1^∞ ≠ 1?

 

What, Is this really indeterminate?

 

1 * 1 * ... * 1 = 1

 

1 * 1 = 1 = a

a * 1 = 1 = a

 

edit: I misunderstood "indeterminate"

Edited October 15, 201114 yr by alan2here

But on the other hand 5+5=10, which is does not contain 3 as a prime factor

An indeterminate form is what happens when you take a limit, where the result cannot be determined by separating the stuff inside the limit into two parts, and taking the limit on each. So saying that [math]1^\infty[/math] is indeterminate means that there are 4 sequences [math]a_n, b_n, c_n, d_n[/math] such that [math]\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} c_n = 1[/math] and [math]\lim_{n \rightarrow \infty} b_n = \lim_{n \rightarrow \infty} d_n = \infty[/math], but

[math]\lim_{n \rightarrow \infty} a_n^{b_n} \neq \lim_{n \rightarrow \infty} c_n^{d_n}[/math]

 

We can see that this is true with the following sequences:

[math]a_n = 1, b_n = n[/math] which has limit 1

[math]c_n = (1 + \frac{1}{n}), d_n = n[/math], which has limit e.

=Uncool-

Edited October 15, 201114 yr by uncool

Thanks. I don't understand much of that but what you'r saying is that indeterminate dosn't mean that it cannot be determined. It's instead to do with limits and the such.

Edited October 15, 201114 yr by alan2here

It's instead to do with limits and the such.

 

Loosely, because you can define and approach the limit in different ways and get different answers the thing is indeterminate.

 

Consider

 

[math]\lim_{x \rightarrow 0} 1^{\frac{1}{x}} = 1[/math],

 

 

which suggests that you should set [math]1^{\infty}=1[/math]. However, to have a really well defined definition we require some robustness in how we define this. So instead consider

 

[math]\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}}[/math].

 

This also looks like how we should define [math]1^{\infty}=1[/math]. However taking the limit gives us

 

[math]\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = e \neq 1[/math].

 

So we see that we do not have a very clear way to define [math]1^{\infty}[/math], so it is indeterminate.

 

Uncool also shows this in a very similar way, he makes more precise statements. This is the correct way to approach indeterminates, in terms of "compound limits".

This seems like how 2 + 2 can seem to = 5, ∞^2 or whatever if the algebra is rearranged correctly, but there is always a good reason why it's not valid.

Edited October 21, 201114 yr by alan2here

Loosely, because you can define and approach the limit in different ways and get different answers the thing is indeterminate.

 

Consider

 

[math]\lim_{x \rightarrow 0} 1^{\frac{1}{x}} = 1[/math],

 

 

which suggests that you should set [math]1^{\infty}=1[/math]. However, to have a really well defined definition we require some robustness in how we define this. So instead consider

 

[math]\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}}[/math].

 

This also looks like how we should define [math]1^{\infty}=1[/math]. However taking the limit gives us

 

[math]\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = e \neq 1[/math].

 

So we see that we do not have a very clear way to define [math]1^{\infty}[/math], so it is indeterminate.

 

Uncool also shows this in a very similar way, he makes more precise statements. This is the correct way to approach indeterminates, in terms of "compound limits".

 

If you were to interpret [math] 1^ \infty[/math] as a cardinal number [math]1^K[/math] where [math]K[/math] is an infinite cardinal then you would be talking about the cardinality of an infinite cartesian product of the singleton set. That cardinality is 1.

 

As you note, if you try to interpret the expression in the context of real numbers, you can make all sorts of interpretations and get about any answer that strikes your fancy.

 

"There's no sense in being precise when you don't even know what you're talking about." -- John von Neumann