Solve Indeterminate Forms with Binomial Expansion

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In summary, the problem is that the person who solved it used a method that is unfamiliar to the person who is trying to solve the problem. They use a binomial expansion. This solution starts with taking the limit as x goes to infinity. Then, they use the binomial theorem to find the value of e. This is the problem: I am having difficulty seeing how binomial expansion could help. (Even in its general form, which you'd have to use, because x isn't restricted to the integers)
  • #1
RadiationX
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I have the solution to this particular problem, the problem is the person who solved it used a method that unfamiliar with. Does anyone know how to solve this problem using Binomial expansion. I know that the answer is [tex] e^k[/tex] where K is some constant. this is the problem:

[tex]\lim_{x\rightarrow\infty}(1 + \frac{k}{x})^x[/tex]
 
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  • #2
I'm having difficulty seeing how binomial expansion could help. (Even in its general form, which you'd have to use, because x isn't restricted to the integers)


You can't solve the problem until you decide upon an acceptable definition of e^k, and possibly have proven some facts about it. (For example, sometimes e^k is defined to be the value of this limit! So, in that case, the proof is trivial)
 
  • #3
Hurkyl said:
I'm having difficulty seeing how binomial expansion could help. (Even in its general form, which you'd have to use, because x isn't restricted to the integers)


You can't solve the problem until you decide upon an acceptable definition of e^k, and possibly have proven some facts about it. (For example, sometimes e^k is defined to be the value of this limit! So, in that case, the proof is trivial)


his solution starts like this:

[tex]1 + x\frac{k}{x} + x\frac{(x-1)}{2!}+\frac{k^2}{x^2}+x\frac{(x-1)(x-2)}{3!}\frac{k^3}{x^3}[/tex]


But i have no idea what this means.
 
  • #4
Here's an artifical way you could do this:

Let the limit be L. Take natural log of both sides. This gives

[itex]\ln L = \ln(\lim_{x\rightarrow\infty}(1 + \frac{k}{x})^x)[/tex]

As the log of a limit is the limit of log for continuous functions, we have

[tex]\ln L = \lim_{x\rightarrow\infty}x\ln((1 + \frac{k}{x}))[/tex]

which is the same as

[tex]\ln L = \lim_{x\rightarrow\infty}k\frac{\ln((1 + \frac{k}{x}))}{k/x}[/tex]

the limit is k and so [itex]L = e^k[/itex].

The theorem we have used here is that the log of a limit of a continuous function is the limit of the log of the function. But there is no better way than to use the definition of e, which is

[tex]e = \lim_{n\rightarrow\infty}(1+\frac{1}{n})^n[/tex]

Cheers
Vivek
 
  • #5
RadiationX said:
his solution starts like this:

[tex]1 + x\frac{k}{x} + x\frac{(x-1)}{2!}+\frac{k^2}{x^2}+x\frac{(x-1)(x-2)}{3!}\frac{k^3}{x^3}[/tex]


But i have no idea what this means.


Okay I get it now. "He" has used a binomial expansion for [itex](1+k/x)^x[/itex]. You should know that x can be negative, fractional or positive integral so you haave to use the binomial theorem for general nonintegral index (as done here). But there is a catch.

The idea used here is that as the order of every term grows, the terms in the brackets partially cancel the the (1/x)^k term. In more precise terms, the at least one of the product terms reduces the power of the denominator to 0. This method looks practical but it is not mathematically correct unless you first prove that the binomial expansion written actually converges to zero for higher order terms as [itex]x\rightarrow\infty[/itex]. This is one of those things which doesn't seem to deserve special proof, but it does.

So I would still think that the argument given by Hurkyl (and me in my first post) are closer to the acceptable definition of e.

Cheers
Vivek

EDIT: By the way, the second term of your original binomial expansion is incorrect. Change the + sign to a product and it'll be okay then.

EDIT#2: A better way to understand the order argument is as follows: a typical term of the the continued product in the numerator is of the form (x-k) for k = 0, 1, 2, ...kmax and for every such product the term in the denominator is x^kmax. Hence, there are as many terms in the denominator (I am breaking x^k as x*x*...*x (k times)). So the order of the numerator = order of denominator in terms of the powers of x. Of course, the order is same but there are still some lurking lower order terms in the numerator after cancellation. They can be reclubbed with the higher order denominator. This intuitively tells you that for large x, the sequence converges to some value. But you need to prove that it does (rigourously) first and find that value. All this is unnecessary if you are allowed to use the proof right out of the box though.
 
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  • #6
Thank you Guys!
 
  • #7
I would like to point out that method of proof has a lot of subtle technicalities. Remember the famous "proof" that [itex]\lim_{n \rightarrow \infty} n / n = 0[/itex]:

Fake proof:
1/1 = 1/1
2/2 = 1/2 + 1/2
3/3 = 1/3 + 1/3 + 1/3
4/4 = 1/4 + 1/4 + 1/4 + 1/4
...
Now, each individual term goes to zero. Therefore, the limit is simply a sum of many zeroes, and is thus zero!


The moral is that you're interchanging the limit operations, and that's something you should only do if you really know what you're doing.

With the binomial expansion method, you have to let the number of terms go to infinity first, and THEN let x go to infinity.

So, the argument you posted really isn't valid unless you apply some advanced machinery that let's you do things the other way around. (Though it is a nice heuristic argument)
 

Related to Solve Indeterminate Forms with Binomial Expansion

1. What is an indeterminate form?

An indeterminate form is a mathematical expression that does not have a definite value when evaluated. This usually occurs when there is a division by zero or an infinity raised to the power of zero.

2. How can I use binomial expansion to solve indeterminate forms?

Binomial expansion is a method for expanding binomial expressions, which can be used to simplify expressions with indeterminate forms. By expanding the expression and cancelling out common terms, the indeterminate form can often be eliminated.

3. Can all indeterminate forms be solved using binomial expansion?

No, not all indeterminate forms can be solved using binomial expansion. Some forms, such as 0/0 or ∞/∞, may require other methods such as L'Hôpital's rule or algebraic manipulation.

4. Is there a specific formula for solving indeterminate forms with binomial expansion?

There is not a specific formula, as the method for solving indeterminate forms with binomial expansion may vary depending on the specific form and expression. It is important to understand the principles of binomial expansion and how to apply them to different forms.

5. Can binomial expansion be used to solve indeterminate forms in all branches of mathematics?

Yes, binomial expansion is a widely applicable mathematical concept and can be used in various branches of mathematics, such as calculus, algebra, and probability.

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