Encountering Euler's Number

A random variable question Suppose values are selected at random from the unit- interval [0,1] until their total exceeds 1 . Let X count the number of values required. What is the expected value of X ? (In other words, what is the mean μX ?)

Computer simulation We have constructed a computer simulation which repeatedly performs this experiment using a random number generating function It shows the average number of randomly chosen values needed to obtain a total larger than 1

Output of 'eulernum.cpp' demo

Euler's Number e = 2.718... This number e has been called the most important number in mathematics It turns up often in probability studies But it was discovered in connection with investigations of compound interest e = limitn→∞ (1 + 1/n)n

Two numerical sequences Consider the two sets L and U defined by L = { (1 + 1/n)n | n = 1, 2, 3, ... } and U = { (1 + 1/n)n+1 | n = 1, 2, 3, ... } These set names come from “lower” and “upper”.

Three theorems to prove Numbers in L form an increasing sequence i.e., (1 + 1/n)n < (1 + 1/(n+1))m+1 , for n > 0 Numbers in U form a decreasing sequence i.e., (1 + 1/n)n+1 > (1 + 1/(n+1))n+2 , for n > 0 Numbers in L are less than numbers in U i.e., (1 + 1/n)n < (1 + 1/n)n+1 , for n > 0

And this “squeeze” theorem 0 < (1 + 1/n)n+1 – (1 + 1/n)n < 4/n , for n > 0 It implies that the distance which separates numbers in L from numbers in U Is smaller than any positive real number (in other words, it's equal to zero)

Furthermore... The sum Sn of these reciprocal factorials Sn = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ... + 1/n! satisfies the inequality (1 + 1/n)n < Sn < (1 + 1/n)n+1

An inequality from algebra Lemma 1: If 0 ≤ a < b, then (n+1)an < (bn+1 – an+1) / (b - a) < (n+1)bn Proof: We use (bn+1 – an+1) = (b - a)( ∑k akbn-k ) and the fact that each of the n+1 summands satisfies an < akbn-k < bn . ▄

Another algebra fact Lemma 2: For any positive integer n , we have (1 + 1/(n+1))2 < (1 + 1/(n+1)) + 1/n Proof: Reexpress both sides in terms of their common denominator n(n+1)2 and then observe that their numerators can readily be compared: n3 + 3n2 + 3n < n3 + 3n2 + 3n + 1 ▄

Recall the Binomial Theorem (1 + 1/n)n = ∑k b(n;k) (1/n)k (1 + 1/n)n+1 = ∑k b(n+1;k) (1/n)k

Timeout for demo and proofs The foregoing slides lay the groundwork for proving the essential facts about Euler's Number But we also can utilize a demo-program that shows us the behavior of the successive terms that lie in the sets L and U