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$\exp(x)$ is usually defined in three different ways:

1) By its Taylor series: $\exp(x)=\sum_{k=0}^{\infty} \frac{x^k}{k!}$

2) By its derivative: $\exp(x)'=\exp(x)$

3) By the limit $\exp(x)=\lim_{N \rightarrow \infty} \left(1+\frac{x}{N} \right)^N$

In textbooks they mostly use the 3rd way, because $\exp(x)$ is introduced before the derivative.


But it just occured to me - we don't need this much information to define this function. There is a very simple definition which allows to recover all the other properties:

4)

$$\lim_{x \rightarrow 0} \exp(x) = \lim_{x \rightarrow 0} (1+x)$$

Now, it's just two terms of Taylor expansion - it doesn't seem like much. Yet we can prove the following properties:

a) It follows that $\lim_{x \rightarrow 0} \exp(a x) = \lim_{x \rightarrow 0} \exp^a(x)$

$$\lim_{x \rightarrow 0} \exp^a(x) = \lim_{x \rightarrow 0} (1+x)^a = \lim_{x \rightarrow 0} (1+a x) = \lim_{x \rightarrow 0} \exp(a x)$$

b) Using this property we recover the 3rd definition

$$ \lim_{N \rightarrow \infty} \left( 1+\frac{x}{N} \right)^N=\lim_{N \rightarrow \infty} \left( \exp \left(\frac{x}{N}\right) \right)^N=\lim_{N \rightarrow \infty} \left( \exp^{\frac{1}{N}}(x) \right)^N=\exp(x) $$

Does this work, or did I make a mistake somewhere?

Edit

I was obviously wrong - there is an infinite number of functions with the property 4. So the definition needs to be:

4) For $x \rightarrow 0$ $$ \exp(x) \approx 1+x$$

For any $x$

$$\exp^a(x) = \exp(a x)$$

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    $\begingroup$ Your definition (4) is satisfied by any function $f(x)$ such that $\lim_{x\to 0} f(x) = 1$, so there is no way that you can prove $f(x)=\exp(x)$ from it... $\endgroup$ – Hans Lundmark Feb 11 '16 at 11:34
  • $\begingroup$ If $a$ is real, how do you define $\text{thing}^a$? $\endgroup$ – Martín-Blas Pérez Pinilla Feb 11 '16 at 11:51
  • $\begingroup$ As $(\exp(x))^a$ $\endgroup$ – Yuriy S Feb 11 '16 at 11:52
  • $\begingroup$ See the edit of my previous comment. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 11 '16 at 11:53
  • $\begingroup$ $\approx$ doesn't have a standard meaning. However, if we define $f(x)\approx g(x)$ to mean $\lim\limits_{x\to0}\frac{f(x)-g(x)}x=0$, then $\exp(x)\approx 1+x$ is actually equivalent to $\exp'(0)=\exp(0)$. Along with the rule $\exp(a)\exp(b)=\exp(a+b)$, this would imply $\exp'(x)=\exp(x)$ (why?), which is the second definition at the top. $\endgroup$ – Akiva Weinberger Feb 11 '16 at 11:55
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$$\lim_{x\to 0} (1+x)$$ is simply equal to $1$, so the equation you wrote is equal to the equation $$\lim_{x\to0} \exp(x) = 1$$

which is most certainly not enough to characterise the exponential function, for example $\cos(x)$.


As for what you proved in points (a) and (b), I don't even know what you were intending to prove so I cannot comment on that.

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  • $\begingroup$ Thanks, I noticed it, see the edit $\endgroup$ – Yuriy S Feb 11 '16 at 11:35
  • $\begingroup$ I wanted to write $\exp(x) \approx 1+x$ for $x \rightarrow 0$ at first, would it be better? $\endgroup$ – Yuriy S Feb 11 '16 at 11:39
  • $\begingroup$ @YuriyS Not really, since the function $1+x$ also satisfies that condition. $\endgroup$ – 5xum Feb 11 '16 at 11:40
  • $\begingroup$ And together with the property $\exp^a(x)=\exp(ax)$? $\endgroup$ – Yuriy S Feb 11 '16 at 11:42
  • $\begingroup$ @YuriyS That might be enough, yes. But I'm not sure. Also, I don't see the benefit of such a definition. $\endgroup$ – 5xum Feb 11 '16 at 11:44

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