MA 146 Steve Sawin

11.8 Power

Examples: Remember a polynomial is a sum of multiples of powers of a variable x 4 − 3x + 2x2 − 7x3 1 + x + x2 + x3 + x4 + x5 1 − x100. It will be handy to also call polynomials things like this where we have (x − a) in place of x 3 + (x − 1)2 − 5(x − 1)3 1 + (x + 2) + (x + 2)2 + (x + 2)3 + (x + 2)4 + (x + 2)5. A is an infinite polynomial like this ∞ X 1 + x + x2 + x3 + ··· = xk k=0 ∞ 1 1 1 1 1 X 1 + (x − 1) + (x − 1)2 + (x − 1)3 + (x − 1)4 + ··· = (x − 1)k 1 1 2 6 24 k! k=0 ∞ X 0 − (x + 1) + 2(x + 1)2 − 3(x + 1)3 + 4(x + 1)4 − · · · = (−1)kk(x + 1)k. k=0

Formulas: The general way to write this is ∞ X k ck(x − a) k=0 so the ck are the coefficients of each power of x and a is the number you are subtracting from each x before applying the exponent (notice if you are adding a number that is the same as subtracting its negative, so a = −1 in the previous example. Most Important: When you write a polynomial you mean it is a rule that I can plug in any number for x and get a number out. When you write a power series you mean it is a rule that I can plug in any number for x and get an infinite series out. So if I plug in x = 3 into ∞ X 1 (x − 1)k k! k=0 I get ∞ ∞ ∞ X 1 X 1 X 2k (3 − 1)k = 2k = k! k! k! k=0 k=0 k=0 which is an example we showed was absolutely convergent in one of our Zoom Meetings. If for a particular value of x the power series converges, then you can think of the power series as a whose value at that x is the value that this series converges to. Our first question is when (i.e., for which values of x) the series converges. Radius and Interval of Convergence

P∞ k Key fact: A power series k=0 ck(x−a) will always converge on an interval centered at a. That is, it converges absolutely for all x in the interval

a − R < x < a + R.

This number R is called the radius of convergence and the interval (a − R, a + R) is called the interval of convergence. The series diverges for every x > a + R and for every x < a − R. It may converge absolutely, converge conditionally or diverge for the boundary points x = a−R and x = a+R. So the interval may be (a−R, a+R), [a − R, a + R], (a − R, a + R], or [a − R, a + R). R can be zero, in which case the interval contains exactly the point a, or it can be ∞, in which case the interval is every real number.

Absolute value and intervals: a deeply important digression. The idiom a−R < x < a + R comes up a lot. Let’s put in numbers. If a = 3 and R = 2 then we are saying 1 = 3 − 2 < x < 3 + 2 = 5, or x ∈ (1, 5). if you take such an x, its diwtance to 3 is less than 2. So this interval is all the points a distance less than 2 from 3. Remember we say that |x − 3| < 2. So another way to say

a − R < x < a + R is to say |x − a| < R.

Finding the interval of convergence: First Example: Find the interval of convergence of

∞ X 1 • (x − 1)k. k3k k=0

1. : Remember the series converges absolutely if the limit < 1 and diverges if the limit > 1. The absolute value does not know if x − 1 is positive or negative.

k+1 k+1 k+1 k+1 (x − 1) /(k + 1)3 |x − 1| /(k + 1)3 lim = lim k→∞ (x − 1)k/k3k k→∞ |x − 1|k /k3k k |x − 1|k+1 3k = lim k→∞ (k + 1) |x − 1|k 3k+1 |x − 1| 1 = lim 1 · < 1 k→∞ 1 3 |x − 1| < 1. 3

2. Interval: Turn this inequality with absolute value into an interval |x − 1| < 1 3 |x − 1| < 3 here’s the radius of convergence −3

so the series converges absolutely when x ∈ (−2, 4) and diverges when x is bigger than 4 or less than −2.

3. The endpoints: Plug in x = −2 and x = 4 into the original series, see if it converges. This will usually entail LCT and AST.

• x = 4

∞ ∞ X 1 X 3k (4 − 1)k = k3k k3k k=0 k=0 ∞ X 1 = k k=0 This is a p-series with p = 1 so it diverges. • x = −2

∞ ∞ X 1 X (−3)k (−2 − 1)k = k3k k3k k=0 k=0 ∞ X (−1)k = k k=0 We just saw that the absolute value of this series diverges, but it is alternating and decreasing with limk 1/k = 0 so by AST this series converges conditionally. Interval of Convergence: Process 1. Ratio Test: Do the ratio test with your series. Note that (x − a)k is an exponential term, and it can be positive or negative, so when you remove the absolute value elsewhere it will remain there as |x − a|k. The ratio test tells you that the series converges when that limit is < 1, so the end result of the ratio test will always look like |x − a| < 1 R (fine print: sometimes R will be 0 or ∞.)

2. Interval: Unless R is 0 or ∞, solve this by

|x − a| < 1 R |x − a| < R −R

so the series converges absolutely when x is in (a − R, a + R) and diverges when it is outside that interval, except we do not know about the endpoints.

3. Endpoints: If R is not 0 or ∞, plug in a−R and a+R for x in the original series. You should see all the exponentials cancel out nicely except one of the two endpoints will be alternating. Decide if each of these endpoints is absolutely convergent, conditionally convergent, or divergent. Report as an interval with ()[] to indicate the status of the endpoints.

Examples: Find the interval and radius of convergence of

∞ X xn • . n! n=0 1. Ratio Test:

n+1 n+1 x /(n + 1)! n! |x| lim = lim n n→∞ xn/n! n→∞ (n + 1)! |x| 1 |x| = lim = 0 < 1 n→∞ n + 1 1 which is true for all x which means the radius of convergence is ∞ and therefore the interval is (−∞, ∞) and we are done. This series is absolutely convergent everywhere. Find the interval and radius of convergence of

∞ X • n!(x + 3)n. n=0 1. Ratio Test:

n+1 n+1 (n + 1)!(x + 3) (n + 1)! |x + 3| lim = lim n n→∞ (x + 3)nn! n→∞ n! |x + 3| n + 1 |x + 3| = lim = ∞ unless x = −3 n→∞ 1 1 watch what happened in that last line carefully! If x = −3 then this expression is limn→∞(n + 1)0 = 0, so the series is absolutely convergent.If x is anything else then of course limn→∞(n + 1) |x + 3| = ∞ > 1 so the series is divergent. The radius of converge is 0, the interval is the single point [−3]. Every power series converges absolutely at a, the center of the series.

Find the interval and radius of convergence of

∞ X k • (x − 1)k. 7k k=0 1. Ratio Test:

k+1 k+1 k k+1 (k + 1)(x − 1) /7 k + 1 7 |x − 1| lim = lim k→∞ k(x − 1)k/7k k→∞ k 7k+1 |x − 1|k 1 1 |x − 1| |x − 1| = lim = < 1 k→∞ 1 7 1 7 so the radius of convergence is 7.

2. Interval: |x − 1| < 1 7 |x − 1| < 7 −7 < x − 1 < 7 −6 < x < 8

so the interval (not counting endpoints) is (−6, 8), an interval centered at 1 and with radius 7 (so total length 14).

3. Endpoints: plug in x = 1 + 7 = 8 and 1 − 7 = −6: • x = 8 ∞ ∞ ∞ X k X k7k X (8 − 1)k = = k 7k 7k k=0 k=0 k=0 which diverges by the Divergence Test. • x = −6 ∞ ∞ ∞ X k X k(−7)k X (−6 − 1)k = = (−1)kk 7k 7k k=0 k=0 k=0 which diverges by the Divergence Test. Thus the interval of convergence is the open interval (−6, 8).