Wiki on e (mathematical constant)

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Discussion Overview

The discussion revolves around the mathematical constant e, particularly its definition and properties as presented in a Wikipedia article. Participants explore the implications of limits and derivatives involving e, as well as its representation through infinite series and sequences.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion regarding a limit involving e, specifically questioning the interpretation of a term leading to an indeterminate form (0/0).
  • Another participant suggests writing e^h as an infinite series, indicating a potential method for understanding e.
  • A third participant clarifies the derivative of a^x using limits, emphasizing that the limit process should not lead to the indeterminate form mentioned and explaining how to properly evaluate it.
  • A fourth participant provides a numerical example of the sequence F(n) = (1+1/n)^n, illustrating how it approaches the value of e as n increases.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the limit involving e, and there are differing views on how to approach the concept of e in relation to limits and derivatives.

Contextual Notes

There are unresolved assumptions regarding the definitions and properties of limits and derivatives, particularly in the context of the indeterminate form mentioned by the first participant.

Pellefant
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The source: http://en.wikipedia.org/wiki/E_(mathematical_constant)


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I can not understand the last peace in that equation,

If e=a it will be e^x*((1-1)/0) ...which means 0/0, that don't make sense :(, what am i missing
 
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Write e^h as an infinite series.
 
What you are quoting from Wikipedia is the "difference quotient" calculation for the deriviative of ax:
[tex]\frac{d}{dx}a^x= \lim_{h\rightarrow 0}\frac{a^{x+h}-a^x}{h}= \lim_{h\rightarrow 0}\frac{a^xe^h-a^x}{h}= a^x \lim_{h\rightarrow 0}\frac{a^h- 1}{h}[/tex]

Now, I'm not at all sure what you mean by "If e=a it will be e^x*((1-1)/0) ...which means 0/0". Whether e= a or not, if you replace h by 0 in the limit, you get (1-1)/0 which does not exist. But that's not how you find limits! What you can do is show that
[tex]\lim_{h\rightarrow 0} \frac{a^h- 1}{h}[/tex]
does, in fact, exist so that the derivative of ax exists and is just that constant times ax. You can then define e to be the value of a such that that limit is 1, giving [itex]d e^x/dx= e^x[/itex].
 
When you are looking at a limit process, you can see how numbers work. For example

F(n) = (1+1/n)^n, give us values F(1) = 2, F(2)=2.25; F(5) = 2.49, F(10) = 2.59; F(100) = 2.70, and as we approach infinity it goes to e=2.71828...
 

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