In math when people want to model population growth or radioactive decay we use exponential functions. In many cases, we use base $e$. My question is, what is the purpose of using base $e$ rather than some other base?
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For starters, sometimes people do use other bases. In radioactivity, $2$ is often used, giving meaning to the term half-life.
To answer your question, there is no reason to pick any particular base unless it is useful. Since 2 and $e$ are useful, we use them. If another number is useful for a specific problem having to do with exponential growth or decay, people use it. The reason that $e$ is useful is because $e^{\lambda x}$ is the solution to $y' = \lambda y$, a fundamental differential equation for this type of problem.
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Mh, the question is about $e$, not about the exponential behavior, which does not rely on a particular base. – Jun 26 '15 at 19:57
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2@YvesDaoust But $2^{\lambda x}$ isn't solution to the mentioned differential equation. Having base $e$ surely makes it nicer, and without it, we would have an additional constant involving natural log, which is base $e$. – Wojowu Jun 26 '15 at 20:03
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Obviously, but my point is that $\lambda$ or $\lambda\ln(2)$ do not make a real difference. They are arbitrary values, that's it. Nature doesn't care that there's an additional coefficient in the equations. You cannot derive the value of $e$ from the lifetime of plutonium. A last argument: $e^{0.35265t}=2^{0.508766t}$. – Jun 26 '15 at 20:16
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2You are absolutely right. Nature doesn't care, it's just why scientists use $e $ by convention. Calculations are easier (even if it is just a tiny bit easier). – Plutoro Jun 26 '15 at 20:53
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The number $e$ appears naturally as the limit of the compounding interest formula. That is $$\left( 1 + \frac{r}{n} \right)^n \simeq e^r$$ when $n$ is large. This makes $e$ a good choice for modeling systems with exponential growth.
In order to expand on the above. The compounding interest formula is one of the routes which led to the discovery of $e$. (The other route was through the natural logarithm, which is related to the area underneath the hyperbola $y=1/x$.) Liebniz was among the earlier mathematicians to study $e$, and he used $b$ to represent this number. I believe the proof demonstrating $$\lim_{n\to\infty}\left( 1 + \frac{r}{n} \right)^n = \sum_{n=0}^\infty \frac{r^n}{n!}$$ may be due to him (I would have to consult with some references I don't have handy at the moment).
There is an excellent book on the history of $e$, named appropriately "e: The story of a number". It's a book written for laymen, and I thoroughly enjoyed reading it.
Joel
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The exponential decay phenomena in nature do not involve $e$, but other basis of arbitrary value (more precisely, there are arbitrary time constants). In a way, $e$ does not exist in nature, neither in the discharge of condensers, nor in the decay of radioactive materials, nor on the spirals of sunflowers, nor in the growth of compound loan interests.
$e$ is the preferred base of mathematicians because it is convenient, in the sense that it is the only base for which $$\left(b^x\right)'=b^x.$$
For other bases, a conversion factor appears. This is the very same reason why radians are used to measure angles.
On the opposite, the base $10$ that we use for our numeration system is deeply grounded in human anatomy, and base $2$ is pretty useful in the theory of information and computer science.
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Other comments about $e^{ct}$ for some constant $c$ illustrate that $e$ isn't a natural base in the real world: $e^c$ can be any number, and you never see $c=1$ elsewhere than in theory. – Jun 26 '15 at 19:54
Next, suppose that $$g(x)=2^x$$ Try to show that the slope of $g(x)$ is less than $g(x)$.
If you can do this, then you can conclude that there is a number ($2\lt e\lt 4$) such that $h(x)=e^x$ where the slope of $h(x)$ is equal to $h(x)$.
– John Joy Jun 26 '15 at 20:56