Giant Lobster
Active Member
By definition, e = lim{n -> infinite} (1 + 1/n)^n
But why is
lim{n -> infinite} (1 - 1/n)^n = 1/e ?
But why is
lim{n -> infinite} (1 - 1/n)^n = 1/e ?
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Can someone tell me why? (1 Viewer)
- Thread starter Giant Lobster
- Start date
Find the general case first.
lim {n-> infinity} (1+a/n)^n
= lim {n -> infinity} e^(ln[(1+a/n)^n])
= lim {n -> infinity} e^[n*ln(1+a/n)]
= e^[ lim {n -> infinity} n*ln(1+a/n) ]
= e^[ lim {n -> infinity} ln(1+a/n) / (1/n) ]
Now the question is how do we find lim {n -> infinity} ln(1+a/n) / (1/n) ? You have to use a certain rule known as L'Hoptial's rule. I'll explain this later.
= e^[ lim {n -> infinity} (-a/n^2)/(1+a/n) / (-1/n^2) ]
= e^[ lim {n -> infinity} a/(1+a/n) ]
= lim {n -> infinity} e^( a/(1+a/n) )
= e^a
so finally:
lim {n-> infinity} (1+a/n)^n = e^a
sub in a = 1 and you get the limit value of e, sub in a = -1, and you get the limit value of 1/e.
--------------------------------------------
L'Hopital's Rule:
if you have two functions f(x) and g(x), both continuous & differentiable (which means f'(x) and g'(x) are continuous too) at x=a, and if f(a) = g(a) = 0 then:
lim {x -> a} f(x)/g(x) = lim {x -> a } f'(x)/g'(x)
This can be proven from first principles of differentiation, but don't bother with it
lim {n-> infinity} (1+a/n)^n
= lim {n -> infinity} e^(ln[(1+a/n)^n])
= lim {n -> infinity} e^[n*ln(1+a/n)]
= e^[ lim {n -> infinity} n*ln(1+a/n) ]
= e^[ lim {n -> infinity} ln(1+a/n) / (1/n) ]
Now the question is how do we find lim {n -> infinity} ln(1+a/n) / (1/n) ? You have to use a certain rule known as L'Hoptial's rule. I'll explain this later.
= e^[ lim {n -> infinity} (-a/n^2)/(1+a/n) / (-1/n^2) ]
= e^[ lim {n -> infinity} a/(1+a/n) ]
= lim {n -> infinity} e^( a/(1+a/n) )
= e^a
so finally:
lim {n-> infinity} (1+a/n)^n = e^a
sub in a = 1 and you get the limit value of e, sub in a = -1, and you get the limit value of 1/e.
--------------------------------------------
L'Hopital's Rule:
if you have two functions f(x) and g(x), both continuous & differentiable (which means f'(x) and g'(x) are continuous too) at x=a, and if f(a) = g(a) = 0 then:
lim {x -> a} f(x)/g(x) = lim {x -> a } f'(x)/g'(x)
This can be proven from first principles of differentiation, but don't bother with it
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Giant Lobster
Active Member
amazing...
Im beginning to realise this e number is fascinating (yeh i sound so newb)
if its too too much truble, can u show me why
lim(n -> infinite) [ sum(k=0 to n) { x^k/k! } ] = e^x
Im beginning to realise this e number is fascinating (yeh i sound so newb)
if its too too much truble, can u show me why
lim(n -> infinite) [ sum(k=0 to n) { x^k/k! } ] = e^x
ok here's a really cheap no frills proof:
let f(x) = sum(k = 0 -> n) x^k/k!
= 1 + x + x^2/2 + x^3/6 + ... + x^n/n!
therefore,
f'(x) = 1 + x + x^2/2 + ... + x^(n-1)/(n-1)!
= f(x) - x^n/n!
f'(x) = f(x) - x^n/n!
let g(x) = lim {n -> infinity} f(x)
so g(x) = 1 + x + x^2/2 + x^3/6 + ....
g'(x) = g(x) - lim {n -> infinity} (x^n/n!)
lim {n -> infinity} x^n/n! = 0, for any finite value of x (the factorial function grows alot faster than an exponential function, you can prove this)
g'(x) = g(x)
so g(x) = C*e^x
but g(0) = 1 => C = 1
g(x) = e^x
but we already know g(x) = lim {n -> infinity} sum{k = 0 to n} x^k/k!
lim {n -> infinity} sum {k = 0 to n} x^k/k! = e^x
-------------------------------------------
This is actually known as the taylor series expansion of the function f(x) = e^x.
You'll actually notice that sum(k=0 to n) x^k/k! is actually a polynomial of nth degree!
Taylor series are a way of approximating a function to a polynomial, and the higher the degree of the polynomial you use the more accurate (in fact to represent a function 100% accurately by a polynomial, your polynomial needs to be of infinite degree!). This is how computers & calculators compute certain values like sin(100) or ln(6) etc. They don't store & retrieve the values, they approximate the function (e.g. sin() ) to a high degree polynomial and then sub & solve. All continuous & differentiable functions (not just f(x) = e^x) have taylor series expansions.
Last edited:
Grey Council
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haven't i always said that wogboy is a genius? Moreover, he is the most darn helpful genius ever.
BTW, wogboy, do you do this sort of stuff in Uni? Your doing engineering of some sort, arent ya?
BTW, wogboy, do you do this sort of stuff in Uni? Your doing engineering of some sort, arent ya?
Giant Lobster
Active Member
thats just bloody amazing...
ive always wondered how those trig functions are calculated electronically
hey, you mentioned its a "no frills" proof; by that do u mean theres a more "correct" proof out there?
and how do u prove that factorial grows the fastest?
i know it goes lnx < x^n < n^x < x! as x -> infinite (n E J)
in fact, how do u prove any one of these? i think it may be possible to prove anything up to n^x using harder 3u methods, but dunno how to treat x!
ive always wondered how those trig functions are calculated electronically
hey, you mentioned its a "no frills" proof; by that do u mean theres a more "correct" proof out there?
and how do u prove that factorial grows the fastest?
i know it goes lnx < x^n < n^x < x! as x -> infinite (n E J)
in fact, how do u prove any one of these? i think it may be possible to prove anything up to n^x using harder 3u methods, but dunno how to treat x!
BlackJack
Vertigo!
These appear in first year first semester maths for us. 'tis quite faskinating.
BlackJack
Vertigo!
This is definitely uni work, unless BoS is feeling particularly evil.
Differential calculus, 'tis.
Differential calculus, 'tis.
Giant Lobster
Active Member
by that do you mean i have to use diff to prove
?
How then do i diff x! ?
Not really, it's no frills because I only proved what you asked for (i.e. that particular infinite sum equal to e^x). I could have gone further and proven the general case for any function having a taylor series expansion and deriving the taylor formula, but nah I can't be bothered. If your still interested then use Google to search for "taylor series".
KeypadSDM
B4nn3d
Man, i've been trying to prove the:
x<sup>n</sup> < n<sup>x</sup>
For 0 < n < x (n, x E Z)
By differentiation, showing that the gradient of n<sup>x</sup> is always greater than the gradient of x<sup>n</sup> and hence that x<sup>n</sup> < n<sup>x</sup>
But I haven't had any hints, so I don't really know how to do it ... Ugh.
x<sup>n</sup> < n<sup>x</sup>
For 0 < n < x (n, x E Z)
By differentiation, showing that the gradient of n<sup>x</sup> is always greater than the gradient of x<sup>n</sup> and hence that x<sup>n</sup> < n<sup>x</sup>
But I haven't had any hints, so I don't really know how to do it ... Ugh.
KeypadSDM
B4nn3d
Sorry, i got the number wrong. It starts working after 4.
Giant Lobster
Active Member
no no... what i said is true for x -> infinite for all finite n