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Function represented by power series

  • Thread starter sandra1
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  • #1
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Homework Statement


If a function f is represented by the power series ∑(k=0 to ∞) a_k(x-a)^k, with a radius of convergence )<R<∞, then f is continuous on the interval (a-R, a+R)


Homework Equations





The Attempt at a Solution



I don't know if my proof is loose at some point or not, because it seems so easy. Is there anything that is obviously not right about my proof? Thanks very much

So choose x=m in (a-R, a+R), the series converges on the interval (a-R,a+R)
then we would have
lim(x->m) ∑(k=0 to ∞) a_k(x-a)^k = ∑(k=0 to ∞) (a_k)lim(x->m) (x-a)^k = ∑(k=0 to ∞) (a_k)(m-a)^k = f(m).

So f is continuous on any arbitrary m on (a-R, a+R) therefore ∑(k=0 to ∞) a_k(x-a)^k is continuous on (a-R, a+R)
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
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It is possible to have series of continuous functions that converges to a non-continuous function. To prove the result, you must use the fact that a power series is uniformly continuous within its radius of convergence.
 
  • #3
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Your proof is invalid, because you cannot assume that

[tex]\lim_{x \to m} \lim_{n \to \infty} \sum_{k=0}^n a_k (x-a)^k = \lim_{n \to \infty} \lim_{x \to m} \sum_{k=0}^n a_k (x-a)^k[/tex].

Consider [itex]x_{m,n}[/itex] = 0 when n < m and 1 when n >= m. Then [itex]\lim_{n \to \infty} \lim_{m \to \infty} x_{m,n} \neq \lim_{m \to \infty} \lim_{n \to \infty} x_{m,n}[/itex].
 
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