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If I have that $e^x$ is defined by the unique continuous function $\mathbb R\to \mathbb R^*$ s.t. $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ and $f(1)=e$, is it possible to prove that $$\lim_{h\to 0}\frac{e^h-1}{h}=1\ \ ?$$

Jam
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Walace
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    How are you defining $e$? – user729424 Feb 02 '20 at 14:41
  • @AndrewOstergaard I think the definition of $f(1)=e$ is sufficient, as $f^{-1}(e)=1$ regardless of the value of $e$. – Jam Feb 02 '20 at 14:42
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    @Jam: if $e$ is not the number we know as $e$, the limit is not $1$. – Martin Argerami Feb 02 '20 at 14:44
  • $e$ is the number $2.7...$ @AndrewOstergaard – Walace Feb 02 '20 at 14:46
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    @Walace: Right, but how is it defined? Without a way of continuing that decimal expansion arbitrarily far out, there has to be some definition for it. – anomaly Feb 02 '20 at 14:47
  • @Walace: if you use $e=2.71$, the limit is not $1$. So you must have a more specific number in mind, and we are asking you what it is. – Martin Argerami Feb 02 '20 at 14:48
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    Existence of the limit is proved here. – cqfd Feb 02 '20 at 14:50
  • @MartinArgerami Yes, you're right. I erroneously thought $\lim\frac{f(a)-1}{h}$ was the inverse of $f$ for all $a=f(1)$. – Jam Feb 02 '20 at 14:53
  • @MartinArgerami: Of course it's not $2.71$, I mean, $e$ is the famous number. What else ? I don't understand your intervention... – Walace Feb 02 '20 at 14:56
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    @Walace Their point is that the decimal expansion is meaningless until you've expressed how each decimal is found. If comes from the usual definition of $e=\lim_{h\to0} (1+h)^{1/h}$ then your proposition pops straight out. – Jam Feb 02 '20 at 14:58
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    @Walace: a fairly common definition of $e$ (the one I was taught when I was a student, for instance), is "the number such that $\lim_{h\to0}\frac{e^h-1}h=1$". You still haven't said what $e$ is in your context. – Martin Argerami Feb 02 '20 at 15:02
  • @MartinArgerami: In my lecture $e=\lim_{n\to \infty }\left(1+\frac{1}{n}\right)^n$. – Walace Feb 02 '20 at 15:15
  • Of course $2^x$ also satisfies $f(x+y)=f(x)f(y)$, but then $\lim_{h\to 0}(2^h-1)/h \ne 1$. So you need more than the functional equation. You need $f(1)=e$, where $e$ is a certain unique number. In fact, you might define $e$ by the choice that yields $\lim_{h\to 0}(e^h-1)/h = 1$. – GEdgar Feb 02 '20 at 15:25

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Note that $$ \left(1+\frac1n\right)^{n}=\sum_{k=0}^n {n\choose k}\frac1{n^k}=\sum_{k=0}^n \frac1{k!}\,\prod_{j=1}^{k-1}\left(1-\frac jn\right). $$ From this one can show that $$ e=\sum_{k=0}^\infty \frac1{k!}. $$ Now define $$g(x)=\sum_{k=0}^\infty \frac{x^k}{k!}.$$ This converges uniformly on any bounded set, so $g$ is in particular continuous. It is also differentiable with $g'=g$ and $g(1)=e$. Fix $y$. Then $h(x)=g(x+y)$ satisfies $h'(x)=h(x)$, $h(0)=g(y)$. This is a solution to the initial value problem $h'=h$, $h(0)=g(y)$, of which $g(y)g(x)$ is also a solution. By the uniqueness of the solution (Picard) we get that $g(x+y)=g(x)g(y)$. So $g$ is a continuos function with $g(x+y)=g(x)g(y)$ and $g(1)=e$. By your hypothesis, $g=f$. As $g$ is differentiable, $$ \lim_{h\to0}\frac{e^h-1}h=\lim_{h\to0}\frac{f(h)-1}h=\lim_{h\to0}\frac{g(h)-g(0)}h=g'(0)=1. $$

Martin Argerami
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    (+1) Flawless solution; works with both choices $$e := \lim_{n\to\infty} \left( 1 + \frac 1 n \right)^n, \qquad \text{or} \qquad e := \sum_{k=0}^\infty \frac 1 {k!}. $$ – giobrach Feb 02 '20 at 15:55