I Propagation of uncertainty in the slope

[Jussi Lundahl]

I'm doing a lab report from electronic spectrum of iodine. I did Birge-Sponer plot from my data. Excel gave it to me a slope y = -2,0698x+133,34. From regression analysis I get uncertainties for slope and intercept.

Slope: $-2,069761731 \pm 0,075075941$
Intercept: $133,3385857 \pm 2,396753622$
But the real equation of the slope is $$\Delta E_{v'} = \tilde{\nu}_e'-2\tilde{\nu}_e'x_e'(v'+1)$$, where
$$\tilde{\nu}_e' = 133,3385857 \pm 2,396753622 $$
$$-2\tilde{\nu}_e'x_e' = -2,069761731 \pm 0,075075941 $$
$$(v'+1)=x $$.

Now the question: How I can calculate uncertainty of $$\tilde{\nu}_e'x_e'$$ and $$x_e'$$?

 

[Dale]

I cleaned up your equations a bit. To do what you want you have to write your desired parameters in terms of your measured parameters (the slope and the intercept).

 

[Jussi Lundahl]

I cleaned up your equations a bit. To do what you want you have to write your desired parameters in terms of your measured parameters (the slope and the intercept).

So, you mean like this:
$$\tilde{\nu}_e' - 2\tilde{\nu}_e'x_e' = 0$$
$$ x_e' = \frac{\tilde{\nu}_e'}{2\tilde{\nu}_e'}$$

and the uncertainty is
$$\delta x_e' = \sqrt{ \left( \frac{\partial \frac{ \tilde{\nu}_e' }{2 \tilde{\nu}_e' }}{\partial \tilde{\nu}_e' } \times \delta \tilde{\nu}_e' \right)^2} $$
$$\delta x_e' = \sqrt{ \left( \frac{1/2}{\partial \tilde{\nu}_e'} \times 2,396753622 \right)^2}$$.

There for $$\delta x_e' =0.$$ Is this right?

 

[Vanadium 50]

Maybe. You haven't told us what your measured parameters are.

And by the way 133.3385857 +/- 2.396753622 should be 133.3 +/- 2.4. Do you understand why?

 

[Jussi Lundahl]

Maybe. You haven't told us what your measured parameters are.

And by the way 133.3385857 +/- 2.396753622 should be 133.3 +/- 2.4. Do you understand why?

Yes, I understand. I just copy/paste the values from excel. :) What do you mean by measured parameters?

I think I figured out this problem...

$$Lets~calculate~x_e':$$
\begin{align*}
&& k= - 2\tilde{\nu}_e'x_e' &= -2,069761731\\ \\
\iff && x_e' &= -\frac{k}{2\tilde{\nu}_e'}\\ \\
\iff && x_e' &= \frac{2,069761731}{2\tilde{\nu}_e'} = 0,007761301
\end{align*}
and the uncertainty is:
\begin{align*}
&&\delta x_e' &= \sqrt{ \left( \frac{\partial x_e'}{\partial \tilde{\nu}_e' } \times \delta \tilde{\nu}_e' \right)^2 + \left(\frac{\partial x_e'}{\partial k} \times \delta k\right)^2} \\ \\
\iff && \delta x_e' &= \sqrt{\left(\frac{k}{2\times (\tilde{\nu}_e')^2} \times \delta \tilde{\nu}_e'\right)^2 + \left(-\frac{1}{2 \times \tilde{\nu}_e'}\times \delta k \right)^2} \\ \\
\iff && \delta x_e' &= \sqrt{ \left( \frac{-2,069761731}{2\times (133,3385857)^2} \times 2,396753622 \right)^2 + \left(-\frac{1}{2\times133,3385857}\times 0,075075941\right)^2} \\ \\
&& &= 0,000314195 \approx 0,001
\end{align*}

 

[WWGD]

So, you mean like this:
$$\tilde{\nu}_e' - 2\tilde{\nu}_e'x_e' = 0$$
$$ x_e' = \frac{\tilde{\nu}_e'}{2\tilde{\nu}_e'}$$

and the uncertainty is
$$\delta x_e' = \sqrt{ \left( \frac{\partial \frac{ \tilde{\nu}_e' }{2 \tilde{\nu}_e' }}{\partial \tilde{\nu}_e' } \times \delta \tilde{\nu}_e' \right)^2} $$
$$\delta x_e' = \sqrt{ \left( \frac{1/2}{\partial \tilde{\nu}_e'} \times 2,396753622 \right)^2}$$.

There for $$\delta x_e' =0.$$ Is this right?

Sorry for necropost. There seems to be a mistake/typo here:

$$ x_e' = \frac{\tilde{\nu}_e'}{2\tilde{\nu}_e'}$$

Terms cancel out and we end up with 1/2.

 

[sophiecentaur]

Yes, I understand. I just copy/paste the values from excel. :

You can format numbers (and anything else) to show them in a variety of ways. Choose the decimal places and save yourself ink!