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Furthermore why is it that $e^x$ is used in exponential modelling? Why aren't other exponential functions used, i.e. $2^x$, etc.?

Thank you!

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Depends on how you define $e$. –  Thomas Andrews Feb 6 '13 at 22:39
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One way to define $\operatorname{e}$ is as the limit of $(1+1/t)^t$ as $t$ tends towards infinity. To understand this definition, consider compound interest added yearly, monthly, weekly, daily, hourly, etc. Adding less and less interest, more and more frequently corresponds to letting $t$ tends towards infinity in $(1+1/t)^t$. Allowing a system to grow and decay on an infinitesimally small time scale gives rise to $\operatorname{e}^t$. This is the definition of $\operatorname{e}$ that I use in my answer below to prove that $d\!\operatorname{e}^x\!/dx=\operatorname{e}^x$. –  Fly by Night Feb 6 '13 at 23:19

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Apply the usual definition of the derivative:

$$\frac{d}{dx}\operatorname{e}^x = \lim_{h \to 0} \frac{\operatorname{e}^{x+h}-\operatorname{e}^x}{h} \, . $$

Next, apply the usual laws of indices:

$$\lim_{h \to 0} \frac{\operatorname{e}^{x+h}-\operatorname{e}^x}{h} = \lim_{h \to 0} \frac{\operatorname{e}^x(\operatorname{e}^h-1)}{h}=\operatorname{e}^x \, \lim_{h \to 0}\frac{\operatorname{e}^h-1}{h} \, . $$

If we're able to show that $(\operatorname{e}^h-1)/h \to 1$ as $h \to 0$ then we're done. This is easier said than done. We need to employ a little trickery. Let us first start by defining

$$t = \frac{1}{\operatorname{e}^h-1}$$

so that $1/t = \operatorname{e}^h-1$, and hence $1 + 1/t = \operatorname{e}^h$. Finally, we have: $h = \ln(1+1/t)$. Hence:

$$\begin{array}{ccc} \lim_{h \to 0} \frac{\operatorname{e}^h-1}{h} &=& \lim_{t \to \infty}\frac{1/t}{\ln(1+1/t)} \\ \\ &=&\lim_{t \to \infty}\frac{1}{t\ln(1+1/t)} \\ \\ &=&\lim_{t \to \infty}\frac{1}{\ln[(1+1/t)^t]} \\ \\ &=&\frac{1}{\ln(\operatorname{e})} \\ \\ &=& 1 \end{array}$$

This proof uses the definition that $\operatorname{e} = \lim_{t \to \infty}(1+1/t)^t$. To appreciate this definition, consider compound interest added yearly, monthly, weekly, daily, hourly, etc. Adding less and less interest, more and more frequently corresponds to letting $t$ tends towards infinity in $(1+1/t)^t$. Allowing a system to grow and decay on an infinitesimally small time scale gives rise to $\operatorname{e}^{\tau}$. This definition also benefits from a lack of circular reasoning: to be able to define $\operatorname{e}^x$ as a power series required us knowing how to differentiate it and get its Taylor series. Either that, or it was an amazingly, infinitely-lucky guess.

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Your last statement is completely invalid. You do not need Taylor series to show that $\sum_{k\geq0} \frac1{k!} x^k$ is a convergent sum for all $x\in\mathbb R$ and that it is its own derivative. It is as well a unique solution of $ka_{k-1}=a_k$ and $a_0=1$ which characterizes a power series with sum $f$ such that $f(x)=1$ and $f'(x)=f(x)$. –  yo' Feb 6 '13 at 23:33
    
@tohecz You don't need it... But how did people get there in the first place? They found the Taylor series and then said "Hey, let's take this as an alternative definition." I would disagree that my statement is "completely invalid". You might disagree, and that's okay, but unless you are the official spokesperson for all of mathematics, you can't dismiss my comments totally. –  Fly by Night Feb 6 '13 at 23:37
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@Assad A limit is different to substituting in the number. As you say: $(\operatorname{e}^0-1)/0$ is indeterminate. A limit is different. What I am saying is that as $h$ gets closer and closer to zero, the number $(\operatorname{e}^h-1)/h$ gets closer and closer to $1$. Try putting smaller and smaller numbers into you calculator and see what happens: $$\frac{\operatorname{e}^{0.1}-1}{0.1} \approx 1.0517$$ $$\frac{\operatorname{e}^{00.1}-1}{00.1} \approx 1.0050$$ $$\frac{\operatorname{e}^{000.1}-1}{000.1} \approx 1.0005$$ This caries on like this. –  Fly by Night Feb 7 '13 at 16:02
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@Assad Think about the very definition of differentiation from first principals: $$\frac{\operatorname{d}}{\operatorname{d}x}f = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \, . $$ If you just substitute $h=0$ then you get $(f(x)-f(x))/0$. You need to look at what happens as $h$ gets smaller and smaller. There are lots of limits like this, for example $(\sin x)/x$ gets closer to $1$ as $x$ gets closer to $0$. Moreover $(\cos x - 1)/x \to 0$ as $x \to 0$. Both are needed to prove that $\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx} \cos x = -\sin x$. –  Fly by Night Feb 7 '13 at 16:09
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@Assad It was just a trick. Making that substitution changed the unknown limit into the form of one we could handle. This wasn't the only possible substitution. There is no reason except that it works. Have you learned to integrate by susbtitution? For example, finding $\int \sqrt{2x+3} \,\operatorname{d}\!x$ can't be done using the standard rule that $\int x^k \, \operatorname{d}\!x = \frac{1}{k+1}x^{k+1}+C$. However, substituting $u = 2x + 3$ allows is to change $\int \sqrt{2x+3} \,\operatorname{d}\!x$ into $\int \frac{1}{2}u^{1/2} \,\operatorname{d}\!u$ which we can handle. –  Fly by Night Feb 7 '13 at 16:37

In some contexts it can be prudent to define the exponential function $y=\exp(x)$ as the solution of the differential equation

$\displaystyle \frac{dy}{dx}=y(x)$ such that $y(0)=1$

so that the exponential function is it's own derivative because you define it as such.

By Picard's Existence Theorem the solution of this differential equation is unique. This seems like cheating but depending on what properties of $e^x$ you want to use it is perfectly valid. For example, this doesn't tell you how to even calculate $\exp(2)=e^2$ without a little more work...

It is used in modelling because if you have a quantity $y=f(x)$ that changes in proportion to it's size:

$\displaystyle \frac{dy}{dx}=ky(x)$,

then the solution is $f(x)=y(0)e^{kx}$. Lots of quantities change in proportion to their size; e.g. radioactive decay.

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If you define $$\exp(x) = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots$$ then if you differentiate term by term you get back what you started with.

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The main reason for using $\exp$ rather than other exponentials is that its derivative (and consequently its Taylor series, antiderivative, etc) is simpler.

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Don't you think it has something to do with the fact that $(1+r/t)^t \to \operatorname{e}^r$ as $t \to \infty$? Continuously applied compound interest and radioactive decay can be modelled this way. –  Fly by Night Feb 6 '13 at 23:39
    
Except that radioactive decay uses the Poisson distribution, which is also built on ${e}^{t}$, but it is rather an observation than arbitrary choice. –  András Hummer Apr 5 '13 at 18:31

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