zakinutuzi

2021-12-31

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

rodclassique4r

Beginner2022-01-01Added 37 answers

Apply the usual definition of the derivative:

$\frac{d}{dx}{e}^{x}=\underset{h\to 0}{lim}\frac{{e}^{x+h}-{e}^{x}}{h}$

Next, apply the usual laws of indices:

$\underset{h\to 0}{lim}\frac{{e}^{x+h}-{e}^{x}}{h}=\underset{h\to 0}{lim}\frac{{e}^{x}({e}^{h}-1)}{h}={e}^{x}\underset{h\to 0}{lim}\frac{{e}^{h}-1}{h}$

If we're able to show that$\frac{{e}^{h}-1}{h}\to 1$ 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

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

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

$\underset{h\to 0}{lim}\frac{{e}^{h}-1}{h}=\underset{t\to \mathrm{\infty}}{lim}\frac{\frac{1}{t}}{\mathrm{ln}(1+\frac{1}{t})}$

$=\underset{t\to \mathrm{\infty}}{lim}\frac{1}{t\mathrm{ln}(1+\frac{1}{t})}$

$\underset{t\to \mathrm{\infty}}{lim}\frac{1}{\mathrm{ln}\left|{(1+\frac{1}{t})}^{t}\right|}$

$=\frac{1}{\mathrm{ln}\left(e\right)}$

$=1$

This proof uses the definition that$e=\underset{t\to \mathrm{\infty}}{lim}{(1+\frac{1}{t})}^{t}$ . 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 $(1+\frac{1}{t})}^{2$ . Allowing a system to grow and decay on an infinitesimally small time scale gives rise to $e}^{\tau$ . This definition also benefits from a lack of circular reasoning: to be able to define $e}^{x$ 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.

Next, apply the usual laws of indices:

If we're able to show that

so that

This proof uses the definition that

Jillian Edgerton

Beginner2022-01-02Added 34 answers

If you define

$\text{exp}\left(x\right)=\sum _{n=0}^{\mathrm{\infty}}\frac{{x}^{n}}{n!}=1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\frac{{x}^{4}}{4!}+\dots$

then if you differentiate term by term you get back what you started with.

then if you differentiate term by term you get back what you started with.

Vasquez

Expert2022-01-09Added 669 answers

In some contexts it can be prudent to define the exponential function

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

It is used in modelling because if you have a quantity

then the solution is

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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