A Introduction of a factor Δℓ when summing equal distants 𝐶ℓ

AI Thread Summary
The discussion centers on understanding the introduction of the factor Δℓ in the standard deviation formula for Cℓ in the context of Legendre expansion. The standard deviation is expressed as σ(Cℓ) = √(2/(2ℓ+1)Δℓ)Cℓ, which differs from the typical form that excludes Δℓ. The need for this factor arises when dealing with a finite number of multipoles, where Δℓ represents the width of the multipole bins, affecting the variance calculation. The reasoning provided suggests that this factor accounts for averaging over the finite multipole range, which is necessary when Cℓ values are not available for every integer ℓ. The conversation emphasizes the importance of understanding this adjustment in the context of numerical and theoretical analysis of angular power spectra.
fab13
Messages
300
Reaction score
7
TL;DR Summary
I would like to demonstrate the formula that defines the standard deviation of a 𝐶ℓ and mostly justify the presence of the factor Δℓ when we have a finite number of 𝐶ℓ's angular power spectrum and associated finite number of multipole ℓ (generated by a code ). I tried at the end of my post to calculate the mean Cℓ for a single bin but it is missing a factor 2. Any help would be really appreciated. This is a problem about the theorical/numerical frontier given the fact I have not Δℓ=1.
Hello,

In the context of Legendre expansion with ##C_\ell## quantities, below the following formula which is the error on a ##C_\ell## :

##\sigma_(C_{\ell})=\sqrt{\frac{2}{(2 \ell+1)\Delta\ell}}\,C_{\ell}\quad(1)##

where ##\Delta\ell## is the width of the multipoles bins used when computing the angular power spectra.

I would like to understand why this factor appears. Normally, we can infer the expression of ##\sigma_(C_{\ell})## and get :

##\sigma_(C_{\ell})=\sqrt{\frac{2}{(2 \ell+1)}}\,C_{\ell}##

That is to say, without having the ##\Delta\ell## factor under the root.

In my case, I have only a finite number of multipoles ##\ell## (60 exactly) in a range between 10 and 5000 and 60 corresponding ##C_\ell##. From my tutor, to compute this standard-deviation, I have to take :

##\Delta\ell=(4990/60)##.

But I would like to understand why I have to use this factor when we have a finite number of equidistant multipoles with associated ##C_\ell## values.

For example, my tutor told me that, ideally, we would have a ##C_\ell## for each ##\ell=1..N##, i.e with ##\ell=1,2,3,4,...N## with the formula :

##\sigma(C_{\ell})^2=\frac{2}{(2 \ell+1)}\,C_{\ell}^{2}##

But he mentionned that, like we have only a finite number of equidistant multipoles ##\ell##, we are obliged to use this factor ##\Delta\ell## in our case (equation##(1)##).

My tutor told me it is like an average on the expression with multipole ##\ell##, i.e on the factor ##\sqrt{\dfrac{2}{(2\ell+1)}}##, but I didn't fully understand this meaning.

Could anyone help me to justify the using of this factor ##\Delta\ell## ?

UPDATE : Someone tried quickly to explain why we introduce this factor ##\Delta\ell## inside the square root, but I am still confused about this demonstration.

Here is his reasoning :

For the interval ##b##: ##b=[\ell, \ell+\Delta\ell]##, one has :

##C_{\ell,b}=\dfrac{1}{\Delta\ell}\,\sum_{\ell'=\ell}^{\ell'+\Delta\ell}\,C_{\ell'}##

If one takes the variance, then :

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell^2}\,\sum_{\ell'}\text{Var}(C_{\ell'})##

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell^2}\,\sum_{\ell'}\dfrac{2}{(2\ell'+1)}\,(C_{\ell'})\quad(2)##

Important step here, one considers the equality (which is actually an approximation) :

##\sum_{\ell'}\dfrac{2}{(2\ell'+1)}\,(C_{\ell'})=\dfrac{2}{(2\ell_{mean}+1)}\,(C_{\ell'_{mean}}^2)\,\Delta\ell\quad(3)##

This way, we have from equation ##(2)## :

##\text{Var}(C_{\ell,b})\simeq \dfrac{1}{\Delta\ell^2}\,\dfrac{2}{(2\ell_{mean}+1)}\,(C_{\ell'_{mean}}^2)\,\Delta\ell##

Finally the expression of the variance for the interval ##b##:

##\text{Var}(C_{\ell,b})\simeq \dfrac{1}{\Delta\ell}\,\dfrac{2}{(2\ell_{mean}+1)}\,(C_{\ell'_{mean}}^2)##

I still didn't understand the step between ##(2)## and ##(3)##.

For me, an average is under the form :

##C_{\ell'_{mean}}=\dfrac{1}{2}\,(C_{\ell'}+C_{\ell'+\Delta\ell})## but I can't see this factor ##\dfrac{1}{2}## in this reasoning.

If someone could explain this critical step between ##(2)## and ##(3)##.
 
Space news on Phys.org
The average ##\bar x## of ##N## different quantities ##x_i## satisfies ##\sum x_i = \bar x N##. This is what is being used.
 
@Orodruin : thanks for your quick answer.

Could you be please more explicit with the notations I took ? your formula is trivial but I have difficulties to apply it with my notations.

Moreover, do you give a reasoning on a single bin of width ##\Delta\ell## or on the total interval ##[\ell_{min},\ell_{max}]## ?

In the formula ##\sum x_i = \bar x N##, does ##N## correspond to ##\Delta\ell## in my formula ?

Thanks in advance, Regards
 
Last edited:
Isn't really there nobody who could explain me my problem of understanding about my last message above ?
 
fab13 said:
Summary:: I would like to demonstrate the formula that defines the standard deviation of a 𝐶ℓ and mostly justify the presence of the factor Δℓ when we have a finite number of 𝐶ℓ's angular power spectrum and associated finite number of multipole ℓ (generated by a code ). I tried at the end of my post to calculate the mean Cℓ for a single bin but it is missing a factor 2. Any help would be really appreciated. This is a problem about the theorical/numerical frontier given the fact I have not Δℓ=1.

For me, an average is under the form :

Cℓmean′=12(Cℓ′+Cℓ′+Δℓ) but I can't see this factor 12 in this reasoning.

If someone could explain this critical step between (2) and (3).
Can you not work it "backwards" to see what N must be in this situation? You have the answer provided to you, It is worrisome that this step eludes you.
 
Hi everyone,

It seems that I may have a "demonstration" about my issue of understanding :

If i take the expression for the bin "##b##" ##[\ell_b,\ell_b+\Delta\ell_b]## :

##C_{\ell,b}=\dfrac{1}{\Delta\ell_b}\,\sum_{\ell=\ell_b}^{\ell_b+\Delta\ell_{b}}\,(C_{\ell})##

Then :

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell_b^2}\,\sum_{\ell=\ell_b}^{\ell_b+\Delta\ell_{b}}\,\text{Var}(C_{\ell})\quad(2)##

By taking ##\Delta\ell=1##, we can have :

##\text{Var}(C_{\ell_b})=\dfrac{1}{\Delta\ell_{b}^2}\,\sum_{\ell=\ell_b}^{\ell+\Delta\ell}\dfrac{2}{(2\ell+1)}\,(C_{\ell}^2)\Delta\ell \simeq \int_{\ell=\ell_{b}}^{\ell_{b}+\Delta\ell_b} \dfrac{1}{\Delta\ell_{b}^2} \dfrac{2}{2 \ell+1}\,C_\ell^2\text{d}\ell \quad(3)##

and this integral can be approximated by rectangular integration method :

##\int_{\ell=\ell_{b}}^{\ell_{b}+\Delta\ell_b} \dfrac{1}{\Delta\ell_{b}^2} \dfrac{2}{2 \ell+1}\,C_\ell^2\text{d}\ell \simeq \dfrac{1}{\Delta\ell_{b}^{2}} \dfrac{2}{(2\ell_{b}+1)} \,C_{\ell_{b}}^{2} \Delta{\ell_{b}} =\quad(4)##

where ##\Delta{\ell_{b}}## is the delta width of bin "##b##" :

then, I could write :

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell_{b}} \dfrac{2}{(2 \ell_{b}+1)}\,C_{\ell_b}^{2} \quad(5)##

Do you think my reasoning is right ?
 
Last edited:
https://en.wikipedia.org/wiki/Recombination_(cosmology) Was a matter density right after the decoupling low enough to consider the vacuum as the actual vacuum, and not the medium through which the light propagates with the speed lower than ##({\epsilon_0\mu_0})^{-1/2}##? I'm asking this in context of the calculation of the observable universe radius, where the time integral of the inverse of the scale factor is multiplied by the constant speed of light ##c##.
The formal paper is here. The Rutgers University news has published a story about an image being closely examined at their New Brunswick campus. Here is an excerpt: Computer modeling of the gravitational lens by Keeton and Eid showed that the four visible foreground galaxies causing the gravitational bending couldn’t explain the details of the five-image pattern. Only with the addition of a large, invisible mass, in this case, a dark matter halo, could the model match the observations...
Hi, I’m pretty new to cosmology and I’m trying to get my head around the Big Bang and the potential infinite extent of the universe as a whole. There’s lots of misleading info out there but this forum and a few others have helped me and I just wanted to check I have the right idea. The Big Bang was the creation of space and time. At this instant t=0 space was infinite in size but the scale factor was zero. I’m picturing it (hopefully correctly) like an excel spreadsheet with infinite...
Back
Top