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Disclaimer: I'm no math expert!

I understand that the constant $$e$$ is expressed as follows:

$$e = \sum_{n=0}^{\infty} \frac1{n!} = 1 + \frac1{1*1} + \frac1{1*2} + ...$$

What would be the implications of defining it as:

$$e = \sum_{n=2}^{\infty}\frac1{1*2} + \frac1{1*3}...$$

I guess algebraically it wouldn't matter much as it just removes 2 from the equation. But are there other cases where the constant is used in such a way that perhaps a value 'higher than one' might be preferred?

To me it feels more logical to have this value 'between zero and one'.

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  • $\begingroup$ I don't fully understand your question. But I can say that $e$ is not a logarithm, it is just a constant. It is the base of the natural logarithm. The series you have listed is actually the series for the exponential $e^x$ for the case of $x=1$. As for the question of the "implication" you mention, the implication is that the sum would no longer be equal to $e$... $\endgroup$
    – graydad
    Commented Mar 14, 2015 at 20:53
  • $\begingroup$ Did you mean $\frac{1}{1 \cdot 2 \cdot 3}$ instead of $\frac{1}{1 \cdot 3}$? Becaues there is no number $x$ such that $x = \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{1 \cdot 4} + \cdots = \sum_{n=2}^\infty \frac{1}{1 \cdot n}$, as your latter form implies. $\endgroup$
    – Freyja
    Commented Mar 14, 2015 at 20:53
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    $\begingroup$ What is the question ? $e\neq \frac{1}{1\cdot 2}+\frac{1}{1\cdot 3}+...$ $\endgroup$
    – Surb
    Commented Mar 14, 2015 at 20:53
  • $\begingroup$ I tried rephrasing the question. Is it more correct now? $\endgroup$
    – Ropstah
    Commented Mar 14, 2015 at 20:58
  • $\begingroup$ I also don't understand the question. We named $e$ for a reason. It's very important. If we redefined it to mean something else then we wouldn't have a name for $e$ anymore, and that would be awful. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 14, 2015 at 21:07

1 Answer 1

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$e$ is nothing but a name for a number, namely $$ e = 2.71828182845904523536028747135266249775724709369995\ldots $$

This number that we call $e$, has a number of interesting properties:

  1. $e^x$ is its own derivative.
  2. The inverse of $e^x$, $\ln x$, has the derivative $\frac{1}{x}$.
  3. $e = \sum_{n=0}^\infty \frac{1}{n!}$
  4. $e = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n$

If we called a chair a microwave, it wouldn't make it able to cook food. Just the same, naming a different number as $e$ wouldn't give it the same properties as above, it would still just be the same number, with its own (often less interesting) properties. (In fact, $e$ is the only number to satisfy have the four properties above.)

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  • $\begingroup$ I saw the 4th method as the definition for e, but it is a property of e. Thanks for clarifying. $\endgroup$
    – Ropstah
    Commented Mar 14, 2015 at 21:19
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    $\begingroup$ A definition of a number is just a property that's unique to a single number. Therefore, all four properties are all definitions of $e$. $\endgroup$
    – Freyja
    Commented Mar 14, 2015 at 21:22
  • $\begingroup$ Nice answer, but the series starts from $n=0$. :) $\endgroup$
    – Vincenzo Oliva
    Commented Mar 14, 2015 at 21:57

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