Do yo know how to solve this equation?
\[ax^2+ bx + c =0\]
Your answer will help me prepare tomorrow’s class
Last class we saw the easy rule for error propagation
\[ \begin{aligned} (x ± Δx) + (y ± Δy) & = (x+y) ± (Δx+Δy)\\ (x ± Δx) - (y ± Δy) & = (x-y) ± (Δx+Δy)\\ (x ± Δx\%) \times (y ± Δy\%)& =xy ± (Δx\% + Δy\%)\\ (x ± Δx\%) ÷ (y ± Δy\%)& =x/y ± (Δx\% + Δy\%) \end{aligned} \]
Here \(Δx\%\) represents the relative uncertainty, that is \(Δx/x\)
We use absolute uncertainty for + and -, and relative uncertainty for ⨉ and ÷
It is easy to get confused with relative errors
Instead of \((x ± Δx\%)\) it is better to write \[x(1± Δx/x)\]
Mathematical notation was invented to make things clear, not confusing
Exercise: Let’s verify the formulas
Remember that we assume that \(Δx/x\) is small
These rules are “pessimistic”. They give the worst case
In general the “errors” can be positive or negative, and they tend to compensate
(This is valid only if the errors are independent)
In this case we can analyze uncertainty using the rules of probabilities
In this case, the value \(Δx\) will represent the standard deviation of the measurement
The standard deviation is the square root of the variance
Then, we combine variances using the rule
“The variance of a sum is the sum of the variances”
(Again, this is valid only if the errors are independent)
\[ \begin{align} (x ± Δx) + (y ± Δy) & = (x+y) ± \sqrt{Δx^2+Δy^2}\\ (x ± Δx) - (y ± Δy) & = (x-y) ± \sqrt{Δx^2+Δy^2}\\ (x ± Δx\%) \times (y ± Δy\%)& =x y ± \sqrt{Δx\%^2+Δy\%^2}\\ \frac{x ± Δx\%}{y ± Δy\%} & =\frac{x}{y} ± \sqrt{Δx\%^2+Δy\%^2} \end{align} \]
When using probabilistic rules we need to multiply the standard deviation by a constant k, associated with the confidence level
In most cases (but not all), the uncertainty follows a Normal distribution. In that case
How much money is there?
(you can make any justified hypothesis)
Eureka!