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<p class="MsoNormal"><span style="color:#1F497D">Steffen, just a note: I should think that another possible worry arises in the approach you’ve outlined by the use of non-orthogonal polynomials. For any expansion of the yield in terms of powers of cos<sup>
</sup></span><span style="color:#1F497D">α</span><span style="color:#1F497D"> , one should use the (orthogonal) Legendre polynomials,
<i>P<sub>n </sub></i>(cos<sup>2</sup></span><span style="color:#1F497D">α), I’d think, for the same reason one uses the (orthogonal) </span><span style="color:#1F497D">cos m</span><span style="color:#1F497D">α</span><span style="color:#1F497D"> terms in the
Fourier moment method</span><span style="color:#1F497D">.</span><span style="color:#1F497D"> Otherwise, since cos<sup>2</sup></span><span style="color:#1F497D">α</span><span style="color:#1F497D"> is not orthogonal to cos<sup>
</sup></span><span style="color:#1F497D">α</span><span style="color:#1F497D">, the cos<sup>2</sup></span><span style="color:#1F497D">α</span><span style="color:#1F497D"> yield term will automatically be picking up pieces both of the zeroth-order
<i>P<sub>0 </sub></i>(cos </span><span style="color:#1F497D">α</span><span style="color:#1F497D">)and the cos<sup>2</sup></span><span style="color:#1F497D">α</span><span style="color:#1F497D"> part of the fourth-order
<i>P<sub>4 </sub></i>(cos </span><span style="color:#1F497D">α</span><span style="color:#1F497D">) term. For a perfect detector (meaning perfect efficiency and 4</span><span style="color:#1F497D">π</span><span style="color:#1F497D"> acceptance), this is not
crucial, but I would worry that it manifestly would be problematic for the real CLAS detector. ---BGR<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span style="font-family:"Book Antiqua",serif;color:blue">Professor Barry G. Ritchie
<br>
Department of Physics <br>
Arizona State University <br>
Tempe, AZ 85287-1504</span></b><b><span style="font-family:"Book Antiqua",serif;color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span style="font-family:"Book Antiqua",serif;color:#1F497D"> <o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span style="font-family:"Book Antiqua",serif;color:blue">Phone: (480) 965-4707
<br>
Fax: (480) 965-7954<o:p></o:p></span></b></p>
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<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"> Frost [mailto:frost-bounces@jlab.org]
<b>On Behalf Of </b>Steffen Strauch<br>
<b>Sent:</b> Wednesday, August 17, 2016 3:17 PM<br>
<b>To:</b> Michael Dugger <dugger@jlab.org><br>
<b>Cc:</b> frost@jlab.org<br>
<b>Subject:</b> Re: [Frost] Question<o:p></o:p></span></p>
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<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Dear Michael,<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">Yes, this is what I originally suggested for the experimentally determined sums over all events.<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">In the meantime, I think I need to revise the expressions and the numerator sum should read:
<b>cos(\alpha)</b> and the denominator sum should read <b>Q cos^2(\alpha)</b>. This and the earlier expressions give the same results for constant background or if the observable does not change over the kinematic bin; like in our simplified examples. Practically,
the differences will be small. However, when there are variations of the observable or of the background over a large bin (like in the double-pion case), the latter expressions give in my opinion results for the observable which are more meaningful.<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">I will update my slides and we can discuss this more during tomorrow’s meeting.<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">Steffen<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">On Aug 17, 2016, at 1:21 PM, Michael Dugger <<a href="mailto:dugger@jlab.org">dugger@jlab.org</a>> wrote:<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal"><br>
Steffen,<br>
<br>
I just want to make sure I understand.<br>
<br>
You weight each numerator event by<br>
<br>
Q cos(\alpha)<br>
<br>
and each denominator event by<br>
<br>
[Q cos(\alpha)]^2 .<br>
<br>
Is this correct?<br>
<br>
Take care,<br>
Michael<br>
<br>
On Wed, 17 Aug 2016, Steffen Strauch wrote:<br>
<br>
<br>
<o:p></o:p></p>
<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
<p class="MsoNormal">Dear Michael,<br>
<br>
It is important that the weights in the moments in the denominator contain an additional factor of Q compared to the numerator.<br>
<br>
I don’t think there is a typo here. If you compare on page 4 the two sets of moment ratios, you see that the first set already contains Q in the integral. Weighing the cos(alpha) moment with an additional factor of Q, gives you the Q^2 in the integral of the
numerator in the second set. The denominator of the first set does not contain any Q in the integral. So, an additional factor of Q^2 in the weight of the moment gives you the Q^2 in the integral of the denominator of the second set.<br>
<br>
For the correct expression that the extracted value is the weighted average of P and not Q*P.<br>
<br>
Thanks,<br>
Steffen<br>
<br>
<br>
<br>
<br>
<o:p></o:p></p>
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<p class="MsoNormal">On Aug 17, 2016, at 12:06 PM, Michael Dugger <<a href="mailto:dugger@jlab.org">dugger@jlab.org</a>> wrote:<br>
<br>
<br>
Steffen,<br>
<br>
On last line of slide 4 and slide 5 of your pdf:<br>
<br>
<a href="https://urldefense.proofpoint.com/v2/url?u=https-3A__www.jlab.org_Hall-2DB_secure_g9_g9-5Fstrauch_mtg_FROST-5Fmeeting-5F2016-5F08-5F18.pdf&d=CwMFaQ&c=AGbYxfJbXK67KfXyGqyv2Ejiz41FqQuZFk4A-1IxfAU&r=NC99X3Muut85jp1nyEEaKzrqGMedseDv3USQMbrzzMU&m=uKTOtlnnx_ZYFBxdgkWlssxnYsmNeTBkCDiI5B_qXYI&s=rNmvyekg4GCUDmS6Ot0NLHq7ce474-Bn80L5NOU7gko&e=">https://www.jlab.org/Hall-B/secure/g9/g9_strauch/mtg/FROST_meeting_2016_08_18.pdf</a><br>
<br>
you show on the left hand side<br>
<br>
Y_{Q cos\alpha}/Y_{Q^2 cos^2 \alpha},<br>
<br>
where there is a factor of Q in the numerator and a factor of Q^2 in the denominator.<br>
<br>
However, in the middle of the line you have Q^2 in both the numerator and denominator. Is there a typo?<br>
<br>
If you have Q^2 in both the numerator and denominator, the expression will not work.<br>
<br>
Take care,<br>
Michael<br>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">_______________________________________________<br>
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