I've used the constant e many times in my finance classes and know how to use it, but I don't understand what e is. To me it's just a number and I use it for discounting.

I looked it up in wikipedia and that page confuses me.

Can anyone explain e to me in kindergarten language?

By definition e is the limit of (1+1/n)^n as n approaches infinity .

Try this out yourself .

Plug a large number for n in your calculator and you'll ~ match the number e you get on your calculator .

If we expand this using the binomial theorem we can represent as a limit of infinite sums .

e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! +....
e ~ 2.718281828....

e is an irrational number much like pi so it cannot be represented as a fraction a/b . There are many interesting properties about e but it's probably worthwhile to start with the definition before you get into other topics .

ok makes sense.

now, and you might not know this, but why do i use it in finance so much? we use it for continuously compounding interest rates. for example $100 at 5% interest continuously compounded for 2 years is $100e^.05*2. so, why e?

Because the definition of e...
...is the formula for compounding interest with the number times you compound limiting to infinity.

Say you compound interest 1 time over a period at rate r.

You get (1+r)^1 = 1+r

Say you compound interest 4 times over a period.

You get (1+r/4)^4

Say you compound monthly.

You get (1+r/12)^12.

Say you compound continuously.

You get lim(n to infinity) (1+r/n)^n = e^r

I forget the proof of the very last equality, but I imagine it's not hard to come by. It should be pretty easy to see where the definition of e comes into play though based on its similarity to the formula for compound interest.

Pretty much what Omaha said but I'll expand some more .

When we talk about compound interest , we have a familiar formula A = P(1+r/n)^(nt)

P = Principal amount
r= annual interest rate
n = the number of compounded periods per annum .
t= t years


The above formula may be re-written as

A= P*[(1+r/n)^(n/r)]^(rt)
Substitute n/r = x

A= P*[1+1/x)^x]^(rt)

So as x becomes large , the quantity (1+1/x)^x approaches e.

A=p*e^(rt)

Example : Find the amount after 3 years if $1000 is invested at an interested rate of 12% per annum compounded continuously .

Solution : Using A=P*e^(rt) , r=0.12 and t=3,
A=1000*e^(0.12*3)
A= $1433.33

I think this definition is arbitrary. e shows up in all sorts of places and you could use any one of them as the starting point to define it. My first introduction to e was by way of the area under the curve 1/x. e is that number such that the area from 1 to e under the curve 1/x is 1. The natural log function, ln, is defined as the area under the curve 1/x so from this definition e is that number such that ln(e)=1. From this definition properties like the one above can be derived.

PairTheBoard

Mathematicians were looking for a function that is equal to its own derivative. They narrowed down the search to functions of the form f(x) = a^x, where a is real.

For a fixed x,

f'(x) = lim (1/h)( f(x+h) - f(x) ) where h--> infinity

a^x = lim (1/h)( a^(x+h) - a^x ) where h---> infinity

factoring out a^x from the right hand side

a^x = a^x * lim (1/h)( a^x - 1) where h--> infinity

1 = lim (1/h) (a^x - 1 ) where h--->infinity

e is defined to be the unique value of a such that the equation above is true. You can massage the equation above and substitute h = 1/n to get the definition provided by previous posters.

e dollars is the amount of money you would have at the end of the year if you put one dollar into a bank that offered 100% interest compounded continuously.

For those who don't know what this means lets change it to a bank that compounded your million dollars every 3.65 days. At a 100% annual interest rate. So every 3.65 days they gave you one percent. After 3.65 days you would have 1,010,000. After 7.30 days you would have 1,020,100. After 10.95 days you would have 1,030,201. At the end of the year you would have just short of "e" million dollars (as opposed to two million with no compounding or 2.25 million if interest was compounded every six months.) The thing is that even though shrinking the time period for compunding makes you more and more money, you run into one of those limit thingies that jason and boris love and you can't get past e.

The more important thing about e concerns making prop bets when poker tournaments redraw. If there are more than a few players left from 20 to a googol, the chances that everybody will draw a new seat is almost exactly one out of e.

oops, in the last 2 lines of my post, it should be a^h not a^x inside the limit =/

Somebody else must have written this post for you.

I don't think this is true. I think Euler was the first one to talk about the constant and he was trying to sum power series. Working out properties of summing power series, he found "e" although of course he didn't call it e and he noticed it was an important constant for summing stuff.

The property that DS mentions about everyone getting a new seat, is a cool probability that Euler discovered while working on what is called the hat problem: If you have n people entering a party and every one leaves his hat at the door to dance. If you give them their hats back randomly, what is the probability that no one got his hat back? The answer is 1/e.

I've never heard it called the "hat problem". Either "matching problem", "problem of recontre", or for the simplest case, counting derangements.

lololololololololol...etc

Oh, and

lololololololololol...etc

Would make a small bet that its known mainly by the words "Euler's problem of the misaddressed letters"

Sure, so long as "mainly" corresponds to what has more (relevant) google hits and/or a wiki entry.

Edit - Spelling Correction

Should be problem of rencontres not recontre

I think you mean that the answer tends to 1/e as n approaches infinity. Lol mathaments. Btw, wiki thinks that Bernoulli was the first to "discover" e, and apparently he did it by considering continuously compounded interest. Of course, this could be wrong.

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

I don't want to spoil it, but in fact, if \sigma is an element of the symmetric group on n letters, p is a polynomial, and \fix\sigma denote the number of fixed points of \sigma, then there are methods to evaluate

\sum_{\sigma\in S_n} p(\fix\sigma)

here is a simple proof of the for the case p = id which generalizes to higher degree polynomials. let \fix_i \sigma = 1 if \sigma fixes the ith place and 0 otherwise.

\sum_{\sigma\in S_n} \fix\sigma =
\sum_{\sigma\in S_n} \sum_i \fix_i\sigma =
\sum_i \sum_{\sigma\in S_n} \fix_i\sigma =
\sum_i (n-1)! =
n(n-1)! =
n!

which is the desired result.

How is this cooler? This is way easier to solve than the original problem, just use indicator functions <---- answer in white.

Then why did you post the answer?

Say \sigma is a permutation of n letters, and V is a vector space of dimension n, with basis e_i. Define a linear map e_i \mapsto e_{\sigma(i)}. The trace of this linear map is equal to the number of fixed points of \sigma. This observation, together with the fact that trace is linear, is basically your argument.

Cooler in that it admits several elegant and simple solutions. To me, easier problems are cooler.

Can't disagree with that =).

That post also jarred my memory and reminded me of an amazingly simple solution to the 1st problem using exponential generating functions. Thanks

sure, this provides an easy solution, but it's essentially a linear technique and thus isn't particularly useful if you want to generalize the problem.

what is

\sum_{\sigma\in S_n} [\fix(\sigma)]^2

Hmm, how about n(1/n)=1!
uA

Obviously. And while I'm mad I just got to this thread so you posted it first, I'm quite happy that boris was nice enough to post a question that perfectly shows why clever amateur will sometimes beat not so clever pros.

Permit me to answer it in a way that everyone will understand.

There is 1000 players in a tournament redrawing for seats. The RIO is paying five thousand dollars to any player who gets his own seat. Each player has a one in a thousand chance of making a thousand dollars. Each player has an EV of $5.

I go around buying up everyone's EV for face value. It cost me five grand. Each purchase is a break even purchase for me. Thus the whole deal is a break even thing for me. Therefore my expected payoff from the Rio (which ranges from zero to 5mil) is the $5000 I paid. If the expected value of my prize is $5000, the expected number of matches is one. And of course this would work for any size tournament.

Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.

And for once, I actually agree with Sklansky rigor. Tournament reseating is the perfect way to conceptualize this problem, for those who already have experience with poker tournaments.