Proving Continuity of Power Series Function

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The discussion centers on proving the continuity of the power series function f(x) = sum(a_n*x^n) within its radius of convergence. Participants emphasize the need to demonstrate that for any |a| < R and given e > 0, there exists d > 0 such that |x-a| < d implies |f(x) - f(a)| < e. A challenge arises in bounding the difference quotient |(f(x) - f(a))/(x-a)| near x = a without prior knowledge of differentiation. The conversation highlights the pedagogical implications of approaching power series before learning differentiation. Ultimately, the need for a rigorous approach to show convergence without differentiation remains a key concern.
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Homework Statement


Show, from the definition of continuity, that the power series function f(x)=sum(a_n*x^n) is continuous for its radius of convergence.

Homework Equations


Definition of continuity

The Attempt at a Solution


Must show that for any |a| < R, given e>0 there exists d>0 such that |x-a|<d => |f(x) - f(a)|.
|f(x)-f(a)| < e.
|f(x) - f(a)| <= |f(x-a)|
Then I get stuck here.
Any help would be appreciated
 
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|f(x)-f(a)| is not less than |f(x-a)|. It's not like f is linear or something. |f(x)-f(a)|=|(f(x)-f(a))/(x-a)|*|x-a|. Now to get a d, you need a bound on |(f(x)-f(a))/(x-a)| near x=a. Hint: doesn't that look like a difference quotient?
 
Hmm we haven't done differentiation yet so I'm not sure how helpful the |(f(x)-f(a))/(x-a)| will be.
 
You are doing power series without having done differentiation!? That's an interesting pedagogical approach. You can still factor (x-a) algebraically from each power of f(x)-f(a), but I'm not sure how you show the rest of it converges without using the differentiability of power series.
 
yeah...
Thanks for the help though
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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