Lecture 5. Power

Audrey Terras May 29, 2010

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∞ n Here we consider convergence and differentiation /integration of power series an(x p) . I will mostly assume p = 0 n=0 − for simplicity. We consider x R mostly. The same arguments should workX for complex numbers and matrices, with a little thought. ∈ ∞ ∞ n n Theorem. a) Suppose an is a of real numbers and anx converges for some x = 0. Then anu { } n=0 6 n=0 converges absolutely for all u such that u < x . X X | | | | ∞ ∞ n n b) If anx diverges, then anu diverges if u > x . n=0 n=0 | | | | X X ∞ n n n Proof. a) anx converges implies lim anx = 0. Since convergent implies bounded, we have anx M for all n. n n=0 →∞ | | ≤ This impliesX if u < x , we can apply the comparison test by writing | | | | n n n n u u anu = anx M . | | xn ≤ xn

∞ ∞ n n u So we can compare the series anu with the convergent geometric series xn . n=0 | | n=0 X X We leave the proof of b) to the reader for extra credit. Hint: Use part a). We are calling the following positive number the because we are really thinking of power series in the where we actually get an open circle of radius R where the power series converges. On the real line where this course lives, we only get an interval.

∞ n Definition 1 Now define the radius of convergence R of anx n=0 X

∞ n R = l.u.b. r r 0, an r converges . ( ≥ | | ) n=0 X

I am using the convention here that R (blackboard bold face font) is the set of real numbers and should not be confused with ordinary capital R. This convention comes to us from Nicolas Bourbaki (the unreal mathematician).

∞ n Corollary. Let S denote the set of real numbers x such that anx converges. Then S must be of the following n=0 X forms: 0 , ( R,R), ( R,R], [ R,R), [ R,R], or ( , ) = R. { } − − − − −∞ ∞

1 ∞ n Formulas for the Radius of Convergence R of a Power Series anx . We assume an = 0, n. n=0 6 ∀ X ∞ an n 1) Assume the limit lim exists. Then it is the radius of convergence R for anx . That is n an+1 →∞ n=0 X a R = lim n . n a →∞ n+1

2) Again, assuming the limit involved exists, we have the formula:

1 1 n = lim an . R n →∞ | |

∞ n an Proof. 1) Use the for of anx . Setting lim = c, we see that the ratio of the n an+1 n=0 →∞ terms in the power series approaches a limit: X

n+1 an+1x an+1 x = x | |, as n a xn a → c → ∞ n n

The ratio test from Lectures Part 4 says the series converges if x < c. and diverges if x > c. Why? (Extra Credit). So c = R, the radius of convergence. This comes from the definition| | of radius of convergence| | as a least upper bound. If x > R, the series diverges and if x < R, the series converges. That is what we needed to show. | | 2) We leave this to the reader| for| Extra Credit.

∞ n 1 Example 1). Our friend the geometric series x = 1 x . The ratio of convergence is 1 by any of the formulas n=0 − since an = 1, for all n. X ∞ 1 n x Example 2). Our best friend the exponential series n! x = e . Use the first formula for the radius of convergence. n=0 Obtain X a 1 (n + 1)! (n + 1) R = lim n = R = lim n! = lim lim = . n a n 1 n n! n 1 ∞ →∞ n+1 →∞ (n+1)! →∞ →∞

This means the series converges everywhere. The nice thing about this example is that the exponential series also converges for all complex numbers and for all nxn matrices. This gives a nice way to get the power series for and cosine out of Euler’sformula

eiθ = cos θ + i sin θ.

∞ Example 3). Here’sa bad one: n!xn. n=0 The radius of convergence is the reciprocalX of that for Example 2. Thus you get R = 0. This series only converges at x = 0.

2 Differentiation and Integration of Power Series

b ∞ We want to legally interchange and limit (or ) so that we can integrate a power series term-by-term at least if we n=0 Za X d ∞ stay inside the radius of convergence. Similarly we want to legally interchange dx and limit (or ), n=0 X 2 Example. Integrate the geometric series to get the power series for log(1 x). This is legal if x < 1 using the corollary of the following theorem. − | |

x x x x 1 ∞ ∞ ∞ tn+1 ∞ xn+1 log(1 x) = dt = tndt = tndt = = . − − 1 t n + 1 n + 1 Z Z n=0 n=0 Z n=0 0 n=0 0 − 0 X X 0 X X

Note that the formula makes sense if x = 1, although the following theorem and corollary do not justify the equality at 1. − − Theorem. (Interchange of limit and integral). Suppose that fn :[a, b] R is continuous n and converges uniformly on [a, b] to f. Then → ∀

b b b

lim fn = lim fn = f. n n →∞Za Za →∞ Za Proof. We already know that f must be continuous on [a, b]. So we can integrate it. See Lectures, Part 3, Thm. 7, p. 7. Using properties of the integral from last quarter, we find that

b b b b b

f fn = (f fn) f fn f fn = f fn (b a). − − ≤ | − | ≤ k − k k − k − Z Z Z Z Z ∞ ∞ a a a a a

ε It follows that given ε > 0, if we take N so that n N, implies f fn < b a (as we can by uniform convergence), then we have ≥ k − k∞ − b b

f fn < ε. − Z Z a a

∞ Corollary. Suppose gn(x)[a, b] R is continuous n and gn(x) converges uniformly on [a, b] to s(x). Then s(x) → ∀ n=0 is continuous on [a, b] and X b b ∞ ∞ gn = gn. n=0 n=0 X Za Za X Proof. Extra Credit Exercise.

The proof of the last theorem was essentially one line. But differentiation requires more effort, as well as more hypotheses. Why should that be? sin(nx) Example. Consider fn(x) = . This sequence converges uniformly to 0 on [ π, π]. Why? However, √n −

0 fn(x) = √n cos(nx).

This has problems converging at all much less to 0! For example, it diverges at x = 0. In fact, it can be shown to diverge everywhere. Suppose x = 0.You just need to see that if x = 0, N such that cos Nx < 1 and thus 6 6 ∃ | | 2 1 cos(2Nx) = 2 cos2 Nx 1 = 1 2 cos2 Nx > . | | − − 2

1 This means n > N (namely n = 2N) such that cos(nx) > 2 . It follows that there is a subsequence satisfying ∃ 1 | | √nk cos(nkx) > 2 √nk and thus diverging. Extra Credit. Find an example to show that we need uniform convergence of f to f in the hypothesis of the preceding n theorem. at each x in [a, b] does not suffi ce for the interchange of integral and limit. 2 n Hint. Look at fn(x) = n x(1 x) , on [0, 1]. −

3 Theorem. (Interchange of Derivative and Limit). Suppose fn is a sequence of continuously differentiable functions on [a, b]. Suppose, also that the sequence of f 0 {converges} uniformly to g on [a, b]. And, finally { n} assume there exists one point x0 [a, b] such that fn(x0) converges. Then ∈ d d g(x) = lim fn(x) = lim fn(x). n dx dx n →∞ →∞ and the convergence of fn(x) is uniform to the f(x) with f 0(x) = g(x), for all x [a, b]. ∈ Proof. By the fundamental theorem of calculus from last quarter, n, cn s.t. ∀ ∃ x

fn(x) = fn0 + cn. (1) Za

Here cn = fn(a). − Let x = x0 and take the limit of formula (1) as n . This says that c = lim cn. → ∞ ∃ n Next take the limit of formula (1) for general x [a, b] as n using the→∞ preceding Theorem (which we can since ∈ → ∞ the sequence of derivatives converges uniformly). This says, the sequence fn converges pointwise (i.e., for each fixed x in [a, b]) to { } x f(x) = g + c. Za

To finish the proof of this theorem, we just need to see fn converges uniformly to f. Well, try this, using our favorite and properties of integrals from last quarter,{ }

x x

0 fn(x) f(x) = f + cn g c | − | n − − Z Z a a x x x

0 0 f g + cn c f g + cn c ≤ n − | − | ≤ n − | − | Za Za Za

0 (b a) f g + cn c . ≤ − n − | − | ∞

So if you insist on giving me an ε > 0, I can certainly find an N (depending only on ε and not on x) such that n > N implies fn(x) f < ε. That is the meaning of uniform convergence. | − |

∞ 0 Corollary. Suppose gn(x)[a, b] R is continuously differentiable, n and gn(x) converges uniformly on [a, b] to → ∀ n=0 X ∞ ∞ s(x). And assume there is one point x0 [a, b] such that gn(x) converges. Then we have gn(x) converges to r(x) ∈ n=0 n=0 X X uniformly on [a, b] and s(x) = r0 (x); i.e., ∞ d d ∞ g (x) = g (x). dx n dx n n=0 n=0 X X Proof. Extra Credit. Deduce this corollary from the preceding theorem. Examples. For our power series, these 2 theorems, allow us to integrate and differentiate term by term on closed subintervals of ( R,R), where R is the radius of convergence. I find this somewhat surprising, since the power series you get by differentiation− term-by-term blows up the term by a factor of n: d a xn = na xn 1. dx n n − It is also a bit surprising that integration does not produce a series that converges in a larger region since

4 x n+1 n x ant dt = an , for n 0. n + 1 ≥ Z0

∞ n Theorem. Suppose f(x) = anx with radius of convergence R. Then for x in ( R,R), we have the following n=0 − facts. X 1) x ∞ xn+1 f(t)dt = a . n n + 1 n=0 Z0 X 2)

∞ n 1 f 0(x) = nanx − . n=0 X The integrated and differentiated series have the same radius of convergence R. Proof. 2) Note that if R is its radius of convergence, a power series converges uniformly on closed subintervals of ( R,R). − So we just have to convince ourselves that the extra factor of n in nan does not affect the radius of convergence. To see this, note that if 0 < x < c < R, we know by the definition of radius of convergence R on the first page of this | | ∞ n n n part of the lectures, that an c < . Thus lim anc = 0. It follows that there is a bound M such that anc M, n n=0 | | ∞ →∞ | | ≤ for all n. X Therefore we have: n 1 n 1 n 1 n x − n n x − n an x − = | | an c M | | . | | | | c c | | ≤ c c     n 1 ∞ n 1 ∞ x − So n an x − can be compared with n |c| . This series converges by the ratio test as the ratios are n=0 | | | | n=0 X X   n x (n + 1) |c| n + 1 x x n 1 = | | | | < 1, as n . x  − n c → c → ∞ n |c|   1) is left to the reader for Extra Credit.

Examples. The power series for ex, sin(x), cos(x) converge absolutely and uniformly on closed and bounded sets in the real line. So they are differentiable and integrable term by term.

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Last quarter we proved Taylor’sformula with remainder:

n 1 (k) − f (a) k f(x) = (x a) + Rn. k! − k=0 X (k) f (c) k We had 2 formulas for the remainder. The easiest to remember is Rn = (x a) , for some c between a and x. If k! − one can show that lim Rn = 0, then the function f(x) is represented by its Taylor series within the radius of convergence. n That is, for x a R,→∞ | − | ∞ f (k)(a) f(x) = (x a)k. k! − k=0 X

5 See Wikipedia (Taylor series article) for animated pictures showing the convergence of the Taylor series to ex. We proved last quarter the our favorite functions: ex, sin(x), cos(x), log(1 x) are represented by their Taylor series within the radius of convergence interval. So we have: − ∞ 1 ex = xn, for all x R; n! n=0 ∈ X ∞ 1 log(1 x) = xn, for x < 1. n − − n=0 | | X x x2 x3 x4 x5 x6 Figure 1 shows the sum of the first 7 terms of the Taylor series for e , namely f(x) = 1 + x + 2 + 6 + 24 + 5 24 + 30 24 in red and ex in blue on the interval [ 5, 5]. ∗ ∗ −

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•5 •4 •3 •2 •1 0 1 2 3 4 5 x

Figure 1: Plot of ex blue and 1st 7 terms of its Taylor series about 0 red.

Another example is the Binomial Series:

∞ α (1 + x)α = xn, (2) n n=0 X   where we define the generalized binomial coeffi cient α α(α 1) (α n + 1) = − ··· − . n n!   As usual, define 0! = 1. When α is a positive integer, this is the ordinary binomial theorem. Otherwise you need to restrict x to have x < 1. Why? Extra| Credit| Exercise. Prove formula (2) using Taylor’sformula from last quarter. Find the radius of convergence. Why is this the binomial theorem when α is a positive integer?

We also saw last quarter that there are functions f(x) not represented by their Taylor series; e.g., f(0) = 0, and for 1 x = 0, define f(x) = e− x2 . For this function, one can show that f (n)(0) = 0, for all n. This means the Taylor series for f(6x) around x = 0 is 0 even though f(x) itself is positive except at x = 0. For x = 0, 6 1 (n) 2 ∞ f (0) e− x = f(x) = xn = 0. n! 6 n=0 X See Figure 2.

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•5 •4 •3 •2 •1 0 1 2 3 4 5 x

1 Figure 2: The plot of f(x) = e− x2 . This function is not represented by its Taylor series about 0 which is 0 for all x.

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