Register to reply

Proof for exponentials

by island-boy
Tags: exponentials, proof
Share this thread:
island-boy
#1
Mar14-06, 02:44 AM
P: 99
hello, I need the proof to show that:
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]

Here's what I was able to come up with so far:
since the derivative of [tex]e^x[/tex] is also [tex]e^x[/tex],

then let f(x) = [tex]e^x[/tex]
thus:
D(f(x)) = [tex]\lim_{n\rightarrow\infty} \frac{f(x+h) - f(x)}{h} = \lim_{n\rightarrow\infty} \frac{e^{x+h} - e^x}{h} = e^x\lim_{n\rightarrow\infty} \frac{e^h - 1}{h}[/tex]

so for the derivative of [tex]e^x[/tex] to equal itself,
[tex]\lim_{n\rightarrow\infty} \frac{e^h - 1}{h} = 1[/tex]

so for small values of h, we can write:
[tex]e^h - 1 = h[/tex]
and so
[tex] e = (1+h)^{1/h}[/tex]

Replacing h by 1/n, we get:
[tex] e = (1 + 1/n)^n[/tex]
As n gets larger and approaches infinity, we get:
[tex]e = \lim_{n\rightarrow\infty} (1+1/n)^n[/tex]

so, how do I get:
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]

Also, is it true that:
[tex](1+x/n)^n \leq e^x[/tex] and
[tex](1-x/n)^n \leq e^{-x}[/tex] for every natural n and every x element of X?
How can I prove this?

thanks!
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
Fermat
#2
Mar14-06, 05:28 AM
HW Helper
P: 878
Quote Quote by island-boy
hello, I need the proof to show that:
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]
...
thanks!
Have you tried the Binomial theorem ?
Hurkyl
#3
Mar14-06, 08:59 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,099
I would've taken the expression for e and raised both sides to the x power.

so for small values of h, we can write:
You really ought to do this more rigorously...

P.S. many of the limits in your limits are wrong.

arildno
#4
Mar14-06, 09:08 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Proof for exponentials

As a hint, set 1/u=x/n, and re-express in terms of u and x.
island-boy
#5
Mar15-06, 12:31 AM
P: 99
Quote Quote by Hurkyl
I would've taken the expression for e and raised both sides to the x power.


You really ought to do this more rigorously...

P.S. many of the limits in your limits are wrong.
which one of the limits are wrong?

I was able to prove that:
[tex] (1 + x/n)^n \leq e^x [/tex] and [tex] (1 - x/n)^n \leq e^{-x} [/tex]

Using the Taylor expansion for [tex]e^x[/tex], I got

[tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...[/tex]
which means
[tex]e^x \geq 1 + x[/tex]
subtituting x by x/n, I get
[tex]e^{x/n} \geq 1 + x/n[/tex]
thus
[tex]e^{(x/n)n} \geq (1 + x/n)^n[/tex]
simplifying, I get:
[tex]e^x \geq (1 + x/n)^n [/tex]

Similarly I used the Taylor Expnansion to prove that
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]

since [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...[/tex]

and using binomial theorem for [tex]\lim_{n\rightarrow\infty} (1+x/n)^n[/tex]
I got:
[tex]\lim_{n\rightarrow\infty}1^n + \lim_{n\rightarrow\infty} x + \lim_{n\rightarrow\infty} \frac{n(n-1)(x^2)}{(n^2)(2!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(x^3)}{(n^3)(3!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(n-3)(x^4)}{(n^4)(4!)} + ...

= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...

=e^x
[/tex]

these correct?
Hurkyl
#6
Mar15-06, 01:19 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,099
You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)
benorin
#7
Mar15-06, 02:40 AM
HW Helper
P: 1,025
See this thread posts # 3 and 12.

I take it that your definition for [tex]e^x[/tex] is the Taylor series about x=0, viz.

[tex]e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/tex]
benorin
#8
Mar15-06, 02:44 AM
HW Helper
P: 1,025
Essentially, this

Quote Quote by benorin
That [tex]\lim_{n\rightarrow\infty} (1+\frac{x}{n})^n = e^x[/tex] may be proven as follows:

by the binomial theorem,

[tex](1+\frac{x}{n})^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}\left( \frac{x}{n}\right) ^{k} = \sum_{k=0}^{n} \frac{n!}{(n-k)!}\frac{1}{n^{k}} \frac{x^k}{k!}\rightarrow \sum_{k=0}^{\infty} \frac{x^k}{k!}=:e^{x}\mbox{ as }n\rightarrow\infty[/tex]

since [tex] \frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty[/tex]

although there are some uniform convergence issues to be handled when taking the limit of the above sum...
and this,

Quote Quote by benorin
This is not rigorous, but it works...

[tex] \frac{n!}{(n-k)!} = \frac{n(n-1)(n-2)\cdots (n-(k-1))(n-k)!}{(n-k)!} =n(n-1)(n-2)\cdots (n-(k-1))[/tex]

counting the number of terms in the above product: it goes n-0 through n-(k-1) so there are (k-1)-0+1 = k terms so we know that the leading term will be nk when the product is expanded, and hence

[tex] \frac{n!}{(n-k)!} = n^k + \mbox{ some polynomial of degree k-1 in }n[/tex]

or rather

[tex] \frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty[/tex]


Register to reply

Related Discussions
Convolution with exponentials Engineering, Comp Sci, & Technology Homework 6
Convolution with exponentials Calculus 0
Integrating exponentials Precalculus Mathematics Homework 3
Derivatives of Exponentials Differential Equations 5
Differentiating exponentials Calculus 2