Methods of calculating e.

[hellomeow]

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

[hellomeow]

I have made a few calculations using the James Bernoulli compounding interest problem.

If the compounding interest was compounded once per micro second, this would be the form that the equation would take:

According to Speedcrunch:
(1+(1/(365days*24hours*60mins*60secs*1000millisec*1000microseconds)))^(365*24*60*60*1000*1000)=
2.71828182845900213728

And once per nanosecond:
(1+(1/(365days*24hours*60mins*60secs*1000millisec*1000microseconds*1000nanoseconds)))^(365*24*60*60*1000*1000*1000)=
2.71828182845904519226

These results seem odd to me, as even on the first calculation after the ninth digit the number exceeds the given value of e. One property of this calculation is that the limit is that:
e =lim x -> ∞ f(x)=(1+1/x)^x
And given that the provided number is hardly infinity, it seems strange that it would not be able to accurately calculate e. I can only assume there is some error with either the method being used, or the calculator SpeedCrunch is not as accurate as I might have hoped.

[hellomeow]

Another interesting thing about e is that it can be arranged as the following calculation based on factorials and squares.

e^x=(x^0/0!)+(x^1/1!)+(x^2/2!)+(x^3/3!)+(x^4/4!)+(x^5/5!).....(x^n/n!)

As an infinite series, again being arranged as an infinite limit.

lim n -> ∞ = e^x = (x^n/n!)+(x^(n-1)/(n-1)!)+(x^(n-2)/(n-2)!)+(x^(n-3)/(n-3))....(x^(n-n)/(n-n)!)

I am thinking of a way to write a script that can provide a calculation of an approximated e based on an arbitrary value of n. It would be redundant to attempt using an infinite n, but a very large n (Hopefully something over Avogadro's Number, let's say) might provide some accuracy provided a good means of calculation using float variables.

[Finn]

don't even think about posting in this thread, daryl

you know why

[hellomeow]

May 20, 2008 23:10:38 GMT -5 Finn said:

don't even think about posting in this thread, daryl

you know why

Daryl is probably my sole collaborator in my endeavor to calculate e, unless you'd also like to collaborate.

[Mikhail Kalashnikov]

Another fairly interesting method is to use an infinite complex fraction which follows the pattern 2+1/(1+1/(2+1/(1+1/(1+1/(4+1/(1+1/(1+1/(6+1/(1+1/(1+1/(8.... Though for the amount of work you must enter into the calculator, it converges rather slowly. In case you couldn't pick up the pattern, you first take 2, then add the inverse of 1 plus the inverse of two plus the inverse of one plus the inverse of one plus the inverse of four plus the inverse of one plus the inverse of one plus the inverse of six.... And so on. So it follows the pattern 2 1 2 1 1 4 1 1 6 1 1 8 1 1 10.... You can construct a series such as this for any number, such as Pi, for instance, to achieve a fairly accurate fraction for any given decimal. To do this, you first take the number, and subtract the first digit, in this case two. Then you take the inverse of that, subtract the first digit, take the inverse of that, subtract the first digit, et cetera. The numbers you subtract are the ones you must insert into this formula. Using this method, for example, I can get the fraction 833719/265381, to correctly identify Pi to 11 decimal places.

Of course, there are easier ways to achieve such things, though it is still interesting to note that some numbers (e) have patterns when using this method, while others (Pi) have none.