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40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
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ID Date Author Type Category Subject
13066   Thu Jun 15 18:56:31 2017 jigyasaUpdateComputer Scripts / ProgramsMC2 Pitch-Yaw offset

A python script to randomly vary the MC2 pitch and yaw offset and correspondingly record the value of MC transmission has been started on Donatella in the control room and should run for a couple of hours overnight.

The script is named MC_TRANS_1.py and is located in my user directory at /users/jigyasa

Apologies for any inconvenience.
Data analysis will follow.

13073   Mon Jun 19 18:41:12 2017 jigyasaUpdateComputer Scripts / ProgramsMC2 Pitch-Yaw offset

The previous run of the script had produced some dubious results!

The script has been modified and now scans the transmission sum for a longer duration to provide a better estimate on the average transmission. The pitch and yaw offsets have been set to the values that were randomly generated in the previous run as this would enable comparison with the current data.

I am starting it on Donatella and it should run for a couple of hours.

Apologies for the inconvenience.

 Quote: A python script to randomly vary the MC2 pitch and yaw offset and correspondingly record the value of MC transmission has been started on Donatella in the control room and should run for a couple of hours overnight. The script is named MC_TRANS_1.py and is located in my user directory at /users/jigyasa
13076   Tue Jun 20 17:44:12 2017 jigyasaUpdateComputer Scripts / ProgramsMC2 Pitch-Yaw offset

The script didn't run properly last night, due to an oversight of variable names! It's been started again and has been running for half an hour now.

Quote:

I am starting it on Donatella and it should run for a couple of hours.

Apologies for the inconvenience.

 Quote: A python script to randomly vary the MC2 pitch and yaw offset and correspondingly record the value of MC transmission has been started on Donatella in the control room and should run for a couple of hours overnight. The script is named MC_TRANS_1.py and is located in my user directory at /users/jigyasa

13084   Tue Jun 27 18:47:49 2017 jigyasaUpdateComputer Scripts / ProgramsMC2 Pitch-Yaw offset

The values generated from the script were analyzed and a 3D scatter plot in addition to a 2D map were plotted.

Yesterday, Rana pointed me to another method of collecting and analyzing the data. So I worked on the code today and have left a script (MC2rerun.py) running on Ottavia which should run overnight.

 Quote: The script didn't run properly last night, due to an oversight of variable names! It's been started again and has been running for half an hour now.

13087   Thu Jun 29 10:04:18 2017 jigyasaUpdateComputer Scripts / ProgramsMC2 Pitch-Yaw offset

The script is being executed again, now.

 Quote: I worked on the code today and have left a script (MC2rerun.py) running on Ottavia which should run overnight.

5905   Wed Nov 16 09:21:56 2011 SureshUpdateIOOMC2 Shifted in Pitch, corrected by adjusting the pitch bias

[Steve, Suresh]

Steve went over to the MC2 walkway and stepped over the barrier to pick up some stuff there.  MC2 stack shifted and MC2 pitch as off.  MC unlocked and could not relock till the MC2 pitch bias was readjusted

Without the WFS the MC to PSL alignment is poor, but it is largely due to a shift in the MC and not a shift in the PSL beam.  We know this 'coz the shift in the DC spot positions on WFS (when the MC is unlocked) is not significant nor is the shift on the C1:IOO-QPD.  When WFS loops are engaged the MC optics are turned to optimise the PSL to MC alignment, but the shift is large at the moment.

(Sorry Mirko your measurement could not be completed.  The MC unlocked in the middle)

Please Note:  If you need to access the blocked off area near MC2 stack, do not step over the barrier.  The disturbance is too great and the MC2 stack will shift.  Instead please move the barrier aside and walk as gently as possible near it, taking care not to touch the MC2 Chamber.

5910   Wed Nov 16 10:53:35 2011 SureshUpdateIOOMC2 Shifted in Pitch, corrected by adjusting the pitch bias

 Quote: [Steve, Suresh]     Steve went over to the MC2 walkway and stepped over the barrier to pick up some stuff there.  MC2 stack shifted and MC2 pitch as off.  MC unlocked and could not relock till the MC2 pitch bias was readjusted previous MC2PIT reading: 3.6235           current MC2PIT reading:  3.9565 Without the WFS the MC to PSL alignment is poor, but it is largely due to a shift in the MC and not a shift in the PSL beam.  We know this 'coz the shift in the DC spot positions on WFS (when the MC is unlocked) is not significant nor is the shift on the C1:IOO-QPD.  When WFS loops are engaged the MC optics are turned to optimise the PSL to MC alignment, but the shift is large at the moment. (Sorry Mirko your measurement could not be completed.  The MC unlocked in the middle) Please Note:  If you need to access the blocked off area near MC2 stack, do not step over the barrier.  The disturbance is too great and the MC2 stack will shift.  Instead please move the barrier aside and walk as gently as possible near it, taking care not to touch the MC2 Chamber.

Apparently the MC2 stack had not finished shifting.   The MC unlocked while Steve was working on the PSL table installing the mirror for IOO_QPD and then it could not relock.  So I moved the MC2 once again in Pitch.  The current status of the sliders is here

Yesterday I fixed the yellow buttons on the MC_ALIGN and MCLOCK screens.  They use the new updatesnap script  .  Could we also add a couple of lines to this script so that eveytime we save a snap shot the various values are written(appended) to a text file?  That way we do not need to depend solely on the conlog, which is quite slow.

5913   Wed Nov 16 17:03:19 2011 KojiUpdateIOOMC2 Shifted in Pitch, corrected by adjusting the pitch bias

MC was not locked for more than 5 hours because of the misalignment.

Noticed that MC2 WFS feedback filters had big outputs (particularly in Pitch).
They were reset to zero.

MC2 was aligned and recovered the lock. Once the WFS is engaged, the transmission returned to the uisual value.

9884   Wed Apr 30 21:16:42 2014 ranaUpdateLSCMC2 Strad

I found the YARM LSC feedback going to MC2 and the MC2 violin mode (at 644.69 Hz) rung up. The existing notch was just a second order Twin-T style notch (so not a good idea) and also not turned on, since it was in the FM4 spot of LSC-MC2 and the vio triggers are ganged between all mirrors and don't touch FM4.

I copied the PRM vio bandstop into FM2 of this bank, deleted the old notch, and tuned the bandstop frequencies a little to get the violin peak into one of the zeros of the elliptic bandstop. Attached are the Y-arm / MCF spectrum with the mode rung up as well as the new filter's TF compared with the old notch.

P.S. I installed http://en.wikipedia.org/wiki/Midnight_Commander on pianosa.

Attachment 2: MC_Y_vio.pdf
Attachment 3: MC2_vio.pdf
16066   Wed Apr 21 15:50:01 2021 Anchal, PacoUpdateSUSMC2 Suspension Optimization summary
MC2 Coil Balancing DC and AC Gains
POS PIT YAW COIL_GAIN (AC balancing)

UL

1.038 1 1 1.05
UR 1.009 1 -1 -1.05
LL 0.913 -1 1 -1.030
LR 0.915 -1 -1 0.995

MC2 Diagonalized input matrix
UL UR LR LL SIDE
POS 0.2464 0.2591 0.2676 0.2548 -0.1312
PIT 1.7342 0.7594 -2.494 -1.5192 -0.0905
YAW 1.2672 -2.0309 -0.9625 2.3356 -0.2926
SIDE 0.1243 -0.1512 -0.1691 0.1064 0.9962

MC2 Suspension Gains
Old gain New Gain
SUSPOS 150 150
SUSPIT 10 30
SUSYAW 10 30

16067   Wed Apr 21 18:49:29 2021 ranaUpdateSUSMC2 Suspension Optimization summary

the POS column should be all 1 for the AC balancing. Where did those non-1 numbers come from?

16071   Thu Apr 22 08:50:21 2021 AnchalUpdateSUSMC2 Suspension Optimization summary

Yes, during the AC balancing, POS column was set to all 1. This table shows the final values after all the steps. The first 3 columns are DC balancing results when output matrix was changed. While the last column is for AC balancing. During AC balancing, the output matrix was kept to ideal position as you suggested.

 Quote: the POS column should be all 1 for the AC balancing. Where did those non-1 numbers come from?

5635   Fri Oct 7 22:48:26 2011 SureshUpdateIOOMC2 Trans QPD spot size and incident power decreased

After centering the spot on the MC2, I started to adjust the spot position on MC_TRANS_QPD to center the beam on it.  I noticed that the spot size was about 3 to 4mm dia. because the 200mm lens was too close to the QPD.  I moved it back and decreased the spot size to about 1mm and the sensitivity to spot position increased.  However, Koji noted that the QPD sectors were near saturation, so I put in a ND=0.3 filter to reduce the incident power on the QPD.

At optimal alignment the current QPD_SUM is around 25k to 26k counts (factor of 2 down). Eventually the gain of the QPD ckts have to be reduced to prevent saturation, for the moment this is temporary fix.

The MC_TRANS_SUM trigger for MC autolocker is working fine no further change was required.

13705   Mon Mar 26 21:25:55 2018 ranaSummaryIOOMC2 Trans table has issues

Gautam, Rana

While at the MC2 table, we noticed that it has some optical problems:

1. There is an ND filter mounted to the beam reducing lens. ND filters are illegal, Steve. It causes too much scattering noise. We should instead have a beamsplitter and a dump.
2. One of those bad U100-AC28 mounts is in use. This is one of those ones with plastic clips that Osamu liked, but the plastic gets in the way of the beam. Needs to be removed.
3. Reflections from the QPD and the PDA255 are not dumped. This causes noise. Bad.
4. The QPD transimpedance should be reduced so that it can handle more light. I don't know what it has now, but its probably 10-100 kOhm.

We estimated that the power in the IMC is (1 W)*Finesse/pi = 500 W. The MC2 Transmission spec is < 10 ppm, so the power on the table is probably ~5 mW. Since the PDA255 has a transimpedance of 10 kOhm and a max output power of 10V, it can handle up to ~1 mW. Probably we can get the QPD to handle 4 mW.

Gautam, Steve  3-27

We measured MC2 transmitted power right at the uncoated window ~2.5 mW  The beam was just a little bigger than the meter.

Attachment 1: 20180326_201929.jpg
674   Tue Jul 15 12:23:22 2008 JenneUpdateGeneralMC2 Watchdog tripped
Alberto, Jenne

Mode Cleaner was unlocked. We checked, and found that MC2's watchdog was tripped. It didn't look like anything bad was going on, so we turned the optic back on, and tried to relock the MC. It looks like the Mode Cleaner is now locked, but the lock bit on the LockMC screen is still red. I don't know what's up.
2208   Mon Nov 9 11:11:19 2009 AlbertoConfigurationLSCMC2 Watchdogs tripped

The MC2 watchdogs tripped. I just restored them.

I also had to relock the MZ and then the Mode Cleaner.

2123   Tue Oct 20 02:14:29 2009 robUpdateIOOMC2 alignment bias changed

the mode cleaner was having trouble locking in a 00 mode, needing several tries.  I changed the MC2 coil biases, and it seems better for now.

16163   Wed May 26 11:45:57 2021 Anchal, PacoConfigurationIMCMC2 analog camera

[Anchal, Paco]

We went near the MC2 area and opened the lid to inspect the GigE and analog video monitors for MC2. Looked like whatever image is coming through the viewport is split into the GigE (for beam tracking) and the analog monitor. We hooked the monitor found on the floor nearby and tweaked the analog video camera around to get a feel for how the "ghost" image of the transmission moves around. It looks like in order to try and remove this "extra spots" we would need to tweak the beam tracking BS. We will consult the beam tracking authorities and return to this.

14774   Thu Jul 18 22:03:00 2019 KruthiUpdateCamerasMC2 and cameras

[Kruthi, Yehonathan, Gautam]

Today evening, Yehonathan and I aligned the MC2 cameras. As of now there are 2 GigEs in the MC2 enclosure. For the temporary GigE (which is the analog camera's place), we are using an ethernet cable connection from the Netgear switch in 1x6. The MC2 was misaligned and the autolocker wasn't able to lock the mode cleaner. So, Gautam disabled the autolocker and manually changed the settings; the autolocker was able to take over eventually.

4674   Tue May 10 00:44:52 2011 valeraUpdateIOOMC2 centering

Kiwamu, Koji, Valera

We centered the beam on MC2 in pitch by moving the MC1,2,3 in the following combination [-9,+3,-7]. This actuation vector mostly moves the spot on MC2 vertically. The attached plot shows the dither before and after the centering. We monitored the demodulated signals and saw the reduction of the MC2 pit response from -1.0 to -0.22 which corresponds to the beam spot position change from 9 to 2 mm. Thus all the spots on MC mirrors are within 2 mm of the center. We estimate based on the distance between the MC1-MC3 of 20 cm, the distance from the center between MC1 and MC3 to the end of the Faraday isolator of 80 cm, and the aperture of the FI of 12 mm, the maximum angle out of MC of 3/200 rad. Which implies the maximum differential spot motion of 3 mm not to be limited by the FI aperture.

Attachment 1: mc2centering.pdf
5782   Wed Nov 2 10:04:05 2011 steveUpdateSUSMC2 chamber anchoring tightened

 Quote: Thinks to do before the NEXT realignment: B,  tie 4 ancher bolts on table legs to the floor C,  tie 4 dog clamps between table and chamber D,  check the locked position of the 4 x 4 positioning screws E,  check bellow protecting 4 tubes are not shorting A,  here is the concrete  slab cut   It reminds me to check the  IFO vacuum envelope dog clamps on the chambers to floor  with torque wrench.

I locked the PMC and MC resonated in TM00 right on. Than I started checking anchoring screws with the torque wrench.

B,  3/8 foor bolts were tight, They were set  to 50 ft/lbs

C,  dog clamp bolts were tight at 80 ft/lbs

D,  horizontal position locks were lose. They were locked by finger.

E,  below covers were reset not to short

The MC is locking in TM01 mode now

13706   Mon Mar 26 21:40:26 2018 gautamSummaryIOOMC2 classical radiation pressure noise

[rana, gautam]

we measured the RIN of the MC2 transmission using the PDA255 I had put on the MC2 trans table sometime ago for ringdowns. Attached are (i) spectra for the RIN, (ii) spectra for the classical rad. pressure noise assuming 500W circulating power and (iii) a tarball of data and code used to generate these plots.

We took a full span measurement (to make sure there aren't any funky high-freq features) and a measurement from DC-800 Hz (where we are looking for excess noise). The DC level of light on the photodiode was 2.76V (measured using o'scope)

I'll add this to the noise budget later. But the measured RIN seems consistent with a 2013 measurement at 100Hz (though the 2013 measurement is using DTT and so doesn't have high frequency information).

Attachment 1: IMC_RIN.pdf
16055   Tue Apr 20 18:19:30 2021 AnchalUpdateSUSMC2 coil balanced at DC

Following up from morning's work, I balanced the coils at DC as well. Attachment 1 is screenshot of striptool in which blue and red traces show ASCYAW and ASCPIT outputs when C1:SUS-MC2_LSC_OFFSET was switched by 500 counts. We see very slight disturbance but no real DC offset shown on PIT and YAW due to position step. This data was taken while nominal F2A filter calculated to balance coils at DC was uploaded

I have uploaded the filters on filter banks 7-10 where FM7 is the nominal filter with Q close to 1 and 8-10 are filters with Q 3, 7 and 10 respectively. The transfer function of these filters can be seen in Attachment 2. Note, that the high frequency gain drops a lot when higher Q filters are used.

These filters are designed such that the total DC gain after the application of coil outputs gains for high frequency balancing (as done in morning 16054) balances the coils at DC.

Since I had access to the complete output matrix that balances the coils to less than 1% cross coupling at high frequencies from 16009, I also did a quick test of DC coil balancing with this kind of high frequency balancing. In this case, I uploaded another set of filters which were made at Q close to 1 and gain such that effective DC gain matrix becomes what I got by balancing in the above case. This set of filter also worked as good as the above filters. This completes the proof that we can also use complete matrix for high frequency coil balancing which can be calculated by a script in 20min and works with DC coil balancing as well. In my opinion, this method is more clear and much faster than toggling values in coil output gains where we have only 4 values to optimize 6 cross-coupling parameters. But don't worry, I'm not wasting time on this and will abandon this effort for now, to be taken up in future.

Next up:

• Tomorrow, we'll finish DC balancing for MC1 and MC3 with the method I practiced today. This should not take much time and should be completed before the meeting.
• I'll also, calculate and upload the F2A filters for MC1 and MC3.
• Next, we'll optimize gains in the suspension damping loops by doing step response test (with TRAMP = 0s). We'll look for decaying response (at MC_F, and WFS sensors) with a few oscillations for each step in POS, PIT, and YAW.

Edit Tue Apr 20 21:25:46 2021 :

Corrected the calculation of filters in case of Q different than $\large \sqrt{G_{DC}}$. There was a bug in the code which I overlooked. I'll correct the filter bank modules tomorrow.

Edit Wed Apr 21 11:06:42 2021 :

I have uploaded the corrected foton filters. Please see attachment 3 for the transfer functions calculated by foton. They match the filters we intended to upload. Only after uploading and closing the foton filter, I realized that the X=7 filter plot (bottom left in attachment 3) does not have dB units on y-axis. It is plotted in linear y-scale (this plot in foton is for phase by default to I guess I forgot to change the scaling when repurposing it for my plot).

Attachment 1: MC2_DC_Coil_Balanced_St.png
Attachment 2: IMC_F2A_Params_MC2.pdf
15224   Tue Feb 25 19:58:06 2020 HangUpdateIOOMC2 coil balancing

[Yehonathan, Hang]

We did some quick DC balancing of the MC2 coil drivers to reduce the l2a coupling. We updated the gains in the C1:SUS-MC2_UL/UR/LR/LLCOIL to be 1, -0.99, 0.937,-0.933, respectively. The previous values were 1, -1, 1, -1.

The procedures are the following:

Lock IMC.

Drive UL+LR and change the gain of LR to zero pitch.

Drive UR+LL and change the gain of LL to zero pitch.

Lastly, drive all 4 coils and change UR & LR together to zero yaw.

We used C1:SUS-MC2_LOCKIN1_OSC to create the excitations at 33 Hz w/ 30,000 cts. The angular error signals were derived from IMC WFSs.

While this time we did things by hand, in the future it should be automated as the procedure is sufficiently straightforward.

15265   Wed Mar 11 16:46:25 2020 HangUpdateIOOMC2 coil balancing

My old scheme was flawed as I used pitch as the readback. The pitch signal could not distinguish the cross-coupling due to coil imbalance and that due to the natural suspension L2P. A new scheme based on yaw alone has been developed and will be integrated into ifo_test. For now we revert the C1:SUS-MC2_UL/UR/LR/LLCOIL gains back to 1, -1, 1, -1.

 Quote: [Yehonathan, Hang] We did some quick DC balancing of the MC2 coil drivers to reduce the l2a coupling. We updated the gains in the C1:SUS-MC2_UL/UR/LR/LLCOIL to be 1, -0.99, 0.937,-0.933, respectively. The previous values were 1, -1, 1, -1. The procedures are the following: Lock IMC. Drive UL+LR and change the gain of LR to zero pitch. Drive UR+LL and change the gain of LL to zero pitch. Lastly, drive all 4 coils and change UR & LR together to zero yaw.  We used C1:SUS-MC2_LOCKIN1_OSC to create the excitations at 33 Hz w/ 30,000 cts. The angular error signals were derived from IMC WFSs. While this time we did things by hand, in the future it should be automated as the procedure is sufficiently straightforward.
15468   Fri Jul 10 15:26:28 2020 gautamUpdateLSCMC2 coils need DC balancing?

I was looking at some signals from last night, see Attachment #1.

• It looks like as the DC control signal to the MC2 suspension increases, the MC transmission decreases.
• I confirmed that the IMC REFL level doesn't correspondingly trend up, but didn't include it here for plot compactness, so I think the cavity length isn't being detuned.
• So the problem is suggestive of some L2A coupling, and the MC2 coil actuators need to be balanced better at DC?
• You can see from the IMC WFS control signals that the WFS servo is presumably trying to counteract this L2A action, but doesn't succeed, probably because the servo isn't tuned correctly.
• This is a problem that is distinct from the drifting TT alignment. So it complicates the alignment situation.
• Ideally, if the dither alignment servos could be made to work for the arm cavities when locked in the PRFPMI config, this wouldn't be so much of a problem, as the TTs would just adjust the beam pointing to match the cavity axes of the arms. But since I haven't managed to get that servo working yet...
• But why should MC2 need such a large DC control signal ever?
• In the PRFPMI lock, the CARM servo is supposed to match the laser frequency to the average length of the two arm cavities.
• The MC2 suspension is used as a frequency actuator in order to realize this matching.
• But, as you can see, the digital CARM control signal picks up a significant DC offset the deeper we go into the lock.
• Can't we offload this DC signal to the laser crystal temperature servo? Is there going to be some weird interaction with the existing slow loop? Or is the idea itself flawed?

Attachment #2 shows some ASC metrics. My conclusion here is that running the PRCL and MICH dither alignment servos (former demodulating REFLDC and latter demodulating ASDC to get an error signal) that running the dither alignment servo and hand tuning the arm ASC loop offsets improves the mode matching to the IFO, because:

1. The arm transmission increases.
2. POPDC increases.
3. ASDC decreases.

The REFLDC behavior needs a bit more interpretation I think, because if the IFO is overcoupled (as I claim it is), then better alignment would at some point actually result in REFLDC increasing.

All the DC signals recorded by the fast system come from the backplane P2 connector of the PD interface boards. According to the schematic, these signals have a voltage gain of 2. The LSC photodiodes themselves have a nominal DC gain of 50 ohms. So, the conversion from power to digital counts is: 0.8 A/W * 50 V/A * 2 * 3276.8 cts/V * whtGain. Inverting, I get 3.8 uW/ct for a whitening gain of 1. This is power measured at the photodiode - optical losses upstream of the photodiode will have to be accounted for separately.

Assuming a modulation depth of 0.2, the 55 MHz sideband power should be ~20 mW. The Schnupp asymmetry is supposed to give us O(1) transmission of this field to the AS port. Then, the SRM will attenuate the field by a factor of 10, so we expect ~2 mW at the AS port. Let's assume 80 % throughput of this field to the AP table, and then there is a 50/50 beamsplitter dividing the light between the AS55 and AS110 photodiodes. So, we expect there to be ~700 uW of power in the TEM00 mode 55 MHz sideband field. This corresponds to 1600 cts according to the above calibration (the ASDC whitening gain is set to 18 dB). The fact that much smaller numbers were seen for ASDC indicates that (i) the schnupp asymmetry is not so perfectly tuned and the actual transmission of the sideband field to the dark port is smaller, or (ii) one or more optical splitting fractions assumed above is wrong. If the former is true, we can still probably infer the contrast defect if we can somehow get an accurate measurement of the sideband transmission to the dark port.

Attachment 1: MC2_balancing.png
Attachment 2: ASDC.png
15470   Sat Jul 11 18:24:30 2020 KojiUpdateLSCMC2 coils need DC balancing?

> Can't we offload this DC signal to the laser crystal temperature servo?
No. PSL already follows the MC length. So this offset is coming from the difference between the MC length and the CARM length.
What you can do is to offload the MC length to the CARM DC if this helps.

15474   Mon Jul 13 11:36:08 2020 ranaUpdateLSCMC2 coils need DC balancing?
1. if IMC REFL is not increasing, I don't think its a mis-alignment. Usually, REFL is a more sensitive indicator of alignment than TRANS since its usually near zero. Maybe the MC2 TRANS PD is not centered or doesn't have enough lens action.
2. to reduce the DC load on MC2, you could have a slow releif drive the ETMs and DC and minimize LSC-MCL
996   Fri Sep 26 09:05:47 2008 steveUpdateSUSMC2 damping restored
The MC2 sus damping was restored.
1300   Fri Feb 13 08:38:03 2009 steveUpdateIOOMC2 damping restored
2207   Mon Nov 9 08:43:16 2009 steveUpdateSUSMC2 damping restored

MC2 damping restored,  MZ locked and the arms are flashing now.

7268   Fri Aug 24 09:21:45 2012 SteveUpdateIOOMC2 damping restored

 Quote: I turned on some filters and gain in the SUS-MC2_MCL filter bank tonight so as suppress the seismic noise influence on MC_F. This may help the MC stay in lock in the daytime. Koji updated the mcdown and mcup scripts to turn the MCL path on and off and to engage the Boost filters at the right time. The attached PNG shows the MCL screen with the filters all ON. In this state the crossover frequency is ~45 Hz. MC_F at low frequencies is reduced by more than 10x. I also think that this may help the X-Arm lock. The number of fringes per second should be 2-3x less.

Attachment 1: 10hrsMC2.png
1607   Tue May 19 15:57:07 2009 steveUpdateIOOMC2 damping restored after EQ

Earthquake mag 4.0 at Lennox, Ca     trips MC2 watchdogs       http://quake.usgs.gov/recenteqs/Quakes/ci10411545.html

See 40m accelerometers as they see it.

Attachment 1: acc.jpg
2033   Thu Oct 1 10:23:00 2009 steveUpdatePSLMC2 damping restored again

The EQ did not change the input beam pointing. All back to normal, except MC2 wachdogs tripped again.

Attachment 1: mc2again.jpg
2034   Thu Oct 1 11:39:47 2009 JenneUpdateSUSMC2 damping restored again

 Quote: The EQ did not change the input beam pointing. All back to normal, except MC2 wachdogs tripped again.

Round 3 for the day of MC2 watchdogs tripping.

13174   Wed Aug 9 11:33:49 2017 gautamUpdateElectronicsMC2 de-whitening

Summary:

The analog de-whitening filters for MC2 are different from those on the other optics (i.e. ITMs and ETMs). They have one complex pole pair @7Hz, Q~sqrt(2), one complex zero pair @50Hz, Q~sqrt(2), one real pole at 2.5kHz, and one real zero @250Hz (with a DC gain of 10dB).

Details:

I took the opportunity last night to measure all 4 de-whitening channel TFs. Measurements and overlaid LISO fits are seen in Attachment #1.

The motivation behind this investigation was that last week, I was unable to lock the IMC to one of the arms. In the past, this has been done simply by routing the control signal of the appropriate arm filter bank (e.g. C1:LSC-YARM_OUT) to MC2 instead of ETMY via the LSC output matrix (if the matrix element to ETMY is 1, the matrix element to MC2 is -1).

Looking at the coil output filter banks on the MC2 suspension MEDM screen (see Attachment #2), the positions of filters in the filter banks is different from that on the other optics. In general, the BIO outputs of the DAC are wired such that disengaging FM9 on the MEDM screen engages the analog de-whitening path. FM10 then has the inverse of the de-whitening filter, such that the overall TF from DAC to optic is unity. But on MC2, these filters occupy FM7 and FM8, and FM9 was originally a 28Hz Elliptic Low-pass filter.

So presumably, I was unable to lock the IMC to an arm because for either configuration of FM9 (ON or OFF), the signal to the optic was being aggressively low-passed. To test this hypothesis, I simply copied the 28Hz elliptic to FM6, put a gain of 1 on FM9, left it engaged (so that the analog path TF is just flat with gain x3), and tried locking the IMC to the arm again - I was successful. See Attachment #3 for comparison of the control signal spectra of the X-arm control signal, with the IMC locked to the Y-arm cavity.

In this test, I also confirmed that toggling FM9 in the coil output filter banks actually switches the analog path on the de-whitening boards.

Since I now have the measurements for individual channels, I am going to re-configure the filter arrangement on MC2 to mirror that on the other optics.

Unrelated to this work: the de-whitening boards used for MC1 and MC3 are D000316, as opposed to D000183 used for all other SOS optics. From the D000316 schematic, it looks like the signals from the AI board are routed to this board via the backplane. I will try squishing this backplane connector in the hope it helps with the glitching MC1 suspension.

GV Aug 13 11:45pm - I've made a DCC page for the MC2 dewhitening board. For now, it has the data from this measurement, but if/when we modify the filter shape, we can keep track of it on this page (for MC2 - for the other suspensions, there are other pages).

Attachment 1: MC2deWhites.pdf
Attachment 2: MC2Coils.png
Attachment 3: MC2stab.pdf
4146   Wed Jan 12 22:33:24 2011 kiwamuUpdateCDSMC2 dewhitening are healthy except for UR

I briefly checked the MC2 analog dewhitening filters.

It turned out that the switching of the dewhitening filters from epics worked correctly except for the UR path.

I couldn't get a healthy transfer function for the UR path probably because the UR monitor at the front panel on either the AI filter or the dewhitening filter maybe broken.

Need a check again.

 Quote:  #4144 Tomorrow I'll take a look at the binary outputs and see why the analog filters aren't actually changing.

7276   Sun Aug 26 11:53:09 2012 JenneUpdateIOOMC2 getting kicked up regularly

We need to re-look at this new MC autolocker stuff, and the new MCL filters.

MC2 is getting kicked up (sometimes the watchdog trips, sometimes it just comes close) pretty regularly.  I'm not sure yet what is causing this, but we need to deal with it since it's pretty obnoxious.

5862   Thu Nov 10 12:28:31 2011 JenneUpdateSUSMC2 is being a little wild...WFS to blame

Mirko and  Den are measuring MC_F, which they will report about later, but I noticed that MC2 is totally crazy right now.  It shouldn't matter that they are doing things (like unplugging the feedback to the PSL's PZT), because we actuate on the laser, not on the MC.  I disabled the MC autolocker before they started working.

Anyhow, somehow MC2 got kicked up (whatever, that happens), but it won't re-damp.  I think it's the WFS.  The yaw output from the WFS is truely crazy.

I have disabled the WFS output / ASC input on the MC SUS screens, and MC2 was then able to damp.  My disabling only the MC2 WFS input at first kicked up MC1 and 3, so I disabled all of the WFS stuff, and all 3 MC mirrors are again happy.

SURESH: FIX ME!  (signed, The WFS)

5877   Fri Nov 11 21:09:30 2011 SureshUpdateSUSMC2 is being a little wild...WFS to blame

 Quote: Mirko and  Den are measuring MC_F, which they will report about later, but I noticed that MC2 is totally crazy right now.  It shouldn't matter that they are doing things (like unplugging the feedback to the PSL's PZT), because we actuate on the laser, not on the MC.  I disabled the MC autolocker before they started working.  Anyhow, somehow MC2 got kicked up (whatever, that happens), but it won't re-damp.  I think it's the WFS.  The yaw output from the WFS is truely crazy.  I have disabled the WFS output / ASC input on the MC SUS screens, and MC2 was then able to damp.  My disabling only the MC2 WFS input at first kicked up MC1 and 3, so I disabled all of the WFS stuff, and all 3 MC mirrors are again happy.  SURESH: FIX ME!  (signed, The WFS)

The Problem

Turning off the WFS servo loops usually should be done using the 'mcwfsoff' script.  The script takes care of switching off the integrators and Clears the History.

'mcdown' and 'mcup' scripts run the 'mcwfsoff' and 'mcwfson' scripts so when the MC unlocks the WFS servos are shutdown and restarted properly.  However if the MC autolocker script is suspended by pressing the Enable/Disable switch in the LOCKMC screen and then the MC unlocks, it results in the WFS servo integrators accumulating large values.  If these values are passed through the ASC filter banks the optic will get a pretty huge kick.

The Solution

I have added some indicators which will let us know if the WFS Servo Filter outputs are larger than +/-1000.  When engaging the WFS loops the user has to take care to Clear History in the servo filter banks if these indicators are steadily Red.   before engaging the WFS Servo loops ensure that the servo filter outputs are zeroes.

Koji and I discussed whether it would be useful to run the 'mcwfsoff' script when the Disable button is pressed in the autolocker.  His recommendation is that we should keep  the autolocker script simple and that user has to be cautious when switching on the WFS servos and when directing the ASC outputs to the suspensions.

1002   Fri Sep 26 19:18:39 2008 JenneUpdateIOOMC2 is having a bad day
As Steve mentioned earlier in today's elog, MC2 keeps ringing up for no clear reason. It is definitely only MC2 that is ringing up, since it's sensors will read several hundreds of counts, while all the other optics are at regular 2 counts and below on the Watchdog screen.

Preliminary investigation results: Around the time of these "kick up" events, the Ranger seismometer does not see any motion, nor does the set of accelerometers under the MC1 chamber. The set of accelerometers under the MC2 chamber do see activity that is at the same time as these events. These events are not caused just by someone walking around, since Rana went inside and clunked around near MC2 while I watched the sensor levels. MC2's watchdog did not trip.

For further investigation: Why is it that only the MC2 accelerometers are seeing the motion? Similarly, why is MC2 the only optic being kicked? Has anyone done anything lately to the MC2 stack?
2457   Mon Dec 28 12:35:57 2009 JenneUpdateSUSMC2 is having a bad day

MC2 is having a bad day, and I'm not yet sure why.  It's to do with the damping though.  When the damping is off, after a little while it will settle to ~30mV or so on the Watchdog screen.  When I enable all of the outputs and then turn on the damping, the optic gets kicked up.  It's like there's a minus sign error somewhere, maybe in a bad burtrestore?  This has been going on since I did my morning bootfest.

It's started to sit down and play nicely now.  Is someone doing magic remotely that is fixing things that I hadn't figured out yet?

2458   Mon Dec 28 12:45:55 2009 KojiUpdateSUSMC2 is having a bad day

The MCL path of MC2 was in a strange state as the filters were activated as if it is in lock even though we had no lock. So I manually ran "mcdown". This reset the filters of the MCL path.

 Quote: MC2 is having a bad day, and I'm not yet sure why.  It's to do with the damping though.  When the damping is off, after a little while it will settle to ~30mV or so on the Watchdog screen.  When I enable all of the outputs and then turn on the damping, the optic gets kicked up.  It's like there's a minus sign error somewhere, maybe in a bad burtrestore?  This has been going on since I did my morning bootfest. It's started to sit down and play nicely now.  Is someone doing magic remotely that is fixing things that I hadn't figured out yet?

14788   Sun Jul 21 02:07:04 2019 KruthiUpdateLoss MeasurementMC2 loss map

I'm running the MC2 loss map scripts on pianosa now. The camera server is throwing an error and is not grabbing snapshots :(

Update: I finished taking the readings for MC2 loss map. I couldn't get the snapshots with the script, so I manually took some 4-5 pictures.

14789   Sun Jul 21 12:54:18 2019 gautamUpdateLoss MeasurementMC2 loss map

Can you please be more specific about what the error is? Is this the usual instability with the camera server code? Or was it something new?

 Quote: The camera server is throwing an error and is not grabbing snapshots :(
14791   Sun Jul 21 17:17:03 2019 KruthiUpdateLoss MeasurementMC2 loss map

The camera server keeps throwing the error: failed to grab frames. Milind suggested that it might a problem with the ethernet cable, so I even unplugged it and connected it again; it still had the same issue. One more thing I noticed was, it does take snapshots sometimes with the terminal command caput C1:CAM-ETMX_SNAP 1, but produces a segmentation fault when ezca.Ezca().write(C1:CAM-ETMX_SNAP, 1) ezca.Ezca().write(CAM-ETMX_SNAP, 1) is used via ipython. When the terminal command also fails to take snapshots, I noticed that the SNAP button on the GigE medm screen remains on and doesn't switch back to OFF like it is supposed to.

Quote:

Can you please be more specific about what the error is? Is this the usual instability with the camera server code? Or was it something new?

 Quote: The camera server is throwing an error and is not grabbing snapshots :(
14792   Sun Jul 21 19:27:25 2019 MilindUpdateLoss MeasurementMC2 loss map

I think ezca.Ezca().write() takes the string "CAM-ETMX_SNAP" as an argument and not C1:CAM-ETMX_SNAP. See this, line 47. Are you sure this is not the problem?

 Quote: The camera server keeps throwing the error: failed to grab frames. Milind suggested that it might a problem with the ethernet cable, so I even unplugged it and connected it again; it still had the same issue. One more thing I noticed was, it does take snapshots sometimes with the terminal command caput C1:CAM-ETMX_SNAP 1, but produces a segmentation fault when ezca.Ezca().write(C1:CAM-ETMX_SNAP, 1) is used via ipython. When the terminal command also fails to take snapshots, I noticed that the SNAP button on the GigE medm screen remains on and doesn't switch back to OFF like it is supposed to.
14796   Mon Jul 22 12:57:35 2019 KruthiUpdateLoss MeasurementMC2 loss map

In my script I have used "CAM-ETMX_SNAP" only; while entering it in the elog I made a mistake, my bad!

Quote:

I think ezca.Ezca().write() takes the string "CAM-ETMX_SNAP" as an argument and not C1:CAM-ETMX_SNAP. See this, line 47. Are you sure this is not the problem?

 Quote: The camera server keeps throwing the error: failed to grab frames. Milind suggested that it might a problem with the ethernet cable, so I even unplugged it and connected it again; it still had the same issue. One more thing I noticed was, it does take snapshots sometimes with the terminal command caput C1:CAM-ETMX_SNAP 1, but produces a segmentation fault when ezca.Ezca().write(C1:CAM-ETMX_SNAP, 1) is used via ipython. When the terminal command also fails to take snapshots, I noticed that the SNAP button on the GigE medm screen remains on and doesn't switch back to OFF like it is supposed to.
9759   Fri Mar 28 20:23:02 2014 ranaSummaryIOOMC2 moved

I aligned MC2 suspension by 0.01 in pit and yaw to align the MC better to the PSL beam. Then I turned the WFS back on. The beams are not centered on the WFS heads.

Nic and Gabriele ought to send their SURF some example code (in April) for how to start redesigning the WFS telescopes so that we can order some optics in early June.

I've also turned on the MC2 TRANS path to gather some data over the weekend on how well or bad it works. Please turn it off on Monday.

9764   Mon Mar 31 11:34:00 2014 manasaSummaryIOOMC2 moved

 Quote: I've also turned on the MC2 TRANS path to gather some data over the weekend on how well or bad it works. Please turn it off on Monday.

MC2_TRANS path in WFS servo turned OFF.

ELOG V3.1.3-

format: plain


text:
# Is the push forward of a sigma-finite measure equivalent to a sigma-finite measure?

Let $X$ and $Y$ be locally compact, second countable spaces, and let $\varphi \colon X \to Y$ be a Borel map. Let $\mu$ be a sigma-finite measure on $X$. In general, the push-forward $\varphi_*\mu$ is not sigma-finite. Does there, however, exist a sigma-finite measure $\nu$ that is mutually absolutely continuous with $\varphi_*\mu$?

• You mean "mutually absolutely continuous with $\varphi_* \mu$", I suppose. Jan 10 '15 at 20:00
• Yes, because $\mu$ is equivalent to a finite measure. Jan 10 '15 at 20:03

Note that $\mu$ is equivalent to a finite measure $\nu$ on $X$. For example, if we can write $X = \bigcup_n X_n$ where $X_n$ is Borel with $0 < \mu(X_n) < \infty$, set $f = \sum_{n=1}^\infty \frac{1}{2^n \mu(X_n)} 1_{X_n}$ and take $d\nu = f\,d\mu$. Then $\nu$ is a probability measure and is mutually absolutely continuous with $\mu$ (since $f>0$ everywhere). That is, $\mu(A) = 0$ iff $\nu(A) = 0$ for all Borel $A \subset X$. Now the pushforward $\varphi_* \nu$ is a finite measure on $Y$, and is mutually absolutely continuous with $\varphi_* \mu$, since for any Borel $B \subset Y$ we have $(\varphi_* \mu)(B) = \mu(\varphi^{-1}(B)) = 0$ iff $(\varphi_* \nu)(B) = \nu(\varphi^{-1}(B)) = 0$.


text:
# Intersecting Chords (Finding Angle Measure)

If two chords intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle .

In the figure,

$m\angle 1=\frac{1}{2}\left(m\stackrel{⌢}{QR}+m\stackrel{⌢}{PS}\right)$ .


text: This is a project that contains the files and scripts to reproduce the all the analyses and figures described in [this manuscript][1], which is a primer on creating synthetic datasets using the synthpop R package. To run this analysis in the browser using RStudio server, click on [this link][2]. Note that this analysis does not contain the scripts to generate Supplementary Figures 1-3. These were not included due to resource constraints of the RStudio server instance. The script to generate Supplementary Figures 1-3 are available on this on this Open Science Framework page. [1]: https://psyarxiv.com/dmfb3/ [2]: https://mybinder.org/v2/gh/dsquintana/synthpop-primer/master?urlpath=rstudio


text:
# Alternatives to percentage for measuring relative difference?

If the value of something changes from $\;a\;$ to $\;b\;$, their relative difference can be expressed as a percentage: $$\newcommand{\upto}{\mathop\nearrow} (D0) \;\;\; a \upto b \;=\; (b-a)/a \color{gray}{\times 100\%}$$

(In this question $\;a\;$, $\;b\;$, and $\;c\;$ are positive (non-zero) real numbers throughout, and $\;a \upto b\;$ is a real number.)

However, percentages are not anti-symmetrical, i.e., the above definition does not satisfy $$(P0) \;\;\; a \upto b \;=\; -(b \upto a)$$ This could be fixed by instead defining $$(D1) \;\;\; a \upto b \;=\; (b-a)/(b+a)$$ (By the way, does this difference measure have a standard name?)

However, neither of the above definitions add up: they don't satisfy the more general property $$(P1) \;\;\; a \upto b \;+\; b \upto c \;=\; a \upto c$$ After some attempts I discovered that $$(D2) \;\;\; a \upto b \;=\; \log_q (b/a)$$ (for an arbitrary base $\;q > 1\;$) does satisfy $(P1)$. And it also satisfies the following other properties of percentage $(D0)$ and of $(D1)$ which seem desirable for any relative difference: \begin{align} (P2) \;\; & a \upto a \;=\; 0 \\ (P3) \;\; & \textit{sgn}(a \upto b) \;=\; \textit{sgn}(b - a) \\ (P4) \;\; & (s \times a) \upto (s \times b) \;=\; a \upto b \text{ for all $\;s > 0\;$} \\ (P5) \;\; & a \upto b \to M \text{ for } a \to 0 \text{ for some $\;M\;$ (which may be $\;\infty\;$)} \\ (P6) \;\; & a \upto b \to N \text{ for } b \to \infty \text{ for some $\;N\;$ (which may be $\;\infty\;$)} \\ \end{align} (I mention $(P2)$ only for completeness; it follows of course from $(P0)$.)

My question: Apart from $(D2)$, what other definitions (if any) satisfy all of the above properties?

• In (p4), do you mean $\forall s>0$, or $\exists s>0$? – TZakrevskiy Sep 19 '13 at 16:46
• @tzakrevskiy Good point, I mean $\forall$. I have corrected this.Thanks! – Marnix Klooster Sep 19 '13 at 17:21

The general (real-valued) solution of axioms P1 and P2 is to take a function $F$ on the set $V$ of values that $a,b,\cdots$ could possibly assume, and define $a \upto b = F(b) - F(a)$.

If $F$ is injective, the converse of P2 is satisfied: $a \upto b = 0$ only for $a=b$.

If $F$ is increasing, P3 is satisfied.

If $F$ is continuous, P5 and P6 are satisfied. (Assume here that the set of values $V$ is an interval, or put the order topology on the set.)

This shows that there is a large family of solutions of the axioms, one for every increasing continuous function on $V$, if you do not assume homogeneity, P4. Homogeneity is a strong requirement that cuts down the space of continuous solutions to the logarithmic ones stated in the question.

Homogeneity of $a \upto b$ is the functional equation $F(sb) - F(sa) = F(b) - F(a)$. Solutions $F$ and $F+c$ are equivalent for constant $c$, and we can assume $F(1)=0$ by adjusting the constant. Taking $a=1$ this is $F(sb)=F(s)+F(b)$ and all continuous solutions (on intervals) are well-known to be multiples of the logarithm.

Assuming that you meant to work with open intervals of values, and $a \upto x$ to be an increasing continuous function of $x$, the not necessarily homogeneous solutions correspond to one parameter groups of homeomorphisms of the interval.

I'll rewrite your conditions in a slightly different way: we look for a function $$f:\Bbb R_+^2\to \Bbb R,$$ such that we have

1) "antisymetry" $$f(a,b)=-f(b,a),$$ 2) "reflexivity" $$f(a,a)=0,$$ 3) homogenity of first degree $$f(sa,sb)=f(a,b)$$ 4) "transitivity" $$f(a,b)+f(b,c)=f(a,c)$$ 5) "monotonicity" if $b>a$, then $$f(a,b)>0.$$

From (3) it's apparent that it's sufficient to define our function only on $\{(a,b)\in R_+^2: a^2+b^2=1\}$; in other words, our function depends only on the ratio $\frac ba$.

Thus, we can say that, in fact, our function is given by $$f(a,b) = g(\ln (b/a)),$$ where $$g:\Bbb R\cup \{\pm\infty\}\to \Bbb R.$$ The conditions rewrite

1) $g(x)=-g(-x)$.

2) $g(0) =0$.

4) $g(\ln (b/a))+g(\ln (c/b))=g(\ln(c/a))$
5) $g(t)>0$ for $t>0$.
We need to use (4), other conditions give nothing significant. It's equivalent to $g(x)+g(y) = g(x+y)$. This is a Cauchy's functional equation, there're some questions on Math.SE on this topic, for example this.
If we suppose that $g$ is continuous at some point, then it's continuous everywhere and is linear, i.e. giving essentialy the same solutions as you've already proposed. Other solutions are not explicit and require axiom of choice. However, wikisays that the graph of such functions is dense in $\Bbb R^2$, but in our case we know that $g(t)>0$ for $t>0$, so these "implicit" functions are ruled out.
We can finally conclude that the only possible case is $g(t) = kt$ for $k>0$, or $f(a,b)=k\ln(b/a)$.


text:
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# A Consumer Advice Web Site Tested a Fuel AdditiveA) Compute the Test Statistic B) Can You Conclude That

Question 39
Multiple Choice
Question 39
Multiple Choice

## A consumer advice web site tested a fuel additive. The distance that 12 cars could travel on five gallons of gasoline was recorded without and then with the additive. The results were as follow.$\begin{array}{l|cccccccccccc}\text { Car } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\hline \text { No Additive } & 176 & 145 & 124 & 179 & 139 & 186 & 180 & 176 & 90 & 159 & 92 & 167 \\\text { With additive } & 186 & 162 & 130 & 179 & 149 & 191 & 193 & 184 & 105 & 159 & 94 & 173\end{array}$a) Compute the test statisticb) Can you conclude that the median distance traveled with the additive difference from the median distance traveled with the additive? Use the $\alpha = 0.05$ level of significance.

A) a) $\mathrm { S } = 15$
b) We can not conclude that the median distances are different.
B) a) $S = 12$
b) We can conclude that the median distances are different.
C) a) $\mathrm { S } = 12$
b) We can not conclude that the median distances are different.
D) a) $\mathrm { S } = 15$
b) We can conclude that the median distances are different.

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text:
# What degenerates mode in a fiber have in common? Do they have the same eigenvalues?

According to the definition degenerates modes in a optical fiber "have the same propagation constants". These modes can be combined to from the linearly polarized modes.

This "constant" should be the $\beta$ constant that appear in the Bessel equation for fields (see e.g. "Optical Fiber Communications - Principles and Practice" - John M. Senior). The eigenvalues that are associated to a mode in a optical fiber in the core and in the cladding are $$\sqrt{n_{core}k^2-\beta^2}\,\,\,,\,\,\,\,\, \sqrt{n_{cladding}k^2-\beta^2}$$ Where $k=\frac{2\pi \nu}{ c}$.

So if $\beta$ is the same for these "degenerates modes" does it mean that these modes also have the same eigenvalues (since they have the same frequency $\nu$)?

This doesn't seem reasonable, since one eigenvalue should correspond to a particular mode and not to two or more modes that can be combined.

So what does the degenerates modes have in common (which constant or parameter)?

• Who says that modes can't share eigenvalues? – Emilio Pisanty Sep 4 '17 at 0:20
• For instance the idealized (amagnetic) hydrogen atom has degenerate modes for all non-s orbitals. It is common for absolute introductory treatments of the eigen-problem to ignore degenerate modes until the basic terminology and phenomenology is developed, but it's an important topic. – dmckee Sep 4 '17 at 0:27
• moreover if the two modes did not have the same eigenvalue it would not be terribly meaningful to combine them to form linearly polarized modes... – ZeroTheHero Sep 4 '17 at 0:48
• @ZeroTheHero In practice they generally don't - the degeneracy is broken by manufacturing imperfevctions. That's why fibers mess with polarization state in the most maddening, environment-dependent ways. – WetSavannaAnimal Sep 4 '17 at 2:13

Degenerate modes share eigenvalues of the linear operator $\mathscr{L} = \nabla^2 + k^2 n^2$, which is the operator whose eigenvalue problem we solve for weak guidance, scalar waveguide theory. Any linear superposition of such modes is also a mode of the waveguide. We change this operator to the exact one when we do vector waveguide theory, but the idea is the same of course.


text:
Matching interval brackets

I'm using vim to edit LaTeX code, and I'm often writing intervals like (a,b] but this messes with other vim features. For instance,

• Jumping with %: if I have something like \phi((a,b]), the final ) will match the second ( and some closing parenthesis later in the code will match the first.
• There's a feature (I'm not sure if it's specific to vimtex or just a feature of vim) where ]] closes any open delimiters, and this makes that behave wrongly - it'll try to close the ) from many lines ago.
• The tsd/csd/dsd features of vimtex which toggle/change/delete surrounding delimiters don't work (I think csd and dsd are extensions of vim-surround but I'm not sure). In particular, either they select the wrong delimiters or do nothing at all.

Currently I'm putting )s in comments to mitigate this. This solves the problem of parentheses later in the file matching wrongly, but matching in the same line doesn't work (and things like tsd from vimtex don't work properly).

Is there any better solution, or a standard solution for this kind of problem? Ideally this would deal with [a,b) also.

• This is a very difficult problem, actually. It is very hard to know how to match parantheses if the start and end paranthesis is of different type. I remember we had a discussion about this at a vimtex issue once. You could try to search the issue threads and see if you find anything, or you could create a new issue. May 23 '18 at 6:29
• @KarlYngveLervåg I think this is the relevant issue? May 23 '18 at 19:31
• Yes, that's relevant. although the topic is indentation. The point is still, as mentioned above, that it is inherently very difficult to get this right for unbalanced parantheses. May 24 '18 at 6:01
• A workaround might be to use the \DeclarePairedDelimiter command from the mathtools package to then just write like \leftopenint{a, b}. I learned about this in this question
– Jan
May 28 '19 at 7:52
• @Jan I quite like this solution! If you'd like to write it as an answer I'm happy to accept it. May 28 '19 at 17:02

1 Answer

A workaround might be to use the \DeclarePairedDelimiter command from the mathtools package to then just write like \leftopenint{a, b}. See the accepted answer to this question.

\documentclass{minimal}
\usepackage{mathtools}
\DeclarePairedDelimiter\leftopenint{(}{]}
\DeclarePairedDelimiter\rghtopenint{[}{)}

\begin{document}
\begin{equation*}
\leftopenint{-\infty, \frac{1}{N} } \cup \rghtopenint*{\frac{1}{N}, \infty}
\end{equation*}
\end{document}

which gives me what is posted below. Note that the * makes the delimiters scale.

Just noted the asymmetric appearance of the brackets in the header also breaks syntax highlighting and the commands like closing of matching environments. I could fix this with outsourcing these definitions to a separate file and then use \include...


text:
## Appendix 8: The Articulation of Theory: Weak Interactions

Radioactivity, the spontaneous decay of a substance, produces alpha particles (positively charged helium nuclei), or beta particles (electrons), or gamma rays (high energy electromagnetic radiation). It was discovered in 1896 by Henri Becquerel. Experimental work on the energy of the electrons emitted in $$\beta$$ decay began in the early twentieth century, and the observed continuous energy spectrum posed a problem. If $$\beta$$ decay were a two-body decay (for example, neutron $$\rightarrow$$ proton $$+$$ electron) then applying the laws of conservation of energy and of conservation of momentum requires that the energy of the electron have a unique value, not a continuous spectrum.[1] Thus, the observed continuous energy spectrum cast doubt on both of these conservation laws. Physicists speculated that perhaps the electrons lost energy in escaping the substance, with different electrons losing different amounts of energy, thus accounting for the energy spectrum. Careful experiments showed that this was not the case so the problem remained. In the early 1930s Pauli suggested that a low-mass neutral particle, named by Fermi as the neutrino, was also emitted in β decay.[2] This solved the problem of the continuous energy spectrum because in a three-body decay ( neutron proton $$+$$ electron$$+$$ neutrino) the energy of the electron was no longer required to be unique. The electron could have a continuous energy spectrum and the conservation laws were saved.[3]

In 1934 Fermi proposed a new theory of $$\beta$$ decay that incorporated this new particle (Fermi 1934). He added a perturbation energy due to the decay interaction to the Hamiltonian describing the nuclear system. Pauli (1933) had previously shown that the perturbation could have only five different forms if the Hamiltonian is to be relativistically invariant. These are $$S$$, the scalar interaction; $$P$$, pseudoscalar; $$V$$, vector; $$A$$, axial vector; and $$T$$, tensor. Fermi knew this but chose, in analogy with electromagnetic theory, to use only the vector interaction. His theory initially received support from the work of Sargent (1932; 1933) and others. There remained, however, the question of whether or not the other forms of the interaction also entered into the Hamiltonian.[4] In this episode we shall see how experiment helped to determine the mathematical form of the weak interaction.

Gamow and Teller (1936) soon proposed a modification of Fermi’s vector theory. Fermi’s theory had originally required a selection rule, the change in $$J = 0$$, where $$J$$ is the angular momentum of the nucleus, and did not include the effects of nuclear spin. Gamow and Teller included nuclear spin and obtained selection rules, change in $$J = 0 \pm 1$$ for allowed transitions, with no $$0-0$$ transitions allowed. The Gamow-Teller modification required either a tensor or an axial vector form of the interaction. Their theory helped to solve some of the difficulties that arose in assigning nuclear spins using only the Fermi selection rule. At then end of the 1930s there was support for Fermi’s theory with some preference for the Gamow-Teller selections rules and the tensor interaction.

The work of Fierz (1937) helped to restrict the allowable forms of the interaction. He showed that if both $$S$$ and $$V$$ interactions were present in the allowed -decay interaction, or both $$A$$ and $$T$$, then there would be an interference term of the form $$1 + a/W$$ in the allowed beta-decay spectrum, where W is the electron energy. This term vanished if the admixtures were not present. The failure to observe these interference terms showed that the decay interaction did not contain both $$S$$ and $$V$$, or both $$A$$ and $$T$$.

The presence of either the $$T$$ or $$A$$ form of the interaction in at least part of the beta-decay interaction was shown by Mayer, Moszkowski, and Nordheim (1951). They found twenty five decays for which the change in $$J$$ was $$0\pm 1$$, with no parity change. These decays could only occur if the $$A$$ or $$T$$ forms were present. Their conclusion depended on the correct assignment of nuclear spins which, although reliable, still retained some uncertainty. Further evidence, which did not depend on knowledge of the nuclear spins, came from an examination of the spectra of unique forbidden transitions.[5] These were n-times forbidden transitions in which the change in nuclear spin was $$n + 1$$. These transitions require the presence of either $$A$$ or $$T$$. In addition, only a single form of the interaction makes any appreciable contribution to the decay. This allows the prediction of the shape of the spectrum for such transitions. Konopinski and Uhlenbeck (1941) showed that for an $$n$$-times forbidden transition the spectrum would be that of an allowed transition multiplied by an energy dependent term $$a_n(W)$$. For a first-forbidden transition $$a_1 = C[(W^2 - m^2 c^4) + (W_o - W)^2]$$. The spectrum for 91Y measured by Langer and Price (1949) (Figure 16) shows the clear presence of either the A or T forms of the interaction. The spectrum requires the energy-dependent correction.

Figure 16. The unique, once-forbidden spectrum of $$\ce{^{91}Y}$$. The best theoretical fit is that which gives a straight line. The Fermi theory alone, $$a_1=1$$, does not give a straight line. The correction factor $$a_1 = C[(W^2 - m_o c^2) + (W_o - W)^2]$$, does give a linear plot. From Konopinski and Langer (1953).

Evidence in favor of the presence of either the $$S$$ or $$V$$ forms of the interaction was provided by Sherr, Muether, and White (1949)and by Sherr and Gerhart (1952). They observed the decay of $$\ce{^{14}O}$$ to an excited state of $$\ce{^{14}N}$$, $$\ce{^{14}N^{\ast}}$$. They argued that both $$\ce{^{14}O}$$ and $$\ce{^{14}N^{\ast}}$$ had spin 0. This required the presence of either $$S$$ or $$V$$ because the decay was forbidden by $$A$$ and $$T$$. (Recall that the Gamow-Teller selection rules specified no 0 to 0 transitions).

Further progress in isolating the particular forms of the interaction was made by examining the spectra of once-forbidden transitions. Here too, interference effects, similar to those predicted by Fierz, were also expected. A. Smith (1951)and Pursey (1951) found that the spectrum for these transitions would contain energy dependent terms of the form $$G_V G_T / W$$, $$G_A G_P / W$$, and $$G_S G_A / W$$, where the $$G$$s are the coupling constants for the various interactions, and $$W$$ is the electron energy. The linear spectrum found for $$\ce{^{147}Pm}$$ demonstrated the absence of these terms (Langer, Motz and Price 1950).

Let us summarize the situation. There were five allowable forms of the decay interaction; $$S$$, $$T$$, $$A$$, $$V$$, $$P$$. The failure to observe Fierz interference showed that the interaction could not contain both $$S$$ and $$V$$ or both $$A$$ and $$T$$. Experiments showing the presence of Gamow-Teller selections rules and on unique forbidden transitions had shown that either $$A$$ or $$T$$ must be present. The decay of $$\ce{^{14}O}$$ to $$\ce{^{14}N^{\ast}}$$ had demonstrated that either $$S$$ or $$V$$ must also be present. This restricted the forms of the interaction to $$STP$$, $$SAP$$, $$VTP$$, or $$VAP$$ or doublets taken from these combinations. The absence of interference terms in the once-forbidden spectra eliminated the $$VT$$, $$SA$$, and $$AP$$ combinations. $$VP$$ was eliminated because it did not allow Gamow-Teller transitions. This left only the $$STP$$ triplet or the $$VA$$ doublet as the possible interactions.

The spectrum of RaE provided the decisive evidence. Petschek and Marshak (1952) analyzed the spectrum of RaE and found that the only interaction that would give a good fit to the spectrum was a combination of $$T$$ and $$P$$. This was, in fact, the only evidence favoring the presence of the P interaction. This led Konopinski and Langer (1953), in their 1953 review article on $$\beta$$ decay to conclude that, “As we shall interpret the evidence here, the correct law must be what is known as an $$STP$$ combination (1953, p. 261).”

Unfortunately, the evidence from the RaE spectrum had led the physics community astray. Petschek and Marshak had noted that their conclusion was quite sensitive to assumptions made in their calculation. “Thus, an error in the finite radius correction of approximately 0.1 percent leads to an error of up to 25% in $$C_{1(T+P)}$$ [the theoretical correction term].” Further theoretical analysis cast doubt on their assumptions, but all of this became moot when $$K$$. Smith[6] measured the spin of RaE and found it to be one, incompatible with the Petschek-Marshak analysis.

The demise of the RaE evidence removed the necessity of including the $$P$$ interaction in the theory of $$\beta$$ decay, and left the decision between the $$STP$$ and $$VA$$ combinations unresolved. The dilemma was resolved by evidence provided by angular-correlation experiments, particularly that from the experiment on $$\ce{^{6}He}$$ by Rustad and Ruby (1953; 1955)

(a) Angular Correlation Experiments. Angular correlation experiments are those in which both the decay electron and the recoil nucleus from $$\beta$$ decay are detected in coincidence. The experiments measured the distribution in angle between the electron and the recoil nucleus for a fixed range of electron energy, or measured the energy spectrum of either the electron or the nucleus at a fixed angle between them. These quantities are quite sensitive to the form of the decay interaction and became decisive pieces of evidence in the search for the form of the decay interaction. Hamilton (1947) calculated the form of the angular distribution expected for both allowed and forbidden decays, assuming only one type of interaction ($$S$$, $$V$$, $$T$$, $$A$$, $$P$$) was present. He found, for allowed transitions, that the angular distributions for the specific forms of the interaction would be different. A more general treatment was given by de Groot and Tolhoek (1950). They found that the general form of the angular distribution for allowed decays depended on the combination of the particular forms of the interactions in the decay Hamiltonian. For single forms their results agreed with those of Hamilton.

Figure 17. Schematic view of the experimental apparatus for the $$\ce{^{6}He}$$ angular correlation experiment of Rustad and Ruby (1953; 1955).

The most important of the experiments performed at this time was the measurement of the angular correlation in the decay of $$\ce{^{6}He}$$. This decay was a pure Gamow-Teller transition and thus was sensitive to the amounts of $$A$$ and $$T$$ present in the decay interaction. The decisive experiment was that of Rustad and Ruby (1953; 1955). This experiment was regarded as establishing that the Gamow-Teller part of the interaction was predominantly tensor. This was the conclusion reached in several review papers on the nature of $$\beta$$ decay. (Ridley 1954; Kofoed-Hansen 1955; Wu 1955). The experimental apparatus is shown in Figure 17. The definition of the decay volume was extremely important. In order to measure the angular correlation one must know the position of the decay so that one can measure the angle between the electron and the recoil nucleus. The decay volume for the helium gas in this experiment was defined by a 180 microgram/cm2 aluminum hemisphere and the pumping diaphragm. Rustad and Ruby (1953) presented two experimental results. The first was the coincidence rate as a function of the angle between the electron and the recoil nucleus for electrons in the energy range (2.5–4.0) mc2. The second was the energy spectrum of the decay electrons with the angle between the electron and the recoil nucleus fixed at 180°. Both results are shown in Figure 18 along with the predicted results for $$A$$ and $$T$$, respectively. The dominance of the tensor interaction is clear. This conclusion was made more emphatic in their 1955 paper which included more details of the experiment and even more data. The later results, shown in Figure 19, clearly demonstrate the superior fit of the tensor interaction.

Figure 19. Coincidence counting rate versus angle between the electron and the recoil nucleus for (a) electrons in the energy range 4.5–5.5 mc2 and (b) electrons in the energy range 5.5–7.5 mc2. From Rustad and Ruby (1955).

The Rustad-Ruby result, along with several others, established that the Gamow-Teller part of the decay interaction was tensor and that the decay interaction was $$STP$$, or $$ST$$, rather than $$VA$$. We have seen clearly in this episode the fruitful interaction between experiment and theory. Theoretical predictions became more precise and were tested experimentally until the form of the weak interaction was found. Fermi’s theory of $$\beta$$ decay had been confirmed. It had also been established that the interaction was a combination of scalar, tensor, and pseudoscalar $$(STP)$$.

Figure 20. Energy spectrum of recoil ions from $$\ce{^{35}A}$$ decay. From (Hermannsfeldt et al. 1958).

(b) Epilogue. It would be nice to report that such a simple, satisfying story, with its happy ending was the last word. It wasn’t. Work continued on angular correlation experiments and the happy agreement was soon destroyed (Franklin 1990, Chapter 3). Things became more complex with the discovery of parity nonconservation in the weak interactions, including $$\beta$$ decay. Sudarshan and Marshak (1958) and Feynman and Gell-Mann (1958) showed that only a $$V-A$$ interaction was compatible with parity nonconservation. If there was to be a single interaction describing all the weak interactions then there was a serious conflict between this work and the Rustad-Ruby result. This led Wu and Schwarzschild (1958) to reexamine and reanalyze the Rustad-Ruby experiment. They found, by calculation and by constructing a physical analogue of the gas system, that a considerable fraction of the helium gas (approximately 12%) was not in the decay volume. This changed the result for the angular correlation considerably and cast doubt on the Rustad-Ruby result.[7] The 6He angular correlation experiment was redone, correcting the problem with the gas target, and the new result is shown in Figure 20 (Hermannsfeldt et al. 1958). It clearly favors A, the axial vector interaction. Once again, physics was both fallible and corrigible. This new result on $$\ce{^{6}He}$$ combined with the discovery of parity nonconservation established that the form of the weak interaction was $$V-A$$.


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# HAPcryst

A HAP extension for crystallographic groups

Version 0.1.13
Released 2020-02-10

This project is maintained by Marc Roeder, The GAP Team

(C) Marc Roeder 2007-2008

HAPcryst -- A GAP4 extension for HAP calculating
resolutions for Bieberbach groups

Marc Roeder
(NUI Galway, Ireland)

This package is an add-on for Graham Ellis' HAP package. HAPcryst
implements some functions for crystallographic groups (namely
OrbitStabilizer-type methods). It is also capable of calculating free
resolutions for Bieberbach groups.

for the text of version 2 or www.http://gnu.org/copyleft/gpl.html

Installation
------------
HAPcryst depends on the following packages:
hap (as HAPcryst is a extension of HAP)
polymaking (which in turn depends on external software)
Cryst
packages installed.
The following packages are not required but suggested:
Carat
CrystCat
GAPDoc

HAPcryst.tar.bz2
HAPcryst.tar.gz
HAPcryst-win.zip
(where  is some version number) to the directory pkg/ of the
GAP home directory. If you do not have permission to do so, create
a directory called gap/pkg in your home directory.

2. change directory to pkg/ and unpack the archive using the according command:
tar -xjf HAPcryst.tar.bz2
tar -xzf HAPcryst.tar.gz
unzip HAPcryst-win.zip
(replace  with the version number)

3. start GAP. If you have created the directory gap/pkg in your home
directory, use "gap -l '/gap;'" where  is the path of
your home directory (use "pwd" to find out what it is, if you don't know)

If GAP returns some error messages, load hap.

5. Run the test file by calling "ReadPackage("hapcryst","tst/testall.g");"

Recommendation:
--------------

Calculating resolutions of Bieberbach groups involves convex hull
computations. polymake by default uses cdd to compute convex
hulls. Experiments suggest that lrs is the more suitable algorithm for
the convex hull computations done in HAPcryst than the standard
cdd. You can change the behaviour of polymake of by editing the file
"/.polymake/prefer.pl" It should contain a section like this
(just make sure lrs is before cdd, the position of beneath_beyond does
not matter):

#########################################
application polytope;

prefer "*.convex_hull  lrs, beneath_beyond, cdd";

Documentation
-------------

The documentation of the package can be found in the doc/ subdirectory.
A second part of the documentation describes datatypes used in HAPcryst.
This can be found in the lib/datatypes/doc subdirectory of the HAPcryst
main directory

Examples
--------

The directory "examples" contains some code that generates pictures
(JavaView applets) of Dirichlet domains of 3 dimensional Bieberbach
groups with coloured orbits (of the group acting on the faces of the
tessellation of R^3).
Just say


text:
# How do you write the quadratic function in standard form y=-1/2(x-6)(x-8)?

Apr 3, 2017

$7 x - {x}^{2} / 2 - 24$

#### Explanation:

Let's first multiply $\left(x - 6\right) \left(x - 8\right)$. We will use FOIL to expand this (first, outer, inner, last).

First: $x \cdot x = {x}^{2}$
Outer: $x \cdot - 8 = - 8 x$
Inner: $x \cdot - 6 = - 6 x$
Last: $- 8 \cdot - 6 = 48$

Add these all up to get ${x}^{2} - 14 x + 48$.

Now, don't forget to multiply each of these terms by $- \frac{1}{2}$ to get $- {x}^{2} / 2 + 7 x - 24$. We can arrange the terms so that the positive term is in the front $7 x - {x}^{2} / 2 - 24$.


text:
# Cube Slicing in Rn

Keith Ball
Proceedings of the American Mathematical Society
Vol. 97, No. 3 (Jul., 1986), pp. 465-473
DOI: 10.2307/2046239
Stable URL: http://www.jstor.org/stable/2046239
Page Count: 9

You are not currently logged in.

Access your personal account or get JSTOR access through your library or other institution:

Preview not available

## Abstract

We prove that every (n - 1)-dimensional section of the unit cube in Rn has volume at most $\sqrt2$. This upper bound is clearly best possible.

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text:
Volume 28, Issue 1
An Existence Result for a Class of Chemically Reacting Systems with Sign-Changing Weights
10.4208/jpde.v28.n1.1

J. Part. Diff. Eq., 28 (2015), pp. 1-8.

Preview Full PDF BiBTex 207 437
• Abstract

We prove the existence of positive solutions for the system

\begin{align*}\begin{cases}-\Delta_{p} u =\lambda a(x){f(v)}{u^{-\alpha}},\qquad x\in \Omega,\\-\Delta_{q} v = \lambda b(x){g(u)}{v^{-\beta}},\qquad x\in \Omega,\\u = v =0, \qquad x\in\partial \Omega,\end{cases}\end{align*}
where $\Delta_{r}z={\rm div}(|\nabla z|^{r-2}\nabla z)$, for $r>1$ denotes the r-Laplacian operator and $\lambda$ is a positive parameter, $\Omega$ is a bounded domain in $\mathbb{R}^{n}$, $n\geq1$ with sufficiently smooth boundary and $\alpha, \beta \in (0,1).$ Here $a(x)$ and $b(x)$ are $C^{1}$ sign-changing

functions that maybe negative near the boundary and $f,g$ are $C^{1}$ nondecreasing functions, such that $f, g :\ [0,\infty)\to [0,\infty);$ $f(s)>0,$ $g(s)>0$ for $s> 0$, $\lim_{s\to\infty}g(s)=\infty$ and

$$\lim_{s\to\infty}\frac{f(Mg(s)^{\frac{1}{q-1}})}{s^{p-1+\alpha}}=0,\qquad \forall M>0.$$

We discuss the existence of positive weak solutions when $f$, $g$, $a(x)$ and $b(x)$ satisfy certain additional conditions. We employ the method of sub-supersolution to obtain our results.

• History

Published online: 2015-03

• Keywords


text:
# Thread: Independent variables and partial differentiation

1. ## Independent variables and partial differentiation

I hope this makes sense... I'm not familiar with the terminology.

There is a vector space (x,y), with a linear transformation between (x, y) and (x', y'):
$x' = g_1(x, y) \\
y' = g_2(x, y)$

There is a set of points (a,b) in (x,y) defined by the following function:
$a = f(b,c)$
Where c is an independent variable.
- What is db/dc? Is it zero? Is it undefined?

The points (a,b) are transformed to (a',b').
- Can we say from the above whether b' and c are independent?
- Is there a rigorous definition of what 'independent' means?
- Does it depend on the particular equation?

Say we use the above to derive an expression like this:
$a' = h(a', b', c)$
- Are b' and c independent variables in that expression?
- Can I simply treat b' as a constant when finding $\frac{\partial a'}{\partial c}$, or do I need to treat it as a function of c?

2. I'm not sure whether I can answer this question in the abstract. So let me see if I can construct a simple example...

Originally Posted by Pete
There is a vector space (x,y), with a linear transformation between (x, y) and (x', y'):
$x' = g_1(x, y) \\
y' = g_2(x, y)$
I'll use

$x' = x + y\\
y'=x-y$

There is a set of points (a,b) in (x,y) defined by the following function:
$a = f(b,c)$
Where c is an independent variable.
I'll take

$a=e^{bc}$

What is db/dc? Is it zero? Is it undefined?
You said c is an independent variable, by which I assume you mean c is independent of b. That is, we can vary b as much as we want to without affecting c, and vice versa.

This suggests to me that db/dc=0.

The points (a,b) are transformed to (a',b').
Using my example

$a=e^{bc} \\
x' = x + y\\
y'=x-y$

so

$a'=a + b = e^{bc} + b\\
b'=a-b = e^{bc} - b$

- Can we say from the above whether b' and c are independent?
It seems clear to me from the example that b' depends on c.

Is there a rigorous definition of what 'independent' means?
Doesn't it mean that you can vary one variable without affecting the other?

Say we use the above to derive an expression like this:
$a' = h(a', b', c)$
Do you mean $a'=h(a,b',c)$? Using my example we have

$a'= e^{bc} + b = b' + 2b = b' + 2\frac{\ln a}{c}$

- Are b' and c independent variables in that expression?
No, because $b'=e^{bc}-b$

- Can I simply treat b' as a constant when finding $\frac{\partial a'}{\partial c}$, or do I need to treat it as a function of c?
Does't the notation $\frac{\partial a'}{\partial c}$ mean that all variables other than c are to be held constant? That would include b', wouldn't it?

3. Thanks James,
Originally Posted by James R
You said c is an independent variable, by which I assume you mean c is independent of b. That is, we can vary b as much as we want to without affecting c, and vice versa.

This suggests to me that db/dc=0.
But the same logic suggests that dc/db=0, ie db/dc is infinite or something.
It seems to me that db/dc=0 would imply that b can't vary:
\begin{align}
\frac{db}{dc} &= 0 \\
\int \frac{db}{dc} dc &= \int 0 dc \\
b + C_1 &= C_2 \\
b &= C
\end{align}

...but I'm not enough of a mathematician to be certain that this proof is valid.

Doesn't it mean that you can vary one variable without affecting the other?
That's my understanding, but I was wondering if there was any formal definition.

Do you mean $a'=h(a,b',c)$? Using my example we have

$a'= e^{bc} + b = b' + 2b = b' + 2\frac{\ln a}{c}$
In the specific case this question is based on, the expression was $a'=h(a',b',c)$, but the key point is that the expression involves b' and c, and doesn't involve b, so your example is good.

No, because $b'=e^{bc}-b$
This is where I get confused.
In that function, it seems that b' is an dependent variable, depending on b and c. (Or are they three interdependent variables?)
But in the function:
$a'= b' + 2\frac{\ln a}{c}$
Where b is not involved, it seems to me that b' is independent of c, because they can vary without affecting the other.
For any given value of c, b' can have any value (because b can have any value).
For any given value of b', c can have any value (because b can have any value).

Does that make sense, or am I right off the track?

Doesn't the notation $\frac{\partial a'}{\partial c}$ mean that all variables other than c are to be held constant? That would include b', wouldn't it?
I think so... but I'm getting confused by b.
We can hold b constant or b' constant, but not both (because that would imply holding c constant).
b is not in the equation, but b' was earlier derived/defined as a function of b and c.
So, do we hold b' constant, because b' is in the equation and b isn't?
Or do we hold b constant, for some other reason?

4. Originally Posted by Pete
But the same logic suggests that dc/db=0, ie db/dc is infinite or something.
It seems to me that db/dc=0 would imply that b can't vary:
\begin{align}
\frac{db}{dc} &= 0 \\
\int \frac{db}{dc} dc &= \int 0 dc \\
b + C_1 &= C_2 \\
b &= C
\end{align}

...but I'm not enough of a mathematician to be certain that this proof is valid.
Consider a horizontal line, which is really what you're describing.

The value of Y (or in this case b) is both constant and independent of the value of X (or in this case c).

Or to put it another way, if you set the variable c to some constant, you make it independent of b.

I'm not sure it neccessarily follows that it's true of all independent variables, however, that the derivative is 0.

5. Originally Posted by Trippy
Consider a horizontal line, which is really what you're describing.

The value of Y (or in this case b) is both constant and independent of the value of X (or in this case c).

Or to put it another way, if you set the variable c to some constant, you make it independent of b.

I'm not sure it necessarily follows that it's true of all independent variables, however, that the derivative is 0.
So, dy/dx=0 implies that y is a constant, right?
So if y is not a constant, then this implies that dy/dx is not zero?

6. Originally Posted by Pete
So, dy/dx=0 implies that y is a constant, right?
So if y is not a constant, then this implies that dy/dx is not zero?
I would have said so, yes.

If you consider

Y=0
Y=6
Y=$\frac{\sqrt{2}}{\pi}$
and Y=999999999999999999999999999999999999999999999999 999999999999999

In all cases $\frac{dy}{dx}=0$

7. Originally Posted by Pete
So, dy/dx=0 implies that y is a constant, right?
So if y is not a constant, then this implies that dy/dx is not zero?
What if y = z, and z is independent of x?

In general, if y=f(x,z), then

$\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial z}\frac{\partial z}{\partial x}$

So, if y=f(x,z)=z, then

$\frac{dy}{dx} = 0 + 1\times \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}$

so, we need to know how z varies with x to find the total derivative of y with respect to x.

If z is independent of x, then it seems to me that we must have $\frac{\partial z}{\partial x} = 0$, and it then follows also that $\frac{dy}{dx}=0$.

8. Originally Posted by James R
What if y = z, and z is independent of x?

In general, if y=f(x,z), then
$\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial z}\frac{\partial z}{\partial x}$
I think that should be
$\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial z}\frac{dz}{dx}$

So, if y=f(x,z)=z, then

$\frac{dy}{dx} = 0 + 1\times \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}$

so, we need to know how z varies with x to find the total derivative of y with respect to x.
Right... but "how z varies with x" implies some relationship between z and x. It doesn't make sense if z and and are independent, because when you vary x, z can do anything.

Maybe I'm just not getting the concept of independence.
It seems to me that independence is not an absolute thing, but is relative to the equation.
eg:
y = x+z
implies y depends on independent variables x and z.
But rewriting it as:
x = y - z
implies that x depends on independent variables y and z.

Is it better in this case to say that x, y, and z are interdependent?

9. Penrose states in his book "The Road to Reality", that for a pair of independent variables, say u and v, expressing that some quantity, say Z, is a function of u and v but independent of v is given by: dZ/dv = 0. It also says that for any value of u, Z is constant in v, and so Z depends only on u.

He also makes the point that this only holds locally, it might not be true for all Z.

10. Originally Posted by arfa brane
Penrose states in his book "The Road to Reality", that for a pair of independent variables, say u and v, expressing that some quantity, say Z, is a function of u and v but independent of v
...so Z is really only a function of u?
... is given by: dZ/dv = 0.
Are you sure it didn't say $\partial Z/\partial v = 0$?

It also says that for any value of u, Z is constant in v, and so Z depends only on u.
That looks very much like a description of $\partial Z/\partial v = 0$.

11. Originally Posted by Pete
Are you sure it didn't say $\partial Z/\partial v = 0$?
Yes, sorry, it actually concerns the Cauchy-Riemann equations. Partial derivatives are of one variable by definition.

Ed: but the principle holds for ordinary differentiation: if Z is constant in v the "slope" dZ/dv is zero, and v is independent.
Originally Posted by Pete
so Z is really only a function of u?
For a function of two variables, Z(u,v), there is no problem with Z being constant in v but still a function of u and v. The v could be contour lines of constant height (latitude) above the equator of a sphere, say, so they have gradients = 0.

12. Originally Posted by Pete
There is a vector space (x,y), with a linear transformation between (x, y) and (x', y'):
$x' = g_1(x, y) \\
y' = g_2(x, y)$

There is a set of points (a,b) in (x,y) defined by the following function:
$a = f(b,c)$
Where c is an independent variable.
- What is db/dc? Is it zero? Is it undefined?
If you fix c, you'll get a curve a=f(b,c) in general. So c cannot be taken as the only independent variable. Rather, we should consider b as a function of a and c. Differentiate the equation a=f(b,c) with respect to c to get

$
0=\frac{\partial f}{\partial b} \frac{\partial b}{\partial c} + \frac{\partial f}{\partial c},
$

so
$
\frac{\partial b}{\partial c} = - \frac{\partial f}{\partial c} \left(\frac{\partial f}{\partial b}\right)^{-1},
$
or in short $b_c = - f_c/f_b$,
provided the latter term can be inverted.
In effect, it is the same as holding a fixed, and defining the implicit function b(c) by a=f(b,c).

Originally Posted by Pete
The points (a,b) are transformed to (a',b').
- Can we say from the above whether b' and c are independent?
- Is there a rigorous definition of what 'independent' means?
- Does it depend on the particular equation?
- Not from the above, bu you can choose to consider b' and c as independent, because you have one equation and 3 variables.
- It means you can freely choose the values of the variables, and still have freedom to satisfy the given equation/relation.
- Yes, but in a weak way. If you have 1 equation involving 3 variables, you can choose 2 of them to be independent, unless some singular situations arise (look up implicit function theorem).

Originally Posted by Pete
Say we use the above to derive an expression like this:
$a' = h(a', b', c)$
- Are b' and c independent variables in that expression?
- Can I simply treat b' as a constant when finding $\frac{\partial a'}{\partial c}$, or do I need to treat it as a function of c?
- You can choose so.
- Yes.

13. Originally Posted by temur
If you fix c, you'll get a curve a=f(b,c) in general. So c cannot be taken as the only independent variable. Rather, we should consider b as a function of a and c. Differentiate the equation a=f(b,c) with respect to c to get

$
0=\frac{\partial f}{\partial b} \frac{\partial b}{\partial c} + \frac{\partial f}{\partial c},
$

so
$
\frac{\partial b}{\partial c} = - \frac{\partial f}{\partial c} \left(\frac{\partial f}{\partial b}\right)^{-1},
$
or in short $b_c = - f_c/f_b$,
provided the latter term can be inverted.
In effect, it is the same as holding a fixed, and defining the implicit function b(c) by a=f(b,c).
Thanks temur.
What about the total derivative $\frac{db}{dc}$?
Does it have a value?
Is it even meaningful?

- Not from the above, bu you can choose to consider b' and c as independent, because you have one equation and 3 variables.
- It means you can freely choose the values of the variables, and still have freedom to satisfy the given equation/relation.
- Yes, but in a weak way. If you have 1 equation involving 3 variables, you can choose 2 of them to be independent, unless some singular situations arise (look up implicit function theorem).
Thanks, I'll exercise due care

One more...
We have:
\begin{align}
x &= f(y,z) \\
\frac{dx}{dy} &= 0
\end{align}

Is this correct:
\begin{align}
\int \frac{dx}{dy} dy &= \int 0 dy \\
x + C_1 &= C_2 \\
x &= C
\end{align}

Or this:
\begin{align}
\int \frac{dx}{dy} dy &= \int 0 dy \\
x + g(z) &= C_2
\end{align}

My understanding is that the use of the expression \frac{dx}{dy} implies that z is a function of y, so x is implicitly a function of only y, and that the first process is correct.

But I'm not certain, and there's disagreement about in another thread (it's a debate thread, so only Tach and I can post there).

14. Originally Posted by Pete
What about the total derivative $\frac{db}{dc}$?
Does it have a value?
Is it even meaningful?
It is not meaningful. In order to define this, you need to specify how a depends on c. The "total" derivative $\frac{db}{dc}$ is taken along a given curve. Without a curve (i.e., a dependence such as a=a(c)), there is no "total" derivative.

Originally Posted by Pete
We have:
\begin{align}
x &= f(y,z) \\
\frac{dx}{dy} &= 0
\end{align}

Is this correct:
\begin{align}
\int \frac{dx}{dy} dy &= \int 0 dy \\
x + C_1 &= C_2 \\
x &= C
\end{align}

Or this:
\begin{align}
\int \frac{dx}{dy} dy &= \int 0 dy \\
x + g(z) &= C_2
\end{align}

My understanding is that the use of the expression \frac{dx}{dy} implies that z is a function of y, so x is implicitly a function of only y, and that the first process is correct.

But I'm not certain, and there's disagreement about in another thread (it's a debate thread, so only Tach and I can post there).
Again you need to know how z depends on y. Otherwise it has no meaning.

15. Right, we can't even talk about $\frac{dx}{dy}$ unless z depends on y...

So, if:
\begin{align}
x &= f(y,z) \\
z &= g(y) \\
\frac{dx}{dy} &= 0
\end{align}

Then is this correct?
\begin{align}
\int \frac{dx}{dy} dy &= \int 0 dy \\
x + C_1 &= C_2 \\
x &= C
\end{align}

Thanks again

16. Originally Posted by Pete
Right, we can't even talk about $\frac{dx}{dy}$ unless z depends on y...

So, if:
\begin{align}
x &= f(y,z) \\
z &= g(y) \\
\frac{dx}{dy} &= 0
\end{align}
These conditions imply only that the function f(y,z) is constant along the curve z=g(y). For example, take g(y)=0. Then any of f(y,z)=1, f(y,z)=zy, f(y,z)=sin(z) satisfy the equation $\frac{dx}{dy} = f_y+f_zg_y=0$.

17. *click*

Thanks!

#### Posting Permissions

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text:
# Algorithm-Self Doubt

97 views
How in a heap there are at most $\lceil \frac{n}{2^{h+1}} \rceil$ nodes of height h.

A heap of size $n$ has atmost $\left \lceil \frac{n}{2^{h+1}} \right \rceil$ nodes with height $h.$

Proof by induction:

Suppose it is true for height $\left ( h-1 \right )$

Let $N_{h}$ be the number of nodes at height $h$ in the $n$ node tree $T.$

Now, the tree $T'$ formed by removing the leaves of $T.$

It has $n'=n-\left \lceil \frac{n}{2} \right \rceil=\left \lfloor \frac{n}{2} \right \rfloor$ nodes.

Let $N'_{h-1}$ denote the number of node at height $\left ( h-1 \right )$ in tree $T'.$

By induction, we have $N_{h}=N'_{h-1}=\left \lceil \frac{n'}{2^{h}} \right \rceil=\left \lceil \frac{n/2}{2^{h}} \right \rceil=\left \lceil \frac{n}{2^{h+1}} \right \rceil$

https://www2.cs.sfu.ca/CourseCentral/307/petra/2009/SLN_2.pdf

0

@srestha-I have one doubt.How have you done $N_h=N'_{h-1}$?

0
Can u chk the link below?

$N'_{h-1}$ denote number of nodes in height $h-1$ in $T'$

## Related questions

1
359 views
What is the best case and worst case of the algorithm? And when will best case and worst case will happen?? int main() { for(i=1 ; i<=n ; i++) { if(n%i == 0) { for(j=1 ; j<=n ; j++) { printf("Hello"); } } } }


text:
# Tag Info

## New answers tagged finite-element

0

I finally did it! You can check the code on my GitLab repo. The function is to do this is the domain_extract which identifies the meshed boundaries to separate. To check a working example, you can check the sphere_meshing.py file, under the Scripts/Meshing directory.

2

Yes. That's all there is to the stability condition. Taking the material properties - shear modulus ($\mu$), bulk modulus ($\kappa$) and density ($\rho$) - into account, the global critical time step is evaluated as the minimum of the critical time step for each element ($\Delta t^e$) $\Delta t^e = CFL * h^e / c_{\kappa}$ where CFL is the Courant-Friedrichs-...

6

Domain decomposition was developed in the late 1990s and early 2000s because it allowed the re-use of sequential PDE solvers: You only have to write a wrapper around it that sends the computed solution to other processors, receives other processors' solutions, and uses these as boundary values for the next iteration. This works reasonably well for the small ...

1

Although I have never used these modules you mentioned, I do believe Comsol is an appropriate choice. My experience with the software was very positive a couple of years ago, mainly because of its support team. Unfortunately I cannot compare it to other alternatives because I have never used another commercial multiphysics FE software. You can parametrize ...

2

This can be solved as follows. If $L_0$ is the initial distance between the two nodes you want to displace, $L$ is the distance between the two nodes in the displaced body, and $d$ is the amount of length change you want to define, the following constraint relation can be defined $$G=L-L_0-d=0$$ Note that $L$ is a nonlinear function of the nodal ...

3

Suppose that we have a piecewise-linear one-dimensional basis. Technically the gradient has no point value defined at the finite element nodes because the derivative is discontinuous at the nodes as you've noticed. However, there are several techniques to approximate the gradient. One example is based on projecting the weak derivative onto a piecewise-linear ...

4

So so many places you have to rewrite. The whole mesh handling (accessing faces and edges from cells, neighbors from cells, ...). Shape functions. Dealing with the question of how the normal vector of a face when seen from one cell matches that on a neighboring cell. You will also likely encounter that 3d problems are always much larger and that solvers, ...

4

Ok, here comes the answer promised in the comment section. Let's start the other way round. In grid methods, one basically selects a number of $N+1$ gridpoints $\{x_k\}_{k=0}^{N}$. As basis functions, one can use Lagrange polynomials constructed over these nodes, each one having polynomial order $N$. For this setup, using Newton-Cotes-like quadrature weights ...

1

I would suggest Introduction to Numerical Methods for Variational Problems by Langtangen and Mardal. Besides that, you could check: scikit-fem. SfePy. I would also suggest our own course Introductory Finite Elements. It has Notebooks with the material and lecture notes; and our own package SolidsPy.

Top 50 recent answers are included


text:
### Are You a Smart Shopper?

In my local town there are three supermarkets which each has a special deal on some products. If you bought all your shopping in one shop, where would be the cheapest?

### Would You Rather?

Would you rather: Have 10% of £5 or 75% of 80p? Be given 60% of 2 pizzas or 26% of 5 pizzas?

### You Never Get a Six

Charlie thinks that a six comes up less often than the other numbers on the dice. Have a look at the results of the test his class did to see if he was right.

# The Magic Number and the Hepta-tree

## The Magic Number and the Hepta-Tree

The Mathemagician needs to break a dreadful spell! Help him to find the magic number that will do the trick!

Here is the hepta-tree hung with silver balls.

Do you think the balls will have similar values?

Try estimating the amount on each ball.

In order to break the spell, firstly find the exact difference between the largest ball and the smallest ball.

Multiply your result by $244$.

Round this to a whole number and that gives us the MAGIC NUMBER!

Give this number to the Mathemagician and the terrible spell will be broken for ever!

### Why do this problem?

This problem gives work on percentages in an interesting and challenging way. It also gives you an opportunity to discuss different ways of estimating and therefore shows learners how expressions which look very different can give very close answers.

### Possible approach

You could start by showing the group the picture of the hepta-tree and invite them to give a quick response as to which ball they think has the highest value and which one the smallest. Ask them to estimate the value of each ball in pairs and encourage several pairs to share their method with everyone. This will bring out the fact that actually the balls have very similar values.

After this they could work in pairs on the problem itself, possibly using this sheet (which lists the expressions on the balls). Learners are likely to need pencil and paper or a mini-whiteboard and pen, and possibly a calculator.

At the end of the lesson encourage the group to discuss the problem. Finding out how each pair approached it could be very useful as learners will gain from seeing what others have done. By sharing calculation strategies, children's understanding of the relationship between percentages, fractions and decimals will be deepened and you will be able to assess how comfortable they are with this area of mathematics.

### Key questions

What do you notice when you estimate the value of each ball?
How could you work out the exact value of this ball?
What do you notice about the numbers you have found?
How do you know which is the highest/lowest number?

### Possible extension

Learners could create more percentages like the ones on the hepta-tree which look different but have very close answers. These could range like the ones in the problem or approximate to another number. It would be interesting to talk to children about how they were coming up with the percentages.

### Possible support

Using this sheet will help children organise their calculations and it may be appropriate for some to use a calculator.