Some math
SuperVar Source
It is interesting that you can do this more efficiently. For example, lets say that "x" ranges from 0 to 3, y from 0 to 4 and z from 0 to 7. Then the number a+4b+20z is unique (or a+(3+1)b+(3+1)(4+1)c, if you prefer.)
You can use a similar method to prove that e is irrational, though it is neither harder nor easier than the normal proof.
You can use a similar method to prove that e is irrational, though it is neither harder nor easier than the normal proof.
There's an easier way to prove natural base e (dots are for space purposes):
e = lim (x + 1) ^ (1 / x)
... x->0
So, as x approaches 0, e approaches ~2.718....
e = lim (x + 1) ^ (1 / x)
... x->0
So, as x approaches 0, e approaches ~2.718....
That's not a very rigorous proof of irrationality, since the decimal might repeat after some huge number of digits.
August 27th 2004, 11:42 AM
AS
Is there any definite value for e? the smaller you make x, the larger the decimal becomes...
I guess I'm not making sense
I guess I'm not making sense
August 27th 2004, 12:26 PM
MiloBones
'Tis a limit, like many well known constants. As an exact value, e is 1/(0!)+1/(1!)+1/(2!)+1/(3!)+..., but I think you could call Merlin's definition exact as well.
I would brag about my 1337 mathematical education, but I'll let it go.
Well, n! = 1 * 2 * ... * n.
One would think 0! therefore would be 0. Not so. Watch:
1! = 1
2! = 2 * 1! ; 2! = 2
3! = 3 * 2! ; 3! = 6
4! = 4 * 3! ; 4! = 24
.....
So, turn that around:
4! = 24
3! = 4!/4 ; 3! = 6
2! = 3!/3 ; 2! = 2
1! = 2!/2 ; 1! = 1
0! = 1!/1 ; 0! = 1
Now, how does that satisfy n!?
1! = 1
n! = n(n-1)! ; n[1, Infinity)
Take our answer from before and check to see that:
0! = 1
n! = n(n-1)! ; n[0, Infinity)
That's why 0! == 1.
One would think 0! therefore would be 0. Not so. Watch:
1! = 1
2! = 2 * 1! ; 2! = 2
3! = 3 * 2! ; 3! = 6
4! = 4 * 3! ; 4! = 24
.....
So, turn that around:
4! = 24
3! = 4!/4 ; 3! = 6
2! = 3!/3 ; 2! = 2
1! = 2!/2 ; 1! = 1
0! = 1!/1 ; 0! = 1
Now, how does that satisfy n!?
1! = 1
n! = n(n-1)! ; n[1, Infinity)
Take our answer from before and check to see that:
0! = 1
n! = n(n-1)! ; n[0, Infinity)
That's why 0! == 1.
August 30th 2004, 05:23 PM
MiloBones
Also, IIRC, the gamma function is 1 at zero (I cant remeber exactly what the gamma funtion is...) But I imagine that it has long been known that 0! = 1.
....Infinity
G(z) = INT x^(z-1) e^(-x) dx
........0
Eek. I had to look this one up.
INT = Integration sign.
MiloBones: G(0) is undefined.
G(z) = INT x^(z-1) e^(-x) dx
........0
Eek. I had to look this one up.
INT = Integration sign.
MiloBones: G(0) is undefined.
August 30th 2004, 06:00 PM
MiloBones
Thats because n! = G(n+1), and gamma is 1 (duh) at 0... my memory is going .