Uncertainty Derivations/Calculation

[tmobilerocks]

Homework Statement

If F = aXⁿ = f +- f +δf where a is a constant, show f = xⁿ and $\frac{δf}{f}$ = $\frac{nδx}{x}$.

X = x +- δx

x refers to the average and δx refers to uncertainty in x.

Homework Equations

The power rule for error propagation shows that the uncertainty is multiplied n times (where n is the power raised).

The Attempt at a Solution

I'm having trouble showing that f = xⁿ. Through the use of algebraic manipulation, I was able to get a(x+δx)ⁿ = f + δf. I then made the assumption to ignore the constant a and by deduction say x = 5 +- 0.5, set f = xⁿ because it is continuously multiplied by whatever the function x is to the nth power. The second part is easier- mainly I just took the differential δf = n*x^n-1δx. This simplifies to $\frac{δf}{f}$ = n$\frac{δx}{x}$

 

[BvU]

Hello t and welcome to PF. If I want to help, I have to be able to read what you've written. Could you proofread your stuff and explain what isn't completely universally known language ?

I recognize some aspects from error propagation theory, but I find your jargon hard to understand.

 

[tmobilerocks]

Hello,

I am supposed to show that f = xⁿ given that F = aXⁿ = f ± δf where a is a constant. X is equal to x ± δx. In plain English, X is equal to the average x, plus or minus the uncertainty in x (δx).

I am having trouble showing that f = xⁿ. I have already derived δf/f using the method shown above.

Thanks again!

 

[BvU]

Yes, it still looks weird that a should disappear from f. Is that really how it is presented to you ?
I mean if F = 1000 X² and X = 5.000 ± 0.001 there is no way that f can be x²

 

[HalfLostAndSearching]

which is the desired result.

Great job on your attempt at solving this problem! You are correct in saying that the power rule for error propagation states that the uncertainty is multiplied n times when raised to the power of n. This means that when we have a function F = aXn, the uncertainty δF can be calculated using the formula δF = n*aXn-1*δX.

To show that f = xn, we can use algebraic manipulation as you mentioned. By substituting F with f and X with x, we get: f = a*xn = f + δf. From here, we can cancel out the f on both sides and we are left with a(x+δx)n = δf. Since a is a constant, we can ignore it and focus on the (x+δx)n term. By using the binomial expansion, we get (x+δx)n = xn + nxn-1*δx + ... + δxn. Since we are only interested in the first two terms, we can ignore the rest and we are left with f = xn + nxn-1*δx, which simplifies to f = xn.

For the second part, you are correct in saying that δf = n*xn-1*δx. From here, we can divide both sides by f to get \frac{δf}{f} = n*xn-1*δx/f. We know that xn = f, so we can substitute this in to get \frac{δf}{f} = n*δx/x, which is the desired result.

Overall, great job on your attempt and explanation of the problem! Keep up the good work.