What Does n^{-1} Really Represent in Mathematical Terms?

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Homework Help Overview

The discussion revolves around the mathematical representation of negative exponents, specifically the notation n^{-1} and its implications in various contexts, including dimensional analysis and physical equations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the meaning of negative exponents, questioning the original poster's interpretation of n^{-1} and n^{-2}. Some attempt to relate these concepts to dimensional analysis and physical equations, while others seek clarification on definitions and implications.

Discussion Status

The discussion is ongoing, with multiple interpretations being explored. Some participants provide clarifications on the definitions of negative exponents, while others express confusion and seek further explanation. There is no explicit consensus on the understanding of the notation.

Contextual Notes

Participants reference specific examples from physics, such as energy and the Dirac operator, to illustrate their points. There is a noted contradiction in the interpretation of dimensional analysis notation, which adds to the complexity of the discussion.

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Homework Statement



The problem is stated in the attempt as a solution

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The Attempt at a Solution



Suppose you have

n^{-1} am I right in thinking this is just \frac{n}{1} so for something like n^{-2} would it be simply \frac{n}{2}?
 
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No. Why would you think that? n^(-1)=1/n, n^(-2)=1/n^2.
 
Because in Dimensional analysis, we tend to use the notation ie. MTl^{-2} which would mean that it is mass times time divided by length squared...


mmm... has it got something to do with things like

\epsilon \mu = c^{-2}

such that

\sqrt{\epsilon \mu} = \frac{1}{c}

I could be totally off, mind my ignorance.
 
I have no idea what you are confused about. The definition of n^(-k) where k>=0 is pretty simple. It's 1/n^k. Could you try and explain again?
 
Let's try this again. Let us say, E is energy.

What would be E^{-2} be?
 
help1please said:
Let's try this again. Let us say, E is energy.

What would be E^{-2} be?

It would be 1/E^2. Since energy has units M*L^2/T^2, E^2 has units M^2*L^4/T^4. So 1/E^2 would have units T^4/(M^2*L^4). Wouldn't it?
 
In fact let's do this another way. I was reading about the dirac operator

D(\psi) = \hbar^2 R^{-2} \psi(x)

How does R^{-2} read to you? Just 1/R^2 like you said before?
 
help1please said:
In fact let's do this another way. I was reading about the dirac operator

D(\psi) = \hbar^2 R^{-2} \psi(x)

How does R^{-2} read to you? Just 1/R^2 like you said before?

If R is a number, then YES, it is just 1/R2. If R is an operator, then NO, it means something different.

RGV
 
R is taken to be a radius or a length.
 
  • #10
help1please said:
Because in Dimensional analysis, we tend to use the notation ie. MTl^{-2} which would mean that it is mass times time divided by length squared...
Yes, "divided by length squared" which contradicts what you wrote! MTl2= MT/l2, Not "MTl/2".


mmm... has it got something to do with things like

\epsilon \mu = c^{-2}

such that

\sqrt{\epsilon \mu} = \frac{1}{c}

I could be totally off, mind my ignorance.
You are making this much too difficult- \epsilon\mu= \frac{1}{c^2} which again contradicts what you said originally.
 

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