Prove that a sub n equal to e

[Tracker]

How do I show that [math] a_n = \left( 1 +\frac{1}{n} \right)^{n} = e [/math]

[insane_alien]

you can't because it is not.

 

counter examples include n=1 and well, any number except infinity

 

your missing some limits.

[Tracker]

Sorry about the confusion I believed that [math] a_n [/math] was one of the standard notions to mean it is an infinite sequence. Assuming that fact how do I prove it?

[NeonBlack]

Here is a hint to get you started:

(Note: If you want to be more rigorous, define a function [math]f(x)=(1+\frac{1}{x})^x[/math]

Assuming this limit exists, set it equal to say L. Take the (natural) log and then see if you can solve the limit.

(You will probably also need L'hopital's rule)

[the tree]

I believed that [math] a_n [/math] was one of the standard notions to mean it is an infinite sequence.

What you're trying to prove is that [math]a_n \to e[/math] (as [math]n \to \infty)[/math], it's fairly important to note that for no natural value of [math]n[/math] is [math]a_n[/math] at all equal to [math]e[/math], since [math]a_n[/math] is always rational.

Assuming this limit exists,

That's a horrible thing to assume, and you certainly don't need to if your going to do this:

Take the (natural) log and then see if you can solve the limit.

That's not enough, but it's start. Given that, you'd also need to show that [math]x_n \to l \Rightarrow e^{x_n} \to e^l[/math].

 

Note: If you want to be more rigorous, define a function [math]f(x)=(1+\frac{1}{x})^x[/math]

How is that more rigorous? You're trying to prove something about the limit of a rational sequence, redefining the problem wouldn't help.

(You will probably also need L'hopital's rule)

I'd hope not.

[NeonBlack]

tree, it may seem stupid to you, but mathematicians can be very uptight (I don't to offend anyone ) If you are going to use l'Hopital's rule, (I don't know of any other way to do it, but their may be) you will need to take derivatives. You can't really take the derivative of [math]a_n[/math] because it is a discrete sequence, but if you have a continuous function f(x) that passes through all of the same points as [math]a_n[/math] then you can use that.

 

Yes, I think you do need to assume that it exists, but there are a number of ways to prove whether or not it does. If a limit diverges, and you try to work with it as if it is a finite number, you may get inconsistent results.

[D H]

Take the (natural) log and then see if you can solve the limit.

This approach assumes the natural logarithm function.

 

Tracker: [math]e=\lim_{n\to\infty}\left(1+\frac 1 n\right)^n[/math] is one of the standard definitions of [math]e[/math]. You must be using some other definition of e and your job is to show that this limit is the same value. So, first things first: What are you using as a definition of e?

 

Hint: Have you tried a binomial expansion of [math]\left(1+\frac 1 n\right)^n[/math]?

[Tracker]

This is where I am at:

 

Using/assuming the Continuity of f(x)

[math]

a_n = \left( 1 +\frac{1}{n} \right)^{n} = e^{x\ln(1 + \frac{1}{x})} [/math]

 

[math]

 

\lim_{x\to\infty} 1 + \frac{1}{x} = 1 [/math] therefore [math] e^{x\ln(1)} = 0

[/math]

 

So I am making a mistake somewhere but I am not sure where.

 

I should mention this is just a practice problem and not a homework assignment, but this seemed the best place to post the question.

 

This approach assumes the natural logarithm function.

 

Tracker: [math]e=\lim_{n\to\infty}\left(1+\frac 1 n\right)^n[/math] is one of the standard definitions of [math]e[/math]. You must be using some other definition of e and your job is to show that this limit is the same value. So, first things first: What are you using as a definition of e?

 

Hint: Have you tried a binomial expansion of [math]\left(1+\frac 1 n\right)^n[/math]?

 

I understand it one of the standard definitions, but I want to prove it for my own knowledge. I am not using anything else as a definition of e. How could I do a binomial expansion as it is an infinite sequence?

 

Here how to do it with L'Hopital's rule:

 

[math]

e^{x\ln(1 + \frac{1}{x})}

[/math]

[math]

\lim_{x\to\infty} [/math][math] \ln(\frac{1 + \frac{1}{x}}{\frac{1}{x}} ) = 1[/math] therefore [math] =e^1 = e[/math]

 

Anyone know how to do it another way?

Edited November 12, 2008 by Tracker
multiple post merged

[the tree]

tree, it may seem stupid to you, but mathematicians can be very uptight (I don't to offend anyone ) If you are going to use l'Hopital's rule, (I don't know of any other way to do it, but their may be) you will need to take derivatives. You can't really take the derivative of [math]a_n[/math] because it is a discrete sequence, but if you have a continuous function f(x) that passes through all of the same points as [math]a_n[/math] then you can use that.

I don't see any reason why you would need L'Hopitals rule. All you need to show is that [math]n\ln(1+\frac{1}{n}) \to 1[/math] and that [math]x_n \to l \Rightarrow e^{x_n} \to e^l[/math], like I said. (the statement is the harder to prove)*

 

There are some cases where squeezing the goalposts can be a useful trick (e.g. if you want to prove that [math]x \in S[/math] then it might be easier to prove that [math]x \in S_1 \subset S[/math]) but you could never ever claim that this makes your proof more rigorous.

 

Yes, I think you do need to assume that it exists, but there are a number of ways to prove whether or not it does.

If you can prove it, which you can, then you are not assuming it exists. Taking the existence of a limit as a given is a very fatal error and it basically the opposite of rigour.

 

* edit PROTIP for this approach it's best to use the definition of e as exp(1) where exp is defined as the function equal to it's derivative and exp(0)=1.

Edited November 12, 2008 by the tree

[NeonBlack]

This approach assumes the natural logarithm function.

 

Is there anything wrong with that? We can also define the natural logarithm as the anti-derivative of 1/x no?

 

tree, you don't understand. When I say "I assume that this exists," it only means that I am not going to show the proof.

 

All you need to show is that [math]n\ln(1+\frac{1}{n}) \to 1[/math]

 

And how would you do that? If you let n go to infinity since ln 1 =0' date=' you have [math']\infty \cdot 0 [/math] which is a good time to bust out l'Hopital

 

By the way: the sequence is not rational. Look at n=2. [math]a_2=\sqrt{\frac{3}{2}}[/math]

 

DH: The binomial expansion way is very nice. I didn't know about that before.

[Tracker]

If you let n go to infinity since ln 1 =0, you have [math]\infty \cdot 0 [/math] which is a good time to bust out l'Hopital

 

DH: The binomial expansion way is very nice. I didn't know about that before.

 

Thank you for the help, as I am sure you noticed I used L'Hopital's rule to solve it. Though how do I do it with binomial expansion?

[the tree]

Is there anything wrong with that? We can also define the natural logarithm as the anti-derivative of 1/x no?

...ish.

 

tree, you don't understand. When I say "I assume that this exists," it only means that I am not going to show the proof.

So what the hell do you say when you mean that you actually are going to assume something? Do you often say the complete opposite to what you mean?

 

And how would you do that? If you let n go to infinity since ln 1 =0, you have [math]\infty \cdot 0 [/math] which is a good time to bust out l'Hopital

Oh easy. From the definition of ln, you can tell that [math]\frac{1}{n+1} < \ln(n+1) - \ln(n) < \frac{1}{n}[/math] (which is how you'd prove that ln tends to zero) then obviously:

[math]\frac{n}{n+1} < n \ln(n+\frac{1}{n} )< \frac{n}{n}[/math]

So by the magic of the sandwich principle you have that:

[math]n \ln(n+\frac{1}{n}) \to 1[/math].

 

Neat, but not hugely difficult.

 

 

 

By the way: the sequence is not rational. Look at n=2. [math]a_2=\sqrt{\frac{3}{2}}[/math]

[math]a_2 = (1+\frac{1}{2})^2 = \frac{3}{2}^2 \not= \sqrt{\frac{3}{2}}[/math]

Edited November 12, 2008 by the tree

[NeonBlack]

 

[math]a_2 = (1+\frac{1}{2})^2 = \frac{3}{2}^2 \not= \sqrt{\frac{3}{2}}[/math]

 

Oops. You're right. I was also thinking about [math]

\lim_{x\to 0} (1 + x)^{\frac{1}{x}}

[/math] and in my head I turned it into [math](1 + \frac{1}{x})^{\frac{1}{x}}[/math]

 

So what the hell do you say when you mean that you actually are going to assume something?

 

I guess it's usually clear based on on context then. But it's not very often that I will do that.

 

Tracker:

The binomial expansion is [math]

1+\frac{n!}{(n-1)!} \frac{1}{n}+\frac{1}{2}\frac{n!}{(n-2)!}\frac{1}{n^2}+\frac{1}{6}\frac{n!}{(n-3)!}\frac{1}{n^3}+...[/math]

[the tree]

I guess it's usually clear based on on context then.

Well in the context of maths, which by an incredible coincidence is the context in question, assuming something means not proving it.

[NeonBlack]

assuming something means not proving it.

 

...means that I am not going to show the proof.

 

I'm sorry I don't understand why you're getting so upset over this.

[ydoaPs]

[math]

 

\lim_{x\to\infty} 1 + \frac{1}{x} = 1 [/math]

 

If you mean [math]\lim_{x\to\infty}(1+\frac{1}{x})=1[/math], then you're wrong. [math]\lim_{x\to\infty}(1+\frac{1}{x})=\lim_{x\to\infty}1+\lim_{x\to\infty}\frac{1}{x}=1+\lim_{x\to\infty}\frac{1}{x}[/math].

 

[math]\lim_{x\to\infty}\frac{1}{x}[/math] does not exist, so [math]\lim_{x\to\infty}(1+\frac{1}{x})[/math] does not exist.

[D H]

So, first things first: What are you using as a definition of e?

I understand it one of the standard definitions, but I want to prove it for my own knowledge. I am not using anything else as a definition of e[/i'].

 

Of course you are. Otherwise you couldn't do this:

Here how to do it with L'Hopital's rule:

 

[math]

e^{x\ln(1 + \frac{1}{x})}

[/math]

[math]

\lim_{x\to\infty} [/math][math] \ln(\frac{1 + \frac{1}{x}}{\frac{1}{x}} ) = 1[/math] therefore [math] =e^1 = e[/math]

 

Anyone know how to do it another way?

You implicitly assumed a whole lot of things there. You assumed

1. [math]a^b=\exp(b\ln a)[/math]
    
    
   
2. [math]\ln\left(1+\frac 1 x\right) \rightarrow \frac 1 x\;\text{as}\;x\to\infty[/math]
    
    
   
3. [math]\exp(1)=e[/math]

That's a lot of unjustified assumptions!

 

 

The "standard" definitions of e include

1. [math]e=\lim_{n\to\infty}\left(1+ \frac 1 n\right)^n[/math], which is the problem at hand
    
    
   
2. [math]e=\lim_{x\to 0}\left(1+x\right)^{1/x}[/math], which is just a restatement of the first definition
    
    
   
3. [math]e=\sum_{n=0}^{\infty}\frac 1 {n!}[/math]
    
    
   
4. The solution to [math]\int_1^{\,e} \frac 1 x\,dx = 1[/math]

The binomial expansion of definition 1 leads directly to definition 3 (see post #13 by Neon Black).

[Mr Skeptic]

tree, it may seem stupid to you, but mathematicians can be very uptight (I don't to offend anyone )

 

I don't to offend you, but you left out a word there. Not that it matters since it is not math, and I am not an English major so I don't have to be uptight about a verb being left out of a sentence since it is clear anyways.

 

If you mean [math]\lim_{x\to\infty}(1+\frac{1}{x})=1[/math], then you're wrong. [math]\lim_{x\to\infty}(1+\frac{1}{x})=\lim_{x\to\infty}1+\lim_{x\to\infty}\frac{1}{x}=1+\lim_{x\to\infty}\frac{1}{x}[/math].

 

[math]\lim_{x\to\infty}\frac{1}{x}[/math] does not exist, so [math]\lim_{x\to\infty}(1+\frac{1}{x})[/math] does not exist.

 

Methinks you confused an infinity with a zero.

[D H]

Methinks you confused an infinity with a zero.

His mistake was here:

[math]\lim_{x\to\infty}(1+\frac{1}{x})=\lim_{x\to\infty}1+\lim_{x\to\infty}\frac{1}{x}[/math]

yourdad: That is completely invalid.

[ydoaPs]

Opps...I did. Don't mind me. I'm crazy on lack of sleep.

[Mr Skeptic]

yourdad: That is completely invalid.

 

Now I'm confused. How's that step invalid?

[D H]

This is not a valid operation:

 

[math]\lim_{x\to a}f(g(x)+h(x)) = f(\lim_{x\to a}g(x))+f(\lim_{x\to a}h(x))[/math]

[Cap'n Refsmmat]

But that wasn't the operation being done. This was:

 

[math]\lim_{x \to a} (g(x) + h(x)) = \lim_{x \to a} g(x) + \lim_{x \to a} h(x)[/math]

 

And as far as I know, that's valid.

 

Unless, of course, to continue the chain, I misread somebody's post as well...

[D H]

Oops, sorry ydop. It was Tracker in post #8 who made the mistake of applying the limit to the function arguments. ydop's mistake was exactly what Mr Skeptic said.