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This is a question that came to my mind as I was reading about power series. Is it possible for two power series to, in some sense, represent the same function, and thus be "identical"? So the definition would go like:

The power series $f(z)=a_0+a_1z+a_2z^2+\ldots$ and $g(z)=b_0+b_1z+b_2z^2+\ldots$ are identical if $f(z)=g(z)$ for all values of $z$ such that $f(z)$ and $g(z)$ converge.

With this definition, if $f(z)$ or $g(z)$ has radius of convergence $0$ and $a_0=b_0$, then they are identical. What if the radius of convergence $R_f$ and $R_g$ are both positive? Does that guarantee that $a_i=b_i$ for all $i\geq 0$?

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Good question! Yes, it does guarantee that $a_i = b_i$. The way to see this is to note that if two functions are the same, then they have the same derivative, the same 2nd derivative, etc. And the derivative of a power series

$$f(x) = a_0 + a_1x + \dotsb$$

is always $f'(0) = a_1$, and the second derivative is always $2a_2$, etc.

Here we've taken the power series to be centered at zero, but the same applies if it is centered somewhere else.

Note that the condition $R_f, R_g>0$ is used in claiming that $f'(0) = a_1$, because if $R_f \not > 0$, there is no derivative for such a function, since it is not defined in an open set. Once we have convergence in $(-r,r)$ for $r>0$, we know that $f$ converges uniformly on compact subsets of $(-r,r)$, and so we can take the derivative term-by-term on $(-r,r)$.

Note as a simple counter-example that $$f(x) = 1 + 2x + 2^2x^2 + 3^3x^3 + \dotsb$$ and $$g(x) = 1 + 3x + 2^2x^2 + 3^3x^3 + \dotsb$$ agree everywhere where they converge, though they are not identical.

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  • $\begingroup$ Hmm... where is the condition that $R_f,R_g>0$ being used here? Because if $\min\{R_f,R_g\}=0$, then it is not necessary that $a_i=b_i$ for $i>0$ (as I mentioned in the post). $\endgroup$ – PJ Miller Aug 27 '13 at 2:48
  • $\begingroup$ I added a bit more. $\endgroup$ – Eric Auld Aug 27 '13 at 2:57
  • $\begingroup$ Thank you for your answer, Eric! I missed the point that the derivative wouldn't exist if $R_f=0$ (or $R_g=0$). $\endgroup$ – PJ Miller Aug 27 '13 at 3:00
  • $\begingroup$ You're very welcome. $\endgroup$ – Eric Auld Aug 27 '13 at 3:03

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