text:
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# Golf Drives

APSTAT-LTU1JT

A local driving range keeps track of the distances its clients hit golf balls. Distances are normally distributed and have a standard deviation of $43\text{ yards}$.

If a $350\text{ yard}$ drive puts an individual in the top $1\%$ of all drives, what is the mean of this distribution?

A

About $250\text{ yards}$.

B

About $450\text{ yards}$.

C

About $350\text{ yards}$.

D

About $295\text{ yards}$.

E

About $268\text{ yards}$.

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# basic rules logarithm of exponential

I am looking for proof of the basic rules of logarithm. I can prove all basic rules except this $$\log_ab^y=y\log_ab$$ how to get this rule using definition of logarithm.

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What definition? There are several equivalent definitions. Most either have it as the inverse exponential function or $\int_1^x\frac{1}{x}dx$ –  CBenni Feb 18 '13 at 15:29
@CBenni: I expect it's just this definition: "$z = \log_a b$ if $a^z = b$". –  Tara B Feb 18 '13 at 15:32
@TaraB well that might be the definition, but I know a lot of professors choose another definition for various (or no) reasons. –  CBenni Feb 18 '13 at 15:38
Yes, I know, I just meant that I expect that in this case what I said is the relevant definition (I was the one who 'upvoted' your comment). –  Tara B Feb 18 '13 at 15:41

If $\log_ab=m\iff a^m=b$

So, $\log_ab^y=n\iff a^n=b^y=(a^m)^y=a^{my}\implies n=my$ if $a\ne0,1$

So, $\log_ab^y=n=my=y\log_ab$

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Just check that $$a^{y \log_{a}(b)} = a^{\log_{a}(b) y} = \left(a^{\log_{a}(b)}\right)^{y} = b^{y}.$$

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Assuming you know, or have proven that $\log_a(xy) = \log_a x + \log_z y$ We use this to prove:

$$\log_ab^y= \log_a \underbrace{(b \cdot b \cdot ...\cdot b)}_{\large y \; times}$$ $$\iff \log_a b^y = \underbrace{\log_a b + \log_a b + \cdots + \log_z b}_{\large y \;\;times}$$ $$\iff \log_ab^y = y\log_z b$$

Else: by definition only

$$a^{y \log_{a}(b)}= a^{\log_{a}(b) y}=\left(a^{\log_{a}(b)}\right)^{y}=b^{y}.$$

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what if $y\notin \mathbb{N}$? –  CBenni Feb 18 '13 at 15:36
@CBenni That's why I included the "else"...by definition... –  amWhy Feb 18 '13 at 15:53


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# What is the slope of the line through (-4,-1) and (2,-7) ?

###### Question:

What is the slope of the line through (-4,-1) and (2,-7) ?

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## 2008 Vol. 25, No. 5

Display Method:
2008, 25(5): 705-708. doi: 10.1007/s00376-008-0705-1
[Abstract](973) [PDF 114KB](1238)
Abstract:
NULL
2008, 25(5): 709-722. doi: 10.1007/s00376-008-0709-x
Abstract:
A global atmospheric general circulation model has been used to perform eleven idealized numerical experiments, i.e., TP00, TP10, ..., TP100, corresponding to different percentages of the Tibetan Plateau altitude. The aim is to explore the sensitivity of East Asian climate to the uplift and expansion of the Tibetan Plateau under the reconstructed boundary conditions for the mid-Pliocene about 3 Ma ago. When the plateau is progressively uplifted, global annual surface temperature is gradually declined and statistically significant cooling signals emerge only in the Northern Hemisphere, especially over and around the Tibetan Plateau, with larger magnitudes over land than over the oceans. On the contrary, annual surface temperature rises notably over Central Asia and most parts of Africa, as well as over northeasternmost Eurasia in the experiments TP60 to TP100. Meanwhile, the plateau uplift also leads to annual precipitation augmentation over the Tibetan Plateau but a reduction in northern Asia, the Indian Peninsula, much of Central Asia, parts of western Asia and the southern portions of northeastern Europe. Additionally, it is found that an East Asian summer monsoon system similar to that of the present initially exists in the TP60 and is gradually intensified with the continued plateau uplift. At 850 hPa the plateau uplift induces an anomalous cyclonic circulation around the Tibetan Plateau in summertime and two anomalous westerly currents respectively located to the south and north of the Tibetan Plateau in wintertime. In the mid-troposphere, similar-to-modern spatial pattern of summertime western North Pacific subtropical high is only exhibited in the experiments TP60 to TP100, and the East Asian trough is steadily deepened in response to the progressive uplift and expansion of the Tibetan Plateau.
2008, 25(5): 723-736. doi: 10.1007/s00376-008-0723-z
[Abstract](1637) [PDF 2153KB](1254)
Abstract:
Various paleoclimate records have shown that the Asian monsoon was punctuated by numerous sub-orbital time-scale events, and these events were coeval with those that happened in the North Atlantic. This study investigates the Asian summer monsoon responses to the Atlantic Ocean forcing by applying an additional freshwater flux into the North Atlantic. The simulated results indicate that the cold North Atlantic and warm South Atlantic induced by the weakened Atlantic thermohaline circulation (THC) due to the freshwater flux lead to significantly suppressed Asian summer monsoon. The authors analyzed the detailed processes of the Atlantic Ocean forcing on the Asian summer monsoon, and found that the atmospheric teleconnection in the eastern and central North Pacific and the atmosphere-ocean interaction in the tropical North Pacific play the most crucial role. Enhanced precipitation in the subtropical North Pacific extends the effects of Atlantic Ocean forcing from the eastern Pacific into the western Pacific, and the atmosphere-ocean interaction in the tropical Pacific and Indian Ocean intensifies the circulation and precipitation anomalies in the Pacific and East Asia.
2008, 25(5): 737-747. doi: 10.1007/s00376-008-0737-6
[Abstract](1004) [PDF 1050KB](1493)
Abstract:
Impacts of initial conditions on cloud-resolving model simulations are investigated using a series of sensitivity experiments. Five experiments with perturbed initial temperature, moisture, and cloud conditions are conducted and compared to the control experiment. The model is forced by the large-scale vertical velocity and zonal wind observed and derived from NCEP/Global Data Assimilation System (GDAS). The results indicate that model predictions of rainfall are much more sensitive to the initial conditions than those of temperature and moisture. Further analyses of the surface rainfall equation and the moisture and cloud hydrometeor budgets reveal that the calculations of vapor condensation and deposition rates in the model account for the large sensitivities in rainfall simulations.
2008, 25(5): 748-756. doi: 10.1007/s00376-008-0748-3
[Abstract](934) [PDF 302KB](1558)
Abstract:
Increased precipitation during the vegetation periods was observed in and further predicted for Inner Mongolia. The changes in the associated soil moisture may affect the biosphere-atmosphere exchange of greenhouse gases. Therefore, we set up an irrigation experiment with one watered (W) and one unwatered plot (UW) at a winter-grazed Leymus chinensis-steppe site in the Xilin River catchment, Inner Mongolia. UW only received the natural precipitation of 2005 (129 mm), whereas W was additionally watered after the precipitation data of 1998 (in total 427 mm). In the 3-hour resolution, we determined nitrous oxide (N2O), methane (CH4) and carbon dioxide (CO2) fluxes at both plots between May and September 2005, using a fully automated, chamber-based measuring system. N2O fluxes in the steppe were very low, with mean emissions (±s.e.) of 0.9±0.5 and 0.7±0.5 μg N m-2 h-1 at W and UW, respectively. The steppe soil always served as a CH4 sink, with mean fluxes of -24.1±3.9 and -31.1±5.3 μg C m-2 h-1 at W and UW. Nighttime mean CO2 emissions were 82.6±8.7 and 26.3±1.7 μg C m-2 h-1 at W and UW, respectively, coinciding with an almost doubled aboveground plant biomass at W. Our results indicate that the ecosystem CO2 respiration responded sensitively to increased water input during the vegetation period, whereas the effects on CH4 and N2O fluxes were weak, most likely due to the high evapotranspiration and the lack of substrate for N2O producing processes. Based on our results, we hypothesize that with the gradual increase of summertime precipitation in Inner Mongolia, ecosystem CO2 respiration will be enhanced and CH4 uptake by the steppe soils will be lightly inhibited.
2008, 25(5): 757-764. doi: 10.1007/s00376-008-0757-2
[Abstract](1013) [PDF 2544KB](1272)
Abstract:
Continuous observation data collected over the whole year of 2004 on a cropland surface in Tongyu, a semi-arid area of northeastern China (44o25'N, 122o52'$E), have been used to investigate the variations of surface albedo and soil thermal parameters, including heat capacity, thermal conductivity and thermal diffusivity, and their relationships to soil moisture. The diurnal variation of surface albedo appears as a U shape curve on sunny days. Surface albedo decreases with the increase of solar elevation angle, and it tends to be a constant when solar elevation angle is larger than 40o. So the daily average surface albedo was computed using the data when solar elevation angle is larger than 40o. Mean daily surface albedo is found to decrease with the increase of soil moisture, showing an exponential dependence on soil moisture. The variations of soil heat capacity are small during Julian days 90--300. Compared with the heat capacity, soil thermal conductivity has very gentle variations during this period, but the soil thermal diffusivity has wide variations during the same period. The soil thermal conductivity is found to increase as a power function of soil moisture. The soil thermal diffusivity increases firstly and then decreases with the increase of soil moisture. 2008, 25(5): 765-777. doi: 10.1007/s00376-008-0765-2 Abstract: This study investigates the characteristics of a heavy snowfall event over the southwestern part of the Korean Peninsula on 4 December 2005. The snowstorm was a type of mesoscale maritime cyclone which resulted from barotropic instability, and diabatic heating from the warm ocean in continental polar air masses. Based on surface observations, radiosonde soundings, MTSAT-1R satellite data and the 10-km grid RDAPS (Regional Assimilation and Prediction System based on the PSU/NCAR MM5) data, the evolution of the mesocyclone is explained by the following dynamics; (1) In the initial stage, the primary role in the cyclogenesis process of the mesocyclone is a barotropic instability in the horizontal shear zone. (2) In the developing stage, the mesocyclone moves and deepens into a baroclinic zone corresponding to the surface heating and moistening. (3) In the mature stage, it is found that the mesocyclone is intensified by the destabilization caused by enhanced low-level heating and condensation, the moisture flux convergence, and the interaction between upper and lower-level potential vorticity anomalies. We suggest that a checklist with stepwise indicators responsible for development be prepared for the forecasting of heavy snowfall over the southwestern part of the Korean Peninsula. 2008, 25(5): 778-790. doi: 10.1007/s00376-008-0778-x Abstract: In mountain areas, radar observations are often contaminated (1) by echoes from high-speed moving vehicles and (2) by point-wise ground clutter under either normal propagation (NP) or anomalous propagation (AP) conditions. Level II data are collected from KMTX (Salt Lake City, Utah) radar to analyze these two types of contamination in the mountain area around the Great Salt Lake. Human experts provide the ground truth" for possible contamination of either type on each individual pixel. Common features are then extracted for contaminated pixels of each type. For example, pixels contaminated by echoes from high-speed moving vehicles are characterized by large radial velocity and spectrum width. Echoes from a moving train tend to have larger velocity and reflectivity but smaller spectrum width than those from moving vehicles on highways. These contaminated pixels are only seen in areas of large terrain gradient (in the radial direction along the radar beam). The same is true for the second type of contamination - point-wise ground clutters. Six quality control (QC) parameters are selected to quantify the extracted features. Histograms are computed for each QC parameter and grouped for contaminated pixels of each type and also for non-contaminated pixels. Based on the computed histograms, a fuzzy logical algorithm is developed for automated detection of contaminated pixels. The algorithm is tested with KMTX radar data under different (clear and rainy) weather conditions. 2008, 25(5): 791-804. doi: 10.1007/s00376-008-0791-0 [Abstract](1053) [PDF 2313KB](1639) Abstract: The characteristics of the upper-level circulation and thermodynamical properties for the period when two distinct cold surges broke out over East Asia during the 2005/06 winter are investigated. From early December 2005 to early January 2006, exceptionally cold weather lasted for approximately one month due to two successive cold surges that took place on 2 December 2005 and 2 January 2006, respectively. This study reveals that both involve the upper-tropospheric circulation, which induces the amplification and expansion of the surface Siberian high toward East Asia, but arose from different causes: the former is caused by the upper-level blocking originated from the North Pacific and the latter is caused by the upper-level wave train across the Eurasian Continent. In particular, it is suggested that the lower-tropospheric anomalous wind caused by upper-level circulation anomalies and a steep meridional temperature gradient amplified by phase-locked annual cycle combined to induce very strong cold advection in East Asia, which resulted in exceptionally cold weather that lasted for several weeks. The present results emphasize that the characteristics of the upper-tropospheric circulation can be considered as important precursors to cold surge occurrences in East Asia. 2008, 25(5): 805-814. doi: 10.1007/s00376-008-0805-y Abstract: A simple approach for estimating the equivalent diffusion for diagnosing tracer transport is proposed. Two different expressions are derived; one is based directly on an analytical solution of the two-dimensional advection-diffusion equation, the other uses the variance of the tracer distribution. To illustrate some features of the equivalent diffusion and possible applications thereof, idealized releases of passive tracers from the nuclear fuel reprocessing plants at Sellafield in the Irish Sea and Dounreay on the northern coast of Scotland have been simulated with a regional isopycnic co-ordinate Ocean General Circulation Model. Both continuous and pulse releases are considered; the former being representative of the actual historical discharges from the reprocessing plants, the latter resembling an accidental scenario. Age tracers are included to calculate the mean time elapsed since the tracers left their source regions. It is found that in the Nordic Seas the age of tracers from Dounreay is approximately 2 years younger than the age from Sellafield. Although tracers from both sources eventually end up along the same transport routes, significant qualitative differences regarding the dispersion properties are found. It is argued that one single parameter, the equivalent horizontal diffusion, which is estimated to be in the range of 20--56 m$^2$s$^{-1}$from Sellafield and 170--485 m$^2$s$^{-1}$from Dounreay, determines these differences. 2008, 25(5): 815-823. doi: 10.1007/s00376-008-0815-9 Abstract: In order to investigate the spatial patterns of the Tropospheric Biennial Oscillation (TBO) on the global scale, the Climate Prediction Center (CPC) Merged Analysis of Precipitation (CMAP) monthly averaged precipitation and the Climate Diagnostics Center (CDC) monthly outgoing long-wave radiation (OLR) and SST are used in conjunction with TBO bandpass-filtering. The results indicate active biennial variability in the tropical eastern-central Pacific regions. It is evident that observations reflect the biennial component of the ENSO rather than the TBO itself. Since some studies have pointed out that the TBO is a broad-scale phenomenon differing from the ENSO, to investigate the pure TBO the ENSO signal must be excluded. The Scale Interaction Experiment-FRCGC (SINTEX-F) coupled general circulation model (CGCM) developed at Japan Frontier Research Center for Global Change (FRCGC) can capture both the ENSO and the biennial signals. Air-sea interactions in the tropical eastern-central Pacific are decoupled to eliminate the effects of ENSO in a experiment by SINTEX-F and the results show that biennial variability still exists even without ENSO. It seems to mean that the TBO and ENSO are independent from each other. Furthermore, the model results indicate that the two key regions are southwest Sumatra and the tropical western Pacific for the TBO cycle. 2008, 25(5): 824-840. doi: 10.1007/s00376-008-0824-8 [Abstract](1007) [PDF 1701KB](1632) Abstract: A study of the circulation in the northern South China Sea (SCS) is carried out with the aid of a three-dimensional, high-resolution regional ocean model. One control and two sensitivity experiments are performed to qualitatively investigate the effects of surface wind forcing, Kuroshio intrusion, and bottom topographic influence on the circulation in the northern SCS. The model results show that a branch of the Kuroshio in the upper layer can intrude into the SCS and have direct influence on the circulation over the continental shelf break in the northern SCS. There are strong southward pressure gradients along a zonal belt largely seaward of the continental slope. The pressure gradients are opposite in the southern and northern parts of the Luzon Strait, indicating inflow and outflow through the strait, respectively. The sensitivity experiments suggest that the Kuroshio intrusion is responsible for generating the imposed pressure head along the shelf break and has no obvious seasonal variations. The lateral forcing through the Luzon Strait and Taiwan Strait can induce the southwestward slope current and the northeastward SCS Warm Current in the northern SCS. Without the lateral forcing, there is no high-pressure-gradient zonal belt seaward of the continental slope. The wind forcing mainly causes the seasonal variation of the circulation in the SCS. The wind-induced water pile-up results in the southward high pressure gradient along the northwestern boundary of the basin. Without the blocking of the plateau around Dongsha Islands, the intruded Kuroshio tends to extend northwest and the SCS branch of the Kuroshio becomes wider and stronger. The analyses presented here are qualitative in nature but should lead to a better understanding of the oceanic responses in the northern SCS to these external influence factors. 2008, 25(5): 841-846. doi: 10.1007/s00376-008-0841-7 [Abstract](945) [PDF 268KB](1472) Abstract: By analyzing the linkage of the Northern Annular Mode (NAM) anomaly to the East Asian jet and the East Asian trough during Stratospheric Sudden Warming (SSW), the influence of SSW on East Asian weather is studied. The results show that the East Asian jet is strengthened and the East Asian trough is deepened during SSW. With the downward propagation of SSW, the strengthened East Asian jet and the East Asian trough would move southward, expand westward and gradually influence the area of north and northeastern China. This implies that the winter monsoon tends to be enhanced over East Asia during SSW. 2008, 25(5): 847-855. doi: 10.1007/s00376-008-0847-1 [Abstract](909) [PDF 467KB](1437) Abstract: Characterization of carbonaceous aerosols including CC (carbonate carbon), OC (organic carbon), and EC (elemental carbon) were investigated at Xi'an, China, near Asian dust source regions in spring 2002. OC varied between 8.2 and 63.7 μg m-3, while EC ranged between 2.4 and 17.2 μ m-3 during the observation period. OC variations followed a similar pattern to EC and the correlation coefficient between OC and EC is 0.89 (n=31). The average percentage of total carbon (TC, sum of CC, OC, and EC) in PM2.5 during dust storm (DS) events was 13.6%, which is lower than that during non-dust storm (NDS) periods (22.7%). CC, OC, and EC accounted for 12.9%, 70.7%, and 16.4% of TC during DS events, respectively. The average ratio of OC/EC was 5.0 in DS events and 3.3 in NDS periods. The OC-EC correlation (R2=0.76, n=6) was good in DS events, while it was stronger (R2=0.90, n=25) in NDS periods. The percentage of water-soluble OC (WSOC) in TC accounted for 15.7%, and varied between 13.3% and 22.3% during DS events. The distribution of eight carbon fractions indicated that local emissions such as motor vehicle exhaust were the dominant contributors to carbonaceous particles. During DS events, soil dust dominated the chemical composition, contributing 69% to the PM2.5 mass, followed by organic matter (12.8%), sulfate (4%), EC (2.2%), and chloride (1.6%). Consequently, CC was mainly entrained by Asian dust. However, even in the atmosphere near Asian dust source regions, OC and EC in atmospheric dust were controlled by local emission rather than the transport of Asian dust. 2008, 25(5): 856-866. doi: 10.1007/s00376-008-0856-0 Abstract: This study investigates the decadal and interannual variability of the Indian Ocean Dipole (IOD). It is found that the long-term IOD index displays a decadal phase variation. Prior to 1920 negative phase dominates, but after 1960 positive phase prevails. Under the warming background of the tropical ocean, a larger warming trend in the western Indian Ocean is responsible for the decadal phase variation of the IOD mode. Due to reduced latent heat loss from the local ocean, the western Indian Ocean warming may be caused by the weakened Indian Ocean westerly summer monsoon. The interannual air-sea coupled IOD mode varies on the background of its decadal variability. During the earlier period (1948--1969), IOD events are characterized by opposing SST anomaly (SSTA) in the western and eastern Indian Ocean, with a single vertical circulation above the equatorial Indian Ocean. But in the later period (1980--2003), with positive IOD dominating, most IOD events have a zonal gradient perturbation on a uniform positive SSTA. However, there are three exceptionally strong positive IOD events (1982, 1994, and 1997), with opposite SSTA in the western and eastern Indian Ocean, accompanied by an El Nino event. Consequently, two anomalous reversed Walker cells are located separately over the Indian Ocean and western-eastern Pacific; the one over the Indian Ocean is much stronger than that during other positive IOD events. 2008, 25(5): 867-884. doi: 10.1007/s00376-008-0867-x [Abstract](986) [PDF 737KB](1409) Abstract: The 21-yr ensemble predictions of model precipitation and circulation in the East Asian and western North Pacific (Asia-Pacific) summer monsoon region (0o--50oN, 100o--150oE) were evaluated in nine different AGCM, used in the Asia-Pacific Economic Cooperation Climate Center (APCC) multi-model ensemble seasonal prediction system. The analysis indicates that the precipitation anomaly patterns of model ensemble predictions are substantially different from the observed counterparts in this region, but the summer monsoon circulations are reasonably predicted. For example, all models can well produce the interannual variability of the western North Pacific monsoon index (WNPMI) defined by 850 hPa winds, but they failed to predict the relationship between WNPMI and precipitation anomalies. The interannual variability of the 500 hPa geopotential height (GPH) can be well predicted by the models in contrast to precipitation anomalies. On the basis of such model performances and the relationship between the interannual variations of 500 hPa GPH and precipitation anomalies, we developed a statistical scheme used to downscale the summer monsoon precipitation anomaly on the basis of EOF and singular value decomposition (SVD). In this scheme, the three leading EOF modes of 500 hPa GPH anomaly fields predicted by the models are firstly corrected by the linear regression between the principal components in each model and observation, respectively. Then, the corrected model GPH is chosen as the predictor to downscale the precipitation anomaly field, which is assembled by the forecasted expansion coefficients of model 500 hPa GPH and the three leading SVD modes of observed precipitation anomaly corresponding to the prediction of model 500 hPa GPH during a 19-year training period. The cross-validated forecasts suggest that this downscaling scheme may have a potential to improve the forecast skill of the precipitation anomaly in the South China Sea, western North Pacific and the East Asia Pacific regions, where the anomaly correlation coefficient (ACC) has been improved by 0.14, corresponding to the reduced RMSE of 10.4% in the conventional multi-model ensemble (MME) forecast. 2008, 25(5): 885-896. doi: 10.1007/s00376-008-0885-8 [Abstract](992) [PDF 761KB](1877) Abstract: Ultraviolet (UV) solar radiation has a significant influence on human health, the environment and climate. A series of measurements, including UV radiation (290--400 nm) and global solar radiation$(R_{\text s})\$, were continuously recorded from August 2004 at the Lhasa and Haibei sites on the Tibetan Plateau. Both observation sites' altitudes are above 3000 m and have similar meteorological conditions. The data from 2005--2006 was used to identify the varying characteristics of UV radiation. It's relation to the clearness index Ks, the relative optical mass mr, and Rs were established. The annual mean values of total daily UV radiation are 0.92 and 0.67 MJ m-2 at Lhasa and Haibei, respectively. The UV radiation in Lhasa represented 4.6\% of the global solar radiation while in Haibei this percentage was 4.2%. In the case of clear days (Ks>0.8), these percentages ranged between 4.0% and 4.5% in Lhasa and between 5.1% and 5.5% in Haibei. In the case of cloudy days (Ks<0.4), these percentages ranged from 4.4% to 6.8% in Lhasa and from 5.1% to 5.5% in Haibei. The maximum values of UV radiation for each relative optical mass diminished exponentially with mr. Thus, for Lhasa and Haibei, UV=46.25 mr-1.29, and UV=51.76 mr-1.42, respectively. The results of this study can be used to obtain more UV radiation data for the study of UV climate characteristics, the effects of UV on ecological processes and the feedback of the thinning of the stratospheric ozone, from more routine measurements Rs data.
2008, 25(5): 897-904. doi: 10.1007/s00376-008-0897-4
Abstract:
A three-dimensional variational method is proposed to simultaneously retrieve the 3-D atmospheric temperature and moisture profiles from satellite radiance measurements. To include both vertical structure and the horizontal patterns of the atmospheric temperature and moisture, an EOF technique is used to decompose the temperature and moisture field in a 3-D space. A number of numerical simulations are conducted and they demonstrate that the 3-D method is less sensitive to the observation errors compared to the 1-D method. When the observation error is more than 2.0 K, to get the best results, the truncation number for the EOF's expansion have to be restricted to 2 in the 1-D method, while it can be set as large as 40 in a 3-D method. This results in the truncation error being reduced and the retrieval accuracy being improved in the 3-D method. Compared to the 1-D method, the rms errors of the 3-D method are reduced by 48\% and 36\% for the temperature and moisture retrievals, respectively. Using the real satellite measured brightness temperatures at 0557 UTC 31 July 2002, the temperature and moisture profiles are retrieved over a region (20o--45oN, 100o--125oE) and compared with 37 collocated radiosonde observations. The results show that the retrieval accuracy with a 3-D method is significantly higher than those with the 1-D method.


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# 11 In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Given,

Initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours = $P(1+\frac{R}{100})^n$

$= 506000(1+\frac{2.5}{100})^2 \approx 531616$

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

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text:
The unit of moment of inertia of an area is

This question was previously asked in
ISRO Scientist Civil 2020 Paper
View all ISRO Scientist CE Papers >
1. kg/m
2. kg/m2
3. m4
4. m3

Answer (Detailed Solution Below)

Option 3 : m4
Free
NPCIL 2020 Scientific Assistant Physics Mini Live Test
1311
30 Questions 30 Marks 40 Mins

Detailed Solution

Moment of inertia of an area or Second moment of area (M.I):

M.I of a body about any axis is defined as the summation of the second moment of all elementary areas about the axis.

I = Σ(A × d2)

Unit: m4 or mm4 or cm4

The following table shows the Second moment of inertia of different shapes

 Shape Figure Moment of Inertia Rectangle $${{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{b}}{{\rm{d}}^3}}}{{12}}$$ $${{\rm{I}}_{{\rm{yy}}}} = \frac{{{\rm{d}}{{\rm{b}}^3}}}{{12}}$$ Triangle $${{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}$$ $${{\rm{I}}_{{\rm{yy}}}} = \frac{{{\rm{h}}{{\rm{b}}^3}}}{{36}}$$ Circle $${{\rm{I}}_{{\rm{xx}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}$$ $${{\rm{I}}_{{\rm{yy}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}$$ Semicircle $${{\rm{I}}_{{\rm{xx}}}} = {\rm{\;}}0.11{{\rm{R}}^4}$$ $${{\rm{I}}_{{\rm{yy}}}} = \frac{{\rm{\pi }}}{8}{{\rm{R}}^4}$$ Quarter circle $${{\rm{I}}_{{\rm{xx}}}} = 0.055{{\rm{R}}^4}$$ $${{\rm{I}}_{{\rm{yy}}}} = 0.055{{\rm{R}}^4}$$


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## Silicon Optoelectronic Integrated Circuits (Springer Series

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Our research addresses several of the most exciting questions in current atomic and molecular sciences, including quantum control of molecular states and processes (Odom, Seideman), tests of astrophysics and fundamental symmetries ( Gabrielse, Odom, Shahriar ), and trapping and cooling of atoms, anti-atoms, and molecules ( Gabrielse, Ketterson, Odom, Shahriar ). So are our loyal friends lying when they insist a photograph looks nothing like us?

## Mathematics for quantum mechanics;: An introductory survey

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Phys9000 even includes sections for trigonometry, chemistry, unit conversions and a generic equation solver. Astronomers, in conjunction with other types of physicists, might be called upon to solve problems connected with space flight navigation and satellite communications. Applied research includes the development of novel high power lasers, parametric oscillators and amplifier; techniques for waveguide, photonic crystal and other photonic devices.

## Next-Generation FTTH Passive Optical Networks: Research

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Cottingham (ed.), Cambridge: Cambridge University Press. We should thus expect the time-reversed versions of their orbits to obey the laws of physics. a book sliding across a table does generate heat from friction as it slows down. which involves heat. Bell & sons Published in 1915, 64 pages Harvey Gould, Jan Tobochnik The Graduate School established this new award to recognize and honor the outstanding contributions that Graduate Assistants provide to students, faculty, departments, administrative units, and the University as a whole.

## MAKING A REFRACTING TELESCOPE

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In a fourth column, calculate the theoretical intensity $$I_0\cos^2\theta$$. His research area is in computational physics, specifically in soft matter physics/materials science. Tuesday, 11 August 3:45 pm - 5:15 pm, Los Angeles Convention Center, Room 404AB UPDATE: here are the slide decks: Scientific and Technical Academy Award for the development of Bullet Physics! That's right - this amazing piece of bright red all-natural (except for having been dyed) material was only found in one single isolated place on God's Green Earth: The Pacific Ocean. (The Pacific Ocean covers about 30% of the Earth's surface.) The same woman also suggested that rutilated quartz - titanium oxide slivers inside quartz (silicon oxide) get there the same way that a straw can penetrate a tree during a tornado.

## High Dielectric Constant Materials: VLSI MOSFET Applications

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Light scattering and scanning of a sharp metal tip illuminated by a focused laser source: Sample contrast and resolution are obtained by means of the locally enhanced tip-sample interaction spatially confined by the tip apex radius. Condensed Matter physicists study how matter arises from a large number of interacting atoms and electrons, and what physical properties it has as a result of these interactions.

## International Conference on Experimental Mechanics 2013 and

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Bending a ray like 2 inward implies bending its imaginary continuation 3 outward. in the same way that raising one end of a seesaw causes the other end to go down. The number of turns per second is the frequency of the light. The proof of this equation is left as an exercise to those readers who are suﬃciently brave and motivated. 111 Discussion Questions A In ﬁgures n/1 and n/2.3 The Lensmaker’s Equation The focal length of a spherical mirror is simply r/2. and another left turn when it exits. of which ﬁgures n/1 and n/2 exhaust only two.

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Because light is an electromagnetic wave, other forms of electromagnetic radiation such as X-rays, microwaves, and radio waves exhibit similar properties.[1] Most optical phenomena can be accounted for using the classical electromagnetic description of light. F. (Prerequisite: M CC 124; M CC 126; PHCC 110 or PHCC 121 or PHCC 141.) Properties of the Milky Way, galaxies, quasars and related objects; special and general relativity, cosmology. Physics Letters A does not publish articles on High Energy and Nuclear Physics.

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In this paper, we show that Inkjet Printing can be successfully applied to external-cavity vertically emitting thin-film organic lasers and can be used to generate a diffraction-limited output beam with an output energy as high as 33.6 μJ with a slope efficiency S of 34%. Prerequisite(s): PHYS 2102 with a grade of C or above, senior standing, and MATH 2171. Nolas) has recently been awarded a Carnegie Fellowship, allowing him to expand on his current research on high-pressure and high-temperature synthesis of carbon materials with superlative properties at the Carnegie Institute of Washington.

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Daliang Li kayak fishing on Quinnipiac river of New Haven. May be repeated for up to 6 hours of credit with permission of the Optics Program Director. (Fall/Spring/Summer) OPTI 6101. While we strive to provide the most comprehensive notes for as many high school textbooks as possible, there are certainly going to be some that we miss. Michel Orrit and colleagues want to build an apparatus for that, with the help of a 0.5 MEuro investment from the Stichting Fundamenteel Onderzoek der Materie (FOM).

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## 8.1 Distance, Midpoint, and the Parabola

### Learning Objectives

1. Apply the distance and midpoint formulas.
2. Graph a parabola using its equation given in standard from.
3. Determine standard form for the equation of a parabola given general form.

## Conic Sections

A conic sectionA curve obtained from the intersection of a right circular cone and a plane. is a curve obtained from the intersection of a right circular cone and a plane. The conic sections are the parabola, circle, ellipse, and hyperbola.

The goal is to sketch these graphs on a rectangular coordinate plane.

## The Distance and Midpoint Formulas

We begin with a review of the distance formulaGiven two points $(x1, y1)$ and $(x2, y2)$, the distance d between them is given by $d=(x2−x1)2+(y2−y1)2$.. Given two points $( x1, y1)$ and $( x2, y2)$ in a rectangular coordinate plane, the distance d between them is given by the distance formula,

$d=(x2−x1)2+(y2−y1)2$

Furthermore, the point that bisects the line segment formed by these two points is called the midpointGiven two points $(x1, y1)$ and $(x2, y2)$, the midpoint is an ordered pair given by $(x1+x22, y1+y22).$ and is given by the formula,

$(x1+x22, y1+y22)$

The midpoint is an ordered pair formed by the average of the x-values and the average of the y-values.

### Example 1

Given $(−2,−5)$ and $(−4,−3)$ calculate the distance and midpoint between them.

Solution:

In this case, we will use the formulas with the following points:

$(x1, y1) (x2, y2)(−2,−5)(−4,−3)$

It is a good practice to include the formula in its general form before substituting values for the variables; this improves readability and reduces the probability of making errors.

$d=(x2−x1)2+(y2−y1)2=[−4−(−2)]2+[−3−(−5)]2=(−4+2)2+(−3+5)2=(−2)2+(2)2=4+4=8=22$

Next determine the midpoint.

$(x1+x22, y1+y22)=(−2+(−4)2, −5+(−3)2)=(−62, −82)=(−3, −4)$

Plotting these points on a graph we have,

Answer: Distance: $22$ units; midpoint: $(−3,−4)$

### Example 2

The diameter of a circle is defined by the two points $(−1,2)$ and $(1,−2).$ Determine the radius of the circle and use it to calculate its area.

Solution:

Find the diameter using the distance formula.

$d=(x2−x1)2+(y2−y1)2=[1−(−1)]2+(−2−2)2=(2)2+(−4)2=4+16=20=25$

Recall that the radius of a circle is one-half of the circle’s diameter. Therefore, if $d=25$ units, then

$r=d2=252=5$

The area of a circle is given by the formula $A=πr2$ and we have

$A=π(5)2=π⋅5=5π$

Area is measured in square units.

Answer: Radius: $5$ units; area: $5π$ square units

Try this! Given $(0,0)$ and $(9,−3)$ calculate the distance and midpoint between them.

Answer: Distance: $310$ units; midpoint: $(92,−32)$

## The Parabola

A parabolaThe set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. is the set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. In other words, if given a line L the directrix, and a point F the focus, then $(x,y)$ is a point on the parabola if the shortest distance from it to the focus and from it to the line is equal as pictured below:

The vertex of the parabola is the point where the shortest distance to the directrix is at a minimum. In addition, a parabola is formed by the intersection of a cone with an oblique plane that is parallel to the side of the cone:

Recall that the graph of a quadratic function, a polynomial function of degree 2, is parabolic. We can write the equation of a parabola in general formThe equation of a parabola written in the form $y=ax2+bx+c$ or $x=ay2+by+c$, where a, b, and c are real numbers and $a≠0.$ or we can write the equation of a parabola in standard formThe equation of a parabola written in the form $y=a(x−h)2+k$ or $x=a(y−k)2+h.$:

$General Form Standard Formy=ax2+bx+c y=a(x−h)2+k$

Here a, b, and c are real numbers, $a≠0.$ Both forms are useful in determining the general shape of the graph. However, in this section we will focus on obtaining standard form, which is often called vertex formThe equation of a parabola written in standard form is often called vertex form. In this form the vertex is apparent: $(h,k).$. Given a quadratic function in standard form, the vertex is $(h,k).$ To see that this is the case, consider graphing $y=(x+3)2+2$ using transformations.

$y=x2 Basic squaring function.y=(x+3)2 Horizontal shift left 3 units.y=(x+3)2+2 Vertical shift up 2 units.$

Use these translations to sketch the graph,

Here we can see that the vertex is $(−3,2).$ This can be determined directly from the equation in standard form,

$y=a(x−h)2+k↓ ↓ y=[x−(−3)]2+2$

Written in this form we can see that the vertex is $(−3,2).$ However, the equation is typically not given in standard form. Transforming general form to standard form, by completing the square, is the main process by which we will sketch all of the conic sections.

### Example 3

Rewrite the equation in standard form and determine the vertex of its graph: $y=x2−8x+15.$

Solution:

Begin by making room for the constant term that completes the square.

$y=x2−8x+15=x2−8x +___+15−___$

The idea is to add and subtract the value that completes the square, $(b2)2$, and then factor. In this case, add and subtract $(b2)2=(−82)2=(−4)2=16.$

$y=x2−8x+15Add and subtract 16.=(x2−8x +16)+15 −16Factor.=(x−4)(x−4)−1=(x−4)2−1$

Adding and subtracting the same value within an expression does not change it. Doing so is equivalent to adding 0. Once the equation is in this form, we can easily determine the vertex.

$y=a(x − h)2+k↓ ↓ y=(x−4)2+(−1)$

Here we have a translation to the right 4 units and down 1 unit. Hence, h = 4 and k = −1.

Answer: $y=(x−4)2−1$; vertex: $(4,−1)$

If there is a leading coefficient other than 1, then begin by factoring out that leading coefficient from the first two terms of the trinomial.

### Example 4

Rewrite the equation in standard form and determine the vertex of the graph: $y=−2x2+12x−16.$

Solution:

Since $a=−2$, factor this out of the first two terms in order to complete the square. Leave room inside the parentheses to add and subtract the value that completes the square.

$y=−2x2+12x−16=−2(x2−6x +___−___)−16$

Now use −6 to determine the value that completes the square. In this case, $(b2)2=(−62)2=(−3)2=9$. Add and subtract 9 and factor as follows:

$y=−2x2+12x−16=−2(x2−6x +___ −___ )−16Add and subtract 9.=−2(x2−6x +9−9)−16 Factor.=−2[(x−3)(x−3)−9]−16=−2[(x−3)2−9]−16Distribute the−2.=−2(x−3)2+18−16=−2(x−3)2+2$

In this form, we can easily determine the vertex.

$y=a(x−h)2+k↓↓y=−2(x−3)2+2$

Here h = 3 and k = 2.

Answer: $y=−2(x−3)2+2$; vertex: $(3,2)$

Make use of both general form and standard form when sketching the graph of a parabola.

### Example 5

Graph: $y=−2x2+12x−16.$

Solution:

From the previous example we have two equivalent forms of this equation,

$General Form Standard Formy=−2x2+12x−16 y= −2(x−3)2+2$

Recall that if the leading coefficient $a>0$ the parabola opens upward and if $a<0$ the parabola opens downward. In this case, $a=−2$ and we conclude the parabola opens downward. Use general form to determine the y-intercept. When $x=0$ we can see that the y-intercept is $(0,−16).$ From the equation in standard form, we can see that the vertex is $(3,2).$ To find the x-intercept we could use either form. In this case, we will use standard form to determine the x-values where $y=0$,

$y=−2(x−3)2+2 Set y =0 and solve.0=−2(x−3)2+2 −2=−2(x−3)21=(x−3)2 Apply the square root property.±1=x−33±1=x$

Here $x=3−1=2$ or $x=3+1=4$ and therefore the x-intercepts are $(2,0)$ and $(4,0).$ Use this information to sketch the graph.

Answer:

So far we have been sketching parabolas that open upward or downward because these graphs represent functions. At this point we extend our study to include parabolas that open right or left. If we take the equation that defines the parabola in the previous example,

$y= −2(x−3)2+2$

and switch the x and y values we obtain

$x= −2(y−3)2+2$

This produces a new graph with symmetry about the line $y=x.$

Note that the resulting graph is not a function. However, it does have the same general parabolic shape that opens left. We can recognize equations of parabolas that open left or right by noticing that they are quadratic in y instead of x. Graphing parabolas that open left or right is similar to graphing parabolas that open upward and downward. In general, we have

In all cases, the vertex is $(h,k).$ Take care to note the placement of h and k in each equation.

### Example 6

Graph: $x=y2+10y+13.$

Solution:

Because the coefficient of $y2$ is positive, $a=1$, we conclude that the graph is a parabola that opens to the right. Furthermore, when $y=0$ it is clear that $x=13$ and therefore the x-intercept is $(13,0).$ Complete the square to obtain standard form. Here we will add and subtract $(b2)2=(102)2=(5)2=25.$

$x=y2+10y+13=y2+10y +25−25+13=(y+5)(y+5)−12=(y+5)2−12$

Therefore,

$x=a (y−k)2+h↓↓x=(y−(−5))2+(−12)$

From this we can see that the vertex $(h,k)=(−12,−5).$ Next use standard form to find the y-intercepts by setting $x=0.$

$x= (y+5)2−120=(y+5)2−1212=(y+5)2±12=y+5±23=y+5−5±23=y$

The y-intercepts are $(0,−5−23)$ and $(0,−5+23).$ Use this information to sketch the graph.

Answer:

### Example 7

Graph: $x= −2y2+4y−5.$

Solution:

Because the coefficient of $y2$ is $a=−2$, we conclude that the graph is a parabola that opens to the left. Furthermore, when $y=0$ it is clear that $x=−5$ and therefore the x-intercept is $(−5,0).$ Begin by factoring out the leading coefficient as follows:

$x=−2y2+4y−5=−2(y2−2y+___−___)−5$

Here we will add and subtract $(b2)2=(−22)2=(−1)2=1.$

$x= −2y2+4y−5=−2(y2−2y +1−1)−5=−2[(y−1)2−1]−5=−2(y−1)2+2−5=−2(y−1)2−3$

Therefore, from vertex form, $x= −2(y−1)2−3$, we can see that the vertex is $(h,k)=(−3,1).$ Because the vertex is at $(−3,1)$ and the parabola opens to the left, we can conclude that there are no y-intercepts. Since we only have two points, choose some y-values and find the corresponding x-values.

$xyx=−2(y−1)2−3−11−1x= −2(−1−1)2−3=−2(−2)2−3=−11−52x= −2(2−1)2−3=−2(1)2−3=−5−113x= −2(3−1)2−3=−2(2)2−3=−11$

Answer:

Try this! Graph: $x= y2−y−6.$

Answer:

### Key Takeaways

• Use the distance formula to determine the distance between any two given points. Use the midpoint formula to determine the midpoint between any two given points.
• A parabola can open upward or downward, in which case, it is a function. In this section, we extend our study of parabolas to include those that open left or right. Such graphs do not represent functions.
• The equation of a parabola that opens upward or downward is quadratic in x, $y=ax2+bx+c.$ If $a>0$, then the parabola opens upward and if $a<0$, then the parabola opens downward.
• The equation of a parabola that opens left or right is quadratic in y, $x=ay2+by+c.$ If $a>0$, then the parabola opens to the right and if $a<0$, then the parabola opens to the left.
• The equation of a parabola in general form $y=ax2+bx+c$ or $x=ay2+by+c$ can be transformed to standard form $y=a(x−h)2+k$ or $x=a(y−k)2+h$ by completing the square.
• When completing the square, ensure that the leading coefficient of the variable grouping is 1 before adding and subtracting the value that completes the square.
• Both general and standard forms are useful when graphing parabolas. Given standard form the vertex is apparent $(h,k).$ To find the x-intercept set $y=0$ and solve for x and to find the y-intercept set $x=0$ and solve for y.

### Part A: The Distance and Midpoint Formulas

Calculate the distance and midpoint between the given two points.

1. $(−1,−3)$ and $(5,−11)$

2. $(−3,2)$ and $(1,−1)$

3. $(4,−2)$ and $(−2,−6)$

4. $(−5,−6)$ and $(−3,−4)$

5. $(10,−1)$ and $(9,6)$

6. $(−6,−4)$ and $(−12,1)$

7. $(0,0)$ and $(2,3)$

8. $(0,0)$ and $(22,−3)$

9. $(5,−3)$ and $(25,−3)$

10. $(310,6)$ and $(10,−56)$

11. $(12,−1)$ and $(−2,32)$

12. $(−43,2)$ and $(−13,−12)$

13. $(15,−95)$ and $(310,−52)$

14. $(−12,43)$ and $(−23,56)$

15. $(a,b)$ and $(0,0)$

16. $(0,0)$ and $(a2,2a)$

Determine the area of a circle whose diameter is defined by the given two points.

1. $(−8,12)$ and $(−6,8)$

2. $(9,5)$ and $(9,−1)$

3. $(7,−8)$ and $(5,−10)$

4. $(0,−5)$ and $(6,1)$

5. $(6,0)$ and $(0,23)$

6. $(0,7)$ and $(5,0)$

Determine the perimeter of the triangle given the coordinates of the vertices.

1. $(5,3)$, $(2,−3)$, and $(8,−3)$

2. $(−3,2)$, $(−4,−1)$, and $(−1,0)$

3. $(3,3)$, $(5,3−23)$, and $(7,3)$

4. $(0,0)$, $(0,22)$, and $(2,0)$

Find a so that the distance d between the points is equal to the given quantity.

1. $(1,2)$ and $(4,a)$; $d=5$ units

2. $(−3,a)$ and $(5,6)$; $d=10$ units

3. $(3,1)$ and $(a,0)$; $d=2$ units

4. $(a,1)$ and $(5,3)$; $d=13$ units

### Part B: The Parabola

Graph. Be sure to find the vertex and all intercepts.

1. $y=x2+3$

2. $y=12(x−4)2$

3. $y=−2(x+1)2−1$

4. $y=−(x−2)2+1$

5. $y=−x2+3$

6. $y=−(x+1)2+5$

7. $x=y2+1$

8. $x=y2−4$

9. $x=(y+2)2$

10. $x=(y−3)2$

11. $x=−y2+2$

12. $x=−(y+1)2$

13. $x=13(y−3)2−1$

14. $x=−13(y+3)2−1$

Rewrite in standard form and give the vertex.

1. $y=x2−6x+18$

2. $y=x2+8x+36$

3. $x=y2+20y+87$

4. $x=y2−10y+21$

5. $y=x2−14x+49$

6. $x=y2+16y+64$

7. $x=2y2−4y+5$

8. $y=3x2−30x+67$

9. $y=6x2+36x+54$

10. $x=3y2+6y−1$

11. $y=2x2−2x−1$

12. $x=5y2+15y+9$

13. $x=−y2+5y−5$

14. $y=−x2+9x−20$

Rewrite in standard form and graph. Be sure to find the vertex and all intercepts.

1. $y=x2−4x−5$

2. $y=x2+6x−16$

3. $y=−x2+12x−32$

4. $y=−x2−10x$

5. $y=2x2+4x+9$

6. $y=3x2−6x+4$

7. $y=−5x2+30x−45$

8. $y=−4x2−16x−16$

9. $x=y2−2y−8$

10. $x=y2+4y+8$

11. $x=y2−2y−3$

12. $x=y2+6y−7$

13. $x=−y2−10y−24$

14. $x=−y2−12y−40$

15. $x=3y2+12y+12$

16. $x=−2y2+12y−18$

17. $x=y2−4y−3$

18. $x=y2+6y+1$

19. $x=−y2+2y+5$

20. $y=2x2−2x+1$

21. $y=−3x2+2x+1$

22. $y=−x2+3x+10$

23. $x=−4y2−4y−5$

24. $x=y2−y+2$

25. $y=x2+5x−1$

26. $y=2x2+6x+3$

27. $x=2y2+10x+12$

28. $x=y2+y−1$

### Part C: Discussion Board

1. Research and discuss real-world applications that involve a parabola.

2. Do all parabolas have x-intercepts? Explain.

3. Do all parabolas have y-intercepts? Explain.

4. Make up your own parabola that opens left or right, write it in general form, and graph it.

### Answers

1. Distance: 10 units; midpoint: $(2,−7)$

2. Distance: $213$ units; midpoint: $(1,−4)$

3. Distance: $52$ units; midpoint: $(192,52)$

4. Distance: $5$ units; midpoint: $(22,32)$

5. Distance: $5$ units; midpoint: $(352,−3)$

6. Distance: $522$ units; midpoint: $(−34,14)$

7. Distance: $22$ units; midpoint: $(14,−4320)$

8. Distance: $a2+b2$ units; midpoint: $(a2,b2)$

9. $5π$ square units

10. $2π$ square units

11. $92π$ square units

12. $6+65$ units

13. 12 units

14. −2, 6

15. 2, 4

1. $y=(x−3)2+9$; vertex: $(3,9)$

2. $x=(y+10)2−13$; vertex: $(−13,−10)$

3. $y=(x−7)2$; vertex: $(7,0)$

4. $x=2(y−1)2+3$; vertex: $(3,1)$

5. $y=6(x+3)2$; vertex: $(−3,0)$

6. $y=2(x−12)2−32$; vertex: $(12,−32)$

7. $x=−(y−52)2+54$; vertex: $(54,52)$

8. $y=(x−2)2−9$;

9. $y=−(x−6)2+4$;

10. $y=2(x+1)2+7$;

11. $y=−5(x−3)2$;

12. $x=(y−1)2−9$;

13. $x=(y−1)2−4$;

14. $x=−(y+5)2+1$;

15. $x=3(y+2)2$;

16. $x=(y−2)2−7$;

17. $x=−(y−1)2+6$;

18. $y=−3(x−13)2+43$;

19. $x=−4(y+12)2−4$;

20. $y=(x+52)2−294$;

21. $x=2(y+52)2−12$;

1. Answer may vary

2. Answer may vary


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# Electric Field

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 Sub Topics Scientist Michael Faraday brought the concept of electric field in the 19th century. He used the patterns of electric field lines that shows the existence of the electric field but these lines are invisible in reality. He explained that around a charge particle there a space in which another charge particle experience a force of attraction or repulsion, this region is called the electric field.

## Electric Field Definition

When electric charges acts in certain region of space. They produce force around it. This is what electric field is! It decides whats the strength the field carries due to these charge interaction. Hence, the electric field is defined as"Electrical force acting per unit charge exerted on the charged body".

## Electric Field Equation

The electric field by a point charge q is given by,

Electric field strength = $\frac{Force}{Charge}$
or
E = $\frac{F}{q}$

Electric field unit :
The electric field strength is expressed in Newton per coulomb (NC-1) or volt per meter (Vm-1).

## Electric Field Lines

Electric field lines (electric flux lines) are the curved or straight paths along which a unit positive charge tends to move in the electric field if it is free to do so. These lines are imaginary but they give us the pictorial visualization of electric field. They show patterns of several lines, which extend between infinity and the source charges.

Pictorial view of electric field lines

## Electric Field Strength

The electric field is a region where force acts due to the interaction between the charges. The strength of the field is measured as
Electric field strength = $\frac{Force}{Charge}$
or
E = $\frac{F}{q}$

The unit of electric field strength is in newton per coulomb (NC-1). It is a vector quantity. It obeys inverse square law. Hence, if a charge q is at a distance r from another charge Q, the force between them is given as

F = $\frac{kqQ}{r^2}$

The charge Q has the electric field strength is given as
E = $\frac{F}{q}$ = $\frac{\frac{kqQ}{r^2}}{q}$ = $\frac{kQ}{r^2}$

This is the formula to calculate the electric field strength.

## Magnitude of electric field

The Magnitude of the electric field is always positive whereas the direction will be in the way the electrical force acts given as
$\vec{E}$ = $\frac{\vec{F}}{q}$The electric field is given as $\vec{E}$ tells that the electric field has always the vector quantity that rely on charge q (positive or negative).

## Electric Field of a Dipole

Let us consider two charges +q and -q separated by a small distance. They form a dipole.
A system which has two equal and opposite charges separated by a small distance is called electric dipole. It is the ratio of product of magnitude of charges to the separation between them given as

E = |q|d

where,
q is the charge and
d is the distance of separation between them.

## Electric Field in a Capacitor

A parallel plate capacitor consists of two conducting metal plates X and Y each of area A separated by distance d. The plates can be of any shape. When the plates are connected to a battery, a charge +q appears on plate X and – q appears on plate Y. On each plate the charge is distributed uniformly. As the distance between the plates is small compared to the area of the plate, the electric field is uniform between the plates. The electric field is non uniform at the outer edge of the plates.

The electric field E acting in a parallel plate capacitors is given by
E = $\frac{\sigma}{\epsilon_o}$ = $\frac{V}{d}$
where $\epsilon_o$ is the permittivity of free space and d is the distance between the plates
The capacitance is
C = $\frac{\epsilon_o A}{d}$Here A is the area of the plate.

## Net Electric Field

The net electric field is the sum total of the charges acting in the given electric field. If q1, q2, q3,...... qn are the charges acting in the given field.

The Net electric field is given by
q = q1 + q2 + q3 + q4 + ...... +qn.

## Calculating electric field

Lets see some examples on calculating electric field:

### Solved Examples

Question 1: Calculate the magnitude of electric charge $\vec{E}$ at a field point 3 m from the point charge 6 nc?
Solution:

Given:
Distance r = 3 m, Charge q = 6 nc

The magnitude of electric field is
E = $\frac{1}{4 \pi \epsilon_0}$ $\frac{q}{r^2}$
= (9 $\times$ 109 Nm2/C2) $\frac{6 \times 10^{-9}}{(3 m)^2}$
= 6 N/C

Question 2: If the charge + 3e and a charge -3 e acts separated by a distance of 4 nm. What is the potential energy of the system?
Solution:

Given:
Charge q = + 3e, Q = - 3e, distance r = 4 $\times$ 10-9 m

The electric potential is given by
U = k $\frac{qQ}{r}$
= (9 $\times$ 109) $\times$ $\frac{3 \times 1.6 \times 10^{-19} \times (-3) \times 1.6 \times 10^{-19}}{4 \times 10^{-9}}$
= - 5.065 $\times$ 10-9 N.


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# Reverse Taylor Series

Almost everyone is familiar with the famous Taylor Series:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$

which, if it converges at more than one point, will converge in some interval about $a$. Has anyone considered the "Reverse" Taylor Series:

$g(x) = \sum_{n=0}^\infty \frac{f^{(n)}(x-a)}{n!} a^n$

I call it reverse, because its what you get for symbolically taking $a \rightarrow (x-a)$. You might be saying that there is absolutely no reason to believe that this series should converge to $f(x)$, but there are two big examples where it does.

For $e^x$:

$g(x) = \sum_{n=0}^\infty \frac{e^{(x-a)}}{n!} a^n = \frac{e^x}{e^a} \sum_{n=0}^\infty \frac{a^n}{n!} = e^x$

For $x^k$:

$g(x) = \sum_{n=0}^k \frac{k(k-1)...(k-n+1)(x-a)^{k-n} a^n}{n!} = \sum_{n=0}^k {k \choose n} (x-a)^{k-n} a^n = ((x-a) + a)^k$

I have yet to find a counter-example for when the Reverse Taylor Series does not give back the original function for an analytic function, and I also have yet to think of a way to prove that the Reverse Taylor Series should converge for a given function. Does anyone have any ideas?

-
If it works for $x^k$, then if you express your real analytic function as a power series, just arguing term by term should give you the convergence. – Willie Wong Dec 23 '10 at 2:37

Your two series are the same thing pointwise when they converge. $g(x)$ is the Taylor series for $f$ expanded around the point $x-a$. So it will converge as long as the Taylor series for $f$ at $x-a$ has a radius of convergence at least $a$.
However, in $f(x)$, moving $x$ around may make the Taylor series fail to converge due to your $x$ moving out of the radius of convergence. Whereas in $g(x)$, moving $x$ around will make the series fail to converge due to moving the point $x - a$ somewhere with a radius of convergence smaller than $a$. That is, once you fix $a$, in the Taylor series for $f$ around $a$, the convergence radius is fixed, independent of $x$. But in the series $g(x)$, the convergence of the series depends on how the radius of convergence at $x-a$ (which depends on $x$) compares with $a$. So that the function $g(x)$ converges at more than one point is insufficient to guarantee that $g(x)$ converges in a small interval about $a$.
Put $y=x-a$. Then, $$g(x) = \sum\limits_{n = 0}^\infty {\frac{{f^{(n)} (y)}}{{n!}}(x - y)^n } = f(x).$$


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# Machine A takes 10 hours to complete a certain job and starts that job

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Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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Machine A takes 10 hours to complete a certain job and starts that job at 9AM. After one hour of working alone, machine A is joined by machine B and together they complete the job at 5PM. How long would it have taken machine B to complete the job if it had worked alone for the entire job?

(A) 15 hours
(B) 18 hours
(C) 20 hours
(D) 24 hours
(E) 35 hours

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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23 Sep 2015, 22:52
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Let us assume total job = 100 units

A finishes 100 units in 10 hrs(given)

Hence A (working rate)= 10 units /hr

Now Given that A works for 1 hr( so 10 units done) then A and B finish total work in 7 hours. Hence A and B finish 90 units in 7 hours. Of these 7x10=70 units were done by A. Hence B did 20 units in 7 hours.

Hence B(working rate)= 20/7 units /hr

Hence B takes 100x7/20 = 35 hours to complete the job.

Answer E.

Regards,
Dom.
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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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24 Sep 2015, 01:14
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A does the job in 10 hrs, so in 1 hr it will complete 1/10 of the total work
When A is joined by B at 10 AM, 9/10 (1 - 1/10)of the total work is left. They both complete this work in 7 hrs.
Therefore, they would have completed the whole work in 70/9 hrs.
From this we can say that:
1/10 + 1/B = 9/70 (where B is the number of hours B takes to complete the job alone)
=> 1/B = 9/70 - 1/10 = 2/70 = 1/35
=> B = 35 hours

Hence the answer is E.
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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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Bunuel wrote:
Machine A takes 10 hours to complete a certain job and starts that job at 9AM. After one hour of working alone, machine A is joined by machine B and together they complete the job at 5PM. How long would it have taken machine B to complete the job if it had worked alone for the entire job?

(A) 15 hours
(B) 18 hours
(C) 20 hours
(D) 24 hours
(E) 35 hours

Kudos for a correct solution.

Let x be the time taken by machine B to complete the work alone. Then it's 1 hour work will be 1/x.
Total work done by machine A and B in 1 hour will be
$$\frac{1}{10}+\frac{1}{x}$$
=$$\frac{x+10}{10x}$$

So time taken to complete whole work will be $$\frac{10x}{x+10}$$

Machine A worked for 1 hour before Machine B joined.
Work done by A in 1 hour is 1/10
work left to do is 9/10
Machine A and B working together take 7 hours to complete 9/10 work.
$$\frac{9}{10}*\frac{10x}{x+10}=7$$

or $$9x=7x+70$$
or $$2x=70$$
or $$x=35$$

Answer:- E
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A alone can finish work in 10 hours..
Therefore a can do 1/10th of work in 1 hour..

A starts at 9 a.m and works alone for 1 hour. A finishes 1/10th of the work.
Remaining work = 1 - (1/10) = 9/10

A and B together then finish the remaining work (9/10) from 10a.m to 5 p.m = 7 hours
therefore in 1 hour together they finish 9/70 work

Adding Rate of A and B

1/A + 1/B = 9/70
1/10 + 1/B = 9/70
B = 35

Answer E
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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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24 Sep 2015, 13:44
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Bunuel wrote:
Machine A takes 10 hours to complete a certain job and starts that job at 9AM. After one hour of working alone, machine A is joined by machine B and together they complete the job at 5PM. How long would it have taken machine B to complete the job if it had worked alone for the entire job?

(A) 15 hours
(B) 18 hours
(C) 20 hours
(D) 24 hours
(E) 35 hours

Another approach:

Working ALONE Machine A takes 10 hours to do the job.
With Machine B's help, the job took 8 hours to complete.
So, Machine B saved Machine A 2 hours of work.

In other words, during the 7 hours that Machine B was helping, Machine B did the work that would have taken Machine A an extra 2 hours to complete.
So, it takes Machine B 7 hours to complete the same amount of work that it takes Machine A to complete in 2 hours.
So, it takes Machine B 7/2 times as long to complete what Machine A can complete

It takes Machine A 10 hours to do the job.
So, the time it takes Machine B to do the job = (10)(7/2) = 35 = E

Cheers,
Brent
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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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27 Sep 2015, 10:03
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Bunuel wrote:
Machine A takes 10 hours to complete a certain job and starts that job at 9AM. After one hour of working alone, machine A is joined by machine B and together they complete the job at 5PM. How long would it have taken machine B to complete the job if it had worked alone for the entire job?

(A) 15 hours
(B) 18 hours
(C) 20 hours
(D) 24 hours
(E) 35 hours

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Using only algebra and the memorized formula for Work problems ($$Work = Rate * Time$$ or $$Rate = \frac{Work}{Time}$$), we can break this problem into 3 parts: Machine A alone from 9 AM to 10 AM , Machines A and B from 10 AM to 5 PM, and then the hypothetical Machine B alone. Let’s set up these three steps to see how we can solve this on the scenic path.

Machine A takes 10 hours to do the job, so each hour it works finishes 1/10th of the total job. From 9 AM to 10 AM, Machine A works alone, so at 10 AM, machine B kicks in and 1/10th of the job is done. Ergo, 9/10th of the job is left to complete for both machines.

From 10 AM to 5 PM, 7 hours pass, during which 9/10th of the job gets completed. Thus we can calculate the rate of the machines working together: $$\frac{9}{10} = Rate_{A+B} * 7$$ hours. $$Rate_{A+B} = \frac{9}{70}$$.

Since we know that $$Rate_A + Rate_B = Rate_{A+B}$$ (i.e. rates are additive), we can leverage the fact that we know 2 of these 3 rates to find the third using basic fraction addition. $$\frac{1}{10} + Rate_B = \frac{9}{70}$$. Putting them all on a common denominator: $$\frac{7}{70} + Rate_B = \frac{9}{70}$$, so $$Rate_B = \frac{2}{70}$$, or 1/35.

Now that we have B’s rate of 1/35, we can easily tell that it would take 35 hours to complete the entire job.

The correct answer is (E).

The algebraic solution works fine and gets you the right answer, but there are many moving parts to keep track of and many opportunities for mistakes. Can we get to the same answer but faster using conceptual understanding and avoid the scenic route entirely?

If machine A does 1/10th of the work in an hour, and it works from 9 to 5, then it works for 8 hours and accomplishes 80% of the job on its own. This means that machine B only accomplishes 20% of the job, and it does so in 7 hours (10 AM to 5 PM). If the machine does 1/5 of the job in 7 hours, it will take (7*5) 35 hours to complete 5/5 of the job. Answer choice E, using almost no math whatsoever, but rather by exploiting the logic of the question.

In general, if you see how to solve a problem via algebra and are confident you can solve it in 3 minutes or less, then by all means go for it. However, you can save some time if you really understand how questions are set up and what they are testing. It may not be possible to come up with a handy shortcut on test day because of nerves and stress, but during your preparation take a look at how problems are solved and see if you can find a more elegant solution. All roads lead to Rome, and the more routes you know, the less likely you are to get stuck in unfamiliar territory.
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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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23 Feb 2016, 11:39
rate of A=1/10=7/70
rate of A+B=9/70
rate of B=2/70=1/35
B will complete job alone in 35 days
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Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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18 Mar 2016, 21:20
Bunuel wrote:
Machine A takes 10 hours to complete a certain job and starts that job at 9AM. After one hour of working alone, machine A is joined by machine B and together they complete the job at 5PM. How long would it have taken machine B to complete the job if it had worked alone for the entire job?

(A) 15 hours
(B) 18 hours
(C) 20 hours
(D) 24 hours
(E) 35 hours

Kudos for a correct solution.

A completes 1/10 part of the work in 1 hour
The remaining work is 9/10; therefore, 7(1/10 + 1/y) = 9/10
Solving y = 35 hours.

Alternatively,
Machine A completes 4/5 of the total work. Remaining 1/5 of the work is completed by Machine B in 7 hours. Therefore, B will take 35 hours to complete the whole work.

Last edited by Mathivanan Palraj on 21 Dec 2017, 05:46, edited 1 time in total.
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Re: Machine A takes 10 hours to complete a certain job and starts that job [#permalink]

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21 Mar 2017, 01:02
Bunuel wrote:
Machine A takes 10 hours to complete a certain job and starts that job at 9AM. After one hour of working alone, machine A is joined by machine B and together they complete the job at 5PM. How long would it have taken machine B to complete the job if it had worked alone for the entire job?

(A) 15 hours
(B) 18 hours
(C) 20 hours
(D) 24 hours
(E) 35 hours

Kudos for a correct solution.

let the total work be x units
rate of work for A = x/10
let the rate for B be y
A worked for 8 hrs and b worked for 7 hrs

therefore
8x/10 +7y=x
7y = 2x/10
y = x/35

thefore time taken by B = x/y = x/x/35 = 35

Option E
Re: Machine A takes 10 hours to complete a certain job and starts that job   [#permalink] 21 Mar 2017, 01:02
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# Machine A takes 10 hours to complete a certain job and starts that job

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Are there poly time algorithms to determine if a graph is almost bipartite?

Given an undirected graph G, we can say that G is almost bipartite if deleting k edges (or vertices) would make it bipartite.

Are there poly time algorithms to determine if a graph is exactly or approximately almost bipartite?

• What do you mean by "approximately almost bipartite"?   – user6973 Aug 28 '16 at 17:01
• It's np-hard for general $k$ because it's basically the max cut problem. I don't think this is research-level – Sasho Nikolov Aug 28 '16 at 18:26
• @RickyDemer I meant that the output could be a 1 + eps approximation of the number of edges or vertices needed to make the graph bipartite for example. I would allow some probability is of error too. – Lembik Aug 28 '16 at 19:18

The vertex version is called "odd cycle transversal"; it's NP-complete but fixed-parameter tractable. See:

Yannakakis, Mihalis (1978), "Node-and edge-deletion NP-complete problems", Proceedings of the 10th ACM Symposium on Theory of Computing (STOC '78), pp. 253–264, doi:10.1145/800133.804355.

Reed, Bruce; Smith, Kaleigh; Vetta, Adrian (2004), "Finding odd cycle transversals", Operations Research Letters, 32 (4): 299–301, doi:10.1016/j.orl.2003.10.009.

Hüffner, Falk (2005), "Algorithm engineering for optimal graph bipartization", Experimental and Efficient Algorithms: 240–252, doi:10.1007/11427186_22.

The edge version has been called "edge bipartization"; it's also NP-complete but fixed-parameter tractable. See:

Guo, Jiong; Gramm, Jens; Hüffner, Falk; Niedermeier, Rolf; Wernicke, Sebastian (2006), "Compression-based fixed-parameter algorithms for feedback vertex set and edge bipartization", JCSS 72 (8): 1386–1396, doi:10.1016/j.jcss.2006.02.001.

Odd cycle transversal has an $O(\sqrt{\log n})$ approximation algorithm, but (assuming the Unique Games Conjecture) no constant-factor approximation; see (references copied from "On Polynomial Kernels for Structural Parameterizations of Odd Cycle Transversal" by Jansen and Kratsch):
Agarwal, Amit, Charikar, Moses, Makarychev, Konstantin, Makarychev, Yury, $O(\sqrt{\log n})$ approximation algorithms for Min UnCut, Min 2CNF deletion, and directed cut problems, STOC'05, pp. 573–581.


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# Why does L’Hopital’s rule fail in calculating $\lim_{x \to \infty} \frac{x}{x+\sin(x)}$?

$$\lim_{x \to \infty} \frac{x}{x+\sin(x)}$$

This is of the indeterminate form of type $\frac{\infty}{\infty}$, so we can apply l’Hopital’s rule:

$$\lim_{x\to\infty}\frac{x}{x+\sin(x)}=\lim_{x\to\infty}\frac{(x)’}{(x+\sin(x))’}=\lim_{x\to\infty}\frac{1}{1+\cos(x)}$$

This limit doesn’t exist, but the initial limit clearly approaches $1$. Where am I wrong?

Your only error — and it’s a common one — is in a subtle misreading of L’Hopital’s rule. What the rule says is that IF the limit of $$f’$$ over $$g’$$ exists then the limit of $$f$$ over $$g$$ also exists and the two limits are the same. It doesn’t say anything if the limit of $$f’$$ over $$g’$$ doesn’t exist.


text:
# American Institute of Mathematical Sciences

• Previous Article
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Second order elliptic operators in $L^2$ with first order degeneration at the boundary and outward pointing drift
March  2015, 14(2): 421-437. doi: 10.3934/cpaa.2015.14.421

## On small perturbation of four-dimensional quasi-periodic system with degenerate equilibrium point

 1 School of Mathematics, Shandong University, Jinan, Shandong 250100, China, China

Received  December 2013 Revised  July 2014 Published  December 2014

This paper focuses on quasi-periodic perturbation of four dimensional nonlinear quasi-periodic system. Using the KAM method, the perturbed system can be reduced to a suitable normal form with zero as equilibrium point by a quasi-periodic transformation. Hence, the perturbed system has a quasi-periodic solution near the equilibrium point.
Citation: Wenhua Qiu, Jianguo Si. On small perturbation of four-dimensional quasi-periodic system with degenerate equilibrium point. Communications on Pure and Applied Analysis, 2015, 14 (2) : 421-437. doi: 10.3934/cpaa.2015.14.421
##### References:
 [1] N. N. Bogoljubov, Y. A. Mitropolskii and A. M. Samoilenko, Methods Of Accelerated Convergence In Nonlinear Mechanics, Springer, New York, 1976 (Russian original: Naukova Dumka, Kiev 1969). [2] H. Her and J. You, Full measure reducibility for generic one-parameter family of quasiperiodic linear systems, J. Dynam. Differential Equations, 20 (2008), 831-866. doi: 10.1007/s10884-008-9113-6. [3] A. Jorba and C. Simó, On quasi-periodic perturbations of elliptic equilibrium points, SIAM J. Math. Anal., 27 (1996), 1704-1737. doi: 10.1137/S0036141094276913. [4] A. Jorba and C. Simó, On the reducibility of linear differential equations with quasiperiodic coefficients, J. Differential Equations, 98 (1992), 111-124. doi: 10.1016/0022-0396(92)90107-X. [5] J. Moser, Convergent series expansion for quasi-periodic motions, Math. Ann., 169 (1967), 136-176. [6] J. Xu, On Quasi-periodic perturbations of hyperbalic-type degeneate equilibrium point of a class of planar system, Discrete and Continous Dynamical Systems, 33 (2013), 2593-2619. doi: doi:10.3934/dcds.2013.33.2593. [7] J. Xu, On small perturbation of two-dimensional quasi-periodic systems with hyperbolic-type degenerate equilibrium point, J. Differential Equations, 250 (2011), 551-571. doi: 10.1016/j.jde.2010.09.030. [8] J. Xu, Persistence of Floquet invariant tori for a class of non-conservative dynamical systems, Proc. Amer. Math. Soc., 135 (2007), 805-814. doi: 10.1090/S0002-9939-06-08529-7. [9] J. Xu and Q. Zheng, On the reducibility of linear differential equations with quasiperiodic coefficients which are degenerate, Proc. Amer. Math. Soc., 126 (1998), 1445-1451. doi: 10.1090/S0002-9939-98-04523-7. [10] J. Xu and S. Jiang, Reducibility for a class of nonlinear quasi-periodic differential equations with degenerate equilibrium point under small perturbation, Ergodic Theory and Dynamical Systems, 31 (2011), 599-611. doi: 10.1017/S0143385709001114.

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##### References:
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# Group Theory (math.GR)

• Feb 28 2017 math.GR math.CT math.DG arXiv:1702.08282v1
We explain that general differential calculus and Lie theory have a common foundation: Lie Calculus is differential calculus, seen from the point of view of Lie theory, by making use of the groupoid concept as link between them. Higher order theory naturally involves higher algebra (n-fold groupoids). Keywords: (conceptual, topological) differential calculus, groupoids, higher algebra($n$-fold groupoids), Lie group, Lie groupoid, tangent groupoid, cubes of rings
• Feb 28 2017 math.GR arXiv:1702.08426v1
In the present article we introduce and study a class of topological reflection spaces that we call Kac-Moody symmetric spaces. These are associated with split real Kac-Moody groups and generalize Riemannian symmetric spaces of non-compact type. Based on preliminary work by the third-named author we observe that in a non-spherical Kac-Moody symmetric space there exist pairs of points that do not lie on a common geodesic; however, any two points can be connected by a chain of geodesic segments. We moreover classify maximal flats in Kac-Moody symmetric spaces and study their intersection patterns, leading to a classification of global and local automorphisms. Furthermore, we define a notion of asymptoticity for certain geodesic rays in a Kac-Moody symmetric space and use it to define a partial boundary. We show that such a partial boundary is simplicially isomorphic to the geometric realization of the twin buildings of the underlying split real Kac-Moody group, and show that every automorphism of the symmetric space is uniquely determined by the induced simplicial automorphism of the partial boundary. Altogether it turns out that Kac-Moody symmetric spaces share several properties with Riemannian symmetric spaces; in many respects they behave similarly to hovels, their non-Archimedean cousins. Some of our methods apply to general topological reflection spaces beyond the Kac-Moody setting.
• Sometimes it is possible to embed an algebraic trapdoor in a block cipher. Building on previous research, in this paper we investigate an especially dangerous algebraic structure, which is called a hidden sum and which is related to some regular subgroups of the affine group. Mixing group theory arguments and cryptographic tools, we pass from characterizing our hidden sums to designing an efficient algorithm to perform the necessary preprocessing for the exploitation of the trapdoor.
• The Markoff group of transformations is a group $\Gamma$ of affine integral morphisms, which is known to act transitively on the set of all positive integer solutions to the equation $x^{2}+y^{2}+z^{2}=xyz$. The fundamental strong approximation conjecture for the Markoff equation states that for every prime $p$, the group $\Gamma$ acts transitively on the set $X^{*}\left(p\right)$ of non-zero solutions to the same equation over $\mathbb{Z}/p\mathbb{Z}$. Recently, Bourgain, Gamburd and Sarnak proved this conjecture for all primes outside a small exceptional set. In the current paper, we study a group of permutations obtained by the action of $\Gamma$ on $X^{*}\left(p\right)$, and show that for most primes, it is the full symmetric or alternating group. We use this result to deduce that $\Gamma$ acts transitively also on the set of non-zero solutions in a big class of composite moduli. Our result is also related to a well-known theorem of Gilman, stating that for any finite non-abelian simple group $G$ and $r\ge3$, the group $\mathrm{Aut}\left(F_{r}\right)$ acts on at least one $T_{r}$-system of $G$ as the alternating or symmetric group. In this language, our main result translates to that for most primes $p$, the group $\mathrm{Aut}\left(F_{2}\right)$ acts on a particular $T_{2}$-system of $\mathrm{PSL}\left(2,p\right)$ as the alternating or symmetric group.
• A regular $t$-balanced Cayley map (RBCM$_t$ for short) on a group $\Gamma$ is an embedding of a Cayley graph on $\Gamma$ into a surface with some special symmetric properties. We propose a reduction method to study RBCM$_t$'s, and as a first practice, we completely classify RBCM$_t$'s for a class of split metacyclic 2-groups.
• We prove that for every $r>0$ if a non-positively curved $(p,q)$-map $M$ contains no flat submaps of radius $r$, then the area of $M$ does not exceed $Crn$ for some constant $C$. This strengthens a theorem of Ivanov and Schupp. We show that an infinite $(p,q)$-map which tessellates the plane is quasi-isometric to the Euclidean plane if and only if the map contains only finitely many non-flat vertices and faces. We also generalize Ivanov and Schupp's result to a much larger class of maps, namely to maps with angle functions.
• Answering a question asked by Agol and Wise, we show that a desired stronger form of Wise's malnormal special quotient theorem does not hold. The counterexamples are generalizations of triangle groups, built using the Ramanujan graphs constructed by Lubotzky--Phillips--Sarnak.
• The role of finite centralizers of involutions in pseudo-finite groups is analyzed. Using basic techniques from infinite group theory, it is shown that a pseudo-finite group admitting a definable involutory automorphism fixing only finitely many elements is finite-by-abelian-by-finite. As a consequence, an alternative proof of the corresponding result for periodic groups due to Hartley and Meixner is given, as well as a gently improvement regarding definable properties. Furthermore, it is shown that any pseudo-finite group has an infinite abelian subgroup, and that in any pseudo-finite group in which the centralizer of any element is finite or has finite index, the FC-center is a finite index definable subgroup.
• Feb 28 2017 cs.FL math.GR arXiv:1702.08083v1
Given a pseudoword over suitable pseudovarieties, we associate to it a labeled linear order determined by the factorizations of the pseudoword. We show that, in the case of the pseudovariety of aperiodic finite semigroups, the pseudoword can be recovered from the labeled linear order.
• This is the second part of a two part work in which we prove that for every finitely generated subgroup $\Gamma < \mathsf{Out}(F_n)$, either $\Gamma$ is virtually abelian or its second bounded cohomology $H^2_b(\Gamma;\mathbb{R})$ contains an embedding of $\ell^1$. Here in Part II we focus on finite lamination subgroups $\Gamma$ --- meaning that the set of all attracting laminations of elements of $\Gamma$ is finite --- and on the construction of hyperbolic actions of those subgroups to which the general theory of Part I is applicable.
• We show that every group in a large family of (not necessarily torsion) spinal groups acting on the ternary rooted tree is of subexponential growth.
• We investigate the action of outer automorphisms of finite groups of Lie type on their irreducible characters. We obtain a definite result for cuspidal characters. As an application we verify the inductive McKay condition for some further infinite families of simple groups at certain primes.
• We note a generalization of Whyte's geometric solution to the von Neumann problem for locally compact groups in terms of Borel and clopen piecewise translations. This strengthens a result of Paterson on the existence of Borel paradoxical decompositions for non-amenable locally compact groups. Along the way, we study the connection between some geometric properties of coarse spaces and certain algebraic characteristics of their wobbling groups.
• We prove that profinite completions of Burnside-type surface group quotients are not virtually prosolvable, in general.
• Feb 28 2017 math.GR arXiv:1702.07761v1
A theorem of Myasnikov and Roman'kov says that any verbally closed subgroup of a finitely generated free group is a retract. We prove that all free (and many virtually free) verbally closed subgroups are retracts in any finitely generated group.

Robert Raussendorf Jan 24 2017 22:29 UTC

Regarding Mark's above comment on the role of the stabilizer states: Yes, all previous works on the subject have used the stabilizer states and Clifford gates as the classical backbone. This is due to the Gottesman-Knill theorem and related results. But is it a given that the free sector in quantum

...(continued)
Planat Jan 24 2017 13:09 UTC

Are you sure? Since we do not propose a conjecture, there is nothing wrong. A class of strange states underlie the pentagons in question. The motivation is to put the magic of computation in the permutation frame, one needs more work to check its relevance.

Mark Howard Jan 24 2017 09:59 UTC

It seems interesting at first sight, but after reading it the motivation is very muddled. It boils down to finding pentagons (which enable KCBS-style proofs of contextuality) within sets of projectors, some of which are stabilizer states and some of which are non-stabilizer states (called magic stat

...(continued)


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# Continuity, and continuity in topology.

Metric spaces:
Neighborhood of a point $a$ is a Set of point $N$, such that $\exists\delta>0:B_\delta(a)\subset N$ ($B_r(x)$ = open ball at x of radius r)

Definition of open set: "A subset $O$ of a metric space is said to be open if it is a neighborhood to each of its points"

(It is now easy to go from this to the metrisable topological spaces)

Continuity:
($a$ is a point, $M$ and $N$ are neighborhoods (of the form $\forall M\exists N$), $M$ is a neigh. to $f(a)$, $N$ of $a$)

$\epsilon-\delta \iff f(B_\delta(a))\subset B_\epsilon(f(a))\iff B_\delta(a)\subset f^{-1}(B_\epsilon(a))\iff f(N)\subset M\iff N\subset f^{-1}(M)\iff f^{-1}(M)\text{ is a neigh. of }a$

This shows the "list" going from the well known $\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon$
requiring only a leap of faith from $|\text{_}|$ to metrics (the ball definition) then it's straight on to the more topological $\forall N_{f(a)}, f^{-1}(N_{f(a)})\text{ is a neigh. of }a$

Topological space:
Given a topological space $(X,J)$ a set $N\subset X$ is a neighborhood of a point $a\in X$ if $N$ contains an open set that contains $a$

Definition of open set: "Thee members of $J$ are called 'open sets'"

Question

The leap of faith ("Trust me, we're not just calling these 'open sets' rather than 'Thingymajig', think of them as open sets" and "Yeah, these are 'neighborhoods' we give them this name because they are like neighborhoods from metric spaces') required is really quite large.

I know it cannot be proved (there's no distance thus no open balls for the general topology) but I'd like some more discussion about accepting these definitions. From what I can tell.

"If we call an open set this, and a neighborhood that, later when we deal with distance and metrics, we can define open sets and neigh. using the metric, and show it has these properties, the that leads to continuity as you think of it"

So it's sort of going "from topology, to metric spaces" in terms of definitions and showing properties, however every book on the subject goes the other way. It builds on metric spaces then suddenly "we call these open sets!" I'd like to be a bit more confident and happy with the transition, recommended reading, examples, anything would be great.

• The notion of a topological space is more general than that of a metric space. Topologies that can be realized by a metric (distance function) are called metrizable. Some topologies are not metrizable, hence the greater generality of topological spaces compared to metric spaces. What precisely is your Question? – hardmath Aug 19 '14 at 0:13

• @AlecTeal There is literally no question mark in the post, so it's a little hard to evaluate whether this is answering the question or not. Anyway, you're right that it's more mathematically sensible to define open sets generally, and then look at special cases like metric spaces, but in practice we usually start with the simpler examples for pedagogical purposes. In this case, students often learn first $\mathbb{Q}$, then $\mathbb{R}$, then $\mathbb{R}^n$, then metric spaces, then topological spaces. Each is a generalization, each has open sets (which we first learn as "open intervals"). – Slade Aug 16 '14 at 14:53


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## A radio is bought for Rs.750 and sold for 4% loss. Find S.P.

C.P. = 750                                          loss = 4%

S.P. = 750 $\times$ $\frac{100 – 4}{100}$

750 $\times$ $\frac{96}{100}$ = 360