Estimating Uncertainty in f(x)=a/x

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Discussion Overview

The discussion revolves around estimating the uncertainty in the function f(x) = a/x, particularly how variations in the input x (denoted as Δx) affect the output f(x). Participants explore the sensitivity of the function to these uncertainties and seek to understand the implications of small fluctuations in x.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant inquires about the proper method to determine how f(x) reacts to uncertainty Δx, questioning the sensitivity of the function to small changes.
  • Another participant asserts that f(x) is sensitive to small Δx, providing a mathematical expression Δf = f '(x)Δx = -(a/x²)Δx to illustrate this sensitivity.
  • A subsequent post seeks clarification on whether the expression Δf = f '(x)Δx is a general definition applicable to any function, regardless of its specific dependence on x.

Areas of Agreement / Disagreement

Participants express differing levels of understanding regarding the sensitivity of f(x) to Δx, with some asserting its sensitivity while others seek clarification on the definitions involved. The discussion does not reach a consensus on the broader implications of the sensitivity.

Contextual Notes

The discussion does not address specific assumptions regarding the nature of Δx or the context in which f(x) is applied, leaving these aspects unresolved.

Niles
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Hi

Say I have a function given by f(x)=a/x, where a is some real constant. I know that there is some uncertainty Δx on x, but I don't know what it is. I just know that it will fluctuate around some value.

What is the proper way of determining how f(x) reacts to the uncertainty Δx? I mean, is there a way to find out if it is very sensitive to even small Δx or not?
 
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f(x) is sensitive to small Δx in the order -a/x2 because Δf = f '(x)Δx = -(a/x2)Δx.
 
EnumaElish said:
f(x) is sensitive to small Δx in the order -a/x2 because Δf = f '(x)Δx = -(a/x2)Δx.

Thanks. Is Δf = f '(x)Δx a definition regardless of how f(x) depends on x?
 
Ok, I think I got it... Thanks!
 

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