Analytic on an interval/expressing with a different power series

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SUMMARY

The discussion focuses on demonstrating that a real power series, defined as f:= ∑ cnxn with a radius of convergence R > 0, can be expressed as another power series centered at b within the interval I = (-R, R). Specifically, it asserts the existence of a power series ∑ dn(x-b)n that converges to f(x) for |x-b| < r - |b|. The participants emphasize the analytic nature of f on I and the need for a rigorous approach to re-center the series.

PREREQUISITES
  • Understanding of power series and their convergence properties
  • Familiarity with the concept of radius of convergence
  • Knowledge of analytic functions and their representation
  • Basic skills in manipulating series and limits
NEXT STEPS
  • Study the properties of power series and their convergence criteria
  • Learn about Taylor series and how to derive them from functions
  • Explore the concept of re-centering power series
  • Investigate the implications of radius of convergence on function behavior
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This discussion is beneficial for mathematics students, particularly those studying real analysis, as well as educators and anyone interested in the properties of power series and analytic functions.

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Homework Statement



Suppose the real power series \sum ^{\infty}_{n=0}c_{n}x^{n} has radius of convergence R > 0. Define f:= \sum ^{\infty}_{n=0}c_{n}x^{n} on I:= (-R, R) and let b \in I. Show that there exists a power series \sum d_{n}(x-b)^{n} that converges to f(x) for |x-b| < r - |b|.

Homework Equations



None that I can think of.

The Attempt at a Solution



I don't even know where to begin. Obviously the function is analytic on the open interval I because it is defined by a power series that converges on I. Intuitively, I understand that the function can be represented by a power series with a different center (in I) and smaller radius of convergence, but I can't think of how to start demonstrating this rigorously.
 
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If f(x) is as defined, what can you say about f(u) where u= x+ b? Notice that x= u- b.
 

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