Can anyone help with peak summation?

  • Thread starter Thread starter physical101
  • Start date Start date
  • Tags Tags
    Peak Summation
physical101
Messages
41
Reaction score
0
Hi there I have read that the area of a peak can be approximated by summing the recorded peak intensities. I can't see how this works?
If you add all the peak intensities together is not just the magnitude of their sum and not the area of the surface the peak overlays?
Someone told me that integration is just a summation of the peak intensities but is it not a summation of the peak areas at an really small interval?
Please help
 
Mathematics news on Phys.org
I suppose you are talking about http://mathworld.wolfram.com/RiemannSum.html" when evaluating the area. Then you are right, you have to take into account the mesh size \Delta x.

Could it be that your mesh size equals 1? Or maybe your are only interested in the relative area of your peak? Because then if you write down the ratio

ratio = peak_Area / total_Area

the mesh size cancels and it is sufficient to add the peak intensitities.
 
Last edited by a moderator:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top