- #1
azaharak
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A while back, one of my undergraduate physics professors gave an argument for why the uncertainty in a function or quantity F is given by
\Delta F = \sqrt{^{N}_{i-1}\sum(\frac{\partial F}{\partial x_{i}})^{2}(\Delta x_{i})^{2}}
He argued to think of a right triangle and think of c=\sqrt{a^{2}+b^{2}}
The uncertainty in the length of side c, would be calculate in a similar method, however it would be
\Delta c =\sqrt{(da)^{2}+(db)^{2}} which via chain rule would be
\Delta c =\sqrt{(\frac{\partial c}{\partial a}\Delta a)^{2}+(\frac{\partial c}{\partial b}\Delta b)^{2}}
He then argued that in a function of several variables, those variables can be thought of as perpendicular to each other in the same way that (a) and (b) are in the right triangle (because they pertain to degrees of freedom), this is why we call for the sum in quadrature.
I know that it can be derived from normal distribution, however is the argument above correct reasoning?
Thank you to all
Alex Z
\Delta F = \sqrt{^{N}_{i-1}\sum(\frac{\partial F}{\partial x_{i}})^{2}(\Delta x_{i})^{2}}
He argued to think of a right triangle and think of c=\sqrt{a^{2}+b^{2}}
The uncertainty in the length of side c, would be calculate in a similar method, however it would be
\Delta c =\sqrt{(da)^{2}+(db)^{2}} which via chain rule would be
\Delta c =\sqrt{(\frac{\partial c}{\partial a}\Delta a)^{2}+(\frac{\partial c}{\partial b}\Delta b)^{2}}
He then argued that in a function of several variables, those variables can be thought of as perpendicular to each other in the same way that (a) and (b) are in the right triangle (because they pertain to degrees of freedom), this is why we call for the sum in quadrature.
I know that it can be derived from normal distribution, however is the argument above correct reasoning?
Thank you to all
Alex Z
