2, x2 3 n = 3, x3 4

11 Notice that n = 0, 1 n = 1, x 2 = x1 1+1 n = 2, x2 3 = x2 2+1 n = 3, x3 4 = x3 3+1 and hence f (x) = 1 + x 2 + x x = x n n + 1.

12 Notice that n = 0, 1 n = 1, x 2 = x1 1+1 n = 2, x2 3 = x2 2+1 n = 3, x3 4 = x3 3+1 and hence f (x) = 1 + x 2 + x x = x n n + 1. Next, we want to determine the interval of convergence. From the theorem we know that the radius of convergence is the same as the original series, hence for f (x) we have R = 1.

13 Notice that n = 0, 1 n = 1, x 2 = x1 1+1 n = 2, x2 3 = x2 2+1 n = 3, x3 4 = x3 3+1 and hence f (x) = 1 + x 2 + x x = x n n + 1. Next, we want to determine the interval of convergence. From the theorem we know that the radius of convergence is the same as the original series, hence for f (x) we have R = 1. However, the interval could be different - so we need to check the endpoints x = 1 and x = 1 for convergence. Do this before clicking next.

14 For x = 1, the series becomes x n n + 1 = = 1 n n n + 1 = = n=1 and this is just the harmonic series (or the p series with p = 1). 1 n

15 For x = 1, the series becomes x n n + 1 = ( 1) n n + 1 = = ( 1) n+1 n=1 and so this is the alternating harmonic series which we know is convergent. n

16 For x = 1, the series becomes x n n + 1 = ( 1) n n + 1 = = ( 1) n+1 n=1 and so this is the alternating harmonic series which we know is convergent. So, for f (x) = x n n + 1 n the interval of convergence is [ 1, 1). Note that this is different from the interval of convergence of f (x) which is [ 1, 1].

17 Next, let s find f (x) dx. x n So let s go back to f (x) =. Like before you may find it n2 n=1 helpful to write out the terms of the series, f (x) = n=1 x n n 2 = x + x x x and integrate it term by term. Then write it in summation form again.

18 f (x) = x + x x x x x f (x) dx = C + x x x x x x = C + x x x x x x

19 f (x) = x + x x x x x f (x) dx = C + x x x x x x = C + x x x x x x Note: the C here is the normal +C that we use for indefinite integration.

20 f (x) = x + x x x x x f (x) dx = C + x x x x x x = C + x x x x x x Note: the C here is the normal +C that we use for indefinite integration. Next, try to write this as a power series using summation notation, ie a n x n.

21 f (x) = x + x x x x x f (x) dx = C + x x x x x x = C + x x x x x x Note: the C here is the normal +C that we use for indefinite integration. Next, try to write this as a power series using summation notation, ie a n x n. This becomes C + n=2 x n n(n 1) 2.

22 f (x) = x + x x x x x f (x) dx = C + x x x x x x = C + x x x x x x Note: the C here is the normal +C that we use for indefinite integration. Next, try to write this as a power series using summation notation, ie a n x n. This becomes C + n=2 x n n(n 1) 2. Next, we will determine the interval of convergence. The theorem

23 First let s consider x = 1. Here our series becomes, C + n=2 x n n(n 1) 2 = C + n=2 1 n n(n 1) 2 = C + n=2 1 n(n 1) 2. Check the convergence or divergence of this before clicking next.

24 First let s consider x = 1. Here our series becomes, C + n=2 x n n(n 1) 2 = C + n=2 1 n n(n 1) 2 = C + n=2 1 n(n 1) 2. Check the convergence or divergence of this before clicking next. Here we will compare the series to 1, which is a convergent p n 3 series with p = 3. To use the regular comparison test we would 1 need to show that 1, which is not true. So instead we n(n 1) 2 n 3 will use the limit comparison test. (or you could use the limit comparison test). Try this before clicking next.

25 lim n 1 n 3 1 n(n 1) 2 n(n 1) 2 = lim n n 3 = 1 > 0 Therefore, by the limit comparison test this series converges. Next we consider the series when x = 1. C + n=2 x n n(n 1) 2 = C + ( 1) n n(n 1) 2. n=2

26 lim n 1 n 3 1 n(n 1) 2 n(n 1) 2 = lim n n 3 = 1 > 0 Therefore, by the limit comparison test this series converges. Next we consider the series when x = 1. C + n=2 x n n(n 1) 2 = C + ( 1) n n(n 1) 2. n=2 Note that we know ( 1) n n(n 1) 2 is convergence, and hence we n=2 can conclude that our series is convergent.

27 lim n 1 n 3 1 n(n 1) 2 n(n 1) 2 = lim n n 3 = 1 > 0 Therefore, by the limit comparison test this series converges. Next we consider the series when x = 1. C + n=2 x n n(n 1) 2 = C + ( 1) n n(n 1) 2. n=2 Note that we know ( 1) n n(n 1) 2 is convergence, and hence we n=2 can conclude that our series is convergent. Putting this all together, the interval of convergence for f (x) dx is [ 1, 1].

28 Section 9.9 Now we move on to the next section, Representation of Functions by Power Series. In this section we will go over techniques for finding a power series that represents a given function.

29 Section 9.9 Now we move on to the next section, Representation of Functions by Power Series. In this section we will go over techniques for finding a power series that represents a given function. For example, suppose you are given cos(x) and you want a power series representation of it. It turns out that the power series representation is ( 1) n x 2n cos(x) =. (2n)! We will study where this and other power series representations came from and use this to find power series representations for other functions.

30 Section 9.9 Consider the following power series x n. Notice that once we fix an x value, then it simply becomes a geometric series and hence we know the values it converges for. This power series is convergent whenever x < 1 and so its interval of convergence is ( 1, 1).

31 Section 9.9 Consider the following power series x n. Notice that once we fix an x value, then it simply becomes a geometric series and hence we know the values it converges for. This power series is convergent whenever x < 1 and so its interval of convergence is ( 1, 1). Further, for a convergent geometric series, we know its sum. In particular, ar n = a 1 r. Thus for x in our interval of convergence, ( 1, 1), we have that x n = 1 1 x. Here we have a power series on the left and a function on the right. So we would call x n a power series representation for the function 1 1 x.

32 Section 9.9 Theorem (Operations with Power Series) If f (x) = a n x n, then for any k R f (kx) = a n (kx) n = a n k n x n

33 Section 9.9 Theorem (Operations with Power Series) If f (x) = a n x n, then for any k R f (kx) = a n (kx) n = a n k n x n For example, suppose we want to find a power series representation ( 1) n x 2n for cos(7x). We know that cos(x) =. This theorem (2n)! says that we can plug 7x in for x in the power series representation for cos(x) to get the power series representation for cos(7x).

34 Section 9.9 Theorem (Operations with Power Series) If f (x) = a n x n, then for any k R f (kx) = a n (kx) n = a n k n x n For example, suppose we want to find a power series representation ( 1) n x 2n for cos(7x). We know that cos(x) =. This theorem (2n)! says that we can plug 7x in for x in the power series representation for cos(x) to get the power series representation for cos(7x). ( 1) n (7x) 2n cos(7x) = (2n)! = ( 1) n 7 2n x 2n (2n)!

35 Section 9.9 Theorem (Operations with Power Series) If f (x) = a n x n, then f (x m ) = a n (x m ) n = a n x mn. For example, suppose we want to find a power series representation 1 1 for. We know that 1 x 2 1 x = x n. This theorem says that we can plug x 2 1 in for x in the power series representation of 1 x 1 to get the power series representation for 1 1 x 2 = (x 2 ) n = 1 x 2, x 2n.

36 Section 9.9 Theorem (Operations with Power Series) If f (x) = a n x n and g(x) = b n x n then f (x) + g(x) = f (x) g(x) = (a n + b n )x n (a n b n )x n Note that with all of this operations with series, it can change the interval of convergence for the new series.

37 Section 9.9 Example (2) Find a power series representation for each of the given functions using 1 1 x = x n and the operations with power series. 1 f (x) = 1 5 x 2 f (x) = 6 x+1 3 f (x) = 7x 2 2x 2 +x 1 The solution for this problem is posted on the course webpage.