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I am confused about the definition of analytical functions on real line, or, to be more precise, how this definition is used. According to Wikipedia

Analytic function is a function that is locally given by a convergent power series.

So it means that in every point on the considered domain there exist an open neighborhood of this point, in which Taylor series converges to function value. However, I often encounter textbooks that write a McLaren series, for example, for an exponent enter image description here

and then write something like "exponent is an analytic function on the entire real line".

  1. Do I understand correctly that the analyticity of a function at one point does not guarantee analyticity at other points on the line?

So for exponent we have analyticity in $0$ and the neighborhood of $0$ is the entire real line. I know (however, I want my proof to be checked) how to prove this for exponent: $\exp(x)=\exp(x-x_0+x_0)=\exp(x_0)\exp(x-x_0),$ so since $\cdot\sum_{i=0}^{\infty}\frac{(x-x_0)^i}{i!}$ converges to $\exp(x-x_0)$ in every neighborhood of $0$ we could write $\exp(x)=\sum_{i=0}^{\infty}\frac{\exp(x_0)(x-x_0)^i}{i!}.$

But examples come to mind, such as the function $\frac{1}{1-x}.$ I know the proof of its analyticity in $0$ and that the neibourhood of $0$ in which Taylor series converges to function value is $(-1,1)$ (which in fact coincides with the convergence radius of the McLaren series). But concerning this example, I have the following questions

  1. Will this function be analytic at other points of the real line? If yes, how can we prove this and how will the neighborhood of analyticity depend on the point? It seems that even the radius of convergence of the Taylor series strongly depends on the point $ {\frac {1}{1-a}}\sum \limits _{k=0}^{\infty }{{\left({\frac {x-a}{1-a}}\right)}^{k}}={\frac {1}{1-a}}\cdot {\frac {1}{1-\left({\frac {x-a}{1-a}}\right)}}={\frac {1}{1-x}},$ so ${\displaystyle \left|{\frac {x-a}{1-a}}\right|<1}$ and, for example, for $a=0.5$ we have $x\in(0,1).$

  2. How is the radius of convergence related to the size of the neighborhood of the point where the function will be analytic? It seems that the vicinity of analyticity must be contained in the interval in which the Taylor power series converges to the value of the function.

  3. Is there any example of any function which is expressed as a McLaren series in some neighborhood of zero, but there are points on the line where it is not analytic?

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1 Answer 1

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If $f(x)=\sum_{n\ge 0} c_n x^n$ converges for $|x|< R$ then it converges absolutely for $|x|<R$ so that for $|a|< R$,

$$f(a+x)=\sum_{n\ge 0}c_n (a+x)^n=\sum_{n\ge 0}c_n \sum_{k=0}^n x^k a^{n-k}{n\choose k}$$ converges absolutely as well for $|x|< R-|a|$, which implies that we can change the order of summation to obtain that

$$f(a+x)= \sum_{k\ge 0}x^k \sum_{n\ge k}c_n a^{n-k}{n\choose k}$$ is given by a power series on the disk $|x|< R-|a|$.

That's why analyticity makes sense, being given by a power series on a disk $U$ implies being given by a power series on each disk contained in $U$.

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  • $\begingroup$ Thanks for your proof! But I still confused about Taylor series. Your argument is suitable for convergence of power series to some value. But still, why is this value is a function value in a given point? I mean, Taylor series not necessarily converge to function value. $\endgroup$
    – Motoko
    Commented Apr 3, 2022 at 13:38
  • $\begingroup$ Unclear what you mean, my $f$ is given by a power series on $|x|<R$ and so does $f(a+x)$ on $|x|<R-|a|$ $\endgroup$
    – reuns
    Commented Apr 3, 2022 at 14:50
  • $\begingroup$ I'm sorry, I didn't read your reply before. Now everything became clear. Thanks! $\endgroup$
    – Motoko
    Commented Apr 3, 2022 at 16:21

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