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On why e^x = lim (1+x/n)^n

by AxiomOfChoice
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AxiomOfChoice
#1
Oct22-09, 06:17 PM
P: 529
Is there someone who can explain why this is true, or point me to an online resource that provides a proof of it?
[tex]
e^x = \lim_{n\to \infty} \left(1 + x/n \right) ^n
[/tex]
I know that in some ways, this is how the exponential function is defined. But any resources you can provide that explain it in more detail would be appreciated. Thanks!
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mathman
#2
Oct22-09, 06:53 PM
Sci Advisor
P: 6,038
I am not sure if this will satisfy you.
However expand (1+x/n)n by the binomial theorem. Take the term by term limit as n -> oo. The result is the power series for ex.
lurflurf
#3
Oct22-09, 08:16 PM
HW Helper
P: 2,263
The reason is when n is large we have
1+x/n~exp(x/n)
we also have (n need not be large)
exp(x/n)^n=exp(x)

The proof depends on ones definition of e^x such as
1)exp'(x)=exp(x) with exp(0)=1
2)exp(x)exp(y)=exp(x+y) with exp'(0)=1
3)exp(x)=1+x+x^2/2!+...+x^n/n!+...
4)exp(log(x))=x log having been previously defined

A common one given 3) is to expand (1+x/n)^n by the binomial theorem and then shown the limit of the sum is the sum of the limits.

Bohrok
#4
Oct22-09, 11:27 PM
P: 867
On why e^x = lim (1+x/n)^n

Let f(x) be the right side, then rewrite it as elnf(x). With ln(f(x)), rearrange it so you can use l'H˘pital's rule (with respect to n), probably use a substitution, and then you can evaluate the limit.
lurflurf
#5
Oct23-09, 02:47 AM
HW Helper
P: 2,263
Why does everybody like l'H˘pital's rule so much? It is not needed here as log'(1)=1 suffices.
elibj123
#6
Oct23-09, 07:34 AM
P: 240
I wrote the proof for you
Attached Thumbnails
ex.jpg  
l'H˘pital
#7
Oct23-09, 12:33 PM
P: 256
Quote Quote by lurflurf View Post
Why does everybody like l'H˘pital's rule so much? It is not needed here as log'(1)=1 suffices.
Are you kidding? l'H˘pital's rule is the best. Made of epic win.

As for the OP, yeah, generally, if you let the limit be a variably, say y, you could ln both sides, and solve it that way.
neelakash
#8
Nov10-09, 07:18 AM
P: 511
Might be I am little late with this post...however,let me say that I could not understand how L Hospital's would have to be implemented to do this...I could follow the binomial expansion argument or the use of standard limit argument.

A second issue...From the texts that deal with symmetry,the limit is also used with operator argument.Like

[tex]\displaystyle\lim_{k\to\infty}\ [\ I_{\ n\times\ n}\ + \frac{\ i \vec{\phi}\ . \mathbb{D} }{k}]^\ {k}[/tex]

[tex] = \ exp\{ \ {\ i \ {\vec{\phi}\ . \mathbb{D} } \}[/tex]
neelakash
#9
Nov10-09, 07:58 AM
P: 511
I am sorry that it took time to fix the Latex symbols...Let me get back to the discussion...For operator valued argument,the identity also holds...My question is will the relation still hold if [tex]\mathbb{D}=\mathbb{D}_{1}\ +\mathbb{D}_{2}[/tex] and the two operators D1 and D2 do not commute...?

I think it will(as for angular momentum matrices,which are generators of rotation),but how to see it?
l'H˘pital
#10
Nov10-09, 09:42 AM
P: 256
Quote Quote by neelakash View Post
Might be I am little late with this post...however,let me say that I could not understand how L Hospital's would have to be implemented to do this...I could follow the binomial expansion argument or the use of standard limit argument.
[itex]
y = \lim_{n\to \infty} \left(1 + x/n \right) ^n
[/itex]

Right? So
[itex]
ln y = ln [ \lim_{n\to \infty} \left(1 + x/n \right) ^n]
[/itex]

Focus on the right, it looks kinda like this ...

[itex]
\lim_{n\to \infty} \left n ln [(1 + x/n \right)]
[/itex]

The n goes to the front because of logarithms.

[itex]
\lim_{n\to \infty} \left (1/n)^-1 ln [(1 + x/n \right)]
[/itex]

If you take the limit, you get [itex]0/0[/itex] So, apply l'H˘pital's rule and you'll get

[itex]
ln y = \lim_{n\to \infty} \left x/(1+x/n) = x [/itex]

So, cancel the ln, and you get as desired,

[itex]

y = e^x [/itex]
neelakash
#11
Nov10-09, 11:59 AM
P: 511
But can you write in general

[itex]

ln [ \lim_{n\to \infty} \left(1 + x/n \right) ^n]=\ lim_{n\to \infty} \left n ln [(1 + x/n \right)]

[/itex]

I vaguely remember,from outside, one can insert ln inside [lim n-->0] with some condition satisfied...I cannot remember what the condition was...
l'H˘pital
#12
Nov10-09, 01:28 PM
P: 256
The condition is about continuity. ln is continuous, so yeah, you can move the limit inside or outside.
neelakash
#13
Nov10-09, 08:54 PM
P: 511
thanks...I forgit it...
mrm0607
#14
Nov13-09, 08:00 PM
P: 7
Hi,

I would like to understand

how l'H˘pital's rule applies. we haven't covered it yet.

Thank you
mrm0607
#15
Nov13-09, 08:58 PM
P: 7
Hi,

We have not covered the l Hopitals rule yet so I am trying to expand (1+x/n)^n to show

as lim(n-->infinity) (1+x/n)^n = e^x for any x> 0

This is what I did so far and after that I am short of lost:

Please read this (n 1) in vertical form ( I do not know how to use Latex)

lim (n-->infinity)(1 + x/n)^n =

lim(x->infinity)[1 + (n 1) (1/n) + (n 2)(1/n^2) + (n 3)(1/n^3) ---------- + --
-- (n k)(1/n^k) + ---+(n n)(1/n^n)]

=1 + 1/1! + 1/2! + 1/3! + ---- 1/k! + --- + 1/n!

now how does this lead to e^x?

May be I did not understand what you meant?

Thank you.


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