Taylor series expansion of a power series.

[Skrew]

If f(x) is a power series on S = (a-r, a+r), we should be able to expand f(x) as a taylor series about any point b within S with radius of convergence min(|b-(a-r)|, |b - (a + r)|)

Does anyone have a proof of this or a link to a proof? I have seen it proved using complex analysis, but I would like to see a proof that uses only concepts from real analysis.

 

[HallsofIvy]

It's really just a substitution. If $f(x)= \sum a_n(x- a)^n$ for x in [a-r, a+r], and b is also in [a-r, a+r], then let y= x- a+ b so that x- a= y- b. Then $\sum a_n(x- a)^n= \sum a_n(y- b)^n$. Simply renaming "y" to "x" gives $\sum a_n(y- b)^n$, a power series centered at x= b. Of course, if the original power series only converged in [a- r, a+ r], and b< a, then the new power series could only converge in some subinterval of [a- r, b+ (b- (a- r))]= [a- r, 2b- a+ r] since a power series always converges in some interval. Similarly, if b> a the new power series could only converge in some subinterval of [2b- a- r, a+r]

 

[Skrew]

It's really just a substitution. If $f(x)= \sum a_n(x- a)^n$ for x in [a-r, a+r], and b is also in [a-r, a+r], then let y= x- a+ b so that x- a= y- b. Then $\sum a_n(x- a)^n= \sum a_n(y- b)^n$. Simply renaming "y" to "x" gives $\sum a_n(y- b)^n$, a power series centered at x= b. Of course, if the original power series only converged in [a- r, a+ r], and b< a, then the new power series could only converge in some subinterval of [a- r, b+ (b- (a- r))]= [a- r, 2b- a+ r] since a power series always converges in some interval. Similarly, if b> a the new power series could only converge in some subinterval of [2b- a- r, a+r]

The subsitution is interesting but your functions are not going to be equal for their common interval of convergence.