- #1

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[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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- Thread starter AxiomOfChoice
- Start date

- #1

- 533

- 1

[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!

- #2

mathman

Science Advisor

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However expand (1+x/n)

- #3

lurflurf

Homework Helper

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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.

- #4

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- #5

lurflurf

Homework Helper

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Why does everybody like l'Hôpital's rule so much? It is not needed here as log'(1)=1 suffices.

- #6

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- #7

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Why does everybody like l'Hôpital's rule so much? It is not needed here as log'(1)=1 suffices.

Are you kidding? l'Hôpital'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.

- #8

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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]

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]

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- #9

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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?

I think it will(as for angular momentum matrices,which are generators of rotation),but how to see it?

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- #10

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[itex]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.

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'Hôpital'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]

- #11

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[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...

- #12

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- #13

- 511

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thanks...I forgit it...

- #14

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Hi,

I would like to understand

how l'Hôpital's rule applies. we haven't covered it yet.

Thank you

I would like to understand

how l'Hôpital's rule applies. we haven't covered it yet.

Thank you

- #15

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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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