E in compound interest

1. Dec 29, 2015

Warp

Originally (as far as I know) the number e (ie. 2.71828...) came up in compound interest calculations.

For example, if you have 1 dollar, and a compound interest of 100% per year, and the interest is continuously calculated, after one year you'll have exactly e dollars.

The generic formula for this is (1+1/n)n, where n is the amount of times compound interest is calculated during the year. As n approaches infinity (which means it's continuously calculated), that formula approaches e.

Is the fact that ex is its own derivative just a coincidence, or is there a correlation with the above?

2. Dec 29, 2015

BvU

The way we were first acquainted with the existence of $e$ was: We are looking for the derivative of $f(x) = a^x$. From visual inspection there must be a real number $a$ such that this derivative is equal to 1 in $(0,1)$.

3. Dec 29, 2015

Warp

That's not what I heard. But either way, it's not really what I asked...

4. Dec 29, 2015

BvU

You are right. I got carried away by memories ...

You could investigate the Taylor series of your expression and see that it yields something looking suspiciously like $f'= f$

 that doesn't make much sense. Dug up a dog-eared yellowish textbook, copyright 1969 , on infinitesimals and found: we call this number $e$.
So the derivative definition gives us $\displaystyle f'(0) = \lim_{h\rightarrow 0}{e^h-1\over h} = 1$ hence $\displaystyle f'(x) = \lim_{h\rightarrow 0}{e^{x+h}-e^x\over h} = e^x \lim_{h\rightarrow 0}{e^h-1\over h}= e^x$.
Not interesting for you, but: a few pages further, on limits:
$f(x)=\log(1+x)\Rightarrow f'(x) = (1+x)^{-1} \Rightarrow \displaystyle f'(0) = \lim_{x\rightarrow 0}{\log(x+1)\over x} = 1$ and now we do have an answer to your question !

Last edited: Dec 29, 2015
5. Dec 29, 2015

SteamKing

Staff Emeritus
Although e is now called Euler's number, it was actually discovered by Jakob Bernoulli (among other mathematicians) before Euler was born when he, Bernoulli, was investigating compound interest:

https://en.wikipedia.org/wiki/E_(mathematical_constant)

Euler's singular contribution was his formula which relates e, i, π, 0, and 1.

6. Dec 29, 2015

Ssnow

Considering the derivative of $f(x)=e^{x}$:

$f'(x)= \lim_{h\rightarrow 0} \frac{e^{x+h}-e^{x}}{h}=\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}\cdot e^{x},$

you can see that the derivative of $e^{x}$ differs from $e^{x}$ by the factor $\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}$. If we write:

$\lim_{h\rightarrow 0}\frac{\lim_{n\rightarrow +\infty}\left(1+\frac{1}{n}\right)^{nh}-1}{h},$

we can examine the sequence $\frac{\left(1+\frac{1}{n}\right)^{nh}-1}{h}$. If $h=1$ then for $n\rightarrow \infty$ we have $e-1$ as limit, because it is a variation of the neper sequence you cited before. The fact that $h\rightarrow 0$ ''deforms'' the result of this sequence that under the limit for $h\rightarrow 0$ tends to $1$. This permit to have the equality between $f'$ and $f$ ... this is the way how the growth of neper sequence is correlated with the derivative of the function $e^{x}$.

7. Dec 29, 2015

MrAnchovy

There is a correlation - this is a direct consequence of the fact that e is its own integral.

8. Dec 29, 2015

Warp

Could you explain in vernacular how?

9. Dec 29, 2015

BvU

Did you read #4 or #6 ?

10. Dec 31, 2015

Staff: Mentor

Or, rather, that $\int e^x dx = e^x$ plus an arbitrary constant.