Real numbers as powers of real exponents

[kulix]

Just a quick question for you guys, I've been unable to find the answer to this. Can all real numbers be written as n^p, where p is a real number?

 

[micromass]

What is n??

Anyway, negative numbers can't be written in that form.

 

[kulix]

Ok, let me restate. Can all positive real numbers x be written as n^p, where n and p are real numbers?

Or

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

 

[micromass]

Yeah. Take n=x and p=1. This even holds for negative numbers, so I guess my first reply wasn't quite correct.

 

[kulix]

Oh, mine wasn't complete, p needs to be different from 1.

The better question;

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

 

[micromass]

Take $n=\sqrt[3]{x}$ and p=3.

 

[kulix]

Oh dear. Let me try this:

let S be a series from 2 to infinity of 1 / (n*sqrt(n^2 -1)).

Can you write S as 1/n^p?

 

[micromass]

if n < n * sqrt(n^2-1) < n^2, does that mean there exists p such that n * sqrt (n^2 -1) = n^p ?

This would force

$$p=\log_n{n\sqrt{n^2-1}}$$

This is ok if the logarithm exists. For that, we need n>0, n≠1 and $n\sqrt{n^2-1}>0$.

 

[dodo]

Also, if you're looking for an expression for p when the number n is fixed, then p is the logarithm (base n) of x. That is, given$$x=n^p$$you take logs on both sides,$$\ln x = \ln (n^p) = p \ln n$$and thus$$p = \frac {\ln x}{\ln n}$$so any positive x can be expressed this way (p will be negative for x<1, zero for x=1 and positive for x>1).

P.S.: Oops, two posts got in the middle while I was writing this. I was intending to continue after post #6.

 

[kulix]

Perfect, thank you!

 

[micromass]

Oh dear. Let me try this:

let S be a series from 2 to infinity of 1 / (n*sqrt(n^2 -1)).

Can you write S as 1/n^p?

This question makes no sense. In the series S, your variable n is the dummy variable and thus ranges over all positive integers. In your last sentence, n has become a fixed number.

 

[kulix]

My bad again, what I meant to say is can you write it as a series of 1/n^p.

 

[micromass]

It seems to me like you just want to check convergence of the series. Why can't you just do a comparison test with

$$\sum C\frac{1}{n^2}$$

where C is a constant.

 

[kulix]

Indeed I do, and yes I could do that. However, I'm interested in knowing if my approach is acceptable.

 

[micromass]

The problem with your approach is that (generally) your p will be dependent on n.

So if you follow Dodo's and my hints, then you won't end up with

$$\sum \frac{1}{n^p}$$

but rather with something in the exponent which is also dependent of n. This makes the situation harder.

 

[kulix]

Unfortunately, I used my approach on an exam, that's why I'm so keen on knowing if it's correct. Would have used the comparison, but alas, I didn't occur to me at the time.

 

[Office_Shredder]

The most straightforward way to determine the accuracy of your solution without waiting to get your exam back is to just post the full question and your full solution. I'm not sure what you would even do after re-writing your series as 1/n^p if you didn't know what the value of p is