My Math Forum Range of e^z

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December 4th, 2017, 02:20 PM   #11
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Quote:
 Originally Posted by zylo In R, if $\displaystyle \lim_{x \rightarrow \infty} f(x) = \infty$ or $\displaystyle L$, then $\displaystyle f(\infty) = \infty$ or $\displaystyle f(\infty) = L$ is clear,transparent, and conventional. The meaning is explicitly implied by the fact that $\displaystyle \infty$ is not a number.

$\displaystyle 1/\infty = 0$

$\displaystyle e^{z}=\lim_{n\rightarrow \infty}P_{n}(z)$ and $\displaystyle P_{n}(z) = 0$ has a solution for all n.

Assumption: If something is true for all n, it is true in the limit as n $\displaystyle \rightarrow \infty$.

If you are dealing with mappings from all of z to all of w, you have to deal with infinity, like it or not.

December 4th, 2017, 02:36 PM   #12
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Quote:
 Originally Posted by zylo Assumption: If something is true for all n, it is true in the limit as n $\displaystyle \rightarrow \infty$.
1/|n| > 0 is true for all non-zero n, but untrue in the limit as n $\rightarrow \infty$.

December 4th, 2017, 02:42 PM   #13
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Quote:
 Originally Posted by zylo Assumption: If something is true for all n, it is true in the limit as n $\displaystyle \rightarrow \infty$.
At least you've got as far as calling this an assumption rather than a self-evident truth. Only a little further to go.
Quote:
 Originally Posted by zylo If you are dealing with mappings from all of z to all of w, you have to deal with infinity, like it or not.
No, because $\infty \not \in \mathbb C$.

 December 4th, 2017, 05:47 PM #14 Senior Member   Joined: Sep 2016 From: USA Posts: 417 Thanks: 231 Math Focus: Dynamical systems, analytic function theory, numerics lol this is so good

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