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If someone with not much mathematics in his luggage asks me: What is so special about $\pi$? then off course I have an answer. Even if $i$ would be the subject (I allready see him gazing at my mysterious smile). But when I am asked about $e$ then I grow silent (or try to change the subject). Please help me out of this. Give me a nice characterization of $e$.


I am informed now that this question is somehow a duplicate of this. I agree with that judgment. Sorry for that. Next time I will first have a closer look at the questions that have allready been asked. Thank you also. I find very nice answers there and advice everyone interested to have a look.

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marked as duplicate by Marc van Leeuwen, Vedran Šego, Dan Rust, Julian Kuelshammer, Tom Oldfield Oct 4 '13 at 9:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$\mathcal{E}$verything $e$bout $e$ $e$s $e$xtraordinary! – dtldarek Oct 4 '13 at 8:06
Consider the hyperbola $xy=1$. Draw the segment from $(0,0)$ to $(1,1)$ (which are geometrically characterized points related to the curve). Draw another segment from $(0,0)$ to another point $(a,b)$ of the hyperbola, with $a>1$. When is the area delimited by the two segments and the arc of hyperbola equal to $1$? Precisely when $a=e$. – egreg Oct 4 '13 at 8:08
Certainly this question has been asked before here. – Marc van Leeuwen Oct 4 '13 at 9:01
@egreg Thank you. If pencil and paper are at hand, then I will use it. – drhab Oct 4 '13 at 11:17
@MarcvanLeeuwen What to do for me? Am I supposed to delete this question because it is a duplicate? – drhab Oct 4 '13 at 11:19
up vote 3 down vote accepted

If you put 1 dollar in a deposit, and the yearly interest rate is 100%, after a year you'll end up with 2 dollars -- that is, if interest is accrued yearly.

If it's accrued twice a year, you'll get 50 cents after 6 months, because yearly percentage is 100%, and half a year has passed, so it's 50% rate for 6 months. After second half a year, you'll actually get 75 cents, as you had 1.50 dollar after a 6 months, and 50% of 1.50 is 0.75, so now you have 2.25 dollars.

Similarly, if the interest is accrued monthly, you'll end up with around $2.63. And if the interest is accrued daily, it's around 2.71 dollars.

Sadly, no matter how often the interest is accrued, you'll never end up with more then e dollars, because the more often the interest is accrued, the closer you balance will be to e at the and of the year.

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Can you show the exact calculation that gives 2.71 dollars? – Vitalij Zadneprovskij Oct 4 '13 at 8:18
Sure: you accrue it 365 times, each time you get $1/365$ of 100%, so it will be initial 1 dollar times (1+1/365) times (1+1/365) and so on 365 times: it works, because increasing some value by 100/365 percent is the same as multiplying it by (1 + 1/365): "1" is for existing value, and 1/365 is for the increase. Thus, it will be exactly $1\cdot(1+\frac{1}{365})^{365}$ – xyzzyz Oct 4 '13 at 8:28
You won't get 2.71, because a bank doesn't deal with parts of a cent ... – Michael Hoppe Oct 4 '13 at 8:49
This is the sort of answer I was looking for. Most other answers deal with the exponential function, but I wanted to avoide that. – drhab Oct 4 '13 at 10:54
@MichaelHoppe: yes, but is it relevant here? You wouldn't put a single dollar on deposit, and you wouldn't get 100% interest rate either. – xyzzyz Oct 5 '13 at 16:52

$$\dfrac d{dx}\left(e^x\right)=e^x$$ $$\int e^xdx=e^x+C$$ It is the only function that does this

Also, $e^x$ can be expressed like this: $$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}......$$

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I believe that $$ e^{i\theta} = \cos \theta + i \sin \theta $$ should be enough! Anyway if you're interested in other things, like differential equations which appear in physics or biology or economics and so on, $e$ is quite important.

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Here is an application of $e$ in economics, which may be accessible to someone without any special education in mathematics.

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The thing that is special is the exponential function $\mathrm{exp}$, which satisfies $$\mathrm{exp}(0)=1,\quad \mathrm{exp'}=\mathrm{exp}.$$ Then of course $e:=\mathrm{exp}(1)$, and because of $\mathrm{exp}(x+y)=\mathrm{exp}(x)\mathrm{exp}(y)$ it makes sense to write $\exp(x)=e^x$.

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