First let s get some practice with determining the interval of convergence of power series.
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.
The last topic to cover in this section is the process of differentiating and integrating power series.
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.
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
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 + 1 + a 2(x c) 3 3 +
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 2 2 2 + x 3 3 3 + x 4 4 2 + and differentiate it term by term. Then write it in summation form again.
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) = 1 + 2x 2 2 + 3x 2 3 2 + 4x 3 4 2 + 5x 4 5 2 + 6x 5 6 2 + f (x) = 1 + x 2 + x 2 3 + x 3 4 + x 5 6 + x 6 6
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) = 1 + 2x 2 2 + 3x 2 3 2 + 4x 3 4 2 + 5x 4 5 2 + 6x 5 6 2 + f (x) = 1 + x 2 + x 2 3 + x 3 4 + x 5 6 + x 6 6 Next, we want to write this in summation form, like a n x n.
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) = 1 + 2x 2 2 + 3x 2 3 2 + 4x 3 4 2 + 5x 4 5 2 + 6x 5 6 2 + f (x) = 1 + x 2 + x 2 3 + x 3 4 + x 5 6 + 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
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 2 3 + x 3 4 + = x n n + 1.
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 2 3 + x 3 4 + = 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.
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 2 3 + x 3 4 + = 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.
For x = 1, the series becomes x n n + 1 = = 1 n n + 1 1 n + 1 = 1 + 1 2 + 1 3 + 1 4 = n=1 and this is just the harmonic series (or the p series with p = 1). 1 n
For x = 1, the series becomes x n n + 1 = ( 1) n n + 1 = 1 1 2 + 1 3 1 4 + = ( 1) n+1 n=1 and so this is the alternating harmonic series which we know is convergent. n
For x = 1, the series becomes x n n + 1 = ( 1) n n + 1 = 1 1 2 + 1 3 1 4 + = ( 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].
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 2 2 2 + x 3 3 2 + x 4 4 2 + and integrate it term by term. Then write it in summation form again.
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) dx = C + x 2 2 + x 3 3 1 2 2 + x 4 4 1 3 2 + x 5 5 1 4 2 + x 6 6 1 5 2 + x 7 7 62 + = C + x 2 2 + x 3 3 2 2 + x 4 4 3 2 + x 5 5 4 2 + x 6 6 5 2 + x 7 7 6 2 +
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) dx = C + x 2 2 + x 3 3 1 2 2 + x 4 4 1 3 2 + x 5 5 1 4 2 + x 6 6 1 5 2 + x 7 7 62 + = C + x 2 2 + x 3 3 2 2 + x 4 4 3 2 + x 5 5 4 2 + x 6 6 5 2 + x 7 7 6 2 + Note: the C here is the normal +C that we use for indefinite integration.
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) dx = C + x 2 2 + x 3 3 1 2 2 + x 4 4 1 3 2 + x 5 5 1 4 2 + x 6 6 1 5 2 + x 7 7 62 + = C + x 2 2 + x 3 3 2 2 + x 4 4 3 2 + x 5 5 4 2 + x 6 6 5 2 + x 7 7 6 2 + 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.
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) dx = C + x 2 2 + x 3 3 1 2 2 + x 4 4 1 3 2 + x 5 5 1 4 2 + x 6 6 1 5 2 + x 7 7 62 + = C + x 2 2 + x 3 3 2 2 + x 4 4 3 2 + x 5 5 4 2 + x 6 6 5 2 + x 7 7 6 2 + 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.
f (x) = x + x 2 2 2 + x 3 3 2 + x 4 4 2 + x 5 5 2 + x 6 6 2 f (x) dx = C + x 2 2 + x 3 3 1 2 2 + x 4 4 1 3 2 + x 5 5 1 4 2 + x 6 6 1 5 2 + x 7 7 62 + = C + x 2 2 + x 3 3 2 2 + x 4 4 3 2 + x 5 5 4 2 + x 6 6 5 2 + x 7 7 6 2 + 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
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.
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.
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
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.
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].
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.
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.
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).
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.
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
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).
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)!
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.
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.
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.