text:
• ### First Sunyaev-Zel'dovich mapping with the NIKA2 camera: Implication of cluster substructures for the pressure profile and mass estimate(1712.09587)

April 6, 2018 astro-ph.CO
The complete characterization of the pressure profile of high-redshift galaxy clusters, from their core to their outskirts, is a major issue for the study of the formation of large-scale structures. It is essential to constrain a potential redshift evolution of both the slope and scatter of the mass-observable scaling relations used in cosmology studies based on cluster statistics. In this paper, we present the first thermal Sunyaev-Zel'dovich (tSZ) mapping of a cluster from the sample of the NIKA2 SZ large program that aims at constraining the redshift evolution of cluster pressure profiles and the tSZ-mass scaling relation. We have observed the galaxy cluster PSZ2 G144.83+25.11 at redshift $z=0.58$ with the NIKA2 camera, a dual-band (150 and 260 GHz) instrument operated at the IRAM 30-meter telescope. We identify a thermal pressure excess in the south-west region of PSZ2 G144.83+25.11 and a high redshift sub-millimeter point source that affect the intracluster medium (ICM) morphology of the cluster. The NIKA2 data are used jointly with tSZ data acquired by the MUSTANG, Bolocam and $Planck$ experiments in order to non-parametrically set the best constraints on the electronic pressure distribution from the cluster core ($\rm{R} \sim 0.02 \rm{R_{500}}$) to its outskirts ($\rm{R} \sim 3 \rm{R_{500}}$). We investigate the impact of the over-pressure region on the shape of the pressure profile and on the constraints on the integrated Compton parameter $\rm{Y_{500}}$. A hydrostatic mass analysis is also performed by combining the tSZ-constrained pressure profile with the deprojected electronic density profile from XMM-$Newton$. This allows us to conclude that the estimates of $\rm{Y_{500}}$ and $\rm{M_{500}}$ obtained from the analysis with and without masking the disturbed ICM region differ by 65 and 79% respectively. (abridged)
• ### Recovering galaxy cluster gas density profiles with XMM-Newton and Chandra(1709.06570)

Sept. 19, 2017 astro-ph.CO
We examine the reconstruction of galaxy cluster radial density profiles obtained from Chandra and XMM X-ray observations, using high quality data for a sample of twelve objects covering a range of morphologies and redshifts. By comparing the results obtained from the two observatories and by varying key aspects of the analysis procedure, we examine the impact of instrumental effects and of differences in the methodology used in the recovery of the density profiles. We find that the final density profile shape is particularly robust. We adapt the photon weighting vignetting correction method developed for XMM for use with Chandra data, and confirm that the resulting Chandra profiles are consistent with those corrected a posteriori for vignetting effects. Profiles obtained from direct deprojection and those derived using parametric models are consistent at the 1% level. At radii larger than $\sim$6", the agreement between Chandra and XMM is better than 1%, confirming an excellent understanding of the XMM PSF. We find no significant energy dependence. The impact of the well-known offset between Chandra and XMM gas temperature determinations on the density profiles is found to be negligible. However, we find an overall normalisation offset in density profiles of the order of $\sim$2.5%, which is linked to absolute flux cross-calibration issues. As a final result, the weighted ratios of Chandra to XMM gas masses computed at R2500 and R500 are r=1.03$\pm$0.01 and r=1.03$\pm$0.03, respectively. Our study confirms that the radial density profiles are robustly recovered, and that any differences between Chandra and XMM can be constrained to the $\sim$ 2.5% level, regardless of the exact data analysis details. These encouraging results open the way for the true combination of X-ray observations of galaxy clusters, fully leveraging the high resolution of Chandra and the high throughput of XMM.
• ### High-resolution SZ imaging of clusters of galaxies with the NIKA2 camera at the IRAM 30-m telescope(1709.01255)

Sept. 7, 2017 astro-ph.CO
The development of precision cosmology with clusters of galaxies requires high-angular resolution Sunyaev-Zel'dovich (SZ) observations. As for now, arcmin resolution SZ observations (e.g. SPT, ACT and Planck) only allowed detailed studies of the intra cluster medium for low redshift clusters (z<0.2). With both a wide field of view (6.5 arcmin) and a high angular resolution (17.7 and 11.2 arcsec at 150 and 260 GHz), the NIKA2 camera installed at the IRAM 30-m telescope (Pico Veleta, Spain), will bring valuable information in the field of SZ imaging of clusters of galaxies. The NIKA2 SZ observation program will allow us to observe a large sample of clusters (50) at redshifts between 0.4 and 0.9. As a pilot study for NIKA2, several clusters of galaxies have been observed with the pathfinder, NIKA, at the IRAM 30-m telescope to cover the various configurations and observation conditions expected for NIKA2.
• ### Mapping the hot gas temperature in galaxy clusters using X-ray and Sunyaev-Zel'dovich imaging(1706.10230)

July 21, 2017 astro-ph.CO
We propose a method to map the temperature distribution of the hot gas in galaxy clusters that uses resolved images of the thermal Sunyaev-Zel'dovich (tSZ) effect in combination with X-ray data. Application to images from the New IRAM KIDs Array (NIKA) and XMM-Newton allows us to measure and determine the spatial distribution of the gas temperature in the merging cluster MACS J0717.5+3745, at $z=0.55$. Despite the complexity of the target object, we find a good morphological agreement between the temperature maps derived from X-ray spectroscopy only -- using XMM-Newton ($T_{\rm XMM}$) and Chandra ($T_{\rm CXO}$) -- and the new gas-mass-weighted tSZ+X-ray imaging method ($T_{\rm SZ}$). We correlate the temperatures from tSZ+X-ray imaging and those from X-ray spectroscopy alone and find that $T_{\rm SZ}$ is higher than $T_{\rm XMM}$ and lower than $T_{\rm CXO}$ by $\sim 10\%$ in both cases. Our results are limited by uncertainties in the geometry of the cluster gas, contamination from kinetic SZ ($\sim 10\%$), and the absolute calibration of the tSZ map ($7\%$). Investigation using a larger sample of clusters would help minimise these effects.
• ### Mapping the kinetic Sunyaev-Zel'dovich effect toward MACS J0717.5+3745 with NIKA(1606.07721)

Dec. 8, 2016 astro-ph.CO
Measurement of the gas velocity distribution in galaxy clusters provides insight into the physics of mergers, through which large scale structures form in the Universe. Velocity estimates within the intracluster medium (ICM) can be obtained via the Sunyaev-Zel'dovich (SZ) effect, but its observation is challenging both in term of sensitivity requirement and control of systematic effects, including the removal of contaminants. In this paper we report resolved observations, at 150 and 260 GHz, of the SZ effect toward the triple merger MACS J0717.5+3745 (z=0.55), using data obtained with the NIKA camera at the IRAM 30m telescope. Assuming that the SZ signal is the sum of a thermal (tSZ) and a kinetic (kSZ) component and by combining the two NIKA bands, we extract for the first time a resolved map of the kSZ signal in a cluster. The kSZ signal is dominated by a dipolar structure that peaks at -5.1 and +3.4 sigma, corresponding to two subclusters moving respectively away and toward us and coincident with the cold dense X-ray core and a hot region undergoing a major merging event. We model the gas electron density and line-of-sight velocity of MACS J0717.5+3745 as four subclusters. Combining NIKA data with X-ray observations from XMM-Newton and Chandra, we fit this model to constrain the gas line-of-sight velocity of each component, and we also derive, for the first time, a velocity map from kSZ data (i.e. that is model-dependent). Our results are consistent with previous constraints on the merger velocities, and thanks to the high angular resolution of our data, we are able to resolve the structure of the gas velocity. Finally, we investigate possible contamination and systematic effects with a special care given to radio and submillimeter galaxies. Among the sources that we detect with NIKA, we find one which is likely to be a high redshift lensed submillimeter galaxy.
• ### Resolving galaxy cluster gas properties at z~1 with XMM-Newton and Chandra(1610.01899)

Nov. 25, 2016 astro-ph.CO
We present a pilot X-ray study of the five most massive ($M_{500}>5 \times 10^{14} M_{\odot}$), distant (z~1), galaxy clusters detected via the Sunyaev-Zeldovich effect. We optimally combine XMM-Newton and Chandra X-ray observations by leveraging the throughput of XMM to obtain spatially-resolved spectroscopy, and the spatial resolution of Chandra to probe the bright inner parts and to detect embedded point sources. Capitalising on the excellent agreement in flux-related measurements, we present a new method to derive the density profiles, constrained in the centre by Chandra and in the outskirts by XMM. We show that the Chandra-XMM combination is fundamental for morphological analysis at these redshifts, the Chandra resolution being required to remove point source contamination, and the XMM sensitivity allowing higher significance detection of faint substructures. The sample is dominated by dynamically disturbed objects. We use the combined Chandra-XMM density profiles and spatially-resolved temperature profiles to investigate thermodynamic quantities including entropy and pressure. From comparison of the scaled profiles with the local REXCESS sample, we find no significant departure from standard self-similar evolution, within the dispersion, at any radius, except for the entropy beyond 0.7$R_{500}$. The baryon mass fraction tends towards the cosmic value, with a weaker dependence on mass than observed in the local Universe. We compare with predictions from numerical simulations. The present pilot study demonstrates the utility and feasibility of spatially-resolved analysis of individual objects at high-redshift through the combination of XMM and Chandra observations. Observations of a larger sample will allow a fuller statistical analysis to be undertaken, in particular of the intrinsic scatter in the structural and scaling properties of the cluster population. (abridged)
• ### High angular resolution SZ observations with NIKA and NIKA2(1605.09549)

May 31, 2016 astro-ph.CO
NIKA2 (New IRAM KID Arrays) is a dual band (150 and 260 GHz) imaging camera based on Kinetic Inductance Detectors (KIDs) and designed to work at the IRAM 30 m telescope (Pico Veleta, Spain). Built on the experience of the NIKA prototype, NIKA2 has been installed at the 30 m focal plane in October 2015 and the commissioning phase is now ongoing. Through the thermal Sunyaev-Zeldovich (tSZ) effect, NIKA2 will image the ionized gas residing in clusters of galaxies with a resolution of 12 and 18 arcsec FWHM (at 150 and 260 GHz, respectively). We report on the recent tSZ measurements with the NIKA camera and discuss the future objectives for the NIKA2 SZ large Program, 300h of observation dedicated to SZ science. With this program we intend to perform a high angular resolution follow-up of a cosmologically-representative sample of clusters belonging to SZ catalogues, with redshift greater than 0.5. The main output of the program will be the study of the redshift evolution of the cluster pressure profile as well as that of the scaling laws relating the cluster global properties.
• ### High-resolution tSZ cartography of clusters of galaxies with NIKA at the IRAM 30-m telescope(1602.07941)

Feb. 25, 2016 astro-ph.CO
The thermal Sunyaev-Zeldovich effect (tSZ) is a powerful probe to study clusters of galaxies and is complementary with respect to X-ray, lensing or optical observations. Previous arcmin resolution tSZ observations ({\it e.g.} SPT, ACT and Planck) only enabled detailed studies of the intra-cluster medium morphology for low redshift clusters ($z < 0.2$). Thus, the development of precision cosmology with clusters requires high angular resolution observations to extend the understanding of galaxy cluster towards high redshift. NIKA2 is a wide-field (6.5 arcmin field of view) dual-band camera, operated at $100 \ {\rm mK}$ and containing $\sim 3300$ KID (Kinetic Inductance Detectors), designed to observe the millimeter sky at 150 and 260 GHz, with an angular resolution of 18 and 12 arcsec respectively. The NIKA2 camera has been installed on the IRAM 30-m telescope (Pico Veleta, Spain) in September 2015. The NIKA2 tSZ observation program will allow us to observe a large sample of clusters (50) at redshift ranging between 0.5 and 1. As a pathfinder for NIKA2, several clusters of galaxies have been observed at the IRAM 30-m telescope with the NIKA prototype to cover the various configurations and observation conditions expected for NIKA2.
• ### High angular resolution Sunyaev-Zel'dovich observations of MACS J1423.8+2404 with NIKA: Multiwavelength analysis(1510.06674)

Feb. 15, 2016 astro-ph.CO
The prototype of the NIKA2 camera, NIKA, is an instrument operating at the IRAM 30-m telescope, which can observe simultaneously at 150 and 260GHz. One of the main goals of NIKA2 is to measure the pressure distribution in galaxy clusters at high resolution using the thermal SZ (tSZ) effect. Such observations have already proved to be an excellent probe of cluster pressure distributions even at high redshifts. However, an important fraction of clusters host submm and/or radio point sources, which can significantly affect the reconstructed signal. Here we report on <20" resolution observations at 150 and 260GHz of the cluster MACSJ1424, which hosts both radio and submm point sources. We examine the morphology of the tSZ signal and compare it to other datasets. The NIKA data are combined with Herschel satellite data to study the SED of the submm point source contaminants. We then perform a joint reconstruction of the intracluster medium (ICM) electronic pressure and density by combining NIKA, Planck, XMM-Newton, and Chandra data, focusing on the impact of the radio and submm sources on the reconstructed pressure profile. We find that large-scale pressure distribution is unaffected by the point sources because of the resolved nature of the NIKA observations. The reconstructed pressure in the inner region is slightly higher when the contribution of point sources are removed. We show that it is not possible to set strong constraints on the central pressure distribution without accurately removing these contaminants. The comparison with X-ray only data shows good agreement for the pressure, temperature, and entropy profiles, which all indicate that MACSJ1424 is a dynamically relaxed cool core system. The present observations illustrate the possibility of measuring these quantities with a relatively small integration time, even at high redshift and without X-ray spectroscopy.
• ### A weak lensing analysis of the PLCK G100.2-30.4 cluster(1505.02887)

May 13, 2015 astro-ph.CO
We present a mass estimate of the Planck-discovered cluster PLCK G100.2-30.4, derived from a weak lensing analysis of deep SUBARU griz images. We perform a careful selection of the background galaxies using the multi-band imaging data, and undertake the weak lensing analysis on the deep (1hr) r-band image. The shape measurement is based on the KSB algorithm; we adopt the PSFex software to model the Point Spread Function (PSF) across the field and correct for this in the shape measurement. The weak lensing analysis is validated through extensive image simulations. We compare the resulting weak lensing mass profile and total mass estimate to those obtained from our re-analysis of XMM-Newton observations, derived under the hypothesis of hydrostatic equilibrium. The total integrated mass profiles are in remarkably good agreement, agreeing within 1$\sigma$ across their common radial range. A mass $M_{500} \sim 7 \times 10^{14} M_\odot$ is derived for the cluster from our weak lensing analysis. Comparing this value to that obtained from our reanalysis of XMM-Newton data, we obtain a bias factor of (1-b) = 0.8 $\pm$ 0.1. This is compatible within 1$\sigma$ with the value of (1-b) obtained by Planck Collaboration XXIV from their calibration of the bias factor using newly-available weak lensing reconstructed masses.
• ### Discovery of large-scale diffuse radio emission and of a new galaxy cluster in the surroundings of MACSJ0520.7-1328(1402.4436)

Feb. 18, 2014 astro-ph.CO, astro-ph.GA
We report the discovery of large-scale diffuse radio emission South-East of the galaxy cluster MACS J0520.7-1328, detected through high sensitivity Giant Metrewave Radio Telescope 323 MHz observations. This emission is dominated by an elongated diffuse radio source and surrounded by other features of lower surface brightness. Patches of these faint sources are marginally detected in a 1.4 GHz image obtained through a re-analysis of archival NVSS data. Interestingly, the elongated radio source coincides with a previously unclassified extended X-ray source. We perform a multi-wavelength analysis based on archival infrared, optical and X-ray Chandra data. We find that this source is a low-temperature (~3.6 keV) cluster of galaxies, with indications of a disturbed dynamical state, located at a redshift that is consistent with the one of the main galaxy cluster MACS J0520.7-132 (z=0.336). We suggest that the diffuse radio emission is associated with the non-thermal components in the intracluster and intergalactic medium in and around the newly detected cluster. We are planning deeper multi-wavelength and multi-frequency radio observations to accurately investigate the dynamical scenario of the two clusters and to address more precisely the nature of the complex radio emission.


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### Four all new exotic particles

LHCb has recently observed four new exotic-like particles in the decay of the $B^+$:
he properties of these structures are consistent with their interpretation as four-quark particles, which are considered as "exotic", (hence the "exotic-like" name in the title), although the details of the four quark $c{\bar c}s{\bar s}$ binding mechanism is still under discussion.
Read also paper 1 and paper 2

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Limits of form 0 x 0 question!

1. Evaluate the limit or determine that it does not exist. (Not sure if I did this correctly or how to do it if I'm wrong)

2. lim(x⇒π/2) tan^2(x) ln(sin(x))
-In text form: the limit as x goes to pi over two, of tangent squared of x, times the natural log of sine of x.

3. My attempt was to change tan^2[x] into sin^2[x]/cos^2[x], where the limit would then give me 0/0, I then used L'Hospital's rule to get the numerator (not sure if I did the derivative correctly with the product rule):
2cos[2x]ln[sin(x)] + -csc^2[x]sin^2[x]
and the denominator as -2cos[2x]
This gave me -1/0 which would mean the limit is zero, and does exist.

Dick
Homework Helper
Ok, so the limit is sin(x)^2*ln(sin(x))/cos(x)^2, where the second factor is the 0/0 part. Just apply l'Hopital to that. What's the derivative of the numerator and the denominator?

Mark44
Mentor
1. Evaluate the limit or determine that it does not exist. (Not sure if I did this correctly or how to do it if I'm wrong)

2. lim(x⇒π/2) tan^2(x) ln(sin(x))
-In text form: the limit as x goes to pi over two, of tangent squared of x, times the natural log of sine of x.

3. My attempt was to change tan^2[x] into sin^2[x]/cos^2[x], where the limit would then give me 0/0, I then used L'Hospital's rule to get the numerator (not sure if I did the derivative correctly with the product rule):
2cos[2x]ln[sin(x)] + -csc^2[x]sin^2[x]
and the denominator as -2cos[2x]
This gave me -1/0 which would mean the limit is zero, and does exist.

You have made a number of mistakes here.
tan^2(x) approaches infinity as x approaches pi/2. It is not of the form [0/0].
A limit of the form -1/0 does not mean that the limit is zero, and BTW, zero does exist.

In your original limit, tan^2(x) * ln(sin(x) is the indeterminate form [infinity * 0]. If you rewrite your limit as ln(sin(x))/cot^2(x), this is of the form [0/0], which you can use L'Hopital's Rule on. Remember when you're using L'Hopital's Rule, you are taking the derivatives separately of the numerator an denominator. It's not the same as the quotient rule for derivative.

2cos[2x]ln[sin(x)] + -csc^2[x]sin^2[x]
and the denominator as -2cos[2x]

I did make a mistake in that -1/0 is zero, that is not true, I was thinking 0/-1. I think if I represented tan^2[x] as 1/cot^2[x] that would give me the form of 0/0 correct? Now I think I can do the derivative of both the top and bottom. My first attempt in changing tangent to sines and cosines would take me down a very un-fun and endless path of derivation.

Mark44
Mentor
I did make a mistake in that -1/0 is zero, that is not true, I was thinking 0/-1. I think if I represented tan^2[x] as 1/cot^2[x] that would give me the form of 0/0 correct?
Yes
Now I think I can do the derivative of both the top and bottom. My first attempt in changing tangent to sines and cosines would take me down a very un-fun and endless path of derivation.
Note that using the technique Dick suggested is not a complicated as you're making it out to be. You only have to use L'H's rule on this part: ln(sin(x))/cos(x)^2, since the limit of the other part is 1.

Alright, I used Dick's method and found the derivative of both the top and bottom which turn out to be:
-csc^2[x] (divided by)
-2cos[2x]
So now I have the form 0/2 correct? Thus the limit is zero?

Mark44
Mentor
No, that's not right.
Starting with ln(sin(x))/cos^2(x), how did you get -csc^2(x)/(-2cos(2x))?

Let's take it in parts:
d/dx(ln(sin(x)) = ?
d/dx(cos^2(x)) = ?

No, that's not right.
Starting with ln(sin(x))/cos^2(x), how did you get -csc^2(x)/(-2cos(2x))?

Let's take it in parts:
d/dx(ln(sin(x)) = ?
d/dx(cos^2(x)) = ?

True, I derived the numerator twice, which is how I got -csc^2(x). (My bad)
After actually writing this down on a piece of paper:
d/dx(ln(sin(x)) is simply cot(x)
d/dx(cos^2(x)) is -2cos(x)sin(x)
but shouldn't the denominator be:
d/dx(cot^2(x)) ? Because tan^2(x) is 1/cot^2(x)

diazona
Homework Helper
but shouldn't the denominator be:
d/dx(cot^2(x)) ? Because tan^2(x) is 1/cot^2(x)
No - as has already been said, the limit of the sin^2(x) part is 1. So you don't need to involve that in the computation with L'Hôpital's rule. And once you take out the sin^2(x) part, what you're left with is ln(sin(x)) in the numerator and cos^2(x) in the denominator.

Dick
Homework Helper
True, I derived the numerator twice, which is how I got -csc^2(x). (My bad)
After actually writing this down on a piece of paper:
d/dx(ln(sin(x)) is simply cot(x)
d/dx(cos^2(x)) is -2cos(x)sin(x)
but shouldn't the denominator be:
d/dx(cot^2(x)) ? Because tan^2(x) is 1/cot^2(x)

You can do it that way. I was just suggesting you split the sin^2(x) off since you know the limit of that is 1 and just concentrate on the ln(sin(x))/cos^2(x) part. You can really do it either way. But when you get to the l'Hopital's result change everything into sines and cosines and see what cancels.

Ah... Ok now I see where you both are coming from. After doing all of that and cancelling sines and cosines as suggested, I got an anwer of -1/2.

Dick


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SectionList
allroots · append · children · printnames · remove · subtree · unique · wholetree

# SectionList¶↑

class SectionList
Syntax:
sl = new SectionList()
Description:
Class for creating and managing a list of sections

SectionList.append()
Syntax:
sl.append()
Description:
append the currently accessed section to the list

SectionList.remove()
Syntax:

n = sl.remove()

n = sl.remove(sectionlist)

Description:
remove the currently accessed section from the list If the argument is present then all the sections in sectionlist are removed from sl. Returns the number of sections removed.

SectionList.children()
Syntax:
sl.children()
Description:
Appends the sections connected to the currently accessed section. Note that this includes children connected at position 0 of parent.

SectionList.subtree()
Syntax:
sl.subtree()
Description:
Appends the subtree of the currently accessed section (including that one).

SectionList.wholetree()
Syntax:
sl.wholetree()
Description:
Appends all sections which have a path to the currently accessed section (including the currently accessed section). The section list has the important property that the sections are in root to leaf order.

SectionList.allroots()
Syntax:
sl.allroots()
Description:
Appends all the root sections. Root sections have no parent section. The number of root sections is the number of real cells in the simulation.

SectionList.unique()
Syntax:
n = sl.unique()
Description:
Removes all duplicates of sections in the SectionList. I.e. ensures that no section appears more than once. Returns the number of sections references that were removed.

SectionList.printnames()
Syntax:
.printnames()
Description:

print the names of the sections in the list.

The normal usage of a section list involves efficiently iterating over all the sections in the list with forsec sectionlist {statement}


text:
2016.06 Release

The 2016.06 release is available on GitHub.

An updated Windows installer and installation guide (PDF) may be via the Nuand Support page.
(more…)

Software defined radios are capable of a surprising number of things. After nearly a year of development, we have developed a method of increasing the bladeRF’s spectrum analysis bandwidth up to 124MHz. This software-only change takes advantage of the bladeRF’s advanced hardware design, and requires only a few additional changes to osmocom-sdr, and GQRX. With 96MHz it is possible to monitor the entirety of the 2.4GHz ISM band and not miss any Bluetooth or WiFi frames.

So how does this work? The bladeRF features a few HDL accelerators that allow it to calibrate and tune the underlying RF front-end several thousand times a second. By performing these tasks so quickly, the bladeRF can take in digitize and stitch together up to 200MHz of additional bandwidth.

How do I get this running? You will have to compile and use gr-osmosdr and GQRX from the Nuand branches, https://github.com/Nuand/gr-osmosdr , and https://github.com/Nuand/gqrx .

$cd ~/software/gnuradio-build/$ wget https://nuand.com/downloads/expiremental-gr-osmosdr.git.tar
$tar xfz experimental-gr-osmosdr.git.tar$ cd ~/software/gnuradio-builds/gr-osmosdr
$mkdir build$ cd build
$cmake -DCMAKE_INSTALL_PREFIX=/opt/gnuradio-3.7.9.2 ../$ make -j8 && sudo make install && sudo ldconfig

To compile and run GQRX

$cd ~/software/gnuradio-build/$ wget https://nuand.com/downloads/expiremental-gqrx.git.tar
$tar xfz experimental-gqrx.git.tar$ cd gqrx
$mkdir build$ cd build
$cmake -DCMAKE_INSTALL_PREFIX=/opt/gnuradio-3.7.9.2 ../$ make -j8 && sudo make install && sudo ldconfig

Monitor the entire 2.4GHz band with one radio.

NB: Screenshots showing 96MHz width represent a configured value below the maximum of 124MHz.

Source release

We are releasing the source to a high performance hardware accelerated ADS-B decoder that is compatible with the bladeRF. The decoder, and all relevant VHDL, C, and MATLAB design files are available. The decoder has a range of >270miles, can resolve many bit errors, and correct for packet collisions; these features that have only been available in commercial decoders until now. The decoder runs on the FPGA instead of the CPU, meaning that even a Raspberry Pi is able to handle the decoder’s maximum performance. The decoder is compatible with the dump1090 visualizer, however it simply passes fully decoded hex packets to dump1090 for displaying flights.

The source code and documentation are being released in conjunction with a multi-part series that explains the applied DSP theory and FPGA implementation of the ADS-B decoder.

The decoder runs on any bladeRF including the bladeRF x40 and bladeRF x115. Any low noise amplifier works with the bladeRF, however there is a bladeRF specific LNA available, the XB-300.

This multi part article will describe the design decisions and implementation of a Software Defined Radio based receiver. The article will focus on ADS-B and Mode-S.

ADS-B, or Automatic Dependent Surveillance – Broadcast, is a wireless protocol used by commercial and private aircraft to identify themselves and broadcast their location in realtime. ADS-B is the digital packet format that sits on top of Mode S, the RF standard ADS-B is broadcasted over.

Why accelerate modems in hardware (with VHDL)?

Hardware acceleration allows for efficient, real-time processing of signals. An FPGA implementation of an algorithm can be designed so that a pipelined design can process a sample per clock cycle. This is not to say that a sample can be processed within one clock cycle of it arriving but rather the pipelined architecture will compute one sample per clock cycle after the pipeline fills up.

Although FPGAs have slower clock rates than processors, FPGAs can massively parallelize tasks. For applied DSP tasks, an FPGA running at 80MHz routinely outperforms an Intel i7 running at 3.5GHz. These capabilities are similar to GPGPUs. A modern GPU running CUDA or OpenCL can outperform a processor at number crunching because of the large number of parallel arithmetic cores in a GPU. So why use an FPGA for applied DSP? One main reason is cost, even the most inexpensive desktop processor setup will cost at least \$150, whereas FPGAs can achieve a comparable performance for a fifth of the price. Additionally, a single chip is much smaller, more low power and power efficient than a complicated GPGPU desktop setup. Although GPGPU solutions in the same price range as FPGAs contains far more on-chip arithmetic processing power than an FPGA, FPGAs do have an added advantage. FPGAs have very a flexible programmable logic fabric that allows for clock-cycle accurate control of the FPGA’s resources. An effective engineer can utilize that level of control to do more with an FPGA running at 80MHz than a GPGPU. In my experience, GPGPUs are better suited for applications where raw computation power is needed and considerations like data latency, processing latency, and power efficiency do not matter.

What design flow is recommended?

Our preferred design flow is to model and simulate in MATLAB, implement and validate the modem using fixed-point arithmetic in C, and ultimately use the C model as a reference to implement the modem in VHDL.

MATLAB is a great tool for prototyping ideas very quickly. A few lines of Matlab can easily capture the essence of an idea that could take hours to implement in C. The goal of the ASD-B core MATLAB script is to simulate the end to end performance of the ADS-B waveform and observe how the modem reacts to expected channel effects (such as fading), and receiver phenomenon (such as DC offsets, clipping, quantization).

It is in this stage of the design that the acquisition, equalization, and decoding algorithms are developed.

Many of the channel effects, and receiver phenomenon are random number processes which means the only way to get a good understanding for the modem’s performance is to simulate many packets with similar perturbations. Some parameters such SNR, and multipath require sweeping, for example sweeping SNR from -5 dB to 3dB with 0.25dB increments is a good idea when working with a waveform that necessitates 1dB of SNR. Simulations where the SNR is swept across such a range is usually referred to as a BER (Bit Error Rate) curve and is one of the fundamental ways of characterizing the performance of a receiver.

Once the receiver architecture is validated by good simulated BER and PER (Packet Error Rate) performance, it is time to implement the modem. Going to VHDL may seem appealing at this point however it is much more time efficient to implement the modem in C. The MATLAB model uses floating point arithmetic and most of the operations are performed on vectors whereas the final VHDL implementation will be a pipelined fixed point arithmetic implementation. There are far too many design decisions that need to be made and validated in going from MATLAB to VHDL, a good intermediary step is to implement the modem in C. C provides 64bit fixed point data types and arithmetic operations, complements of the underlying x86 processor, that perfectly emulate how the VHDL implementation will work. The C implementation will work as an easy to debug implementation and good reference point for how the VHDL should behave.

The final translation to VHDL from a fixed-point C implementation is not without challenges. The greatest challenge in implementing the modem in VHDL is, as almost always, meeting timing closure. Almost all additions can fit in one clock cycle. Some multiplications can fit in one clock system, others may not. And almost no divisions can happen in less than N/2 clock cycles (where N is the number of bits in the dividend). These simple heuristics can be used as a rule of thumb for designing the arithmetic pipelines but there will still likely be timing problems that will have to be resolved.

Stage Pros Cons Notes
MATLAB Very efficient way of prototyping algorithms Many existing DSP and Telecomm tools Not real-time Can use a lot of RAM and CPU time The receiver architecture and techniques should be known before progressing.
C Can be real-time Memory and CPU efficient Provides bit level accurate simulations with VHDL Memory management, file IO, and all signal processing functions have to be implemented The sample-based fixed point implementation should be designed here. The modem should be able to match the vector-based floating point MATLAB reference modem.
VHDL Very fast, efficient, and customizable Very hard to debug interactively Design is no longer entirely a digital design, some physical effects of silicon can cause the design to misbehave Design defensively against operations that might cause problems with timing closure

The next article will begin diving in to the MATLAB code.


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# Solving differential equation using power series representation

1. Apr 28, 2010

### thepatient

1. The problem statement, all variables and given/known data
The problem is:
(x2 - 4) y′′ + 3xy′ + y = 0, y(0) = 4, y′(0) = 1

2. Relevant equations
Existence of power series:
y = $$\sum c(x-x0)^n$$

or

y = (x-x0)^r$$\sum c(x-x0)^n$$

3. The attempt at a solution
I know the point x=2 is an ordinary point of the differential equation, since:
(x^2 - 4) y′′ + 3xy′ + y = 0

y′′ + 3xy′/(x^2-4) + y/(x^2-4)=0

P(x) = 3x/(x^2-4) Q(x) = 1/(x^2 - 4)

x can't equal to +-2.

But then the thing is that, our teacher has only given us problems where 0 is the regular singular point, but it seems like in this problem has x=+-2 as the regular singular points. Would that mean that there exists two li solutions in a form of a power series where:

y = (x-2)^r$$\sum c(x-2)^n$$

y = (x+2)^r$$\sum d(x+2)^n$$

(c and d being constants)

Then each one would have a different set of indicial roots which would give two separate r values in which there would be two different series for each y?

Or should I not be doing the method of Frobenius?

I already tried setting up the sum using the method of frobenius using only x=2 as the ordinary point (because I assumed I should only use the positive term), but for the indicial equation, I obtained two indicial roots of r=0,0, which left me thinking maybe I'm doing this completely wrong and don't know what I'm doing. :( This is probably the most confusing part of this class... Can anyone give me some guidance on what I should do? :(
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 29, 2010

### vela

Staff Emeritus
I think you should get the same indicial equation regardless of which point you expand about. I found two different roots to the indicial equation for your differential equation.

The method of Frobenius will get you at least one solution. In certain cases, you'll get two. If you have a double root, you'll get only one solution. If the roots are distinct and differ by a non-integer, you'll get two solutions. If they differ by an integer, the larger root will yield one solution, and the smaller root may or may not yield a solution.

3. Apr 29, 2010

### HallsofIvy

Staff Emeritus
I can see no reason for expanding about x= 2, especially because x= 2 is a singular point. Because you are given initial conditions y(0)= 4 and y'(0)= 1 at x= 0 write your solution as $y= \sum a_nx^n$ and don't worry the fact that about x= 2 and x= -2 are singular points. Of course, $y(0)= a_0= 4$ and $y'(0)= a_1= 1$

The only time you need to use Frobenius is when your initial values are given at a regular singular point.

4. Apr 29, 2010

### thepatient

I think I did it. :] Took me like 5 hours last night, but then I found out the only thing the teacher needed was the recursion formula. I got from the equation, a_n+2 = (a(n) (n+1)/4(n+2). a(0) = 4, a(1) = 1. I took so long because I was trying to find the entire solution using a series, which came out as something like: y = sum((2n)!!/4^n(2n+1)!! or something like that. Two different ones, that one was for the odds, and I had another one similar to the evens.


text:
Winamp & SHOUTcast Forums WA5 crashes instantly when loading a modern Skin

 11th October 2004, 07:12 #41 madcom Junior Member   Join Date: Oct 2004 Posts: 2 Ok, this is what i think about the bug..: I use windows 98 too, i think that hardware and current proccesses/? have nothing do with the bug, in my computer somtimes winamp crashes: "gen_ff.dll" and sometimes the OS kernel("kernel32.dll"). I readed in some german forums that someone had to rename: "tooltips-elements.xml" to anything else to bypass the xml bug. I also had to disable some ModernSkins options to make it run, like: "WindowShade scale follows Normal Window Scale" disable "enable tooltips" < this one is probably the bug. Remember that Windows 98 has not all XP and 2000 window options like "opacity", etc.. My point?: - rename tooltips-elements.xml to tooltips-elements1.xml - disable "enable tooltips" - configure your winamp correctly I still can't make The Default Winamp Modern Skin. run, it crashes kernel32.dll: ------------------------ 015f:bff9d709. Registros: EAX=c002f8d0 CS=015f EIP=bff9d709 EFLGS=00010216 EBX=00000000 SS=0167 ESP=0063fd88 EBP=00640024 ECX=00000000 DS=0167 ESI=00860f80 FS=3787 EDX=7803a6e8 ES=0167 EDI=004542a0 GS=0000 Bytes en CS:EIP: ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ,02x ------------------------- I hope i could be of help. Regards.
 11th October 2004, 22:19 #42 lazerfisk Member   Join Date: Apr 2002 Location: Karlstad, Sweden Posts: 57 Go through your winamp.ini... Try renaming it to winamp2.ini and start it, try to load a modern skin now. This worked on my computer, however my WinAmp has been upgraded since version 2.x so it was huge when i had a look at it. Didn't feel like putting time into tracking it down, but i'm pretty sure it's one of the options in there... Cheers, Christopher
 12th October 2004, 02:41 #43 madcom Junior Member   Join Date: Oct 2004 Posts: 2 renaming winamp.ini to winamp2.ini did not work for me.. the funny thing is.. i can run any modern skin except for the default modern skin.
 27th December 2004, 14:13 #44 scubed Junior Member   Join Date: Dec 2004 Posts: 2 Resource leak I am using Windows 98 SE (2222) and have just installed WinAmp 5.08. I also use LiteStep. I selected the new skin and had the problem already described in this thread. I changed the skin and now WinAmp starts just fine. However, once a song starts playing, it eats all of my system resources rapidly. Quickly rendering the entire machine unusable and forcing me to reboot. System resources starts at 90%. After running WinAmp, it declines to 66% and steadily drops further until hitting 0%. So, WinAmp appears to have a really bad resource leak. I hope that this can be looked into and fixed. Thank you.
 27th December 2004, 15:35 #45 scubed Junior Member   Join Date: Dec 2004 Posts: 2 Clarification This computer used to run WinAmp 2 just fine. When I returned, running WinAmp 2 had a resource leak, so I updated it, but the problem persisted, exhausting resources in only a few minutes. Given that it used to work, the problem is probably not with WinAmp itself directly. Are there any programs that it does not get along with or anything? What libraries does it rely on that might be broken that would cause this to happen?
 28th December 2004, 14:35 #46 Thantelius Junior Member   Join Date: Dec 2004 Posts: 1 Under this path C:\Program Files\Winamp\Plugins i've deleted the files gen_ml.dll and gen_ml.ini and the problem has been solved.
 29th December 2004, 00:47 #47 lazerfisk Member   Join Date: Apr 2002 Location: Karlstad, Sweden Posts: 57 It sounds very odd that the medialibrary should have anything to do with the crashes. I suspect it relates to another plugin on your system. I did a fresh install of WA5 and went from there, and it worked. The media library is the same in the old installation as in the new one. However, my old one was full of various plugins. My best bet is that it's a general purpose plugin that's causing this.
 3rd March 2005, 14:07 #48 leexgx Junior Member   Join Date: Mar 2005 Posts: 1 i been haveing the same problems with my system with WA5 crashing on start (send an report error) some times it start some time it would crash if it did work (only in clssic mode it would start), if i started millk drop then it whould fall over after 1 min or so i had to use the reset button on me shuttle i found out it was my Nvidia GF4 3D prophet II Ultra card that was doing it replaced it with an Nvidia fx5200 and problem has gone away windows XP amd 2800+ nForce 2 mobo fx5200 (<< is now) and with the crative sound card problems it seems to only happen on VIA chip set in my case as from going from VIA to nForce chip set my sound card has been working with 0 problems (nForce 2 rules and so does nForce 4)
 3rd March 2005, 14:58 #49 JonnyMac Moderator     Join Date: Dec 2000 Posts: 14,384 To anyone using the DEE3 plug-in. Conflicts with the plug-in and modern skins have been reported here and here. This issue can occur even though DEE3 is not set as the current/active output plug-in. Please do not PM me for tech support. Any request for tech support through PM will be ignored. Read the Stickies ---> | | | | <--- Knowledge is power
 4th March 2005, 09:30 #50 lazerfisk Member   Join Date: Apr 2002 Location: Karlstad, Sweden Posts: 57 I can confirm that dsp_dee.dll was found in the plugin folder of my previous winamp installation. However, i believe that it was removed to another folder in order to attempt to clean up the install and make the modern skins work No preferences appear to be present in winamp.ini for this plugin. Cheers, Christopher
 4th April 2005, 00:20 #51 MiG29 Junior Member   Join Date: Apr 2005 Posts: 2 Guess i have the same problem.Im using 5.08 ,with Win Me, the system is AMD K6III500, 128 RAM,Vortex 2 sound card and when i switch from classic skin to modern skin(the default) or any other skin it just crashes.I get a blue screen with "Error: 0D : 0000: 0000306A".However ive noticed that i get this thing only after i install P2P programs(i got Kazaa then Bearshare).After i do a system restore(uninstalling the P2P things doesnt help),everything is back to normal.Do these programs come with anything which interfere with cause it worked just fine untill i got these things installed. I havent got any plugins besides the ones which come with
4th April 2005, 01:01   #52
JonnyMac
Moderator

Join Date: Dec 2000
Posts: 14,384
Quote:
 Originally posted by MiG29 Do these programs come with anything which interfere with cause it worked just fine untill i got these things installed.
Yes – Spyware, adware and other junk like that...
File Sharing Programs: Including list of known spyware

 4th April 2005, 09:58 #53 MiG29 Junior Member   Join Date: Apr 2005 Posts: 2 I know about the spyware,adware and other junk.But i had P2P before and nothing ever happend.Now with 5.08 things just go wrong.
 4th April 2005, 14:39 #54 DJ Egg TechoratorWinamp & SHOUTcast Team   Join Date: Jun 2000 Posts: 35,817 Because the malware bundled with those evil p2p apps has gotten worse (ie. more destructive). You get all you deserve by installing them. The problems won't occur if you only use the safe/recommended p2p apps which come with no bundled spyware/parasites/hijackers :/ If you require further help, start your own thread in the Tech Support forum, and include a HijackThis log. Playlist -|- Twitter
7th April 2005, 16:06   #55
lazerfisk
Member

Join Date: Apr 2002
Posts: 57
Quote:
 Originally posted by DJ Egg Because the malware bundled with those evil p2p apps has gotten worse (ie. more destructive). You get all you deserve by installing them. The problems won't occur if you only use the safe/recommended p2p apps which come with no bundled spyware/parasites/hijackers :/ If you require further help, start your own thread in the Tech Support forum, and include a HijackThis log.
Um, a clean install of WinAmp into a new folder does help for me...

 7th April 2005, 18:36 #56 DJ Egg TechoratorWinamp & SHOUTcast Team   Join Date: Jun 2000 Posts: 35,817 Good for you Playlist -|- Twitter
 8th May 2005, 12:32 #57 chic Junior Member   Join Date: May 2005 Posts: 3 Just installed Winamp 5.09 and encountered said problem.. think I'm probably the first for this new version... now fervently wishing I didn't delete my old Winamp installer =( Here goes... AppName: winamp.exe AppVer: 5.0.0.9 ModName: msvcrt.dll ModVer: 7.0.2600.0 Offset: 00033ace System Specs: Windows XP Pro v2002(no SP) P4 2.4GHz NVidia GeForce4 Ti 4200 Avance Ac'97 Audio ------------------ ------------------------- Chronological recap of events that transpired: 1) Uninstalled my old version 5.04 and installed 5.09, but never successful managed to launch it no matter what i tried 2) Was so desperate that i even resorted to deleting Nullsoft registry key under HKLM after i uninstalled & re-installed for the umpteenth time 3) Even went to manually delete winamp file types under "folder options" just to make sure any reference to winamp is completely wiped out before re-installing, but to no avail 4) Installed Winamp Lite in a moment of desperation, & discovered with glee that it works! 5) In a moment of inspiration, decided to attempt 1 last time to reinstall full version with classic skin selected, n presto! winamp was launched successfully. 6) Switched to modern skin, and everything seems to work fine. Utter ecstasy!!! 7) Exited Winamp (with modern skin still intact) and re-launched Winamp - the dreaded error box pops up again! =( Your prompt assistance would be most appreciated. I can't live without my winamp!!!
 9th May 2005, 22:02 #59 chic Junior Member   Join Date: May 2005 Posts: 3 heyz DJ Egg, Thanx for the links to the old versions.. u're a messiah Yeah, I did install Sonic Engine, but think that wasn't of consequence, since winamp (attempts to) launch automatically after the registration window n there was no way i could have done a reboot before starting up winamp for the 1st time. However, just to b sure, i rebooted after i got my 1st crash, but there were no miracles =P My take is that it definitely has to do with some conflict with the modern skin, since the classic skin works juz fine.. Anyway hope u dudes can iron this out in the next release.. hopefully sooner than later... Cheerios!
 10th May 2005, 00:01 #60 DJ Egg TechoratorWinamp & SHOUTcast Team   Join Date: Jun 2000 Posts: 35,817 We've not actually been able to reproduce it yet, heh. But there's lots of fixes in 5.091, so hopefully the msvcrt.dll error will be covered by one of them....? Playlist -|- Twitter
 11th May 2005, 07:30 #61 chic Junior Member   Join Date: May 2005 Posts: 3 If it's any help, i installed 5.09 on my brand new company laptop, and though it installed properly (thankfully), whenever i launch it, the main window freezes momentarily for abt 20 secs (no exaggeration there =P) b4 i can start working the player eg. playing or configuring.. Either winamp consumes enormous amts of memory or CPU resources when it starts up, or there's some grave incompatibility with certain systems =P Well all the best in bug eradication!!!
 11th May 2005, 13:04 #62 DJ Egg TechoratorWinamp & SHOUTcast Team   Join Date: Jun 2000 Posts: 35,817 Yes, but that behaviour doesn't exist with Winamp 5.08e or earlier, yes? (please confirm). I do believe that the delayed startup in 5.09 has something to do with the new dynamic Online Media check in the Media Library, which is hard-coded in 5.09 to connect to winamp.com and check for new channels/content at Winamp startup. In 5.091 (internal dev/beta build) this behaviour is changed so that the Online Media updater only checks for new channels/content on Winamp exit, and there will be an option in "Prefs > Media Library > Library Options tab > Modify Preferences button" to disable this feature (though, note, if you do disable this feature, then you will always be stuck with the same channels, and they will never be updated). Playlist -|- Twitter
 26th May 2005, 07:43 #63 BluDragon Junior Member   Join Date: May 2005 Posts: 1 I am having a similar problem to the above. I can only use the "Winamp Modern" Skin. If I try to use any other modern skin it displays the default skin. I can't remember what version I started to have the problem, I think it was about 5.06 or maybe earlier. I have had the same problem with every version since, including 5.091. I have tried all the fixes listed above including doing a clean install, and playing around with the modern skin options, all with no effect. My sys specs are: WinXP Pro sp1 with all updates (my system wont accept sp2) AthlonXP 3000+ 512Mb ram MSI KT6 Delta mainboard [rant]And please don't get on my back about sp2 because I have tried 5 different methods of installing it, and every time I had to format it and try again. Including my legit XP Pro, I also tried borrowing XP Home, an XP Pro with Sp2 pre-installed ... *starts rapidly walking in small circles muttering something about a sledge hammer*[/rant]
26th May 2005, 10:08   #64
mark
Forum King

Join Date: Jul 2002
Location: Norn Ir'nd, leek...
Posts: 6,287
Quote:
 Originally posted by DJ Egg In 5.091 (internal dev/beta build) this behaviour is changed so that the Online Media updater only checks for new channels/content on Winamp exit, and there will be an option in "Prefs > Media Library > Library Options tab > Modify Preferences button" to disable this feature (though, note, if you do disable this feature, then you will always be stuck with the same channels, and they will never be updated).
Why not simply add a "check for new content now" button somewhere?

 26th May 2005, 10:10 #65 lazerfisk Member   Join Date: Apr 2002 Location: Karlstad, Sweden Posts: 57 ServicePack 2 has got quite a few things in common with your mother in law. For example, they both cause problem, both show their existance when you need it the least. And so on What does the errormessagebox say? What crashes and where?
26th May 2005, 10:22   #66
lazerfisk
Member

Join Date: Apr 2002
Posts: 57
Quote:
 Originally posted by chic If it's any help, i installed 5.09 on my brand new company laptop, and though it installed properly (thankfully), whenever i launch it, the main window freezes momentarily for abt 20 secs (no exaggeration there =P) b4 i can start working the player eg. playing or configuring.. Either winamp consumes enormous amts of memory or CPU resources when it starts up, or there's some grave incompatibility with certain systems =P Well all the best in bug eradication!!!
Chic, one of many problems may cause this. For example, the windows network is pretty unreliable. If windows goes to shout HELLO down your network cable, your computer is pretty much gonna lock until it gets a reply or windows finally realizes that there is no computer in the other end. This is *ONLY* for the Windows Network and nothing else. So, if you were connected to your corporate domain and had something online, or you just had a mapped drive you tried to play a song from, this could be the cause of it.

Using Process Explorer (www.sysinternals.net) you can see what thread is behind it all. Could be handy to know

Cheers

 Winamp & SHOUTcast Forums WA5 crashes instantly when loading a modern Skin


text:
## 2. Example

In this example, the analysis time is 10 s, and a step load is applied during the first second. The structure is fixed by means of an ebc (essential boundary condition) while the load is applied by means of a nbc (natural boundary condition), e.g. a pressure load. The Rayleigh damping coefficients have been calculated from a free-vibration analysis.

(duration=10.)
case 1
analysis                   dynamic_linear
stage                      2 sfactor (duration)
end

stage 2
ebc            1
nbc            1 sfunction 't<1?1:0'
step_size      (0.001*duration)
rayleigh_alpha +6.4e-02
rayleigh_beta  +9.0e-06
end

end
The analysis time of 10 s is realized by having two case; case 1 references stage 2 (multiple stages are possible) of duration 10. The step load is realized with the expression 't<1?1:0', which means that for t (the analysis time, going from 0 to 10) smaller than 1, the load is applied.


text:
## Thinking Mathematically (6th Edition)

The statement makes sense. A Venn diagram may be used to show that $(A \cup B)'$ is not equal to $A' \cup B'$. by shading regions.


text:
Home » Math Theory » Fractions » Simplifying Fractions

# Simplifying Fractions

## Introduction

We have learnt that a fraction is a number representing a part of a whole. The whole may be a single or a group of objects.  Let us now learn the simplification of fractions.  But, before that, we must recall some basic terms related to fractions that are integral to understanding the process of simplifying them.

Like Fractions – Fractions that have the same denominator are called like fractions. For example, the fractions, $\frac{4}{9}, \frac{13}{9}, \frac{1}{9}$ and $\frac{5}{9}$ are all like fractions having the common denominator 9.

Unlike Fractions – Fractions with different denominators are called unlike fractions. For example, $\frac{3}{7}$ and $\frac{5}{8}$ are unlike fractions as both the fractions have different denominators.

Proper Fractions – Fractions with the numerator less than the denominator are called proper fractions. For example, consider the fractions $\frac{1}{2}$ and $\frac{2}{3}$. In both these fractions, the numerators less than their respective denominators. Hence they are proper fractions.

Improper Fractions – Fractions with the numerator either equal or greater than the denominator are called improper fractions. For example, consider the fractions $\frac{5}{2}$ and $\frac{7}{3}$. In both these fractions, the numerators are greater than their respective denominators. Hence they are improper fractions.

Mixed Fractions – A combination of a whole number and a proper fraction is called a mixed fraction. Example 4$\frac{2}{3}$ is a mixed fraction.

Equivalent fractions – Fractions having the same value are called equivalent fractions. For example, let us consider the fraction $\frac{3}{5}$. Now, if we multiply both the numerator and the denominator of  $\frac{3}{5}$ by 2, we will get, $\frac{3}{5} = \frac{3 x 2}{5 x 2}$ = $\frac{6}{10}$. Here, $\frac{3}{5}$ and $\frac{6}{10}$ are equivalent fractions as the simplest of the fraction $\frac{6}{10}$ is $\frac{3}{5}$.

Thus, an equivalent fraction of a given fraction can be obtained by multiplying its numerator and the denominator by the same number ( other than 0 ).

Now, let us understand how to simplify fractions.

## Reducing a Fraction to its Simplest Form

As the name suggests, a fraction is said to be in the simplest form if the numerator ad denominator has no common factor other than 1 or we say that these are co-primes.

Let us understand this by an example –

Suppose we have a fraction $\frac{24}{56}$ and we want to reduce it to its lowest form.

First of all, we will write the factors of both the numerator and the denominator. We have

The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24

The factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56

From above we can see that the Highest Common Factor (HCF) of 24 and 56 is 8. Therefore we will divide both the numerator and the denominator by 8 to make the numbers a pair of co-primes.

We will get

$\frac{24 ÷8}{56 ÷8} = \frac{3}{7}$

The fraction $\frac{24}{56}$ has now been reduced to $\frac{3}{7}$. 3 and 7 are prime numbers and do not have any common factors between them. Hence we can say that the simplest form of the fraction  $\frac{24}{56}$ is  $\frac{3}{7}$.

We do not need to reduce a fraction to its simplest every time. There may be instances where the numerator and the denominator of a fraction are already in a form such that they do not have any factors in common, though they might not be prime numbers. For example, the numbers 8 and 81 both are not prime numbers, yet they do “not have any factor in common. Therefore, we put them in a fraction, say  $\frac{8}{81}$, they are already in the lowest form. Let us consider another example –

Suppose we have a fraction $\frac{15}{34}$ and we want to reduce it to its lowest form.

First of all, we will write the factors of both the numerator and the denominator. We have

The factors of 15 are 1, 3, 5 and 15

The factors of 34 are 1, 2, 17 and 34

We can see that 15 and 34 have no common factor between them other than 1. Hence the said fraction is““““` said to be in its simplest form.

## How to Simplify Fractions

Let us now understand the steps involved in reducing a fraction to its simplest form. In order to express a given fraction in the standard form, the following steps should be followed –

Step 1 – Check whether the given number is in the form of a fraction. If the fraction is in mixed form, we first convert it into an improper fraction.

Step 2 – Find the greatest common divisor (GCD) of the absolute values of the numerator and the denominator.

Step 3 – Divide the numerator and the denominator of the given fraction by the GCD (HCF) obtained in step III. The fraction so obtained is the standard form of the given fraction.

Let us understand the above steps through an example.

Example

Reduce $\frac{144}{180}$ to lowest terms

Solution

We have been given the fraction $\frac{144}{180}$ and we are required to reduce it to lowest terms. We can clearly see that both the numerators as well as the denominator are even numbers which means that they would certainly have some common factors. Therefore, our first steps are to find the common factors between them and eliminate them. Hence, in order to reduce it to the lowest terms, let us first find the Highest Common Factor ( H. C. F ) of both the numerator and the denominator. For finding the H. C. F of the two numbers, we will have to find their factors. Therefore,

The factors of 144 are – 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 114. Similarly,

The factors of 180 are – 1, 2, 3, 4, 5, 6, 10, 12, 15, 18, 30, 36, 45, 60, 90 and 180

From above we can see that the common factors of 144 and 180 are 1, 2, 3, 4, 6, 12, 18 and 36. This means that the H. C. F of 144 and 180 is 36.

Now, we shall divide both the numerator as well as the denominator by 36 to get

$\frac{144}{180} = \frac{144 ÷36}{180 ÷36} = 45$

Hence, the simplified form of $\frac{144}{180}$ will be 45

### Alternative Method

The above process of finding factors first to find the H. C F and then reducing the fraction can be quite cumbersome for large numbers. Therefore, an alternative to this method is to start reducing the fractions by dividing both the numerator as well as the denominator by the common factors as we spot them.

Let us reduce the fraction $\frac{144}{180}$ through this method. As stated above, we can clearly see that both the numerators as well as the denominator are even numbers which means that they would certainly have some common factors. The first common factor that is easily identifiable is 2. Therefore, let us divide both the numerator as well as the denominator of the given fraction by 2 to get

$\frac{144}{180} = \frac{144 ÷2}{180 ÷2} = \frac{72}{90}$

Again, we can see that still both the numerators as well as the denominator are even numbers which means that they would certainly have some common factors. Let us again divide both the numerator as well as the denominator of the given fraction by 2 to get

$\frac{72}{90} = \frac{72 ÷2}{90 ÷2} = \frac{36}{45}$

Now, we can see that while the numerator is an even number, we have an odd number at the denominator which means that the numerator and the denominator do not have the factor 2 in common anymore. Let us check for the number 3. We can see that both 36 and 45 have 3 as a common factor. Therefore, this fraction can further be reduced by dividing both the numerator as well as the denominator by 3. We will now get,

$\frac{36}{45} = \frac{36 ÷3}{45 ÷3} = \frac{12}{15}$

Again, we can see that while the numerator is an even number, we have an odd number at the denominator. Also, we know that both 12 and 15 have 3 as a common factor. Therefore, this fraction can further be reduced by dividing both the numerator as well as the denominator by 3. We will now get,

$\frac{12}{15} = \frac{12 ÷3}{15 ÷3} = \frac{4}{5}$

Now the fraction obtained is $\frac{4}{5}$ where there no common factors left between the numerator and the denominator. Hence, the simplified form of is $\frac{4}{5}$ which is the same answer that we got using the H. C F method.

## Important points regarding simplification of fractions

Given below are some important points that need to be considered while simplification of fractions or reducing them to their lowest form.

1. The numerator and denominator of a fraction are called its terms. If we simplify a fraction, then we are reducing it to the lowest terms.
2. Reducing a fraction to the lowest terms will not change its value; the reduced fraction will be an equivalent fraction. All we need to do is divide the numerator and the denominator by the same nonzero whole number.
3. If the fraction is in mixed form, we first convert it into an improper fraction.
4. The result of all four operations on fractions, i.e. addition, subtraction, multiplication and division need to be checked whether the result obtained is in simplified form or not. If not, the fraction should be reduced to its simplest form.

## Solved Examples

Example 1 Simplify the fraction { 18 + ( 2 ½ + 4 / 5 ) } of 1 / 1000

Solution We have been given the fraction { 18 + ( 2 ½ + 4 / 5 ) } of 1 / 1000and we are required to simplify it.

We can see that we have a mixed term in the given fraction. Therefore, let us first convert it into an improper fraction.

Now, we know that 2 ½ = 5 / 2. Substituting this value in the given fraction, we have,

{ 18 + ( 2 ½ + 4 / 5 ) } of 1 / 1000

= { 18 + ( 5 / 2 + 4 / 5 ) } of 1 / 1000

Now, let us solve the innermost bracket of the given fraction. We can see that there are two fractions in the bracket which are unlike terms. We know that in order to add two unlike fractions, we will first have to find their L. C. M. therefore, we have,

5 / 2 + 4 / 5 = [ (5 x 5 ) + ( 4 x 2 ) ] / 10 = ( 25 + 8 ) / 10  = 33 / 10

Substituting this value in the given fraction, we have,

{ 18 + ( 5 / 2 + 4 / 5 ) } of 1 / 1000

= { 18 + 33 / 10 } of 1 / 1000

Again, let us solve the remaining bracket of the given fraction. We can see that there are two fractions in the bracket which are unlike terms. Again, finding their L. C. M, we will get,

18 + 33 / 10 = ( ( 18 x 10 ) + 33 ) / 10 = ( 180 + 33 ) / 10 = 213 / 10

Substituting this value in the given fraction, we have,

{ 18 + 33 / 10 } of 1 / 1000

= 213 / 10 of 1 / 1000

= ( 213 / 10 ) x  ( 1 / 1000 )

= 213 / 10000 or 0.0213

Hence, the simplified form of the fraction { 18 + ( 2 ½ + 4 / 5 ) } of 1 / 1000 will be 213 / 10000 or 0.0213

Example 2 Find the value of 15$\frac{3}{7}$ ÷ 1$\frac{23}{49}$

Solution We have been given two fractions 15$\frac{3}{7}$ and 1$\frac{23}{49}$ and we want to find the value of 15$\frac{3}{7}$ ÷ 1$\frac{23}{49}$.

We can see that the fractions given here are both mixed fractions. Therefore, we will first convert them into proper fractions. We will get –

15$\frac{3}{7} = \frac{15 x 7+3}{7} = \frac{105+3}{7} = \frac{108}{7}$

Similarly,

1$\frac{23}{49} = \frac{49 x 1+23}{49} = \frac{49+23}{49} = \frac{72}{49}$

Now, we will replace the mixed fractions in the given question with the improper fractions we have obtained, we will get,

15$\frac{3}{7}\:÷\:1\frac{23}{49} = \frac{108}{7}\: ÷\: \frac{72}{49}$

Here, $\frac{72}{49}$ is the divisor. Therefore we will take its reciprocal. We will get $\frac{49}{72}$

Now, we will replace the divisor with its reciprocal and change the sign of division between the two fractions to the sign of multiplication. We will get,

$\frac{108}{7} x \frac{49}{72}$

Now, this is a case of multiplication of fractions, we will see if the numerators and the denominator have anything in common. We will get,

$\frac{108 3}{7 1} x \frac{49 7}{72 2} = \frac{3 x 7}{2} = \frac{21}{2}$

Hence, 15$\frac{3}{7} ÷ 1\frac{23}{49} = \frac{21}{2}$

Example 3 Find the value of $\frac{18}{20} – \frac{21}{25}$ and simplify the result obtained.

Solution We need to find $\frac{18}{20} – \frac{21}{25}$

The denominators of the above fractions are different; therefore, we will find their L.C.M first.

L.C.M of 20 and 25 = 100

Now, we will convert the given fractions into equivalent fractions with denominator 100.

$\frac{21}{25} = \frac{21 x 4}{25 x 4} = \frac{84}{100}$

$\frac{18}{20} = \frac{18 x 5}{20 x 5} = \frac{90}{100}$

Now, we have two fractions, $\frac{84}{100}$ and $\frac{90}{100}$ who have a common denominator 100. We will now find the difference between the numerators to obtain their difference. Therefore,

$\frac{18}{20} – \frac{21}{25} = \frac{90}{100} – \frac{84}{100} = \frac{90-84}{100} = \frac{6}{100}$

We can see that the above fraction can be reduced to its simplest form if we divide both the numerator and the denominator by 2. We will get,

$\frac{6}{100} = \frac{6 ÷2}{100 ÷2} = \frac{3}{50}$

Hence, $\frac{18}{20} – \frac{21}{25} = \frac{3}{50}$

Example 4 Simplify $\frac{126}{90}$ to its lowest form

Solution We have been given the fraction $\frac{126}{90}$ and we are required to simplify it. We can clearly see that both the numerators as well as the denominator are even numbers which means that they would certainly have some common factors. The first common factor that is easily identifiable is 2. Therefore, let us divide both the numerator as well as the denominator of the given fraction by 2 to get,

$\frac{126}{90} = \frac{126÷2}{90 ÷2} = \frac{63}{45}$

Now, we can see that both the numerator and the denominator have 3 in common. Therefore, let us divide them by 3 to get,

$\frac{63}{45} = \frac{63 ÷3}{45 ÷3} = \frac{21}{15}$

Again, we can see that both the numerator and the denominator have 3 in common. Therefore, let us divide them by 3 to get,

$\frac{21}{15} = \frac{21 ÷3}{15 ÷3} = \frac{7}{5}$

Hence, the simplified form of the fraction $\frac{126}{90}$ is $\frac{7}{5}$

## Key Facts and Summary

1. Fractions that have the same denominator are called like fractions.
2. Fractions with different denominators are called unlike fractions.
3. Fractions with the numerator less than the denominator are called proper fractions.
4. Fractions with the numerator either equal or greater than the denominator are called improper fractions.
5. A combination of a whole number and a proper fraction is called a mixed fraction.
6. Fractions having the same value are called equivalent fractions.
7. A fraction is said to be in the simplest form if the numerator ad denominator has no common factor other than 1 or we say that these are co-primes.
8. In order to reduce a fraction to its lowest terms, we divide the numerator and the denominator by their H. C. F.
9. The numerator and denominator of a fraction are called its terms. If we simplify a fraction, then we are reducing it to the lowest terms.
10. Reducing a fraction to the lowest terms will not change its value; the reduced fraction will be an equivalent fraction. All we need to do is divide the numerator and the denominator by the same nonzero whole number.


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# Existence of Maximal Normal p-subgroup of a finite group.

Let $G$ be a finite group and let $P$ be sylow p subgroup of $G$. Prove the following :

$1)$ If $M$ is a normal p-subgroup of G then $M\leq P$.

$2)$ There is a normal p-subgroup of G that contains all normal p-subgroups of G.

what i have done sofar is :

For 1, by Sylow theorem, Given a sylow p-subgroup $P$ and any p-subgroup, $Q$ there exists $g\in G$ such that $gQg^{-1}\leq P$

Inparticular, for Normal p-subgroup $Q$, we have $Q=gQg^{-1}\leq P$ i.e., $Q\leq P$ for all Normal subgroups.So, I guess $(1)$ is done.

for $(2)$, I have done only a particular case : particular case of sylow p-subgroup being unique,

In that case sylow p-subgroup is Normal and we apply $(1)$ to conclude that there exists a sylow p-subgroup (and hence a Normal p-subgroup whose existence is already known) which contains all Normal p-subgroups.

I am not very sure about the way to see when sylow p-subgroup is not unique.

I would be thankful if someone can give me a hint.

Thank You,.

-
Well, a normal $p$-subgroup is contained in all Sylow $p$-subgroups. So you might want to look at the intersection of all Sylow $p$-subgroups. – Daniel Fischer Aug 31 '13 at 14:28
Let $|G|=p^nm,~~(p,m)=1$, so $|P|=p^n$ and $|M|=p^kt$ in which $k\leq n,~~t\mid m$. $M$ is normal in the group so we can speak about the subgroup $MP$. We have: $$|MP|=\frac{p^n\times p^kt}{p^s}\Big|~ p^nm,~~s\leq k$$ – S. Snape Aug 31 '13 at 14:32


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Tuesday, August 6, 2013

TV Guide Picks The Greatest TV Families of All Time

I think of all the lists TV Guide has done lately, this is one that I don't have as many quibbles with.  To be fair, picking a favorite TV family is much like picking your favorite character out of a book.  There are so many to pick from, that it's hard to narrow it down, and it's hard to not leave anyone off.  This was a list they did back in June, and even after all this time, I can't get mad about the families they left off.  Now I can get annoyed by some they chose to include, The Walshes for one, but other than a few of those types of issues, I'm generally okay with the list.  I do want to give TV Guide a shout out for one aspect of the list, not one reality show family made the list.  Thank the lord they don't think the Kardashians are one of the best TV families.  As usual , I will post the families that should have made the list, but didn't.

General note for all the lists TV Guide has been doing, the first ten are in order, the rest are in alphabetical order.

2.  The Huxtables from The Cosby Show
3.  The Sopranos from The Sopranos
4.  The Bunkers from All in the Family
5.  The Barones from Everybody Loves Raymond
6.  The Conners from Roseanne
7.  The Ewings from Dallas
8.  The Cleavers from Leave it to Beaver

10.  The Waltons from The Waltons
11.  The Andersons from Father Knows Best
12.  The Arnolds from The Wonder Years
13.  The Bankses from The Fresh Prince of Bel-Air
14.  The Bellamys from Upstairs, Downstairs
15.  The Bluths from Arrested Development
16.  The Bradfords from Eight is Enough
17.  The Bundys from Married... With Children
18.  The Camdens from 7th Heaven
19.  The Carringtons from Dynasty
20.  The Clampetts from The Beverly Hillbillies
21.  The Cohens from The O.C.
22.  The Corbetts from The Courtship of Eddie's Father
23.  The Crawleys from Downton Abbey
24.  The Cunninghams from Happy Days

25.  The Davises from Family Affair
26.  The Douglases from My Three Sons
27.  The Evanses from Good Times
28.  The Fishers from Six Feet Under
29.  The Flintstones from The Flintstones
30.  The Formans from That '70s Show
31.  The Gavins from Rescue Me
32.  The Gilmores from Gilmore Girls

33.  The Hecks from The Middle
34.  The Heffernans from The King of Queens
35.  The Ingallses from Little House on the Prairie

36.  The Jeffersons from The Jeffersons
37.  The Keatons from Family Ties
38.  The Lanes from The Patty Duke Show
39.  The Lawrences from Family
40.  The Lopezes from George Lopez
41.  Malcolm's family from Malcolm in the Middle
42.  The Munsters from The Munsters
43.  The Nelsons from The Adventures of Ozzies and Harriet
44.  The Partridges from The Partridge Family
45.  The Petries from The Dick Van Dyke Show
46.  The Pritchetts and the Dumphys from Modern Family
47.  The Ricardos I Love Lucy

48.  The Romanos from One Day at a Time
49.  The Sallingers from Party of Five
50.  The Sanfords from Sandford and Son
51.  The Simpsons from The Simpsons
52.  The Stones from The Donna Reed Show
53.  The Tanners from Full House
54.  The Taylors from The Andy Griffith Show
55.  The Taylors from Friday Night Lights
56.  The Taylors from Home Improvement
57.  The Tates and the Campbells from Soap
58.  The Walshes from Beverly Hills, 90210
59.  The Winslows from Family Matters
60.  The Whites from Breaking Bad

So that's their list, and now that I've actually typed it up, I'm really thinking a lot of these choices are pretty lame.  Most of them I'm sure have their supporters but for a select few, I could care less about them.  Even some of the pictures I used are of families that are just sort of blah for me.  There were a few cases where I would have shown a picture, The Huxtables, but it was right after the one I shared of The Bradys.  It seems like a lot of these families are very similar to each other.  I would liked to have seen a livelier bunch of families, maybe even a wider definition of what makes a family.  I would think that the group of Friends or even Seinfeld would qualify, though I'm not a huge fan of either show.

That being said, I'm not sure the families I would have liked to have been included, would make this list any more diverse, but here they are anyway.

I would have liked to see The Stephenses from Bewitched make the list.  Between a mortal husband, a witch for a wife, two magical kids, and a truly wicked mother-in-law, they are the perfect family.  Other sitcoms families that should have made the list are the Seavers from Growing Pains, the Strattons from Silver Spoons, the Tanners from Alf, and the Papadopolises from Webster.

I would also like to nominate two Western families to the list; the Cartwrights from Bonanza and the Barkleys from The Big Valley.  One other family from a drama, that in my opinion should have been in the top ten, The Summerses from Buffy the Vampire Slayer.

My last comment, where are the families form the daytime soap operas.  How about the Buchanans, Kramers, or the Vegas from One Life to Live.  We could also have had the Martins, Kanes, and Chandlers from All My Children.  And don't froget the Quartermaines, Spencers, or the Corinthos clan from General Hospital.

So those are my thoughts, what do you guys make of the list?

Ryan, you took me back a long way with the list of TV families. I have seen many of these, others I'm only familiar with. I'm glad EVERYBODY LOVES RAYMOND is in the top ten as I enjoyed the sitcom a lot. Not the greatest of TV families but the Hennessys of 8 SIMPLE RULES seem to be fairly popular. It's currently being telecast on Indian television.

bermudaonion said...

I don't watch enough TV to know if this is a good list or not.

Teddyree said...

Some of my favourites up there from when I was a kid ... The Brady Bunch, Flintstones, The Ingallses, The Adams family.

but I agree Ryan, the Stephensons from Bewitched should have made the list & what about the Drummond family from Different Strokes. Glad the Crawley's are on the list but I would have liked to see the Walker family from Brothers and Sisters up there too.

Lisa said...

It looks like they tried very hard to get roughly equal numbers of families from each of the past five decades. I think they've also worked to ensure some ethnic blend is present, particularly where one family represents the first significant family on tv for a particular ethnicity. Had they worked for less balance, they might have come up with a stronger list.

Unknown said...

Best TV family, beyond any debate: the Morgans from Dexter.

Michelle Stockard Miller said...

These lists really bother me because they rarely get it right. I mean, did you see the top 100 lists from Entertainment Weekly in books, TV, Film, music, etc.? What a joke! I wanted to throw the magazine across the room, especially in the TV and music categories. I tend to agree with you, Ryan...meh.


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# Challenging Integral [indefinite]

Integrate: $$\int \frac{x^2+n(n-1)}{(x\sin x+n\cos x )^2}dx$$

I've been beating my head around this problem for quite some time now, but I've got nowhere. I'd request the person writing the solution to please explain his thought process because I would like to learn how to appraoch such Integrals in the future.

Using $\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$

$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$ and $\displaystyle \cos \phi = \frac{n}{\sqrt{x^2+n^2}}$ and $\displaystyle \tan \phi = \frac{x}{n}\Rightarrow \phi = \tan^{-1}\left(\frac{x}{n}\right)$

So Integral is $$\displaystyle = \int \sec^2(x-\phi)\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx$$

Now Let $$\displaystyle (x-\phi) = y\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)=y$$. Then $$\displaystyle \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx = dy$$

So Integral is $$\displaystyle \int \sec^2(y)dy = \tan y +\mathbb{C} = \tan\left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)+\mathbb{C}$$

So $$\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx = \left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathcal{C}= \frac{n\sin x-x\cos x}{n\cos x+x\sin x}+\mathcal{C}$$

• Nice solution and explanation. This method seems much more intuitive to me then the other solution. – Plopperzz Oct 4 '16 at 6:03

Using the identities $n^2+x^2=(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2$ and

$n=\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)$, we can rewrite the integral as

$\displaystyle \int\frac{(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2-[\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)]}{(n\cos x+x\sin x)^2} dx$

$\displaystyle=\int\frac{(n\cos x+x\sin x)[n\cos x+x\sin x-\cos x]-(n\sin x-x\cos x)[-n\sin x+x\cos x+\sin x]}{(n\cos x+x\sin x)^2} dx$

$\displaystyle=\int\bigg(\frac{n\sin x-x\cos x}{n\cos x+x\sin x}\bigg)^{\prime} dx$

$\;\;\;\;\displaystyle=\frac{n\sin x-x\cos x}{n\cos x+x\sin x}.$

Hint: Multiply and Divide the Integrand by $x^{2n-2}$.