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Define exponential function $\exp$ as follows: $$\exp:\Bbb{R}\rightarrow(0,\infty)$$ $$\begin{align}i) \ \forall x,y\in \Bbb{R}:\exp(x+y)=\exp(x)\cdot\exp(y)\end{align}$$ $$\begin{align}ii) \ \forall x\in \Bbb{R}:\exp(x)\ge x+1 \end{align}$$ Book says these 4 properties can be derived using this definition.

$$(1) \ \exp \text{is increasing on } \Bbb{R} $$ $$(2) \ \exp(0)=1$$ $$(3) \ \forall x\in \Bbb{R}: \forall n \in \Bbb{Z}: \exp(nx)=(\exp(x))^n$$ $$(4) \ \forall x\in \Bbb{R},x>0:\exp(x)>1$$ The only thing i managed to do myself was to prove $(3)$ by mathematical induction for $n\in\Bbb{N}$. I also tried to prove the monotonicity directly from definition, but can't seem to find a starting point. Would you please give me hint on where to start with proofs of these? Another thing I am interested in is a proof of uniqueness of this definition. How do i prove, assuming only these two definitions $i),ii)$ that $\exp$ function defined like this is unique?

EDIT: For the proof of $(2)$ i have following idea: $$\exp(0)=\exp{(1+(-1))}=\exp(1)\cdot\exp(-1)$$ and by $(3)$ (not proven yet for negative integers) we can continue $$\exp(1)\cdot\exp(-1)=\frac{\exp(1)}{\exp(1)}=1$$ EDIT 2: What remains is to show that $(3)$ holds for negative values of $n$.

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    $\begingroup$ From (i), (1) and (4) are equivalent. $\endgroup$ – Lord Shark the Unknown Apr 16 '18 at 21:19
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By your condition (ii), $\exp(x) > 1$ for all $x > 0$. Hence if $y > x$, we have

$$\exp(y) = \exp(x + (y - x)) = \exp(x) \cdot \exp(y - x) > \exp(x) \cdot 1$$

giving monotonicity. The fact that $\exp(0) = 1$ follows from setting $x = y = 0$ and knowing that $\exp(0) > 0$.


As for uniqueness, the standard technique is to take two candidates and compare them. For example, if $\exp_2$ is another function satisfying the conditions, study $\exp - \exp_2$ and make sure it's the zero function. Or something more natural here is to take $\exp / \exp_2$ and show it's identically one.

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    $\begingroup$ @MichalDvořák I fixed a typo in my answer. It should have been about $x = y = 0$. $\endgroup$ – user296602 Apr 16 '18 at 21:31
  • $\begingroup$ I would be interested in an elementary argument showing that it is indeed fixed by the two conditions above. I really do think you need at least some real analysis here. $\endgroup$ – Thomas Bakx Apr 16 '18 at 21:57
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I have never come across this definition of the exponential. Where did you find this? Apart from that, let me help you:

Note that we can take $x=y=0$ in i) to find $\mbox{exp}(0)=1$ (it cannot be zero because of ii) as you can see). Also, if we have (4) then we also have (1) by using (i) again. In fact, $\mbox{exp}$ is then strictly increasing. So it remains to prove (4).

If there is an $x>0$ such that $\mbox{exp}(x)=1$, then by (3), which we already proved, we obtain a set of arbitrarily large real numbers $nx$ for which $\mbox{exp}(nx)=1$. This must contradict ii).

As for uniqueness, I think you need something a little more sophisticated. Can you prove that $\mbox{exp}$ is differentiable with derivative equal to itself, using the properties you established? If this is the case, you know that its derivative at $0$ must equal one because otherwise property (ii) cannot hold. Observe that this property ii) is actually very restrictive, because for example if we only use (1)-(4) as assumptions, any function of the form $x \mapsto a^x$ would fit the bill for positive $a$.

Edit: as an elaboration, if we know that the function is differentiable with derivative equal to itself, then the quotient of any two functions satisfying i) and ii) must have zero derivative by application of the quotient rule, so that it is identically $1$.

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    $\begingroup$ I've found that in scripts from Faculty of Mathematics and Physics from the Charles university of Prague, unfortunately, not available in English version $\endgroup$ – Michal Dvořák Apr 16 '18 at 22:03
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    $\begingroup$ If i can prove that $\exp$ is its own derivative, isn't only thing i could show by the quotient rule that, assuming we have two different $\exp$'s $$\bigg(\frac{\exp_1(x)}{\exp_2(x)}\bigg)'=0$$ Can i actually assume that $\frac{\exp_1(x)}{\exp_2(x)}=1$? What makes sense to me is to assume that $\frac{\exp_1(x)}{\exp_2(x)}=c$ for some $c\in\Bbb{R}$ Also unfortunately, i can't seem to prove the derivative, i still need to know somehow shot, that $$\lim_{h\to 0}\frac{\exp(h)-1}{h}=1$$, which i have no idea how to show withouht knowing the limit definition for e and logarithms. $\endgroup$ – Michal Dvořák Apr 16 '18 at 22:20
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    $\begingroup$ As for the quotient rule: true, but both functions must satisfy i) and ii), so both also satisfy (2). This should tell you what the value of $c$ is. For the derivative, you indeed need to calculate the limit you mentioned. Note that by using ii) twice, you can show that $\mbox{exp}(h) \geq 1+h$ and $\mbox{exp}(-h) \geq 1-h$, or by (3) this is $\mbox{exp}(h) \leq \frac{1}{1-h}$. Now apply the squeeze theorem for limits. $\endgroup$ – Thomas Bakx Apr 16 '18 at 22:22
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    $\begingroup$ Oh, I'm sorry. You should apply i) to find $\mbox{exp}(h)\mbox{exp}(-h)=1$, which yields $\mbox{exp}(-h) = \frac{1}{\mbox{exp}(h)}$. Also I should have been a little more careful: the inequality $\mbox{exp}(h) \leq \frac{1}{1-h}$ only holds for $h<1$ of course, but we are only interested in values of $h$ close to zero anyways. $\endgroup$ – Thomas Bakx Apr 16 '18 at 22:33
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    $\begingroup$ This is not so hard to see: we have, in your notation, that $\mbox{exp}_1(0) = \mbox{exp}_2(0) = 1$. We saw that the quotient is constant, so it equals its value at $0$. It follows that $c = \frac{1}{1}=1$. $\endgroup$ – Thomas Bakx Apr 16 '18 at 22:43

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