### Post by hellomeow on May 19, 2008 19:21:30 GMT -5

It has come to my attention that e is quite the interesting number, and I've been spending the last few hours experimenting with means by which to calculate e to the best I can. Essentially, e is the value by which raised to any value of x will create the curve at which x forms the tangent line line at any point of x on a function, to give you a visual aid. Also note that the consequential notation for log_e is ln:

Now there are a few fun facts about e, it was originally discovered by James Bernoulli as a compounded interest problem of the form $1 compounded over time with an interest return of 100%, ie one year would resemble $1(1+1/1a)^1a Simply enough if it was compounded per annum it would return 2, however, as the period of compounding decreases, the approximation begins to reach a certain constant.

To provide an example, let's say we wish to compound this value over a period of 365 days. Then we would have $1(1+1/365d)^365 or $2.71456482... etc etc. However, this can be limited even further, for example let's try a millisecond, ignoring that if this were an actually amount of money the few pennies earned would become inconsequential due to rounding.

$1(1+1/(365d*24h*60min*60sec*1000ms))^(365d*24h*60min*60*1000ms) = 2.71828167 (According to Google calculator :/)

Essentially, there is a limit by which this function can be expressed as lim x ->∞ f(x)=(1+1/x)^x. Basically, as x reaches infinity, the approximation of the constant becomes more accurate. Basically, I want to test various calculators for values that can approximate the constant, which if you haven't guessed yet happens to be e I will periodically update this thread with various calculations as accurate as I can get, at very compounding periods. Anyone else is free to experiment as well.

Here is a reference. e = 2.7182818261984928651

Now there are a few fun facts about e, it was originally discovered by James Bernoulli as a compounded interest problem of the form $1 compounded over time with an interest return of 100%, ie one year would resemble $1(1+1/1a)^1a Simply enough if it was compounded per annum it would return 2, however, as the period of compounding decreases, the approximation begins to reach a certain constant.

To provide an example, let's say we wish to compound this value over a period of 365 days. Then we would have $1(1+1/365d)^365 or $2.71456482... etc etc. However, this can be limited even further, for example let's try a millisecond, ignoring that if this were an actually amount of money the few pennies earned would become inconsequential due to rounding.

$1(1+1/(365d*24h*60min*60sec*1000ms))^(365d*24h*60min*60*1000ms) = 2.71828167 (According to Google calculator :/)

Essentially, there is a limit by which this function can be expressed as lim x ->∞ f(x)=(1+1/x)^x. Basically, as x reaches infinity, the approximation of the constant becomes more accurate. Basically, I want to test various calculators for values that can approximate the constant, which if you haven't guessed yet happens to be e I will periodically update this thread with various calculations as accurate as I can get, at very compounding periods. Anyone else is free to experiment as well.

Here is a reference. e = 2.7182818261984928651