proof for exponentials

island-boy

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!

Fermat

Have you tried the Binomial theorem ?

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.

arildno

As a hint, set 1/u=x/n, and re-express in terms of u and x.

island-boy

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

You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)

benorin

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

Essentially, this

and this,