proof for exponentials


by island-boy
Tags: exponentials, proof
island-boy
island-boy is offline
#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!
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Fermat
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#2
Mar14-06, 05:28 AM
HW Helper
P: 879
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
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#3
Mar14-06, 08:59 AM
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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
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#4
Mar14-06, 09:08 AM
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proof for exponentials


As a hint, set 1/u=x/n, and re-express in terms of u and x.
island-boy
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#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
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#6
Mar15-06, 01:19 AM
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You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)
benorin
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#7
Mar15-06, 02:40 AM
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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
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#8
Mar15-06, 02:44 AM
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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]


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