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I am dealing with $f(x) = \log_3(2x-1)$ and I already found out the Taylor series expansion at $a = 1$ as $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2)^n(x-1)^n}{n\log(3)}$. My problem is I need to find all possible values of a so that Taylor series expansion at a is well-defined. What does it mean if the Taylor series expansion is well-defined? Do I need to solve for the internal of convergence?

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  • $\begingroup$ I guess you need to check if radius of convergence $R>0$. I suppose that Taylor series expansion of $e^{-1/x^2}$ which is $0$ at $a=0$ can also be considered as not well defined even though its radius of convergence is obviously infinite, but we are not in this situation here. $\endgroup$
    – zwim
    Jun 10 at 9:47
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Using, as you did, natural logarithms and computing the successive derivatives, around $x=a$, you have $$\log_e(3)\,f(x)=\log(2a-1)-\sum_{n=1}^\infty \frac{\left(\frac{1}{2}-a\right)^{-n}}{n} (x-a)^n$$ So, the logarithm gives a first condition.

Now, for the radius of convergence $$b_n=\frac{\left(\frac{1}{2}-a\right)^{-n}}{n}\implies \frac{b_n}{b_{n+1}}=\frac{ n+1}{n}\left(\frac{1}{2}-a\right)\quad \to \left(\frac{1}{2}-a\right)$$ I am sure that you can conclude.

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