- #1
Caspian
- 15
- 0
Ok, this isn't a homework question -- more out of curiosity. But it seems so trivial that I hate to post it under "General Physics"
We all know the standard formula for error propagation:
[tex]\sigma_f = \sqrt{\dfrac{\partial f}{\partial x}^2 \sigma_x^2 + \dfrac{\partial f}{\partial y}^2 \sigma_y^2[/tex]
Now, let [tex]f = x^2[/tex]. We get [tex]\sigma_f = \sqrt{(2x)^2 \sigma_x^2}[/tex]
Now, let [tex]f = x \cdot x[/tex]. We get [tex]\sigma_f = \sqrt{x^2 \sigma_x^2 + x^2 \sigma_x^2} = \sqrt{2 x^2 \sigma_x^2}[/tex].
This says that the error of [tex]x \cdot x[/tex] equals [tex]\sqrt{2}[/tex] times the error of [tex]x^2[/tex]!
I'm baffled at this... does anyone know why this is true? I've never seen a derivation of the standard error propagation formula... does the derivation assume that the two variables are not equal? (btw, If someone knows where to find the derivation to the formula, I would be very happy to see it)
Thanks!
We all know the standard formula for error propagation:
[tex]\sigma_f = \sqrt{\dfrac{\partial f}{\partial x}^2 \sigma_x^2 + \dfrac{\partial f}{\partial y}^2 \sigma_y^2[/tex]
Now, let [tex]f = x^2[/tex]. We get [tex]\sigma_f = \sqrt{(2x)^2 \sigma_x^2}[/tex]
Now, let [tex]f = x \cdot x[/tex]. We get [tex]\sigma_f = \sqrt{x^2 \sigma_x^2 + x^2 \sigma_x^2} = \sqrt{2 x^2 \sigma_x^2}[/tex].
This says that the error of [tex]x \cdot x[/tex] equals [tex]\sqrt{2}[/tex] times the error of [tex]x^2[/tex]!
I'm baffled at this... does anyone know why this is true? I've never seen a derivation of the standard error propagation formula... does the derivation assume that the two variables are not equal? (btw, If someone knows where to find the derivation to the formula, I would be very happy to see it)
Thanks!