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In this proof I saw in a stack exchange answer and quite a few other places too, the first step is to substitute $x$ with $i\theta$ in the Taylor expansion of $e^x$. What I don't get is how can we plug in complex values, knowing that the Taylor series loses accuracy at values largely deviating from $0$(here)? How do complex numbers satisfy this condition? How can we even check if it's deviating from $0$, if it's not even real? Is the defining feature of Taylor series different for complex numbers?

AltercatingCurrent
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  • What's the problem with $\theta i$ not being real? For any complex number $z$, we have $\displaystyle e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$. – José Carlos Santos Aug 06 '22 at 18:50
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    It doesn't loose accuracy if you use the whole series. – callculus42 Aug 06 '22 at 18:50
  • @callculus42 how so? Isn't the series calculated "around" specific values? As in, doesn't the Taylored polynomial graph "wrap" around the graph of the actual function only about a specific value, and deviate as we go further from it? – AltercatingCurrent Aug 06 '22 at 18:51
  • Yes, the series is developed around $x=a$. But when you input a value $x_0$, then you get the exact value of $f(x_0)$ – callculus42 Aug 06 '22 at 18:54
  • @callculus42 yeah I know that. But how is putting $x=i\theta$ even comparable to $a=0$(around which the expansion of $e^x$ is computed) in the proof? – AltercatingCurrent Aug 06 '22 at 18:56
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    @AltercatingCurrent: "the Taylor series loses accuracy at values largely deviating from $0$" ... A Taylor polynomial (that is, a finite truncation of a Taylor series) loses accuracy, but you can always overcome that by using a longer polynomial. Indeed, a common exercise is to use a formula for the remainder to determine how long a polynomial you need to keep accuracy within acceptable bounds across a given range. In any case, "in the limit", the infinitely-long Taylor series provides perfect accuracy. – Blue Aug 06 '22 at 18:56
  • @Blue oh, i didn't think of that before! And now that you say it, it does make sense. However it feels unintuitive to plug in complex values, considering the graphical significance of the Taylor series. – AltercatingCurrent Aug 06 '22 at 18:58
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    @AltercatingCurrent: That said, it's not at all obvious that one can substitute complex values into a series derived from real analysis and get a reasonable result. The details of this kind of thing are covered in complex analysis courses; prior to that, getting at Euler's Formula from this substitution is really more of a heuristic argument than a formal proof. – Blue Aug 06 '22 at 19:01
  • @Blue ah, I see. Thanks to you and everyone else who commented! – AltercatingCurrent Aug 06 '22 at 19:18
  • The radius of convergence does matter. Happily, the exponential series converges everywhere and fairly simply can the basic convergence theory from real analysis be carried over to complex analysis. So the map $\theta\mapsto\exp(i\theta)$ is a well-defined complex function (issues of complex analyticity, differentiability etc. are irrelevant here) that, for $\theta$ real, produces a series which may be split into its real and imaginary parts, cosine and sine (whose series also converge everywhere). – FShrike Aug 06 '22 at 19:51
  • But e.g. if you thought you can take the Maclaurin series for $\log(1-x)$ and allow arbitrary complex entries, you certainly would get a convergence problem (but it would be entirely “accurate” within the open unit disc centred at one) – FShrike Aug 06 '22 at 19:53

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