Accuracy of f(x) when x is Accurate to 6%”

  • Thread starter Thread starter huan.conchito
  • Start date Start date
  • Tags Tags
    Accuracy
AI Thread Summary
The discussion focuses on determining the accuracy of the function f(x) = x^{1/4} when x has a 6% accuracy. It is established that the percent error in f(x) is approximately a quarter of the percent error in x, leading to a maximum error of about ±1.5%. The calculations show that the minimum value of f(x) is approximately 0.9847f(x) and the maximum is about 1.01467f(x), indicating an asymmetrical error distribution. The importance of confirming that f(x) is monotone increasing within the relevant domain is also highlighted, as this affects the accuracy assessment. Overall, the analysis demonstrates that while calculus can be used, simpler methods may yield exact answers more efficiently.
huan.conchito
Messages
44
Reaction score
0
Suppose f(x) =x^{1/4} . If x is accurate to within 6%, within what percent is f(x) accurate ?

i know that %percent error = delta F/ f(x) = f'(x)*deltaX / f(x)
what i don't know is how to get delta f out this this information
 
Last edited:
Mathematics news on Phys.org
anyone know ?
 
I think I ran into this in calc, but not quite sure, I believe you can differentiate dy/dx, then you could plug in the %error for dx, then solve for dy. That probably isn't it but its my best guess at the moment so late at night for me right now :-p
 
\Delta f = \frac{df}{dx}|_{x_0}\Delta x
\Delta f = \frac{1}{4}x_0^{-3/4}\Delta x=\frac{1}{4}\frac{x_0^{1/4}\Delta x_p}{100}

where \Delta x_p is your percent error in x and x0 is the value of x at which you're finding the error. To get the percent error in f, you just evaluate:

\Delta f_p=100\frac{\Delta f}{f(x_0)}=100\frac{\Delta f}{x_0^{1/4}}=\frac{1}{4}\Delta x_p

In other words, it's just a quarter of the percent error in x. If you're just looking for the raw error in f, it depends on the value you're evaluating at and it's given by the second equation above.
 
Ah that is indeed it, if you used the method I said before, you would actually need to plug in .06x, not just .06, then solve for dy and divide by the function leaving you with 1.5% I believe.
 
Honestly, there is no need to use calculus for this, unless a calculator is forbidden. You can get an exact answer much more easily directly with a calculator.

f(x) = x^{\frac{1}{4}}

We're given that the maximum error in x is 6 %, meaning that x varies from a min of 0.94x to a max of 1.06x

Hence,

min(f(x)) = (0.94x)^{\frac{1}{4}} which approximately equals to 0.9847(x^{\frac{1}{4}}) = 0.9847f(x), meaning a percentage error of - 1.53 %

and

max(f(x)) = (1.06x)^{\frac{1}{4}} which approx. 1.01467f(x), meaning a percentage error of + 1.47 %

So you can see the relative error in f(x) is assymetrical for a symmetrical relative error in x, and goes from a maximum of 1.53 % on the negative side to a max of 1.47 % on the positive side.

To be rigorous, you actually need to establish that f(x) is monotone increasing for the domain in question (true in this case). But it wouldn't work, for instance, in cases where there is a local maximum or minimum to f(x) around the values in consideration. And example of this would be in computing the relative error in f(x) = (x-1)^2 for a 5 % error in x where x is known to be close to 1.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top