Error analysis in log space: How to handle exponents in regression analysis?

VINAYBAR
Messages
5
Reaction score
0

Homework Statement



Maybe, I am a little stupid, but I just can't understand what I am doing wrong here. I have 2 quantities which I have to convert to log (base 10) and perform a regression analysis.

Quantity=1.235e9
error=3.4475e8

Homework Equations



I have to express the above quantity as log (base 10) with the error. The catch is I do not want to count the exponent i.e. just take log of the quantity and the error directly. I know, that if I take the log with the exponent then there is no problem, but I need to compare some results and hence don't want to do it.

The Attempt at a Solution



So, I took the log of 1.235=0.0917

The error in log space by the standard formula is 0.12. Now, if I count the powers then the quantity is 9.0917 and the error is unchanged. If I use my regression program with these values, then I get a decent fit. But, if I use it with a value of 0.0917 and error 0.12 then my fits are all messed up.

If something is unclear, please let me know and I can supply you with more information.
 
Physics news on Phys.org
Why are you excluding the exponent? If you exclude then it will be a total different number
For example 1 can be written as 0.01e2
log of 1 is 0 and log of 0.01 is -2.
So both are not same. You should not avoid exponents while taking log because it is integral part of that number.
 
I have to ignore the exponent because after my regression is complete, I want to compare it to existing data. The previous data set does the calculation by ignoring the exponent i.e. (log (value))*(10^9) which is why I am ignoring the exponent as well.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top