e is...
the number such that the exponential function e^x has a slope of 1 at x = 0.
What does it mean? Well, lots of regular functions f(x) have slope 1 at x=0. However, e is the base of an exponential, so if we take a derivative (using the definition of derivative):
f'(x) = e^(x+h) - e^(x)/h {lim h->0}
= e^x(e^h - 1)/h
= e^x * lim(h->0) e^h - 1/h
= e^x
(recall that all of the above have lim h->0)
Thus, e^x is its own derivative!
We could do this in reverse, and try to find the base b that satisfies this property by trial and error. We need to find the base that grows just fast enough (not too slow (like 2) or not too fast (like 3)).
e is the function that equals its own derivative.
e is the exponential that can express this idea:
Infinitely large exponential of an infinitely small change
lim (x->0) = (1 + x)^(1/x)
(UNFINISHED)
Look at definition of limits in calculus. Choose any epsilon > 0 which is the distance we are considering. you can find some delta > 0
basically, limit for any definition of precision that you want. whatever your measuring instruments choose, then use that. odometer example :).
From slashdot:
What are you thinking? They didn't just sit down and guess numbers ("hmm, 2.9? no, smaller than that. 2.6? nope..."). e isn't just some magic smear of decimal places. Read that comment the other guy posted. e comes into play in a _lot_ of different, seemingly unrelated fields.
Also, here's a definition of e that you might find interesting:
Think of the number one. What's 1^2? What's 1^9? Still 1.
How about 1.1^2? Just a little bigger than 1.1
How about 1.01^2? Still bigger than the starting 1.01, but by a lot less.
What about (1 + 1/1000)^2?
What about (1 + 1/1000)^500?
The bigger the exponent, the more the number grows
but the smaller the initial number is compared to 1, the less the number grows
What happens if the base number is infinitely close to 1, but the exponent is infinitely large?
The limit as x approaches infinity of (1+1/x)^x = ...
e
Interested in e now?
E -- fight between exponent and base. You can't have 1 as a base (1^n = 1). So you take an infinitesiaml base, raise it to the infinite power. What do you get? e :)
see why it's useful for calculus?
e - most cost-effective base to use?
e is not a fudge factor. It can be thought of as sort of a fundamental relationship between multiplication and exponentiation. It's a direct result of our basic arithmetic operations and treatments of infinite sums. It's really important, and possibly more mathematically pure than p .
e is its own derivative.
physical connection with e: bell curves
try something with 50-50 odds (1/2). try 2 times. chance of winning 3/4 (HH HT TH TT)
3 times at 1/3 chance, 8/27
4 times at 1/4, 81/246
take to limit... N times at 1/N ... get 1/e
e that base such that gradient of curve d/dx lg(n) = lg(n)
let f(x) = log(x) (base 10)...
compute derivative...
find that derivative = c/x, where c is a constant (log e)
e is the base that makes this constant 1.
(don't have to carry it through)
reason for using radians not degrees
in radians, d/dx sin = cos
degrees, d/dx sin = 57.3 * cos
e for something that grows continually - calculus... infinitesimal growths (not discrete ones, like 2^x)
exponentiation: rate of change depends on magnitude
