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If analytic continuation gives the wrong answer sometimes, at least under typical/basic reasoning, like when it assigns a divergent series a finite value, then why do we trust it to give us values like $e^{i\pi}$? Why is this the chosen way to give values to expressions outside of their domain?

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  • $\begingroup$ How do you want to define $e^{i\pi}$? $\endgroup$ – Robert Israel Mar 4 at 21:28
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    $\begingroup$ Analytic continuation does not "give the wrong values" -- ever. If you think it does, then you are misinterpreting it. $\endgroup$ – saulspatz Mar 4 at 21:34
  • $\begingroup$ You seem to misunderstand what analytic continuation means. For example $f(z)=\frac{1}{1-z}=\sum_{k=0}^\infty z^k$, for $|z|\lt 1$. For the domain of equality, there is no problem, but for $|z|\ge 1$, the analytic continuation gives sensible results, but the series diverges. $\endgroup$ – herb steinberg Mar 4 at 21:53
  • $\begingroup$ @saulspatz so along the lines of what herb said, iff it converges, it always converges on the "right value". I didn't know that but it makes sense. To answer the rest of the question, how come this type of continuation gives the perferred value to the rest of the function though, the one that we consider to be the ACTUAL value? $\endgroup$ – Benjamin Thoburn Mar 4 at 22:33
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    $\begingroup$ Because there is typically only one way to extend a function so that it is analytic in the larger larger domain. (This is true if the larger domain is "simply connected.") So for example, if we want to extend the sine function from the real line to the complex plane, there's only one way to do it, if we require that the extended function be differentiable. Since no one would want to extend the sine function in a way that loses differentiability, there's only one "right" way to do it, and that is the analytic continuation. $\endgroup$ – saulspatz Mar 4 at 23:16

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