8.4: Boundary Issues and Abel’s Theorem
( \newcommand{\kernel}{\mathrm{null}\,}\)
- Explain the Abel's theorem
Summarizing our results, we see that any power series ∑anxn has a radius of convergence r such that ∑anxn converges absolutely when |x|<r and diverges when |x|>r. Furthermore, the convergence is uniform on any closed interval [−b,b]⊂(−r,r) which tells us that whatever the power series converges to must be a continuous function on (−r,r). Lastly, if
f(x)=∞∑n=0anxn
for x∈(−r,r), then
f′(x)=∞∑n=0annxn−1
for x∈(−r,r) and
∫xt=0f(t)dt=∞∑n=0anxn+1n+1
for x∈(−r,r).
Thus power series are very well behaved within their interval of convergence, and our cavalier approach from Chapter 2 is justified, EXCEPT for one issue. If you go back to Exercise Q1 of Chapter 2, you see that we used the geometric series to obtain the series,
arctanx=∞∑n=0(−1)n12n+1x2n+1.
We substituted x=1 into this to obtain π4=∑∞n=0(−1)n12n+1. Unfortunately, our integration was only guaranteed on a closed subinterval of the interval (−1,1) where the convergence was uniform and we substituted in x=1. We “danced on the boundary” in other places as well, including when we said that
π4=∫1x=0√1−x2dx=1+∞∑n=1(∏n−1j=0(12−j)n!)((−1)n2n+1)
The fact is that for a power series ∑anxn with radius of convergence r, we know what happens for x with |x|<r and x with |x|>r. We never talked about what happens for x with |x|=r. That is because there is no systematic approach to this boundary problem. For example, consider the three series
∞∑n=0xn,∞∑n=0xn+1n+1,∞∑n=0xn+2(n+1)(n+2)
They are all related in that we started with the geometric series and integrated twice, thus they all have radius of convergence equal to 1. Their behavior on the boundary, i.e., when x=±1, is another story. The first series diverges when x=±1, the third series converges when x=±1. The second series converges when x=−1 and diverges when x=1.
Even with the unpredictability of a power series at the endpoints of its interval of convergence, the Weierstrass-M test does give us some hope of uniform convergence.
Suppose the power series ∑anxn has radius of convergence r and the series ∑anrn converges absolutely. Then ∑anxn converges uniformly on [−r,r].
- Hint
-
For |x|≤r, |anxn|≤|anrn|.
Unfortunately, this result doesn’t apply to the integrals we mentioned as the convergence at the endpoints is not absolute. Nonetheless, the integrations we performed in Chapter 2 are still legitimate. This is due to the following theorem by Abel which extends uniform convergence to the endpoints of the interval of convergence even if the convergence at an endpoint is only conditional. Abel did not use the term uniform convergence, as it hadn’t been defined yet, but the ideas involved are his.
Suppose the power series ∑anxn has radius of convergence r and the series ∑anrn converges. Then ∑anxn converges uniformly on [0,r].
The proof of this is not intuitive, but involves a clever technique known as Abel’s Partial Summation Formula.
Let a1,a2,...,an,b1,b2,...,bn be real numbers and let Am=∑mk=1ak. Then
a1b1+a2b2+···+anbn=n−1∑j=1Aj(bj−bj+1)+Anbn
Prove Lemma 8.4.1.
- Hint
-
For j>1, aj=Aj−Aj−1.
Let a1,a2,...,an,b1,b2,...,bn be real numbers with b1≥b2≥...≥bn≥0 and let Am=∑mk=1ak. Suppose |Am|≤B for all m. Then |∑nj=1ajbj|≤B⋅b1.
Prove Lemma 8.4.2.
Prove Theorem 8.4.1.
- Hint
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Let ϵ>0. Since ∑∞n=0anrn converges then by the Cauchy Criterion, there exists N such that if m>n>N then |∑mk=n+1akrk|<ϵ2. Let 0≤x≤r. By Lemma 8.4.2,
|m∑k=n+1akxk|=|m∑k=n+1akrk(xr)k|≤ϵ2(xr)n+1≤ϵ2
Thus for 0≤x≤r, n>N,
|∞∑k=n+1akxk|=limn→∞|m∑k=n+1akxk|≤ϵ2<ϵ
Suppose the power series ∑anxn has radius of convergence r and the series ∑an(−r)n converges. Then ∑anxn converges uniformly on [−r,0].
Prove Corollary 8.4.1.
- Hint
-
Consider ∑an(−x)n.