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# 12.8 Proof of the Differentiation Theorem

12.73   Lemma. The power series has radius of convergence equal to .

12.74   Exercise. Prove lemma 12.73. (We proved this lemma earlier using the differentiation theorem. Since we need this result to prove the differentiation theorem, we now want a proof that does not use the differentiation theorem.)

12.75   Lemma. Let be a power series. Then the two series and have the same radius of convergence.

Proof: I'll show that for all .

a) If converges, then converges.
b) If converges and , then converges.
a) follows from the comparison test, since

To prove b), suppose converges and . By lemma 12.73, converges, and hence is bounded. Choose such that

Then , and

By the comparison test, converges.

12.76   Corollary. and have the same radius of convergence.

Proof: Use the lemma twice.

12.77   Theorem. Let be a power series with positive radius of convergence. Let for all in the disc of convergence for and let be the function corresponding to the formal derivative of . Then is differentiable on its disc of convergence, and for all in the disc of convergence.

Proof: Let be a point in the disc of convergence, and let be a different point in the disc. Then

Let

Then

and since

the theorem will follow if we can show that is continuous at .

In the calculation below, I quietly use the following facts:

a) When , .
b) When , .

 (12.78)

Let the radius of convergence of our power series be , and let . Then

Let . Then , and

(Here I've used the fact that for .) Thus

We noticed in the corollary to lemma 12.75 that the series has radius of convergence , and hence converges to a limit , and by (12.78),

If is a sequence in such that , then

for all large , and by the null-times bounded theorem and comparison theorem for null sequences, . Hence, is continuous at .

Next: 12.9 Some XVIII-th Century Up: 12. Power Series Previous: 12.7 Special Values of   Index