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Trigonometric Parallaxes: G9.62+0.20

# Trigonometric Parallaxes of Massive Star-Forming Regions. Vii. g9.62+0.20 and the Expanding 3 Kpc-Arm

A. Sanna11affiliation: Dipartimento di Fisica, Universitá degli Studi di Cagliari, S.P. Monserrato-Sestu km 0.7, I-09042 Cagliari, Italy 22affiliation: INAF, Osservatorio Astronomico di Cagliari, Loc. Poggio dei Pini, Str. 54, 09012 Capoterra (CA), Italy 33affiliation: Harvard-Smithsonian Center for Astrophysics, 60 Garden Street, Cambridge, MA 02138, USA , M. J. Reid33affiliation: Harvard-Smithsonian Center for Astrophysics, 60 Garden Street, Cambridge, MA 02138, USA , L. Moscadelli44affiliation: INAF, Osservatorio Astrofisico di Arcetri, Largo E. Fermi 5, 50125 Firenze, Italy , T. M. Dame33affiliation: Harvard-Smithsonian Center for Astrophysics, 60 Garden Street, Cambridge, MA 02138, USA , K. M. Menten55affiliation: Max-Planck-Institut für Radioastronomie, Auf dem Hügel 69, 53121 Bonn, Germany , A. Brunthaler55affiliation: Max-Planck-Institut für Radioastronomie, Auf dem Hügel 69, 53121 Bonn, Germany , X. W. Zheng66affiliation: Department of Astronomy, Nanjing University, Nanjing 210093, China and Y. Xu77affiliation: Purple Mountain Observatory, Chinese Academy of Sciences, Nanjing 210093, China
###### Abstract

We report a trigonometric parallax of 12 GHz methanol masers associated with the massive star forming region (catalog G9.62+0.20), corresponding to a distance of  kpc. With a local standard of rest velocity of about 2 km s, the region’s kinematic distances of 0.5 and 16 kpc differ greatly from the distance derived here. Our measurement of the peculiar motion of the star forming region shows a very large deviation from a circular Galactic orbit: 41 km s radially outward from the Galactic center and 60 km s counter to Galactic rotation. The combination of its radial velocity and distance places (catalog G9.62+0.20) in the inner region of the Galaxy close to the Expanding Near 3 kpc-Arm, where the bulge/bar potential has strong gravitational influence. We also map the distribution of 12 GHz methanol masers, locate them with respect to a hypercompact H II region, and compare our data with the periodic flare phenomenon reported previously for this source.

astrometry — Galaxy: fundamental parameters — Galaxy: kinematics and dynamics — masers — stars: formation — techniques: high angular resolution

## 1 Introduction

(catalog G9.62+0.20) is a massive star-forming complex harboring a number of H II regions in different evolutionary phases (e.g. Garay et al. 1993; Hofner et al. 1996; Testi et al. 2000; Hofner et al. 2001; De Buizer et al. 2003). Several masing species are associated with component (catalog G9.62+0.20) E, which has been classified as a candidate hypercompact (HC) H II region (Kurtz & Franco, 2002). Both 6.7 and 12.2 GHz methanol masers have been detected toward this object (Phillips et al. 1998; Caswell et al. 1995), as well as 22.2 GHz water (Hofner & Churchwell, 1996) and 1.665 and 1.667 GHz hydroxyl (Caswell, 1998) masers. Furthermore (catalog G9.62+0.20) has been claimed to be the most luminous 6.7 GHz methanol maser known in the Galaxy and both its 6.7 and 12.2 GHz methanol emissions show periodic flaring (Goedhart et al. 2003, 2005).

(catalog G9.62+0.20) is in the first Galactic quadrant, projected 9.6 degrees from the Galactic center. While its Local Standard of Rest (LSR) velocity is near zero, 21-cm wavelength H I spectra toward its H II region(s) show clouds of absorbing gas with LSR velocities out to 55 km s(Hofner et al., 1994). This effectively rules out the near kinematic distance of 0.5 kpc 111We adopt a “revised” kinematic distance using the prescription of Reid et al. (2009b), assuming (catalog G9.62+0.20) has an LSR velocity of +2.0 km s.. The apparent cutoff in H I absorption above 55 km s suggests that the source is not at the far distance either (16 kpc), but rather lies deep in the inner Galaxy on the near side of the tangent point. However, this argument may partly be affected by the decrease of cold and dense H I inside the Galactic gas hole (Galactocentric radius  kpc), since this gas component is the one responsible for well-defined H I absorption lines.

In order to determine the true distance and reliably locate (catalog G9.62+0.20) in the Galaxy, we conducted multi-epoch, phase referencing, VLBA observations of its 12 GHz methanol masers. In this paper, we report a measurement of trigonometric parallax based on our spectral line interferometric observations. In addition to the parallax, we accurately determine the proper motion of the masers with respect to background quasars. Combining the coordinates, distance, LSR velocity and proper motions yields the full 3-dimensional location and kinematics of the massive star-forming region. This complete information clearly associates (catalog G9.62+0.20) with the complex inner region of the Galaxy where the central bar has strong gravitational influence.

## 2 Observations and Data Analysis

Using the NRAO222The National Radio Astronomy Observatory is a facility of the National Science Foundation operated under cooperative agreement by Associated Universities, Inc. Very Long Baseline Array (VLBA), we observed the high-mass star forming region (HMSFR) (catalog G9.62+0.20) in the  E CHOH transition (rest frequency 12178.597 GHz). Background information about our VLBA program to measure parallaxes for HMSFRs is given in Reid et al. (2009a), hereafter called Paper I. We observed under program BR100A at 5 epochs: 2005 March 25 and October 8, 2006 March 16 and October 4, 2007 March 16. These dates were chosen to sample the peaks of the parallax signature in Right Ascension, since the amplitude of the Declination parallax signature is small and those position measurements are more susceptible to systematic errors at the low Declination of this source.

Paper I describes the general observational strategy and data calibration procedures. The maser observations were centered at the systemic velocity (V) of +2.0 km s, also traced with the CHCN  line emission by Hofner et al. (1996). Spectral resolution was 0.38 km s. Two fringe finders, J1743–0350 and J1800+3848 from the ICRF catalog, were observed for single-band delay and instrumental phase-offset calibration. For background position references, we observed 4 calibrators, J1803–2030, J1804–2147, J1808–2124 and J1811–2055, from the extragalactic radio survey by Xu et al. (2006). Only the first and last calibrators were detected with our VLBA measurements (see Table 1 and Figure 1). We selected the LSR velocity channel at +0.1 km s as the phase reference, corresponding to the peak intensity of the bright and compact maser spot F1, following the labeling of Minier et al. (2002) (see Figure 1 and 3). Table 1 summarizes source properties.

The absolute positions of the two background sources were not known with sufficient accuracy to be used to determine the absolute position of the maser emission. However, the a priori position of the maser reference spot (F1) was correct to about 20 mas, as indicated by the low fringe rates of the interferometer phases. After calibration, we imaged the phase reference maser channel and continuum sources using the AIPS task IMAGR with a circular restoring beam of at each epoch, equal to the east-west naturally weighted dirty beam size, which was typically at a P.A. of east of north, since the parallax information comes essentially from the east-west position shifts. The intensity peak of the phase reference spot varied by less than % across our observations spanning 2 yr (see Figure 4).

Figure 1 shows the images of the background sources, J1803–2030 and J1811–2055, and the phase reference maser spot from the middle epoch. For J1803–2030 the small deviations from a pointlike image are most likely caused by unmodeled short-term atmospheric fluctuations and follow the dirty beam shape (cf. Brunthaler et al. 2009). The calibrator J1811–2055 turned out to have an elongated-structure, with a possible extended halo. When imaging this source, we used a larger circular restoring beam of . This beam size was set to maximize the brightness sensitivity in the image. For parallax purposes, we fitted only the peak of the emission. At the first epoch (2005 March 25) only 4 (BR, FD, NL, OV) VLBA antennas were usable and the dynamic range of the images was about 7 times worse than for the subsequent 4 epochs. This precluded the use of the extended source J1811–2055 at this epoch. We measured the positions of the maser and the background sources by fitting elliptical Gaussian brightness distributions to the detected emission (AIPS task JMFIT).

## 3 Results

We measured the parallax and the proper motion of 12 GHz methanol masers by modeling the position differences between the reference maser spot and each calibrator versus time. The model is the sum of the parallax sinusoid, including the effects of the ellipticity of the Earth’s orbit, and a linear motion in each coordinate. Since systematic errors usually dominate over random noise, we adopted an “a posteriori” estimate of the errors from the fit itself. We assigned independent “error floors” to the east and north position offsets and added them in quadrature to the formal position fitting uncertainties. This procedure was iterated, adjusting error floors to achieve a reduced per degree of freedom near unity in each coordinate (see Paper I for a detailed discussion). We down-weighted the data from the first epoch (2005 March 25) by a factor 7, equal to the dynamic range degradation of the background source images. We can also put an upper limit on the shift over time of the centroid position of the quasars due to a possible jet component evolution. The small residuals for the parallax fits relative to the two quasars, and the similar parallax results, suggest that such effects are less than about 50 as.

Table 2 and Figure 2 show the results of the parallax and proper motion fit for each background source. While the post-fit residuals for the fit using J1803–2030 were about half that of the fit using J1811–2055, the J1803–2030 fit is dominated by data from only three epochs. As such, there are effectively near-zero degrees of freedom and the formal errors are not reliable. Since we could not reliably determine the relative weights for the two parallax data sets, we calculated an un-weighted average, instead of the combined fit procedure applied in previous papers (e.g. Paper I). The measured parallax of (catalog G9.62+0.20) obtained in this manner is  mas, corresponding to a distance of  kpc.

Our new distance indicates that the luminosity of the brightest 6.7 GHz methanol maser observed in the Galaxy has been substantially underestimated in the past. The near kinematic distance (2.0 kpc) adopted by Phillips et al. (1998) is only 38% of our measured distance implying a 7 times brighter source. This kinematic distance was derived by Forster & Caswell (1989) based on OH masers velocities, assuming the Schmidt (1965) Galactic rotation model and a Sun–Galactic center distance (R) of 10 kpc. Adopting the IAU value of the distance to the Galactic center, R kpc, our measured distance translates to a Galactocentric radius of 3.5 kpc. The measured mean proper motion is  mas yr toward the east and  mas yr toward the north; at our measured distance these values correspond to  km sand  km s eastward and northward, respectively. Completing the kinematic information, we assume a LSR velocity of  km s for the maser complex.

We compare our measurements with a simple rotation model of the Galaxy, converting these values from the equatorial heliocentric reference frame to a rotating Galactic reference frame. This approach subtracts a circular velocity component at the position of the source and yields a peculiar motion vector with respect to circular rotation (Reid et al., 2009b). Assuming negligible internal motions of CHOH masers (e.g. Moscadelli et al. 2002, 2009) and adopting the IAU value for the distance to the Galactic center (R kpc), the rotation velocity at the solar circle ( km s), and the Hipparcos measurements of the solar motion (Dehnen & Binney, 1998), the peculiar velocity components of (catalog G9.62+0.20) are  km s, where U, V and W are directed toward the Galactic center, in the direction of Galactic rotation and toward the North Galactic Pole, respectively. Using instead the best-fit values of R kpc and  km s, obtained from 16 parallax and proper motion measurements (Reid et al., 2009b), the peculiar velocity components are  km s. Quoted uncertainties are based on proper motion and measurement errors and also a 7 km s uncertainty in each peculiar velocity component, owing to expected Virial motions of a star within a high mass star forming region.

## 4 Galactic Location and Peculiar Motion

In the literature there are two estimates of distance to the source (catalog G9.62+0.20) consistent with our direct measurement. Scoville et al. (1987) derived a distance of 4.7 kpc for the H II complex (catalog G9.62+0.20), based on the association to a CO cloud with a LSR velocity of +4 km s and following the expanding ring model of the 3 kpc-Arm given by Bania (1980). Hofner et al. (1994) showed that there is HI absorption over the LSR velocity range of -5 to 55 km s and suggested a kinematic distance of 5.7 kpc. Our trigonometric parallax places (catalog G9.62+0.20) at a distance close to 5.2 kpc. The combination of distance and Galactic coordinates locates (catalog G9.62+0.20) in the inner region of the Galaxy within the so-called “molecular ring”, which appears as a complex structure in CO longitude-velocity maps (e.g. Dame et al. 2001; Rodriguez-Fernandez & Combes 2008). Based on the observations of distinct tangents points, traced by the early CO surveys in the southern Milky Way (e.g. Robinson et al. 1984; Bronfman et al. 1989), the inner edge of the molecular ring is thought to be composed of the Norma and Expanding 3 kpc arms (e.g. Figure 5 in Bronfman et al. 2000; Figure 3 in Dame et al. 2001). Due to observational difficulties detecting and following the Norma Arm in CO emission at positive longitudes, the question to which arm, if any, does the Norma Arm connect in the first quadrant is still controversial (e.g. Russeil 2003; Vallée 2008). Therefore in the following, we focus our comparison of the spatial and velocity properties of (catalog G9.62+0.20) with the well-studied Expanding 3 kpc-Arm.

Since the LSR velocity for a nearly circular Galactic orbit at the longitude and distance of (catalog G9.62+0.20) would be  km s, about 50 km s greater than observed, (catalog G9.62+0.20) is clearly highly kinematically anomalous. We can gain a more complete picture of the region by comparing the Galactic location (,b) and the LSR velocity (+2 km s) of (catalog G9.62+0.20) with respect to the distribution of CO toward this region. Using the longitude-velocity and latitude-velocity CO  line maps of Bitran et al. (1997), we find the HMSFR (catalog G9.62+0.20) associated with the edge of a CO clump (cutoff at 0.5 K). Dame & Thaddeus (2008) locate the Expanding (Near) 3 kpc-Arm (hereafter simply the 3 kpc-Arm) below the plane of the Milky Way at Galactic longitude , whereas (catalog G9.62+0.20) is slightly above the plane. The LSR velocity of the 3 kpc-Arm in this direction is expected to be  km s, blueshifted by about  km s with respect to a circular orbit at  km s. We find (catalog G9.62+0.20) to have a  km s, which, while close to the CO velocity at this longitude, is shifted by about 15 km s with respect to CO in the 3 kpc-Arm. Thus, (catalog G9.62+0.20) may not be part of the 3 kpc-Arm as previously proposed and, perhaps, could belong to a nearby structure such as the Norma Arm. Either way, this source, and presumably its associated arm, has a large peculiar motion with a  km s component radially outward and a  km s component counter to Galactic rotation, rendering kinematic distances useless.

The spatial and velocity properties of the HMSFR (catalog G9.62+0.20) and the 3 kpc-Arm, although not strictly associated, suggest they could participate in a similar dynamical anomaly. The large peculiar motion of the 3 kpc-Arm might be induced by the gravitational influence of a central bar(s) potential, as explored by many authors. Habing et al. (2006) studied the distribution of maser stars in the inner Milky Way with respect to the ISM distribution and compared their spatial properties with a set of orbits derived by combining an axisymmetric potential plus a weak rotating bar. They found an overall agreement between maser stars and gas in both space and velocity, moving inward along the Galactic plane with almost circular orbits that become more and more elongated inside the molecular ring with radial motion increasingly important. Following Sevenster (1999), they suggested the 3 kpc-Arm to be the locus of ballistic orbits originating at the corotation radius of the bar. On the contrary, Rodriguez-Fernandez & Combes (2008) modeling the dynamics of the Milky Way gas flow in the presence of a potential from two nested bars, suggested by the Two Micron All Sky Survey (2MASS) data, reproduced the longitude-velocity features of the 3 kpc-Arm as a spiral density wave maximum surrounding the primary bar. Further parallax and 3-D velocity measurements of HMSFRs sampling the inner region of the Galaxy will help to constraint the dynamics of this complex region.

As a final remark, we note that two supernova remnants (SNRs), G9.8+0.6 and G9.7-0.0, are projected to within a half degree of (catalog G9.62+0.20). G9.8+0.6 has distance estimates of 8.8 kpc (Stupar et al., 2007) and 10.4 kpc (Case & Bhattacharya, 1998), both based on the -D relation, which may not be reliable (e.g. Green 2004), whereas the G9.7-0.0 has a near kinematic distance estimate (4.7 kpc) based on an associated OH maser velocity (+43 km s) detected toward the SNR (Hewitt & Yusef-Zadeh, 2009). However, the association of (catalog G9.62+0.20) and G9.7-0.0 is questionable given their about 40 km s different radial velocity.

Y.X. was supported by Chinese NSF through grants NSF 10673024, NSF 10733030, NSF 10703010, and NSF 10621303.

We are very grateful to Testi and collaborators for providing the VLA continuum map of G9.62+0.20. Facilities: VLBA.

## Appendix A Appendix material: 12 GHz Methanol Masers

Our measurements allow us to locate the distribution of the individual maser spots with respect to the HCH II component (catalog G9.62+0.20) E. Figure 3 presents the absolute positions of the 12 GHz maser spots superposed on the 22 GHz continuum emission map of Testi et al. (2000). For this data taken in the BnA-configuration, the VLA has a synthesized HPBW of at a P.A. of east of north and is insufficient to resolve the emission of the HCH II region E. The accuracy in the absolute position of the VLA map is better than 0.2 arcsec. The spectral index of component E at radio frequencies, from 8.4 to 110 GHz, is 0.95 (Franco et al., 2000). We find that individual 12 GHz methanol maser spot velocities and flux densities, as well as their overall distribution, change little over a time span of about 8 yr, from 1998 November (Minier et al., 2002) to our observations in 2007 March. We detected all nine distinct clusters imaged by Minier et al. (2000, 2002) within a region of about  mas, labeled from A to I, and resolved them into 20 distinct spots/knots. The maser emission generally appears as compact cores of high brightness embedded in extended low brightness structures (Figure 3 edge panels).

Analysis of the light curves of the brightest channel of each spot is consistent with the timing of the flares measured by Goedhart et al. (2003). Light curve data of 12 GHz methanol maser emission in G9.62+0.20 E at our 5 epochs are plotted in Figure 4. Brightness (Jy beam) and integrated flux density (Jy) of the brightest channel of each maser feature are estimated using the task IMSTAT. Selected BOXes for IMSTAT purposes are the same per spot at every epoch. Note that flux densities from the first epoch should be taken as lower limits because of the loss of extended emission due to the lack of short-baselines at that epoch (cf. visibility amplitude vs. uv-distance diagrams in Minier et al. 2002). Epoch 2 falls just before the start of the flare predicted by the periodicity and phasing reported by Goedhart et al. (2003); epoch 3 and 4 fall in the expected quiescent periods. Interestingly, Epoch 5 (2007 March) falls during the expected flaring time and our data for features B and C1 (which displayed the strongest flares previously; cf. Figure 7 in Goedhart et al. 2005) show the greatest increases of flux density.

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## August 21, 2008

### Homotopical Algebra with Applications to Mathematical Physics

#### Posted by Urs Schreiber

guest post by Jim Stasheff

We are organizing a special session at the 2009 Spring Southeastern Section Meeting in Raleigh, NC, during April 4-5, 2009:

Session Title: Homotopical Algebra with Applications to Mathematical Physics

Organizers:

Jim Stasheff, UNC Chapel Hill, Emeritus, jds at math.upenn.edu

Various aspects of homotopy algebras such as $A_\infty$, $L_\infty$, and open-closed homotopy algebras are currently the object of research by a large number of mathematicians. These algebras are closely linked to a variety of topics in mathematical physics, for example, open string theory, closed string theory, gauge theory. This session will provide a forum for reports on current developments in both areas – algebraic structures as well as physical implications.

We particularly encourage graduate students to report on their current research on these topics. We hope to have at least one lecture that will present the overall picture that relates these subjects. If you are interested in giving such a survey, please let us know.

If you are interested, even provisionally, please respond to both of us. The official deadline for submitting an abstract to the AMS is —.

Posted at August 21, 2008 8:19 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1772


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# Pump A can empty a pool in A minutes, and pump B can empty

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Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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Updated on: 24 Jan 2014, 06:19
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Difficulty:

75% (hard)

Question Stats:

62% (02:31) correct 38% (02:31) wrong based on 380 sessions

### HideShow timer Statistics

Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

A. $$\frac{A+B-1}{2}$$

B. $$\frac{A(B+1)}{A+B}$$

C. $$\frac{AB}{(A+B)}$$

D. $$\frac{AB}{(A+B)} -1$$

E. $$\frac{A(B-1)}{(A+B)}$$

Originally posted by gmatgambler on 24 Jan 2014, 06:13.
Last edited by Bunuel on 24 Jan 2014, 06:19, edited 2 times in total.
Edited the question.
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Posts: 50543
Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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24 Jan 2014, 06:29
6
6
gmatgambler wrote:
Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

A. $$\frac{A+B-1}{2}$$

B. $$\frac{A(B+1)}{A+B}$$

C. $$\frac{AB}{(A+B)}$$

D. $$\frac{AB}{(A+B)} -1$$

E. $$\frac{A(B-1)}{(A+B)}$$

The rate of pump A is $$\frac{1}{A}$$ pool/minute;
The rate of pump B is $$\frac{1}{B}$$ pool/minute.

Their combined rate is $$\frac{1}{A} + \frac{1}{B} = \frac{A+B}{AB}$$ pool/minute.

In one minute that A works alone it does (rate)*(time) = $$\frac{1}{A} * 1= \frac{1}{A}$$ part of the job, thus $$1-\frac{1}{A}=\frac{A-1}{A}$$ part of the job is remaining to be done by A and B together.

To do $$\frac{A-1}{A}$$ of the job both pumps working together will need (time) = (job)/(combined rate) = $$\frac{A-1}{A}*\frac{AB}{A+B}=\frac{(A-1)B}{A+B}$$.

So, the total time is $$1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B}$$.

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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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24 Jan 2014, 06:35
1
1
Bunuel wrote:
gmatgambler wrote:
Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

A. $$\frac{A+B-1}{2}$$

B. $$\frac{A(B+1)}{A+B}$$

C. $$\frac{AB}{(A+B)}$$

D. $$\frac{AB}{(A+B)} -1$$

E. $$\frac{A(B-1)}{(A+B)}$$

The rate of pump A is $$\frac{1}{A}$$ pool/minute;
The rate of pump B is $$\frac{1}{B}$$ pool/minute.

Their combined rate is $$\frac{1}{A} + \frac{1}{B} = \frac{A+B}{AB}$$ pool/minute.

In one minute that A works alone it does (rate)*(time) = $$\frac{1}{A} * 1= \frac{1}{A}$$ part of the job, thus $$1-\frac{1}{A}=\frac{A-1}{A}$$ part of the job is remaining to be done by A and B together.

To do $$\frac{A-1}{A}$$ of the job both pumps working together will need (time) = (job)/(combined rate) = $$\frac{A-1}{A}*\frac{AB}{A+B}=\frac{(A-1)B}{A+B}$$.

So, the total time is $$1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B}$$.

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Hope this helps.
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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30 Mar 2014, 14:32
2
I did this question using smart numbers

A = 6
B = 3

Therefore in 1 minute A did 1/6 of the job.
5/6 of the job remains

Together A and B work at a rate of 1/2.

Thus 1/2 (x) = 5/6

x = 5/3 minutes

Now we need to add the minute worked by A so total of 5/3 + 1 = 8/3 minutes

Replacing we can see that in (B)

(6)(4) / 9 = 8/3

Therefore it is the correct answer choice

Gimme some freaking Kudos!
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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30 Mar 2014, 19:24
jlgdr wrote:
I did this question using smart numbers

A = 6
B = 3

Therefore in 1 minute A did 1/6 of the job.
5/6 of the job remains

Together A and B work at a rate of 1/2.

Thus 1/2 (x) = 5/6

x = 5/3 minutes

Now we need to add the minute worked by A so total of 5/3 + 1 = 8/3 minutes

Replacing we can see that in (B)

(6)(4) / 9 = 8/3

Therefore it is the correct answer choice

Gimme some freaking Kudos!
Cheers
J

Taking numbers is a great strategy but try to take easier numbers. Say, 1, 0 etc as long as you meet all conditions mentioned.
I would take A = B = 1 (since there are no symmetric options in A and B such as A/(A+B) and B(A + B), I can take both A and B equal)
Now I need to do no calculations. A starts and works for a minute. In that time, the pool is empty. B joins but has nothing to do. Total time taken to empty the pool is 1 min.
Only option (B) gives 1.
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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01 Apr 2014, 09:32
1
2
Method 1:
The time taken by both pumps working together to do the work = AB/(A+B) minutes
Now, since Pump A has already worked for 1 minute, 1*(1/A) of the work has already been done.
Time taken to do the remaining work = [AB/(A+B)]*(1-1/A) = B(A-1)/(A+B) minutes
Total time taken from the start = [B(A-1)/(A+B)] + 1 = A*(B+1)/(A+B)
Option (B).

Method 2:
Let t be the time taken from the start. Then pump A works for t minutes to finish the work and pump B for (t-1) minutes
=> t(1/A) + (t-1)(1/B) = 1
=> t = A*(B+1)/(A+B)
Option (B).

Method 3:
We know that the time taken to do the work by both pumps working together right from the beginning is AB/(A+B) minutes.
Lets flip the question around to say that the two pumps worked together to do the work, except pump B was off in the first minute. So we must add the time it would take the two pumps to do the work that pump B would have done in this one minute to the total time. In this one minute, pump B would have done 1*(1/B) of the work = 1/B of the work.
Total time taken is therefore = AB/(A+B) + AB*(1/B)/(A+B)
= A*(B+1)/(A+B)
Option (B).
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Solve Work Problem using Numerical approach  [#permalink]

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Updated on: 18 Apr 2014, 00:25
Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

(a) A+B-1/2

(b) A(B+1)/A+B

(c) AB/A+B

(d) AB/A+B -1/1

(e) A(B-1)/A+B

-----------------------------------------------

METHOD ONE: Algebraic approach

This approach to this question involves some tricky algebra.

Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.

In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of

The time it will take the two pumps, working at the combined rate, to drain this, is:

That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)

METHOD TWO: Numerical approach

Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.

Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.

Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?

Add the first minute for total time.

If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?

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Originally posted by royQV on 17 Apr 2014, 09:57.
Last edited by royQV on 18 Apr 2014, 00:25, edited 1 time in total.
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Re: Solve Work Problem using Numerical approach  [#permalink]

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17 Apr 2014, 18:05
1
royQV wrote:
Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

METHOD ONE: Algebraic approach

This approach to this question involves some tricky algebra.

Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.

In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of

The time it will take the two pumps, working at the combined rate, to drain this, is:

That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)

METHOD TWO: Numerical approach

Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.

Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.

Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?

Add the first minute for total time.

If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?

Can you please post the OA?

Rate of Pump A $$= \frac{1}{a}$$

Rate of Pump B $$= \frac{1}{b}$$

Work done by Pump A in 1 min $$= \frac{1}{a} * 1 = \frac{1}{a}$$

Work remaining (To be done by both pumps) $$= 1 - \frac{1}{a} = \frac{a-1}{a}$$

Combined rate of Pump A & B$$= \frac{1}{a} + \frac{1}{b}$$

Time required to complete the remaining work = t

$$\frac{a+b}{ab} * t = \frac{a-1}{a}$$

$$t = \frac{b(a-1)}{a+b}$$

Total time required

= $$\frac{b(a-1)}{a+b} + 1$$

$$= \frac{a(b-1)}{a+b}$$
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Re: Solve Work Problem using Numerical approach  [#permalink]

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18 Apr 2014, 01:30
royQV wrote:
Q. Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

(a) A+B-1/2

(b) A(B+1)/A+B

(c) AB/A+B

(d) AB/A+B -1/1

(e) A(B-1)/A+B

-----------------------------------------------

METHOD ONE: Algebraic approach

This approach to this question involves some tricky algebra.

Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.

In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of

The time it will take the two pumps, working at the combined rate, to drain this, is:

That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)

METHOD TWO: Numerical approach

Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.

Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.

Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?

Add the first minute for total time.

If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?

Merging similar topics. Please refer to the discussion above.

Hope it helps.

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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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07 Aug 2017, 22:24
gmatgambler wrote:
Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

A. $$\frac{A+B-1}{2}$$

B. $$\frac{A(B+1)}{A+B}$$

C. $$\frac{AB}{(A+B)}$$

D. $$\frac{AB}{(A+B)} -1$$

E. $$\frac{A(B-1)}{(A+B)}$$

Since A empties a pool in A minutes. In 1 minute A empties 1/A fraction of the pool.
B empties a pool in B minutes. In 1 minute B empties 1/B fraction of the pool.

So, 1/A + (1/A + 1/B)n = 1 where is n is the time in minutes for which A and B worked together.
(A+b)/AB *n = 1- 1/A
n = AB/(A+B) - B /(A+B)= (AB- B)/(A+B)

SO, total time taken to empty a pool = 1+ (AB- B)/(A+B) = (A+B+ AB-B)/(A+B) = A(B+1)/(A+B)

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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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17 Aug 2017, 06:49
The fastest time in which tank can be emptied is by working together which is AB /(A+B)

But in this case it would be greater than AB/(A+B) since they have not worked together all time.

out of option B , C , D & E only option "B" is greater than AB /(A+B)
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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# Effect on output voltage for buck converter in discontinuous mode

In a dc-dc switch mode buck converter, why the output voltage is different from the desired value when the inductor current is discontinuous? How can the voltage be adjusted to the required value? What would be the effect of open-circuiting the load terminals on the output voltage? Will the capacitor keep on charging until it blows out if the load is removed?

The average value of the output of a buck converter in discontinuous mode is still the same as continuous mode. However, the ripple voltage increases.

This happens on light loads when the switcher cannot produce pulses of low enough duty cycle and it transfers a little too much energy per switching cycle for the light load.

This means the average output voltage starts to rise too high and the control system shuts down the switcher for several cycles resulting in a slightly lower than normal output voltage. Until the load starts to take more power this situation persists.

Discontinuous mode also happens when the input voltage to the buck rises too high.

If the buck converter is operating at 100kHz and the inductor is 10uH. Let's also say that the supply is 12V and duty cycle is 50%. The "on" pulse of the mosfet will charge the inductor with current and this current is determined by input voltage, output voltage and inductance. Let's say the output voltage is 5V - this means the voltage across the inductor is 7V when the mosfet is "on".

V = L$\dfrac{di}{dt} \therefore 7 = 10\times 10^{-6} \times\dfrac{di}{dt} \therefore\dfrac{di}{dt} =\space$700,000

Because dt is 50% of 10us we can calculate I, which is $700000\times 5\times 10^{-6}$ = 3.5A.

The energy transferred per cycle is therefore $\dfrac{L\times 3.5^2}{2}$ = 57.8uJ.

This transfers 100k times per second $\therefore$ the power to the load is $57.8\mu J \times 100000$ = 5.78W.

If the load resistor is too high to take that power at 5V then the simple fact is that the output voltage will rise and the buck convertor will enter discontinuous mode unless the control loop reduces the duty cycle.

This duty cycle reduction will of course happen because it would be a poor buck converter that couldn't produce less than 50% duty but, it will have a minimum value and at this point, if the energy-load equation isn't in equilibrium the converter will enter discontinuous mode.

• What if the output voltage becomes higher? – Anusha Oct 13 '13 at 10:55
• The energy into the inductor may become too much for the load and the output rises because of minimum duty cycle of switcher. – Andy aka Oct 13 '13 at 11:11
• I've added a question above. Can you please answer it? – Anusha Oct 13 '13 at 14:53
• If load is removed, energy in the inductor will directly charge the capacitor, voltage will rise and continu to rise unless the control system forces discontinuous operation. Then voltage will remain steady because switcher is inhibited. Cap voltage will slowly self- discharge due to leakage and eventually, switcher will restart for one or two cycles and process repeats. – Andy aka Oct 13 '13 at 15:16

The average voltage of a buck converter in DCM is more than CCM, because the inductor current will become zero before switching period, say 100 $\mu$sec switching time and duty_on(d)=0.5, duty_off(d2)=0.3(because of DCM). So inductor current becomes zero after 80 $\mu$sec, then diode will turn off (current flowing through the diode is zero). After diode is turned off the capacitor voltage will appear across the load. By this way the average voltage of buck converter in DCM is mode ($V_o/V_g = d/(d+d2)$).


text:
# How do you solve the following a - b(mod5) = a + b(mod5)?

attempted solution:

a - b(mod5) = a + b(mod5)

2a(mod5) = 0

a = 5

5 + b mod 5 = 5 - b mod 5

b mod 5 = -b mod 5

b = 0

-
Almost: $5\bmod 5=0$, so you should have $a=0$, not $a=5$, if you’re working in the usual residue system $\{0,1,2,3,4\}$. If you’re working over the integers, and the starting point is actually $a-b\equiv a+b\pmod 5$, then the solution is that $a$ and $b$ can be any multiples of $5$. –  Brian M. Scott Nov 30 '12 at 0:18
@BrianM.Scott: In the usual residue system, can't $a$ be anything? $3-5\equiv 3+5 \pmod 5$ –  Ross Millikan Nov 30 '12 at 0:36
Is this a single equation or a system of 2 equations? Where does that $2a\equiv0\pmod5$ come from? –  Mike Nov 30 '12 at 1:08
so it's the other way around a = 0 and b = 5?> in R? is 0 a multiple of 5? –  internetlearning Nov 30 '12 at 2:15
what's the step by step solution? –  internetlearning Nov 30 '12 at 2:15

If you're working in the usual residue system of $\{0, 1, 2, 3, 4\}$, and $b = 0$, $a$ can be any number $n$ in the set, that is, $a-0 = a+0 \pmod 5 \iff a = a \pmod 5$ which is true for any $a$. Solutions of the form $(a, b)$ are given by $\{(a, 0): a \in \{0, 1, 2, 3, 4\}\}$

Similarly, if you're working over the integers, then provided $b$ is any integer multiple of $5$ (that is, provided $b\equiv 0 \bmod 5),$ then $a-b\equiv a+b\pmod 5 \iff a - 5k \equiv a + 5k \pmod 5$, which is true for all integer values of $a$, $k \in \mathbb{Z}$. Solutions of the form $(a, b)$ are given by $\{(a, 5k): a, k\in \mathbb{Z}\}$.

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what is a residue system? –  internetlearning Nov 30 '12 at 2:13
Is simply means limiting the integers of concern, when considering $\mod n$, to $\{0, 1, 2, ...., n-1\}$. But if you are looking for all integer solutions for a and b, then the second part of the answer will give you that. –  amWhy Nov 30 '12 at 2:19
why is b any integer multiple of 5? –  internetlearning Nov 30 '12 at 2:21
if you subtract (a - b) - (a + b) = 0 (mod 5), you get a-a - b - b= -2b = 0 (mod 5). That happens if and only if b is a multiple of 5: $b\in \{...-10, -5, 0, 5, 10, ...\}$. –  amWhy Nov 30 '12 at 2:32

From $a-b\equiv a+b \pmod 5$ you can derive $2b \equiv 0 \pmod 5$, so $b \equiv 0 \pmod 5$, but $a$ can be anything-it cancels. Try $a=3, b=5$, for example.

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## söndag 23 oktober 2011

### Do Living Physicists Support DLR/Backradiation?

The inventor of the idea of DLR/backradiation supporting CO2 alarmism: Pierre Provost, 1791.

In Section 5.2 of Mathematical Physics of Blackbody Radiation I recall the following statement from Planck's The Theory of Heat Radiation:
• We shall assume that the radiation in one direction is completely independent of the radiation in a different direction, even opposite.
• But the empirical principal law that the emission of any volume-element depends entirely on what takes places inside of this element holds true in all cases (Prevost's principle). A body A at $100^{\circ} C$ emits toward a body B at $0^{\circ} C$ exactly the same amount of radiation as toward an equally large and similarly situated body B' at $1000^{\circ} C$. The fact that the body A is cooled by B and heated by B' is due entirely to the fact that B is weaker, B' stronger emitter than A.
This statement can be seen as the origin of the idea underlying CMB and DLR that the radiation from a blackbody is independent of the temperature the environment or background. Planck refers as support to Prevost's exchange principle from 1791:
• Absolute equilibrium of free heat is the state of this fluid in a portion of space which receives as much of it as it lets escape. Relative equilibrium of free heat is the state of this fluid in two portions of space which receive from each other equal quantities of heat, and which moreover are in absolute equilibrium, or experience precisely equal changes.
• The heat of several portions of space at the same temperature, and next to one another, is at the same time in the two species of equilibrium.
Mathematical Physics of Blackbody Radiation I give a new analysis of blackbody radiation challenging the idea of Provost/Plank of two-way radiative exchange of heat between two bodies with the hotter winning the game of net heat transfer as a stronger emitter.

So what is true physics:
1. Is radiative heat transfer one-way from hot to cold, as my analysis suggests?
2. Is radiative heat transfer two-way with the hotter being the stronger, as Provost claimed in 1791?
Does it matter? I say yes, because the stability properties can be vastly different in one-way and two-way transfer. It can be the difference between a harmless climate sensitivity of 0.3 C and and alarming of 3 C.

To empirically distinguish between 1. and 2. is difficult since only net heat transfer is directly observable. To refer to two-way transfer as an empirical principal law, as Planck does, is questionable.

The question is open. It would be interesting to see new independent support of 2. from active physicists living today. One can argue that the 2nd law of thermodynamics supports 1. but not 2. The 2nd law was unknown to Provost.

A scientist putting forward a certain statement as true often refers to some other scientist as being the responsible, typically the scientist who first gave the evidence of the truth. If that
first scientist is dead and cannot answer questions, which is often the case (as in Planck's reference to Provost), a scientific controversy may arise from different interpretations of what the now dead scientist actually meant or did. Therefore science should be carried by living scientists who carry the responsibility and are willing to answer questions.

In the present case there is a confusion between
which makes the debate confusing, but there seems to be no living physicist willing to answer the questions and carry the responsibility.

#### 18 kommentarer:

1. Hi Claes!

I am trying to understand your arguments considering non background radiation, but I haven't succeded so far.

I will suggest a simple test situation. Take to boxes, cages, of similar conveniant size, strong enough (steel and glass) to endure beeing evaporated of air. Let's hang a metal ball in a wool thread from the top inside each case. Let evrything have living room temperature. Then we evacuate the air in both cages, so there will be no "temperature" inside unless the inner walls and the ball, and no conduction beetween the balls and the surroundings. Then vi put cage A inside a freezer where the walls and the air has a temperature of -20C, and let cage B stay outside in the living room temperature.

Let's check them up the next day with the following hypothesis:

H0: The two balls will have the same temperature.

H1: One of the balls (cage B will be my guess) will be considerably colder.

Which one will be your favorit, and why?

2. A will be cooler. What is this exp supposed to test?

3. Thank's for your answer, A will be my guess too (wrongly written in my first post).

But you have only adressed the first half of my question. The other half is WHY. I think A will be colder because it recieves less heat radiation from the surroundings, (the only way to exchange energy in this case) but enough to never get colder than -20C. So both balls will be in equilibrium with the surroundings due to the radiation from the surroundings, i.e. "back radiation" is real physics.

Or, have I missed something of the traditional explanation or of your interpretattion?

4. So, if I understand it right, the question is does body A at temperature T_A radiate independently of body B at temperature T_B?

Why isn't this a testable configuration?

If one sets up an experiment where body A sees body B , but one screens body B so that the detector only sees body A, shouldn't this be able to test the hypothesis that body A radiates independently of body B by conducting the test with and without body B present?

Sincerely,
Dol

5. (continuation of my previous comment)

Thinking about it, an even better experiment would be the same setting. And then to introduce first a body B with temperature T_B >> T_A. And then repeat the experiment with T_B << T_A.

Sincerely,
Dol

6. Claes,
Is the silvering in the thermos flask not supporting 2? Not sure how efficient silvering works but if through reflection body B may look as warm as body A is this not supporting the idea that radiation goes in all directions equally? Body B looks warmer as it reflects incoming radiation and not due to changes in temperature difference between A and B
Lars

7. I do not see the beauty of this exp. A is radiating to a background of lower temp
and thus cools off, while B keeps it's temp. Where is the beef?

8. Why not let the surroundings be body B and screen the detector so it only sees radiation from body A?

9. Well, lets alter the situation a bit. While A is still put into the freezer, B is put outside the room where the surrounding temperature is at 0C. Then they both will have to cool, and after sufficient time they both will enter 0C. (And A will still reach -20). Will the rate of temperature change be the same for both balls, or do you think, as I do, that the rate between 20C and 0C will be higher for ball A, A will reach 0C sooner. My explaination for this will be that B during the process will absorb more heat radiation from the surroundings than A. Do you have an other opinion?

I think this example has enough beauty to deserve an answer based upon physics rather than sighing over the estetichs. The point is I am really trying to understand your point of view here, and I am trying to clarify the situation with some basic question.

10. B will not absorb more heat than A, just radiate away less because the background of B is warmer than that of A. Elementary heat transfer according to True-SB. No reason to refer to any False-SB.

11. I'm looking at your Computational Blackbody Radiation-document and are trying to understand you arguments for arriving at the Plank's law and there is one thing I really need your explanation and justification for.

In section 7.1, p. 181 (PDF p. 17) you write "viewing physics as a form of analog computation with finite precision.

First of all, can you elaborate exactly what you mean with physics in this context?

And secondly what is your experimental/theoretical arguments that this is the right view?

Sincerely,
Dol

12. One more question, are you working with c=1 units?

Sincerely,
Dol

13. It is natural to view physical processes as cause-effect systems or input-output
mechanism and the procedure from input to output can be viewed as a form of (analog) computation. Finite precision computation can be seen as a small dissipative diffusion (averaging-smoothing) effect causing damping or cut-off of
high frequencies which is seen in blackbody radition.

14. Claes wrotez
Finite precision computation can be seen as a small dissipative diffusion (averaging-smoothing) effect causing damping or cut-off of high frequencies which is seen in blackbody radition.

But this then looks like an ad-hoc justification to get the right spectra.

Is there any ab initio justification for taking this view?

Sincerely,
Dol

15. Thank's Claes for a straight forward answer. It's a bit hard to imagine that a body not will radiate against a warmer body. Your thoughts are challenging. I'll reread some of your post on the topic, and continue to follow your blog.

W.k.r.

16. Good luck Trond.

17. IceskaterFinland says, Radiation emitted from materials results in cooling where the radiation is emitted.

You can easily show that a warm cooling objects surface rises in temperature when the cooling rate is reduced by the presence of an introduced midrange temperature object or if you allow the hotter object to increase the temperature of the nearby cold object.

No laws of thermodynamics are broken by this

The backradiation is part of a cooling process.

The second law of thermodynamics applies to the impossibility of constructing heat engines that use cooling to do work. Obviously no work is being done since both objects are cooling when viewed in terms of their total heat content.


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# PUMac 2015 Algebra A Problem 3

Algebra Level 4

Find the sum of the non-repeated roots of the polynomial $$P(x) = x^6-5x^5-4x^4-5x^3+8x^2+7x+7$$.

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# is there a formula for an innovation outlier regressor in time series intervention analysis

For an innovative outlier, the equations I got from a paper by Tsay, http://docplayer.net/12080848-Outliers-level-shifts-and-variance-changes-in-time-series.html, are these:

$$y_t = f(t) +z(t)\\ f(t) = w_0 \frac{\theta(B)}{\phi(B)}\epsilon_t^d\\ \epsilon_t^d = \begin{array}{cc} \{ & \begin{array}{cc} 1 & t = d \\ 0 & t \neq d \end{array} \end{array}$$

and, $$\frac{\theta(B)}{\phi(B)}$$ are polynomials in a backshift operator, representing an arma model for a time series $$z(t)$$, while $$y_t$$ is a new time series with the innovation taken into account.

I am trying to figure out how to work this into a regressor for different values of $$\frac{\theta(B)}{\phi(B)}$$.

For example, I tried to work out this simple case, but it is a ton of work.

if $$\frac{\theta(B)}{\phi(B)} = \frac{1}{1-\phi_1B - \phi_2B^2}$$:

$$y_t = \frac{w_0 \epsilon_t ^ d}{1 - \phi_1B - \phi_2B ^ 2} + z_t\\ (1 - \phi_1 B - \phi_2 B ^ 2) y_t = w_0 \epsilon_t ^ d + (1-\phi_1 B - \phi_2 B ^ 2) z_t$$ case 1:

$$t < d \rightarrow y_d = z_d$$

case 2:

$$t = d : \\ y_d - \phi_1 y_{d - 1} - \phi_2 y_{d - 2} = w_0 \epsilon_d^d + (1-\phi_1 B - \phi_2 B ^ 2) z_t \rightarrow \\ \rightarrow y_d - \phi_1 z_{d - 1} - \phi_2 z_{d - 2} = w_0 + (1-\phi_1 B - \phi_2 B ^ 2) z_t \rightarrow \\ \rightarrow y_d = w_0 + z_d$$

case 3:

$$t = d + 1\\ y_{d + 1} - \phi_1 y_{d} - \phi_2 y_{d - 1} = w_0 + (1-\phi_1 B - \phi_2 B ^ 2) z_{d + 1} \rightarrow \\ \rightarrow y_{d+1} - \phi_1 (w_0 + z_d) - \phi_2 z_{d - 1} = (1-\phi_1 B - \phi_2 B ^ 2) z_{d + 1} \rightarrow \\ \rightarrow y_{d+1} = \phi_1 w_0 + z_{d+1}$$

case 4:

$$t = d + 2\\ y_{d + 2} - \phi_1 y_{d+1} - \phi_2 y_{d} = w_0 + (1-\phi_1 B - \phi_2 B ^ 2) z_{d + 2} \rightarrow \\ \rightarrow y_{d + 2} - \phi_1 (\phi_1 w_0 + z_{d+1}) - \phi_2 (w_0 + z_d) = (1-\phi_1 B - \phi_2 B ^ 2) z_{d + 2} \rightarrow \\ \rightarrow y_{d+2} = \phi_1 ^2 w_0 + \phi_2 w_0 + z_{d+2}$$

case 5:

$$t = d + 3 \rightarrow y_{d+3} = \phi_1 ^ 3 w_0 + 2 \phi_1 \phi_2 w_0 + z_{d+3}$$

case 6:

$$t = d + 4 \rightarrow y_{d+4} = \phi_1 ^ 4 w_0 + 3 \phi_1^2 \phi_2 w_0 + \phi_2 ^2 w_0 + z_{d+3}$$

My question is... is there some formula, where given any arma model $$\theta(B)$$ and $$\phi(B)$$, I can create a regressor with the terms solved above?

For example, in the above case I would have a regressor with $$0$$'s until $$t = d$$, and then the above terms :

$$[0, 0, ..., 0, 1, \phi_1, \phi_1 ^2 + \phi_2, \phi_1 ^ 3 + 2 \phi_1 \phi_2, \phi_1 ^ 4 + 3 \phi_1^2 \phi_2 + \phi_2 ^2, ...]$$

The idea here is that using this regressor, and knowledge of the ARMA terms, I could find $$w_0$$ with a linear regression.

Or, is there no real formula, and I just have to write something recursive in R?