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 January 17th, 2018, 03:37 AM #11 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra Essentially your argument is one of context and the OP implies through the usr of R that we are talking about real numbers, where $0^0$ is undefined. In other contexts different definitions or conventions do apply. The example of $\displaystyle e^x=\sum_{n=0}\frac{x^n}{n!}$ is a rather amusing one. It's an expression packed with conventions and notational obfuscation.
January 17th, 2018, 08:38 AM   #12
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Quote:
 Originally Posted by v8archie Essentially your argument is one of context and the OP implies through the usr of R that we are talking about real numbers, where $0^0$ is undefined. In other contexts different definitions or conventions do apply.
I never read anywhere in the OP that he's talking about the real numbers only. Sure $R$ can be a real number, it's the exponent that must be natural.

Quote:
 The example of $\displaystyle e^x=\sum_{n=0}\frac{x^n}{n!}$ is a rather amusing one. It's an expression packed with conventions and notational obfuscation.
But all of math is packed with convention and notational obfucation!!!! Some like to define math as that which is invariant under notation.

January 17th, 2018, 09:16 AM   #13
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 Originally Posted by topsquark Hmmmmm.... Mathematica does not like this one. See here. -Dan
If you plot $\left(\dfrac 1 x\right)^{\frac 1 x}$

from 1 to some big number the graph certainly tends towards 1.

 January 17th, 2018, 10:20 AM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra $\displaystyle \left( \frac1x \right)^{\frac1{\ln x}}$ doesn't
 January 17th, 2018, 10:25 AM #15 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond $\mathcal{O}(x)\gg\mathcal{O}(\log x)$ Last edited by greg1313; January 18th, 2018 at 06:53 AM.
January 17th, 2018, 12:44 PM   #16
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Quote:
 Originally Posted by topsquark Hmmmmm.... Mathematica does not like this one.
One can just do this or this.

January 17th, 2018, 12:47 PM   #17
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 Originally Posted by greg1313 $\mathcal{O}(x)\gg\mathcal{O}(\log x)$
Yes. More to the point $$x^{-\frac1{\ln x}}=e^{\ln{x^{-\frac1{\ln x}}}}=e^{-\frac1{\ln x}\ln x}=e^{-1}$$

Last edited by greg1313; January 18th, 2018 at 06:53 AM.

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