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# IO Processors: Writing Your Own Software¶
There are a number of steps required before you can start writing your own software for an IOP (IO Processor). This document will describe the IOP architecture, and how to set up and build the required software projects to allow you to write your own application for the MicroBlaze inside an IOP. Xilinx® SDK projects can be created manually using the SDK GUI, or software can be built using a Makefile flow.
## IO Processors¶
As seen previously, an IOP can be used as a flexible controller for different types of external peripherals. The ARM® Cortex®-A9 is an application processor, which runs Pynq and Jupyter notebook on a Linux OS. This scenario is not well suited to real-time applications, which is a common requirement for an embedded systems. In the base overlay there are three IOPs. As well as acting as a flexible controller, an IOP can be used as dedicated real-time controller.
IOPs can also be used standalone to offload some processing from the main processor. However, note that the MicroBlaze processor inside an IOP in the base overlay is running at 100 MHz, compared to the Dual-Core ARM Cortex-A9 running at 650 MHz. The clock speed, and different processor architectures and features should be taken into account when offloading pure application code. e.g. Vector processing on the ARM Cortex-A9 Neon processing unit will be much more efficient than running on the MicroBlaze. The MicroBlaze is most appropriate for low-level, background, or real-time applications.
There are two types of IOP, a Pmod IOP and an Arduino IOP.
Previous sections showed the similarities between the Pmod IOP and Arduino IOP. Each IOP contains a Xilinx MicroBlaze processor, a Debug module, and one or more of the following functional units and interface peripherals:
The Arduino also includes a UART, and XADC.
The interface peripherals are connected to a Configurable Switch. The switch is different for the Pmod and the Arduino IOPs. The Pmod configurable switch connects to a Pmod port, and the Arduino configurable switch connects to an Arduino interface connector.
Pmod IOP:
The IOP’s configurable switch can be used to route signals between the physical interface, and the available internal peripherals in the IOP sub-system.
## Software requirements¶
A MicroBlaze cross-compiler is required to build software for the MicroBlaze inside an IOP. Xilinx SDK contains the MicroBlaze cross-compiler and was used to build all Pmod device drivers released with Pynq and is available for free. It should be noted that Pynq ships with precompiled IOP executables to support various peripherals (see Pynq Modules), but that full source code is available from the project GitHub. Xilinx software is only needed if you intend to build your own IOP applications/peripheral drivers. A free, fully functional, version of the Xilinx tools is available for Pynq if required (see the free Xilinx Vivado WebPack for more details).
The current Pynq release is built using Vivado and SDK 2016.1. it is recommended to use the same version to rebuild existing Vivado and SDK projects. If you only intend to build software, you will only need to install SDK. The full Vivado and SDK installation is only required to design new overlays.
You can use the Vivado HLx Web Install Client and select SDK and/or Vivado during the installation.
Pynq also support building of bitstreams from SDSoC. SDSoC is currently a separate download and installation from the main Vivado/SDK software.
## Compiling projects¶
Software executables run on the MicroBlaze inside the IOP. Code for the MicroBlaze can be written in C or C++ and compiled using the Xilinx SDK (Software Development Kit).
You can pull or clone the Pynq GitHub repository, and all the driver source and project files can be found in <GitHub Repository>\Pynq-Z1\sdk, (Where <GitHub Repository> is the location of the PYNQ repository).
These projects are considered SDK Application projects and contain the top level application. Each SDK project requires a BSP project (Board Support Package), and a hardware platform project. See below for more details. Software libraries are included in a Board Support Package (BSP) project, and the BSP is linked to from the application project.
All Application projects can be compiled from the command line using Makefiles, or imported into the SDK GUI.
You can also use existing projects as a starting point to create your own project.
### HDF file¶
Before an Application project and BSP can be created or compiled in SDK, a Hardware Platform project is required. A Hardware Platform defines the peripherals in the IOP subsystem, and the memory map of the system, and is used by the BSP to build software libraries to support the underlying hardware.
A Hardware Description File (.hdf), created by Vivado, is used to create the Hardware Platfrom project in SDK.
A precompiled .hdf file is provided, so it is not necessary to run Vivado to generate a .hdf file:
<GitHub Repository>/Pynq-Z1/sdk/
### Board Support Package¶
The BSP (Board Support Package) contains software libraries and drivers to support the underlying peripherals in the system.
A BSP must be linked to a Hardware Platform, as this is where the peripherals in the system are defined. An Application Project is then linked to a BSP, and can use the libraries available in the BSP.
### Building the projects¶
A Makefile to automatically create and build the Hardware Platform and the BSP can be found in the same location as the .hdf file.
<GitHub Repository>/Pynq-Z1/sdk/makefile
Application projects for peripherals that ship with Pynq (e.g. Pmods and Grove peripherals) can also be found in the same location. Each project is contained in a separate folder.
The Makefile uses the .hdf file to create the Hardware Platform. The BSP can then be created. The application projects will also be compiled automatically as part of this process.
The Makefile requires SDK to be installed, and can be run from Windows, or Linux.
To run make from Windows, open SDK, and choose a temporary workspace (make sure this path is external to the downloaded GitHub repository). From the Xilinx Tools menu, select Launch Shell
In Linux, open a terminal, and source the SDK tools.
From either the Windows Shell, or the Linux terminal, navigate to the sdk folder in your local copy of the GitHub repository:
cd to <GitHub Repository>/Pynq-Z1/sdk and run make
This will create the Hardware Platform Project (hw_def), and the Board Support Package (bsp), and then link and build all the application projects.
If you examine the Makefile, you can see how the MBBINS variable at the top of the makefile is used to compile the application projects. If you want to add your own custom project to the build process, you need to add the project name to the MBBINS variable, and save the project in the same location as the other application projects.
Individual projects can be built by navigating to the <project directory>/Debug and running make.
### Binary files¶
Compiling code produces an executable file (.elf) which needs to be converted to binary format (.bin) to be downloaded to, and run on, an IOP.
A .bin file can be generated from a .elf by running the following command from the SDK shell:
mb-objcopy -O binary <inputfile>.elf <outputfile>.bin
This is done automatically by the makefile for the existing application projects. The makefile will also copy all .bin files into the <GitHub Repository>/Pynq-Z1/sdk/bin folder.
### Creating your own Application project¶
Using the Makefile flow, you can use an existing project as a starting point for your own project.
Copy and rename the project, and modify or replace the .c file in the src/ with your C code. The generated .bin file will have the same base name as your C file.
e.g. if your C code is my_peripheral.c, the generated .elf and .bin will be my_peripheral.elf and my_peripheral.bin.
We encourage the following naming convention for applications <pmod|grove|arduino>_<peripheral>
You will need to update references from the old project name to your new project name in <project directory>/Debug/makefile and <project directory>/Debug/src/subdir.mk
If you want your project to build in the main Makefile, you should also append the .bin name of your project to the MBBINS variable at the top of the makefile.
If you are using the SDK GUI, you can import the Hardware Platform, BSP, and any application projects into your SDK workspace.
The SDK GUI can be used to build and debug your code.
## IOP Memory¶
The IOP instruction and data memory is implemented in a dual port Block RAM, with one port connected to the IOP, and the other to the ARM processor. This allows an executable binary file to be written from the ARM (i.e. the Pynq environment) to the IOP instruction memory. The IOP can also be reset from Pynq, allowing the IOP to start executing the new program. The IOP data memory is also used as a mailbox for communication and data exchanges between the Pynq environment and the IOP.
### Memory map¶
The IOP memory is 64KB of shared data and instruction memory. Instruction memory for the IOP starts at address 0x0. Pynq and the application running on the IOP can write to anywhere in the shared memory space (although care should be taken not to write to the instruction memory unintentionally as this will corrupt the running application).
When building the MicroBlaze project, the compiler will only ensure that the application and allocated stack and heap fit into the BRAM. For communication between the ARM and the MicroBlaze, a part of the shared memory space must also be reserved within the MicroBlaze address space.
There is no memory management in the IOP. You must ensure the application, including stack and heap, do not overflow into the defined data area. Remember that declaring a stack and heap size only allocates space to the stack and heap. No boundary is created, so if sufficient space was not allocated, the stack and heap may overflow.
If you need to modify the stack and heap for an application, the linker script can be found in the <project>/src/ directory.
It is recommended to follow the convention for data communication between the two processors via MAILBOX. These MAILBOX values are defined in the header file.
Instruction and data memory start 0x0 Instruction and data memory size 0xf000 Shared mailbox memory start 0xf000 Shared mailbox memory size 0x1000 Shared mailbox Command Address 0xfffc
The following example explains how Python could initiate a read from a peripheral connected to an IOP.
1. Python writes a read command (e.g. 0x3) to the mailbox command address (0xfffc).
2. MicroBlaze application checks the command address, and reads and decodes the command.
3. MicroBlaze performs a read from the peripheral and places the data at the mailbox base address (0xf000).
4. MicroBlaze writes 0x0 to the mailbox command address (0xfffc) to confirm transaction is complete.
5. Python checks the command address (0xfffc), and sees that the MicroBlaze has written 0x0, indicating the read is complete, and data is available.
6. Python reads the data in the mailbox base address (0xf000), completing the read.
## Controlling the Pmod IOP Switch¶
There are 8 data pins on a Pmod port, that can be connected to any of 16 internal peripheral pins (8x GPIO, 2x SPI, 4x IIC, 2x Timer).
Each pin can be configured by writing a 4 bit value to the corresponding place in the IOP Switch configuration register. The following function, part of the provided pmod_io_switch_v1_0 driver (pmod.h) can be used to configure the switch.
void config_pmod_switch(char pin0, char pin1, char pin2, char pin3, char pin4, \
char pin5, char pin6, char pin7);
While each parameter is a “char” only the lower 4-bits are currently used to configure each pin.
Switch mappings used for IOP Switch configuration:
Pin Value
GPIO_0 0x0
GPIO_1 0x1
GPIO_2 0x2
GPIO_3 0x3
GPIO_4 0x4
GPIO_5 0x5
GPIO_6 0x6
GPIO_7 0x7
SCL 0x8
SDA 0x9
SPICLK 0xa
MISO 0xb
MOSI 0xc
SS 0xd
PWM 0xe
TIMER 0xf
For example, to connect the physical pins GPIO 0-7 to the internal GPIO_0 - GPIO_7:
config_pmod_switch(GPIO_0, GPIO_1, GPIO_2, GPIO_3, GPIO_4, \
GPIO_5, GPIO_6, GPIO_7);
From Python all the constants and addresses for the IOP can be found in:
<GitHub Repository>/python/pynq/iop/iop_const.py
Note that if two or more pins are connected to the same signal, the pins are OR’d together internally. This is not recommended and should not be done unintentionally.
Any application that uses the Pmod driver should also call pmod_init() at the beginning of the application.
## Running code on different IOPs¶
The shared memory is the only connection between the ARM and the IOPs in the base overlay. The shared memory of a MicroBlaze is mapped to the ARM address space. Some example mappings are shown below to highlight the address translation between MicroBlaze and ARM’s memory spaces.
IOP Base Address MicroBlaze Address Space ARM Equivalent Address Space
0x4000_0000 0x0000_0000 - 0x0000_ffff 0x4000_0000 - 0x4000_ffff
0x4200_0000 0x0000_0000 - 0x0000_ffff 0x4200_0000 - 0x4200_ffff
0x4400_0000 0x0000_0000 - 0x0000_ffff 0x4400_0000 - 0x4400_ffff
Note that each MicroBlaze has the same range for its address space. However, the location of each IOPs address space in the ARM memory map is different for each IOP. As the address space is the same for each IOP, any binary compiled for one Pmod IOP will work on another Pmod IOP.
e.g. if IOP1 exists at 0x4000_0000, and IOP2 (a second instance of an IOP) exists at 0x4200_0000, the same binary can run on IOP1 by writing the binary from python to the 0x4000_0000 address space, and on IOP2 by writing to the 0x4200_0000.
## IOP Application Example¶
Taking Pmod ALS as an example IOP driver (used to control the PMOD light sensor), first open the pmod_als.c file:
<GitHub Repository>/Pynq-Z1/sdk/pmod_als/src/pmod_als.c
Note that the pmod.h header file is included.
Some COMMANDS are defined by the user. These values can be chosen to be any value, but must correspond with the Python part of the driver.
By convention, 0x0 is reserved for no command/idle/acknowledge, and IOP commands can be any non-zero value.
The ALS peripheral has as SPI interface. Note the user defined function get_sample() which calls an SPI function spi_transfer() call defined in pmod.h.
In main() notice config_pmod_switch() is called to initialize the switch with a static configuration. This means that if you want to use this code with a different pin configuration, the C code must be modified and recompiled.
Next, the while(1) loop is entered. In this loop the IOP continually checks the MAILBOX_CMD_ADDR for a non-zero command. Once a command is received from Python, the command is decoded, and executed.
Taking the first case, reading a single value:
case READ_SINGLE_VALUE:
MAILBOX_DATA(0) = get_sample();
get_sample() is called and a value returned to the first position (0) of the MAILBOX_DATA.
MAILBOX_CMD_ADDR is reset to zero to acknowledge to the ARM processor that the operation is complete and data is available in the mailbox.
### Examine Python Code¶
With the IOP Driver written, the Python class can be built that will communicate with that IOP.
<GitHub Repository>/python/pynq/iop/pmod_als.py
First the MMIO, request_iop, iop_const, PMODA and PMODB are imported.
import time
from pynq import MMIO
from pynq.iop import request_iop
from pynq.iop import iop_const
from pynq.iop import PMODA
from pynq.iop import PMODB
ALS_PROGRAM = "pmod_als.bin"
The MicroBlaze binary for the IOP is also declared. This is the application executable, and will be loaded into the IOP instruction memory.
The ALS class and an initialization method are defined:
class Pmod_ALS(object):
def __init__(self, if_id):
The initialization function for the module requires an IOP index. For Grove peripherals and the StickIt connector, the StickIt port number can also be used for initialization. The __init__ is called when a module is instantiated. e.g. from Python:
from pynq.pmods import Pmod_ALS
als = Pmod_ALS(PMODB)
Looking further into the initialization method, the _iop.request_iop() call instantiates an instance of an IOP on the specified pmod_id and loads the MicroBlaze executable (ALS_PROGRAM) into the instruction memory of the appropriate MicroBlaze.
self.iop = request_iop(if_id, PMOD_ALS_PROGRAM)
An MMIO class is also instantiated to enable read and write to the shared memory.
self.mmio = self.iop.mmio
Finally, the iop.start() call pulls the IOP out of reset. After this, the IOP will be running the als.bin executable.
self.iop.start()
### Example of Python Class Runtime Methods¶
The read method in the Pmod_ALS class will simply read an ALS sample and return that value to the caller. The following steps demonstrate a Python to MicroBlaze read transaction specfic to the ALS class.
def read(self):
First, the command is written to the MicroBlaze shared memory using mmio.write(). In this case the value 0x3 represents a read command. This value is user defined in the Python code, and must match the value the C program running on the IOP expects for the same function.
self.mmio.write(iop_const.MAILBOX_OFFSET+
iop_const.MAILBOX_PY2IOP_CMD_OFFSET, 3)
When the IOP is finished, it will write 0x0 to the command area. The Python code now uses mmio.read() to check if the command is still pending (in this case, when the 0x3 value is still present at the CMD_OFFSET). While the command is pending, the Python class blocks.
while (self.mmio.read(iop_const.MAILBOX_OFFSET+
iop_const.MAILBOX_PY2IOP_CMD_OFFSET) == 3):
pass
Once the command is no longer 0x3, i.e. the acknowledge has been received, the result is read from the DATA area of the shared memory MAILBOX_OFFSET using mmio.read().
return self.mmio.read(iop_const.MAILBOX_OFFSET)
Notice the iop_const values are used in these function calls, values that are predefined in iop_const.py.
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## G = Dic3×C2×C10order 240 = 24·3·5
### Direct product of C2×C10 and Dic3
Series: Derived Chief Lower central Upper central
Derived series C1 — C3 — Dic3×C2×C10
Chief series C1 — C3 — C6 — C30 — C5×Dic3 — C10×Dic3 — Dic3×C2×C10
Lower central C3 — Dic3×C2×C10
Upper central C1 — C22×C10
Generators and relations for Dic3×C2×C10
G = < a,b,c,d | a2=b10=c6=1, d2=c3, ab=ba, ac=ca, ad=da, bc=cb, bd=db, dcd-1=c-1 >
Subgroups: 152 in 108 conjugacy classes, 86 normal (14 characteristic)
C1, C2, C2 [×6], C3, C4 [×4], C22 [×7], C5, C6, C6 [×6], C2×C4 [×6], C23, C10, C10 [×6], Dic3 [×4], C2×C6 [×7], C15, C22×C4, C20 [×4], C2×C10 [×7], C2×Dic3 [×6], C22×C6, C30, C30 [×6], C2×C20 [×6], C22×C10, C22×Dic3, C5×Dic3 [×4], C2×C30 [×7], C22×C20, C10×Dic3 [×6], C22×C30, Dic3×C2×C10
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C5, S3, C2×C4 [×6], C23, C10 [×7], Dic3 [×4], D6 [×3], C22×C4, C20 [×4], C2×C10 [×7], C2×Dic3 [×6], C22×S3, C5×S3, C2×C20 [×6], C22×C10, C22×Dic3, C5×Dic3 [×4], S3×C10 [×3], C22×C20, C10×Dic3 [×6], S3×C2×C10, Dic3×C2×C10
Smallest permutation representation of Dic3×C2×C10
Regular action on 240 points
Generators in S240
(1 112)(2 113)(3 114)(4 115)(5 116)(6 117)(7 118)(8 119)(9 120)(10 111)(11 192)(12 193)(13 194)(14 195)(15 196)(16 197)(17 198)(18 199)(19 200)(20 191)(21 171)(22 172)(23 173)(24 174)(25 175)(26 176)(27 177)(28 178)(29 179)(30 180)(31 185)(32 186)(33 187)(34 188)(35 189)(36 190)(37 181)(38 182)(39 183)(40 184)(41 228)(42 229)(43 230)(44 221)(45 222)(46 223)(47 224)(48 225)(49 226)(50 227)(51 201)(52 202)(53 203)(54 204)(55 205)(56 206)(57 207)(58 208)(59 209)(60 210)(61 165)(62 166)(63 167)(64 168)(65 169)(66 170)(67 161)(68 162)(69 163)(70 164)(71 148)(72 149)(73 150)(74 141)(75 142)(76 143)(77 144)(78 145)(79 146)(80 147)(81 153)(82 154)(83 155)(84 156)(85 157)(86 158)(87 159)(88 160)(89 151)(90 152)(91 132)(92 133)(93 134)(94 135)(95 136)(96 137)(97 138)(98 139)(99 140)(100 131)(101 127)(102 128)(103 129)(104 130)(105 121)(106 122)(107 123)(108 124)(109 125)(110 126)(211 240)(212 231)(213 232)(214 233)(215 234)(216 235)(217 236)(218 237)(219 238)(220 239)
(1 2 3 4 5 6 7 8 9 10)(11 12 13 14 15 16 17 18 19 20)(21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48 49 50)(51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70)(71 72 73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99 100)(101 102 103 104 105 106 107 108 109 110)(111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130)(131 132 133 134 135 136 137 138 139 140)(141 142 143 144 145 146 147 148 149 150)(151 152 153 154 155 156 157 158 159 160)(161 162 163 164 165 166 167 168 169 170)(171 172 173 174 175 176 177 178 179 180)(181 182 183 184 185 186 187 188 189 190)(191 192 193 194 195 196 197 198 199 200)(201 202 203 204 205 206 207 208 209 210)(211 212 213 214 215 216 217 218 219 220)(221 222 223 224 225 226 227 228 229 230)(231 232 233 234 235 236 237 238 239 240)
(1 106 95 81 62 76)(2 107 96 82 63 77)(3 108 97 83 64 78)(4 109 98 84 65 79)(5 110 99 85 66 80)(6 101 100 86 67 71)(7 102 91 87 68 72)(8 103 92 88 69 73)(9 104 93 89 70 74)(10 105 94 90 61 75)(11 34 28 59 45 238)(12 35 29 60 46 239)(13 36 30 51 47 240)(14 37 21 52 48 231)(15 38 22 53 49 232)(16 39 23 54 50 233)(17 40 24 55 41 234)(18 31 25 56 42 235)(19 32 26 57 43 236)(20 33 27 58 44 237)(111 121 135 152 165 142)(112 122 136 153 166 143)(113 123 137 154 167 144)(114 124 138 155 168 145)(115 125 139 156 169 146)(116 126 140 157 170 147)(117 127 131 158 161 148)(118 128 132 159 162 149)(119 129 133 160 163 150)(120 130 134 151 164 141)(171 202 225 212 195 181)(172 203 226 213 196 182)(173 204 227 214 197 183)(174 205 228 215 198 184)(175 206 229 216 199 185)(176 207 230 217 200 186)(177 208 221 218 191 187)(178 209 222 219 192 188)(179 210 223 220 193 189)(180 201 224 211 194 190)
(1 218 81 177)(2 219 82 178)(3 220 83 179)(4 211 84 180)(5 212 85 171)(6 213 86 172)(7 214 87 173)(8 215 88 174)(9 216 89 175)(10 217 90 176)(11 137 59 144)(12 138 60 145)(13 139 51 146)(14 140 52 147)(15 131 53 148)(16 132 54 149)(17 133 55 150)(18 134 56 141)(19 135 57 142)(20 136 58 143)(21 116 231 157)(22 117 232 158)(23 118 233 159)(24 119 234 160)(25 120 235 151)(26 111 236 152)(27 112 237 153)(28 113 238 154)(29 114 239 155)(30 115 240 156)(31 130 42 164)(32 121 43 165)(33 122 44 166)(34 123 45 167)(35 124 46 168)(36 125 47 169)(37 126 48 170)(38 127 49 161)(39 128 50 162)(40 129 41 163)(61 186 105 230)(62 187 106 221)(63 188 107 222)(64 189 108 223)(65 190 109 224)(66 181 110 225)(67 182 101 226)(68 183 102 227)(69 184 103 228)(70 185 104 229)(71 196 100 203)(72 197 91 204)(73 198 92 205)(74 199 93 206)(75 200 94 207)(76 191 95 208)(77 192 96 209)(78 193 97 210)(79 194 98 201)(80 195 99 202)
G:=sub<Sym(240)| (1,112)(2,113)(3,114)(4,115)(5,116)(6,117)(7,118)(8,119)(9,120)(10,111)(11,192)(12,193)(13,194)(14,195)(15,196)(16,197)(17,198)(18,199)(19,200)(20,191)(21,171)(22,172)(23,173)(24,174)(25,175)(26,176)(27,177)(28,178)(29,179)(30,180)(31,185)(32,186)(33,187)(34,188)(35,189)(36,190)(37,181)(38,182)(39,183)(40,184)(41,228)(42,229)(43,230)(44,221)(45,222)(46,223)(47,224)(48,225)(49,226)(50,227)(51,201)(52,202)(53,203)(54,204)(55,205)(56,206)(57,207)(58,208)(59,209)(60,210)(61,165)(62,166)(63,167)(64,168)(65,169)(66,170)(67,161)(68,162)(69,163)(70,164)(71,148)(72,149)(73,150)(74,141)(75,142)(76,143)(77,144)(78,145)(79,146)(80,147)(81,153)(82,154)(83,155)(84,156)(85,157)(86,158)(87,159)(88,160)(89,151)(90,152)(91,132)(92,133)(93,134)(94,135)(95,136)(96,137)(97,138)(98,139)(99,140)(100,131)(101,127)(102,128)(103,129)(104,130)(105,121)(106,122)(107,123)(108,124)(109,125)(110,126)(211,240)(212,231)(213,232)(214,233)(215,234)(216,235)(217,236)(218,237)(219,238)(220,239), (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130)(131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190)(191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230)(231,232,233,234,235,236,237,238,239,240), (1,106,95,81,62,76)(2,107,96,82,63,77)(3,108,97,83,64,78)(4,109,98,84,65,79)(5,110,99,85,66,80)(6,101,100,86,67,71)(7,102,91,87,68,72)(8,103,92,88,69,73)(9,104,93,89,70,74)(10,105,94,90,61,75)(11,34,28,59,45,238)(12,35,29,60,46,239)(13,36,30,51,47,240)(14,37,21,52,48,231)(15,38,22,53,49,232)(16,39,23,54,50,233)(17,40,24,55,41,234)(18,31,25,56,42,235)(19,32,26,57,43,236)(20,33,27,58,44,237)(111,121,135,152,165,142)(112,122,136,153,166,143)(113,123,137,154,167,144)(114,124,138,155,168,145)(115,125,139,156,169,146)(116,126,140,157,170,147)(117,127,131,158,161,148)(118,128,132,159,162,149)(119,129,133,160,163,150)(120,130,134,151,164,141)(171,202,225,212,195,181)(172,203,226,213,196,182)(173,204,227,214,197,183)(174,205,228,215,198,184)(175,206,229,216,199,185)(176,207,230,217,200,186)(177,208,221,218,191,187)(178,209,222,219,192,188)(179,210,223,220,193,189)(180,201,224,211,194,190), (1,218,81,177)(2,219,82,178)(3,220,83,179)(4,211,84,180)(5,212,85,171)(6,213,86,172)(7,214,87,173)(8,215,88,174)(9,216,89,175)(10,217,90,176)(11,137,59,144)(12,138,60,145)(13,139,51,146)(14,140,52,147)(15,131,53,148)(16,132,54,149)(17,133,55,150)(18,134,56,141)(19,135,57,142)(20,136,58,143)(21,116,231,157)(22,117,232,158)(23,118,233,159)(24,119,234,160)(25,120,235,151)(26,111,236,152)(27,112,237,153)(28,113,238,154)(29,114,239,155)(30,115,240,156)(31,130,42,164)(32,121,43,165)(33,122,44,166)(34,123,45,167)(35,124,46,168)(36,125,47,169)(37,126,48,170)(38,127,49,161)(39,128,50,162)(40,129,41,163)(61,186,105,230)(62,187,106,221)(63,188,107,222)(64,189,108,223)(65,190,109,224)(66,181,110,225)(67,182,101,226)(68,183,102,227)(69,184,103,228)(70,185,104,229)(71,196,100,203)(72,197,91,204)(73,198,92,205)(74,199,93,206)(75,200,94,207)(76,191,95,208)(77,192,96,209)(78,193,97,210)(79,194,98,201)(80,195,99,202)>;
G:=Group( (1,112)(2,113)(3,114)(4,115)(5,116)(6,117)(7,118)(8,119)(9,120)(10,111)(11,192)(12,193)(13,194)(14,195)(15,196)(16,197)(17,198)(18,199)(19,200)(20,191)(21,171)(22,172)(23,173)(24,174)(25,175)(26,176)(27,177)(28,178)(29,179)(30,180)(31,185)(32,186)(33,187)(34,188)(35,189)(36,190)(37,181)(38,182)(39,183)(40,184)(41,228)(42,229)(43,230)(44,221)(45,222)(46,223)(47,224)(48,225)(49,226)(50,227)(51,201)(52,202)(53,203)(54,204)(55,205)(56,206)(57,207)(58,208)(59,209)(60,210)(61,165)(62,166)(63,167)(64,168)(65,169)(66,170)(67,161)(68,162)(69,163)(70,164)(71,148)(72,149)(73,150)(74,141)(75,142)(76,143)(77,144)(78,145)(79,146)(80,147)(81,153)(82,154)(83,155)(84,156)(85,157)(86,158)(87,159)(88,160)(89,151)(90,152)(91,132)(92,133)(93,134)(94,135)(95,136)(96,137)(97,138)(98,139)(99,140)(100,131)(101,127)(102,128)(103,129)(104,130)(105,121)(106,122)(107,123)(108,124)(109,125)(110,126)(211,240)(212,231)(213,232)(214,233)(215,234)(216,235)(217,236)(218,237)(219,238)(220,239), (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130)(131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190)(191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230)(231,232,233,234,235,236,237,238,239,240), (1,106,95,81,62,76)(2,107,96,82,63,77)(3,108,97,83,64,78)(4,109,98,84,65,79)(5,110,99,85,66,80)(6,101,100,86,67,71)(7,102,91,87,68,72)(8,103,92,88,69,73)(9,104,93,89,70,74)(10,105,94,90,61,75)(11,34,28,59,45,238)(12,35,29,60,46,239)(13,36,30,51,47,240)(14,37,21,52,48,231)(15,38,22,53,49,232)(16,39,23,54,50,233)(17,40,24,55,41,234)(18,31,25,56,42,235)(19,32,26,57,43,236)(20,33,27,58,44,237)(111,121,135,152,165,142)(112,122,136,153,166,143)(113,123,137,154,167,144)(114,124,138,155,168,145)(115,125,139,156,169,146)(116,126,140,157,170,147)(117,127,131,158,161,148)(118,128,132,159,162,149)(119,129,133,160,163,150)(120,130,134,151,164,141)(171,202,225,212,195,181)(172,203,226,213,196,182)(173,204,227,214,197,183)(174,205,228,215,198,184)(175,206,229,216,199,185)(176,207,230,217,200,186)(177,208,221,218,191,187)(178,209,222,219,192,188)(179,210,223,220,193,189)(180,201,224,211,194,190), (1,218,81,177)(2,219,82,178)(3,220,83,179)(4,211,84,180)(5,212,85,171)(6,213,86,172)(7,214,87,173)(8,215,88,174)(9,216,89,175)(10,217,90,176)(11,137,59,144)(12,138,60,145)(13,139,51,146)(14,140,52,147)(15,131,53,148)(16,132,54,149)(17,133,55,150)(18,134,56,141)(19,135,57,142)(20,136,58,143)(21,116,231,157)(22,117,232,158)(23,118,233,159)(24,119,234,160)(25,120,235,151)(26,111,236,152)(27,112,237,153)(28,113,238,154)(29,114,239,155)(30,115,240,156)(31,130,42,164)(32,121,43,165)(33,122,44,166)(34,123,45,167)(35,124,46,168)(36,125,47,169)(37,126,48,170)(38,127,49,161)(39,128,50,162)(40,129,41,163)(61,186,105,230)(62,187,106,221)(63,188,107,222)(64,189,108,223)(65,190,109,224)(66,181,110,225)(67,182,101,226)(68,183,102,227)(69,184,103,228)(70,185,104,229)(71,196,100,203)(72,197,91,204)(73,198,92,205)(74,199,93,206)(75,200,94,207)(76,191,95,208)(77,192,96,209)(78,193,97,210)(79,194,98,201)(80,195,99,202) );
G=PermutationGroup([(1,112),(2,113),(3,114),(4,115),(5,116),(6,117),(7,118),(8,119),(9,120),(10,111),(11,192),(12,193),(13,194),(14,195),(15,196),(16,197),(17,198),(18,199),(19,200),(20,191),(21,171),(22,172),(23,173),(24,174),(25,175),(26,176),(27,177),(28,178),(29,179),(30,180),(31,185),(32,186),(33,187),(34,188),(35,189),(36,190),(37,181),(38,182),(39,183),(40,184),(41,228),(42,229),(43,230),(44,221),(45,222),(46,223),(47,224),(48,225),(49,226),(50,227),(51,201),(52,202),(53,203),(54,204),(55,205),(56,206),(57,207),(58,208),(59,209),(60,210),(61,165),(62,166),(63,167),(64,168),(65,169),(66,170),(67,161),(68,162),(69,163),(70,164),(71,148),(72,149),(73,150),(74,141),(75,142),(76,143),(77,144),(78,145),(79,146),(80,147),(81,153),(82,154),(83,155),(84,156),(85,157),(86,158),(87,159),(88,160),(89,151),(90,152),(91,132),(92,133),(93,134),(94,135),(95,136),(96,137),(97,138),(98,139),(99,140),(100,131),(101,127),(102,128),(103,129),(104,130),(105,121),(106,122),(107,123),(108,124),(109,125),(110,126),(211,240),(212,231),(213,232),(214,233),(215,234),(216,235),(217,236),(218,237),(219,238),(220,239)], [(1,2,3,4,5,6,7,8,9,10),(11,12,13,14,15,16,17,18,19,20),(21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48,49,50),(51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70),(71,72,73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99,100),(101,102,103,104,105,106,107,108,109,110),(111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130),(131,132,133,134,135,136,137,138,139,140),(141,142,143,144,145,146,147,148,149,150),(151,152,153,154,155,156,157,158,159,160),(161,162,163,164,165,166,167,168,169,170),(171,172,173,174,175,176,177,178,179,180),(181,182,183,184,185,186,187,188,189,190),(191,192,193,194,195,196,197,198,199,200),(201,202,203,204,205,206,207,208,209,210),(211,212,213,214,215,216,217,218,219,220),(221,222,223,224,225,226,227,228,229,230),(231,232,233,234,235,236,237,238,239,240)], [(1,106,95,81,62,76),(2,107,96,82,63,77),(3,108,97,83,64,78),(4,109,98,84,65,79),(5,110,99,85,66,80),(6,101,100,86,67,71),(7,102,91,87,68,72),(8,103,92,88,69,73),(9,104,93,89,70,74),(10,105,94,90,61,75),(11,34,28,59,45,238),(12,35,29,60,46,239),(13,36,30,51,47,240),(14,37,21,52,48,231),(15,38,22,53,49,232),(16,39,23,54,50,233),(17,40,24,55,41,234),(18,31,25,56,42,235),(19,32,26,57,43,236),(20,33,27,58,44,237),(111,121,135,152,165,142),(112,122,136,153,166,143),(113,123,137,154,167,144),(114,124,138,155,168,145),(115,125,139,156,169,146),(116,126,140,157,170,147),(117,127,131,158,161,148),(118,128,132,159,162,149),(119,129,133,160,163,150),(120,130,134,151,164,141),(171,202,225,212,195,181),(172,203,226,213,196,182),(173,204,227,214,197,183),(174,205,228,215,198,184),(175,206,229,216,199,185),(176,207,230,217,200,186),(177,208,221,218,191,187),(178,209,222,219,192,188),(179,210,223,220,193,189),(180,201,224,211,194,190)], [(1,218,81,177),(2,219,82,178),(3,220,83,179),(4,211,84,180),(5,212,85,171),(6,213,86,172),(7,214,87,173),(8,215,88,174),(9,216,89,175),(10,217,90,176),(11,137,59,144),(12,138,60,145),(13,139,51,146),(14,140,52,147),(15,131,53,148),(16,132,54,149),(17,133,55,150),(18,134,56,141),(19,135,57,142),(20,136,58,143),(21,116,231,157),(22,117,232,158),(23,118,233,159),(24,119,234,160),(25,120,235,151),(26,111,236,152),(27,112,237,153),(28,113,238,154),(29,114,239,155),(30,115,240,156),(31,130,42,164),(32,121,43,165),(33,122,44,166),(34,123,45,167),(35,124,46,168),(36,125,47,169),(37,126,48,170),(38,127,49,161),(39,128,50,162),(40,129,41,163),(61,186,105,230),(62,187,106,221),(63,188,107,222),(64,189,108,223),(65,190,109,224),(66,181,110,225),(67,182,101,226),(68,183,102,227),(69,184,103,228),(70,185,104,229),(71,196,100,203),(72,197,91,204),(73,198,92,205),(74,199,93,206),(75,200,94,207),(76,191,95,208),(77,192,96,209),(78,193,97,210),(79,194,98,201),(80,195,99,202)])
Dic3×C2×C10 is a maximal subgroup of
C30.24C42 C23.26(S3×D5) (C2×C30).D4 C10.(C2×D12) C1528(C4×D4) (C2×C6)⋊D20 (C2×C10)⋊8Dic6 S3×C22×C20
120 conjugacy classes
class 1 2A ··· 2G 3 4A ··· 4H 5A 5B 5C 5D 6A ··· 6G 10A ··· 10AB 15A 15B 15C 15D 20A ··· 20AF 30A ··· 30AB order 1 2 ··· 2 3 4 ··· 4 5 5 5 5 6 ··· 6 10 ··· 10 15 15 15 15 20 ··· 20 30 ··· 30 size 1 1 ··· 1 2 3 ··· 3 1 1 1 1 2 ··· 2 1 ··· 1 2 2 2 2 3 ··· 3 2 ··· 2
120 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 2 2 type + + + + - + image C1 C2 C2 C4 C5 C10 C10 C20 S3 Dic3 D6 C5×S3 C5×Dic3 S3×C10 kernel Dic3×C2×C10 C10×Dic3 C22×C30 C2×C30 C22×Dic3 C2×Dic3 C22×C6 C2×C6 C22×C10 C2×C10 C2×C10 C23 C22 C22 # reps 1 6 1 8 4 24 4 32 1 4 3 4 16 12
Matrix representation of Dic3×C2×C10 in GL4(𝔽61) generated by
1 0 0 0 0 60 0 0 0 0 1 0 0 0 0 1
,
60 0 0 0 0 60 0 0 0 0 9 0 0 0 0 9
,
1 0 0 0 0 1 0 0 0 0 1 60 0 0 1 0
,
60 0 0 0 0 1 0 0 0 0 57 31 0 0 27 4
G:=sub<GL(4,GF(61))| [1,0,0,0,0,60,0,0,0,0,1,0,0,0,0,1],[60,0,0,0,0,60,0,0,0,0,9,0,0,0,0,9],[1,0,0,0,0,1,0,0,0,0,1,1,0,0,60,0],[60,0,0,0,0,1,0,0,0,0,57,27,0,0,31,4] >;
Dic3×C2×C10 in GAP, Magma, Sage, TeX
{\rm Dic}_3\times C_2\times C_{10}
% in TeX
G:=Group("Dic3xC2xC10");
// GroupNames label
G:=SmallGroup(240,173);
// by ID
G=gap.SmallGroup(240,173);
# by ID
G:=PCGroup([6,-2,-2,-2,-5,-2,-3,240,5765]);
// Polycyclic
G:=Group<a,b,c,d|a^2=b^10=c^6=1,d^2=c^3,a*b=b*a,a*c=c*a,a*d=d*a,b*c=c*b,b*d=d*b,d*c*d^-1=c^-1>;
// generators/relations
×
𝔽
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{}
|
# A square base
A solid right pyramid has a square base. The length of the base edge is 4 centimeters and the height of the pyramid is 3 centimeters. What is the volume of the pyramid?
Result
V = 16 cm3
#### Solution:
$a = 4 \ cm \ \\ h = 3 \ cm \ \\ \ \\ S_{ 1 } = a^2 = 4^2 = 16 \ cm^2 \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ S_{ 1 } \cdot \ h = \dfrac{ 1 }{ 3 } \cdot \ 16 \cdot \ 3 = 16 = 16 \ cm^3$
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## A DC programming approach for feature selection in support vector machines learning.(English)Zbl 1284.90057
Summary: Feature selection consists of choosing a subset of available features that capture the relevant properties of the data. In supervised pattern classification, a good choice of features is fundamental for building compact and accurate classifiers. In this paper, we develop an efficient feature selection method using the zero-norm $$l_0$$ in the context of support vector machines (SVMs). Discontinuity at the origin for $$l_0$$ makes the solution of the corresponding optimization problem difficult to solve. To overcome this drawback, we use a robust DC (difference of convex functions) programming approach which is a general framework for non-convex continuous optimisation. We consider an appropriate continuous approximation to $$l_0$$ such that the resulting problem can be formulated as a DC program. Our DC algorithm (DCA) has a finite convergence and requires solving one linear program at each iteration. Computational experiments on standard datasets including challenging feature-selection problems of the NIPS 2003 feature selection challenge and gene selection for cancer classification show that the proposed method is promising: while it suppresses up to more than 99% of the features, it can provide a good classification. Moreover, the comparative results illustrate the superiority of the proposed approach over standard methods such as classical SVMs and feature selection concave.
### MSC:
90C26 Nonconvex programming, global optimization 62-07 Data analysis (statistics) (MSC2010) 68T05 Learning and adaptive systems in artificial intelligence
### Keywords:
feature selection; SVM; nonconvex optimisation; DC programming; DCA
Full Text:
### References:
[1] Amaldi E, Kann V (1998) On the approximability of minimizing non zero variables or unsatisfied relations in linear systems. Theor Comput Sci 209: 237–260 · Zbl 0915.68072 [2] Bradley PS, Mangasarian OL (1998) Feature selection via concave minimization and support vector machines. In: Proceeding of international conference on machina learning ICML’98 [3] Boser B, Guyon I, Vapnik VN (1992) A training algorithm for optimal margin classifiers. In: Proceedings of the fifth annual workshop on computational learning theory, Kaufmann, San Mateo, pp 144–152 [4] Cortes C, Vapnik VN (1995) Support vector networks. Mach Learn 20(3): 273–297 · Zbl 0831.68098 [5] Cristianini N., Shawe-Taylor J (2000) Introduction to support vector machines. Cambridge University Press, Cambridge · Zbl 0994.68074 [6] Golub TR, Slonim DK, Tamayo P, Huard C, Gaasenbeek M, Mesirov JP, Coller H, Loh ML, Downing JR, Caligiuri MA, Bloomfield CD, Lander ES (1999) Molecular classifcation of cancer: class discovery and class prediction by gene expression monitoring. Science 286: 531–537 [7] Guyon I, Weston J, Barnhill S, Vapnik VN (2002) Gene selection for cancer classification using support vector machines. Mach Learn 46(1–3): 389–422 · Zbl 0998.68111 [8] Guyon I, Gunn S, Nikravesh M, Zadeh LA (2006) Feature extraction, foundations and applications. Springer, Berlin · Zbl 1114.68059 [9] Hastie T, Rosset S, Tibshirani R, Zhu J (2004) The entire regularization path for the support vector machine. J Mach Learn Res 5: 1391–1415 · Zbl 1222.68213 [10] Krause N, Singer Y (2004) Leveraging the margin more carefully. In: Proceedings of the twenty-first international conference on machine learning ICML 2004. Banff, Alberta, 63 pp. ISBN:1-58113-828-5 [11] Le Thi HA (1997) Contribution á l’optimisation non convexe et l’optimisation globale: Théorie, Algorithmes et Applications, Habilitation á Diriger des Recherches, Université de Rouen [12] Le Thi HA, Pham Dinh T (1997) Solving a class of linearly constrained indefinite quadratic problems by DC algorithms. J Global Optim 11(3): 253–285 · Zbl 0905.90131 [13] Le Thi HA, Pham Dinh T (2005) The DC (difference of convex functions) programming and DCA revisited with DC models of real world nonconvex optimization problems. Ann Oper Res 133: 23–46 · Zbl 1116.90122 [14] Le Thi HA, Belghiti T, Pham Dinh T (2006) A new efficient algorithm based on DC programming and DCA for clustering. J Global Optim 37: 593–608 · Zbl 1198.90327 [15] Le Thi HA, Le Hoai M, Pham Dinh T (2007) Optimization based DC programming and DCA for Hierarchical Clustering. Eur J Oper Res 183: 1067–1085 · Zbl 1149.90117 [16] Le Thi HA, Mamadou T, Pham Dinh T (2008) DC Programming approach for solving a class of nonconvex programs dealing with zero-norm. In: Modelling, computation and optimization in information systems and management science. CCIS 14. Springer, Heidelberg, pp 348–357 [17] Liu Y, Shen X, Doss H (2005) Multicategory {$$\psi$$}-learning and support vector machine: computational tools. J Computat Graph Stat 14: 219–236 [18] Liu Y, Shen X (2006) Multicategory{$$\psi$$}-learning. J Am Stat Assoc 101: 500–509 · Zbl 1119.62341 [19] Neumann J, Schnörr C, Steidl G (2005) Combined SVM-based feature selection and classification. Mach Learn 61(1–3): 129–150 · Zbl 1137.90643 [20] Pham Dinh T, Le Thi HA (1998) DC optimization algorithms for solving the trust region subproblem. SIAM J Optim 8: 476–505 · Zbl 0913.65054 [21] Rakotomamonjy A (2003) Variable selection using SVM-based criteria. J Mach Learn Res 3: 1357–1370 · Zbl 1102.68583 [22] Ronan C, Fabian S, Jason W, Lé B (2006) Trading convexity for scalability. In: Proceedings of the 23rd international conference on machine learning ICML 2006. Pittsburgh, pp 201–208. ISBN:1-59593-383-2 [23] Singh D, Febbo GPG, Ross K, Jackson DG, Manola J, Ladd C, Tamayo P, Renshaw AA, D’Amico AV, Richie JP, Lander ES, Loda M, Kantoff PW, Golub TR, Sellers WR (2002) Gene expression correlates of clinical prostate cancer behavior Copyright © 2002 Cell Press, Cancer Cell, vol 1, pp 203–209, March 2002 [24] Shen KQ, Ong CJ, Li XP, Wilder-Smith EPV (2008) Feature selection via sensitivity analysis of SVM probabilistic outputs. Mach Learn 70: 1–20 · Zbl 1470.68175 [25] Schmidt M, Fung G, Rosales R (2007) Fast optimization methods for L1 regularization: a comparative study and two new approaches. In: Proceedings of machine learning: ECML 2007, Lecture notes in computer science, vol 4701/2007, pp 286–297 [26] Van’t Veer L et al (2002) Gene expression profiling predicts clinical outcome of breast cancer. Nature 415: 530–536 [27] Vapnik VN (1995) The nature of statistical learning theory. Springer, New York · Zbl 0833.62008 [28] Weston J, Elisseeff A, Scholkopf B, Tipping M (2003) Use of the zero-norm with linear models and kernel methods. J Mach Learn Res 3: 1439–1461 · Zbl 1102.68605 [29] Yuille AL, Rangarajan A (2003) The convex concave procedure. Neural Computat 15(4):915–936. ISSN:0899-7667 · Zbl 1022.68112
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# Bullet misses the enemy
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3 replies to this topic
### #1FantasyVII Members - Reputation: 559
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Posted 10 October 2012 - 03:40 PM
Hello everyone,
I have one tower and one enemy. the tower job is to shoot bullets at the enemy and each bullet will follow the enemy until it hits him.
However the problem is when the enemy is faster than the bullet, the bullet misses the enemy and tries to him the enemy but can't and it will go off screen.
look at 0:05 and 0:08 and 0:09 see how the bullet misses the enemy and tries to hit him and can't.
So what i want is for the bullet to keep following the enemy no matter what until it hits him.
class Bullet
{
void CalculateBulletPosition()
{
DeltaPosition.X = Enemy.EnemyRectangle.X - Tower.Position.X;
DeltaPosition.Y = Enemy.EnemyRectangle.Y - Tower.Position.Y;
double VectorLength = Math.Sqrt((DeltaPosition.X * DeltaPosition.X) + (DeltaPosition.Y * DeltaPosition.Y));
Direction.X = (float)DeltaPosition.X / (float)VectorLength;
Direction.Y = (float)DeltaPosition.Y / (float)VectorLength;
}
public void Update()
{
CalculateBulletPosition();
Velocity.X += Direction.X * Speed;
Velocity.Y += Direction.Y * Speed;
BulletRectangle = new Rectangle((int)(Velocity.X + BulletPosition.X), (int)(Velocity.Y + BulletPosition.Y), Texture.Width, Texture.Height);
}
public void Draw(SpriteBatch spriteBatch)
{
spriteBatch.Draw(Texture, BulletRectangle, Color.White);
}
}
Edited by FantasyVII, 10 October 2012 - 03:42 PM.
### #2HappyCoder Members - Reputation: 3058
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Posted 10 October 2012 - 05:04 PM
Try having the bullet go to where the enemy will be. This will mean you have to take into account the velocity of the bullet and the velocity of the player.
I also noticed you calculate the bullets new direction based on the difference between the enemies position and the towers position (enemy.pos - tower.pos). What if you did the difference between the bullets position and the enemies position. (bullet.pos - enemy.pos)
### #3JTippetts Moderators - Reputation: 10030
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Posted 10 October 2012 - 07:46 PM
http://www.gamedev.n...ng-their-shots/
FLeBlanc runs through the math to calculate the point you need to aim at using a quadratic equation, assuming there is a point you can aim at that will possibly hit the target. Note that sometimes there just is no solution.
Edit:
Eh, just re-read and noticed you want tracking bullets, rather than target leading. In this case, it becomes a sort of tradeoff. How tightly do you want them to track? The way you have it currently, you calculate a trajectory for the bullet based on the vector from the tower to the enemy, but then you add this vector to the existing velocity vector. This will give you an accurate heading initially, but as the trajectory diverges away from a straight line the new velocity vector has a harder and harder time trying to correct out the old, now-incorrect velocity with the new heading, so the bullet just goes more and more awry.
If you want the bullet to tightly follow the enemy, then each update just replace the old trajectory with the newly calculate trajectory. However, this could result in some pretty tight curves. If you want to limit the curves, then you can calculate some vector that lies in between the old velocity and the new velocity whose angle relative to the original velocity is no greater than some limit, A. Then set the velocity to this new vector. This way, the bullet will loop around and eventually spiral in on the target, without sudden and drastic changes in direction.
Edited by JTippetts, 10 October 2012 - 07:54 PM.
### #4FantasyVII Members - Reputation: 559
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Posted 11 October 2012 - 02:28 AM
Try having the bullet go to where the enemy will be. This will mean you have to take into account the velocity of the bullet and the velocity of the player.
I also noticed you calculate the bullets new direction based on the difference between the enemies position and the towers position (enemy.pos - tower.pos). What if you did the difference between the bullets position and the enemies position. (bullet.pos - enemy.pos)
http://www.gamedev.n...ng-their-shots/
FLeBlanc runs through the math to calculate the point you need to aim at using a quadratic equation, assuming there is a point you can aim at that will possibly hit the target. Note that sometimes there just is no solution.
Edit:
Eh, just re-read and noticed you want tracking bullets, rather than target leading. In this case, it becomes a sort of tradeoff. How tightly do you want them to track? The way you have it currently, you calculate a trajectory for the bullet based on the vector from the tower to the enemy, but then you add this vector to the existing velocity vector. This will give you an accurate heading initially, but as the trajectory diverges away from a straight line the new velocity vector has a harder and harder time trying to correct out the old, now-incorrect velocity with the new heading, so the bullet just goes more and more awry.
If you want the bullet to tightly follow the enemy, then each update just replace the old trajectory with the newly calculate trajectory. However, this could result in some pretty tight curves. If you want to limit the curves, then you can calculate some vector that lies in between the old velocity and the new velocity whose angle relative to the original velocity is no greater than some limit, A. Then set the velocity to this new vector. This way, the bullet will loop around and eventually spiral in on the target, without sudden and drastic changes in direction.
thank you so very much. I really appreciate your help.
here is the code if anyone is interested.
I know this code is not clean. going to clean it up right now.
class Bullet
{
public Vector2 Position;
Vector2 DeltaPosition;
Vector2 Direction;
Vector2 Velocity;
void CalculateBulletPosition(Enemy enemy)
{
DeltaPosition.X = (enemy.EnemyRectangle.X) - Position.X;
DeltaPosition.Y = (enemy.EnemyRectangle.Y) - Position.Y;
double VectorLength = Math.Sqrt((DeltaPosition.X * DeltaPosition.X) + (DeltaPosition.Y * DeltaPosition.Y));
Direction.X = (float)DeltaPosition.X / (float)VectorLength;
Direction.Y = (float)DeltaPosition.Y / (float)VectorLength;
}
void CalculatePositionBetweenEnemyAndCurrentBulletPosition(Enemy enemy)
{
DeltaPosition.X = (enemy.EnemyRectangle.X) - (Velocity.X + Position.X);
DeltaPosition.Y = (enemy.EnemyRectangle.Y) - (Velocity.Y + Position.Y);
double VectorLength = Math.Sqrt((DeltaPosition.X * DeltaPosition.X) + (DeltaPosition.Y * DeltaPosition.Y));
Direction.X = (float)DeltaPosition.X / (float)VectorLength;
Direction.Y = (float)DeltaPosition.Y / (float)VectorLength;
}
public void Update(Enemy enemy, bool Towershoot)
{
if (Towershoot)
{
CalculateBulletPosition(enemy);
Towershoot= false;
}
if (Towershoot== false)
{
CalculatePositionBetweenEnemyAndCurrentBulletPosition(enemy);
}
Velocity.X += Direction.X * Speed;
Velocity.Y += Direction.Y * Speed;
BulletRectangle = new Rectangle((int)(Velocity.X + Position.X ), (int)(Velocity.Y + Position.Y ), Texture.Width, Texture.Height);
}
public void Draw(SpriteBatch spriteBatch)
{
spriteBatch.Draw(Texture, BulletRectangle, Color.White);
}
}
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# zbMATH — the first resource for mathematics
Rolle’s theorem fails in $$\ell_ 2$$. (English) Zbl 0888.46017
M. Furi and M. Martelli suggested multidimensional analogs of the Rolle theorem [Am. Math. Monthly 102, No. 3, 243-249 (1995; Zbl 0856.26009)]. The author shows by example that these analogs fail in Hilbert space.
##### MSC:
46G05 Derivatives of functions in infinite-dimensional spaces 46B45 Banach sequence spaces 26B05 Continuity and differentiation questions
Rolle’s theorem
Full Text:
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# Domain of function
• June 6th 2006, 03:35 PM
starbrite
Algebra question
1/ x^2-10x - 8
I know I have to solve for zero and factor it but when I do it just comes out 1, -5 but the awnser is -1 ,5 ... I just dont understand why ? Sorry math is the subject I struggle with most .Thank you
• June 6th 2006, 03:45 PM
ThePerfectHacker
Quote:
Originally Posted by starbrite
1/ x^2-10x - 8
I know I have to solve for zero and factor it but when I do it just comes out 1, -5 but the awnser is -1 ,5 ... I just dont understand why ? Sorry math is the subject I struggle with most .Thank you
Where is the equation!
$\frac{1}{x^2-10x-8}=??$
• June 6th 2006, 03:48 PM
starbrite
oh sorry its f(x)
I have to find the domain of the function . Thanks
• June 6th 2006, 03:55 PM
ThePerfectHacker
Quote:
Originally Posted by starbrite
oh sorry its f(x)
I have to find the domain of the function . Thanks
It math course they teach you whenever they write a function they always assume all possible values that 'x' can take since you have a fraction,
$\frac{1}{x^2-10x-8}$
It can be any value EXCEPT when the denominator is zero.
Thus,
$x^2-10x-8\not =0$
Since this is not factorable use quadradic formula,
$x=\frac{10\pm \sqrt{100+64}}{2\cdot 1}$
Simplfy,
$x=\frac{10\pm 2\sqrt{41}}{2}$
Thus,
$x=5\pm \sqrt{41}$
---
To make things clearly, let me explain,
$\sqrt{100+64}=\sqrt{164}=\sqrt{4\cdot 41}=2\sqrt{41}$
And, in the final step since the numbers shared a common factor namely 2 I divided it out. Important "you cannot cancel out of a sum" meaning you cannot cancel a 2 with the 10 and leave the other 2. You must to cancel both of them whenever you deal with add/subt.
• June 6th 2006, 04:08 PM
starbrite
oh ok thank u so much
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# Frequency and wavelength!
#### waywardtigerlily
A resonant circuit in a radio receiver is tuned to a certian station when the inductor has a value of 0.151mH and the capacitor has a value of 21.2pF. Calculate the frequency of the radio station.
Frequency is 2.81 MHz ( I got this one right)
B) what is the wavelength of the radiowaves emited by the radio station?
I was told to do this just take 1/f,but this gave me a wrong answer. Is there another way of going about it?
Related Introductory Physics Homework Help News on Phys.org
#### MathStudent
1\f is the period,
for wavelength you want to use the equation for all waves
$$v = f \lambda$$
where v is the speed of propogation of the wave
#### whozum
For this problem, $$v = c$$. All electromagnetic waves travel at the speed of light, so:
$$f = c/\lambda$$
#### MathStudent
Well there goes the chance of the OP doing any of the work for himself.
#### whozum
A lot of people dont know that, its not a question of doing work, its just a fact he needs.
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# How do you solve 3^(x-2)=27?
May 23, 2015
${3}^{x - 2} = 27$
27 can be written as ${3}^{3}$
${3}^{x - 2} = {3}^{3}$
at this stage exponents can be equated
$x - 2 = 3$
color(blue)( x = 5 is the solution for the expression.
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Volume 340 - The 39th International Conference on High Energy Physics (ICHEP2018) - Parallel: Astro-particle Physics and Cosmology
The LHCf experiment: recent physics results
E. Berti,* O. Adriani, L. Bonechi, M. Bongi, R. D'Alessandro, M. Haguenauer, Y. Itow, K. Kasahara, Y. Makino, K. Masuda, H. Menjo, Y. Muraki, K. Ohashi, P. Papini, S. Ricciarini, T. Sako, N. Sakurai, K. Sato, M. Shinoda, T. Suzuki, T. Tamura, A. Tiberio, S. Torii, A. Tricomi, W.C. Turner, M. Ueno, Q.D. Zhou
*corresponding author
Full text: pdf
Published on: August 02, 2019
Abstract
The main aim of the LHCf experiment is to provide precise measurements of the production spectra relative to neutral particles produced in the very forward region by high energy proton-proton and proton-ion collisions. This information is necessary in order to test and tune hadronic interaction models used by ground-based cosmic rays experiments. In order to reach this goal, LHCf makes use of two small sampling calorimeters installed in the LHC tunnel at $\pm 140$ m from IP1, able to detect neutral particles having pseudo-rapidity $\eta > 8.4$. After the operations relative to p-p collisions at $\mathrm{\sqrt{s} = 13~TeV}$, the collaboration focused the activity on the study of forward photons and neutrons production at this energy. In this paper, we discuss the recent results relative to the photons analysis, regarding, in particular, the inclusive production cross section, the role of diffractive and non-diffractive processes in forward photons production (investigated thanks to the LHCf-ATLAS joint analysis) and the forward differential energy flow.
DOI: https://doi.org/10.22323/1.340.0207
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
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J. KIMS Technol > Volume 20(5); 2017 > Article
Journal of the Korea Institute of Military Science and Technology 2017;20(5):608-619. DOI: https://doi.org/10.9766/KIMST.2017.20.5.608
Analysis of the Effects of Laser Shock Peening under Initial Tensile Residual Stress Using Numerical Analysis Method Juhee Kim, Jongwoo Lee, Samhyeun Yoo Department of Mechanical and System Engineering, Korea Military Academy 수치해석기법을 이용한 초기 인장잔류응력에 대한 레이저 충격 피닝 효과 분석 김주희, 이종우, 유삼현 육군사관학교 기계시스템공학과 Abstract In this paper, the effects of parameters related to the residual stress induced due to laser shock peening process to determine mitigation of the initial tensile residual stresses are discussed, such as the maximum pressure, pressure pulse duration, laser spot size and number of laser shots. In order to estimate the influence of the initial tensile residual stresses, which is generated by welding in 35CD4 50HRC steel alloy, the initial condition option was employed in the finite element code. It is found that $2{ imes}HEL$ maximum pressure and a certain range of the pressure pulse duration time can produce maximum mitigation effects near the surface and depth, regardless of the magnitudes of tensile residual stess. But plastically affected depth increase with increasing maximum pressure and pressure pulse duration time. For the laser spot size, maximum compressive residual stresses have almost constant values. But LSP is more effective with increasing the magnitudes of tensile residual stress. For the multiple LSP, magnitudes of compressive residual stresses and plastically affected depths are found to increase with increasing number of laser shots, but the effect is less pronounced for more laser shots. And to conclude, even though the initial tensile residual stresses such as weld residual stress field are existed, LSP is enough to make the surface and depth reinforcement effects. Key Words: Laser Shock Peening, Residual Stress, Initial Tensile Residual Stress, Dynamic Yield Stress
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# barium as a mineralizer in cement clinker formation
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The effect of barium on the formation of clinker phases was studied (by XRD Rietveld analysis and by the microscopic point counting method), as well as the rate of alite formation under isothermal conditions. Furthermore, the ability of barium to become a part of clinker minerals was studied by SEM with EDS.
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### Heat Liberation of Barium Cements as a Background
[10] V.E. Kaushansky, O.N. Valyaeva, The barium-containing waste as mineralizer of clinker burning process, Cement and its use, 3 (2002) 31-32. [11] Guo Xiangyang, Wang Shoude, Lu Lingchao and Wang Hui, Influence of barium oxide on the composition and performance of alite-rich Portland cement, Adv. Cem. Res. 24 (2012): 139-144.
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1998-8-1 Fluxes and mineralizers improve burnability and quality of clinker, solving particularly the free lime problem faced by cement plants. The purpose of the present paper is to highlight the role of CaF 2 and CaSO 4 as mineralizers with industrial raw mix in VSK technology in conserving energy and improving the quality of clinker.
### [PDF] Properties of Portland Cement Clinker Using
2014-2-4 It was evaluated that when cement clinker is produced, the chloric component of polysilicon acted as a mineralizer in the firing process. In addition, the physical features of the produced cement were measured. The setting time of the produced cement was reduced as the amount of content of polysilicon sludge increased.
### Mineralizers in cement manufacture: Fluorspar
2021-4-16 the cement raw material, improve the percentage of the principal phase, C. 3. S. CONCLUSION The addition of 0.75% of this mineralizer in clinker raw mixture permits to decrease the free lime content in the clinker to 1.25% for burning temperature of 1300°C. These very satisfactory results are due to the mineralogical
### Optimal fluorite/gypsum mineralizer ratio in Portland
Clinker formation was studied mainly on the . basis of free lime content in the burned samples. This is the most widely used procedure because the calcium oxide, initially formed by CaCO. 3. dissocia-tion, is gradually consumed by the clinker phases (20). The free lime was measured with the ethylene method; optical microscopy; Differential Scanning
### The Influence of Barium Sulphate and Barium Carbonate
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### Effect of barium oxide on the formation and
2009-7-9 Formation and coexistence of tricalcium silicate (C 3 S) and calcium sulphoaluminate (C 4 A 3 $$\bar S$$) minerals in Portland cement clinker containing calcium sulphoaluminate were investigated.The f-CaO content, mineral composite and formation of mineral in the clinker were analyzed respectively by chemical analysis, differential scanning calorimetry(DSC) and X-ray diffraction.
### The Influence of Barium Sulphate and Barium
The work is focused on influence of barium ions on the formation of Portland clinker and its properties. As a source of barium, barium sulphate and barium carbonate were added to raw meal. Clinkers and cements with a different amount of barium were prepared and the effect of barium on the clinker and cement properties respectively was studied.
### [PDF] Properties of Portland Cement Clinker Using
2014-2-4 This study reviewed the usability of sludge, a material that is additionally created when polysilicon (a solar light material) is produced, as the raw material for cement clinker. It was evaluated that when cement clinker is produced, the chloric component of polysilicon acted as a mineralizer in the firing process. In addition, the physical features of the produced cement were measured.
### barium por le crusher dakdekker-nijkerk.nl
barium as a mineralizer in cement clinker formation. Advances in Cement Research Welcome to ICE Virtual,The clinker composition and performance of alite-rich Portland cement with different contents of barium oxide and calcium barium sulfoaluminate (abbr C275B125A3S¯) were investigated by orthogonal testing The results show that alite and C275B125A3S¯ mineral can coexist in a clinker
### AN OVERVIEW OF INDUSTRIAL WASTES AS FUEL AND
2019-6-4 Keywords industry waste, clinker, mineralizer I. INTRODUCTION The cement production has undergone development from 2000 year. The annual global cement production has reached 2.8 billion tons and is expected to increase to some 4 billion tons per year (Schneider et al., 2011). The typical electrical energy consumption of a modern cement plant is
### Influence of Nano-Barium Sulfate Agglomeration on
2015-11-27 hydrous clinker phases and the formation of hydration products 34].[Nano-barium sulfate filled cement pastes had the highest combined water contents as compared to OPC and limestone filled samples. Moreover, n- the i crease of the nano-barium sulfate content resulted in higher combined water contents at all addition levels, while Figure 1.
### Cement Clinker Based on Industrial Waste Materials
2020-6-8 Three samples of clinker were obtained (CL1, CL2 and CL3) based on the content of free lime consistent with the maximum value. The first type of clinker CL1, from RM1, was produced at the lowest temperature (about 1000°C). Due to the high content of CS in RM1 (59%) which acted as a mineralizer, the liquid phase formation seems to take
### MINERALIZER FOR CALCIUM SULFOALUMINATE
1. A method for producing a ternesite or calcium sulfoaluminate (belite, ferrite) ternesite clinker with improved formation and/or thermal stability of ternesite in a single sintering step above 1200° C. comprising the following steps providing a raw meal comprising at least sources of CaO, SiO 2, Al 2 O 3, and SO 3 2, Al 2 O 3, and SO 3
### Evaluation of portland cement clinker with optical
2020-11-30 Clinker microscopy is a powerful tool for the evaluation of clinker and cement properties. Microstructural inves-tigations yield important information on phase distribution and the conditions of the phase formation. The correct understanding of clinker microstructure is crucial for an accurate evaluation of raw material, fuel or process parame
### The Influence of Barium Sulphate and Barium
The work is focused on influence of barium ions on the formation of Portland clinker and its properties. As a source of barium, barium sulphate and barium carbonate were added to raw meal. Clinkers and cements with a different amount of barium were prepared and the effect of barium on the clinker and cement properties respectively was studied.
### AN OVERVIEW OF INDUSTRIAL WASTES AS FUEL AND
2019-6-4 Keywords industry waste, clinker, mineralizer I. INTRODUCTION The cement production has undergone development from 2000 year. The annual global cement production has reached 2.8 billion tons and is expected to increase to some 4 billion tons per year (Schneider et al., 2011). The typical electrical energy consumption of a modern cement plant is
### barium por le crusher dakdekker-nijkerk.nl
barium as a mineralizer in cement clinker formation. Advances in Cement Research Welcome to ICE Virtual,The clinker composition and performance of alite-rich Portland cement with different contents of barium oxide and calcium barium sulfoaluminate (abbr C275B125A3S¯) were investigated by orthogonal testing The results show that alite and C275B125A3S¯ mineral can coexist in a clinker
### Study the Critical Role of Admixtures in Cement
2013-12-24 clinker. Irradiating the powdered samples took place in the range between 3 and 75 degrees. We use amplified electronic scales of four decimals, which weighed 0,5 gr of ground clinker. We used 50 ml ethylene glycol and HCl 0.1 N and 1 N accordingly. 3) Secondary electron microscopy (SEM) analyses of two clinker samples
### Berlinite substitution in the cement clinker (Journal
2017-2-15 OSTI.GOV Journal Article: Berlinite substitution in the cement clinker. Berlinite substitution in the cement clinker. Full Record; Other Related Research
### MINERALIZER FOR CALCIUM SULFOALUMINATE
The present invention relates to a method for the production calcium sulfoaluminate (belite, ferrite) ternesite clinker using fluxes/mineralizers comprising the following steps providing a raw meal comprising at least sources of CaO, SiO 2, Al 2 O 3, SO 3 sintering the raw meal in a kiln at >1200 to 1400 °C to provide a clinker, cooling the clinker, wherein a mineralizer comprising at
### Cement Clinker based on industrial waste materials
Calcium silicate • Cement clinker • Coal fly ash • Industrial waste • Sewage sludge ash. Introduction. Ordinary Portland Cement (OPC) has been a priority material of construction used in many countries throughout the world. In 2017, approximatively 4,100 million metric tons of cement
### Method and plant for manufacturing mineralized
We claim: 1. A method for preparing mineralized Portland cement clinker in a kiln system, said method comprising the steps of preheating a rawmix feedstock in a preheater, calcining the preheated feedstock in a calciner, burning the calcined feedstock in the kiln to form clinker, and then cooling the clinker in a cooler, said method further comprising adding a mineralizer to the feedstock such
### Evaluation of portland cement clinker with optical
2020-11-30 Clinker microscopy is a powerful tool for the evaluation of clinker and cement properties. Microstructural inves-tigations yield important information on phase distribution and the conditions of the phase formation. The correct understanding of clinker microstructure is crucial for an accurate evaluation of raw material, fuel or process parame
### WO2016206785A1 Mineralizer for calcium
The present invention relates to a method for the production calcium sulfoaluminate (belite, ferrite) ternesite clinker using fluxes/mineralizers comprising the following steps: providing a raw meal comprising at least sources of CaO, SiO 2,AI 2 O 3,SO 3,sintering the raw meal in a kiln at >1200 to 1400 °C to provide a clinker, cooling the clinker, wherein a mineralizer comprising at
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# Tag Info
6
tl;dr- Quantum computers can't really help us to simulate the whole universe as the universe is likely vastly more complex than even quantum mechanics can capture, plus we can't even begin to guess how big it is or many other basic fundamental features. In short, simulating the whole universe is beyond sci-fi. We can't really simulate the entire universe, ...
4
It seems this problem is open. Watrous [J. Comp. Sys. Sci. 59, (pp. 281-326), 1999] proved that any space $s$ bounded quantum Turing Machine (for space constructible $s(n)>\Omega(\log n)$) can be simulated by deterministic Turing machine with $O(s^2)$ space. With the assumption $\mathsf{P \neq SC}$ (where $\mathsf{SC \subseteq P}$ is defined as ...
3
It's not so much a matter of big data, but that of saving data. Quantum storage is still (much like the rest of the field) in its infancy. (Take what I write with a grain of salt. It's likely to change rapidly.) There are a few theories on how quantum computers might be able to hold "memory". One of these is using nuclear spin. E.g. using long-lived ...
3
This doesn't exactly answer your question, but it may aid you in understanding the problem and possibly the solution: In their paper "Breaking the 49-Qubit Barrier in the Simulation of Quantum Circuits" (arXiv:1710.05867), the authors describe simulating a 49-qubit and a 56-qubit quantum computer. According to the paper, they required 4.5 Terrabytes of RAM ...
Only top voted, non community-wiki answers of a minimum length are eligible
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基于动态重叠网格方法的尾翼对螺旋桨滑流的影响
1. 1. 中国空气动力研究与发展中心 计算空气动力研究所, 绵阳 621000;
2. 航空工业第一飞机设计研究院, 西安 710089
• 收稿日期:2018-05-18 修回日期:2018-06-04 出版日期:2019-04-15 发布日期:2018-07-04
• 通讯作者: 陈波 E-mail:chenbo01010401@163.com
Influence of tail wing on propeller slipstream based on dynamic overlapping grid method
MIAO Tao1, CHEN Bo1, MA Shuai1, YANG Xiaochuan1, DING Xingzhi2
1. 1. Computational Aerodynamics Institute, China Aerodynamics Research and Development Center, Mianyang 621000, China;
• Received:2018-05-18 Revised:2018-06-04 Online:2019-04-15 Published:2018-07-04
Abstract: The slipstream produced by the propeller airplane will significantly interfere with the swept components. The study of the effect of the tail wing on the slipstream will help to decouple the slipstream and tail wing disturbances. The dynamic overlapping grid method is used to simulate the rotation of the propeller fixed axis. By solving the three-dimensional Unsteady Reynolds Average Navier-Stokes(URANS) equations, the state of a propeller airplane with tail wing with/without slipstream is simulated numerically. The correctness of the calculation method is verified by the experimental results. On this basis, calculations of the slipstream of the configuration with/without tail wing are carried out. The results show that the lift and drag aerodynamics variation law are almost the same after deducting of the aerodynamic force of the tail wing. The pitching moment is different due to the modification of the fuselage rear body. By comparing the slice velocity distribution clouds of the configurations with/without tail wing, variation curves of the down wash angle and the side wash angle at different spatial positions and with different thrust coefficients, it is found that the effect of the tail wing on the slipstream is only limited to the surroundings, and the rules of disturbance of the tail wing with different thrust coefficients are also very similar. Based on the study, it can be concluded that at the preliminary and selection stages of the airplane design, the mutual interference between the propeller slipstream and the tail wing can be simplified to be a one-way effect of the slipstream on the tail wing, and the effect of the tail wing on the slipstream can be ignored.
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# Forums | Mahara Community
## Support / Mahara 15.04 and CAS plugin
Posts: 112
11 July 2015, 1:55 AM
Hi
We planned to upgrade our Mahara from 1.9 to 15.04 next week. But we stopped our upgrade process since we discovered that the CAS plugin doesn't work with 15.04.
Who is using the CAS plugin ? How did / will you manage your MAHARA upgrade to 15.04 with the CAS authentification ?
Thanks
Best regards,
Emilie
Posts: 515
##### Re: Mahara 15.04 and CAS plugin
11 July 2015, 2:52 PM
Hi Emilie,
I had to get CAS working in 15.04 for a client. The problem is due to how the session is handled in Mahara 15.04 (it is slightly different to before).
So I had to adjust all the $_SESSION calls to$SESSION
Cheers
Robert
Posts: 112
##### Re: Mahara 15.04 and CAS plugin
15 July 2015, 8:46 PM
Hi Robert
Thanks a lot for your feedback !! :-)
Could you publish your plugin version for 15.04 on the plugin page ? i am sure other institutions will appreciate to have it available for download .
As you use this CAS plugin for your clients, will you also maintain it for future releases ?
The CAS authentification is key for our Mahara instance. Do you think we could ask Catalyst to have it included in Mahara core ?
Cheers
Emilie
Posts: 3866
##### Re: Mahara 15.04 and CAS plugin
18 July 2015, 4:50 AM
Hello Emilie,
Robert suggested his changes to Patrick so they can be incorporated into the actual plugin. For the meantime, they can be found in his repository. See https://wiki.mahara.org/wiki/Plugins for the links to both plugins. We are not planning on taking over the plugin maintenance, but prefer to push changes to the actual plugin so that things don't get confusing (I still remember the different versions we had for the assignmen submission plugin which got pretty tricky ;-) ).
Cheers
Kristina
Posts: 112
##### Re: Mahara 15.04 and CAS plugin
05 April 2016, 2:49 AM
Hi Robert,
Do you know if CAS plugin works for Mahara 16.04 ?
Best regards
Emilie Lenel
Posts: 3866
##### Re: Mahara 15.04 and CAS plugin
06 April 2016, 3:18 PM
Hello Emilie,
We haven't tried it on 16.04 yet, but it does work on 15.10. I believe there weren't that many differences.
Cheers
Kristina
Posts: 112
##### Re: Mahara 15.04 and CAS plugin
06 April 2016, 7:06 PM
Hello Kristina
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Can I optimize band-pass FFT filter with skipping values instead of zeroing them when decimating?
I currently use very simple band-pass filter: FFT->Keeping needed, zeroing not needed->iFFT->decimation with skipping
This works perfectly for my needs (I apply envelope after filter so no problem with slight deformation of real signal) but can I optimize it with decimation in the middle?
For example, I have pack of 4096 samples for my 1kHz signal, and I want to keep 0.5..2Hz range:
1. I FFT 4096 samples and get 4096 complex with resolution of 1000/4096/2 ~ 0.5Hz
2. I zero first sample, keep 3 samples, zero all other samples
3. iFFT 4096 samples, getting 4096 samples
4. I take 4 samples (with indices 0, 1000, 2000, 3000, 4000)
5. Envelope them with simple detector
Result is totally fine for my needs, but when I need 100 frequency zones I have to iFFT 4096 complex 100 times - can I somehow reduce sampling rate not in step 4 but in step 2 to save processing time and ram with cheaper/faster iFFT?
I tried iFFT on reduced to 4 complex samples array but result was total failure
• What 100 frequency zones do you need? Jun 22 at 23:51
Zeroing bins as a method of filtering is not recommended unless only those distinct frequencies need to be rejected. This is the frequency sampling approach to filter design and results in very large errors for any frequencies that may exist in between those bins that were zeroed compared to optimized methods for filter design.
If there is any interest in "optimizing" and being faster/cheaper then consider other filter design approaches which are just as simple and provide much better performance for a given number of coefficients. My favored approach is to use the least squares algorithm (firls in Matlab/Octave and Python scipy.signal) for optimum filter design (in the least squared sense) or if there is any concern with peak error instead of best overall rms error, then consider the Parks-Mclellan (firpm in Octave and Python scipy.signal) for optimum design in the peak error sense.
If computational cheapness is needed for determining the filter coefficients themselves, then consider using a Kaiser window for the window design method of filter design. For a comparison on these approaches please see: FIR Filter Design: Window vs Parks McClellan and Least Squares
Not zeroing bins is further detailed in:
Why is it a bad idea to filter by zeroing out FFT bins?
For a .5 Hz to 2 Hz bandpass filter with a signal sampled at 1 KHz sampling rate, my suggestion for optimized performance is to decimate to from 1 KHz to 10 Hz and then use a least square filter for the bandpass. The length of the filter is driven by the fractional transition band between the bandpass frequencies and bandstop frequencies. With a ratio of 10 Hz sampling this should be fairly wide resulting in a highly efficient structure overall with less processing than even one of the FFTs.
As a quick example of the dramatic efficiency difference (and I suspect that a series of 6 half-band filters for decimating from 1 KHz to doing the final selection filter at 15.575 Hz, or even a CIC filter with a 3 tap droop compensator would be even more efficient than this, but this was quick to do and demonstrates the dramatic efficiency improvement over a 4096 FFT and IFFT operation) I propose the following as shown below. The OP's FFT processing could still be done if desired instead of the shaping filter with a considerable improvement in efficiency:
Each filter is a linear phase filter and therefore requires half the multipliers for the given number of taps. The above will provide over 55 dB of rejection from 2.3 Hz to 1 KHz, and from DC to 0.2 Hz with the desired passband of 0.5 Hz to 2 Hz, and only requires $$31+31+51 = 113$$ total multipliers!
Even better performance can be achieved if desired with a modest increase in multipliers. For 4096 input samples there will be 409 output samples at the 10 Hz rate after the overall decimation by 100 with the filter delay of 51 samples (so ~350 samples will represent the settled desired output).
This is compared to $$2N \log_2( N) = 2(4096)12 = 98,304$$ multipliers for one radix-2 FFT (and two are required in the OP's initial approach!). Another approach for the IDFT computation of a small number of samples as the OP desires is the the Goertzel algorithm but even that would take $$2N+4$$ multipliers per output sample so for the five output samples desired by the OP would require $$40,980$$ multipliers.
If an FFT is desired instead of the shaping filter and continuing with the OP's initial strategy, then this would still require approximately 3,550 multipliers for each FFT.
The optimized filter designs using multiband decimation filters are shown below for the 10:1 filter (100 Hz to 10 Hz is shown, but the 1KHz to 100 Hz would be the identical filter). Note that rejection is concentrated to be only at the alias frequency locations that matter (rejection is maximized at the locations that would alias to the desired 0.5 to 2 Hz band):
And the final shaping filter with the given number of taps used has the following frequency response:
This has room for further optimization but shows quickly what can be done with proper decimation and filtering approaches rather than filtering through the frequency sampling method of zeroing FFT bins. The same decimation operations could have been done and then FFT's performed on the 10 Hz signal which would also have provided a dramatic improvement; reducing 98,304 multipliers to 3550 for each of the two FFT operations, but the FIR approach was shown to provide what would be better performance with only 113 total multipliers! For filtering applications the least squares is an optimized algorithm that significantly outperforms FFT sampling approaches (except for the cases when we truly only care about the frequencies that are on the FFT bin centers).
If there is a need for multiple different bands each with similar bandwidth but at a different similar frequency, then polyphase filter approaches combined with IFFT blocks can make for very efficient channelizer topologies. For more details on that see this write-up from fred harris titled "Filter Banks for Software Defined Radios":
https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_book_ch_6_text_5_61615.pdf
• Thanks! I understand downsides of zeroing, but envelope filter and adaptive gain works like a charm fixing most of them (I don't need to keep signal form, just looking for pikes in time-domain). I compared results with Chebyshev and Butterworth filters and they seem 90-95% identical on real signal. FFT also has a nice benefit of good semi-hardware accelerated realization in CMSIS, that makes it fast enough, just looking for optimization ways :-) Jun 22 at 19:18
• I wouldn't recommend Chebyshev and Butterworth filters for digital design either (that was in the 60's before they came up with optimum filter design directly in digital rather than mapping from the analog). So if you want optimized ways, please look closer at my suggestions and then see if you can do better. Avoid all that extra processing with envelope filters and adaptive gain! This is all worked out for you already and simple to use/ implement. Jun 22 at 19:26
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MOS
Seminar
On the motives of moduli spaces of Higgs bundles
Heinloth, J (Amsterdam)
Tuesday 15 March 2011, 15:00-16:00
Satellite
Abstract
We will explain an approach to the computation of the cohomology of moduli spaces of Higgs bundles on surfaces, that is closely related to the argument of Harder-Narasimhan for moduli spaces of vector bundles and give an application to the middle cohomology of the moduli space of SL_n Higgs bundles. This is joint work with O. Garcia-Prada and A. Schmitt.
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# Redux Tutorial For Beginners
KrunalLathiya
72.2K views
## Redux Tutorial For Beginners
Redux is quite an excellent State Managment Framework usually used with React.js library. In Single Page Application, data management at client side is far more complicated than just imagine. Now, you are familiar that, ReactJS is relying on the State of the application. However, In React.js state management is possible, but when the application gets bigger and bigger, unwanted errors and data changes are detected, and which module has changed which state and which view is updated, all these matters get complex, and we feel like, we trapped in our application. Facebook gives the solution. Its developer has created one State management pattern called Flux.
## Flux
It complements React’s composable view components by utilizing the unidirectional data flow. When the users interact with views, the views propagates actions through a central dispatcher, to the various stores that hold the application data and logic, which updates all of our views that are affected.
Flux has so many stores, and each store is using different small part of the state or data of our application. In other words, each different module has its store.
## Flux Data Flow
1. The user interacts with the view and view triggers an action.
2. Action dispatch the corresponding function and then that function change the store.
3. When store updates its data, the subscriber views are automatically updated. We do not need to modify the data in different modules manually.
It is a unidirectional data flow. When the application gets bigger and bigger, then multiple stores manage the data.
When multiple stores are there, the condition of our application looks like above figure, but the data flow is Unidirectional.
Flux has Multiple Stores.
## Redux
Redux is the predictable state container for JavaScript applications. Redux is also following the Unidirectional flow, but it is entirely different from Flux. Flux has multiple stores.
Redux has a Single Store.
Redux can not have multiple stores. The store is divided into various state objects. So all we need is to maintain the single store, or we can say the only source of truth.
## Three Principles Of Redux
1. Single source of truth.
3. Changes are made with pure functions.
It is the state of our whole application is stored in an object within a single store. There is an only way to change the state is to emit an action, an object describing what happened. To specify how actions transform the state, you write pure reducers.
## Actions
Actions are payloads of information that send data from your application to your store. You send them to the store using store.dispatch()
Actions are plain JavaScript objects. Actions must have a type property that indicates the type of action being performed. Types should typically be defined as string constants.
import { ADD_TODO, REMOVE_TODO } from '../actionTypes'
## Action Creators
Action creators are exactly the functions that create actions.
function addTodo(text) {
return {
text
}
}
## Reducers
Actions describe the fact that something happened but don’t specify how the application’s state changes in response. That is the job of reducers.
## Handling Actions
This is called a reducer because it is the type of function you would pass to Array.prototype.reduce(reducer, ?initialValue). It is essential that the reducer stay pure.
Following are the things you should never do inside a reducer:
1. Mutate reducer’s arguments;
2. Perform side effects like database calls, API calls, and routing transitions;
3. Call non-pure functions, e.g., Date.now() or Math.random()
## Store
A store is an object that brings them together. A store has the following responsibilities:
1. Holds application state;
3. Allows state to be updated via dispatch(action);
4. Registers listeners via subscribe(listener);
5. It handles unregistering of listeners via the function returned by subscribe(listener)
t is important to note that you will only have a single store in a Redux application. If you want to split your data handling logic, you will use reducer composition instead of many stores.
import { createStore } from 'redux'
import todoApp from './reducers'
let store = createStore(todoApp)
## Redux Tutorial With Example
Create one project folder and in that create one file called package.json. Copy the following code into it.
{
"name": "reduxapp",
"version": "1.0.0",
"description": "",
"main": "index.js",
"scripts": {
"start": "webpack-dev-server"
},
"author": "KRUNAL LATHIYA",
"devDependencies": {
"babel-core": "^6.24.0",
"babel-preset-es2015": "^6.24.0",
"babel-preset-react": "^6.23.0",
"babel-preset-stage-3": "^6.22.0",
"webpack": "^2.3.2",
"webpack-dev-server": "^2.4.2"
},
"dependencies": {
"react": "^15.4.2",
"react-dom": "^15.4.2",
"react-redux": "^5.0.6",
"redux": "^3.7.2"
}
}
Switch to a terminal and type the following command.
npm install
Next step will be to create the webpack.config.js file in the root folder.
// webpack.config.js
module.exports = {
entry: './src/main.js',
output: {
filename: 'bundle.js'
},
module: {
{
test: /\.js\$/,
exclude: /node_modules/
}
]
},
devServer: {
port: 3000
}
};
Also, create one file called in root called .babelrc
{
"presets": ["es2015", "react", "stage-3"]
}
Make one file in the root called index.html
<!DOCTYPE html>
<html>
<meta charset="utf-8">
<title>Redux Tutorial 2017</title>
<body>
<div id="root"></div>
</body>
</html>
Go to a terminal and type this command.
npm start
Server will start at this URL: http://localhost:3000
// main.js
import React from 'react';
import { render } from 'react-dom';
import { Provider } from 'react-redux';
import App from './components/App';
render(
<Provider>
<App />
</Provider>,
document.getElementById('root')
)
I have included the react particular dependency, but also I have included react-redux. It provides us a store through our entire application. So our App element is wrapped around Provider.
## Step 3: Create components directory inside src.
Make one directory and inside make our most important component file called App.js
// App.js
import React from 'react';
const App = () => {
<div className="container">
App Component
</div>
}
export default App;
App.js the component where our other components are included. We are creating the simple counter application. However, we need to define first how many components are required to complete the application.
There are mainly two types of components when you are dealing with React and Redux.
1. Smart Component
2. Dumb Component
## Smart Component
The smart component is the kind of component, which directly interacts with the state of our application. It has access to the store and it can either dispatch the actions or get the current state of our application. It is the smart components because when the store is changed, by default, it subscribes the new state and changes the view according to it. In our application, there are three smart components.
1. Counter.js
3. RemoveCounter.js
All these smart components are put in the containers folder, which I will create later in this article. Container components only contain that components that are smart.
## Dumb Component
App.js is the Dumb component, it includes the child component but, it does not interact the store. So we put that component inside components folder.
## Step 4: Create container dir. Inside src folder.
We need to create three container components inside this directory as I have mentioned before. We are creating AddCounter.js component.
// AddComponent.js
import React, { Component } from 'react';
import { connect } from 'react-redux';
import { addCounter } from '../actions';
import { bindActionCreators } from 'redux';
constructor(props) {
super(props);
}
render() {
return (
<div className="container">
<form>
<div className="field is-grouped">
<div className="control">
<button className="button is-primary"
</button>
</div>
</div>
</form>
</div>
)
}
}
function mapDispatchToProps(dispatch) {
return { actions: bindActionCreators(addCounter, dispatch) }
}
Here, I need to explain lots of things so let us get started.
If AddCounter component is container component, then it connects to the store remember, that is why it is a smart component.
So the last line of the code is that we are connecting our component to the Redux store.
When the user clicks the button according to Redux principle, it dispatches an action, so we need to pass dispatch function as a property to this component.
Now, it dispatches the action, and in our scenario, it named as addCounter(). So addCounter return an action.
## Create an Action
Make one folder inside src called actions and in that create one file called index.js
// index.js
import * as actionType from './ActionType';
export const addCounter = () => ({
});
So as I mentioned, It returns an object describes our actions.
Here, I have also included one more file called ActionType.js
This file is needed because all the action names are constant and if we create one file, which only exports the naming constant then it is easy for us to define any action without any typos because actions names are on the strings.
## Make ActionType.js
This file is inside actions directory.
// ActionType.js
export const REMOVE_COUNTER = 'REMOVE_COUNTER';
Now, coming to the point, we are at AddCounter.js file, right?
// AddCounter.js
import React, { Component } from 'react';
import { connect } from 'react-redux';
import { addCounter } from '../actions';
import { bindActionCreators } from 'redux';
constructor(props) {
super(props);
}
render() {
return (
<div className="container">
<form>
<div className="field is-grouped">
<div className="control">
<button className="button is-primary"
</button>
</div>
</div>
</form>
</div>
)
}
}
function mapDispatchToProps(dispatch) {
return { actions: bindActionCreators(addCounter, dispatch) }
}
Function mapDispatchToProps is needed because we need to pass dispatch as a property to our component and also we need to bind the actions with this component.
## Step 5: Create reducer directory inside src and make one file
Make one file called counterReducer.js inside reducer directory.
// counterReducer.js
import * as actionType from '../actions/ActionType';
const counterReducer = (state = 0, action) => {
let newState;
switch (action.type) {
return newState = state + action.payload;
case actionType.REMOVE_COUNTER:
return newState = state - action.payload;
default:
return state
}
}
export default counterReducer;
Please note that I have described both cases in here
1. Increment (for add +1 in counter state)
2. Decrement (decrease one from counter state)
If you carefully see the above cases then, I have not modified the state directly. See, I have defined a new variable and then assign new state in that variable and return that variable.
So, it is a pure function, which is not mutating any store state, just take the old state value and add that old value plus new value and assign it to the variable and return the new state of our application. This is the central principle of Redux after all.
Make one file inside reducer directory called index.js
// index.js
import { combineReducers } from 'redux';
import counterReducer from './counterReducer';
const counterApp = combineReducers({
counterReducer
})
export default counterApp
Now, this is our single and final store. This is the store we include in our main.js file. combineReducer() function as the name suggests to combine all reducers in one store and return as a global application state object.
Now, our main.js file will look like this. I have passed the store to our entire application. Also Included the reducers.
// main.js
import React from 'react';
import { render } from 'react-dom';
import { createStore } from 'redux';
import { Provider } from 'react-redux';
import App from './components/App';
import reducer from './reducers';
const store = createStore(reducer);
render(
<Provider store={store}>
<App />
</Provider>,
document.getElementById('root')
)
## Step 6: Create same smart component for RemoveCounter.js
Make RemoveCounter.js inside container directory.
// RemoveCounter.js
import React, { Component } from 'react';
import { connect } from 'react-redux';
import { removeCounter } from '../actions';
import { bindActionCreators } from 'redux';
class RemoveCounter extends Component {
constructor(props) {
super(props);
}
render() {
return (
<div className="container">
<form>
<div className="field is-grouped">
<div className="control">
<button className="button is-primary"
onClick={(e) => {e.preventDefault();this.props.dispatch(removeCounter())}}>
Remove
</button>
</div>
</div>
</form>
</div>
)
}
}
function mapDispatchToProps(dispatch) {
return { actions: bindActionCreators(removeCounter, dispatch) }
}
export default connect(mapDispatchToProps)(RemoveCounter);
Now, switch to src >> actions >> index.js and add new action for removeCounter. So our final file looks like this.
// index.js
import * as actionType from './ActionType';
export const addCounter = () => ({
});
export const removeCounter = () => ({
type: actionType.REMOVE_COUNTER,
});
We have already defined the decrement reducer.
// counterReducer.js
import * as actionType from '../actions/ActionType';
const counterReducer = (state = 0, action) => {
let newState;
switch (action.type) {
return newState = state + action.payload;
case actionType.REMOVE_COUNTER:
return newState = state - action.payload;
default:
return state
}
}
export default counterReducer;
## Step 7: Now final smart component remains is Counter.js
// Counter.js
import React, { Component } from 'react';
import { connect } from 'react-redux';
class Counter extends Component {
constructor(props){
super(props);
}
render(){
return (
<div className="cotainer">
<h1>
{this.props.count}
</h1>
</div>
</div>
)
}
}
function mapStateToProps(state){
return {
count: state.counterReducer,
};
}
export default connect(mapStateToProps)(Counter);
This is the smart component, so we need to connect it with Redux store. So we have connected it with the store and fetched the latest state from the store and display it.
mapStateToProps maps the state to the props of current component and shows the data as a property of the component.
## Step 8: Include all the components into the App.js file.
// App.js
import React from 'react';
import Counter from '../containers/Counter';
import RemoveCounter from '../containers/RemoveCounter';
const App = () => {
return (
<div className="container">
<Counter></Counter><br />
<div className="columns">
<div className="column is-11">
</div>
<div className="column auto">
<RemoveCounter></RemoveCounter>
</div>
</div>
</div>
)
}
export default App;
npm start
If you switch to this URL: http://localhost:3000. You can see a counter application.
Redux Tutorial For Beginners
|
{}
|
# Defining an abbreviated notation for referencing glossary entries
I have defined a command for adding a new symbol to the glossaries as follows:
\newcommand{\NewSymb}[4]{
\newglossaryentry{#1}{
name=#2,
description={\nopostdesc #3},
sort=#4,
type=los,
nonumberlist=true
}
}
which can then be used for example like this:
\NewSymb{s:lattice}{\ensuremath{\Lambda}}{A Lattice}{lattice}
Now, in the text I can refer to this entry via
\gls[hyper=false]{s:lattice}
but this is a bit too long for my taste. Therefore I have defined a command via
\newcommand{\LA}{{\gls[hyper=false]{s:lattice}}}
but I was wondering if it is somehow possible to integrate this \newcommand into the \NewSymb command, such that, after executing, say
\NewSymb{s:lattice}{\ensuremath{\Lambda}}{A Lattice}{lattice}{LA}
I automatically have access to an abbreviated command \LA. The name of the abbreviated command would be provided as a parameter of the NewSymb command.
-
Please enclose your inline code with . – xport Jun 2 '11 at 19:23
\newcommand{\NewSymb}[5]{
\newglossaryentry{#1}{
name=#2,
description={\nopostdesc #3},
sort=#4,
type=los,
nonumberlist=true
}
\newcommand#5{\gls[hyper=false]{#1}}
}
\NewSymb{s:lattice}{\ensuremath{\Lambda}}{A Lattice}{lattice}{\LA}
The fifth argument is the control sequence for calling \gls. It would be possible to express the argument without the backslash, but it's overkill.
It's possible to avoid the check made by \newcommand by changing it into \def. However it's a very risky behavior to redefine existing commands. For example, try to redefine \fi. An acceptable compromise might be
\newglossaryentry{#2}{
name=#3,
description={\nopostdesc #4},
sort=#5,
type=los,
nonumberlist=true
}
#1#6{\gls[hyper=false]{#2}}
}
after which
\NewSymb{s:lattice}{\ensuremath{\Lambda}}{A Lattice}{lattice}{\LA}
would work as before. But
\NewSymb[\renewcommand]{x}{y}{z}{X}{\fi}
would redefine \fi`. (DON'T do it!)
-
works like a charm! thanks!! – TriSSSe Jun 2 '11 at 19:34
@Christian In order that the question doesn't remain among the unanswered ones, it's necessary to mark an answer as accepted. There's no hurry, however. – egreg Jun 2 '11 at 19:40
@egreg: Actually, one upvote is sufficient. ;-) – lockstep Jun 2 '11 at 19:41
@lockstep: nice to know. – egreg Jun 2 '11 at 20:00
@Christian: It's very dangerous to automatically redefine commands. See the edited answer. – egreg Jun 4 '11 at 8:45
|
{}
|
# Can any matrix be expressed as the product of two vectors?
• I
## Main Question or Discussion Point
For example, does this always hold true?
M_ab = v_a × w_b
If not, where does it break down?
hilbert2
Gold Member
Let's say there are a column vector $A = \begin{bmatrix}a \\ b\end{bmatrix}$ and row vector $B = \begin{bmatrix}c & d\end{bmatrix}$ that have
$AB = \begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$.
Is this possible, knowing that $xy = 0$ for real numbers $x,y$ implies that either $x=0$ or $y=0$ ?
Let's say there are a column vector $A = \begin{bmatrix}a \\ b\end{bmatrix}$ and row vector $B = \begin{bmatrix}c & d\end{bmatrix}$ that have
$AB = \begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$.
Is this possible, knowing that $xy = 0$ for real numbers $x,y$ implies that either $x=0$ or $y=0$ ?
Thanks. I was trying to think of a counter example. This is very obvious.
RPinPA
Homework Helper
No, such a matrix has rank 1. Using the notation that $\mathbf w^T$ is the row vector of $\mathbf w$, then row 1 of M is $v_1 \mathbf w^T$, row 2 is $v_2 \mathbf w^T$, etc. Every row is a multiple of every other row. And every column is a multiple of every other column, as all have the form $w_j \mathbf v$.
And in particular, any invertible matrix 2x2 or larger is going to be a counterexample.
What is true is that you can express any matrix M of rank n as a sum of n rank-1 matrices $\sum_{i=1}^n {\mathbf v_i \mathbf w_i^T}$.
Last edited:
fresh_42
Mentor
For example, does this always hold true?
M_ab = v_a × w_b
If not, where does it break down?
It can always be done by a linear combination of those where $\operatorname{rank}M$ is the minimal length of it.
Stephen Tashi
For example, does this always hold true?
M_ab = v_a × w_b
If not, where does it break down?
You didn't say how you define the "product of two vectors". Lets assume you are thinking of this type of example.
v_a = (a1, a2)
v_b = (b1,b,2,b3)
The "product" is a table of data with 2 rows and 3 columns. The (i,j) entry of the table is (a_i)(b_j).
It would be nice if all data tables were so simple! A person who could do multiplication wouldn't need the body of the table.
Thinking about the various complicated data tables that we encounter makes it clear that not all data tables (matrices) have such a simple structure.
However, as @fresh_42 indicates in post #6, all data tables can be written as linear combinations of such simple data tables. This is a remarkable and important fact. It's a concrete interpretation of the "singular value decomposition" of a matrix.
hilbert2
Gold Member
Also, I guess the determinant of this kind of matrices (if they're square) is always zero, as the columns are multiples of each other. This is a very limiting property for a matrix.
fresh_42
Mentor
Also, I guess the determinant of this kind of matrices (if they're square) is always zero, as the columns are multiples of each other. This is a very limiting property for a matrix.
Sure, they are rank one matrices.
The question gets interesting, if we ask for the minimal length of linear combinations of $x\otimes y \otimes z$ to represent a given bilinear mapping, e.g. matrix multiplication. If we define the matrix exponent $\omega := \min\{\,\gamma\,|\,(A;B) \longmapsto A\cdot B = \sum_{i}^Rx_i(A) \otimes y_i(B) \otimes Z_i \,\wedge \, R=O(n^\gamma)\,\}$ then $2\leq \omega \leq 2.3727$ and we do not know how close we can come to the lower bound.
Two is the lower and three the trivial upper bound. I find this fascinating, as it somehow contains the question whether there are intrinsically difficult problems out there, or whether we just haven't found the right clue - similar to NP=P and ERH.
hilbert2
The question gets interesting, if we ask for the minimal length of linear combinations of $x\otimes y \otimes z$ to represent a given bilinear mapping, e.g. matrix multiplication. If we define the matrix exponent $\omega := \min\{\,\gamma\,|\,(A;B) \longmapsto A\cdot B = \sum_{i}^Rx_i(A) \otimes y_i(B) \otimes Z_i \,\wedge \, R=O(n^\gamma)\,\}$ then $2\leq \omega \leq 2.3727$ and we do not know how close we can come to the lower bound.
|
{}
|
Tommy Butler > File-Util-3.39 > File::Util
File-Util-3.39.tar.gz
Dependencies
Annotate this POD (1)
Website
# CPAN RT
Open 0
View/Report Bugs
Module Version: 3.39 Source Latest Release: File-Util-4.161950
# NAME
File::Util - Easy, versatile, portable file handling
# DESCRIPTION
File::Util provides a comprehensive toolbox of utilities to automate all kinds of common tasks on file / directories. Its purpose is to do so in the most portable manner possible so that users of this module won't have to worry about whether their programs will work on other OSes and machines.
# SYNOPSIS
use File::Util;
my $f = File::Util->new(); my$content = $f->load_file('foo.txt');$content =~ s/this/that/g;
$f->write_file( file => 'bar.txt', content =>$content,
);
$f->write_file( file => 'file.bin', content =>$binary_content, '--binmode'
);
my @lines = $f->load_file('randomquote.txt', '--as-lines'); my$line = int rand scalar @lines;
print $lines[$line ];
my @files = $f->list_dir('/var/tmp', qw/ --files-only --recurse /); my @textfiles =$f->list_dir('/var/tmp', '--pattern=\.txt$'); if ($f->can_write('wibble.log') ) {
my $HANDLE =$f->open_handle(
file => 'wibble.log',
mode => 'append'
);
print $HANDLE "Hello World! It's ", scalar localtime; close$HANDLE
}
my $log_line_count =$f->line_count('/var/log/httpd/access_log');
print "My file has a bitmask of " . $f->bitmask('my.file'); print "My file is a " . join(', ',$f->file_type('my.file')) . " file."
warn 'This file is binary!' if $f->isbin('my.file'); print "My file was last modified on " . scalar localtime$f->last_modified('my.file');
# ...and _lots_ more
# INSTALLATION
To install this module type the following at the command prompt:
perl Build.PL
perl Build
perl Build test
sudo perl Build install
On Windows systems, the "sudo" part of the command may be omitted, but you will need to run the rest of the install command with Administrative privileges
Exporter
# EXPORTED SYMBOLS
Exports nothing by default. File::Util respects your namespace.
## EXPORT_OK
The following symbols comprise @File::Util::EXPORT_OK), and as such are available for import to your namespace only upon request.
atomize_path (see atomize_path)
bitmask (see bitmask)
can_flock (see can_flock)
can_read (see can_read)
can_write (see can_write)
created (see created)
ebcdic (see ebcdic)
escape_filename (see escape_filename)
existent (see existent)
file_type (see file_type)
isbin (see isbin)
last_access (see last_access)
last_changed (see last_changed)
last_modified (see last_modified)
NL (see NL)
needs_binmode (see needs_binmode)
return_path (see return_path)
size (see size)
SL (see SL)
strip_path (see strip_path)
valid_filename (see valid_filename)
Note: Symbols in @Class::OOorNO::EXPORT_OK are also available for import.
## EXPORT_TAGS
:all (exports all of @File::Util::EXPORT_OK)
# METHODS
Note: In the past, some of the methods listed would state that they were autoloaded methods. This mechanism has been changed. Only the error handling and help messages are AutoLoad'ed now. (see AutoLoader.) if you want to know more about AutoLoading in Perl. See the CHANGES file distributed with File::Util for an explanation of why this change was made.
Methods listed in alphabetical order.
## atomize_path
Syntax: atomize_path( [file/path or file_name] )
This method is used internally by File::Util to portably handle absolute filenames on different platforms, but it can be a useful tool for you as well.
This method takes a single string as its argument. The string is expected to be a fully-qualified (absolute) or relative path to a file or directory. It carefully splits the string into three parts: The root of the path, the rest of the path, and the final file/directory named in the string.
Depending on the input, the root and/or path may be empty strings. The following table can serve as a guide in what to expect from atomize_path()
+-------------------------+----------+--------------------+----------------+
| INPUT | ROOT | PATH-COMPONENT | FILE/DIR |
+-------------------------+----------+--------------------+----------------+
| C:\foo\bar\baz.txt | C:\ | foo\bar | baz.txt |
| /foo/bar/baz.txt | / | foo/bar | baz.txt |
| ./a/b/c/d/e/f/g.txt | | ./a/b/c/d/e/f | g.txt |
| :a:b:c:d:e:f:g.txt | : | a:b:c:d:e:f | g.txt |
| ../wibble/wombat.ini | | ../wibble | wombat.ini |
| ..\woot\noot.doc | | ..\woot | noot.doc |
| ../../zoot.conf | | ../.. | zoot.conf |
| /root | / | | root |
| /etc/sudoers | / | etc | sudoers |
| / | / | | |
| D:\ | D:\ | | |
| D:\autorun.inf | D:\ | | autorun.inf |
+-------------------------+----------+--------------------+----------------+
## bitmask
Syntax: bitmask( [file name] )
Gets the bitmask of the named file, provided the file exists. If the file exists, the bitmask of the named file is returned in four digit octal notation e.g.- 0644. Otherwise, returns undef if the file does not exist.
## can_flock
Syntax: can_flock
Returns 1 if the current system claims to support flock() and if the Perl process can successfully call it. (see "flock" in perlfunc.) Unless both of these conditions are true a zero value (0) is returned. This is a constant method. It accepts no arguments and will always return the same value for the system on which it is executed.
Note: Perl will try to support or emulate flock whenever it can via available system calls, namely flock; lockf; or with fcntl.
## can_read
Syntax: can_read( [file name] )
Returns 1 if the named file (or directory) is readable by your program according to the applied permissions of the file system on which the file resides. Otherwise a value of undef is returned.
This works the same as Perl's built-in -r file test operator, (see "-X" in perlfunc), it's just easier for some people to remember.
## can_write
Syntax: can_write( [file name] )
Returns 1 if the named file (or directory) is writable by your program according to the applied permissions of the file system on which the file resides. Otherwise a value of undef is returned.
This works the same as Perl's built-in -w file test operator, (see "-X" in perlfunc), it's just easier for some people to remember.
## created
Syntax: created( [file name] )
Returns the time of creation for the named file in non-leap seconds since whatever your system considers to be the epoch. Suitable for feeding to Perl's built-in functions "gmtime" and "localtime". (see "time" in perlfunc.)
## ebcdic
Syntax: ebcdic
Returns 1 if the machine on which the code is running uses EBCDIC, or returns 0 if not. (see perlebcdic.) This is a constant method. It accepts no arguments and will always return the same value for the system on which it is executed.
## escape_filename
Syntax: escape_filename( [string], [escape char] )
Returns it's argument in an escaped form that is suitable for use as a filename. Illegal characters (i.e.- any type of newline character, tab, vtab, and the following / | * " ? < : > \), are replaced with [escape char] or "_" if no [escape char] is specified. Returns an empty string if no arguments are provided.
## existent
Syntax: existent( [file name] )
Returns 1 if the named file (or directory) exists. Otherwise a value of undef is returned.
This works the same as Perl's built-in -e file test operator, (see "-X" in perlfunc), it's just easier for some people to remember.
## file_type
Syntax: file_type( [file name] )
Returns a list of keywords corresponding to each of Perl's built in file tests (those specific to file types) for which the named file returns true. (see "-X" in perlfunc.)
The keywords and their definitions appear below; the order of keywords returned is the same as the order in which the are listed here:
PLAIN File is a plain file.
TEXT File is a text file.
BINARY File is a binary file.
DIRECTORY File is a directory.
SYMLINK File is a symbolic link.
PIPE File is a named pipe (FIFO).
SOCKET File is a socket.
BLOCK File is a block special file.
CHARACTER File is a character special file.
## flock_rules
Syntax: flock_rules( [keyword list] )
Sets I/O race condition policy, or tells File::Util how it should handle race conditions created when a file can't be locked because it is already locked somewhere else (usually by another process).
An empty call to this method returns a list of keywords representing the rules that are currently in effect for the object.
Otherwise, a call should include a list with array containing your chosen directive keywords in order of precedence. The rules will be applied in cascading order when a File::Util object attempts to lock a file, so if the actions specified by the first rule don't result in success, the second rule is applied, and so on.
Recognized keywords:
NOBLOCKEX
tries to get an exclusive lock on the file without blocking (waiting)
NOBLOCKSH
tries to get a shared lock on the file without blocking
BLOCKEX
waits to try getting an exclusive lock
BLOCKSH
waits to try getting a shared lock
FAIL
dies with stack trace
WARN
warn()s about the error with a stack trace and returns undef
IGNORE
ignores the failure to get an exclusive lock
UNDEF
returns undef
ZERO
returns 0
Examples:
ex- flock_rules( qw/ NOBLOCKEX FAIL / );
This is the default policy. When in effect, the File::Util object will first attempt to get a non-blocking exclusive lock on the file. If that attempt fails the File::Util object will call die() with a detailed error message and a stack trace.
ex- flock_rules( qw/ NOBLOCKEX BLOCKEX FAIL / );
The File::Util object will first attempt to get a non-blocking exclusive lock on the file. If that attempt fails it falls back to the second policy rule "BLOCKEX" and tries again to get an exclusive lock on the file, but this time by blocking (waiting for its turn). If that second attempt fails, the File::Util object will fail with a detailed error message and a stack trace.
ex- flock_rules( qw/ BLOCKEX IGNORE / );
The File::Util object will first attempt to get a file non-blocking lock on the file. If that attempt fails it will ignore the error, and go on to open the file anyway and no failures will occur or warings be issued.
## isbin
Syntax: isbin( [file name] )
Returns 1 if the named file (or directory) exists. Otherwise a value of undef is returned, indicating that the named file either does not exist or is of another file type.
This works the same as Perl's built-in -B file test operator, (see "-X" in perlfunc), it's just easier for some people to remember.
## last_access
Syntax: last_access( [file name] )
Returns the last accessed time for the named file in non-leap seconds since whatever your system considers to be the epoch. Suitable for feeding to Perl's built-in functions "gmtime" and "localtime". (see "time" in perlfunc.)
## last_changed
Syntax: last_changed( [file name] )
Returns the inode change time for the named file in non-leap seconds since whatever your system considers to be the epoch. Suitable for feeding to Perl's built-in functions "gmtime" and "localtime". (see "time" in perlfunc.)
## last_modified
Syntax: last_modified( [file name] )
Returns the last modified time for the named file in non-leap seconds since whatever your system considers to be the epoch. Suitable for feeding to Perl's built-in functions "gmtime" and "localtime". (see "time" in perlfunc.)
## line_count
Syntax: line_count( [file name] )
Returns the number of lines in the named file. Fails with an error if the named file does not exist.
## list_dir
Syntax: list_dir( [directory name] , [--opts] )
Returns alphabetically sorted all file names in the directory specified if it exists. Fails with an error message if no such directory is found, or the directory is inaccessible.
The behavior of this method has changed slightly after version 3.29. If running with the --fatals-as-warning flag, the previous behavior was to abort immediately. This is not the case anymore. If running with the --fatals-as-warning flag, list_dir() will still emit a warning when it encounters an otherwise fatal error, but it will also return whatever directory contents it is able to successfully access.
Flags accepted by list_dir()
--dirs-only
return only directory contents which are directories
--files-only
return only directory contents which are files
--no-fsdots
do not include "." and ".." in the list of directory contents
--pattern
return only files/directories matching pattern provided. argument should be plain text string. It will be converted to a perl regex and passed to CORE::grep as the method scans through directory listings for a match.
(ex- '--pattern=\.txt$' returns all file/directory names ending in ".txt". It will match "foo.txt", but not "foo.txt.gz" because of the "$" anchor in the regular expression passed in.)
or for the opposite effect, '--pattern=.*(?<!\.txt)$' returns all file/directory names that don't end in ".txt" --with-paths Include file paths with the contents of the directory list, relative to the directory named in the call. --recurse Recurse subdirectories --follow Recurse subdirectories, same as --recurse --dirs-as-ref When returning directory listing, include first a reference to the list of subdirectories found, followed by anything else returned by the call. --files-as-ref When returning directory listing, include last a reference to the list of files found, preceded by a list of subdirectories found (or preceded by a list reference to subdirectories found if --dirs-as-ref was also used). --as-ref Return a pair list references: the first is a reference to any subdirectories found by the call, the second is a reference to any files found by the call. --sl-after-dirs Append a directory separator ("/, "\", or ":" depending on your system) to all directories found by the call. Useful in visual displays for quick differentiation between subdirectories and files. --ignore-case Items returned by the call to this method are sorted alphabetically by default, so "Zoo.txt" comes before "alligator.txt" because the alphabetical sort is case-sensitive. This is also the way directories are listed at the system level on most operating systems. If you'd like the directory contents returned by this method to be sorted without regard to case , use this flag. --count-only Returns a single value: an integer reflecting the number of items found in the directory after applying the filter criteria specified by any other flags (ie- "--dirs-only", "--recurse", etc.) that may have been passed in as well. ## load_dir Syntax: load_dir( [directory name] , [--ds-type] ) Returns a data structure containing the contents of each file present in the named directory. The type of data structure returned is determined by the optional data-type switch. Only one option may be used for a given call to this method. Recognized options are listed below. Flags accepted by load_dir() --as-list Causes the method to return a list comprised of the contents loaded from each file (in case-sensitive order) located in the named directory. --as-listref Same as above, except an array reference to the list of items is returned rather than the list itself. --as-hashref *(default) Implicit. If no option is passed in, the default behavior is to return a reference to an anonymous hash whose keys are the names of each file in the specified directory; the hash values for contain the contents of the file represented by its corresponding key. Note: This method does not distinguish between plain files and other file types such as binaries, FIFOs, sockets, etc. Restrictions imposed by the current "read limit" (see the readlimit()) entry below will be applied to the files opened by this method as well. Adjust the readlimit as necessary. my$files = $fu->load_dir('directory/to/load/'); The above code creates an anonymous hash reference that is stored in the variable named "$files". The keys and values of the hash referenced by "$files" would resemble those of the following code snippet (given that the files in the named directory were the files 'a.txt', 'b.html', 'c.dat', and 'd.conf') my($files) =
{
'a.txt' => "the contents of file a.txt",
'b.html' => "the contents of file b.html",
'c.dat' => "the contents of file c.dat",
'd.conf' => "the contents of file d.conf",
};
## load_file
Syntax: load_file( [file name] , [--opts] )
OR: load_file( 'FH' => [file handle reference] , [--opts] )
If [file name] is passed, returns the contents of [file name] in a string. If a [file handle reference] is passed instead, the filehandle will be CORE::read() and the data obtained by the read will be returned in a string.
If you desire the contents of the file (or file handle data) in a list of lines instead of a single string, this can be accomplished through the use of the --as-lines flag (see below).
Flags accepted by load_file()
--as-lines
If this flag is passed then your call to load_file will return an ordered list of strings, each of which is a line from the file [file name]. The lines are returned in the order they are read, from the beginning of the file to the end.
This is not the default behavior. The default behavior is for load_file to return a single string containing the entire contents of the file, including line break characters.
--no-lock
By default this method will attempt to get a lock on the file while it is being read, following whatever rules are in place for the flock policy established either by default (implicitly) or changed by you in a call to File::Util::flock_rules() (see the flock_rules()) entry below.
This method will not try to get a lock on the file if the File::Util object was created with the option --no-lock or if the method was called with the option --no-lock.
This method will automatically call binmode() on binary files for you. If you pass in a filehandle instead of a file name you do not get this automatic check performed for you. In such a case, you'll have to call binmode() on the filehandle yourself. Once you pass a filehandle to this method it has no way of telling if the file opened to that filehandle is binary or not.
Notes: This method does not distinguish between plain files and other file types such as binaries, FIFOs, sockets, etc.
Restrictions imposed by the current "read limit" (see the readlimit()) entry below will be applied to the files opened by this method as well. Adjust the readlimit as necessary.
## make_dir
Syntax: make_dir( [new directory name] , [bitmask], [--opts] )
Attempts to create (recursively) a directory as [new directory name] with the [bitmask] provided. The bitmask is an optional argument and defaults to 0777. If specified, the bitmask must be supplied in the form required by the native perl umask function. see "umask" in perlfunc for more information about the format of the bitmask argument.
As mentioned above, the recursive creation of directories is transparently handled for you. This means that if the name of the directory you pass in contains a parent directory that does not exist, the parent directory(ies) will be created for you automatically and silently in order to create the final directory in the [new directory name].
Simply put, if [new directory] is "/path/to/directory" and the directory "/path/to" does not exist, the directory "/path/to" will be created and the "/path/to/directory" directory will be created thereafter. All directories created will be created with the [bitmask] you specify, or with the default of 0777.
Upon successful creation of the [new directory name], the [new directory name] is returned to the caller.
Flags accepted by make_dir()
--if-not-exists
If this flag is passed in then make_dir will not attempt to create the directory if it already exists. Rather it will return the name of the directory as it normally would if the directory did not exist previous to calling this method.
If a call to this method is made without the --if-not-exists flag and the directory specified as [new directory name] does in fact exist, an error will result as it is impossible to create a directory that already exists.
## max_dives
Syntax: max_dives( [integer] )
When called without any arguments, this method returns an integer reflecting the current number of times the File::Util object will dive into the subdirectories it discovers when recursively listing directory contents from a call to File::Util::list_dir(). The default is 1000. If the number is exceeded, the File::Util object will fail with a diagnostic error message.
When called with an argument, it sets the maximum number of times a File::Util object will recurse into subdirectories before failing with an error message.
This method can only be called with a numeric integer value. Passing a bad argument to this method will cause it to fail with an error message.
(see list_dir)
## needs_binmode
Syntax: needs_binmode
Returns 1 if the machine on which the code is running requires that binmode() (a built-in function) be called on open file handles, or returns 0 if not. (see "binmode" in perlfunc.) This is a constant method. It accepts no arguments and will always return the same value for the system on which it is executed.
## new
Syntax: new( ['parameters' => 'values', etc], [--flags] )
This is the File::Util constructor method. eg- It returns a new File::Util object reference when you call it. It recognizes various parameters and flags that govern the behavior of the new File::Util object.
Parameters accepted by new()
use_flock => true/false value
Optionally specify this option to the File::Util::new method instruct the new object that it should never attempt to use flock() in it's I/O operations. The default is to use flock() when available on your system. Specify this option with a true or false value, true to use flock(), false to not use it.
Optionally specify this option to the File::Util::new method to instruct the new object that it should never attempt to open and read in a file greater than the number of bytes you specify. Obviously this argument can only be a numeric integer value, otherwise it will be silently ignored. The default readlimit for File::Util objects is 52428800 bytes (50 megabytes).
max_dives => positive integer
Optionally specify this option to the File::Util::new method to instruct the new object to set the maximum number of times it will recurse into subdirectories while performing directory listing operations before failing with an error message. This argument can only be a numeric integer value, otherwise it will be silently ignored.
Flags accepted by new()
--fatals-as-warning
Directive to instruct the new File::Util object that when any call to one of its methods results in a fatal error that it should return undef instead of the value(s) that would normally be returned by the call, and to send an error message to STDERR as well.
--fatals-as-status
Directive to instruct the new File::Util object that when any call to one of its methods results in a fatal error that it should return undef instead of the value(s) that would normally be returned by the call.
--fatals-as-errmsg
Directive to instruct the new File::Util object that when any call to one of its methods results in a fatal error that it should return an error message instead of the value(s) that would normally be returned by the call.
## open_handle
Syntax: open_handle( file => [file name], [--opts] )
OR: open_handle( file => [file name], mode => [mode], [--opts] )
OR: open_handle( file => [file name], mode => [mode], bitmask => [bitmask], [--opts] )
OR: open_handle( file => [file name], mode => [mode], bitmask => [bitmask], dbitmask => [bitmask], [--opts] )
Attempts to get a unique open file handle on [file name] in [mode] mode. Returns the file handle if successful or generates a fatal error with a diagnostic message if the operation fails.
You will need to remember to call close() on the filehandle yourself, at your own discretion. Leaving filehandles open is not a good practice, and is not recommended. see "close" in perlfunc).
Once you have the file handle you would use it as you would use any file handle. Remember that unless you specifically turn file locking off when the File::Util object is created (see (see new) or by using the --no-lock flag when calling open_handle, that file locking is going to automagically be handled for you behind the scenes, so long as your OS supports file locking of any kind at all. Great! It's very convenient for you to not have to worry about portably taking care of file locking between one application and the next; by using File::Util in all of them, you know that you're covered.
A slight inconvenience for the price of a larger set of features (compare write_file to this method) you will have to release the file lock on the open handle yourself. File::Util can't manage it for you anymore once it hands the handle over to you. At that point, it's all yours. In order to release the file lock on your file handle, call unlock_open_handle() on it. Otherwise the lock will remain for the life of your process. If you don't want to use the free portable file locking, remember the --no-lock flag, which will turn off file locking for your open handle. Seldom, however, should you ever opt to not use file locking unless you really know what you are doing.
If the file does not yet exist it will be created, and it will be created with a bitmask of [bitmask] if you specify a file creation bitmask using the 'bitmask' option, otherwise the file will be created with the default bitmask of 0777.
Any non-existent directories in the path preceding the actual file name will be automatically (and silently - no warnings) created for you and any new directories will be created with a bitmask of [dbitmask], provided you specify a directory creation bitmask with the 'dbitmask' option.
If specified, the directory creation bitmask [dbitmask] must be supplied in the form required by the native perl umask function.
If there is an error while trying to create any preceding directories, the failure results in a fatal error with a diagnostic error message. If all directories preceding the name of the file already exist, the dbitmask argument has no effect and is silently ignored.
Native Perl open modes
The default behavior of open_handle() is to open file handles using Perl's native open() (see "open" in perlfunc). Unless you use the --use-sysopen flag, the following modes and only these modes are valid.
'mode' => 'read'
[file name] is opened in read-only mode. If the file does not yet exist then a fatal error will occur with a diagnostic help message to help you troubleshoot the problem.
'mode' => 'write' (this is the default mode)
[file name] is created if it does not yet exist. If [file name] already exists then its contents are overwritten with the new content provided.
'mode' => 'append'
[file name] is created if it does not yet exist. If [file name] already exists its contents will be preserved and the new content you provide will be appended to the end of the file.
System level open modes ("open a la C")
Optionally you can ask File::Util to open your handle using CORE::sysopen instead of using the native Perl CORE::open(). This is accomplished by passing in the --use-sysopen flag. Using this feature opens up more possibilities as far as the open modes you can choose from, but also carries with it a few caveats so you have to be careful, just as you'd have to be a little more careful when using sysopen() anyway.
Specifically you need to remember that when using this feature you must NOT mix different types of I/O when working with the file handle. You can't go opening file handles with sysopen() and print to them as you normally would print to a file handle. You have to use syswrite() instead. The same applies here. If you get a sysopen()'d filehandle from open_handle() it is imperative that you use syswrite() on it. You'll also need to use sysseek() and other type of sys* commands on the filehandle instead of their native Perl equivalents.
That said, here are the different modes you can choose from to get a file handle when using the --use-sysopen flag. Remember that these won't work unless you use the flag, and will generate an error if you try using them without it. The standard 'read', 'write', and 'append' modes are already available to you by default. These are the extended modes:
'mode' => 'rwcreate'
[file name] is opened in read-write mode, and will be created for you if it does not already exist.
'mode' => 'rwupdate'
[file name] is opened for you in read-write mode, but must already exist. If it does not exist, a fatal error will result and a diagnostic help message will be printed out to help you troubleshoot the problem.
'mode' => 'rwclobber'
[file name] is opened for you in read-write mode. If the file already exists it's contents will be "clobbered" or wiped out. The file will then be empty and you will be working with the then-truncated file. This can not be undone. Once you call open_handle() using this option, your file WILL be wiped out. If the file does not exist yet, it will be created for you.
'mode' => 'rwappend'
[file name] will be opened for you in read-write mode ready for appending. The file's contents will not be wiped out; they will be preserved and you will be working in append fashion. You will only be able to write starting at the end of the file. If the file does not exist, it will be created for you.
Remember to use sysread() and not plain read() when reading those sysopen()'d filehandles!
Flags accepted by open_handle()
--binmode
Makes sure that CORE::binmode() is called on the filehandle when your content is written. This is useful for times when the content you are writing to file is a binary stream. (see "binmode" in perlfunc).
--no-lock
By default this method will attempt to get a lock on the file while it is being read, following whatever rules are in place for the flock policy established either by default (implicitly) or changed by you in a call to File::Util::flock_rules() (see the flock_rules()) entry below.
This method will not try to get a lock on the file if the File::Util object was created with the option --no-lock or if this method is called with the option --no-lock.
--use-sysopen
Instead of opening the file using Perl's native open() command, File::Util will open the file with the sysopen() command. You will have to remember that your filehandle is a sysopen()'d one, and that you will not be able to use native Perl I/O functions on it. You will have to use the sys* equivalents. See perlopentut for a more in-depth explanation of why you can't mix native Perl I/O with system I/O.
## readlimit
Syntax: readlimit( [integer] )
By default, the largest size file that File::Util will read into memory and return via the load_file is 52428800 byptes (50 megabytes).
This value can be modified by calling this method with an integer value reflecting the new limit you want to impose, in bytes. For example, if you want to set the limit to 10 megabytes, call the method with an argument of 10485760.
If this method is called without an argument, the read limit currently in force for the File::Util object will be returned.
## return_path
Syntax: return_path( [string] )
Takes the file path from the file name provided and returns it such that "/foo/bar/baz.txt" is returned "/foo/bar".
## size
Syntax: size( [file name] )
Returns the file size of [file name] in bytes. Returns 0 if the file is empty, returns undef if the file does not exist.
## strip_path
Syntax: strip_path( [string] )
Strips the file path from the file name provided and returns the file name only.
## touch
Syntax: touch( [file name] )
Behaves like the *nix touch command; Updates the access and modification times of the specified file to the current time. If the file does not exist, File::Util tries to create it empty. This method will fail with a fatal error if system permissions deny alterations to or creation of the file.
Returns 1 if successful. If unsuccessful, fails with a descriptive error message about what went wrong.
## trunc
Syntax: trunc( [file name] )
Truncates [file name] (i.e.- wipes out, or "clobbers" the contents of the specified file. Returns 1 if successful. If unsuccessful, fails with a descriptive error message about what went wrong.
## unlock_open_handle
Syntax: unlock_open_handle([file handle])
Release the flock on a file handle you opened with open_handle.
Returns true on success, false on failure. Will not raise a fatal error if the unlock operation fails. You can capture the return value from your call to this method and die() if you so desire. Failure is not ever very likely, or File::Util wouldn't have been able to get a portable lock on the file in the first place.
If File::Util wasn't able to ever lock the file due to limitations of your operating system, a call to this method will return a true value.
If file locking has been disabled on the file handle via the --no-lock flag at the time open_handle was called, or if file locking was disabled using the use_flock method, or if file locking was disabled on the entire File::Util object at the time of its creation (see new()), calling this method will have no effect and a true value will be returned.
## use_flock
Syntax: use_flock( [true / false value] )
When called without any arguments, this method returns a true or false value to reflect the current use of flock() within the File::Util object.
When called with a true or false value as its single argument, this method will tell the File::Util object whether or not it should attempt to use flock() in its I/O operations. A true value indicates that the File::Util object will use flock() if available, a false value indicates that it will not. The default is to use flock() when available on your system.
## write_file
Syntax: write_file( file' => [file name], 'content' => [string], [--opts] )
OR: write_file( file => [file name], content => [string], mode => [mode], [--opts] )
OR: write_file( file => [file name], content => [string], mode => [mode], bitmask => [bitmask], [--opts] )
OR: write_file( file => [file name], content => [string], mode => [mode], bitmask => [bitmask], dbitmask => [bitmask], [--opts] )
Attempts to write [string] to [file name] in mode [mode]. If the file does not yet exist it will be created, and it will be created with a bitmask of [bitmask] if you specify a file creation bitmask using the 'bitmask' option, otherwise the file will be created with the default bitmask of 0777.
[string] should be a string or a scalar variable containing a string. The string can be any type of data, such as a binary stream, or ascii text with line breaks, etc. Be sure to pass in the --binmode flag for binary streams.
Returns 1 if successful or fails (fatal) with an error message if not successful.
Any non-existent directories in the path preceding the actual file name will be automatically (and silently - no warnings) created for you and any new directories will be created with a bitmask of [dbitmask], provided you specify a directory creation bitmask with the 'dbitmask' option.
If specified, the directory creation bitmask [dbitmask] must be supplied in the form required by the native perl umask function.
If there is an error while trying to create any preceding directories, the failure results in a fatal error with a diagnostic error message. If all directories preceding the name of the file already exist, the dbitmask argument has no effect and is silently ignored.
'mode' => 'write' (this is the default mode)
[file name] is created if it does not yet exist. If [file name] already exists then its contents are overwritten with the new content provided.
'mode' => 'append'
[file name] is created if it does not yet exist. If [file name] already exists its contents will be preserved and the new content you provide will be appended to the end of the file.
Flags accepted by write_file()
--binmode
Makes sure that CORE::binmode() is called on the filehandle when your content is written. This is useful for times when the content you are writing to file is a binary stream.
--empty-writes-OK
Allows you to call this method without providing a content argument (it lets you create an empty file without warning you or failing. Be advised that if you use this flag, it will have the same effect as truncating a file that already has content in it (i.e.- it will "clobber" non-empty files)
--no-lock
By default this method will attempt to get a lock on the file while it is being read, following whatever rules are in place for the flock policy established either by default (implicitly) or changed by you in a call to File::Util::flock_rules() (see the flock_rules()) entry below.
This method will not try to get a lock on the file if the File::Util object was created with the option --no-lock or if this method is called with the option --no-lock.
## valid_filename
Syntax: valid_filename( [string] )
For the given string, returns 1 if the string is a legal file name for the system on which the program is running, or returns undef if it is not. This method does not test for the validity of file paths! It tests for the validity of file names only. (It is used internally to check beforehand if a file name is useable when creating new files, but is also a public method available for external use.)
# CONSTANTS
## NL
Syntax: NL
Returns the correct new line character (or character sequence) for the system on which your program runs.
## SL
Syntax: SL
Returns the correct directory path separator for the system on which your program runs.
## OS
Syntax: OS
Returns the File::Util keyword for the operating system FAMILY it detected. The keyword for the detected operating system will be one of the following, derived from the conents of $^O, or if $^O can not be found, from the contents of $Config::Config{osname} (see native Config library), or if that doesn't contain a recognizable value, finally falls back to UNIX. Generally speaking, Linux operating systems are going to be detected as UNIX. This isn't a bug. The OS FAMILY to which it belongs uses UNIX style filesystem conventions and line endings, which are the relevant things to file handling operations. UNIX Specifics: OS name =~ /^(?:darwin|bsdos)/i CYGWIN Specifics: OS name =~ /^cygwin/i WINDOWS Specifics: OS name =~ /^MSWin/i VMS Specifics: OS name =~ /^vms/i DOS Specifics: OS name =~ /^dos/i MACINTOSH Specifics: OS name =~ /^MacOS/i EPOC Specifics: OS name =~ /^epoc/i OS2 Specifics: OS name =~ /^os2/i # PREREQUISITES Perl 5.006 or better Exception::Handler v1.00_0 or better # EXAMPLES ## Get the names of all files and subdirectories in a directory use File::Util; my$f = File::Util->new();
# option --no-fsdots excludes "." and ".." from the list
my @dirs_and_files = $f->list_dir('/foo', qw/--no-fsdots --count-only/); ## Get the names of files and subdirs in a directory as separate array refs use File::Util; my$f = File::Util->new();
my( $dirs,$files ) = $f->list_dir('/foo', '--as-ref'); -OR- my($dirs, $files ) =$f->list_dir('.', qw/--dirs-as-ref --files-as-ref/);
## Get the contents of a file in a string
use File::Util;
my @contents = $f->load_file('filename','--as-lines'); ## Get an open file handle for reading use File::Util; my$f = File::Util->new();
my $fh =$f->open_handle(
file => 'new_filename',
);
## Get an open file handle for writing
use File::Util;
my $f = File::Util->new(); my$fh = $f->open_handle( file => 'new_filename', mode => 'write' ); ## Write to a new or existing file use File::Util; my$content = 'Pathelogically Eclectic Rubbish Lister';
my $f = File::Util->new();$f->write_file( file => 'a new file.txt', content => $content ); # optionally specify a creation bitmask when writing to a new file$f->write_file(
file => 'a new file.txt',
content => $content ); ## Append to a new or existing file use File::Util; my$content = 'Pathelogically Eclectic Rubbish Lister';
my $f = File::Util->new();$f->write_file(
file => 'a new file.txt',
mode => 'append',
print $f->valid_filename("foo?+/bar~@/#baz.txt") ? 'ok' : 'bad'; ## Get the number of lines in a file use File::Util; my$f = File::Util->new();
my $linecount =$f->line_count('foo.txt');
## Strip the path from a file name
use File::Util;
my $f = File::Util->new(); # On Windows # (prints "hosts") my$path = $f->strip_path('C:\WINDOWS\system32\drivers\etc\hosts'); # On Linux/Unix # (prints "perl") print$f->strip_path('/usr/bin/perl');
# On a Mac
# (prints "baz")
print $f->strip_path('foo:bar:baz'); ## Get the path preceding a file name use File::Util; my$f = File::Util->new();
# On Windows
# (prints "C:\WINDOWS\system32\drivers\etc")
my $path =$f->return_path('C:\WINDOWS\system32\drivers\etc\hosts');
# On Linux/Unix
# (prints "/usr/bin")
print $f->return_path('/usr/bin/perl'); # On a Mac # (prints "foo:bar") print$f->return_path('foo:bar:baz');
## Find out if the host system can use flock
use File::Util qw( can_flock );
print can_flock;
-OR-
print File::Util->can_flock;
-OR-
my $f = File::Util->new(); print$f->can_flock;
## Find out if the host system needs to call binmode on binary files
use File::Util qw( needs_binmode );
print needs_binmode;
-OR-
use File::Util;
print File::Util->needs_binmode;
-OR-
use File::Util;
my $f = File::Util->new(); print$f->needs_binmode;
## Find out if a file can be opened for read (based on file permissions)
use File::Util;
my $f = File::Util->new(); my$is_readable = $f->can_read('foo.txt'); ## Find out if a file can be opened for write (based on file permissions) use File::Util; my$f = File::Util->new();
my $is_writable =$f->can_write('foo.txt');
## Escape illegal characters in a potential file name (and its path)
use File::Util;
my $f = File::Util->new(); # prints "C__WINDOWS_system32_drivers_etc_hosts" print$f->escape_filename('C:\WINDOWS\system32\drivers\etc\hosts');
# prints "baz)__@^"
# (strips the file path from the file name, then escapes it
print $f->escape_filename( '/foo/bar/baz)?*@^', '--strip-path' ); # prints "_foo_!_@so~me#illegal$_file&(name"
# (yes, that is a legal filename)
print $f->escape_filename(q[\foo*!_@so~me#illegal$*file&(name]);
## Find out if the host system uses EBCDIC
use File::Util qw( ebcdic );
print ebcdic;
-OR-
use File::Util;
print File::Util->ebcdic;
-OR-
use File::Util;
my $f = File::Util->new(); print$f->ebcdic;
## Get the type(s) of an existent file
use File::Util qw( file_type );
print file_type('foo.exe');
-OR-
use File::Util;
print File::Util->file_type('bar.txt');
-OR-
use File::Util;
my $f = File::Util->new(); print$f->file_type('/dev/null');
## Get the bitmask of an existent file
use File::Util qw( bitmask );
-OR-
use File::Util;
-OR-
use File::Util;
my $f = File::Util->new(); print$f->bitmask('/dev/null');
## Get time of creation for a file
use File::Util qw( created );
print scalar localtime created('/usr/bin/exim');
-OR-
use File::Util;
print scalar localtime File::Util->created('C:\COMMAND.COM');
-OR-
use File::Util;
my $f = File::Util->new(); print scalar localtime$f->created('/bin/less');
## Get the last access time for a file
use File::Util qw( last_access );
print scalar localtime last_access('/usr/bin/exim');
-OR-
use File::Util;
print scalar localtime File::Util->last_access('C:\COMMAND.COM');
-OR-
use File::Util;
my $f = File::Util->new(); print scalar localtime$f->last_access('/bin/less');
## Get the inode change time for a file
use File::Util qw( last_changed );
print scalar localtime last_changed('/usr/bin/vim');
-OR-
use File::Util;
print scalar localtime File::Util->last_changed('C:\COMMAND.COM');
-OR-
use File::Util;
my $f = File::Util->new(); print scalar localtime$f->last_changed('/bin/cpio');
use File::Util qw( last_modified );
print scalar localtime last_modified('/usr/bin/exim');
-OR-
use File::Util;
print scalar localtime File::Util->last_modified('C:\COMMAND.COM');
-OR-
use File::Util;
my $f = File::Util->new(); print scalar localtime$f->last_modified('/bin/less');
## Make a new directory, recursively if neccessary
use File::Util;
my $f = File::Util->new();$f->make_dir('/var/tmp/tempfiles/foo/bar/');
# optionally specify a creation bitmask to be used in directory creations
print $f->NL; # EXAMPLES (Full Programs) ## Batch File Rename # Code changes the file suffix of all files in a directory ending in # *.foo so that they afterward end in *.bar use strict; use vars qw($dir );
use File::Util qw( NL SL );
my $f = File::Util->new(); my$dir = '../wibble';
my $old = 'foo'; my$new = 'bar';
my @files = $f->list_dir($dir, '--files-only');
foreach ( @files ) {
# don't change the file suffix unless it is *.foo
if ($_ =~ /\.$old$/o) { my$newname = $_;$newname =~ s/\.$old/\.$new/;
if (rename($dir . SL .$_, $dir . SL .$newname)) {
print qq($_ ->$newname), NL
}
else { warn <<__ERR__ }
Couldn't rename "$_" to "$newname"!
__ERR__
}
else { print <<__NOCHANGE__ }
File retained as "$_" __NOCHANGE__ } ## Recursively remove a directory and all its contents # This code removes a directory and everything in it use strict; # always use File::Util qw( NL ); my$f = File::Util->new();
my $removedir = '/path/to/directory/youwanttodelete'; my @gonners =$f->list_dir($removedir, '--follow'); # remove directory and everything in it my($a, $b ); @gonners = reverse sort { length$a <=> length $b } @gonners; foreach ( @gonners,$removedir ) {
print "Removing $_ ..." . NL; -d$_ ? rmdir($_) || die$! : unlink($_) || die$!;
}
print 'Done. w00T!', NL x 2;
## Wrap the lines in a file at 72 columns, then save it
# This code opens a file, wraps its lines, and saves the file with
# the newly formatted content
use strict; # always
use File::Util qw( NL );
use Text::Wrap qw( wrap );
$Text::Wrap::columns = 72; # wrap text at this many columns my$f = File::Util->new();
my $textfile = 'myreport.txt'; # file to wrap and save$f->write_file(
filename => $textfile, content => wrap('', '',$f->load_file($textfile)) ); print 'Done.', NL x 2; ## Read and increment a counter file, then save it # This code opens a file, reads a number value, increments it, # then saves the newly incremented value back to the file use strict; # always use File::Util; my$f = File::Util->new();
my $counterfile = 'counter.txt'; # if the counter file doesn't exist, let's make one if ( !$f->existent( $counterfile ) ) {$f->touch($counterfile); } my$count = $f->load_file($counterfile );
# convert textual number to in-memory int type, -this will default
# to a zero if it encounters non-numerical or empty content
chomp $count; # strip off any trailing lines$count =~ s/[^[:digit:]]//g; # remove non-numeric data
$count = 0 if "$count" eq ''; # set count to 0 if empty string
$count = int$count; # numberify $count print 'Count value from file: ' .$f->load_file($counterfile),$f->NL;
$count++; # increment the counter value by 1 # save the incremented count back to the counter file$f->write_file( filename => $counterfile, content =>$count);
# verify that "it worked"
print 'Count is now: ' . $f->load_file($counterfile), $f->NL; print 'Done.',$f->NL x 2;
## Batch Search & Replace
# Code does a batch find or search and replace for all files in a given
# directory, recursively or non-recursively based on choices set forth
# in the code.
use strict;
use File::Util qw( NL SL );
# will get search pattern from file named below
use constant SFILE => './sr/searchfor';
# will get replace pattern from file named below
use constant RFILE => './sr/replacewith';
# will perform batch operation in directory named below
use constant INDIR => '/foo/bar/baz';
# specify whether the operation will do a find or a search and replace
use constant RMODE => [qw| read-only write |]->[1];
# set the options for the search (will or will not recurse, etc)
my @opts = [qw/ --files-only --with-paths --recurse /]->[0,1];
# create new File::Util object, set File::Util to send a warning for
# fatal errors instead of dieing
my $f = File::Util->new('--fatals-as-warning'); my$rstr = $f->load_file(RFILE); my$spat = quotemeta $f->load_file(SFILE);$spat = qr/$spat/; my$gsbt = 0;
my $action = RMODE eq 'read-only' ? 'detections' : 'substitutions'; my @files =$f->list_dir(INDIR, @opts);
for (my $i = 0;$i < @files; ++$i) { next if$f->isbin( $files[$i] );
my $sbt = 0; my$file = $f->load_file($files[$i] );$file =~ s/$spat/++$sbt;++$gsbt;$rstr/ge;
$f->write_file( file =>$files[$i], content =>$file)
if RMODE eq 'write';
print $sbt ? qq($sbt $action in$files[$i]) . NL : ''; } print( NL . <<__DONE__ . NL x 2 ) and exit;$gsbt $action in${\scalar(@files)} files.
__DONE__
## Pretty-Print A Directory Recursively
use strict;
use vars qw( $a$b );
use File::Util qw( NL );
my $ind = ''; my$f = File::Util->new();
my @o = qw(
--with-paths
--sl-after-dirs
--no-fsdots
--files-as-ref
--dirs-as-ref
);
my $filetree = {}; my$treetrunk = '/var/';
my( $subdirs,$sfiles ) = $f->list_dir($treetrunk, @o);
$filetree = [{$treetrunk => [ sort({ uc $a cmp uc$b } @$subdirs, @$sfiles) ]
}];
descend( $filetree->[0]{$treetrunk }, scalar(@$subdirs) ); walk( @$filetree );
sub descend {
my( $parent,$dirnum ) = @_;
for (my $i = 0;$i < $dirnum; ++$i) {
my $current =$parent->[$i]; next unless -d$current;
my( $subdirs,$sfiles ) = $f->list_dir($current, @o);
map { $_ =$f->strip_path($_) } @$sfiles;
splice(@$parent,$i,1,{
$current => [ sort({ uc$a cmp uc $b } @$subdirs, @$sfiles) ] }); descend($parent->[$i]{$current }, scalar @$subdirs ); } return$parent;
}
sub walk {
my $dir = shift(@_); foreach (@{ [ %$dir ]->[1] }) {
my $mem =$_;
if (ref $mem eq 'HASH') { print$ind . $f->strip_path([ %$mem ]->[0]) . '/', NL;
$ind .= ' ' x 3; walk($mem );
$ind = substr($ind, 3 );
} else { print $ind .$mem, NL }
}
}
# BUGS
Send bug reports and patches to the CPAN Bug Tracker for File::Util at https://rt.cpan.org/Dist/Display.html?Name=File%3A%3AUtil
# RESOURCES
If you want to get help, contact the authors (links below in the AUTHORS section)
I fully endorse http://www.perlmonks.org as an excellent source of help with Perl in general.
# CONTRIBUTING
The project website for File::Util is at https://github.com/tommybutler/file-util/wiki
The git repository for File::Util is on Github at https://github.com/tommybutler/file-util
Clone it at git://github.com/tommybutler/file-util.git
This project was a private endeavor for too long so don't hesitate to pitch in. I want to say I very much appreciate the emails, bug reports, patches, and all those who contribute their time and talents as CPAN testers.
# AUTHORS
Tommy Butler http://www.atrixnet.com/contact
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# Asymptotic notation for a function
I have a question about asymptotic notation in some proof of a claim which I don't fully understand. In the proof there is a part in which they show that there is a function $f:\Bbb{N} \to [0,1]$ such that $$(1-g^2(n))^n \le f(n) \le (1-g^2(n))^n e^{\frac{ng^3(n)}{2(1-g^2(n))}}$$
where $g(n) = \frac{1}{log(n)}$, and from it they conclude that $f(n) = (1-g^2(n))^{n(1-o(1))}$. What does it mean? and why is it true?
It means that $f(n)$ is equal to $(1-g^2(n))^{n(1-h(x))}$
where $h(x)$ is a function such that $\lim\limits_{g\to 0} = 0$.
This is of course the case since we have $0\leq -nh(x) \leq \log_{1-g^2(n)}(e^{\frac{ng^3(n)}{2(1-g^2(n))}})=\frac{\frac{ng^3(n)}{2(1-g^2(n))}}{\log(1-g^2(n))}=\frac{n}{2(1-1/\log(n)^2)\log(1-1/\log(n)^2)1/log(n)^ 3}$
dividing by $-n$ gives:
$0\geq h(x)\geq - \frac{1}{2(1-1/\log(n)^2)\log(1-1/\log(n)^2)1/log(n)^ 3}$.
Which implies $h(x)$ goes to zero as desired.
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# Interfacial Effect on the Deformation Mechanism of Bulk Nanolaminated Graphene-Al Composites
### Abstract
Uniaxial tensile tests were carried on graphene (reduced graphene oxide, RGO)–Al laminated composites with Al lamella thicknesses varying from 1 $μ$m down to 200 nm. It was found that there was a transition in plastic deformation mechanism, from a Hall–Petch-typed mechanism at 500 nm and 1 $μ$m Al lamella thicknesses, to confined layer slip (CLS) of dislocations at the 200 nm Al lamella thickness. Moreover, the strengthening effect of RGO was only demonstrated in composite having 200-nm-thick Al lamellas, which can be rationalized by the enhanced barrier for dislocation de-pinning processes in the CLS mechanism.
Type
Publication
Metallurgical and Materials Transactions A: Physical Metallurgy and Materials Science
##### Shmuel Osovski
###### Asst. Prof.
Thought is a strenuous art - few practice it, and then only at rare times.
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# What happens to the space group of a crystal when introducing a non-trivial basis?
I am trying to understand crystallography and the space groups of crystals, but I have one major question bugging me. The book I am using adresses different lattice symmetries and applications of group theory. More specifically, the electron energy bands are characterised by the irreducible representations of the point group of the Bravais lattice. By basis I mean the atoms associated with a lattice point.
Is this characterisation altered or does it break down in any way when a non-trivial basis is used?
I would suspect the answer to be along the lines: "The form of the bands are changed, but the characterisation is conserved...", but this is only guesswork.
• Are you asking a question about like, finding good quantum numbers for such systems? The point groups are obviously altered, but you can still have an underlying Bravais lattice. The species / unit cells however will have different (or even just the trival) point groups - hence, distinct space groups. – daydreamer Oct 14 '20 at 15:43
• The energy bands of a crystal can be characterised by the irreducible representations of the space group of the Bravais lattice. My question is then, is this classification still valid with a non-trivial basis? I will edit my question to make it more clear – B. Brekke Oct 14 '20 at 16:21
• I see no other reason but extreme luck/coincidence for the presence of non-trivial basis not altering the wave functions symmetries in general cases. If you put distinct basis, the point group changes. Find the new group, compute its invariants and proceed in the usual Wigner-lesque way to find the finite-dimensional irreps, or just the nicest irreps possible. Group Theory always gives you good quantum numbers – daydreamer Oct 14 '20 at 16:40
• Although I'm not very experienced, I have doubt in this answer. The reason is that compound materials are mentioned several times in the discussion of specific simple lattice symmetries in my book. Also, if I were to find a new symmetry respecting the basis, I would very often end up with only the trivial symmetry group rendering the method quite useless. Maybe I didn't understand your comment, however, I am still not sure of what is going on – B. Brekke Oct 14 '20 at 17:03
By adding atoms to the crystal basis, the space group will generally change and so will the symmetry classification of electronic states $$|\psi_{nk}\rangle$$. To be a bit more complete, let's look at exactly how states are classified and labeled.
If we have a symmorphic space group $$\mathbb{S}$$, then for a given $$k$$ we can define its little cogroup $$\mathbb{G}_k$$ (group of the wavevector) by $$\mathbb{G}_k = \{\:(R|0)\in \mathbb{S} \:\:\:|\:\:\: Rk\equiv k \:\}$$ where $$Rk\equiv k$$ means equivalence up to a reciprocal lattice vector. In a crystal, electronic states $$|\psi_{nk}\rangle$$ are classified by the irreducible representations of the little cogroup of $$k$$. Each high symmetry line in the Brillouin Zone has a single little cogroup associated with it and the irreps will then label the bands along that line.
## Example
FCC with one atom basis
The basic FCC lattic has space group $$Fm\bar{3}m$$ (#225). If we look at the high symmetry line $$\Delta:\Gamma\to X$$, we can find (from this source) that the little cogroup is $$4mm$$ ($$C_{4v}$$). Then looking at the irreps here, we see there are 5 of them including one 2D irrep
FCC with two atom basis
A zincblende crystal is FCC but has a two atom basis (like CdTe). It has space group $$F\bar{4}3m$$ (#216). Looking at the same high symmetry line $$\Delta:\Gamma\to X$$, the little cogroup in this case is $$2mm$$ ($$C_{2v}$$) (as found here). $$C_{2v}$$ has only 4 irreps and they are all 1D. So the symmetry classification is clearly different.
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# Even though the direct costs of malpractice disputes amounts
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Re: sc- less than Vs lower than [#permalink] 10 Dec 2010, 11:42
concept tested here is : S-V agreement +lower/less
subject is the direct costs so verb has to be singular..-a,b out
uncountables use less/more not lower so option C
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Re: sc- less than Vs lower than [#permalink] 10 Dec 2010, 14:12
first of all the sentence is in past tense. So "amounted to" is right usage. this eliminates A and B.
"amounted to a lower sum" - redundant. eliminate E
"amounted to lower"- amounted is uncountable ,usage of "less" is correct.
Answer is C
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Re: sc- less than Vs lower than [#permalink] 11 Dec 2010, 22:26
C.
A and B are incorrect: use amounts (singular) for direct costs (plural)
Quote:
C : less is used for countable nouns. lower than should be used for non countable nouns
Well, after posting the answer I went through my notes and reference material.
I found the reasoning that I used in the quoted part to be wrong.
less than is used for non countable nouns and lower than for countable nouns
In this sentence 1% of xx billion $= money. Money is uncountable -> fewer than some money - worng (fewer is used for countable nouns) lower than some money - wrong less than some money - correct The confusion is created as x$ is countable. But it is countable only if it refers to $bills (actual paper notes). Looking at the sentence (and 99% of the time in GMAT) x$ is referring to money and not actual $bills. hence it becomes uncountable. Manager Joined: 31 May 2010 Posts: 99 Followers: 1 Kudos [?]: 29 [0], given: 25 Re: sc- less than Vs lower than [#permalink] 13 Dec 2010, 18:02 Its C Redundacy - Amount & SUM so A and E out Parallelism lead 'amonted; with 'spent' As its uncountable - so 'Less' not 'Lower' _________________ Kudos if any of my post helps you !!! Manager Joined: 13 Jul 2010 Posts: 169 Followers: 1 Kudos [?]: 27 [0], given: 7 Re: sc- less than Vs lower than [#permalink] 13 Dec 2010, 18:17 Agreed with C. Amounted needs to be in the past tense to fit the setting of last year and money is uncountable so requires the descriptive less rather than lower. Senior Manager Joined: 19 Oct 2010 Posts: 274 Location: India GMAT 1: 560 Q36 V31 GPA: 3 Followers: 6 Kudos [?]: 42 [0], given: 27 Re: sc- less than Vs lower than [#permalink] 22 Jan 2011, 01:20 USCTrojan2006 wrote: Should be less than, with amount in the past tense form. this is the simplest explanation why it's C _________________ petrifiedbutstanding Manager Joined: 04 Sep 2010 Posts: 84 Followers: 1 Kudos [?]: 5 [0], given: 11 Re: sc- less than Vs lower than [#permalink] 26 May 2011, 04:58 spriya wrote: Even though the direct costs of malpractice disputes amounts to a sum lower than one percent of the$541 billion the nation spent on health care last year, doctors say fear of lawsuits plays major role in health-care inflation.
(A) amounts to a sum lower
(B) amounts to less
(C) amounted to less
(D) amounted to lower
(E) amounted to a lower sum
costs is plural so A and B are out E has redundant "sum so it is also out
Between Less and Lower ;
I will go for Less as it is for Costs
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Re: sc- less than Vs lower than [#permalink] 26 May 2011, 09:53
+1 C
"Lower" can be only an adjective.
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Re: sc- less than Vs lower than [#permalink] 26 May 2011, 10:01
Note the difference in lower vs. less
i scored lower than my sister
vs.
i scored 1% less than my sister
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Re: Even though the direct costs of malpractice disputes amounts [#permalink] 14 Dec 2012, 18:40
Percent is uncountable unless the clearly specified how much is the money it should be considered Countable but when percent is mentioned it requires uncountable usage i.e. less
B and C : B is eliminated since it is singular subject the costs is plural hene answer is C.
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Re: Even though the direct costs of malpractice disputes amounts [#permalink] 03 Jan 2013, 23:31
I got the OA but not sure if my method was corrrect. Basically I substituted in the ellipsis-ed (!) word into the second half of the sentence:
" but (whom) were eventually discovered" --> wrong because they (not them) were eventually discovered
"but (who) were eventually discovered" --> great!
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Re: Even though the direct costs of malpractice disputes amounts [#permalink] 25 Jan 2014, 06:41
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Re: Even though the direct costs of malpractice disputes amounts [#permalink] 25 Jan 2014, 06:41
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# A rectangle with its two sides as 'x' and 'y' units respectively is
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A rectangle with its two sides as 'x' and 'y' units respectively is [#permalink]
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21 Jul 2018, 21:02
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A rectangle with its two sides as 'x' and 'y' units respectively is rotated along its side 'x' to form a cylinder C1, such that the circumference of base of C1 is 'x' and the height of C1 is 'y'. Another rectangle with its two sides as 'p' and 'q' units respectively is rotated along its side 'p' to form another cylinder C2, which now has circumference of base as 'p' and height as 'q'. Is the volume of C1 lesser than that of C2?
(1) Ratio of x to p is 3:2
(2) Ratio of y to q is 2:3
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Re: A rectangle with its two sides as 'x' and 'y' units respectively is [#permalink]
### Show Tags
21 Jul 2018, 21:43
1
Volume of C1, V1 = πr1^2h1
Volume of C2, V2 = πr2^2h2
We need both r1,r2 and h1,h2 to find out which volume is greater.
Statement 1:
Circumference of C1 = X, 2πr1=X
Circumference of C2 = P, 2πr2=P
Ratio of X:P = 3:2
Consider X = 3c and P = 2c, where C is constant
Circumference C1,
2πr1 = 3c --> r1 = 3c/2π
Circumference C2,
2πr2 = 2c --> r2 = 2c/2π --> C/π
Volume V1 = πr1^2h1, we still miss h1 here so insufficient.
Statement 2:
Ratio of y to q is 2:3
Height of C1, y = 2h
Height of C1, q= 3h, where h is constant
We don't have r1 and r2 here hence insufficient.
Statement 1&2:
We have both r1,r2 from (1) and h1,h2 from (2).
V1/V2 = π*(3c/2π)^2*2h/π*(c/π)^2*3h
Reducing the above equation we get
V1/V2 = 9/4
Hence V1 is not lesser than V2. Sufficient
Method 2:
We can solve this without using any calculations. We know that to find volume of a cylinder we require both height and radius. This tell us that each statement alone is not sufficient and we require both statement 1 and statement 2 to find the volume and determine which volume is greater. Hence C (But to be sure you can solve and find volumes)
Please give kudos if you like the explanation.
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A rectangle with its two sides as 'x' and 'y' units respectively is [#permalink]
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21 Jul 2018, 21:50
amanvermagmat wrote:
A rectangle with its two sides as 'x' and 'y' units respectively is rotated along its side 'x' to form a cylinder C1, such that the circumference of base of C1 is 'x' and the height of C1 is 'y'. Another rectangle with its two sides as 'p' and 'q' units respectively is rotated along its side 'p' to form another cylinder C2, which now has circumference of base as 'p' and height as 'q'. Is the volume of C1 lesser than that of C2?
(1) Ratio of x to p is 3:2
(2) Ratio of y to q is 2:3
Volume is dependent on both the variables for a cylinder so each statement just giving ratio of one side is insufficient
Combined both should be sufficient as the volume of both cylinders would be in terms of the product of these two variables X&y for one and p&q for other..
Therefore we will get the ratio of two volumes ...
Although we can answer C with out further work but let us see how algebraically we get the answer..
Volume of cylinder = $$πr^2h$$
For cylinder with base X....2πr=X.....r=X/2π and H =y
$$V_x=π*(\frac{X}{2π})^2*y$$.....
Similarly for other one $$V_p=π*(\frac{p}{2π})^2*q$$..
$$\frac{V_x}{V_p}=\frac{x^2y}{p^2q}=(\frac{X}{p)})^2*\frac{y}{q}=(9/4)*(2/3)=3/2$$
So V_x is more
Sufficient
C
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Re: A rectangle with its two sides as 'x' and 'y' units respectively is [#permalink]
### Show Tags
21 Jul 2018, 21:56
A rectangle with its two sides as 'x' and 'y' units respectively is rotated along its side 'x' to form a cylinder C1, such that the circumference of base of C1 is 'x' and the height of C1 is 'y'.
The circumference of base of C1 is 'x'
height of C1 is 'y'
=> $$2\pi r_{1} = x$$
=> $$r_{1} = \frac{x}{2\pi}$$
volume of this cylinder C1 is $$\pi (r_{1})^2 y$$ = $$\pi \frac{x^2}{4(\pi)^2}y$$ = $$\frac{x^2 y}{4\pi}$$
Another rectangle with its two sides as 'p' and 'q' units respectively is rotated along its side 'p' to form another cylinder C2, which now has circumference of base as 'p' and height as 'q'.
The circumference of base of C2 is 'p'
height of C2 is 'q'
=> $$2\pi r_{2} = p$$
=> $$r_{2} = \frac{p}{2\pi}$$
volume of this cylinder C2 is $$\pi (r_{2})^2 q$$ = $$\pi \frac{p^2}{4(\pi)^2}$$ = $$\frac{p^2 q}{4\pi}$$
We need to compare the volume of cylinders C1 and C2
=> we need to compare $$\frac{x^2 y}{4\pi}$$ and $$\frac{p^2 q}{4\pi}$$
=> We need to compare $$x^2 y$$ and $$p^2 q$$
Statement 1
Ratio of x to p is 3:2 => $$\sqrt{x^2 /p^2} = \frac{9}{4}$$
This doesn't tell anything about y and q
Statement 1 is insufficient
Statement 2
Ratio of y to q is 2:3 => $$\sqrt{y/q} = \frac{2}{3}$$
This doesn't tell anything about x and p
Statement 2 is insufficient
Combining statements 1 and 2
=> $$\frac{C1 volume}{C2 volume}$$ = $$\frac{x^2}{p^2} * \frac{t}{q}$$
= $$\frac{9}{4} * \frac{2}{3} = \frac{3}{2}$$ > 1
Since the $$\frac{C1 volume}{C2 volume}$$ > 1 => C1 volume is > C2 volume
Hence option C
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Re: A rectangle with its two sides as 'x' and 'y' units respectively is &nbs [#permalink] 21 Jul 2018, 21:56
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# Vizing theorem
Let $G$ be a graph (assumed to be finite, undirected and loop-less), and let $\mu ( G )$ be the maximum number of edges joining every pair of vertices. A simple graph will mean a graph without parallel edges. The chromatic index $\chi ^ { \prime } ( G )$ is the least number of colours needed to colour the edges of $G$ in such a way that no two adjacent edges are assigned the same colour (cf. also Graph colouring). If $\Delta ( G )$ is the maximum degree of $G$, then, obviously, $\Delta ( G ) \leq \chi ^ { \prime } ( G )$. The origins of chromatic graph theory may be traced back to 1852, with the birth of the four-colour problem. The first paper on edge colourings appeared 1880, when P.G. Tait [a18] made the important observation that the four-colour conjecture is equivalent to the statement that every $3$-regular planar graph has chromatic index $3$. But after this, little was done. In 1916, D. König [a13] proved that if $G$ is a bipartite graph (cf. also Graph, bipartite), then $\chi ^ { \prime } ( G ) = \Delta ( G )$. The first non-trivial upper bound has been obtained by C.E. Shannon [a17] in 1949, when he proved $\chi ^ { \prime } ( G ) \leq 3 \Delta ( G ) / 2$ for an arbitrary graph $G$. The great breakthrough occurred in 1964, however, when V.G. Vizing [a19] gave the following surprisingly strong result, now known as Vizing's theorem: If $G$ is a graph, then
\begin{equation*} \Delta ( G ) \leq \chi ^ { \prime } ( G ) \leq \Delta ( G ) + \mu ( G ). \end{equation*}
In particular, $\Delta ( G ) \leq \chi ^ { \prime } ( G ) \leq \Delta ( G ) + 1$ if $G$ is simple.
The cornerstone of Vizing's proof is a brilliant recolouring technique. Up to now (1999) all further proofs of his theorem are based more or less on this method (see, for example, [a5], [a10], and [a22]). In addition, the proof of Vizing's theorem can be used to obtain a polynomial-time algorithm to colour the edges of every graph $G$ with $\Delta ( G ) + \mu ( G )$ colours. Nowadays, graphs are normally divided into two classes: those with $\chi ^ { \prime } ( G ) = \Delta ( G )$ are called class-1 graphs, and the others class-2 graphs. Class-2 graphs are relatively scarce, even among the simple graphs [a9]. In view of König's result, bipartite graphs are of class 1; however, the general question whether a given graph is class 1 or class 2 is very difficult, and it is known as the classification problem. This problem is extremely important and has received wide interest. The difficulty of the classification problem lies in the fact that it is $\cal N P$-complete (cf. [a12]; Complexity theory), and that its solution would imply the four-colour theorem. Very little progress has been made for general graphs, and hence it is natural to consider this problem for special families. For regular graphs the so-called one-factorization conjecture goes back to the 1950s: Every simple $r$-regular graph with $2 n$ vertices is class 1 (i.e. $G$ is $1$-factorizable) when $r \geq n$. This conjecture is still open (1999). The best partial solutions go back to A.G. Chetwynd and A.J.W. Hilton [a7] and T. Niessen and L. Volkmann [a15]. They showed independently that the $1$-factorization conjecture is valid if $r \geq ( \sqrt { 7 } - 1 ) n \approx 1.647 n$. Appreciable progress has also been achieved in the particular case of planar graphs. On the one hand, if $\Delta ( G ) \leq 5$, then a planar graph $G$ can lie in either class 1 or class 2. On the other hand, Vizing [a20] has proved that every simple planar graph $G$ with $\Delta ( G ) \geq 8$ is necessarily of class 1. However, the problem of determining what happens when $\Delta ( G )$ is either $6$ or $7$ remains open, and has led [a20] to the planar graph conjecture: Every simple planar graph of maximum degree $6$ or $7$ is of class 1.
The total chromatic number $\chi _ { T } ( G )$ is the least number of colours needed to colour the elements (edges and vertices) of a graph $G$ so that adjacent elements are coloured differently. The centre of interest lies in the still (1999) unproven total colouring conjecture of Vizing [a20] and M. Behzad [a3] that $\chi _ { T } ( G ) \leq \Delta ( G ) + 2$ for a simple graph $G$. A.J.W. Hilton and H.R. Hind [a11] proved this conjecture by vertex colouring methods when $\Delta ( G ) \geq 3 n / 4$, where $n$ is the number of vertices of $G$. Also, Vizing's theorem is a powerful tool in attacking this conjecture. For example, K.H. Chew [a6] has given a new proof of the Hilton–Hind result by transforming the total colouring problem to an edge-colouring problem. Using probabilistic methods, recently M. Molloy and B. Reed [a2] have obtained the following very deep theorem: There is an absolute constant $C$ such that $\chi _ { T } ( G ) \leq \Delta ( G ) + C$ for every simple graph $G$. See [a23] for a thorough treatment of this topic.
In order to give reasonable upper bounds for $\chi _ { T } ( G )$, Vizing [a21] introduced the concept of list colouring: If $\chi _ { l } ^ { \prime } ( G )$ is the list-chromatic index of a graph $G$, then, obviously, $\chi ^ { \prime } ( G ) \leq \chi _ { l } ^ { \prime } ( G )$. Vizing [a21] even posed the still open (1999) list colouring conjecture: Every graph $G$ satisfies $\chi ^ { \prime } ( G ) = \chi _ { l } ^ { \prime } ( G )$. With a surprisingly short proof, F. Galvin [a1] showed that the list colouring conjecture is true for bipartite graphs. A.V. Kostochka [a14] proved that if all cycles in a simple graph $G$ are sufficiently long relative to $\Delta ( G )$, then $\chi _ { l } ^ { \prime } ( G ) \leq \Delta ( G ) + 1$. If the list colouring conjecture is true, then, according to Vizing's theorem, the same bound would hold for any simple graph.
The determination of the chromatic index can be transformed into a problem dealing with the chromatic number $\chi ( G )$. Namely, from the definition it is immediate that $\chi ^ { \prime } ( G ) = \chi ( L ( G ) )$, where $L ( G )$ is the line graph of the simple graph $G$. If, in addition, $\omega ( G )$ is the clique number (cf. also Fixed-point property) of $G$, then Vizing's theorem implies
\begin{equation*} \chi ( L ( G ) ) \leq \omega ( L ( G ) ) + 1. \end{equation*}
Therefore, this bound is called [a16] the Vizing bound. L.W. Beineke [a4] characterized line graphs by nine forbidden induced subgraphs. Using only four forbidden induced subgraphs of Beineke's nine graphs, S.A. Choudom [a8] determined two superclasses of line graphs for which the Vizing bound is also valid. Thus, since $\Delta ( G ) = \omega ( L ( G ) )$ if $G$ is not a triangle, Choudom's result extends Vizing's theorem. The most general contribution in this area was found by H. Randerath [a16]: If $G$ is a simple graph containing neither a complete $5$-graph with an edge missing nor a claw $K _ { 1,3 }$ with one subdivided edge as an induced subgraph, then $\chi ( G )$ is equal to $\omega ( G )$ or $\omega ( G ) + 1$.
#### References
[a1] F. Galvin, "The list chromatic index of a bipartite multigraph" J. Combin. Th. B , 68 (1995) pp. 153–158 [a2] M. Molloy, B. Reed, "A bound on the total chromatic number" Combinatorica , 18 (1998) pp. 241–280 [a3] M. Behzad, "Graphs and their chromatic numbers" Doctoral Thesis Michigan State Univ. (1965) [a4] L.W. Beineke, "Derived graphs and digraphs" , Beiträge zur Graphentheorie , Teubner (1968) pp. 17–33 [a5] K.H. Chew, "On Vizing's theorem, adjacency lemma and fan argument generalized to multigraphs" Discrete Math. , 171 (1997) pp. 283–286 [a6] K.H. Chew, "Total chromatic number of graphs of high maximum degree" J. Combin. Math. Combin. Comput. , 18 (1995) pp. 245–254 [a7] A.G. Chetwynd, A.J.W. Hilton, "$1$-factorizing regular graphs of high degree: an improved bound" Discrete Math. , 75 (1989) pp. 103–112 [a8] S.A. Choudom, "Chromatic bound for a class of graphs" Quart. J. Math. , 28 (1977) pp. 257–270 [a9] P. Erdös, R.J. Wilson, "On the chromatic index of almost all graphs" J. Combin. Th. , 23 B (1977) pp. 255–257 [a10] M.K. Goldberg, "Edge-coloring of multigraphs: recoloring technique" J. Graph Theory , 8 (1984) pp. 123–137 [a11] A.J.W. Hilton, H.R. Hind, "The total chromatic number of graphs having large maximum degree" Discrete Math. , 117 (1993) pp. 127–140 [a12] I. Holyer, "The NP-completeness of edge-coloring" SIAM J. Comput. , 10 (1981) pp. 718–720 [a13] D. König, "Über Graphen und ihre Anwendung auf Determinantentheorie und Mengenlehre" Math. Ann. , 77 (1916) pp. 453–465 [a14] A.V. Kostochka, "List edge chromatic number of graphs with large girth" Discrete Math. , 101 (1992) pp. 189–201 [a15] T. Niessen, L. Volkmann, "Class $1$ conditions depending on the minimum degree and the number of vertices of maximum degree" J. Graph Theory , 14 (1990) pp. 225–246 [a16] H. Randerath, "The Vizing bound for the chromatic number based on forbidden pairs" Doctoral Thesis RWTH Aachen (1998) [a17] C.E. Shannon, "A theorem on coloring the lines of a network" J. Math. Phys. , 28 (1949) pp. 148–151 [a18] P.G. Tait, "On the colouring of maps" Proc. R. Soc. Edinburgh , 10 (1880) pp. 501–503; 729 [a19] V.G. Vizing, "On an estimate of the chromatic class of a $p$-graph" Diskret. Anal. , 3 (1964) pp. 25–30 (In Russian) [a20] V.G. Vizing, "Critical graphs with a given chromatic class" Diskret. Anal. , 5 (1965) pp. 9–17 (In Russian) [a21] V.G. Vizing, "Vertex colouring with given colours" Diskret. Anal. , 29 (1976) pp. 3–10 (In Russian) [a22] L. Volkmann, "Fundamente der Graphentheorie" , Springer (1996) [a23] H.P. Yap, "Total colourings of graphs" , Lecture Notes Math. , 1623 , Springer (1996)
How to Cite This Entry:
Vizing theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Vizing_theorem&oldid=51407
This article was adapted from an original article by Lutz Volkmann (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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# Importance of the Time Step Value for the Accuracy of a Transient CFD Simulation
Overview
My understanding is that one should use a time step $\Delta t < \frac{h}{v}$ (where h - smallest mesh element, v - velocity) to get an accurate result.
But how important is this really for the accuracy of the simulation? Is it as important as having an independent mesh?
Is there even such a thing as a time step independent solution? Can a very small $\Delta t$ actually be bad for the accuracy of the solution?
I am running computational optimisation, where speed is important. Just how much am I justified to use $\Delta t > \frac{h}{v}$?
Also, I am running a transient simulation, where $v$ changes from zero to 60 m/s. Should I just set it to the smallest $\Delta t \approx 0.0007$ s (I can't dynamically change $\Delta t$)?.
Problem Details
I am using an Euler-Euler model (in Fluent™) to simulate particle-air interaction in a fluidised bed.
• Hard limits are usually for explicit solvers. For implicit solvers just run a number of test cases on a small (2D) problem to see how different the solution is with increasing delta_t. While you're at it you can also test the effect of mesh resolution on the solution. – stali Jun 19 '13 at 21:21
It depends on your problem and your ODE solver/time discretization. If you have a hyperbolic PDE and want to solve it with an explicit method, then you need the time step restriction (called the Courant-Friedrichs-Lewy/CFL condition) or your numerical solution will typically oscillate and may grow to $\pm\infty$.
On the other hand, if you have a parabolic problem and an implicit time discretization, then you don't need the restriction.
You will have to tell more about your problem for us to be able to give a more detailed answer.
• In case of linear hyperbolic problems, large time step algorithms can be used to update the solution. That way Courant number much greater than 1 are used. – Subodh Jun 19 '13 at 14:42
• But only if you have an implicit method. I should have been clearer in my answer. (I edited it for that now.) – Wolfgang Bangerth Jun 19 '13 at 20:52
• To be honest, I am having difficulty establishing which method Fluent™ is using. It seems that it can do both. The user guide for Fluent v14 (p. 1250) implies that for the multiphase Euler-Euler simulation an explicit method is used. I like the suggestion by @stali do simply do some testing (although he does suggest this for an implicit solver). The solver is stable with all the time steps, but there is quite a big difference in the solution (and computational cost) between $\Delta t=0.0007$ and $\Delta t=0.0021$. – A.L. Verminburger Jun 20 '13 at 8:52
• Implicit steps exceeding CFL 1 are not time-accurate. For pure advection, there are no faster processes, so they're not of much use for non-steady simulation. Implicit stepping is more interesting when you are trying to follow a quasi-equilibrium process that appears when much faster processes are nearly balanced (e.g., slow-moving vortices in low-Mach shallow-water or gas dynamics). – Jed Brown Jun 20 '13 at 19:46
• @Wolfgang Bangerth Would you say that a CFL number $C_{max}=1$, although required for the explicit scheme is also a good rough guide for the implicit scheme? Or is testing on a simple 2D problem the only way to determine the appropriate $\Delta t$ when using an implicit scheme? – A.L. Verminburger Jun 22 '13 at 14:46
There are two factors that are influenced by time step size and the choice of scheme: accuracy and stability.
Accuracy is typically measured by the "local error" or "consistency error" of the scheme. You want to choose your time step such that this error is balanced with a comparable error of the space discretization. That would be a good balance for accuracy.
Unfortunately, most timestepping schemes also change the dynamics of your system, which is usually subsumed under the term stability. This question goes beyond explicit or implicit. And this goes both ways: a perfectly stable solution can be converted to an explosion if you use the wrong method with a large timestep. And the opposite holds: if you use a method that is too stable, turbulent, instationary flow might be turned into honey. I know of simulations where a single backward Euler step every 100 Crank-Nicolson steps made an oscillatory solution stationary.
The terms used to categorize the dynamics of timestepping schemes are A-, L-, and B-stability. As far as I know, only the Crank-Nicolson scheme and Gauss-collocation methods preserve the essential dynamics, but even for those, certain features of your solution may be amplified or suppressed in an unphysical way, if your timestep is too large.
If you want to be able to predict these effects, you have to know your scheme. Else, you are stuck with test examples, or with computing everything with at least two time step sizes
It is really important! If you do not have an appropriate time step, you can not achieve mesh independence.
Use my personal experience: I investigate CFD of fluid-structure interaction using FEM. I was trying to do the mesh independence study to make sure that mesh does not affect the accuracy of my simulations. However, when I refine the mesh, the simulation results even diverges further. At last, I figured out that I forgot to adjust my time step accordingly.
When you reduce the size of element, you are recommended to reduce the time step correspondingly. Otherwise, you may have trouble.
Also, if you got time and computational resources, you may just run some test cases to see how the $\Delta t$ affects your specific cases, which I think should be the most reliable method.
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Example 13.37.2. Let $\mathcal{A}$ be an abelian category. Let $\ldots \to A_2 \to A_1 \to A_0$ be a chain complex in $\mathcal{A}$. Then we can consider the objects
$X_ n = A_ n \quad \text{and}\quad Y_ n = (A_ n \to A_{n - 1} \to \ldots \to A_0)[-n]$
of $D(\mathcal{A})$. With the evident canonical maps $Y_ n \to X_ n$ and $Y_0 \to Y_1[1] \to Y_2[2] \to \ldots$ the distinguished triangles $Y_ n \to X_ n \to Y_{n - 1} \to Y_ n[1]$ define a Postnikov system as in Definition 13.37.1 for $\ldots \to X_2 \to X_1 \to X_0$. Here we are using the obvious extension of Postnikov systems for an infinite complex of $D(\mathcal{A})$. Finally, if colimits over $\mathbf{N}$ exist and are exact in $\mathcal{A}$ then
$\text{hocolim} Y_ n[n] = (\ldots \to A_2 \to A_1 \to A_0 \to 0 \to \ldots )$
in $D(\mathcal{A})$. This follows immediately from Lemma 13.31.7.
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## Algebra: A Combined Approach (4th Edition)
Use the formula $v=\frac{-b}{2a}$ This equation works for both sideways and normal parabolas. to get the x intercept of the vertex, then substitute that into the original equation into x, and solve for the y point of the vertex. In this case the vertex is (0,0).
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The Students of a Vidyalaya Were Asked to Participate in a Competition for Making And Decorating Pen Holders in the Shape of a Cylinder with a Base, Using Cardboard. Each Pen Holder Was to Be - Mathematics
The students of a Vidyalaya were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution
Radius of circular end of cylinder pen holder = 3 cm
Height of pen holder = 10.5 cm
Surface area of 1 pen holder ="CSA of penholder+Area of base of SA of 1 penholder"= 2pirh+pir^2
2pirh+pir^2
= 2xx3.14xx3xx10.5+3.14138
= 132xx1.5+198/7cm
=198+198/7cm^2
=1584/7cm^2
Area of car board sheet used by 1 competitor = 1584/7 cm^2
Area of car board sheet used by 35 competitors=1584/735cm^2=7920cm^2
Concept: Surface Area of Cylinder
Is there an error in this question or solution?
APPEARS IN
RD Sharma Mathematics for Class 9
Chapter 19 Surface Areas and Volume of a Circular Cylinder
Exercise 19.1 | Q 9 | Page 8
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:
We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$).
If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = {t}$ \lbrace t\rbrace$for some$t \in [0,1]$. Then we must have$c_1(t) = c_2(t)$. 1 How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in$\mathbb{R}$is non-empty. Here goes the proof: We construct recursively a nested sequence$I_j := [a_j, b_j]$of closed intervals for$j \geq 0$. Let$I_0 := [0,1]$. For every$j \geq 0$, construct$I_{j+1}$as follows: let$m_j$be the midpoint of$I_j$. If the curves intersect at$t = m_j$, then we are done, so stop the sequence. Otherwise set$I_{j+1}$to be$[a_j, m_j]$or$[m_j, b_j]$depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that$c_1$is to the "left" of$c_2$at$t = a_j$and to the "right" of$c_2$at$t = b_j$). If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that$\cap_{j=0}^\infty I_j = {t}$for some$t \in [0,1]$. Then we must have$c_1(t) = c_2(t)\$.
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# Nonlinear shooting example calculation
How can we use the nonlinear shooting method with $h = 0.25$ to approx the solution to $y'' = 2y^3$, $-1 \leq x\leq 0$, $y(-1) = 1/2$, and $y(0) = 1/3$.
I tried to convert this to a first order system but the RK $4$ didnt look good. Also, I needed help by doing it via hand. Can someone please show or at least provide a setup?
We are given $(1)$:
$y'' = 2y^3$, $-1 \leq x\leq 0$, $y(-1) = 1/2$, and $y(0) = 1/3$, $h = 0.25.$
We are asked to use the nonlinear shooting method (I am not 100% sure of which algorithm you are using because you mention both Runge-Kutta 4-th and a linearized system, but do not mention if you are using the secant or Newton's with that - anyway, you can work those details).
From $(1)$, we have an exact solution (for comparison purposes) of:
$$y(x) = \frac{1}{x+3}.$$
Of course, we know $y(-1)$ and it should match what we get at the start and $y(0)$, which should match what we get at the end of the algorithm.
With the nonlinear shooting method, we need to approximate two IVPs from $y'' = f(x, y, y') = 2 y^3$, that are given by:
• $\displaystyle y'' = 2y^3$, $-1 \leq x\leq 0$, $y(-1) = 1/2$, and $y(0) = 1/3$, and
• $\displaystyle z'' = \frac{\partial f}{\partial y} z + \frac{\partial f}{\partial y'} z' = (6 y^2)z + (0)z' = 6y^2z$, $-1 \leq x \leq 0$, $z(-1) =0$, and $z'(-1) = 1.$
We are given:
$\displaystyle h = \frac{b-a}{N} = \frac{0 - (-1)}{N} = \frac{1}{N} = 0.25 = \frac{1}{4} \rightarrow N = 4.$
That is what you need for the Nonlinear Shooting Method.
If you further want to linearize this system as a first order system, you could write:
$y_1' = y_2$ with $y_1(-1) = \frac{1}{2}$, and
$y_2' = 2y_1^3$, with $y_2(-1) = t_k$,
where $y_1 = y$ and $y_2 = y'.$
If you are interested, you can actually see this example in Section 6.8.2 of Introduction To Numerical Analysis Using MATLAB by Rizwan Butt for the RK plus secant method Note that he has different limits, but this can help validate your approach to manual calculations if you don't want to code it up.
• Nice - I am lost when it comes to numerical methods...!! But I've got you to learn from; thanks! ;-) +1 – Namaste Apr 14 '13 at 0:41
• @BabakS. Always appreciate the support! :-) – Amzoti Apr 14 '13 at 5:16
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## Reboot and a paper
May 25, 2009
Over the last few days I received several spam comments on this blog (written using Cyrillic alphabet, incidentally). That made me visit this blog for the first time in several months, to turn off comments on old posts. Inadvertently, that also made me consider rebooting this. (Spam can be useful!) At the least, I can use this as a space where I put down comments about interesting papers I read. I might also write about some of the things I am working on.
***
Let me begin by discussing a paper of mine that just got accepted at SMC 2009. The title of the paper is ‘Classes of Optimal Network Topologies under Multiple Efficiency and Robustness Constraints’. In this paper, we have looked at the problem of designing optimal network topologies under efficiency, robustness and cost trade-offs. Efficiency, robustness and cost are defined in terms of structural properties: diameter, average path length, closeness centrality (efficiency); degree centrality, node betweenness, connectivity (robustness); number of edges (cost). A slider $\alpha$ is used to indicate the emphasis on robustness (on a scale of [0, 1]), and thus decides the trade-off between efficiency and robustness. Another parameter $\beta$ (in [0, 1]) decides how cheap or expensive edges are. For example, if $\beta$ is 0, we might make use of the full budget allocated, whereas if $\beta$ is 1, we want to “squeeze out” the most cost-effective topology by removing superfluous edges. Finally, there is also a constraint on the maximum permissible degree (indegree and outdegree, in case of digraphs) on a node in the resulting graph. Maximum degree can be thought of as a constraint on the “bookkeeping” (for example, size of DHT finger tables) a node has to do, or as constraints on maximum inflow and outflow through a node.
Different efficiency and robustness metrics are useful in different application contexts. For example, in case of traffic flow networks, congestion is a major challenge. Hence, node betweenness is a useful robustness metric, along with diameter as the efficiency metric which relates to upper bounds on the communication delay. In a Network Centric Warfare (NCW) scenario, it might be important to have alternate communication paths in case targeted attacks occur on nodes or links. Connectivity is a useful robustness metric there. And so on. Thus, what we do next is an ‘exploration of the parameter space’ involving efficiency and robustness metrics, $\alpha$, $\beta$, cost and maximum degree. We conduct genetic algorithm based experiments to evolve the “fittest” or the “most optimal” topologies under given trade-offs.
The main results in the paper are the following:
• Two prominent classes of topologies that emerge as optimal: (1) star-like or scale-free networks with small diameters, low cost, high resilience to random failures and very low resilience to targeted attacks (2) Circular skip lists (CSL) which are highly resilient in the face of both random failures and targeted attacks, have small diameters under moderate costs.
• Further, we observe that star-like networks are optimal only when both the following design requirements are satisfied: (1) very low emphasis on robustness (or very high emphasis on efficiency) and (2) severe restrictions on cost. In all other cases, CSL are optimal topologies in terms of balancing efficiency, robustness and cost.
Thus, we observe a sharp transition from star-like networks to CSL as emphasis on robustness ($\alpha$) increases or cost restrictions become less severe.
• Circular skip lists are highly homogeneous wrt many structural properties such as degree, betweenness, closeness, pairwise connectivity, pairwise path lengths. Further, they have structural motifs such as Hamiltonicity, (near) optimal connectivity, low diameter, which are optimal under varied application requirements. We argue that CSLs are potential underpinnings for optimal network design in terms of balancing efficiency, robustness and cost.
## Graph Powers
August 7, 2008
Earlier I had talked about multiplying the adjacency matrix of a graph by itself and how it leads to some interesting results. Multiplying the adjacency matrix by itself essentially corresponds to raising the graph by some integer exponent while maintaining the same number of nodes.
The powers of graphs are interesting structures. The p^th power, G^p of a graph G is obtained by adding edges between nodes in G that are separated by a pathlength (greater than 1 and) less than or equal to p. Specifically, the square of a graph is the graph resulting from joining all nodes that are separated by a pathlength 2. A cube of a graph is obtained by short-circuiting all 2 length and 3 length paths in the graph.
An important consequence of raising the graph to a power is diameter reduction. If G has diameter d, then graph G^p has diameter \ceil{d/p}. There are several other interesting properties of graph powers that we can observe, which are important in the design of optimal topologies. I’d be interested in knowing your ideas.
## A note on Hamiltonian circuits (and Uncle Paul)
June 22, 2008
Sometime back I read a short paper by Chvatal and Erdos that presents a couple of sufficiency conditions for hamiltonicity.
One of the results there is – If G is an s-connected graph with at least 3 vertices, and has a maximum independence number of s, then G has a hamiltonian circuit.
It is an interesting result connecting robustness and hamiltonicity. A graph is s-connected if there are s edge independent paths between any two nodes in the graph. In other words, it is the size of the minimum edge cut i.e. the minimum number of edges whose deletion increases the number of components in the graph.
The maximum independence number of a graph is the size of the biggest independent set. An independent set is a set of vertices of the graph such that no two vertices in the set are adjacent. In other words, there is no edge between any pair of vertices in the independent set. The bigger the independence number, the easier it is to fragment the network.
As you can see both these concepts can be used to measure the robustness of a graph. Also, it is not unnatural that connectivity and hamiltonicity are related. Intuitively, the more independent paths, the greater the chances of a graph being hamiltonian. And, a hamiltonian circuit is at least 2-connected: a circle is the simplest hamiltonian circuit and it is 2-connected.
***
Well, but the point of this post is something else really. The paper I referred to in the beginning is just a 3-page note. And it has an interesting footnote on the 1st page. The footnote says: This note was written in Professor Richard K. Guy’s car on the way from Pullman to Spokane, Wash. The authors wish to express their gratitude to Mrs. Guy for smooth driving.
## Seminar on Complex Networks
May 14, 2008
A series of seminars on Complex Systems has been scheduled through the summer over here. I had been asked to give the first seminar. I gave a talk yesterday. I gave an overview of the philosophy, design and analyses of complex networks. Among other things, I talked about how several graph theoretic measures can be used in both design and analyses of complex networks. There was a lot of lively discussion. The total seminar spanned about 2.5 hours, which probably means the seminar went off quite well. In case you are interested, here are the slides I had used.
## Graph invariants, “identifiers” and reconstruction
March 25, 2008
Graph invariants are properties associated with a graph that do not change across all possible isomorphisms of the graph. In other words, if a set of graphs are isomorphic, then they all have the same values for certain properties which are called invariants. Examples of graph invariants are the following:
• no of nodes
• no of edges
• the degree distribution (or degree sequence)
• vertex (edge) chromatic number: the minimum no. of colours needed to colour all vertices (edges) such that adjacent vertices (edges) do not have the same colour
• vertex (edge) covering number: the minimum no. of vertices (edges) needed to cover all edges (vertices)
As you can see, the invariants are numbers. A graph invariant describes a graph in terms of a simple number. Given a graph, it has a unique number of nodes, number of edges, degree sequence etc.. They find application in chemistry where large chemical compounds are indexed based on these numbers.
It should be noted that the incidence and adjacency matrices are not graph invariants. This is because, when isomorphism is computed, if for every pair of adjacent nodes in graph-1, you can find a pair of adjacent nodes in graph-2, and vice versa, the two graphs are isomorphic; labels are not important.
Also, for a given graph, there are unique invariants. However, the converse need not hold. For example, consider the invariant degree sequence. Shown below are two graphs that have the same degree sequence, but are non-isomorphic.
***
What I am interested in finding what can be called graph “identifiers”. Graph identifiers are properties or metrics that uniquely identify a graph. Examples of identifiers are adjacency and incidence matrices. Given an adjacency matrix, there is only one graph corresponding to it. Similarly, are there more identifiers? It need not be one property; it can be a combination of a set of properties like the no. of edges, degree sequence, diameter, centrality sequence and so on. What is perhaps also interesting is to measure the “uniqueness” of sets of properties. Uniqueness indicates in percentage terms “the extent” to which a property or a set of properties identify a graph. For instance, the adjacency matrix has a uniqueness value of 1. Again, this can be measured in terms of, say edit distances.
I am not sure there is much work in this direction. Do you think this is an interesting problem?
A consequence of this is graph reconstruction. If we have a set of properties that identify a graph, it is possible to construct an isomorphic graph using those properties. It can be used to recover networks that are subject to disruptions. (My usage of reconstruction is similar to that in literature, but slightly different.)
## More on Regularity Preservation
March 16, 2008
As sri rightly pointed out in the comments section of the previous post, clique insertion is just one way to extend an r-regular graph into a bigger r-regular graph. We should be able to extend the graph by inserting any (r – 2)-regular graph (provided the insertion is possible — refer sri’s comment). In fact, clique insertion is the minimal case: the smallest connected (r-2)-regular graph has r -1 nodes.
There are interesting questions here. Can we insert any (r – 2)-regular graph? What is the procedure to do so? Are there some (r – 2)-regular graphs that cannot be inserted?
Let’s use some notations for ease of reference. Let Ro be the r-regular graph we start with, Ri be the (r – 2)-regular graph we want to insert and Rn be the graph resultant of the insertion.
A simple but important thing to be noted here is that, pairs of nodes that are adjacent in Ri, cannot be inserted in a single edge of Ro because that will result in an overlapping edge (we don’t allow multiedges between nodes); adjacent nodes in Ri have to be inserted in different edges. Even for small values of r, this becomes pretty complicated to visualize. I tried to insertions for small graphs, and it was possible to do. But I don’t know if we can always avoid creating a multigraph.
But then in the above discussion we have assumed Ri to be a connected graph. What if Ri is disconnected? If Ri is disconnected we can easily prove that there is no bound on the size of Ri that can be inserted. The simplest way to show this is by saying that our Ri has an arbitrarily large number of (r – 2)-cliques as disconnected components (disconnected edges, disconnected cycles, disconnected cubic graphs etc.). Now, you can either imagine this as adding an arbitrary number of cliques (like in the previous post) or as adding one large disconnected graph.
Of course, cliques are just one way to insert. We can simply insert nodes and add edges between the new nodes in such a way that all the new nodes have a degree of r.
Note that the resulting graph Rn is, of course, a connected graph.
***
Now, since we are talking about p2p networks, what is more important is the minimum number of nodes and edges that we need to add when nodes join or leave. Or more generally, the minimum change in topology that is required. When a new node comes in, let’s say we need to accommodate it without disrupting symmetry; so we may need to add dummy nodes and edges. The thing is – in a p2p network, most of the time, unrelated nodes join or leave unrelated parts of the network. So, essentially, we are dealing with disconnected components.
## Regular Graphs: “Clique Insertion”
March 12, 2008
After talking about the basics of regular graphs, let me now introduce something more interesting. When we make changes to a regular graph topology, like addition or removal of edges or nodes, the resulting topology might not remain regular. An interesting question then is the following: are there ways to modify a given regular graph topology, while maintaining its regularity?
Here I introduce a simple operation called “clique insertion” that “extends” a graph (or makes a graph bigger by adding more nodes) while preserving the regularity of the graph. The basic idea of this is due to sri. Later on I developed it a bit more and now it looks like it is a useful result.
Given an r-regular graph of n nodes, the graph can be extended to an r-regular graph of n + (r – 1) nodes by a process of “node insertions” (or “clique insertion”).
Explanation: Let us start with the minimally connected non-trivial regular graph: a circle with n nodes. We can see that by “inserting” a node between any two nodes, we can extend the graph into a bigger graph while preserving the regularity. In other words, cut an arbitrary edge into two, and make their loose ends incident on a newly added node. We can go on inserting any number of nodes in this fashion. We just get bigger circles, all of which have a regularity of 2.
It is pretty simple to see in a circle. Now, let’s go to a slightly more complicated topology. Consider a 3-regular graph. We can get a bigger 3-regular graph by inserting 2 nodes (as explained above), and adding another edge between the 2 nodes.
It is equivalent to inserting a clique of exactly r – 1 (where r is the regularity) nodes in the existing graph (refer the figure below). For an arbitrary n and arbitrary r, we can construct an n + (r – 1) node graph by inserting exactly r – 1 nodes, and adding edges among them to form a r – 1 node clique.
In the figure, we have a 4-regular graph (incidentally, it is the complete graph K5). We now do a “clique insertion” operation on the graph to get a new 4-regular graph of 8 nodes. The nodes marked in brown form the clique.
The above result can be extended thus: given an r-regular graph of n nodes, we can construct a graph of n + m * (r – 1), where m is a positive integer. So, we can easily construct arbitrarily large regular graphs of a given regularity r by inserting any number of cliques of size exactly r – 1.
Usefulness: This has an important connotation in distributed systems, where nodes joining and failures are frequent. Several distributed hash tables (DHTs) require a symmetric degree distribution. They also need to accommodate for node joining and leaving.
• For one, this result can be developed to be a measure the amount of disruption in symmetry that can be expected when changes in topology occur.
• Also, it can be used to design topology changes so as to preserve symmetry. Perhaps, by using dummy nodes.
Question: Is there a similar way to make a graph smaller while preserving the regularity?
That was one example of preserving a structural property of a graph. It is an interesting to find out if we can have a whole set of such operators. We can look at several other properties as well: diameter, average path length, centrality distributions, connectivity etc.. I shall define this problem more formally in another post later on.
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Wednesday, May 29, 2019
No global symmetries in QG 2019
Peter F. was intrigued by an article in Phys.Org,
Researchers demonstrate constraints on symmetries from holography
about a paper by Hiroši Ooguri and Daniel Harlow that was just published in a good enough PRL. Note that these two men have written two similar papers together.
The heart of the main Harlow-Ooguri argument. The global transformation would be a current integrated or transformations composed of boundary regions $$R_i$$ but each of them only affects a nearby grey wedge, leaving e.g. the green bulk non-transformed. It follows that the boundary currents or operators responsible for symmetries cannot correspond to global symmetries in the bulk.
Well, the main claim, as we promised in 2017, is that the symmetries in the bulk of quantum gravity have to be local, gauge symmetries. No global symmetries are allowed. And this statement also extends to non-Abelian and discrete symmetries, they argue. They praise their own paper as the most general, rigorous, and conceptual proof of a "similar" kind of a result.
Now, the statement that quantum gravity doesn't allow the global symmetries has been a part of the lore for decades – I guess that many sharp physicists have been convinced about it before Daniel Harlow was born. I have surely believed it from the 1990s. My adviser Tom Banks would make sure that I did believe it after 1997-1998 because he has always been a physicist who liked to get familiar and refine all similar lore.
Fundamentally, gauge symmetries are preferred in general relativity because general relativity wants "everything to be local". Note that even the existence of a gravitational field or acceleration depends on the locally chosen coordinate system. Also, gauge symmetries may be obtained from diffeomorphisms involving extra dimensions, using the Kaluza-Klein trick. That's why gauge symmetries are naturally "siblings" of the diffeomorphisms of general relativity, while adding global symmetries would be like trying to restore whole pieces of special relativity within general relativity.
The spirit of their proof may be found on Page 6 (8/175) of this long paper. They prove the qualitative statements about the character of symmetries within theories that have both the AdS bulk description and the dual boundary CFT description. They assume that there is a global symmetry and there must be some currents as well as fields that create charged objects.
They need to assume some kind of locality for the "wedges": the operators corresponding to bulk objects near (or very close to) the boundary must be written as operators acting within a (very small) region of the boundary CFT. With this assumption (which might of course be incorrect in principle – local bulk objects could be created by Wilson-line operators going all over the boundary along some paths etc.), the proof of the most basic theorem is trivial. If there is a generator of the global symmetry within the CFT, then it is an integral over the boundary – or a sum over terms from different regions. But each region is only capable of transforming objects in some "wedge" which is near the boundary.
All of them avoid the points that are very distant from the boundary – such as the "center" of the AdS space – which means that the fields at the AdS center don't get transformed by the "global" symmetry transformation. It follows that the symmetry transformation wasn't really "global" in the first place. The assumptions cannot be satisfied at the same moment.
It's great. I haven't written the most precise arguments for the "lore" that I knew after the AdS/CFT was proposed in late 1997 but I guess that they could be virtually identical. It's good that someone writes these things explicitly but I would distrust the statement that their explanations are really new.
At any rate, I believe they are correct and they are effectively relevant for all of quantum gravity. The fact that the proof took place in the AdS space is hardly a limitation. Gravity – even when it is quantum gravity – should still be local in some conceptual sense. It means that its basic properties (such as the existence of symmetries, global or local, and gauge fields) doesn't depend on the asymptotic behavior of the spacetime or the events in very distant points.
It means that if global symmetries are banned locally in the AdS spaces with a CFT dual, they are almost certainly banned in any other spaces of quantum gravity, including those that are asymptotically dS. This assertion is a reasonable expectation. I don't have a full-blown proof, of course. I can't have a full-blown proof because I can't even rigorously define general concepts such as a "general quantum gravity in an asymptotically dS space" – well, we can't define quantum gravity in any de Sitter space terribly well.
Nevertheless, papers like that represent clear progress. Physicists are finding a mixture of rigorous, reliable results and results that are just "rather persuasive" but that nevertheless shape the expectations and move them in certain directions, directions that are increasingly justified by the evidence.
The absence of global symmetries has consequences. For example, the proton seems stable because it's the lightest object with the baryon charge $$B=+1$$. The baryon charge is a generator of a $$U(1)$$ symmetry, in analogy with the electric charge (the lightest particles with a nonzero electric charge, electrons and positrons, are stable because they are lightest particles with a nonzero electric charge).
OK, is the baryon $$U(1)$$ a global or local symmetry? Because the protons live in a spacetime that also has quantum gravity, it can't be a global symmetry. So it must be either a local symmetry, or a non-existent or broken symmetry. If it's a broken symmetry, then the conservation law for $$B$$ is violated in general, so the proton may decay to a positron, photons, and perhaps other stuff.
If the baryon $$U(1)$$ is a local symmetry, then there should be gauge bosons – baryon-charge counterparts of photons – that mediate a new force. Such a new force would cripple the tests of the equivalence principle (there would be new $$B$$ forces between electrically neutral pieces of matter, and those wouldn't be proportional to the masses) unless the gauge coupling $$g_B$$ is tiny.
However, even the loophole with a tiny $$g_B$$ may be basically excluded by a quantum gravity-based, swampland-like reasoning, namely by the weak gravity conjecture. Coupling constants shouldn't be too low, our conjecture states. In effect, the weak gravity conjecture prevents you from "fraudulently circumventing" the ban on global symmetries by considering local symmetries with tiny couplings. Indeed, this was a top motivation of the whole conjecture that I/we added to that paper.
To summarize, quantum gravity-based formal arguments make it very likely that despite the non-observation of any proton decay so far, proton is ultimately unstable and may decay although the lifetime may be much longer than the current bound.
You can see that quantum gravity – as exemplified by string/M-theory – increases the predictivity of the framework of physics. It generally bans some things (global symmetries in this case) and it also prevents you or Nature from applying "cheap tricks" to circumvent the bans. This is an ability of quantum gravity that the critics of physics rarely understand. They think that any rule can be circumvented by a trick, they may always find a loophole, fool Nature in any way, which is why they think it's right to ignore any formal finding in string theory or theoretical physics in general. But Nature, as soon as the consistency of quantum gravity is added to the picture, doesn't really want to be pißed upon.
In other words, whenever we find a principle or a no-go theorem, Nature doesn't obey it just theoretically. It also obeys it in practice. Your method to circumvent the ban unavoidably predicts new effects whose observability is nonzero – the observability is implied to be greater than a lower bound that Nature guarantees not to be "arbitrarily small" in general. Qualitative lessons of quantum gravity matter even in practice.
You know, the knowledge of probably correct answers to questions such as "should the fundamental theory allow global symmetry" is very important and, in some deep sense, primary for a meaningful research program in fundamental physics. Do you remember Feynman's Cargo Cult Caltech commencement speech? A man named Young did some careful experiments – those are analogous to this swampland-like research of quantum gravity – that were ignored by the sloppy researchers who just let their rats run through the mazes chaotically, without a proper quality control. Feynman wrote:
They just went right on running rats in the same old way, and paid no attention to the great discoveries of Mr. Young, and his papers are not referred to, because he didn’t discover anything about the rats. In fact, he discovered all the things you have to do to discover something about rats. But not paying attention to experiments like that is a characteristic of Cargo Cult Science.
Young discovered things like "you have to be careful not to allow the rats to sniff something" etc. if you want to know how the rats get their information etc. Rules like that are necessary for you to do all the experiments properly. Similarly, the stringy or swampland-like research also determines answers to fundamental questions such as "should you play with hypothetical theories that have global symmetries?". And the answer seems to be No: global symmetries shouldn't be there at all.
In this analogy, the "work" on new unifying theories or theories of quantum gravity that ignores the stringy research is a cargo cult science. People doing things like that are just letting their rats recklessly run in mazes – but they may keep on doing serious mistakes (such as the inclusion of global symmetries) that render their life-long research totally worthless.
On one hand, the stringy research is "analogous to any other legitimate type of research". On the other hand, it is often more fundamental and defining or redefining the rules of the game. In this sense, the string theory research may be considered "meta-research" or "metaphysical research". But that doesn't make it less relevant for science. On the contrary, it makes it more relevant for science because every piece of scientific research assumes some metaphysics and it's damn important for this metaphysics not to be completely wrong.
The search for the deepest theories in Nature is a very advanced activity that is unavoidably detached from the people's everyday intuition. Even the most obvious metaphysical assumptions may be wrong. And that's why every person who approaches these questions as a real scientist must be ready to question and revise any assumptions, however obvious, and he must prefer the answers to the fundamental questions that are supported by the evidence, not those that are supported by some prejudices or intuition based on the everyday life. Much of the formal research in quantum gravity and/or string/M-theory is a search for the rules of the game that Nature Herself respects, for the refinements of the rules of the scientific method.
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## I'm out - for now
I just clicked on the 'Cancel Subscription' button, this means I wont be renewing GDNet+ at this time. Why? for many reasons, most of which are personal, no I'm not getting out of games (not yet), no I don't hate GameDev, I quite enjoy it, actually. I'm just trying to evaluate where all my money, or lack thereof is going. I work in the gaming industry, and I haven't been able to afford an xbox360... and I've been wanting a new computer of a decent laptop for a while now but somehow, I never have enough money. So I believe I need to reduce my expenses as much as possible, then identify where the money is going and then decide where I want to use it. Plus, I have more bills now than I used to, hell, the gas bill up here is like \$80...
Anyway, I'll be back (will my journal get deleted? that would be crappy), hopefully soon.
Cya.
## It's getting ridiculous
This just in! Terrorists were planning on using clothes as explosives in airplanes, from now on, everyone must fly naked!
Ok, so it's not an actual headline, but at the rate it's going I won't be surprised when full cavity searches are "standard procedure", hell, they already fondle old ladies, cause everyone knows that grandma can be one mean terrorist.
When I was a kid, I used to love flying in an airplane, now it's one of the most deplorable things you can do, even worse now that I won't be able to fight with damn dehydration by bringing my own water or gatorade in, I'll have to drink the 0.42 oz. of soda they give you. Screw that, if you're getting on a plane, demand that the airline have a large bottle of chilled gatorade waiting for you inside the plane, I think if consumers begin to demand stuff, they may get it.
We should all as the world community use the money we would've otherwise spent at the duty free and fund research into teleportation. Anyway, enough ranting, I gotta go to work
## Je suis fatigue
So between the baby and the new job I don't really get a chance to do much lately. Work is going great, there's plenty of it and I'm having fun, though I am tired; I haven't been working what I consider long hours yet, but I've definitely been doing a big of OT, which is fine because I really like the game I'm working on (did you see EGM's cover this month?)
I've been thinking, but I can't seem to decide... I would like to make some more money on the side, sempesoft.com is not generating as much as I thought it would (or hardly any at all), so should I bother making a small puzzle game for the pocketpc, it would be something I could make fairly quickly, and sell it, or should I continue with SGE and the game I've been planning.. which I also plan to sell one day, but it'll take more time to get there. It's a tough decision, because I have so little time to spend on this (although it's getting better, we're getting the hang of the baby).
Anyway, that is all for now, it's 8am and I gotta get ready to go to work. Oh yea, I never though it would be this f'n hot in Montreal, it's been so hot and muggy that I'm actually looking forward to winter.
Finally, a few days ago I ran into a programming issue that I wanted to share, check it out here: site I could definitely use some opinions, I haven't been able to spend much time figuring out what's going on.
Oh, and the french lessons are going well, hence the title (btw, e does not work on the title)... hehe
ce tout, au revoir
## Welcome to the world of tomorrow!!
I neglected to write a post about a great milestone in my life. I am now the proud father of a little baby boy, who I hope will be impressed when his brain develops enough to understand what his dad does for a living. His name is Luka and he was born on June 26th, which happens to also be my birthday, he has a really cool birthdate 06/26/06.
Here he is, notice how his hands have perfect keyboard posture, he's asking "is that a laptop?"
He was not thrilled when I told him I couldn't bring the xbox to the hospital.
His mom is healthy and practically recuperated and he's doing great, and we're all sleeping very little, but it's ok.
## Happiest place on earth?
Guatemala is among the top 10 happiest places on earth according to this:
http://www.happyplanetindex.org/list.htm
I guess it would be for me if there was a game development or large software development industry (which I predict there will be in 10-15 years), and if there wasn't as much violence and terrible governments that keep the country poor and underdeveloped. It's also not the case that there is so much violence, there probably is the same amount of violence in other places in the world (LA), but it's in a much smaller country/city, so you're affected more directly.
It's a pretty nice place, the weather is awesome most of the time, really nice tourist places to go to, very welcoming people, good schools, lots of cultural diversity and a city that's growing too fast for its own good.
Anyway, I gotta go make games now... I've lingered here too long.
## bleh
I was able to get the mesh to draw properly, but I stopped right before I started converting the skeleton and animation data and I havent been able to pick it up. I had some time this weekend, but I just couldn't get myself to do it, so I decided to rest, I need it, my wife is going to give birth any day now and we're all anxiously awaiting the big moment.
Anyway, I'm cooking, so I gotta go check up on the sauteed shrimp & potatoes, it's my own recipe, yum.
cya.
So I'm working on getting importing models with full skeletal animation support. I've been doing this for the past couple of days, but mostly I had just been evaluating my options. First I gave Cal3D a try, setting it up was a pain; I'm not fond of DLL's, not for something that's not a plugin to an existing application, I know they're useful in some cases, but if I can avoid them I'm happy, I actually forgot what was the ultimate reason why I decided against it, I didn't like the way it was coded, I didn't like how a model was broken up into many files, just too many negatives. In the end, I decided to go with Microsoft, I mean, I'm already using Direct3D, so why not use their file format?
It took me about 20 minutes to get a hacked in animated character (that completely bypassed my entire rendering architecture), and now I'm working on actually making it conform to the architecture; here's the first thing I saw when I compiled:
This is actually better than I expected, I expected my computer to die a horrible death, but instead I got that! So it's all good, it just means somewhere the data is getting read or written incorrectly and it's jacking up the indices, or the vertices. Shouldn't be too bad to figure out.
Alright, I'm going to keep at it because it's pretty exciting.
Cool, just a few minutes after writing this post I found my error, I forgot to account for the index buffer stride, so after quickly fixing that:
Now, I know that looks weird, but, that's just because I have only converted the geometry over, I completely ignored all the transforms, so every "submesh" is being drawn at the origin (everything has the identity matrix as its transform), so now, it's only a matter of getting the right transforms for each submesh, then I'll have to also get the frame data for the animations and get the bone information as well.
If all goes well, I'll have an untextured, skeletal animated model tomorrow, making it textured is the easiest part, all I need to do is load the textures into the manager and save the handle in the submesh's material, it's so easy, I'll leave it for last.
## Beware of zombies!
I thought this was pretty funny, so I decided to share it...
I can never go near that file again, I do not wish for it to eat my brains.
*edit: I had the wrong image before...
## wth gd?
Ok, so something bothers me about gamedev. When I released EasyShots, I sent an email to announce it in the news section, I know it's not strictly a gamedev tool, but *I* use it to take screenshots of my works in progress, so I thought gamedev readers might be interested; my news never got posted, I wasn't happy, but I thought that perhaps gamedev is just trying to seem like a more _industry_ portal, I mean most of the news that get posted are industry news; but anyway, I didnt even get a response saying "sorry, we wont run your news, we are trying to stick to industry type things", which I would've understood... but I did not complain, I just kindly let it slip.
But now... I saw that currently there are only two prices posted for 4E5, so I thought I'd give up a license of EasyShots to every contestant; I sent an email to the link they have on this page with my offer, this was on either friday or saturday... maybe I'm not as patient as I should be, but at least send a "we're busy but we'll get to you..." reply.
My offer still stands, albeit "unofficially" for now, if any of the contestants is interested, send me a PM and provided that you can with a link to your entry and I'll happilly hook you up.
EasyShots is great for taking IOTD screenshots, btw.
## So good
It's so nice when things actually work. This weekend has been incredibly productive, I ported everything to VS2k5, I'm using a new project organization which is working very nicely, I used to have a separate project for each component, like Renderer, Game, Input, Network, System, etc. Now I got rid of that, I have 1 project, SGE and all my files live under it, and I create a separate project for an executable, so now I can do stuff like, work on my game, but if I want to test out a feature, I can create a project for say, AnimationTest, and it will create its own executable, or maybe to join the 4E5 contest.. hmmm...
The contest sounds interesting, and although I haven't seen the movie or read the book, it made me think of the Da Vinci Code...
Anyway, I've got good inertia going, and the weekend is almost over (who was the genius who carved in stone that 2 days were enough?)
Ok, more later, gotta while (IsWeekend()) { drink coffee; write code; }
## Well this is a first
So I'm in the middle of porting my old code to VS 2k5, and I'm running into this:
when I use std::sort
I guess it's possible that I was always doing something wrong, but it used to work well. It also could be that something from my port is broken and that this is a red herring, but who knows. Has anyone run into this?
Here's the sort callback:
** code removed because it was embarrasingly dumb **
nevermind, I fixed it
## Finally!
I was FINALLY able to get all my projects in their right locations with all the perforce bindings working correctly, at least it seems more likely that it worked this time. It seems that everytime I think they're working when I come back at night and start messing with the projects, they're not quite right. I do think I got it right this time, that's the one thing I don't like about perforce, it's not always easy to get the projects working correctly, but once they are, it's so damn cool.
So, I've been thinking a lot about it, and I think I've finally accepted what I always knew, but didn't really care. I've been concentrating too much on writing an engine, which is fun and all, but it's not going anywhere. This is perhaps because the goal is elusive.. a moving target even, the closer I got, the newer more advanced thing I wanted to add... there was a lot of work, but not enough direction, no vision.
So I've decided that I'm going to take what I have, do a quick evaluation of what stays and what goes (I already have a pretty good idea what that is) and I'm going to move in the direction of writing a game.
On the game design side of things I already have an idea, an idea I like quite a bit, so I'm working on evolving that idea into a design. Currently I have a page with a story overview (I'm keeping the story light, but I do want some story elements) and a couple of mindmaps with ideas and structure.
Hopefully, I'll be able to keep this simple and finally evolve SGE into something more real.
## Fresh start, again
Today I bought a 250gb HD, it was at a pretty good price, I'm pretty happy about it, the only thing that I don't like is that I had to redo everything I had done since last week again, I'm just now finishing up installing my software again and I still have to setup my projects and redo my perforce bindings and all that fun stuff. The great thing about it is that now I have a separate 80gb HD that I'll put in one of those external casings and I'll use it to backup my project to it nightly.
In other news, I started up a GD group for Montreal, there wasn't one and there's like 130 people from Montreal registered. It might be fun, I look forward to people joining, it might be fun to do a gathering and meet some of the people here in real life.
Alright, I still have a lot to install, so I better get going.
## Not much to report
I've been so tired the past few days... it's starting to affect me... I don't want to do much of anything, I spend half the day fighting to stay awake... it's not work either, work has been great, it's the amount of activities that I've had after work that's killing me. I've been attending childbirth classes with my wife, but since she's due real soon, we had to take several a week, when normally it would be one a week, and each of these classes lasts 3 hours. We're still unpacking stuff and cleaning up our new apartment, which sounds easy but it's surprisingly more work than you'd expect, add to this the fact that I now do a lot more excercise than I used to, I now walk a few blocks to the metro station, then walk through a mall to get to work.
I was pissed off yesterday, I had left work, I was feeling shitty, I got on the metro and I stood holding on to one of the center poles, reading Lord of the Rings: The Fellowship of the Ring (yea, I hadn't read it until now, I'm a bad geek), when an asshole that was sitting right next to me decides it would probably be funny to light a firecracker inside the metro, from the corner of my eye I saw the bright flash for a split second and then the sound hit my left ear, immediately I went quite deaf and could only hear that loud ringing in my ear. The doors opened right at that moment, so I walked out and got on an adjoining cart so that I could continue to enjoy my book.
tabarnak!
Today I formatted my C: partition, reinstalled all my software (well, I'm in the process of doing that) and next I'll be cleaning up my development drives and preparing for a new wave or renewed enthusiasm and desire to take SGE to the finish line. DX9 SDK April 2006 release just finished installing, yay!
I have a ton of plans on how I'm going to move forward, I haven't cleaned them up into human readable form yet, currently it's a combination of unintelligible streaks of ink in a notebook and deceiving diagrams, but after I'm done getting all my software back into shape, I'll have something nice and clean to work with.
Anyway, things seem to be going a bit back to normal, and some sort of a routine is starting to develop, which is fine, it'll make planning a bit easier... at least till until a month from now, then there'll be a few weeks of disarray and reorganization, but I don't think it will be so bad... sleep deprivation is a friend, not food.
Ok, I'm getting tired, so I better get back to installing shit.
## Busy, busy
I keep wanting to post here, and go back to programming at home a bit, but I've been really busy, not so much at work, the work is cool, maybe a bit slow since we're in post-E3 mode, but nice. But Im still unpacking, and the last two nights I've gone to pre-natal classes with my wife (we're having a baby REAL soon), so we have tons to do, like setup the nursery and finish unpacking and cleaning up.
In addition, my computer is going real slow, so I think it's time for a good ole format of the C: drive, but that's time consuming and I have to do it in one sitting, I don't like leaving the computer crippled or vulnerable by not doing it correctly, perhaps I'll get a chance this weekend, Monday is a holiday here in Montreal, so it's a perfect opportunity.
And now I gotta run off to work, yay!
lates.
## EA Montreal
It's been a while, I know... I'm currently at my new job at EA Montreal and it's awesome. I've moved into an apartment, but I haven't set up my computer or subscribed to an ISP yet, so I can't update from home.
Anyway, Read this it's a bit about EA Montreal and the project I'll be working on. This game is going to be awesome.
## It's official
My work permit has been approved and this pretty much closes the deal. I'm moving to Montreal as soon as next week to start working at EA. I can't wait, I'm very excited to go back to EA, while some people complained, I enjoyed it enormously and I'm sure I'll continue doing so.
My posting frequency may diminish, although I intend to find out EA's policy on "blogging" to see whether I can post more programming stuff than I have so far.
So that's the news I have for now, I gotta go pack stuff like crazy, make backups of my HD (I never trust those xray machines, even though they seem to be safe) and do a whole bunch of stuff, man, I got tons to do, I better go do it.
## Pulled the switch
It's done! I've pulled the switch and the new Sphere Games site is now live.
www.spheregames.com
For sentimental value, I left a copy of the old site here:
www2.spheregames.com
at least for a while, and in case something catastrophic happens I can revert back to it.
## a game?!
Ok, so I'm FINALLY working on a game project. I started yesterday with some real basic stuff. It's going to be a small puzzle game in C# that I'm going to try and sell similar to how I'm selling EasyShots and all that.
I figured, I already have 3 programs out there, instead of using my time and energy on a fourth one (which is already in development), I'll give making a game a shot.
So here's a bit of what I know:
- Will use C# with .Net framework 2.0
- 2D puzzle/action game
- Uses DirectX (Direct3D for 2D, no HW stuff)
- Single player
- PC standalone & web version (web version will be a port, no details right now)
Here's what I have:
- A prototype game design
- A basic 2D drawing library (at 'alpha')
- A photoshop mockup of the game interface
- A high level design of the game mechanics
And that's it... I'm still thinking and writing down ideas because I currently don't know where to begin, and to me that means I have not properly planned and design the game, so opening up visual studio and programming blindly would cause this project to quickly make its way to the "incomplete" bin, instead I'm going to go plan more... way more.
## xRC is now freeware
Some of you might remember xRC, a little program I made that maps joystick input to keyboard input. It's a very simple program that I wrote so that I could use my xbox wireless controller to change channels on my TV tuner. I finally got around to yanking out the shareware type stuff and made it freeware, the only thing I did to it was put some ads for EasyShots, Explore++ and SleepyTime on it, in hopes that if someone liked xRC they'll try out the other programs too.
It's a little strange because I've only mentioned xRC here, and just that one time, and even then it gets downloaded about once a week... which means people either read my old posts, or that somehow google is pointing them towards it, either case it's pretty cool.
So here it is:
xRC (freeware) Map joystick buttons to keyboard input.
## New Release: SleepyTime
Hello!
I just released a new program, it's called SleepyTime, it allows you to schedule when you want your computer to shutdown, reboot, log off, suspend or hibernate. It allows you to set a specific time or to set a minute timer. It's incredibly useful if you don't want your computer to stay on all night after you've gone to sleep, or if you want your computer to reboot after a long process is complete.
You can check it out here: SleepyTime
I also updated EasyShots, here are the latest updates:
- JPEG quality is now set to the highest possible
- Screenshots can now be resized in the edit window
- Resized or cropped image size is now properly updated in the thumbnail list
- Edit window will not get set to a size larger than the desktop resolution upon zoom
- Crop & capture box dimensions are now white text over blue background, improves visibility
- Optimized area capture
- Images can now be dragged into the thumbnail list
- Thumbnails for small screenshots are no longer zoomed in
- The + and - keys now control zoom in the edit window
- Optimized screenshot capture and thumbnail list process
## Sphere Games Website
Today, in just a couple of hours I added the ability to for people to add comments to my "news" postings on Sphere Games essentially turning it into a type of blog.
I think that's the last bit of functionality I want to have working, now I just need to work on the content, which I think the heavy part is done (the gallery was a lot of work), and then it's just a matter of evolving the site over time.
Tomorrow I will very likely upload the newest update to EasyShots, it has couple of new features, a couple of fixes and a couple of optimizations, it's the best version yet =) I'll post here when the update goes live.
## .Net 2.0 bug?
I'm currently updating EasyShots a bit, and in doing so I added a new feature. I now allow images to be dragged into EasyShots, once you do, you can manage them, edit, send them and everything that EasyShots lets you do. It worked great, that is until I tried to drag an animated GIF into it.
Dragging the GIF into it wasn't a problem, when I went into edit mode the program threw an exception deep within .Net code, particularily inside the code responsible for switching the GIF's frames.
So after a while of debugging, I came across this piece of code:
FileStream fs = File.Open(mImageFilename, FileMode.Open, FileAccess.Read, FileShare.Read);mBmp = new Bitmap(fs);fs.Close();
So, I thought, could it be possible that in creating the bitmap in that way I lose valuable information?
I replaced that code with this:
mBmp = (Bitmap)Bitmap.FromFile(mImageFilename);
and this worked. So I wonder if it's a bug, or perhaps it's documented somewhere in MSDN (I didn't look) that creating a bitmap from a stream does not guarantee the same results as using the Bitmap class to load the image directly.
But nothing is free... the second approach may fix animated GIFs, but it appears to leave my file open, so when I try to update it, I get an exception that says "Unable to write file, another progress is using the file".
So I guess it's actually two bugs, first using streams to open files loses a GIF's structure, and using Bitmap.FromFile does not close the file after itself.
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# Search for B^0 Meson Decays to \pi^0 K^0_SK^0_S, \eta K^0_S K^0_S, and \eta^{\prime}K^0_SK^0_S
Abstract : We describe searches for $B^0$ meson decays to the charmless final states $\pi^0 K^0_SK^0_S$, $\eta K^0_S K^0_S$, and $\eta^{\prime}K^0_SK^0_S$. The data sample corresponds to $467 \times 10^{6}$ $B\bar{B}$ pairs produced in $e^+e^-$ annihilation and collected with the BABAR detector at the SLAC National Accelerator Laboratory. We find no significant signals and determine the 90% confidence level upper limits on the branching fractions, in units of $10^{-7}$, $\cal{B}(B^0 \ra \pi^0K^0_SK^0_S) <9$, $\cal{B}(B^0 \ra \eta K^0_SK^0_S) <10$, and $\cal{B}(B^0 \ra \eta^{\prime}K^0_SK^0_S) <20$.
Document type :
Journal articles
http://hal.in2p3.fr/in2p3-00418404
Contributor : Swarna Bassava <>
Submitted on : Friday, September 18, 2009 - 2:03:15 PM
Last modification on : Tuesday, December 22, 2020 - 12:40:02 PM
### Citation
B. Aubert, Y. Karyotakis, J. P. Lees, V. Poireau, E. Prencipe, et al.. Search for B^0 Meson Decays to \pi^0 K^0_SK^0_S, \eta K^0_S K^0_S, and \eta^{\prime}K^0_SK^0_S. Physical Review D, American Physical Society, 2009, 80, pp.011101. ⟨10.1103/PhysRevD.80.011101⟩. ⟨in2p3-00418404⟩
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## samtab Group Title Solve X/-3 = -15. please help me:( 45 -45 5 -5 one year ago one year ago
1. satellite73
$\frac{x}{-3}=-15\iff x=-15\times (-3)$
2. samtab
what? srry i dont understand
3. strenesmee
you have to multiply both sides with -3 to leave x alone
4. samtab
o its 5?
5. strenesmee
if you multiply -15 with -3 you get 45 not 5 you will not divide, just multiply
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# Reducing key shares in Damgård-Dupont threshold RSA
I'm working on understanding and implementing
Damgård, I., & Dupont, K. (2005). Efficient Threshold RSA Signatures with General Moduli and No Extra Assumptions. Public Key Cryptography-PKC 2005, 346–361. doi:10.1007/978-3-540-30580-4_24. Retrieved from http://www.iacr.org/cryptodb/data/paper.php?pubkey=3431
On page 7 (356), step 2 of the key share generation process is
$D$ chooses a random polynomial $f(x)$ of degree at most $t$ with integer coefficients, such that $f(0) = d$; $$f(x) = d + c_1x + \ldots + c_tx^t$$ where the $c_i$'s are random independent integers chosen from the interval $[0\ldots\Delta{}n2^t2^L]$, where $L$ is a secondary security parameter. The secret share of the $i$'th server is $s_i = f(i)$. With the given choice of coefficients, it can be shown that, if we compare the distribution of any $t$ shares resulting from sharing $d$ with the one sharing resulting from sharing any other $d^\prime$, the statistical distance between the two is at most $2^{-L}$.
($n$ is the standard RSA modulus; $d$ the standard RSA secret exponent; $l$ is the number of secret shares/servers; $\Delta = l!$; $t$ is the maximum number of corrupted servers/shares we want to accept)
The given formula for secret shares makes them numerically rather larger than the original secret (by a factor of something like $l^t \cdot l! \cdot 2^{tL}$). I think I can reduce them $\text{mod } \phi(n)$, since the further calculation of partial signatures/decryptions uses the shares in a standard RSA operation (with some extra multipliers): $x^{2\Delta{}s_i} \text{ mod } n$.
Question 1a: Will this modulo operation break the functionality of the scheme? I did some experiments, and think it will work, but have no firm basis. From all the papers I've read I seem to understand that doing operations $\text{mod } \phi(n)$ in the exponents is always right, but I don't understand why and thus can't apply this understanding here.
Question 1b: Will this modulo operation break the security of the scheme? I have an intuition that one of the options might be bad, but can't tell which. Points I've considered:
• Without modulo, it's standard Shamir secret sharing, so an attacker with up to $t$ shares shouldn't learn anything.
• $\phi(n)$ is a secret, so applying it might give the attacker something he wouldn't otherwise have.
Question 2: The last sentence of the section I quoted leaves me at a loss. The authors cite "M. Koprowski: Threshold Integer Secret Sharing, manuscript, 2003", but I couldn't find a copy of that online. What is the "distribution of shares" and "statistical distance" and why is it important here? Shouldn't Shamir sharing unconditionally hide all information about $d$ ($=c_0$) if all $c_i$ are from the same number space? I assume that the last part of the quoted sentence means that I should set $L$ to something like 128 for 128 bit security?
• If I am understanding the protocol, this isn't shamir secret sharing. In shamir secret sharing, the coefficients are drawn uniformly at random from a finite field and the evaluation of the polynomial is computed in the finite field too. In this, the coefficients are chosen from an interval and the shares are not computed within a field. – mikeazo Feb 3 '15 at 20:55
• So, re 2, you aren't guaranteed to "unconditionally hide all information about d". So, proving that the statistical distance between the two distributions of shares resulting from a sharing of $d$ and any other $d'$ is important in proving security. – mikeazo Feb 3 '15 at 20:57
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# Hamiltonian cycle on graphs without small cycles
While answering this question on cstheory, I (informally) proved on the fly the following theorem:
Theorem: For any fixed $l \geq 3$ the Hamiltonian cycle probem remains NP-complete even if restricted to planar bipartite undirected graphs of maximum degree 3 that don't contain cycles of length $\leq l$.
It seems very unlikely that it has not already appeared somewhere.
But it allows to settle many Hamiltonian cycle/path problems on graphclasses.org that are marked as "Unknown to ISGCI" (see for example this one); indeed a direct corollary is that Hamiltonian cycle and path problems are still NP-complete if restricted to $(H_1,...,H_k)\text{-free}$ graphs, where each of the $H_i$ contains at least one cycle.
Can you give me a reference of the paper/book where it appeared?
(then I'll contact people at graphclasses.org)
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At least these discussions helped for new results in graphclasses.org so please inform graphclasses about unknown to them result - The Contact link gives a form, email address is optional. – joro Jun 18 '14 at 8:02
@joro: I already contacted them, yesterday (I also gave them my email). I'll wait a few days and see if they update the status of those problems. – Marzio De Biasi Jun 18 '14 at 8:42
I heard they don't update the database very often and reply with "thanks" after updating the DB and they are quite responsive. – joro Jun 18 '14 at 9:04
@joro: I think they updated the database (they are very collaborative and polite) – Marzio De Biasi Jun 23 '14 at 7:27
The result is stated in the paper Two New Classes of Hamiltonian Graphs by Arkin, Mitchell and Polinshchuk.
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This unpublished manuscript by Hougardy, Emden-Weinert and Kreuter in 1997 provided a simple proof for the following result which is much stronger than the result pointed out in Kristoffer Arnsfelt Hansen's answer:
For any given rational number $0\le r <1/2$, the Hamiltonian cycle probem remains NP-complete even if restricted to bipartite planar $n$-vertex graphs of maximum degree 3 and girth $\ge n^r$.
The manuscript contains also similar results for other problems such as Dominating set, Max cut, VFS, etc.
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Ok, thanks! I forgot to mention that my proof works for planar undirected bipartite graphs of max-degree 3 ... so the Hourgardy et al. paper is stronger ... but not much stronger :-) :-). I'll probably accept Kristoffer's answer because he posted it first. – Marzio De Biasi Jun 17 '14 at 17:00
No problem :-)) – vb le Jun 17 '14 at 17:13
@MarzioDeBiasi, I think the strongness is about the size of a girth. your proof is about fixed number, accepted answer is for some f(n) which is less than sqrt and this answer is more general than all of them. (IMHO restriction to the graph is not very important here) – Saeed Jun 17 '14 at 21:26
The paper contains other NP-hard problems, it will be an answer to the linked question about cyclic graphs. – joro Jun 18 '14 at 9:08
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# Probability regarding getting a full house
I am trying to calculate the probability of getting a full house on a standard 5-card deal.
I am comfortable with how combinations, permutations and the fundamental principle of counting, but I am not sure how this problem works.
I understand that we need to avoid situations such as having a 4 of a kind or a flush, so can someone help me out ?
I would deeply appreciate it if you could give me another example like this as well.
• Umm, what exactly is a full house? – Lord Soth Dec 4 '13 at 22:48
You'll need to count how many different combinations of cards will result in a full house.
There are $13 \times 12$ ways to choose the particular two values of the cards that will make up your hand (i.e. kings over eights).
For each of these combinations, there are $_{4} C_{3} = 4$ combinations for the three of a kind, and $_{4} C_{2} = 6$ combinations for the pair.
Overall, there are $_{52} C_{5} = 2598960$ card combinations.
Hence the probability is $$\frac{13 \times 12 \times 4 \times 6}{2598960} \approx 0.00144.$$
• Thank you. I don't know why I was thinking that full house can come with a flush. – hyg17 Dec 4 '13 at 23:24
Hint: You need to count the number of full houses. They do not overlap with flushes or four of a kinds. How many choices for the rank of the triplet? How many choices for which cards of that rank? How many choices for the rank of the pair? It can't be the same as the triplet. Then how many choices for the cards of the pair, given the rank?
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# Directional derivative - conventions
1. Sep 2, 2005
### rachmaninoff
"directional derivative" - conventions
So the "directional derivative" in vector analysis - the differential rate of increase of a scalar function φ in a vector direction - is by convention* defined starting from the vector gradient (the 'del' operator): $$\nabla \phi \cdot \hat{u}= \vec{\nabla} \phi(\vec{r}) | _{r_0} \cdot \hat{u}$$
(the directional derivative of φ in the u-direction).
So it's not really fundamental - it's defined in terms of del-φ, which is 'more fundamental' I guess. Big deal.
*"by convention" meaning, in every one of a dozen reference books I looked through...
For example, in cartesian coordinates, a directional derivative (at r) might look like:
$$\vec{\nabla} \phi_r \cdot \hat{u} = \left(\hat{i}\frac{\partial{\phi}}{\partial{x}}| _r + \hat{j} \frac{\partial{\phi}}{\partial{y}} | _r + \hat{k}\frac{\partial{\phi}}{\partial{z}} | _r \right) \cdot \left(u_x \hat{i} + u_y \hat{j} + u_z \hat{k} \right)$$
$$=\frac{\partial{\phi}}{\partial{x}}u_x +\frac{\partial{\phi}}{\partial{y}}u_y + \frac{\partial{\phi}}{\partial{z}}u_z$$
So, the vector u is decomposed into some coordinate axes, and the derivatives of the scalar function are found with respect to those three axes, and things are scalar-multiplied and added together. Your definition of the directional derivatives goes through a coordinate system, of your preference - through the definitions of the del operator.
But why not define this directly, in the usual manner for defining a derivative:
$$\lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}$$
Which is the same thing, without those coordinate decompositions,
$$d \phi = \frac{\partial{\phi}}{\partial{x}}dx + \frac{\partial{\phi}}{\partial{y}}dy + \frac{\partial{\phi}}{\partial{z}}dz$$
$$\mbox{and, } d \vec{u} = \hat{i} d(u_x) + \hat{j}d(u_y) + \hat{k}d(u_z)??$$
Because $d \phi$ looks well-defined to me. And so does $d \vec{u} = \hat{u} d \lambda$. So then, why don't we just say, without ambiguity,
$$\frac{d \phi}{d \vec{u}} \right| _r \equiv \lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}$$
(with the normalized direction $$\hat{u} = \vec{u} / \| \vec{u} \|$$ ) ?!
So as to eliminate any reference to the coordinate system?
I mean, when we're taking the usual partial derivatives, with respect to say the "x" axis, that's just a 'directional' derivative in the $\hat{x}$ direction. So, can we not generalize this to any (normalized) vector direction, and so define the derivative of a scalar function with respect to a vector?
That's my question - is this notation rigorous enough?
Last edited by a moderator: Sep 2, 2005
2. Sep 2, 2005
### rachmaninoff
Just to clarify, by "notation", I mean the idea of writing
$$\frac{d \phi}{d \vec{u}}$$
3. Sep 2, 2005
### Hurkyl
Staff Emeritus
As far as I know, the actual definition of the directional derivative is given as, for a unit vector u:
$$(\nabla_{\vec{u}} f)(\vec{a}) = \lim_{h \rightarrow 0} \frac{f(\vec{a} + h \vec{u}) - f(\vec{a})}{h}$$
So that it may actually defined even if the gradient is not.
4. Sep 2, 2005
### amcavoy
I don't know if this helps, but... (2 variables):
Let u=<a,b> so that the line going through u is given by the following:
$$x=x_0+at$$
$$y=y_0+bt$$
$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$
Now from above, it is clear that $\frac{dx}{dt}=a;\quad\frac{dy}{dt}=b$.
Substituting results in the following:
$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}=\frac{\partial f}{\partial x}a+\frac{\partial f}{\partial y}b=\nabla f\cdot \vec{u}$$
5. Sep 2, 2005
### Castilla
Apmcavoy, a bit more exactly...
Not $$\frac {\partial f}{\partial t}$$ but $$\frac {d f}{d t},$$ because in your example the only independent variable is $$t.$$
6. Sep 2, 2005
### rachmaninoff
Can you direct me to a reference which develops this?
7. Sep 2, 2005
### Hurkyl
Staff Emeritus
Advanced Calculus (Buck) defines it exactly as I have, except he uses the notation Dβ for the directional derivative in the direction β.
But frankly, I would expect just about any calculus text book that ever deals with vectors to have this definition in it.
8. Sep 3, 2005
### arildno
I've never seen a vector calculus book worth its salt lacking Hurkyl's definition of the directional derivitave.
In my opinion, it wouldn't be worth its salt if it lacked a definition like that..
9. Sep 3, 2005
### TD
I agree with Hurkyl, the directional derivative can intrinsicly be defined using the limit-definition, which seems fundamental enough. It's done like that in my Analysis course as well.
If you accept that as definition, it's not so hard to prove that the directional derivative can be calculated using the dot product with the gradient, so it follows as an equivalent way of calculating it.
This definition is given on Wikipedia: Directional Derivative as well.
10. Sep 3, 2005
### lurflurf
I prefer the notation
$$(u\cdot\nabla) f$$
for several reasons
first less confusing, if parenthesis are omited we do not confuse it with the gradient of an inner product, nor become confused should u depend on spatial variables. It is also less confusing should we take a directional derivative of a vector, or a higher order directional derivative.
$$D_u^n f =(u\cdot\nabla)^n f$$
I do not know what you mean by the definition of gradient has reference to the coordinate system. It need not. In any case various operators such as divergence, curl ect. should not be seen as functions of the del operator. Those sugestive notations should only remind us of a way the limit process the operator represents can be realized. i.e.
$$T(\nabla,a)=\lim_{V\rightarrow 0}\frac{1}{V}\oint T(n,a)ds$$
where T(nabla,a) is some combination of nabla and a,
and the integral is over the surface of a simple closed curve having volume V, and n is an outwardly pointed unit normal vector.
thus
$$(u\cdot\nabla)a=\lim_{V\rightarrow 0}\frac{1}{V}\oint (u\cdot n)a \ ds$$
$$(u\cdot\nabla)\times a=\lim_{V\rightarrow 0}\frac{1}{V}\oint (u\cdot n)\times a \ ds$$
and so forth
We can chose a particular curve to get a differential expression such as a cube.
of course we could aways use
$$(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b$$
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Koley & Roy (2018) – Estimating kinetic temperature from H I 21 cm absorption studies: correction for the turbulence broadening
Neutral hydrogen 21 cm transition is a useful tracer of the neutral interstellar medium. However, inferring physical condition from the observed 21 cm absorption and/or emission spectra is often not
Stanimirovic, Heiles & Kanekar (2007) – Properties of the Thinnest Cold HI clouds in the Diffuse Interstellar Medium
We have obtained deep HI observations in the direction of 22 continuum sources without previously detected cold neutral medium (CNM). 18 CNM clouds were detected with a typical HI column
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Grains are determinant for setting the chemical, physical, dynamical, and radiative properties of all the media in which they are present. Their influence depends on the grain composition, size, and
Aannestad & Purcell (1973) – Interstellar Grains
Not Available Aannestad, Per A.; Purcell, Edward M. 1973, Annual Review of Astronomy and Astrophysics, 11, 309 http://adsabs.harvard.edu/abs/1973ARA%26A..11..309A
Mohan, Dwarakanath & Srinivasan (2004) – A High Galactic Latitude HI 21cm-line Absorption Survey using the GMRT: I. Observations and Spectra
We have used the Giant Meterwave Radio Telescope (GMRT) to measure the Galactic HI 21-cm line absorption towards 102 extragalactic radio continuum sources, located at high (|b| > 15?) Galactic
Heiles (1997) – Tiny-Scale Atomic Structure and the Cold Neutral Medium
We consider the tiny-scale atomic structure (TSAS) on scales of tens of astronomical units, which has been detected by 21 cm absorption lines against quasars with VLBI techniques, against pulsars
Roy et al. (2013) – The temperature of the diffuse H I in the Milky Way – I. High resolution H I-21 cm absorption studies
We have carried out deep, high velocity resolution, interferometric Galactic H I-21 cm absorption spectroscopy towards 32 compact extragalactic radio sources with the Giant Metrewave Radio Telescope (GMRT) and the
Stanimirovic et al. (2010) – Arecibo Multi-epoch H I Absorption Measurements Against Pulsars: Tiny-scale Atomic Structure
We present results from multi-epoch neutral hydrogen (H I) absorption observations of six bright pulsars with the Arecibo telescope. Moving through the interstellar medium (ISM) with transverse velocities of 10-150
Heiles & Troland (2003) – The Millennium Arecibo 21 Centimeter Absorption-Line Survey. II. Properties of the Warm and Cold Neutral Media
We use the Gaussian fit results of Paper I to investigate the properties of interstellar H I in the solar neighborhood. The warm and cold neutral media (WNM and CNM)
Colgan, Salpeter & Terzian (1988) – H I emission-absorption studies with high-velocity resolution
The properties of interstellar H I are investigated using high-velocity resolution, Arecibo emission-absorption spectra in the direction of 40 radio continuum sources, including 21 at low Galactic latitude. Comparison of
Payne, Salpeter & Terzian (1983) – Interpretation of neutral hydrogen spin temperature measurements
Arecibo 3-arcsec beam neutral hydrogen emission-absorption observations are analyzed, and spin temperature results in the direction of ‘small’ and ‘large’ extragalactic sources are found to be statistically similar. The variations
Murray et al. (2015) – The 21-SPONGE HI Absorption Survey I: Techniques and Initial Results
We present methods and results from “21 cm Spectral Line Observations of Neutral Gas with the EVLA” (21-SPONGE), a large survey for Galactic neutral hydrogen (H i) absorption with the
Braun & Kanekar (2005) – Tiny H I clouds in the local ISM
We report deep, high spectral resolution WSRT H I 21 cm observations of four high latitude compact radio sources, that have revealed a new population of tiny, discrete clouds in
Mohan, Dwarakanath & Srinivasan (2004) – A High Galactic Latitude HI 21cm-line Absorption Survey using the GMRT: II. Results and Interpretation
We have carried out a sensitive high-latitude (|b| > 15?) HI 21 cm-line absorption survey towards 102 sources using the GMRT. With a 3sigma detection limit in optical depth of
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# If $a^b\equiv a^c\equiv 1\pmod{m}$ and $g=\gcd(b,c)$ prove $a^g\equiv 1\pmod{m}$
Let $$b,c,$$ and $$m$$ be positive integers and suppose $$a$$ is relatively prime to $$m$$. Furthermore, assume $$a^b\equiv a^c\equiv 1\pmod{m}$$ and that $$g=\gcd(b,c)$$.
I know that since $$g=\gcd(b,c)$$ it follows that $$g|b$$ and $$g|c$$. So one could write $$b=gx$$ and $$c=gy$$ for some $$x,\;y\;\epsilon\;\mathbb Z$$. I then have $$a^b\equiv a^{gx}\equiv (a^{g})^x\equiv 1\pmod{m}$$. I don't quite understand how to prove $$a^g\equiv 1\pmod{m}$$. I'm wondering if I should be utilizing the fact that the order of $$a$$ divides any power, $$k$$, such that $$a^k\equiv 1\pmod{m}$$. Or if I was on the right track just need to manipulate $$b$$ and $$c$$ slightly differently. Any help would be much appreciated.
Hint $$\,\ a^{\large b}\equiv 1\equiv a^{\large c}\iff {\rm ord}\,a\mid b,c\iff {\rm ord}\,a\mid \gcd(b,c)\iff a^{\large \gcd(b,c)}\equiv 1$$
• See here for the ${\rm ord}$ law used above, and see here for the $\gcd$ law used. We could instead use the Bezout identity for the gcd, but that is less conceptual (and less general). – Bill Dubuque Feb 23 at 22:59
• Can I use the fact that g=bx+cy to rewrite this as $a^g \equiv a^{bx+cy}\equiv a^{bx}a^{cy}\equiv (a^b)^x(a^c)^y\equiv 1^x*1^y \equiv 1 \pmod{m}$ – joseph Feb 23 at 23:10
• @mjoseph Yes, that's the Bezout-based method that I alluded to above. Note that some exponents are negative so you need to remark that $a$ is invertible because $\ldots$ – Bill Dubuque Feb 23 at 23:14
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## Wikipedia - 6.022×10^23 (en) Wikipedia - Amedeo Avogadro (pt) Wikipedia - Atomic mass unit (en) Wikipedia - Avogadro constant (en) Wikipedia - Chimica (it) Wikipedia - Costante di Avogadro (it) Wikipedia - Discusión:Constante de Avogadro (es) Wikipedia - Discussioni progetto:Chimica (it) Wikipedia - Hằng số Avogadro (vi) Wikipedia - Numărul lui Avogadro (ro) Wikipedia - Talk:Avogadro constant (en) Wikipedia - ثابت أفوجادرو (ar) Wikipedia - عدد آووگادرو (fa) Wikipedia - アボガドロ定数 (ja) Avogadro constant, $$N_{\text{A}}$$, $$L$$
https://doi.org/10.1351/goldbook.A00543
The Avogadro constant, $$N_{\text{A}} = 6.022\ 140\ 76 \times 10^{23}\ \text{mol}^{-1}$$, is a proportionality constant between the quantity @[email protected] ($$n$$, with unit mole) and the quantity for counting @[email protected] ($$N$$ with unit one, symbol 1), $$N = N_{\text{A}}\ n$$.
Notes:
1. In the revision of the International System of Units (SI) of 2019 the Avogadro constant became one of the seven defining constants with the exact value $$6.022\ 140\ 76 \times 10^{23}\ \text{mol}^{-1}$$.
2. The Avogadro number is the exact numerical value of the Avogadro constant, $$6.022\ 140\ 76 \times 10^{23}$$
3. While the symbol $$N_{\text{A}}$$ is used to honour Amedeo Avogadro, the symbol $$L$$ is used in honour of Josef Loschmidt.
Sources:
BIPM, Le Système international d’unités (SI), SI Brochure, 9e ed. (2019), p. 17 [Terms] [Book]
BIPM, The International System of Units, SI Brochure, 9th ed. (2019), p. 129 [Terms] [Book]
Green Book, 3rd ed. [Terms] [Book]
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# Which interrogative pronoun should replace the blank space in the following sentence?: With _____ did you go fishing today?
Mar 15, 2018
whom
#### Explanation:
Use whom after a preposition. A preposition is a word that tells you where something is in relation to something else. Other prepositions are to, from, by, of, above, inside, etc. “With” is your preposition in this sentence. Saying “with who?” is not correct.
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# Plugging a circle into real polynomial
I have written some code (attached below) that generates a random real polynomial $P$ degree and coefficient within some range. I then plotted and looked at $im(P(S^1))$ with $S$ being the unit circle in the complex plane.
To my surprise I got pictures with some interesting properties. (Interesting at least for me, (pictures are viewable below))
I noticed:
1. The figure is connected (not too surprising)
2. The figure is self-intersecting itself and an intersection seems to be always crossed exactly twice. (Might be wrong, considering rounding errors etc.)
3. The intersection points $z_i$ seem to have Im$(z_i) = 0$
Point 1. follows directly from S being compact and $P$ continuous. However I find it harder to justify 2 and 3, especially I can make such a claim, maybe I were just lucky with my numbers. Therefore I would appreciate it, if someone could clarify points 2 and 3 to me, if those statements are correct and especially why. As always thanks in advance.
'''
Created on 16 Sep 2017
@author: Imago
'''
import pylab
import cmath as c
import matplotlib.cm as cm
import matplotlib.pyplot as plt
import numpy as np
import random as r
NUMBER_OF_POINTS = 0.0001
N = NUMBER_OF_POINTS
# generate a random polynominal with degree deg, and integer coefficients in range (min, max)
def grp(min, max, deg):
l = list()
for i in range(deg):
l.append(r.randint(min, max))
return np.poly1d(np.array(l))
# give me Re(z), Im(z)
def split(z):
return complex(z).real, complex(z).imag
# my polynominal
f = grp(-3, 3, 10)
print('Polynominal')
print(f)
# interval of numbers between 0 and 1.
I = np.arange(0, 1, N)
# skip the next 6 lines, if you not want to expande the code
X = list()
Y = list()
n = 1
k = 0
X.append(list())
Y.append(list())
# create the points for plotting
for x in I:
z = R * np.exp(x * 2 * np.pi * 1j)
v = f(z)
X[k].append(complex(v).real) # k = 0
Y[k].append(complex(v).imag)
# colour, plot and show the figure
colors = iter(cm.rainbow(np.linspace(0, 1, n))) # n = 1
for c in colors :
plt.scatter(X[k], Y[k], c)
k = k + 1
plt.show()
• Beautiful and engaging stuff stuff! A couple of remarks:: (i.): in the text, you refer to looking at $im P(S)$; I guess that means the image or $S$ under $P$, right? I first thought it meant the imaginary part of $P(S)$, but that doesn't seem right . . . ; (ii.) it might be very helpful to those of us who find your work engaging if you reported the degrees and coefficients of the polynomials you used; it looks to me at first glance like your codes reports this information. Looking forward to hearing form you again . . . I may have more questions later on . . . Cheers! – Robert Lewis Sep 16 '17 at 20:44
Ad 2: A polynomial with threefold intersection would be $$f(x)=(x^3-1)x.$$
Ad 3: It is clear that the image is symmetric with respect to the $x$-axis, which automatically causes intersection points there as soon as the curve crosses the $x$-axis non-perpendicularly. Nevertheless, $f(x)=x^2$ can be said to pass twice hrough each of its point (even though this might be considered cheating). Maybe try $$f(x)=x^3+(x^{9}-1)x$$
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# Overview
Logistic regression is used for classification. The independent variable of the data is quantitative and the dependent variable of the data is binary (0 or 1) i.e. in a class or not. Instead of modeling the response (0 or 1), the logistic model’s dependent variable is the probability that a data point belongs in the class.
Replace this text with your caption
The logistic function is fit to the data. Regerring to the plot above, the model (blue curve) can take in the balance and output the probability of default. Notice that it does tell you if the person defaults or not; it tells you the probability of default. In order to perform classification, you must choose a cutoff, say 0.5, where all values above are predicted as default and all below are predicted as not default.
# Starting Out
1. Introducion to Statistical Learning (James 2013) - section 4.3
2. Advanced Data Analysis from an Elementary Point of View (Dobson 2001) - chapter 7
3. An Introduction to Generalized Linear Models - chapter 7
# More Detail
## What function are we fitting?
Where $$y\in\lbrace0,1\rbrace$$, we are fitting the logistic function,
${p(y=1;\beta)} = \frac{1}{1 + e^{-(\beta_0+\beta_1x)}}$
This is the probability that $$y=1$$ parameterized by $$\beta_i$$, also written as $$p(x)$$ since the probability defends on $$x$$. In order to fit this function to the data, we must find the parameters $$\beta_i$$ that maximize the likelihood function, $$L(\beta;y)$$. The likelihood function is algebraically the same as the joint probability density function $$f(y;\beta)$$ except that the change in notation reflects a shift of emphasis from the random variables $$\textit{y}_i$$, with parameters $$\beta$$ fixed, to parameters $$\beta$$ with $$\textit{y}_i$$ fixed. The likelihood function is given by
${L(\beta)} = \prod_{i=1}^np(x_i)^{y_i}(1-p(x_i))^{1-y_i}$
We could substitute the equation for p(x) from above but we will not here. The form of the likelihood function, which is the same as the distribution of the response, reflects the fact that we assume the response follows a binomial distribution. Finding the parameters $$\beta_i$$ that maximizes the likelihood function involves numerical analysis that are outside the scope of this paper and are not a concern for a modeler.
## Making Predictions
As stated in the overview, logistic regression gives a model whose output is the probability of y=1. To actually predict if the instance is in the class or not, the user must choose a cutoff value, say 0.5. Then we can say that $$y = 1$$ for $$p(x)>0.5$$. In the case of only one predictor variable, this corresponds to choosing a cutoff x value. With two predictors, a line is chosen that divides the values in and out of the class. And so on for higher dimensions. This is called the decision boundary, and the classification becomes clear.
## Multiple Predictors
Logistic regression can be used when there are n predictor variables, called multiple logistic regression. The curve that is fit to the data now takes the form
${p(y=1;\beta)} = \frac{1}{1 + e^{-(\beta_0+\beta_1x_1+\cdots+\beta_nx_n)}}$
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# The number of U.S. domestic trips, in millions, t years after 1994 can be estimated by p(t) = 8.8t + 954. what do the numbers 8.8 and 954 signify?
Jun 27, 2017
I tried this:
#### Explanation:
$954$ will be the initial number (in millions) of trips at, say, $t = 1994$.
$8.8$ will be in millions/year representing the increase of number of trips each year, so that after $1$ year you'll have:
$8.8 \cdot 1 + 954 = 962.8$ millions trips;
and after $2$ years:
$8.8 \cdot 2 + 954 = 971.6$ millions trips;
etc.
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# Green function of the first quadrant
Find the Green function of the first quadrant $$x_1>0, x_2>0$$.
HINT: Use the Green function of the half space $$\Omega:=\left\{x\in\mathbb{R}^n : x_n > 0\right\}$$ which is given by\begin{align*} G_n(x,y)&:=E_n(x_1-y_1,\cdots,x_{n-1}-y_{n-1},x_n-y_n)\\ &\quad\ -E_n(x_1-y_1,\cdots,x_{n-1}-y_{n-1},x_n+y_n). \end{align*}
How I can find the Green function of the first quadrant? The difference to the half space is that I have another boundary…
For $$1 \leqslant k \leqslant n$$, define$$τ_k(x_1, \cdots, x_n) = (x_1, \cdots, x_{k - 1}, -x_k, x_{k + 1}, \cdots, x_n). \quad \forall x_1, \cdots, x_n \in \mathbb{R}$$ And for $$1 \leqslant k_1 < \cdots < k_m \leqslant n$$, define$$τ_{k_1, \cdots, k_m} = τ_{k_1} \circ \cdots \circ τ_{k_m}.$$ Thus the Green function of $$\{(x_1, \cdots, x_n) \in \mathbb{R}^n \mid x_n > 0\}$$ is $$G_n(x, y) = E_n(x - y) - E_n(x - τ_n(y))$$. Now define\begin{align*} H_n(x, y) &= (E_n(x - y) - E_n(x - τ_1(y))) - (E_n(x - τ_2(y)) - E_n(x - τ_1(τ_2(y))))\\ &= E_n(x - y) - E_n(x - τ_1(y)) - E_n(x - τ_2(y)) + E_n(x - τ_{1, 2}(y)) \end{align*} for $$x, y \in D = \{(x_1, \cdots, x_n) \in \mathbb{R}^n \mid x_1, x_2 > 0\}$$. It is easy to see that$$Δ_x H_n(x, y) = -δ(x - y). \quad \forall x, y \in D$$ Note that for any $$1 \leqslant k \leqslant n$$ and $$x, y \in \mathbb{R}^n$$,$$τ_k(τ_k(x)) = x, \quad τ_k(x + y) = τ_k(x) + τ_k(y), \quad E_n(x) = E_n(-x) = E_n(τ_k(x)),$$ thus\begin{align*} H_n(x, y) &= E_n(x - y) - E_n(x - τ_1(y)) - E_n(x - τ_2(y)) + E_n(x - τ_{1, 2}(y))\\ &= E_n(x - y) - E_n(τ_1(x) - y) - E_n(τ_2(x) - y) + E_n(τ_{1, 2}(x) - y). \end{align*}
For $$x^* \in \partial D \cap \{x_1 = 0\}$$, because $$τ_1(x) → τ_1(x^*) = x^*,\ τ_2(x) → τ_2(x^*)\ (x → x^*,\ x \in D)$$, then $$τ_{1, 2}(x) = τ_2(τ_1(x)) → τ_2(x^*)\ (x → x^*,\ x \in D)$$, which implies $$\lim\limits_{\substack{x → x^*\\x \in D}} H_n(x, y) = 0$$. Analogously, $$\lim\limits_{\substack{x → x^*\\x \in D}} H_n(x, y) = 0$$ for $$x^* \in \partial D \cap \{x_2 = 0\}$$. Thus, $$H_n(x, y)|_{x \in \partial D} = 0$$. Therefore, $$H_n(x, y)$$ is the Green function of $$D$$.
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# Harmonic Oscillator
## Homework Statement
I am having issues with d) and would like to know if I did the a, b, and c correctly. I have tried to look online for explanation but with no success.
A harmonic oscillator executes motion according to the equation x=(12.4cm)cos( (34.4 rad /s)t+ π/5 ) .
a) Determine the amplitude of the oscillation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
d) Determine when the object is at its equilibrium position.
max v=Aw
T=1/f
## The Attempt at a Solution
a) Determine the amplitude of the oscillation.
Amplitude would just be 12.4cm, we can take it straight out from the equation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
Here we know that 2π rad is one turn so 34.4 rad is 5.48 turns.
So we have 5.48 turns per second.
T=1/f=1/5.48=0.183secs
d) Determine when the object is at its equilibrium position.
I know that the object is in equilibrium when x=0, so
0=(12.4cm)cos( (34.4 rad /s)t+ π/5 )
I may be missing some algebra skills, but I even try to compute it online and it wields no solution. What am I doing wrong?
gneill
Mentor
Your work on parts (a) through (c) looks fine.
For part (d), consider what the argument of the cosine function needs to be for the cosine to be zero. (Hint: there are many such angles)
So cos( (34.4 rad /s)t+ π/5 ) need to equal 0?
(34.4rad/s)t+π/5 has to be equal to π/2 or 3π/2?
That is going to wield t= 0.02739s and 0.1187s, does not feel right though because a period takes 0.183secs.
gneill
Mentor
So cos( (34.4 rad /s)t+ π/5 ) need to equal 0?
(34.4rad/s)t+π/5 has to be equal to π/2 or 3π/2?
Yes.
In fact, the cosine will be zero every time its argument is equivalent to π/2 or 3π/2. You should be able to write it as a function of n, where n = 0,1,2,.... Or you can solve for the first instance (n = 0 so that the argument is π/2) and then it will happen every half period after that.
alex91alex91alex
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# how to sort a multi-level pandas data-frame by a particular column?
I want to sort a multi-index pandas dataframe by a column but don't want the entire dataframe to be sorted at once. But rather want to sort by one of the indices. Here is an example of what I mean: Below is an example of a multi-index dataframe.
first second
bar one 0.361041
two 0.476720
baz one 0.565781
two 0.848519
foo one 0.405524
two 0.882497
qux one 0.488229
two 0.303862
What I want to do is to get the following:
first second
bar one 0.476720
two 0.361041
baz one 0.848519
two 0.565781
foo one 0.882497
two 0.405524
qux one 0.488229
two 0.303862
These are generated by hand to show what I want. Notice that the second dataframe is not completely sorted. But within each multi-index, it is sorted in descending order. I have a large dataframe. Is there a simpler way of do it (such as a function) instead of grouping the dataframe based on the indices, and then concatenate the individually sorted dataframes.
• I'm not sure if this is a sorting problem, or a data manipulation problem. Your third column doesn't match up anymore to the second one. Is that what you want, or is that a typo? – S van Balen Feb 27 '20 at 9:07
# TO BE EDITTED
There would definitely be more optimal ways for addressing this problem. But this is a simple approach.
import pandas as pd
df = pd.DataFrame({'first': ['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
'second': ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two'],
'value': [0.361041, 0.476720, 0.565781, 0.848519, 0.405524, 0.882497, 0.488229, 0.303862]})
first second value
0 bar one 0.361041
1 bar two 0.476720
2 baz one 0.565781
3 baz two 0.848519
4 foo one 0.405524
5 foo two 0.882497
6 qux one 0.488229
7 qux two 0.303862
import itertools
#storing unique values of the first column.
unique = list(df['first'].unique())
#unique
#The following line, gives the row indices of the dataframe with specific value on the index ('first') column.
indices =[df.index[df['first'] == unique[i]].tolist() for i in range(len(unique))]
#Then this line sorts each pair (it can be more than 2) of the elements from the value column.
sorted_values = []
for [i,j] in indices:
sorted_values.append(sorted([df['value'].iloc[i],df['value'].iloc[j]]))
#The following line returns the row index of a specific value of 'value' column.
#df['value']
[df['value'] == 0.361041].index.tolist()
#This ravels the sorted_list and gives the goal order of the values in 'value' column
values = []
values = list(itertools.chain(*sorted_values))
#values
#output
[0.361041, 0.47672, 0.565781, 0.848519, 0.405524, 0.882497, 0.303862, 0.488229]
Now that we have the actual order of elements in the value column, we can rebuild the whole dataframe based on this column.
Here is one approach...
import pandas as pd
df = pd.DataFrame({'first': ['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
'second': ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two'],
'value': [0.361041, 0.476720, 0.565781, 0.848519, 0.405524, 0.882497, 0.488229, 0.303862]})
df = df.set_index(['first', 'second'])
Let us say, sort (descending) by level 1 which is column 'second'.
df.sort_index(level=[1],ascending=[False])
Check out more on Pandas sort here
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## Files in this item
FilesDescriptionFormat
application/pdf
9114399.pdf (6MB)
(no description provided)PDF
## Description
Title: Graphs on which dihedral, quaternion, and abelian groups act vertex and/or edge transitively and applications to tensor products Author(s): Sanders, Robin Sue Doctoral Committee Chair(s): Weichsel, Paul M. Department / Program: Mathematics Discipline: Mathematics Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Mathematics Abstract: The graphs on which dihedral, quaternion, and abelian groups act vertex and/or edge transitivity are completely characterized. The vertex transitive graphs belong to one of three families--the well known circulant graphs, the metacirculant graphs constructured by Alspach and Parsons, and a family constructed using a generalization of Alspach's and Parsons' construction. If one of the selected groups acts both vertex and edge transitively on a graph, then it is shown that the graph is the disjoint union of some number of copies of a given cycle. The graphs on which one of the selected groups acts edge transitively but not vertex transitively fall into two broad--disjoint copies of a complete bipartite graph and disjoint copies of a "pseudo-cycle," a graph which is related to the tensor product of a complete bipartite graph and an even cycle.The full automorphism group of a pseudo-cycle is then found. It is shown that this group depends upon several parameters used in creating the pseudo-cycle, and in the most general case, the automorphism group is not quite what one might expect it to be. Next, these results are used to find the full automorphism group of the tensor product of any cycle with a complete bipartite graph. Finally, it is shown that the technique used to find the full automorphism group of a pseudo-cycle can be used to find the automorphism groups of certain tensor products. In particular, conditions on a graph $\Gamma$ are found that insure that the full automorphism groups of the tensor products of $\Gamma$ with the complete graph $K\sb{n}$, the complete graph with loops $K\sbsp{n}{\*}$, and the complete bipartite graph $K\sb{n,n}$ are as small as possible. Issue Date: 1990 Type: Text Language: English URI: http://hdl.handle.net/2142/19384 Rights Information: Copyright 1990 Sanders, Robin Sue Date Available in IDEALS: 2011-05-07 Identifier in Online Catalog: AAI9114399 OCLC Identifier: (UMI)AAI9114399
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# Basic Skills in Higher Mathematics Robert Glen Adviser in Mathematics Mathematics 1(H) Outcome 3.
## Presentation on theme: "Basic Skills in Higher Mathematics Robert Glen Adviser in Mathematics Mathematics 1(H) Outcome 3."— Presentation transcript:
Basic Skills in Higher Mathematics Robert Glen Adviser in Mathematics Mathematics 1(H) Outcome 3
Mathematics 1(Higher) Outcome 2 Use basic differentiation Differentiation f (x) f (x)f (x) f (x) dy dx
Mathematics 1(Higher) Outcome 2 Use basic differentiation PC Index f (x) f (x)f (x) f (x) dy dx PC(a) Basic differentiation PC(b) Gradient of a tangent PC(c) Stationary points Click on the PC you want
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x dy dx f (x) f (x)f (x) f (x) PC(a) - Basic differentiatio n
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x PC(a) - Basic differentiation 1 Simple functions 2 Simple functions multiplied by a constant 3 Negative indices 4 Fractional indices 6 Sums of functions (simple cases) 7 Sums of functions (negative indices) Click on the section you want 8 Sums of functions ( algebraic fractions) 5 Negative and fractional indices with constant
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 1 Simple functions
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Every function f(x) has a related function called the derived function. The derived function is written f (x) “f “f dash x”x” The derived function is also called the derivative. To find the derivative of a function you differentiate the function. Some examples f(x) x 3 x 6 x 3 3x 2 6x 5 1 0 f (x)f (x) x 10 10x 9 (= x 1 ) (= 3x 0 )
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Rule No. 1 for differentiation If f(x) = x n, then f (x) = nx n -1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = x 4 2 f(x) = x 5 3 g(x) = x 8 4 h(x) = x 2 5 f(x) = x 12 6 f(x) = x Here are the answers 1 f (x) = 4x 3 2 f (x) = 5x 4 3 g(x) = 8x 7 4 h(x) = 2x 5 f (x) = 12x 11 6 f (x) = 1 7 f(x) = 5 7 f (x) = 0
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Continue with Section 2 End of Section 1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 2 Simple functions multiplied by a constant
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples f(x) 2x 3 3x 6 2x 10 5x 2 3x 2 2 10x 9 3 6x 5 5 1 6x 2 18x 5 20x 9 5 f (x)f (x)
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Rule No. 2 for differentiation If f(x) = ax n, then f (x) = anx n -1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = 3x 4 2 f(x) = 2x 5 3 g(x) = ½ x 8 4 h(x) = 5x 2 5 f(x) = ¼ x 12 6 f(x) = 8x Here are the answers 1 f (x) = 12x 3 2 f (x) = 10x 4 3 g(x) = 4x 7 4 h(x) = 10x 5 f (x) = 3x 11 6 f (x) = 8 7 f(x) = 10 6 f (x) = 0
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Continue with Section 3 End of Section 2
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 3 Negative indices
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Rule No. 1 for differentiation If f(x) = x n, then f (x) = nx n -1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 1 f(x) = = x -3 f (x) = -3 -4-3-2 0 -5 -3 -4 Note: This is an example of using Rule No.1 with a negative index. = x -4 x ?x ?
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 2 f(x) = = x -1 f (x) = -1 -2 0 -2 Note: This is an example of using Rule No.1 with a negative index. = x -2 x ?x ?
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = x -2 2 f(x) = x -4 3 g(x) = 4 h(x) = 5 f(x) = Here are the answers 1 f (x) = -2x -3 2 f (x) = -4x -5 3 g(x) = -5x -6 = 4 h(x) = -1x -2 = 5 f (x) = -10x -11 =
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Continue with Section 4 End of Section 3
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 4 Fractional indices
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Rule No. 1 for differentiation If f(x) = x n, then f (x) = nx n -1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 3 f(x) = f (x) = 0 1 Note: This is an example of using Rule No.1 with a fractional index. = x ?x ?
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 4 f(x) = f (x) = 0 1 Note: This is an example of using Rule No.1 with a fractional index. = x ?x ?
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = 2 f(x) = 3 g(x) = 4 h(x) = Here are the answers 1 f (x) = 2 f (x) = 3 g(x) = 4 h(x) =
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Continue with Section 5 End of Section 4
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 5 Negative and fractional indices with constant
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Rule No. 2 for differentiation If f(x) = ax n, then f (x) = anx n -1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 5 f(x) = = 2x -3 f (x) = -6 -4-3-2 0 -5 -3 -4 Note: This is an example of using Rule No.2 with a negative index. = x -4 x ?x ?
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 6 f(x) = = 5x -1 f (x) = -2 0 -2 Note: This is an example of using Rule No.2 with a negative index. = x -2 x ?x ? -5
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = 3x -2 2 f(x) = 5x -4 3 g(x) = 4 h(x) = 5 f(x) = Here are the answers 1 f (x) = -6x -3 2 f (x) = -20x -5 3 g(x) = -10x -6 = 4 h(x) = -2x -2 = 5 f (x) = -30x -11 =
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 7 f(x) = f (x) = 0 1 Note: This is an example of using Rule No.2 with a fractional index. = x ?x ? 6 ½ = 3 3
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples 8 f(x) = f (x) = 0 1 Note: This is an example of using Rule No.2 with a fractional index. = x ?x ? 9 2/3 = 6
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = 2 f(x) = 3 g(x) = 4 h(x) = Here are the answers 1 f (x) = 2 f (x) = 3 g(x) = 4 h(x) =
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x When a function is given in the form of an equation, the derivative is written in the form 1 f(x) = 3x 2 f (x) = 6x y = 3x 2 = f (x) 2 f(x) = 4x -3 f (x) = -12x -4 y = 4x -3 = 6x = -12x -4 Examples
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Continue with Section 6 End of Section 5
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 6 Sums of functions (simple cases)
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples f(x) = 5x 2 - 3x + 1 f (x) = g(x)g(x)h(x)h(x)k(x)k(x) - 3+ 010x = 10x - 3 g(x) = (x + 3)(x - 2) = x 2 + x - 6 g (x) = 2x2x + 1 1 2
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Rule No. 3 for differentiation If f(x) = g(x) + h(x) + k(x) +……..., then f (x) = g (x) + h (x) +k (x) +..
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = x 2 + 7x - 3 2 f(x) = 3x 2 - 4x + 10 3 g(x) = x(x 2 + x - 5) 4 h(x) = 4x 3 - 10x 2 5 f(x) = x 5 (7 - 5x 2 ) 6 f(x) = x 3 + x 2 + x + 1 Here are the answers 1 f (x) = 2x + 7 2 f (x) = 6x - 4 3 g(x) = 3x 2 + 2x - 5 4 h(x) = 12x 2 - 20x 5 f (x) = 35x 4 - 35x 6 6 f (x) = 3x 2 + 2x + 1 7 f(x) = ¼ x 4 +½ x 2 7 f (x) = x 3 + x
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Now do Section A2 on page 33 of the Basic Skills booklet End of Section 6
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 7 Sums of functions (negative indices)
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples f(x) = f (x) = = - - 3x = 2x -2 - 5x -1 - 3x -4x -3 + 5x -2 - 3 - + y = + - 1 = x -2 - x -1 + 1 = -2x -3 + x -2 + 0 = - + 1 2
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = 2 y = 3 y = 4 g(x) = Here are the answers 1 f (x) = 2 3 4 g(x) = - 2x + x 2 - x - + 3x 2 + - - - 2 = - + 2x - 1 = - + + 6x - - +
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Continue with Section 8 End of Section 7
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x f (x) f (x)f (x) f (x) dy dx PC(a) - Basic differentiation 8 Sums of functions (algebraic fractions)
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x More examples f(x) = 1 = + - = x f (x) = 1 3x -2 = 1 + 2 y = = + - = 1 = 0 - 3x -2 + 4x -3 = - + + 0 + 5 - 3x -1 + + 3x -1 - 2x -2
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Differentiate each of these functions. 1 f(x) = 2 y = 3 g(x) = 4 y = Here are the answers 1 f (x) = 2x - 2 2 3 g(x) = 1 + 4 = 1 = -+
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(a) Differentiate a function reducible to a sum of powers of x Now do Section A3 on page 33 of the Basic Skills booklet End of Section PC(a)
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation dy dx f (x) f (x)f (x) f (x) PC(b) - Gradient of a tangent
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Rule No. 4 for differentiation The gradient of a tangent to the curve y = f (x) is f (x) or
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Find the gradient of the tangent to the curve y = x2 x2 - 5x at each of the points A and B.B. Gradient of tangent = = 2x When x = 3, = 2(3) = 1 When x = -1, = 2(-1) = -7 y x A (3, -6) B (-1, 6) y = x 2 - 5x So gradient of tangent at A is 1 and gradient of tangent at B is -7 m = 1 m = -7 A B Example 1 - 5
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Example 1 (continued) y x A (3, -6) B (-1, 6) y = x 2 - 5x m = 1 m = -7 m = 1 Point on line is (3, -6) Equation of tan is y - b = m(x m(x - a)a) y y y = Find the equation of the tangent to the curve y = x 2 - 5x at each of the points A and B. Tangent at A - (-6) = 1 (x - 3) + 6 = x - 3 x- 9 y = x - 9
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Find the equation of the tangent to the curve y = x 2 - 5x at each of the points A and B. y x A (3, -6) B (-1, 6) y = x 2 - 5x m = 1 m = -7 Tangent at B m = -7 Point on line is (-1, 6) Example 1 (continued) Equation of tan is y - b = m(x m(x - a)a) y - y - 6 = y = NB 3 negatives 6 -7 (x - (-1)) = -7x -7x - 1 y = x - 9 y = -7x - 1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Find the gradient of the tangent to the curve y = x3 x3 - 5x + 3 at each of the points P and Q.Q. Gradient of tangent = = 3x 2 - 5 When x = 1, = 3(1) 2 - 5 = -2 When x = -2, = 3(-2) 2 - 5 = 7 y x P (1, -1) Q (-2, 5) So gradient of tangent at P is -2 and gradient of tangent at Q is 7 m = -2 m = 7 P Q Example 2 y = x 3 - 5x + 3
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Example 2 (continued) y x m = -2 Point on line is (1, -1) Equation of tan is y - b = m(x m(x - a)a) y - y y = y = x 3 - 5x + 3 P (1, -1) Q (-2, 5) m = -2 m = 7 Tangent at P Find the equation of the tangent to the curve y = x 3 - 5x + 3 at each of the points P and Q. (-1) = -2 (x - 1) + 1 = -2x+ 2 -2x+ 1 y = -2x + 1
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Example 2 (continued) y x Tangent at Q m = 7 Point on line is (-2, 5) Equation of tan is y - b = m(x m(x - a)a) y - y - 5 = y = Find the equation of the tangent to the curve y = x 3 - 5x + 3 at each of the points P and Q. y = x 3 - 5x + 3 P (1, -1) Q (-2, 5) m = -2 m = 7 5=7 (x - (-2)) 7x7x + 14 7x7x+ 19 y = -2x + 1 y = 7x + 19
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation 1 Find the gradient of the tangent to the curve y = x 2 + 5 at each of the points A and B.B. Find the equation of each tangent. y x A (2, 9) B (-4, 21) Tangent at A m = 4 Equation is y = 4x 4x + 1 Tangent at B m = -8 Equation is y = -8x - 11 Answers y = x 2 + 5 = 2x
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation 2 Find the gradient of the tangent to the curve y = (3 + x)(3 - x)x) at each of the points E and F.F. Find the equation of each tangent. y x F Tangent at E m = -4 Equation is y = -4x + 13 Tangent at F (-3, 0) m = 6 Equation is y = 6x 6x + 18 Answers y = (3 + x)(3 - x) E (2, 5) = -2x
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation 3 Find the gradient of the tangent to the curve y = x4 x4 - 5x 2 + 4 at each of the points A, B and C.C. Find the equation of each tangent. y x A (0, 4) B (-2, 0) Tangent at A m = 0 Equation is y = 4 Tangent at B m = -12 Equation is y = -12x - 24 Answers C (1, 0) Tangent at C m = -6 Equation is y = -6x + 6 y = x 4 - 5x 2 + 4 = 4x 3 - 10x
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Now do Sections B1and B2 on page 37 of the Basic Skills booklet End of Section PC(b)
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. dy dx f (x) f (x)f (x) f (x) PC(c) - Stationary points
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. Rule No. 5 for differentiation Stationary points occur when f (x) = 0 or = 0
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. To determine the nature of a stationary point, find out the gradient of the tangent before and after the stationary point. before after y x + ve - ve = 0 x before after + 0 - slope This is a maximum turning point
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. To determine the nature of a stationary point, find out the gradient of the tangent before and after the stationary point. before after y x - ve + ve = 0 x before after - 0 + slope This is a minimum turning point
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. To determine the nature of a stationary point, find out the gradient of the tangent before and after the stationary point. x before after - 0 slope This is a minimum turning point x before after + 0 - slope This is a maximum turning point +
Mathematics 1(Higher) Outcome 3 Use basic differentiation y = x 2 - 6x + 5 = 2x For stationary values, = 0 ie 2x - 6 = 0 x = 3 When x = 3, (3) 2 = 9 = -4 So a stationary point occurs at (3, -4). y = Find the stationary point on the curve with equation y = x2 x2 - 6x + 5. Example 1 Using differentiation determine its nature. PC(c) Determine the coordinates of the stationary points on a curve……. -6(3) + 5 -18 + 5 - 6
Mathematics 1(Higher) Outcome 3 Use basic differentiation Find the stationary point on the curve with equation y = x 2 - 6x + 5. Example 1 (continued) A stationary point occurs at (3, -4). x 3 - - 0 slope (3, -4) is a minimum turning point. Using differentiation determine its nature. 3 3 + = 2x - 6 + PC(c) Determine the coordinates of the stationary points on a curve…….
Mathematics 1(Higher) Outcome 3 Use basic differentiation = x 2 For stationary values, = 0 ie x 2 - 2x - 3 = 0 ( )( ) = 0 When x = 3, 1/3(3) 3 So stationary points occur at (3, 1) and (-1, ) y = Find the stationary points on the curve with equation y = 1/3x 3 - x2 x2 - 3x + 10. Example 2 Using differentiation determine their nature. - 2x - 3 x = When x = -1, y = 1/3(-1) 3 PC(c) Determine the coordinates of the stationary points on a curve……. x = x - 3 x + 1 -32-32 -3(3) +10 = -(-1) 2 -3(-1)+10 = 3,
Mathematics 1(Higher) Outcome 3 Use basic differentiation Find the stationary points on the curve with equation y = 1/3x 3 - x 2 - 3x + 10. Example 2 Using differentiation determine their nature. x 3 - - 0 slope (3, 1) is a minimum turning point. 3 3 + + A stationary point occurs at (3, 1). x2x2 - 2x - 3 = PC(c) Determine the coordinates of the stationary points on a curve…….
Mathematics 1(Higher) Outcome 3 Use basic differentiation Find the stationary points on the curve with equation y = 1/3x 3 - x 2 - 3x + 10. Example 2 Using differentiation determine their nature. x -1 - + 0 slope (-1, ) is a maximum turning point. -1 + - A stationary point occurs at (-1, ). x2x2 - 2x - 3 = PC(c) Determine the coordinates of the stationary points on a curve…….
Mathematics 1(Higher) Outcome 3 Use basic differentiation The graph of the curve y = 1/3x 3 - x2 x2 - 3x + 10 looks something like this: Example 2 y x 2 4 6 0 1 234 8 10 12 -4 -3 -2 5 (3, 1) (-1, ) Maximum turning point. Minimum turning point. -5 y = 1/3x 3 - x2 x2 - 3x + 10 PC(c) Determine the coordinates of the stationary points on a curve…….
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 1 Find the stationary point on the curve y = x 2 + 3. Using differentiation determine its nature. Solution = 2x2x For stationary values, = 0 ie 2x = 0 x = 0 When x = 0, (0) 2 + 3 y = = 0 + 3 So a stationary point occurs at (0, 3). = 3
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 1 Find the stationary point on the curve y = x 2 + 3. Using differentiation determine its nature. Solution (continued) = 2x2x A stationary point occurs at (0, 3). x 0 - - 0 slope 0 0 + + (0, 3) is a minimum turning point.
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 2 Find the stationary point on the curve y = 16 - x 2. Using differentiation determine its nature. Solution = -2x For stationary values, = 0 ie -2x = 0 x = 0 When x = 0, 16 - (0) 2 y = = 16 - 0 So a stationary point occurs at (0, 16). = 16
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. Solution (continued) = -2x A stationary point occurs at (0, 16). x 0 - + 0 slope 0 0 + - (0, 16) is a maximum turning point. 2 Find the stationary point on the curve y = 16 - x 2. Using differentiation determine its nature.
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 3 Find the stationary point on the curve y = (x + 5)(x - 3). Using differentiation determine its nature. Solution = 2x2x For stationary values, = 0 ie 2x + 2 = 0 x = -1 When x = -1, (-1) 2 y = = 1 So a stationary point occurs at (-1, -16). = -16 y = x 2 + 2x - 15 + 2 - 15 +2(-1) - 2- 15
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 3 Find the stationary point on the curve y = (x + 5)( x - 3). Using differentiation determine its nature. Solution (continued) = 2x + 2 A stationary point occurs at (-1, -16). x -1 - - 0 slope -1 + + (-1, -16) is a minimum turning point.
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 4 Find the stationary points on the curve y = 1/3 x 3 - x 2 - 15x. Using differentiation determine their nature. Solution For stationary values, = 0 = x2x2 - 2x- 15 ie ( )( ) = 0 x - 5 x + 3 x = When x = 5,5, y = 1/3(5) 3 - (5) 2 - 15(5) = When x = -3, y = 1/3(-3) 3 -(-3) 2 -15(-3) = 27 So stationary points occur at (-3, 27) and (5, ) 5, -3
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 4 Find the stationary points on the curve y = 1/3 x 3 - x 2 - 15x. Using differentiation determine their nature. Solution (continued) = A stationary point occurs at (-3, 27). x2x2 - 2x- 15 x -3 - + 0 slope -3 -3 + - (-3, 27) is a maximum turning point.
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(c) Determine the coordinates of the stationary points on a curve……. 4 Find the stationary points on the curve y = 1/3 x 3 - x 2 - 15x. Using differentiation determine their nature. Solution (continued) = A stationary point occurs at (5, ). x2x2 - 2x- 15 x 5 - slope 5 5 + (-3, ) is a minimum turning point. - 0 +
Mathematics 1(Higher) Outcome 3 Use basic differentiation PC(b) Determine the gradient of a tangent to a curve by differentiation Now do Sections C1and C2 on page 43 of the Basic Skills booklet End of Section PC(c)
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# Calculus 1 : How to find local maximum graphing functions of curves
## Example Questions
### Example Question #11 : Local Maximum
A ball is tossed into the air and its height is determined by the function . What is the maximum height of the ball (in meters)?
Explanation:
To determine the maximum height of the ball, the first derivative of the function must be found and set equal to zero.
Test values to the left and right to confirm if this point is a maximum.
Because the derivative goes from positive to negative, we know the original function goes from increasing to decreasing and thus this point is a maximum.
Finally, to find the height of the ball at this point, plug in into the original function.
### Example Question #3 : Finding Maximums
What are the -coordinate of the local maximum on the graph of the function
?
Explanation:
To find maxima and minima, find the coordinates of the points where the derivative is undefined or equal to zero. The derivative of p(x) is
Next set the derivative equal to zero and solve for x:
Finally we need to test the critical points in the original equation to determine which is a maximum.
Since the value of the function is greatest at x = -3, that is the x-coordinate of the maximum.
### Example Question #4 : Finding Maximums
Find the local maximum of the function .
There are none.
Explanation:
To find the local maximum, first find the first derivative of the function.
.
Then find all values of x for which the derivative equals 0 or is undefined. The derivative equals 0 when x=0 and is never undefined because the denominator is always greater than 0. Then, by picking points less than and greater than 0, we see that the function is increasing less than 0 and increasing greater than 0.
Therefore, it is a local maximum.
### Example Question #12 : Local Maximum
Given a graph with equation , find the local maximum(s) in the graph if there are any.
No local maxima.
Explanation:
In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of where there are critical points. Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of in order to see whether or not the slope increases/decreases as is approaches or leaves that value.
In order to take the derivative of equation, the power rule must be applied, .
Taking the derivative, we find that the equation becomes
Setting the equation equal to zero and solving for , we find that the critical points of this graph are . In order to determine whether or not these points are maxima or minima, we plug in numbers larger/smaller as well as inbetween the two critical points to find the behavior of the slope.
For this problem I will choose the numbers -5,0 and 1. These numbers are choosen since is approximately -4 and is approximately 0.25. By choosing numbers that are larger and smaller then the critical points as well as a number inbetween the two critical points, we can see the behavior of the slope throughout the entire graph.
Plugging in those numbers into , we find that makes positive, makes negative, and makes positive. This means that the slope is positive approaching and negative leaving it. The slope is negative approaching and positive leaving it. For a critical point to be the local maxima, its slope must be positive approaching it and negative leaving it, therefore is the local maxima in this graph.
### Example Question #11 : Curves
Which of the following are local maxima of the function on the interval ?
Explanation:
and we need to find maxima. So lets take the first derivative by the product rule where . In this case, we can set . Next, we can find the derivatives of each . Now we assemble by the product rule above to get:
.
Now we must find critical points (where , endpoints, or undefined values).
The domain of the function is all real numbers and the endpoints are and . So now we just need to find where . This will be equal to when and where . There is no value of we can choose such that so lets focus on . This is true where . We know that , so maybe we can use this to find values for which . We know that in the second quadrant, sine is positive and cosine is negative. We can check and verify that . The next value will occur in the 4th quadrant, when sine and cosine flip signs. So we can characterize these values that satisfy the equation as where
Now we need to determine which of these critical points are maxima. so, the function is increasing from and we can show this is not a maxima. Since we're going to be testing a lot of points, let's take the second derivative and use the second derivative test. If , we have a max (think concave up). So
At , as (is in the second quadrant, where only sine is positive). Therefore, is a local maximum. The next value, , is in the fourth quadrant where cosine is positive, so the second derivative should be positive, meaning it should be a local minimum. The next point, should be another maximum, but it lies outside the domain. So the last endpoint, , is the last local maximum on the interval.
### Example Question #5 : Finding Maximums
Find the local maxima of the following function:
There are no local maxima
There are no local maxima
Explanation:
To find the local maximum of the function, we must find the point at which the first derivative changes from positive to negative. To do this, we first must find the first derivative:
We found the derivative using the following rule:
Now, we must find the critical point(s), the point(s) at which the first derivative is equal to zero:
Now, we make our intervals over which to analyze the sign of the first derivative:
Over the first interval, the firt derivative is positive, and over the second interval, the first derivative is positive. Because the first derivative doesn't change from positive to negative, there are no local maxima.
### Example Question #14 : Local Maximum
What is the local maximum of the following function in the interval ?
Explanation:
To find the local extrema of a function you must find the zeros of the derivative. The derivative of is , so we can calculate the derivative of our polynomial and pull out a constant, getting:
We can use the quadratic formula or just guess some easy values, and figure out that the zeros of this polynomial are x=6 and x=2. 2 is the only one of these in the interval stated, so we need to check and see if it's a maximum or a minimum. We have to plug at least three values into our function to see this, so we'll choose x=1, 2, 3.
We can see that as x approaches 2, f(x) is increasing, and as it moves away from 2, f(x) decreases.
So must be our local maximum.
### Example Question #15 : Local Maximum
What is the -coordinate of the maximum for the function ?
Explanation:
To find the maximum, you must first find the derivative of the function, which is We found this derivative using the power rule which states,
.
Then, you need to set that equal to zero to get your critical points.
It will be helpful to factor it:
.
Then, set each expression to 0 to get those points:
.
Then, put these points on a number line to then test the sign of the slope in between each point.
To the left of , plug in a value (say, 0) into the derivative. You get a positive value. In between and 1, the value is negative. To the right of 1, the value is positive. Then, to find the max, look for where the signs change from positive to negative.
That is occurring at .
### Example Question #16 : Local Maximum
What is the maximum of over the interval ?
Explanation:
To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, .
Given the function, and applying the power rule we find the following derivative.
Check the -value at each endpoint and when the first derivative is zero, namely
The largest value is .
### Example Question #11 : Derivative As A Function
Find the -value where the local maximum occurs on
.
Explanation:
To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,
.
Given the function, and applying the quotient rule we find the following derivative.
when and when , which indicates that has a local maximum at .
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G = Q8×C3×C15order 360 = 23·32·5
Direct product of C3×C15 and Q8
direct product, metacyclic, nilpotent (class 2), monomial
Aliases: Q8×C3×C15, C60.13C6, C12.5C30, C10.7C62, C4.(C3×C30), C6.9(C2×C30), C20.3(C3×C6), C2.2(C6×C30), (C3×C60).11C2, (C3×C12).5C10, C30.32(C2×C6), (C3×C30).58C22, (C3×C6).17(C2×C10), SmallGroup(360,117)
Series: Derived Chief Lower central Upper central
Derived series C1 — C2 — Q8×C3×C15
Chief series C1 — C2 — C10 — C30 — C3×C30 — C3×C60 — Q8×C3×C15
Lower central C1 — C2 — Q8×C3×C15
Upper central C1 — C3×C30 — Q8×C3×C15
Generators and relations for Q8×C3×C15
G = < a,b,c,d | a3=b15=c4=1, d2=c2, ab=ba, ac=ca, ad=da, bc=cb, bd=db, dcd-1=c-1 >
Subgroups: 72, all normal (12 characteristic)
C1, C2, C3, C4, C5, C6, Q8, C32, C10, C12, C15, C3×C6, C20, C3×Q8, C30, C3×C12, C5×Q8, C3×C15, C60, Q8×C32, C3×C30, Q8×C15, C3×C60, Q8×C3×C15
Quotients: C1, C2, C3, C22, C5, C6, Q8, C32, C10, C2×C6, C15, C3×C6, C2×C10, C3×Q8, C30, C62, C5×Q8, C3×C15, C2×C30, Q8×C32, C3×C30, Q8×C15, C6×C30, Q8×C3×C15
Smallest permutation representation of Q8×C3×C15
Regular action on 360 points
Generators in S360
(1 25 253)(2 26 254)(3 27 255)(4 28 241)(5 29 242)(6 30 243)(7 16 244)(8 17 245)(9 18 246)(10 19 247)(11 20 248)(12 21 249)(13 22 250)(14 23 251)(15 24 252)(31 222 237)(32 223 238)(33 224 239)(34 225 240)(35 211 226)(36 212 227)(37 213 228)(38 214 229)(39 215 230)(40 216 231)(41 217 232)(42 218 233)(43 219 234)(44 220 235)(45 221 236)(46 357 195)(47 358 181)(48 359 182)(49 360 183)(50 346 184)(51 347 185)(52 348 186)(53 349 187)(54 350 188)(55 351 189)(56 352 190)(57 353 191)(58 354 192)(59 355 193)(60 356 194)(61 297 138)(62 298 139)(63 299 140)(64 300 141)(65 286 142)(66 287 143)(67 288 144)(68 289 145)(69 290 146)(70 291 147)(71 292 148)(72 293 149)(73 294 150)(74 295 136)(75 296 137)(76 197 158)(77 198 159)(78 199 160)(79 200 161)(80 201 162)(81 202 163)(82 203 164)(83 204 165)(84 205 151)(85 206 152)(86 207 153)(87 208 154)(88 209 155)(89 210 156)(90 196 157)(91 337 272)(92 338 273)(93 339 274)(94 340 275)(95 341 276)(96 342 277)(97 343 278)(98 344 279)(99 345 280)(100 331 281)(101 332 282)(102 333 283)(103 334 284)(104 335 285)(105 336 271)(106 322 304)(107 323 305)(108 324 306)(109 325 307)(110 326 308)(111 327 309)(112 328 310)(113 329 311)(114 330 312)(115 316 313)(116 317 314)(117 318 315)(118 319 301)(119 320 302)(120 321 303)(121 257 169)(122 258 170)(123 259 171)(124 260 172)(125 261 173)(126 262 174)(127 263 175)(128 264 176)(129 265 177)(130 266 178)(131 267 179)(132 268 180)(133 269 166)(134 270 167)(135 256 168)
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15)(16 17 18 19 20 21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40 41 42 43 44 45)(46 47 48 49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72 73 74 75)(76 77 78 79 80 81 82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99 100 101 102 103 104 105)(106 107 108 109 110 111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130 131 132 133 134 135)(136 137 138 139 140 141 142 143 144 145 146 147 148 149 150)(151 152 153 154 155 156 157 158 159 160 161 162 163 164 165)(166 167 168 169 170 171 172 173 174 175 176 177 178 179 180)(181 182 183 184 185 186 187 188 189 190 191 192 193 194 195)(196 197 198 199 200 201 202 203 204 205 206 207 208 209 210)(211 212 213 214 215 216 217 218 219 220 221 222 223 224 225)(226 227 228 229 230 231 232 233 234 235 236 237 238 239 240)(241 242 243 244 245 246 247 248 249 250 251 252 253 254 255)(256 257 258 259 260 261 262 263 264 265 266 267 268 269 270)(271 272 273 274 275 276 277 278 279 280 281 282 283 284 285)(286 287 288 289 290 291 292 293 294 295 296 297 298 299 300)(301 302 303 304 305 306 307 308 309 310 311 312 313 314 315)(316 317 318 319 320 321 322 323 324 325 326 327 328 329 330)(331 332 333 334 335 336 337 338 339 340 341 342 343 344 345)(346 347 348 349 350 351 352 353 354 355 356 357 358 359 360)
(1 225 347 319)(2 211 348 320)(3 212 349 321)(4 213 350 322)(5 214 351 323)(6 215 352 324)(7 216 353 325)(8 217 354 326)(9 218 355 327)(10 219 356 328)(11 220 357 329)(12 221 358 330)(13 222 359 316)(14 223 360 317)(15 224 346 318)(16 231 191 307)(17 232 192 308)(18 233 193 309)(19 234 194 310)(20 235 195 311)(21 236 181 312)(22 237 182 313)(23 238 183 314)(24 239 184 315)(25 240 185 301)(26 226 186 302)(27 227 187 303)(28 228 188 304)(29 229 189 305)(30 230 190 306)(31 48 115 250)(32 49 116 251)(33 50 117 252)(34 51 118 253)(35 52 119 254)(36 53 120 255)(37 54 106 241)(38 55 107 242)(39 56 108 243)(40 57 109 244)(41 58 110 245)(42 59 111 246)(43 60 112 247)(44 46 113 248)(45 47 114 249)(61 151 121 101)(62 152 122 102)(63 153 123 103)(64 154 124 104)(65 155 125 105)(66 156 126 91)(67 157 127 92)(68 158 128 93)(69 159 129 94)(70 160 130 95)(71 161 131 96)(72 162 132 97)(73 163 133 98)(74 164 134 99)(75 165 135 100)(76 264 339 289)(77 265 340 290)(78 266 341 291)(79 267 342 292)(80 268 343 293)(81 269 344 294)(82 270 345 295)(83 256 331 296)(84 257 332 297)(85 258 333 298)(86 259 334 299)(87 260 335 300)(88 261 336 286)(89 262 337 287)(90 263 338 288)(136 203 167 280)(137 204 168 281)(138 205 169 282)(139 206 170 283)(140 207 171 284)(141 208 172 285)(142 209 173 271)(143 210 174 272)(144 196 175 273)(145 197 176 274)(146 198 177 275)(147 199 178 276)(148 200 179 277)(149 201 180 278)(150 202 166 279)
(1 132 347 72)(2 133 348 73)(3 134 349 74)(4 135 350 75)(5 121 351 61)(6 122 352 62)(7 123 353 63)(8 124 354 64)(9 125 355 65)(10 126 356 66)(11 127 357 67)(12 128 358 68)(13 129 359 69)(14 130 360 70)(15 131 346 71)(16 259 191 299)(17 260 192 300)(18 261 193 286)(19 262 194 287)(20 263 195 288)(21 264 181 289)(22 265 182 290)(23 266 183 291)(24 267 184 292)(25 268 185 293)(26 269 186 294)(27 270 187 295)(28 256 188 296)(29 257 189 297)(30 258 190 298)(31 198 115 275)(32 199 116 276)(33 200 117 277)(34 201 118 278)(35 202 119 279)(36 203 120 280)(37 204 106 281)(38 205 107 282)(39 206 108 283)(40 207 109 284)(41 208 110 285)(42 209 111 271)(43 210 112 272)(44 196 113 273)(45 197 114 274)(46 144 248 175)(47 145 249 176)(48 146 250 177)(49 147 251 178)(50 148 252 179)(51 149 253 180)(52 150 254 166)(53 136 255 167)(54 137 241 168)(55 138 242 169)(56 139 243 170)(57 140 244 171)(58 141 245 172)(59 142 246 173)(60 143 247 174)(76 312 339 236)(77 313 340 237)(78 314 341 238)(79 315 342 239)(80 301 343 240)(81 302 344 226)(82 303 345 227)(83 304 331 228)(84 305 332 229)(85 306 333 230)(86 307 334 231)(87 308 335 232)(88 309 336 233)(89 310 337 234)(90 311 338 235)(91 219 156 328)(92 220 157 329)(93 221 158 330)(94 222 159 316)(95 223 160 317)(96 224 161 318)(97 225 162 319)(98 211 163 320)(99 212 164 321)(100 213 165 322)(101 214 151 323)(102 215 152 324)(103 216 153 325)(104 217 154 326)(105 218 155 327)
G:=sub<Sym(360)| (1,25,253)(2,26,254)(3,27,255)(4,28,241)(5,29,242)(6,30,243)(7,16,244)(8,17,245)(9,18,246)(10,19,247)(11,20,248)(12,21,249)(13,22,250)(14,23,251)(15,24,252)(31,222,237)(32,223,238)(33,224,239)(34,225,240)(35,211,226)(36,212,227)(37,213,228)(38,214,229)(39,215,230)(40,216,231)(41,217,232)(42,218,233)(43,219,234)(44,220,235)(45,221,236)(46,357,195)(47,358,181)(48,359,182)(49,360,183)(50,346,184)(51,347,185)(52,348,186)(53,349,187)(54,350,188)(55,351,189)(56,352,190)(57,353,191)(58,354,192)(59,355,193)(60,356,194)(61,297,138)(62,298,139)(63,299,140)(64,300,141)(65,286,142)(66,287,143)(67,288,144)(68,289,145)(69,290,146)(70,291,147)(71,292,148)(72,293,149)(73,294,150)(74,295,136)(75,296,137)(76,197,158)(77,198,159)(78,199,160)(79,200,161)(80,201,162)(81,202,163)(82,203,164)(83,204,165)(84,205,151)(85,206,152)(86,207,153)(87,208,154)(88,209,155)(89,210,156)(90,196,157)(91,337,272)(92,338,273)(93,339,274)(94,340,275)(95,341,276)(96,342,277)(97,343,278)(98,344,279)(99,345,280)(100,331,281)(101,332,282)(102,333,283)(103,334,284)(104,335,285)(105,336,271)(106,322,304)(107,323,305)(108,324,306)(109,325,307)(110,326,308)(111,327,309)(112,328,310)(113,329,311)(114,330,312)(115,316,313)(116,317,314)(117,318,315)(118,319,301)(119,320,302)(120,321,303)(121,257,169)(122,258,170)(123,259,171)(124,260,172)(125,261,173)(126,262,174)(127,263,175)(128,264,176)(129,265,177)(130,266,178)(131,267,179)(132,268,180)(133,269,166)(134,270,167)(135,256,168), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)(16,17,18,19,20,21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75)(76,77,78,79,80,81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100,101,102,103,104,105)(106,107,108,109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132,133,134,135)(136,137,138,139,140,141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160,161,162,163,164,165)(166,167,168,169,170,171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190,191,192,193,194,195)(196,197,198,199,200,201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220,221,222,223,224,225)(226,227,228,229,230,231,232,233,234,235,236,237,238,239,240)(241,242,243,244,245,246,247,248,249,250,251,252,253,254,255)(256,257,258,259,260,261,262,263,264,265,266,267,268,269,270)(271,272,273,274,275,276,277,278,279,280,281,282,283,284,285)(286,287,288,289,290,291,292,293,294,295,296,297,298,299,300)(301,302,303,304,305,306,307,308,309,310,311,312,313,314,315)(316,317,318,319,320,321,322,323,324,325,326,327,328,329,330)(331,332,333,334,335,336,337,338,339,340,341,342,343,344,345)(346,347,348,349,350,351,352,353,354,355,356,357,358,359,360), (1,225,347,319)(2,211,348,320)(3,212,349,321)(4,213,350,322)(5,214,351,323)(6,215,352,324)(7,216,353,325)(8,217,354,326)(9,218,355,327)(10,219,356,328)(11,220,357,329)(12,221,358,330)(13,222,359,316)(14,223,360,317)(15,224,346,318)(16,231,191,307)(17,232,192,308)(18,233,193,309)(19,234,194,310)(20,235,195,311)(21,236,181,312)(22,237,182,313)(23,238,183,314)(24,239,184,315)(25,240,185,301)(26,226,186,302)(27,227,187,303)(28,228,188,304)(29,229,189,305)(30,230,190,306)(31,48,115,250)(32,49,116,251)(33,50,117,252)(34,51,118,253)(35,52,119,254)(36,53,120,255)(37,54,106,241)(38,55,107,242)(39,56,108,243)(40,57,109,244)(41,58,110,245)(42,59,111,246)(43,60,112,247)(44,46,113,248)(45,47,114,249)(61,151,121,101)(62,152,122,102)(63,153,123,103)(64,154,124,104)(65,155,125,105)(66,156,126,91)(67,157,127,92)(68,158,128,93)(69,159,129,94)(70,160,130,95)(71,161,131,96)(72,162,132,97)(73,163,133,98)(74,164,134,99)(75,165,135,100)(76,264,339,289)(77,265,340,290)(78,266,341,291)(79,267,342,292)(80,268,343,293)(81,269,344,294)(82,270,345,295)(83,256,331,296)(84,257,332,297)(85,258,333,298)(86,259,334,299)(87,260,335,300)(88,261,336,286)(89,262,337,287)(90,263,338,288)(136,203,167,280)(137,204,168,281)(138,205,169,282)(139,206,170,283)(140,207,171,284)(141,208,172,285)(142,209,173,271)(143,210,174,272)(144,196,175,273)(145,197,176,274)(146,198,177,275)(147,199,178,276)(148,200,179,277)(149,201,180,278)(150,202,166,279), (1,132,347,72)(2,133,348,73)(3,134,349,74)(4,135,350,75)(5,121,351,61)(6,122,352,62)(7,123,353,63)(8,124,354,64)(9,125,355,65)(10,126,356,66)(11,127,357,67)(12,128,358,68)(13,129,359,69)(14,130,360,70)(15,131,346,71)(16,259,191,299)(17,260,192,300)(18,261,193,286)(19,262,194,287)(20,263,195,288)(21,264,181,289)(22,265,182,290)(23,266,183,291)(24,267,184,292)(25,268,185,293)(26,269,186,294)(27,270,187,295)(28,256,188,296)(29,257,189,297)(30,258,190,298)(31,198,115,275)(32,199,116,276)(33,200,117,277)(34,201,118,278)(35,202,119,279)(36,203,120,280)(37,204,106,281)(38,205,107,282)(39,206,108,283)(40,207,109,284)(41,208,110,285)(42,209,111,271)(43,210,112,272)(44,196,113,273)(45,197,114,274)(46,144,248,175)(47,145,249,176)(48,146,250,177)(49,147,251,178)(50,148,252,179)(51,149,253,180)(52,150,254,166)(53,136,255,167)(54,137,241,168)(55,138,242,169)(56,139,243,170)(57,140,244,171)(58,141,245,172)(59,142,246,173)(60,143,247,174)(76,312,339,236)(77,313,340,237)(78,314,341,238)(79,315,342,239)(80,301,343,240)(81,302,344,226)(82,303,345,227)(83,304,331,228)(84,305,332,229)(85,306,333,230)(86,307,334,231)(87,308,335,232)(88,309,336,233)(89,310,337,234)(90,311,338,235)(91,219,156,328)(92,220,157,329)(93,221,158,330)(94,222,159,316)(95,223,160,317)(96,224,161,318)(97,225,162,319)(98,211,163,320)(99,212,164,321)(100,213,165,322)(101,214,151,323)(102,215,152,324)(103,216,153,325)(104,217,154,326)(105,218,155,327)>;
G:=Group( (1,25,253)(2,26,254)(3,27,255)(4,28,241)(5,29,242)(6,30,243)(7,16,244)(8,17,245)(9,18,246)(10,19,247)(11,20,248)(12,21,249)(13,22,250)(14,23,251)(15,24,252)(31,222,237)(32,223,238)(33,224,239)(34,225,240)(35,211,226)(36,212,227)(37,213,228)(38,214,229)(39,215,230)(40,216,231)(41,217,232)(42,218,233)(43,219,234)(44,220,235)(45,221,236)(46,357,195)(47,358,181)(48,359,182)(49,360,183)(50,346,184)(51,347,185)(52,348,186)(53,349,187)(54,350,188)(55,351,189)(56,352,190)(57,353,191)(58,354,192)(59,355,193)(60,356,194)(61,297,138)(62,298,139)(63,299,140)(64,300,141)(65,286,142)(66,287,143)(67,288,144)(68,289,145)(69,290,146)(70,291,147)(71,292,148)(72,293,149)(73,294,150)(74,295,136)(75,296,137)(76,197,158)(77,198,159)(78,199,160)(79,200,161)(80,201,162)(81,202,163)(82,203,164)(83,204,165)(84,205,151)(85,206,152)(86,207,153)(87,208,154)(88,209,155)(89,210,156)(90,196,157)(91,337,272)(92,338,273)(93,339,274)(94,340,275)(95,341,276)(96,342,277)(97,343,278)(98,344,279)(99,345,280)(100,331,281)(101,332,282)(102,333,283)(103,334,284)(104,335,285)(105,336,271)(106,322,304)(107,323,305)(108,324,306)(109,325,307)(110,326,308)(111,327,309)(112,328,310)(113,329,311)(114,330,312)(115,316,313)(116,317,314)(117,318,315)(118,319,301)(119,320,302)(120,321,303)(121,257,169)(122,258,170)(123,259,171)(124,260,172)(125,261,173)(126,262,174)(127,263,175)(128,264,176)(129,265,177)(130,266,178)(131,267,179)(132,268,180)(133,269,166)(134,270,167)(135,256,168), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)(16,17,18,19,20,21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75)(76,77,78,79,80,81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100,101,102,103,104,105)(106,107,108,109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132,133,134,135)(136,137,138,139,140,141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160,161,162,163,164,165)(166,167,168,169,170,171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190,191,192,193,194,195)(196,197,198,199,200,201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220,221,222,223,224,225)(226,227,228,229,230,231,232,233,234,235,236,237,238,239,240)(241,242,243,244,245,246,247,248,249,250,251,252,253,254,255)(256,257,258,259,260,261,262,263,264,265,266,267,268,269,270)(271,272,273,274,275,276,277,278,279,280,281,282,283,284,285)(286,287,288,289,290,291,292,293,294,295,296,297,298,299,300)(301,302,303,304,305,306,307,308,309,310,311,312,313,314,315)(316,317,318,319,320,321,322,323,324,325,326,327,328,329,330)(331,332,333,334,335,336,337,338,339,340,341,342,343,344,345)(346,347,348,349,350,351,352,353,354,355,356,357,358,359,360), (1,225,347,319)(2,211,348,320)(3,212,349,321)(4,213,350,322)(5,214,351,323)(6,215,352,324)(7,216,353,325)(8,217,354,326)(9,218,355,327)(10,219,356,328)(11,220,357,329)(12,221,358,330)(13,222,359,316)(14,223,360,317)(15,224,346,318)(16,231,191,307)(17,232,192,308)(18,233,193,309)(19,234,194,310)(20,235,195,311)(21,236,181,312)(22,237,182,313)(23,238,183,314)(24,239,184,315)(25,240,185,301)(26,226,186,302)(27,227,187,303)(28,228,188,304)(29,229,189,305)(30,230,190,306)(31,48,115,250)(32,49,116,251)(33,50,117,252)(34,51,118,253)(35,52,119,254)(36,53,120,255)(37,54,106,241)(38,55,107,242)(39,56,108,243)(40,57,109,244)(41,58,110,245)(42,59,111,246)(43,60,112,247)(44,46,113,248)(45,47,114,249)(61,151,121,101)(62,152,122,102)(63,153,123,103)(64,154,124,104)(65,155,125,105)(66,156,126,91)(67,157,127,92)(68,158,128,93)(69,159,129,94)(70,160,130,95)(71,161,131,96)(72,162,132,97)(73,163,133,98)(74,164,134,99)(75,165,135,100)(76,264,339,289)(77,265,340,290)(78,266,341,291)(79,267,342,292)(80,268,343,293)(81,269,344,294)(82,270,345,295)(83,256,331,296)(84,257,332,297)(85,258,333,298)(86,259,334,299)(87,260,335,300)(88,261,336,286)(89,262,337,287)(90,263,338,288)(136,203,167,280)(137,204,168,281)(138,205,169,282)(139,206,170,283)(140,207,171,284)(141,208,172,285)(142,209,173,271)(143,210,174,272)(144,196,175,273)(145,197,176,274)(146,198,177,275)(147,199,178,276)(148,200,179,277)(149,201,180,278)(150,202,166,279), (1,132,347,72)(2,133,348,73)(3,134,349,74)(4,135,350,75)(5,121,351,61)(6,122,352,62)(7,123,353,63)(8,124,354,64)(9,125,355,65)(10,126,356,66)(11,127,357,67)(12,128,358,68)(13,129,359,69)(14,130,360,70)(15,131,346,71)(16,259,191,299)(17,260,192,300)(18,261,193,286)(19,262,194,287)(20,263,195,288)(21,264,181,289)(22,265,182,290)(23,266,183,291)(24,267,184,292)(25,268,185,293)(26,269,186,294)(27,270,187,295)(28,256,188,296)(29,257,189,297)(30,258,190,298)(31,198,115,275)(32,199,116,276)(33,200,117,277)(34,201,118,278)(35,202,119,279)(36,203,120,280)(37,204,106,281)(38,205,107,282)(39,206,108,283)(40,207,109,284)(41,208,110,285)(42,209,111,271)(43,210,112,272)(44,196,113,273)(45,197,114,274)(46,144,248,175)(47,145,249,176)(48,146,250,177)(49,147,251,178)(50,148,252,179)(51,149,253,180)(52,150,254,166)(53,136,255,167)(54,137,241,168)(55,138,242,169)(56,139,243,170)(57,140,244,171)(58,141,245,172)(59,142,246,173)(60,143,247,174)(76,312,339,236)(77,313,340,237)(78,314,341,238)(79,315,342,239)(80,301,343,240)(81,302,344,226)(82,303,345,227)(83,304,331,228)(84,305,332,229)(85,306,333,230)(86,307,334,231)(87,308,335,232)(88,309,336,233)(89,310,337,234)(90,311,338,235)(91,219,156,328)(92,220,157,329)(93,221,158,330)(94,222,159,316)(95,223,160,317)(96,224,161,318)(97,225,162,319)(98,211,163,320)(99,212,164,321)(100,213,165,322)(101,214,151,323)(102,215,152,324)(103,216,153,325)(104,217,154,326)(105,218,155,327) );
G=PermutationGroup([[(1,25,253),(2,26,254),(3,27,255),(4,28,241),(5,29,242),(6,30,243),(7,16,244),(8,17,245),(9,18,246),(10,19,247),(11,20,248),(12,21,249),(13,22,250),(14,23,251),(15,24,252),(31,222,237),(32,223,238),(33,224,239),(34,225,240),(35,211,226),(36,212,227),(37,213,228),(38,214,229),(39,215,230),(40,216,231),(41,217,232),(42,218,233),(43,219,234),(44,220,235),(45,221,236),(46,357,195),(47,358,181),(48,359,182),(49,360,183),(50,346,184),(51,347,185),(52,348,186),(53,349,187),(54,350,188),(55,351,189),(56,352,190),(57,353,191),(58,354,192),(59,355,193),(60,356,194),(61,297,138),(62,298,139),(63,299,140),(64,300,141),(65,286,142),(66,287,143),(67,288,144),(68,289,145),(69,290,146),(70,291,147),(71,292,148),(72,293,149),(73,294,150),(74,295,136),(75,296,137),(76,197,158),(77,198,159),(78,199,160),(79,200,161),(80,201,162),(81,202,163),(82,203,164),(83,204,165),(84,205,151),(85,206,152),(86,207,153),(87,208,154),(88,209,155),(89,210,156),(90,196,157),(91,337,272),(92,338,273),(93,339,274),(94,340,275),(95,341,276),(96,342,277),(97,343,278),(98,344,279),(99,345,280),(100,331,281),(101,332,282),(102,333,283),(103,334,284),(104,335,285),(105,336,271),(106,322,304),(107,323,305),(108,324,306),(109,325,307),(110,326,308),(111,327,309),(112,328,310),(113,329,311),(114,330,312),(115,316,313),(116,317,314),(117,318,315),(118,319,301),(119,320,302),(120,321,303),(121,257,169),(122,258,170),(123,259,171),(124,260,172),(125,261,173),(126,262,174),(127,263,175),(128,264,176),(129,265,177),(130,266,178),(131,267,179),(132,268,180),(133,269,166),(134,270,167),(135,256,168)], [(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15),(16,17,18,19,20,21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45),(46,47,48,49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75),(76,77,78,79,80,81,82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99,100,101,102,103,104,105),(106,107,108,109,110,111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130,131,132,133,134,135),(136,137,138,139,140,141,142,143,144,145,146,147,148,149,150),(151,152,153,154,155,156,157,158,159,160,161,162,163,164,165),(166,167,168,169,170,171,172,173,174,175,176,177,178,179,180),(181,182,183,184,185,186,187,188,189,190,191,192,193,194,195),(196,197,198,199,200,201,202,203,204,205,206,207,208,209,210),(211,212,213,214,215,216,217,218,219,220,221,222,223,224,225),(226,227,228,229,230,231,232,233,234,235,236,237,238,239,240),(241,242,243,244,245,246,247,248,249,250,251,252,253,254,255),(256,257,258,259,260,261,262,263,264,265,266,267,268,269,270),(271,272,273,274,275,276,277,278,279,280,281,282,283,284,285),(286,287,288,289,290,291,292,293,294,295,296,297,298,299,300),(301,302,303,304,305,306,307,308,309,310,311,312,313,314,315),(316,317,318,319,320,321,322,323,324,325,326,327,328,329,330),(331,332,333,334,335,336,337,338,339,340,341,342,343,344,345),(346,347,348,349,350,351,352,353,354,355,356,357,358,359,360)], [(1,225,347,319),(2,211,348,320),(3,212,349,321),(4,213,350,322),(5,214,351,323),(6,215,352,324),(7,216,353,325),(8,217,354,326),(9,218,355,327),(10,219,356,328),(11,220,357,329),(12,221,358,330),(13,222,359,316),(14,223,360,317),(15,224,346,318),(16,231,191,307),(17,232,192,308),(18,233,193,309),(19,234,194,310),(20,235,195,311),(21,236,181,312),(22,237,182,313),(23,238,183,314),(24,239,184,315),(25,240,185,301),(26,226,186,302),(27,227,187,303),(28,228,188,304),(29,229,189,305),(30,230,190,306),(31,48,115,250),(32,49,116,251),(33,50,117,252),(34,51,118,253),(35,52,119,254),(36,53,120,255),(37,54,106,241),(38,55,107,242),(39,56,108,243),(40,57,109,244),(41,58,110,245),(42,59,111,246),(43,60,112,247),(44,46,113,248),(45,47,114,249),(61,151,121,101),(62,152,122,102),(63,153,123,103),(64,154,124,104),(65,155,125,105),(66,156,126,91),(67,157,127,92),(68,158,128,93),(69,159,129,94),(70,160,130,95),(71,161,131,96),(72,162,132,97),(73,163,133,98),(74,164,134,99),(75,165,135,100),(76,264,339,289),(77,265,340,290),(78,266,341,291),(79,267,342,292),(80,268,343,293),(81,269,344,294),(82,270,345,295),(83,256,331,296),(84,257,332,297),(85,258,333,298),(86,259,334,299),(87,260,335,300),(88,261,336,286),(89,262,337,287),(90,263,338,288),(136,203,167,280),(137,204,168,281),(138,205,169,282),(139,206,170,283),(140,207,171,284),(141,208,172,285),(142,209,173,271),(143,210,174,272),(144,196,175,273),(145,197,176,274),(146,198,177,275),(147,199,178,276),(148,200,179,277),(149,201,180,278),(150,202,166,279)], [(1,132,347,72),(2,133,348,73),(3,134,349,74),(4,135,350,75),(5,121,351,61),(6,122,352,62),(7,123,353,63),(8,124,354,64),(9,125,355,65),(10,126,356,66),(11,127,357,67),(12,128,358,68),(13,129,359,69),(14,130,360,70),(15,131,346,71),(16,259,191,299),(17,260,192,300),(18,261,193,286),(19,262,194,287),(20,263,195,288),(21,264,181,289),(22,265,182,290),(23,266,183,291),(24,267,184,292),(25,268,185,293),(26,269,186,294),(27,270,187,295),(28,256,188,296),(29,257,189,297),(30,258,190,298),(31,198,115,275),(32,199,116,276),(33,200,117,277),(34,201,118,278),(35,202,119,279),(36,203,120,280),(37,204,106,281),(38,205,107,282),(39,206,108,283),(40,207,109,284),(41,208,110,285),(42,209,111,271),(43,210,112,272),(44,196,113,273),(45,197,114,274),(46,144,248,175),(47,145,249,176),(48,146,250,177),(49,147,251,178),(50,148,252,179),(51,149,253,180),(52,150,254,166),(53,136,255,167),(54,137,241,168),(55,138,242,169),(56,139,243,170),(57,140,244,171),(58,141,245,172),(59,142,246,173),(60,143,247,174),(76,312,339,236),(77,313,340,237),(78,314,341,238),(79,315,342,239),(80,301,343,240),(81,302,344,226),(82,303,345,227),(83,304,331,228),(84,305,332,229),(85,306,333,230),(86,307,334,231),(87,308,335,232),(88,309,336,233),(89,310,337,234),(90,311,338,235),(91,219,156,328),(92,220,157,329),(93,221,158,330),(94,222,159,316),(95,223,160,317),(96,224,161,318),(97,225,162,319),(98,211,163,320),(99,212,164,321),(100,213,165,322),(101,214,151,323),(102,215,152,324),(103,216,153,325),(104,217,154,326),(105,218,155,327)]])
225 conjugacy classes
class 1 2 3A ··· 3H 4A 4B 4C 5A 5B 5C 5D 6A ··· 6H 10A 10B 10C 10D 12A ··· 12X 15A ··· 15AF 20A ··· 20L 30A ··· 30AF 60A ··· 60CR order 1 2 3 ··· 3 4 4 4 5 5 5 5 6 ··· 6 10 10 10 10 12 ··· 12 15 ··· 15 20 ··· 20 30 ··· 30 60 ··· 60 size 1 1 1 ··· 1 2 2 2 1 1 1 1 1 ··· 1 1 1 1 1 2 ··· 2 1 ··· 1 2 ··· 2 1 ··· 1 2 ··· 2
225 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 type + + - image C1 C2 C3 C5 C6 C10 C15 C30 Q8 C3×Q8 C5×Q8 Q8×C15 kernel Q8×C3×C15 C3×C60 Q8×C15 Q8×C32 C60 C3×C12 C3×Q8 C12 C3×C15 C15 C32 C3 # reps 1 3 8 4 24 12 32 96 1 8 4 32
Matrix representation of Q8×C3×C15 in GL3(𝔽61) generated by
47 0 0 0 47 0 0 0 47
,
13 0 0 0 25 0 0 0 25
,
1 0 0 0 0 1 0 60 0
,
60 0 0 0 42 2 0 2 19
G:=sub<GL(3,GF(61))| [47,0,0,0,47,0,0,0,47],[13,0,0,0,25,0,0,0,25],[1,0,0,0,0,60,0,1,0],[60,0,0,0,42,2,0,2,19] >;
Q8×C3×C15 in GAP, Magma, Sage, TeX
Q_8\times C_3\times C_{15}
% in TeX
G:=Group("Q8xC3xC15");
// GroupNames label
G:=SmallGroup(360,117);
// by ID
G=gap.SmallGroup(360,117);
# by ID
G:=PCGroup([6,-2,-2,-3,-3,-5,-2,1080,2185,1087]);
// Polycyclic
G:=Group<a,b,c,d|a^3=b^15=c^4=1,d^2=c^2,a*b=b*a,a*c=c*a,a*d=d*a,b*c=c*b,b*d=d*b,d*c*d^-1=c^-1>;
// generators/relations
×
𝔽
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{}
|
# cs.DC-distributed_parallel_and_cluster_computing
## Strassen’s Algorithm
Recall that an -by- matrix is a two-dimensional array
## What Is An Efficient Parallel Algorithm?
Most algorithms we encounter in the study of computer science are sequential, i.e., designed with the assumption that only one computational operation can be...
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{}
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# Help with some Resistance Questions
1. Nov 7, 2004
### Hollysmoke
Okay we just start electricity in class and almost instantly we got a small lab. I did part of it, but I am having trouble with some of the questions and have no idea where to start from.
The first question is Determine the R value of a resistor that uses 4 m. of copper wire with a thickness of 1/3x the one used in the lab.
Now I have to find the ratio, but how? Also the resistor board used is 1 meter. I'm not sure, but would it be
R2= Resisitivity x 4 m. / Pie (4r)^2
R1= Resistivity x 1 meter / Pie r^2
How do I find the ratio?
2nd question was "Determine how much wider the diameter of the thick nichrome wire is in comparison to the thin nichrome wire".
How would I start such a question?
3rd question was "If the resistivity of nichrome is 100x10^-8 ohms/meter, determine the resistivity of copper"
How would I start to solve this question also?
Thanks (I'm not looking for an answer, Just how to start solving it)
2. Nov 8, 2004
### maverick280857
How do you find the ratio? I'm sorry but I don't quite understand your problem. You seem to know all the parameters. What is the problem? If the thickness is an issue then the thickness refers to the diameter (twice the radius). And you have used
$$R = \rho\frac{L}{A}$$
so where are you stuck?
Cheers
Vivek
3. Nov 8, 2004
### Hollysmoke
Doesn't matter, I figured it out and I got it right ^^
4. Nov 8, 2004
### jai6638
ummm can u please temme how u did it?? havent done electricity for a while now and cant seem to remember how to do it.. what is the value of RHO ( the P symbol ) in this case??
thanks
Last edited: Nov 8, 2004
5. Nov 8, 2004
### Hollysmoke
okay so we have the resistance of the copper from the lab, which we use
R1=P x L / A
Then we have the changes, which is 4x length and 1/3 the thickness. So, since we want the value of the resistor, we do:
R2=P x 4L / pie (1/3r^)2 <--- this is pie r^2 because the wire is circular
Since the resistance of the wire was 0.4 ohms (I'm doing this all off by memory BTW because I handed in my lab) then all we have to do is find the ratio. Since we have 1/3 in the brackets, we square to get 1/9, then the ratio is 4 / 1/9, which is 1/36. So 0.4 x 1/36 = 1.1 x 10^-2 ohms.
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{}
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# Swartzchild metric and free fall
1. Jul 15, 2007
### daniel_i_l
I just finished reading Black Holes by E. Taylor and J. Wheeler. Throughout the book they use the SC metric for the the metric near a massive object(for radial motion only:
$$d \tau^2 = \left( 1 - \frac{ 2M }{r} \right) dt^2 - \frac{dr^2}{\left( 1 - \frac{ 2M }{r} \right)}$$
Where dt is measured very far away from the massive body, r is measured as the circumference of a sphere whose center is in the middle and whose outer shell reaches the point divided by two times pi. M,t and r are mearured in meters.
Then, in order to calculate to path of a particle near a massive object in between two events they use the metric to find the path that give the maximal proper time (the Principal Of Maximal Aging).
Now my question is, to use POMA you need to have initial and final events and then calculate the path between them via the POMA. But I want to prove that the SC metric also predicts that an object at rest next to a massive object will start falling in. To do that shouldn't I start at t0 and r0 and then calculate which r1 will give the maximal proper time for a giver t1? Shouldn't this r1 be smaller than r0 - meaning that in order to maximize proper time the object should start falling towards the massive object? I tried to do this but I got that for any r1=/=r0 the proper time is smaller for a giver t1 then if r1=r0. Is that right? If so, how does the SC metric predict that an object that starts at rest next to a massive body will move towards it?
Thanks.
Last edited: Jul 15, 2007
2. Jul 15, 2007
### StatusX
That principle applies to the shortest path between two points, so you have to specify both the starting and ending locations completely, one coordinate of the final position isn't enough. More specifically, by leaving out this coordinate, you're not imposing the condition that the particle is initially at rest.
Last edited: Jul 15, 2007
3. Jul 15, 2007
### daniel_i_l
Then how can I use the metric to show that a particle released from rest near a massive object starts falling?
Thanks.
4. Jul 15, 2007
### StatusX
You can derive the geodesic equation, which determines the local conditions on a path so that it maximizes proper time globally. All the solutions can be verified to satisfy the maximal time conditions between their endpoints, but being local, they allow you to calculate the path of a particle knowing only its initial conditions, ie, position and velocity.
More visually, if you were to draw the paths maximizing time between (t0,r0) and (t1,r) for all r, these would all have different slopes through (t0,r0), and the condition that the particle is initial at rest picks out one of these, and so one final r.
Last edited: Jul 15, 2007
5. Jul 15, 2007
### daniel_i_l
Can you explain what you mean by:
"determines the local conditions on a path so that it maximizes proper time globally"?
How would I go about deriving this equation? Can you give me some starting points?
Thanks.
6. Jul 15, 2007
### StatusX
Did you see my edit? I think that's a more satisfying explanation. With that picture in mind (ie, a drawing of all possible geodesics passing though each point), note that if a given path maximizes proper time between its endpoints, it also maximizes proper time between any two points lying on it, since otherwise we could replace this section with a longer path and make the overall proper time bigger.
So being a geodesic is not just a global property, it requires that a certain condition must be satisfied between any two points, no matter how close they are. In particular, as you move the endpoint closer and closer to the starting point, you determine how any geodesic passing through a point must evolve given only its initial direction. It turns out that the condition specifices its second derivative with respect to proper time.
The derivation uses calculus of variations. It's very similar to the derivation of Lagrange's equations, and if you've seen these, it's not that hard to derive it yourself. I can get you started if you want, but it should be in any GR textbook.
Last edited: Jul 15, 2007
7. Jul 15, 2007
### daniel_i_l
I've never seen the derivation of the Lagrange's equations (the only quantitive use of the calculus of variations I've seen was in Feynman's Lectures on physics where he discussed the principal of least action - which is awfully similar to the POMA now that I think about it). But can you get me started please?
(Is it basically like this: http://www.eftaylor.com/pub/newton_mechanics.html ?)
And is this the general idea: Get an expression for the path from (t0,r0) to
(t1,r) as a variable of r and find the one whose slope at r0 is zero?
Thanks!
Last edited: Jul 15, 2007
8. Jul 15, 2007
### StatusX
That site presents a derivation of Lagrange's laws that's a little different than the usual one. If you take the time to go through it, it might be a little more intuitive than the standard derivation, but it takes a while to get there, and I think it would be more trouble than it's worth for the geodesic equation.
I've started trying to do the derivation a few times here, but I can't see a way to do without being confusing in less than 2 or 3 pages. I'd really suggest finding a book on GR, there are plenty of good ones. I was planning on writing up something on calculus of variations for the tutorial section here pretty soon, and if I do I'll be sure to include a derivation of the geodesic equation.
It's not a bad picture to have, but when deriving the equations this isn't what's being done directly. Although the way they do it on that site seems to be doing quantitatively what I described above, of imposing the condition that the path must be a maximum over every subpath, and using this condition on infinitesimal sections to derive differential equations governing the path. (ie, doing what you suggested for t1=t0+dt, r=r0+dr)
Last edited: Jul 15, 2007
9. Jul 15, 2007
### pervect
Staff Emeritus
If you just want to work out orbits (which will be geodesics) in the Schwarzschild space-time, the easy approach is to take advantage of the existence of a conserved energy and angular momentum.
Any orbit will occupy a plane, so you can eliminate one coordinate right off. Usually one sets $\theta=0$ for the equatorial plane.
Then you have two coordinates, r and $\phi$. So you need two equations.
The geodesic equations will be equivalent to
d/dt (E) = 0
d/dt (L) = 0
For massive particles one usually uses $\tilde{E}$, which is the energy per unit mass of the orbiting particle, rather than E, and also $\tilde{L}$
I would expect that "Exploring black holes" would cover some of the same material, you can also look at http://www.fourmilab.ch/gravitation/orbits/
Last edited: Jul 16, 2007
10. Jul 16, 2007
### daniel_i_l
Thanks everyone! I'll read those links and see if I can prove it. I'll post my conclusions if anything interesting comes out:)
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# Clifford algebras with inverse?
1. Sep 21, 2010
### mnb96
Hello,
Let's assume we have a Clifford algebra of one of the following two kinds: $$\mathcal{C}\ell_{n,0}$$ or $$\mathcal{C}\ell_{0,n}$$.
Is it possible to say which algebras will admit a multiplicative inverse for each multivector element?
2. Sep 21, 2010
$$Cl_{0,1}=\bf{C}$$
$$Cl_{0,2}=\bf{H}$$
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# Ten New Year’s Algebra Puzzles For 2019
Alright, the holidays are coming to an end and the second half of the school year is about to begin. Let’s work off some of the rust! Here are a few puzzles to give to your students as warm-ups or challenges depending on where they are.
### Puzzle #1
Find all quadratics such that solutions to $$(ax + b)(cx + d) = 0$$ are integer with $$a,b,c,d$$ all unique and chosen only from $$\{2,0,1,9\}$$
Some gotchas and things to note:
• $$(0x + 2)(1x + 9)$$ isn’t a quadratic! Or at the least, you can have a nice debate about this.
• I leave it to you to decide if you will accept $$(2x + 9)(1x + 0)$$ as a quadratic that satisfies the question above (one of the roots is integer, but the second isn’t).
• Your students be warned: $$(2x + 0)(1x + 9)$$ is the same as $$(1x + 9)(2x + 0)$$!
### Puzzle #2
How many distinct quadratics can be made of the form $$(ax + b)(cx + d)$$ where $$a,b,c,d$$ are all unique and chosen only from $$\{2,0,1,9\}$$?
The answer isn’t $$4! = 24$$ nor is the answer $$\binom{4}{2} = 12$$.
### Puzzle #3
This one depends on Puzzle #2.
If $$a,b,c,d$$ were chosen uniformly randomly without replacement from $$\{2,0,1,9\}$$ what is the probability that $$(ax + b)(cx + d)$$ is a quadratic? What if $$a,b,c,d$$ were chosen uniformly randomly with replacement?
### Puzzle #4
Without using a calculator determine which is larger, $$201^{9}$$ or $$20^{19}$$.
### Puzzle #5
Without using a calculator determine which is larger, $$2^{910}$$ or $$9^{210}$$.
### Puzzle #6
Only using the digits $$2,0,1,9$$ exactly once and assigning them to $$a,b,c,d$$ what’s the largest number that can be formed when written as $$a^{bcd}$$ or $$ab^{cd}$$ or $$abc^{d}$$?
Some points of clarification if there is a sense of ambiguity:
• You can only construct numbers of the three forms given above. The task is to find the construction that will yield the largest value.
• When I write “$$abc$$” I mean digit concatentation, not multiplication. So, if $$a = 2, b = 0, c = 9$$ then $$abc = 209$$
### Puzzle #7
Got fraction troubles? Try this
Choose $$a,b,c,d$$ uniquely from $$\{2,0,1,9\}$$. Which subtraction(s) of the form $$\displaystyle \Big|\frac{a}{b} – \frac{c}{d}\Big|$$ is (are) closest to ONE?
### Puzzle #8
How many four-digit prime numbers can be created using $$\{2,0,1,9\}$$ exactly once? And what are they, if there are any? No calculator allowed!
### Puzzle #9
Try this alphanumeric puzzle for the new year!
### Puzzle #10
Construct two lines, one with positive slope and one with negative slope, such that their intersection is at the point $$(20,19)$$.
Do you enjoy this blog?
Consider supporting with a contribution!
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# Logistic Regression with R
October 21, 2012
By
(This article was first published on Tatvic Blog » R, and kindly contributed to R-bloggers)
## Logistic Regression
In my first blog post, I have explained about the what is regression? And how linear regression model is generated in R? In this post, I will explain what is logistic regression? And how the logistic regression model is generated in R?
Let’s first understand logistic regression. Logistic regression is one of the type of regression and it is used to predict outcome of the categorical dependent variable. (i.e. categorical variable has limited number of categorical values) based on the one or more independent variables. For example, if you would like to predict who will win the next T20 world cup, based on player’s strength and other details. It is a prediction done with categorical variable. Logistic regression can be binomial or multinomial.
In the binomial or binary logistic regression, the outcome can have only two possible types of values (e.g. “Yes” or “No”, “Success” or “Failure”). Multinomial logistic refers to cases where the outcome can have three or more possible types of values (e.g., “good” vs. “very good” vs. “best” ). Generally outcome is coded as “0″ and “1″ in binary logistic regression. We will use binary logistic regression in the rest of the part of the blog. Now, we will look at how the logistic regression model is generated in R.
## Logistic regression in R
To fit logistic regression model, glm() function is used in R which is similar to lm(), but glm() includes additional parameters. The format is
Here, Y is dependent variable and X1, X2 and X3 are independent variables. Function includes additional parameter family and it has value binomial(link=”logit”) which means the probability distribution of regression model is binomial and link function is logit (Refer book R in Action for more information). Let’s generate a simple model. Suppose we want to predict whether a student will get admission based on his two exam scores. For this problem we have a historical data from previous applicants which can be used as the training data set to build a model. The data set contains the following parameters.
1. exam_1- Exam-1 score
2. exam_2- Exam-2 score
In the above parameters, parameter admitted has value 1 or 0 for each observation. Now, we will generate a model that can predict, will student get admission based on two exam scores? For a given problem, admitted is considered as dependent variable, exam_1 and exam_2 are considered as independent variables. The R code for the model is given as below.
>Model_1<-glm(admitted ~ exam_1 +exam_2, family = binomial("logit"), data=data)
After generating the model, let’s try to predict using this model. Suppose we have two exam marks of a student, 60 of exam_1 and 85 of exam_2. We will predict that will student get admission? Following is R code for predicting probability of student to get admission.
>in_frame<-data.frame(exam_1=60,exam_2=86)
>predict(Model_1,in_frame, type="response")
Output
0.9894302
Here, the output is given as a probability score which has value in range 0 to 1. If the probability score is greater than 0.5 then it is considered as TRUE. If the probability score is less than or equal to 0.5 then it is considered as FALSE. In our case 1 or 0 will be considered as the output to decide, will student get admission or not? if it is 1 then student will get admission otherwise not. So I have used round() function to convert probability score to 0 or 1. It is as below.
>round(predict(Model_1, in_frame, type="response"))
Output
1
Output is 1 means a student will get admission. We can also predict for other observations in the above manner. Finally we understood what is logistic regression? And how it works in R? If you want to do the same exercise, Click here for R code and sample data set of above example. In the next blog, we will discuss about a specific problem for Google Analytics data and see how to use logistic regression into?
Would you like to understand the value of predictive analysis when applied on web analytics data to help improve your understanding relationship between different variables? We think you may like to watch our Webinar – How to perform predictive analysis on your web analytics tool data. Watch the Replay now!
### Amar Gondaliya
Amar is data modeling engineer at Tatvic. He is focused on building predictive model based on available data using R, hadoop and Google Prediction API. Google Plus Profile: : Amar Gondaliya
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The security to the murdered president’s “what am I?” riddle party [Part TREE(3)]
François Hollande, famous Cordon Bleu chef that he (like all French presidents) is, has invited a select few guests to his party, where he will be showing off the secret cheese-making process that doubles as a plan of attack against Britain. Your task as a UK spy (who happens to also be an Argentinian agent) is to steal this recipe for disaster. Entrance to the château wherein lies the party, however, is overseen by two guards. One guard always lies, and the other always tells the truth.
You arrive early and hide in some bushes. Guests begin approaching the door, and it becomes clear that there is some kind of password involved.
• One guest arrives by himself. “1”, says a guard. “6”, replies the guest. He is permitted entrance.
• Two more guests show up. “3” is prompted; “2” is the response, and it is apparently correct.
• A group of seven is next to request entry, and the guards see fit to ask them each in turn.
• “2!” “7.”
• “11!” “6.”
• “20!” “4.”
• “4!” “2.”
• “6!” “3.”
The group had been doing fine until that last person, whose throat is promptly slit, along with those of what look to be his wife and son. “You were just off by one!”, says one of the guards, who looks like the sort of person who would tell the truth – but then again, it’s hard to tell with them. You’re not in Argentina anymore.
You think you’ve figured out the pattern, though, and are about to take your turn when another guest arrives, late. The challenge is given – “26” – but before the guest can give an answer, a loud gunshot is heard. Someone shouts out the window that the president is dead! It’s a coup d’état! You must get to the plans before the assassin, but one guard is still performing his duty. You are given the number five as a challenge; what is your response?
• What a story........ looks like Homeland in medieval times..;) – Debanjan Chakraborty Nov 13 '14 at 7:45
• Are the Argentina references and the president's murder hints or just extra red-herring details? – Rand al'Thor Nov 15 '14 at 18:57
• @randal'thor: References to other puzzles on the site. The Argentina ones don’t matter, no. – Ry- Nov 16 '14 at 16:36
• To clarify "until that last person" - does this mean that responding "3" to the guard's "6" is incorrect? Or is the interaction of this last person omitted from the list? – Leo Nov 19 '14 at 21:20
• @Leo: Yes, it means that responding “3” to “6” is incorrect. – Ry- Nov 20 '14 at 21:05
6
The rule is
If the number that the guard tells you is n, respond with the number of letters in the nth word of The French National Anthem (in French, of course).
Here is the first stanza:
French lyrics
Allons enfants de la Patrie,
Le jour de gloire est arrivé !
Contre nous de la tyrannie,
L'étendard sanglant est levé, (bis)
Entendez-vous dans les campagnes
Mugir ces féroces soldats ?
Ils viennent jusque dans nos bras
Égorger nos fils, nos compagnes !
All of the words match, and the incorrect answer (off by one) fits as well!
• The response to 3 was 2, not 6. So no typo. – mmking Jun 12 '15 at 21:21
• This is it! Well done. – Ry- Jun 12 '15 at 22:28
• I wish I could upvote this 20 times, so glad it finally got answered! Great job JLee – alexmc Jun 13 '15 at 4:17
• @alexmc Thanks! I had never seen this puzzle til yesterday. I don't know how I missed it. I just got lucky. The French National Anthem (which I didn't even know existed til I Googled it) was my first idea. Surprising I got it, because I'm usually better at creating puzzles, rather than solving them. – JLee Jun 13 '15 at 11:49
I'm guessing the answer is 0. All the answers are to have an implicit relation with number 5.
1 - 6 = 5 | 3 + 2 = 5| 2 - 7 = 5| 11 - 6 = 5| 20 / 4 = 5
But the last two was not so. As the last person is found wrong but the one before that is not means that guest was questioned by the guard who speaks lies and the last guest was questioned by the guard who speaks truth.
“You were just off by one!”, says one of the guards who looks like the sort of person who would tell the truth – but then again, it’s hard to tell with them.
This means that what the guard told was a lie.
• And the 4 and 2? – QuyNguyen2013 Nov 21 '14 at 14:01
• Also, the respond to answering '3' to '6!' was that its only one off. 6 and 2 do not make 5 in any sense and neither do 6 and 4 (unless you take the average but that is a stretch). – Leo Nov 21 '14 at 14:06
• @quy : That might have been asked by the guard who lies. – mani_nz Nov 21 '14 at 14:48
• @Leo: '3' to '6!' was that its only one off, said by a guard who could be lying. Read it again. It says he could tell the truth but not sure – mani_nz Nov 21 '14 at 14:50
• @mani_nz the OP already said in a previous comment that the lie/truth aspect has no bearing on the puzzle and that it is just a reference. In the event that it DID have a bearing on the puzzle we could just pick and choose which Q&A examples we want to use to fit an operation <- this would not make sense as it would make the solution too loose. – Leo Nov 21 '14 at 15:10
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Learning
is
simple
and
fun!
Register now! Browse subjects
### Theory:
Parallelogram:
A parallelogram is a quadrilateral (four-sided figure) in which the opposite sides are parallel.
In the figure, the side $$AB$$ is parallel to $$CD$$, and $$AD$$ is parallel to $$BC$$.
Important!
• The opposite sides are congruent.
• The two diagonals bisect each other.
• The opposite angles are congruent.
• The consecutive angles are supplementary.
Area of the parallelogram:
The area of any parallelogram is calculated using the formula $$A = b \times h$$ square units.
Where $$h$$ represents the height of the parallelogram and $$b$$ represents the base of the parallelogram.
Example:
If the height and the base of the parallelogram are $$12$$ $$cm$$ and $$6$$ $$cm$$, then find its area.
Solution:
Given $$h$$ $$=$$ $$12$$ $$cm$$ and $$b$$ $$=$$ $$6$$ $$cm$$.
Area of the parallelogram, $$A$$ $$=$$ $$b \times h$$
$$=$$ $$12 \times 6$$
$$=$$ $$72$$ $$cm^2$$
Ways to construct a parallelogram:
A parallelogram can be constructed if one of the following four measurements are given.
(i) Two adjacent sides and one angle.
(ii) Two adjacent sides and one diagonal.
(iii) Two diagonals and one included angle.
(iv) One side, one diagonal and one angle.
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# A topology on the set of all complete theories of a first order language
Let $L$ be a first order language and $\mathcal T$ be the collection of all complete $L$ theories. For each $L$-sentence (closed $L$-formula) $\phi$, define $B_\phi :=\{T \in \mathcal T : \phi \in T\}$. Let $\mathcal B:=\{ B_\phi : \phi \in T \}$ .
I can see that if $T \in B_{\phi_1} \cap B_{\phi_2}$, then $\phi_1,\phi_2 \in T$, then for every model $M$ of $T$, $M \vDash \phi_1, M \vDash \phi_2$ [since $T$ is complete and $\phi_1, \phi_2$ are $L$-sentences ] , and then $M \vDash \phi_1 \land \phi_2$ , so $T \vDash \phi_1 \land \phi_2$ , so $T\in B_{\phi_1 \land \phi_2} \subseteq B_{\phi_1} \cap B_{\phi_2}$.
Thus $\mathcal B$ forms a base for a topology on $\mathcal T$ .
How to show that this topology is compact and Hausdorff ?
Throughout I presume that "complete theory" means "complete consistent theory," and that a theory is complete if it contains (not merely proves) each sentence or its negation.
Hausdorffness and compactness of the topology reflect basic properties of first-order logic: namely, the behavior of negation and the finiteness of proofs, respectively.
EDIT: Before going further, based on your comment below I think it's worth stating explicitly the following: if $T$ is complete (and consistent), then for any sentence $\varphi$ exactly one of $\varphi$ and $\neg\varphi$ is in $T$. In particular, if $T$ is complete and we know $\varphi\not\in T$, then we know $\neg\varphi\in T$.
• Towards Hausdorffness: Suppose $T_0, T_1$ are distinct complete theories. Then there is some $\varphi$ such that $\varphi\in T_0$ and $\neg\varphi\in T_1$ (why?). Do you see a corresponding pair of basic open sets which appropriately separate $T_0$ and $T_1$?
• Towards compactness: It will probably be easier, first, to consider basic open covers - that is, covers by basic opens (not just by opens). Suppose $\mathcal{C}=\{B_\varphi:\varphi\in \Gamma\}$ (for some set of sentences $\Gamma$) is a basic open cover. Then every theory $T$ is in some $B_\varphi$. Now think about the set $\check{\Gamma}=\{\neg\varphi: \varphi\in\Gamma\}$. If $\mathcal{C}$ has no finite subcover, then $\check{\Gamma}$ is finitely consistent (why?) and hence actually consistent, since proofs are only finitely long. Consider a complete theory $T\supseteq\check{\Gamma}$; do you see a problem with this theory as a point in our space? HINT: think about the "cover" $\mathcal{C}$ ...
Of course, it might feel a bit odd that we're not using the compactness theorem in proving the compactness of this space. Ultimately this comes down to the precise way we defined the space. "Complete (consistent) theory" is a proof-theoretic property, and doesn't invoke models in any way, so there's no need to bring semantic notions into the picture.
We could, however, have considered the space whose points are theories of structures - that is, sets of sentences of the form $Th(\mathcal{M})$ for $\mathcal{M}$ a structure (equivalently, complete satisfiable theories). Clearly every such set is complete and consistent, but the converse requires the full strength of the completeness theorem. To show that this space is compact, we can either use the completeness theorem to reduce it to the situation above ... or use the (weaker) compactness theorem to (following otherwise the argument above) go from "$\check{\Gamma}$ is finitely satisfiable" to "$\check{\Gamma}$ is satisfiable," and then let $T=Th(\mathcal{M})$ for some $\mathcal{M}\models\check{\Gamma}$.
The space of complete satisfiable theories can also be thought of as the $T_0$-ification](https://en.wikipedia.org/wiki/Kolmogorov_space) of the (proper class sized) space of all structures, where the basic open sets (well, classes) are those of the form $\{\mathcal{M}: \mathcal{M}\models\varphi\}$ for $\varphi$ a sentence. (If you don't like the set/class issue here, by the Lowenheim-Skolem theorem we can safely restrict attention to structures whose underlying set is a subset of $\mathbb{N}$, and there's only set-many of these; note, however, that this fix depends on the specific properties of first-order logic.)
• how can we find $\phi$ with $\phi \in T_0, \neg \phi \in T_1$ ? – user Apr 29 '18 at 16:50
• @users Well, $T_0\not=T_1$; so either there's some $\varphi$ in $T_0\setminus T_1$, or there's some $\varphi\in T_1\setminus T_0$. Let's look at the first possibility: since $\varphi\not\in T_1$ and $T_1$ is complete, what can you say about $\neg\varphi$ and $T_1$? The other possibility is the same but flipped. (See also my edit.) – Noah Schweber Apr 29 '18 at 16:52
• Yes thank you ... complete theories are indeed decidable – user Apr 29 '18 at 16:54
• @users Are you familiar with Zorn's lemma? If so just think about a maximal consistent theory containing $\check{\Gamma}$ - why does such a thing exist (Zorn) and why must such a thing be complete (if not, you could make it bigger)? If not, I think it's a good idea to first think about the case when the language is countable. We list all the sentences as $\psi_n$ ($n\in\mathbb{N}$), and build an extension of $\check{\Gamma}$ via a "greedy algorithm:" throw in $\psi_0$ if that's consistent, otherwise throw in $\neg\psi_0$; now throw in $\psi_1$ if that's consistent, otherwise $\neg\psi_1$; etc. – Noah Schweber Apr 29 '18 at 20:30
• This process clearly produces a complete theory containing $\check{\Gamma}$, and doesn't need any set theory. The fully general case, however, does require (a weak form of) Zorn's lemma, though, so ultimately you're not going to escape the creeping claws of set theory entirely. – Noah Schweber Apr 29 '18 at 20:30
To prove that the space is Hausdorff, consider two $L$-theories $T_1\neq T_2$. Since they are different and complete, there is $\phi$ such that $\phi\in T_1$ but $\phi\not\in T_2$, which implies $\neg\phi\in T_2$. So the open sets $B_{\phi}$ and $B_{\neg\phi}$ witness that the space is Hausdorff.
Suppose that $\mathcal{D}:=\{B_{\phi_i}:i\in I\}$ is an open cover of the space (we can assume that all $B_i$ are basic open sets). If $I$ is finite, we have nothing to do, so suppose $I=\omega$ (there are no other possibilities since the space is second-countable). Suppose for a contradiction that for all $I'$ finite subset of $\omega$ $\{B_{\phi_i}:i\in I'\}$ is not a covering of $\mathcal{T}$. Then, there is an $L$-structure satisfying $\bigwedge_{i\in I'}\neg\phi_i$, which, by compactness (not surprisingly!) means that the theory $T$ containing all the $\neg\phi_i$ is satisfiable. But then, if $T'$ is any completion of $T$, $T'$ cannot be in $\bigcup \mathcal{D}$, a contradiction.
• why is the space second countable ? I don't see it unless $L$ is countable ... – user Apr 29 '18 at 17:01
• I was actually assuming $L$ to be countable. But it does not matter. I am not using the fact that $I$ is $\omega$, you may just replace that sentence with "$I$ is infinite". – Leo163 Apr 29 '18 at 17:07
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# Dispersive Flies Optimisation
## What is DFO?
Dispersive Flies Optimisation Research (DFO) [1] is a swarm intelligence algorithm. That is, it is an algorithm where individual agents follow simple instructions to search for a solution in in a matrix representing an N-dimensional search space [2]. The agents are set up so that they can communicate with one another and adapt their search trajectory based on the behaviour of their neighbours.
The algorithm aims to exploit the emergent adaptive intelligence of the swarm to find an optimal solution to a given search problem. Its behaviour is inspired by the swarming behaviour of flies hovering over food sources and it has been shown to find optimal or “good enough” solutions [3] faster than randomness and of the standard versions of the well-known Particle Swarm Optimisation, Genetic Algorithm (GA) as well as Differential Evolution (DE) algorithms on an extended set of benchmarks over three performance measures of error, efficiency and reliability [4].
The algorithm was first proposed by Mohammad Majid al-Rifaie, a computing lecturer at Goldsmiths, University of London, in his paper published in 2014.
## How does it work?
Just like flies hovering over food, animals, your face, or p– ehm — feces, the algorithm creates a swarm of agents are initially scattered at random, but that eventually converge towards that agent which has found the most interesting feature in the space they are in.
The algorithm does not centrally coordinate the agents’ movements but lets communication between neighbouring flies decide where each of them will move towards. The agents are also programmed to disperse if a certain stochastic function exceeds a preset output threshold, so as to mimic flies running away from a threat and -possibly- finding an even more interesting point of interest.
As described in its original paper,
DFO consists of two tightly connected mechanisms, one is the formation of the swarms and the other is its breaking or weakening.
Unlike real life flies, the agents in the algorithm are not just hunting for food across a 3D search space but live in a matrix made up of all the dimensions that we want. This is because the position vector of each virtual fly represents the variables of a problem that we are trying to solve and for which we don’t have a mathematical formula to compute the solution(s). An example of this could be the identification of certain patterns in a data set, or finding a viable (if not optimal) solution to unsolved path-finding problems. It’s been shown, for example, that it is possible to achieve a good automatic identification of microcalcifications on mammographs using DFO as a method of image analysis[6].
## The hunt
Nature has equipped flies with a nervous system that scans their sensory world in search for salient markers that can the food. When a fly finds something interesting, its flight behaviour is picked up by neighbouring flies, which then converge towards the position of the first fly.
In DFO, the search for food is carried out by a fitness function that tests how good the position values of the fly are at solving the problem at hand –Remember: the position vector of each fly is the set of variables whose variance describe (within reason) all the possible cases in our problem. We could then say that the position of the fly in the global search space is the fly’s attempt (hypothesis) at solving the problem by trial and error.
Once a fly has checked how good its location is as compared to a certain reference fitness value, it scans its two closest neighbours to see whether they have found a better set of variables (position vectors) and it then moves across the search space on the basis of the knowledge it has. The algorithm also stores the position of the fly that has found the best solution so far and uses it to randomly drags other flies towards is direction.
Finally, each fly runs a dispersive stochastic method that may push them to fly away in a random direction. This is to simulate the arrival of a threat in the real world and it helps the algorithm avoid getting stuck on a local optima.
The mathematical description of the fly’s position update method (for each coordinate) is as follows:
$x_{i}^{t=n} = x_{\text{i BN}}^{t=n-1} + U(0,1)\cdot(x_{\text{i SB}}^{t=n-1} - x_i^{t=n-1})$
Where:
• $x_{i} ^ {t=n}$ is the dimension, or coordinate, ” i ” of the fly’s position at time ” n “ that we want to find.
• $x_{\text{i BN}} ^ {t=n-1}$ is the same coordinate but in the current position of the best of the two neighbouring flies
• $x_{\text{i SB}}^{t=n-1}$ is as above but of the best fly in the swarm (elitist approach)
• $U(0,1)$ is the denomination for a a random number between 0 an 1 as chosen by a normal distribution function
If, however, the random function responsible for mimicking the arrival of a threat outputs a value higher than a certain threshold, the update function will randomise the specific fly’s coordinate as below:
$x_{i}^{t=n} = x_{\text{i min}} + U(0,1)\cdot(x_{\text{i max}} - x_{\text{i min}})$
Where $x_{\text{i min}}$ is the minimum value that the dimension, or coordinate, ” i ” of the fly’s position can take in the whole space and $x_{\text{i max}}$ is the maximum.
Below we can see an example for a 2 dimensional problem-space where the current fly we are examining updates its position. We assumed that the random number generating function returned 0.3 for the first dimension and 0.8 for the second dimension –you can ignore the top left and bottom right drawings on the picture.
The overall behaviour is therefore expected to be that of convergence (of the swarm) towards optimal points. Flies will in fact tend to go gravitate (hover) in the direction of those other flies that are in better positions as evaluated by a given fitness function.
In the above visualisation, flies are hovering towards a better and better positions. However, they are also getting stuck in a local optima.
With DFO we hope that by exploring other areas at random every time the dispersion threshold is met, the swarm will eventually converge to a good enough if not globally optimal solution of the problem within a reasonable amount of time.
## Possible Applications
We have mentioned that DFO has been used with positive results for searching an image-function (search space) in order to find certain patterns [6]. Being a promising search algorithm within the so-called Swarm Intelligence (SO) branch of computer science research, DFO could also work well in data mining, where Ant Colony Optimization (ACO) and Particle Swarm Optimization (PSO) have already been deployed with interesting results [7].
As a Music Computing student, I personally would be interested to see if DFO could be used to come up with musical structures that can be considered musical by our standards. To achieve this I think that one are of the musical experience should be selected as the focus, for example the melodic/harmonic structure of a piece of music.
Last year I have exploited with some interesting results a Genetic Algorithm coupled with a tree-structured grammar of Western notes to automatically generate melodic phrases. Since it’s been shown that Swarm Intelligence can be applied as an extension of Lyndenmayer grammars [8], it would be interesting to test out DFO’s performance when asked to come up with meaningful structures.
As always with perceptual experiences and aesthetic art, one of the hardest problems with the above application would be to come up with a meaningful fitness function and a correct set of problem variables and algorithm parameters. In this respect though I think that a possible approach could be the coupling up of DFO with some Deep Learning [9] strategies. This way we might try to approximate the “arousal” and “linking” function of certain melodic/harmonic patterns based on the average response by some chosen population sample.
Another area I am interested in at the moment is detecting meaningful patterns in a stream of data, such as extracting meaningful information from noisy market charts (as in ForEx, Stocks, Bitcoin, etc) or video analysis in a moving image. Again I can see DFO being a good candidate for exploring the data-field in search for specific patterns. Those patterns might then be quickly pointed out to the user as well as used for making some sort of automatic informed decision. Particle Swarm Optimization (PSO) has already produced some interesting results when applied to Computer Vision [10], so DFO may also provide a promising problem-solving vehicle in those files.
## Implementing DFO
I implemented a version of DFO in C++ that is now available on my gitHub. This is basically a porting of the original java code as created by Mohammad Majid al-Rifaie, which has helped me understand how the different elements of the algorithm can be coded as different classes (in a OOP fashion) that can then be set to explore any problem’s space-function that we like.
The repository includes a README and the code is thoroughly commented.
Using the algorithm I then went on and tried to tackle the following minimisation problem:
fitness function = 12/x + 18/y * xy
x & y are positive
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# Lupita Nyong'o Trailers
## 12 Years A Slave - Featurettes
Year 2014 | Genre Dramas
Synopsis - Director Steve McQueen joins the stars of '12 Years A Slave' to praise the immense level of acting skill that went into creating the movie. Among those actors were main...
Directed by
Starring , , , , , Lupita Nyong'o, , , , , , Adepero Oduye, , , , , , , , , Anwan Glover, Mustafa Harris, Ashley Dyke, Adepero Oduye
## Non-stop - Trailer
Year 2013 | Genres Action, Adventure
Synopsis - Bill Marks is a U.S. federal air marshal who ironically can't stand plane journeys. His hatred for flying is only about to get a lot worse when an anonymous person...
Directed by Jaume Collet-Serra
Starring , , , , Lupita Nyong'o, , , , , , , , , , O.T. Fagbenle
## 12 Years A Slave - Trailer
Year 2013 | Genre Dramas
Synopsis - Solomon Northup was a well-educated man from a successful family living in upstate New York with his wife and three children. He was categorised as a free black man and...
Directed by
Starring , , , , , Lupita Nyong'o, , , , , , Adepero Oduye, , , , , , , , , Anwan Glover, Mustafa Harris, Ashley Dyke, Adepero Oduye
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# American Institute of Mathematical Sciences
May 2019, 18(3): 999-1021. doi: 10.3934/cpaa.2019049
## Large time behavior of solutions of local and nonlocal nondegenerate Hamilton-Jacobi equations with Ornstein-Uhlenbeck operator
1 Laboratoire de Mathématiques de l'INSA de Rouen, 685 Avenue de l'Université, 76800 Saint-Étienne-du-Rouvray, France 2 On leave from IRMAR, Université de Rennes 1, France
Received June 2017 Revised March 2018 Published November 2018
We study the well-posedness of second order Hamilton-Jacobi equations with an Ornstein-Uhlenbeck operator in $\mathbb{R}^N$ and $\mathbb{R}^N× [0, +∞).$ As applications, we solve the associated ergodic problem associated to the stationary equation and obtain the large time behavior of the solutions of the evolution equation when it is nondegenerate. These results are some generalizations of the ones obtained by Fujita, Ishii & Loreti 2006 [19] by considering more general diffusion matrices or nonlocal operators of integro-differential type and general sublinear Hamiltonians. Our work uses as a key ingredient the a-priori Lipschitz estimates obtained in Chasseigne, Ley & Nguyen 2017 [10].
Citation: Thi Tuyen Nguyen. Large time behavior of solutions of local and nonlocal nondegenerate Hamilton-Jacobi equations with Ornstein-Uhlenbeck operator. Communications on Pure & Applied Analysis, 2019, 18 (3) : 999-1021. doi: 10.3934/cpaa.2019049
##### References:
show all references
##### References:
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2019 Impact Factor: 1.105
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# Tag Info
0
Yes you would actualy see the car with its speed added with your speed.
1
Your understanding is correct. And switching between different perspectives (what physicists would call different inertial frames of reference) like that is a very useful tool in physics, because it turns out that the laws of physics have the same form no matter which inertial frame of reference a problem is described in. For more information, see ...
2
Rough impressions can be misleading. The other car really is moving 200 km/h from your point of view. One thing to keep in mind is that you tend to perceive motion more readily when it is closer to you. This is at least partly due to the fact that you really only can see angular speed across your field of view (like degrees per second). To convert this into ...
3
Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...
1
Nothing happens to the initial angular momentum. It is simply irrelevant. Let's imagine that the pendulum is placed on a non-rotating body (or at the earth's equator). When the string is cut, the pendulum falls straight to the center. The path it traces is a straight line. In the non-equatorial case, the pendulum structure has some rotation. At the ...
0
You do have to define the velocity with respect to something. The idea (I think) behind defining wrt fixed stars is that the Earth is in motion wrt the stars, too. Barring friction, the moment the rope is burned, the pendulum is only under the influence of gravity and the restraining force of the suspension rope -- analogous to swinging a weight around ...
1
Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$ With $x= h \cot \theta$, ...
1
I've heard that inertial frames are frames in which Newton's laws hold. The modern view of Newton's first law is that it defines the concept of an inertial frame. It also, at least conceptually, provides a mechanism for testing whether a frame of reference is an inertial frame. Suppose you know that no forces act on some particle. If that particle ...
1
1)Definition: An inertial frame of reference is a frame of reference where Newton's first law applies (uniform motion if without external force). Now if we have other frame of references that are moving relative to this inertial frame with uniform relative velocities, then all the others are also called inertial frame of references. 2)Transformation between ...
4
I believe the difference comes from the fact that forces can do different amounts of work in different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work ...
0
Physicist created momentum as the property of the system that is conserved if on the system act only internal forces. A force is defined internal in the system if, the force itself and its reaction are applied on the system. From the previous hypotesis you can demostrate that momentum is defined as $$\vec{p} = \sum m_i\vec{\upsilon}_i$$ The last example ...
0
An elastic collision means that the over all kinetic energy of the entire system before and after the collision is the same. So the ball can bounce off the wall, and the wall can recoil in such a manner that you have an elastic collision.
3
Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$\sum \mathbf p_{i}=\sum\mathbf p_f$$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...
2
It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...
1
Linear momentum will be conserved when the Lagrangian that describes your system is unchanged by translations in space. This is a consequence of Noether's theorem, and it's as close as you're going to get to a fundamental explanation for the conservation of momentum. In general a Lagrangian written in the coordinate system of an accelerating observer is not ...
0
Okay, let's say you have a friend staying at home measuring the time and observing you with a telescope - the whole situation will be described from his point of view. Namely, we will use his measure of distance $x$ and the time $t$ he measures with the stopwatch in his hand. Special relativity tells us that he has a means of calculating your "proper ...
0
My apologies that I can't follow your question enough to answer everything, in particular I can't see how you project your images. If you extended them in what look like straight lines to you, you'll get radial lines expanding from your position, but it seemed like you wanted a Cartesian grid. However I do want to answer the part about strain and its ...
5
The key is the coriolis force. The coriolis force is $F_c = -2m\Omega \times v$. Here $\Omega$ is the rotation of the frame of reference and $v$ is the linear speed of the satellite. If you do the calculations, left as an exercies for the reader, you'll get the missing force. In case 2 the coriolis force is 0, because the velocity $v$ has to be used in ...
0
I'm afraid your statement that magnetic field at a point can have only on(e) "constant value" is not true, and you will have to learn a bit about special relativity to understand why. There is a classic thought experiment, which is to imagine the electromagnetic fields due to a charged particle at rest in what we'll call the stationary frame. Clearly, the ...
0
Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix $a$ that relates the two sets of axes is simply the identity matrix. $dG'$ is a vector, so $a_{ji}dG_j' = dG_i'$ is simply equivalent to the statement that with $I$ the identity matrix and $V$ an arbitrary vector, $IV=V$. Answer ...
0
Planes of simultaneity in special relativity don't really mean much of anything. The real physical structure of spacetime is in the light cones. The takeaway from "relativity of simultaneity" is not that there are "different time orderings for different observers", but rather that there is no meaningful time ordering for spacelike separated events. They ...
11
What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ...
1
"Most probably in reality there are some extremely complex laws and equations which makes this question more complicated." Not really. The equations are rather straightforward. Let's measure velocities in units of the speed of light and let's denote the velocity of $B$ as observed by $A$ as $v_{BA}$, the velocity of $A$ as observed by $C$ (the ...
10
It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ...
1
That light has a fixed velocity in vacuum comes from observations . In order to fit the data Lorenz transformation were imposed on the rigorous mathematical model for electromagnetism, Maxwell's equations. It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to ...
4
In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no". There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the ...
1
"How do we know that clocks slow down relative to each other?" Experimentally. This has been observed many times in the lab. The same answer is true for ANYTHING in physics and science in general. We only know that it is true, because we have experimental evidence for it.
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No, since by the principle of relativity: A body in constant velocity motion cannot determine whether it is in motion in a certain direction or whether everything else is in motion in the other direction. No physical experiment can determine this hence for all purposes a body in motion will simply claim that it is at rest while the other body is in motion. ...
0
But magnetic field at that point can have only on "CONSTANT VALUE". Not true. If a particle is acted on by some combination of electrical and magnetic forces in one frame of reference, then in another frame of reference, it will be a different combination of electrical and magnetic forces. It's possible to have a force that's purely electrical in one ...
0
If you set your reference frame to be fixed to the swing (let's call the origin the pivot point of the swing), you are now dealing with a rotating reference frame. In a rotating reference frame, all objects observe a centrifugal force that pulls them outwards from the origin. For any stationary object in your rotating reference frame (the seat on the ...
0
Your reasoning should be improved. In an inertial system there are no centrifugal force. You may in that system speak about a centripetal force, which is not a real force but an expression for how much net force is needed for the object to perform a circular motion. You probably mean that the expression for the centripetal force should be equal to the ...
1
The crucial point here is, that a reference frame where photons are at rest simply cannot be defined in special relativity: There is no Lorentz transformation which transforms you from a given inertial system into a reference frame or a space-time where some photon is at rest. This, however, means that concepts such as travelled distance and proper time do ...
0
Photons are not considered as observers which would be able to observe their proper time. But if you would do so, their hypothetical proper time would be zero. You obtain this result by multiplying the time observed by any real observer with reciprocal gamma (which is 0 for v=c), see http://en.wikipedia.org/wiki/Time_dilation. However, you may not forget ...
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Why do i have the sense that this can have a meaningful physical interpretation (even if velocities are both $c$)? First lets assume that both velocities are $c$ or $1$ in the question's notation. Then the relativistic velocity addition formula simply gives: $1_A \oplus 1_B = 1_B \oplus 1_A = 1_{tot}$, correct? Yes, does the fact that this is mathematicaly ...
2
To make this easy, I will assume that all speeds are small compared to the speed of light $c$. The motion is in one dimension, so the trajectory of the space vehicle will be the following: $$x(t) = x_0 + v_0 t + \frac12 a t^2$$ We can easily set $x_0 = 0$ if the put the origin of the coordinate system at the start point where the space vehicle started. If ...
4
Your friend is correct. The acceleration of the projectile is determined by the thrust its rocket motor can produce. If the acceleration of the projectile is $2g$ then the thrust of its motor would be $2mg$. But launching the projectile from your space vehicle can't increase it's thrust. You can increase its initial velocity, but once the projectile has ...
0
You are right. Torques are not free vectors, the same way that linear velocity is not a free vector and it is associated with a specific point. When the point moves, the components of torque and velocity change according to the same transformation law \begin{align} \vec{v}_B & = \vec{v}_A + (\vec{r}_{A}-\vec{r}_B) \times \vec{\omega} \\ \vec{\tau}_B ... 1 Torque is defined as the cross product of two vectors: r and F. r points from the pivot point to the point where the force F is applied. Both vectors are not bound and stay the same vectors no matter where you put them. However, if you combine them via mathematical operations, you have to stick to their rules (here: T1=r2*F3-r3*F2 etc...). There are no ... 2 Some key reading if you want to understand this stuff is chapters two to five of the IERS Technical Note 36, the IERS Conventions (2010). It's not just the J2000/FK5 frame (aka the EME2000 frame) that is associated with some epoch date. Every Earth-centered inertial frame has some epoch date. There are two fundamental reasons why this must be the case: ... 0 A simple pendulum would be a good experiment to detect both non-inertial frames - rotating and linearly accelerating: If you know weight of the pendulum in inertial frame, in rotating frame, its weight would decrease because of radially outward centrifugal force acting on it. (If earth stops rotating, our weight would increase!) Think: what would happen if ... 4 That is exactly the point: if the field is in a vacuum with respect to observer A, and observer B accelerates uniformly with respect to A, then B will observe a field state with nonzero particle content. It doesn't really matter whether you talk about different observers on different frames of reference, or of a single observer who 'switches' their frame of ... 2 Yes, I can think of two ways to do this and there may be more. In a rotating frame the acceleration is a function of distance from the pivot. If B has sufficiently precise instruments the variation of acceleration with position will be detectable. However if B is confined to a very small space, or doesn't have precise enough instruments the variation of ... 1 The friend rotating and experiencing the centrifugal force may observe several effects that his linearly accelerating friend doesn't: the acceleration at different points of the box is slightly different i.e. the apparent gravitational field is non-uniform there is the extra Coriolis force acting on objects that are moving relatively to the rotating frame ... 1 Where's the flaw? Here's one: Let γ be the squareroot of 1−v²/c² but, in fact,\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$But, there is a flaw in your reasoning too so correcting the error in the formula for \gamma will not get you to the correct answer. Let's work it out with coordinates to see explicitly what's going on. Let the ... 0 Your question exposes the importance of defining notions in physics unambiguously and universally in terms of "How to measure?". As Einstein put it explicitly (however, referring specificly only to the notion of "simultaneity", and unfortunately only as late as 1917): "We thus require a definition of simultaneity such that this definition supplies us with ... 2 For Special Relativity (SR) i think the Michelson-Morley experiment is compatible and provides a verification of SR principle (some other formulations are also compatible with the experiment). Quantum Field Theory and especially the Dirac prediction and verification of positron is also a verification of SR (and many other expreriments in this context) For ... 0 I don't have the reputation to comment, so I'll comment about David Hammen's (accepted) answer here. His conclusion is correct, but mentions "Those formulae do imply a singularity for the clock that is closest to you" in reference to the forumale he thought you were referring to. He also mentions "In between, you'll get a nice continuous change from faster ... 1 Actually it's not that difficult (but a neat problem), there's only one crucial step in the development that I will show you, but let's start from a bit earlier. Let's first write out the two forces on interest here (in terms of magnitudes, as we know already they're on parallel trajectories): The coulomb repulsion between the charges:$$F_C = ...
1
A more particle-based view of zakk and Moonraker's observation that photons experience zero time between emission and absorption is this: The postulated event that a massless particle (say a photon) could emit another massless particle (say another photon) in the retrograde direction — or in any direction for that matter — is always zero. A photon from its ...
0
I want to check that I get it right. [...] [...] we would be associating time with objects and not with the points of space itself. In physics, the word "time" is used in various related but quite different meanings (which I describe in some detail below). Therefore, if you care to "get it right" (which is of course commendable) you should avoid ...
Top 50 recent answers are included
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### Six Is the Sum
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### (w)holy Numbers
A church hymn book contains 700 hymns. The numbers of the hymns are displayed by combining special small single-digit boards. What is the minimum number of small boards that is needed?
### ABC
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
# Number Detective
### Why do this problem?
This problem offers opportunities to reinforce the language and characteristics of numbers such as odd, even, multiple. However it is also a problem which can be tackled in a systematic way, or, more elegantly, through insightful reading of the clues. Comparing the two methods can be a useful exercise in considering what mathematical thinking looks like.
### Possible approach
Put three two-digit numbers, such as $93$, $56$, $75$ on the board and ask the children which is the odd one out and why. Listen for explanations which include descriptions of place value, odds and evens, and perhaps multiples.
Say that you are going to choose one of them secretly and they can ask just one or two questions to see if they can find which one it is. For each correct guess ask the children to justify their answer.
Put $23$, $45$, $62$, $101$, $94$ on the board and tell then you have chosen one of these. What questions would be good ones to ask and why? Make a record of the questions on the board so that the children can refer to them. Again, ask the children to justify any conclusions they come to.
Then offer the problem to pairs of children. Say that you are interested in the mystery number but also how they know and you will be asking for a description of what they did and in what order, so they may like to keep some notes.
Bring the children back together and ask a few pairs to describe their method. Listen for those who go through the clues in order eliminating the impossible numbers, and explain that working systematically is a very important skill for a mathematician.
But also look out for the children who have scanned through the clues to find one that is more useful and saves some work - for example
'The sum of the two digits is a multiple of five.'
which means the magic number must be 46 or 64. After that it only needs one more carefully chosen clue to distinguish between them:
'The digit in the tens place is greater that the digit in the unit (or ones) place.'
Bring out the importance of scanning through to see which clues might be the most useful.
### Key questions
Which question shall we ask first? Why?
What does 'multiple of five' mean?
### Possible extension
An obvious extension is for the children to make up their own examples for each other. They could repeat the format of the given problem, or play 'what's my number?' with a partner, where they try to find out what the mystery number is in the minimum number of guesses. Give an opportunity for them to describe why some questions are more useful than others.
### Possible support
Sometimes it can be confusing to be given too much information at one time. Write the clues and the numbers out on separate cards. Spread the number cards out and offer the clues one at a time, encouraging discussion of the characteristics of each number in response to the clue.
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On Quantum Chosen-Ciphertext Attacks and Learning with Errors
Large-scale quantum computing is a significant threat to classical public-key cryptography. In strong "quantum access" security models, numerous symmetric-key cryptosystems are also vulnerable. We consider classical encryption in a model which grants the adversary quantum oracle access to encryption and decryption, but where the latter is restricted to non-adaptive (i.e., pre-challenge) queries only. We define this model formally using appropriate notions of ciphertext indistinguishability and semantic security (which are equivalent by standard arguments) and call it QCCA1 in analogy to the classical CCA1 security model. Using a bound on quantum random-access codes, we show that the standard PRF- and PRP-based encryption schemes are QCCA1-secure when instantiated with quantum-secure primitives. We then revisit standard IND-CPA-secure Learning with Errors (LWE) encryption and show that leaking just one quantum decryption query (and no other queries or leakage of any kind) allows the adversary to recover the full secret key with constant success probability. In the classical setting, by contrast, recovering the key uses a linear number of decryption queries, and this is optimal. The algorithm at the core of our attack is a (large-modulus version of) the well-known Bernstein-Vazirani algorithm. We emphasize that our results should *not* be interpreted as a weakness of more »
Authors:
; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10106032
Journal Name:
Theory of Quantum Computing, Communication, and Cryptography 2019
3. One of the primary research challenges in Attribute-Based Encryption (ABE) is constructing and proving cryptosystems that are adaptively secure. To date the main paradigm for achieving adaptive security in ABE is dual system encryption. However, almost all such solutions in bilinear groups rely on (variants of) either the subgroup decision problem over composite order groups or the decision linear assumption. Both of these assumptions are decisional rather than search assumptions and the target of the assumption is a source or bilinear group element. This is in contrast to earlier selectively secure ABE systems which can be proven secure from either the decisional or search Bilinear Diffie-Hellman assumption. In this work we make progress on closing this gap by giving a new ABE construction for the subset functionality and prove security under the Search Bilinear Diffie-Hellman assumption. We first provide a framework for proving adaptive security in Attribute-Based Encryption systems. We introduce a concept of ABE with deletable attributes where any party can take a ciphertext encrypted under the attribute string and modify it into a ciphertext encrypted under any string where is derived by replacing any bits of with symbols (i.e. deleting" attributes of ). The semantics of the systemmore »
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# Which part of a massless wire will break at the breaking stress?
Consider an ideal massless wire of length $$L$$, uniform cross-sectional area $$A$$, Young's modulus $$Y$$ suspended from the ceiling, with a load of weight $$W$$ suspended at the end. There is no variation in the value of acceleration due to gravity $$g$$ along the length of the wire. Any other factors that vary in real life are also constant here.
When the weight is just enough to produce the breaking stress in the wire, the wire breaks (fractures) at a point.
Which point will the wire break at? Is it entirely random (anywhere along the length) or is there a definite point (like the midpoint)?
• From my knowledge of tensile testing, the location most likely to fracture would be at the jaw that holds the wire. For that reason, dog bone shaped test specimens are typically used where the wide ends are secured by the jaws making fracture less likely there, and with the narrow section in between being the subject of the test. Apr 14 at 15:34
Since nothing in the question identifies any point on the wire as particularly remarkable—the stress is perfectly uniform—we can't select any particular point as a location of predicted failure. We might as well assume the failure location to be random.
In practice, the top connection point would draw attention, for at least several reasons:
1. No material is truly massless, so the stress (arising from the suspended weight plus the weight of the wire) would be highest there.
2. Connection points can involve additional stresses because of applied constraints; if the connection is rigid, for example, then the connection point must bear an axial load plus a bending moment.
3. Connection points can involve discontinuities that can create stress concentrations.
4. Connection points can feature additional surface area that may increases susceptibility to corrosion or crack propagation.
As supported by the relevant physics models, it's therefore good engineering practice to build up the connection point cross-section area, and reduce attachment constraints (use a lubricated hinge, for example), and constrain the motion of the hanging object, and protect the top end from environmental damage, and so on, to address the possible failure modes listed above.
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JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $${3 \over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :
A
$$C' = {{3 + K} \over {4K}}{C_0}$$
B
$$C' = {{4 + K} \over {3}}{C_0}$$
C
$$C' = {{4K} \over {K + 3}}{C_0}$$
D
$$C' = {{4} \over {3 + K}}{C_0}$$
2
JEE Main 2021 (Online) 26th February Morning Shift
MCQ (Single Correct Answer)
+4
-1
Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is $${{15} \over 4}$$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $${{{C_2}} \over {{C_1}}}$$.
A
$${{15} \over {11}}$$
B
No Solutions
C
$${{29} \over {15}}$$
D
$${{15} \over {4}}$$
3
JEE Main 2021 (Online) 25th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle '$$\alpha$$' with the plates. It leaves the plates at angle '$$\beta$$' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be :
A
$${{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$$
B
$${{\cos \beta } \over {\cos \alpha }}$$
C
$${{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}$$
D
$${{\cos \beta } \over {\sin \alpha }}$$
4
JEE Main 2021 (Online) 24th February Morning Shift
MCQ (Single Correct Answer)
+4
-1
Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :
A
4 : 1
B
1 : 2
C
2 : 1
D
1 : 4
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dc.contributor.author Narayan, R. dc.contributor.author McKinney, J. C. dc.contributor.author Farmer, A. J. dc.date.accessioned 2019-09-22T14:23:54Z dc.date.issued 2007 dc.identifier.citation Narayan, R., J. C. McKinney, and A. J. Farmer. 2007. “Self-Similar Force-Free Wind from an Accretion Disc.” Monthly Notices of the Royal Astronomical Society 375 (2): 548–66. https://doi.org/10.1111/j.1365-2966.2006.11272.x. dc.identifier.issn 0035-8711 dc.identifier.issn 1365-2966 dc.identifier.uri http://nrs.harvard.edu/urn-3:HUL.InstRepos:41384883 * dc.description.abstract We consider a self-similar force-free wind flowing out of an infinitely thin disc located in the equatorial plane. On the disc plane, we assume that the magnetic stream function P scales as P proportional to R(nu), where R is the cylindrical radius. We also assume that the azimuthal velocity in the disc is constant: v(phi) = Mc, where M < 1 is a constant. For each choice of the parameters nu and M, we find an infinite number of solutions that are physically well-behaved and have fluid velocity <= c throughout the domain of interest. Among these solutions, we show via physical arguments and time-dependent numerical simulations that the minimum-torque solution, i.e. the solution with the smallest amount of toroidal field, is the one picked by a real system. For nu >= 1, the Lorentz factor of the outflow increases along a field line as gamma approximate to M(z/R(fp))((2-nu)/2) approximate to R/R(A), where R(fp) is the radius of the foot-point of the field line on the disc and R(A) = R(fp)/M is the cylindrical radius at which the field line crosses the Alfven surface or the light cylinder. For nu < 1, the Lorentz factor follows the same scaling for z/R(fp) < M(-1/(1-nu)), but at larger distances it grows more slowly: gamma (z/R(fp))(nu/2). For either regime of nu, the dependence of gamma on M shows that the rotation of the disc plays a strong role in jet acceleration. On the other hand, the poloidal shape of a field line is given by z/R(fp) approximate to (R/R(fp))(2/(2-nu)) and is independent of M. Thus rotation has neither a collimating nor a decollimating effect on field lines, suggesting that relativistic astrophysical jets are not collimated by the rotational winding up of the magnetic field. dc.language.iso en_US dc.publisher Oxford University Press dash.license LAA dc.title Self-similar force-free wind from an accretion disc dc.type Journal Article dc.description.version Version of Record dc.relation.journal Monthly Notices of the Royal Astronomical Society dash.depositing.author Narayan, Ramesh::dc7afe5d74d62c7b451015317ea2ccbe::600 dc.date.available 2019-09-22T14:23:54Z dash.workflow.comments 1Science Serial ID 66200 dc.identifier.doi 10.1111/j.1365-2966.2006.11272.x dash.source.volume 375;2 dash.source.page 548-566
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# How to design model in Django using GeoDjango?
I am using Postgresql 9.0 and PostGIS 2.0. I am using Django 1.3 and GeoDjango.I have model with id,latitude,longitude and placename. All the places are in USA.
My requirement is following -
User will give his address. We use his address to get latitude and longitude, and then find places within 'x' miles.
So, should I use PointField in the model? Also, how do I convert users latitude\longitude into geometry object to use dwithin?
-
Your model must be like this:
# Import django module
from django.contrib.gis.db import models
class Point(models.Model):
placename = models.CharField(max_length=50)
geom = models.PointField(srid=4326)
objects = models.GeoManager()
def __unicode__(self):
return '%s %s %s' % (self.name, self.geom.x, self.geom.y)
from django.contrib.gis.geos import *
from django.http import HttpResponse
from django.contrib.gis.measure import D # D is a shortcut for Distance
pnt = fromstr(str(po), srid=4326)
qs = Point.objects.filter(point__distance_lte=(pnt, D(km=5)))
return HttpResponse(your choose for returning object)
it will give you objects in 5 km square.
Geodjango Distance Lookups
Availability: PostGIS, Oracle, SpatiaLite
The following distance lookups are available:
distance_lt
distance_lte
distance_gt
distance_gte
dwithin
Dont forget, qs is in array and it filter all objects for your request that you have to loop for getting all objects one by one..
Now you can convert your result to geojson or anything for your app...
i hope it helps you....
-
I have already done this but getting error when firing distance queries – carlsen Apr 26 '12 at 6:28
What errors are you getting? – blah238 Apr 26 '12 at 6:34
first of all, you have to make geometry column in your db from your x and y. because of spaial query, you will need this...and give me some detail about your error... – Aragon Apr 26 '12 at 6:46
Very helpful, thanks! – Mark Bell Mar 23 '13 at 9:31
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# Solving Radical Equations and Inequalities
This section covers:
Now that we’ve learned about Quadratics and Factoring and did some work with square roots, we can go back and revisit solving radical equations and inequalities, with special emphasis on square root functions. We solved some radical equations in the Solving Exponential and Radical Equations portion in the Exponents and Radicals in Algebra section, but now we can work with more complicated equations where we can multiply binomials to find the answers!
For working with inverse functions of radicals, see the Inverses of Functions section. For factoring and solving with Exponents, see the Exponential Functions section.
# Parent Graphs of Radical Functions
First of all, let’s see what some basic radical function graphs look like. The first set of graphs are the quadratics and the square root functions; since the square root function “undoes” the quadratic function, it makes sense that it looks like a quadratic on its side. But the important thing to note about the simplest form of the square root function $$y=\sqrt{x}$$ is that the range ($$y$$) by definition is only positive; thus we only see “half” of a sideways parabola. The domain ($$x$$) is always positive, too, since we can’t take the square root of a negative number.
Quadratic Function Square Root Function Domain: $$\left( {-\infty ,\infty } \right)\text{ or }\mathbb{R}$$ Range: $$\left[ {0,\text{ }\infty } \right)$$ Domain: $$\left[ {0,\text{ }\infty } \right)$$ Range: $$\left[ {0,\text{ }\infty } \right)$$
Remember also that another way to write $$y=\sqrt{x}$$ is $$y={{x}^{{\frac{1}{2}}}}$$.
See how, since $${{2}^{2}}=4$$, a point on the quadratic graph is $$\left( {2,4} \right)$$? Similarly, since $$\sqrt{4}=2$$, a point on the square root graph is $$\left( {4,2} \right)$$.
Next, we have the cubic (raising something to the 3rd power) and cube root function graphs. Since cube roots can be both positive and negative, the domain and range of both graphs is the set of real numbers.
Cubic Function Cube Root Function Domain: $$\left( {-\infty ,\infty } \right)\text{ or }\mathbb{R}$$ Range: $$\left( {-\infty ,\infty } \right)\text{ or }\mathbb{R}$$ Domain: $$\left( {-\infty ,\infty } \right)\text{ or }\mathbb{R}$$ Range: $$\left( {-\infty ,\infty } \right)\text{ or }\mathbb{R}$$
Remember also that another way to write $$y=\sqrt[3]{x}$$ is $$y={{x}^{{\frac{1}{3}}}}$$.
See how, since $${{2}^{3}}=8$$, a point on the cube function graph is $$\left( {2,8} \right)$$ (and so is $$\left( {-2,-8} \right)$$)? Similarly, since $$\sqrt[3]{8}=2$$, a point on the cube root graph is $$\left( {8,2} \right)$$. Note that $$\left( {-8,-2} \right)$$ is also a point on this graph, since $$\sqrt[3]{-8}=-2$$.
We’ll talk a little later in the Inverses of Functions section that the quadratic and square root functions are “opposites” or inverses of each other. The cubic and cube root functions are also inverses of each other.
The generic equation for a transformation with vertical stretch $$a$$, horizontal stretch $$b$$, horizontal shift $$h$$, and vertical shift $$k$$ is $$\displaystyle f\left( x \right)=a\,\sqrt[n]{{\frac{1}{b}\left( {x-h} \right)}}+k$$ for radical functions.
Remember these rules:
When functions are transformed on the outside of the $$f(x)$$ part, you move the function up and down and do “regular” math, as we’ll see in the examples below. These are vertical transformations or translations.
When transformations are made on the inside of the $$f(x)$$ part, you move the function back and forth (but do the opposite math – basically since if you were to isolate the $$x$$, you’d move everything to the other side). These are horizontal transformations or translations.
When there is a negative sign outside the parentheses, the function is reflected (flipped) across the $$x$$-axis; when there is a negative sign inside the parentheses, the function is reflected across the $$y$$-axis.
Here are some examples, using t-charts:
T-chart
Graph
$$\displaystyle y=-\frac{1}{2}\sqrt{{x+2}}+4$$
Parent function:
$$y=\sqrt{x}$$
For square root functions, use 0, 1, and 4 for the $$x$$ values of the parent function.
x – 2 x y $$-\frac{1}{2}y+4$$ –2 0 0 4 –1 1 1 3.5 2 4 2 3
Domain: $$\left[ {-2,\infty } \right)$$
Range: $$\left( {-\infty ,4} \right]$$
Vertical compression by factor of $$\frac{1}{2}$$ (over 1, down $$\frac{1}{2}$$), reflect over the $$x$$-axis, translate 2 units right, 4 units up.
$$y=\sqrt[3]{{-2\left( {x-1} \right)}}+2$$
Parent function:
$$y=\sqrt[3]{x}$$
For cube root functions, use –1, 0, and 1 for the $$x$$ values of the parent function.
$$-\frac{1}{2}x+1$$ x y y + 2 $$1\frac{1}{2}$$ –1 –1 1 1 0 0 2 $$\frac{1}{2}$$ 1 1 3
Domain: $$\left( {-\infty ,\infty } \right)$$
Range: $$\left( {-\infty ,\infty } \right)$$
Horizontal compression by a factor of $$\frac{1}{2}$$ (over $$\frac{1}{2}$$, down 2), reflect over the $$y$$-axis, translate 1 unit right, 2 units up.
Now let’s solve some problems with square root functions. With even radicals, we have to make sure that our answers never produce a negative number underneath the square root (even radical) sign. Also, if we raise both sides to an even exponent (like squaring), we need to check our answers, since some solutions may not work. Both of these conditions can produce extraneous solutions (solutions that don’t work), since even exponents can be a little quirky.
The main idea in solving these is to get rid of the radical signs by raising both sides to that exponent. For example, with square roots, we have to square both sides. If we have two square roots, it’s easiest to have them separated so when we square both sides, it’s not as complicated. Sometimes we have to take the square of each side more than once, after we’ve FOILED one or both sides (see last example in next set of examples).
Also remember that you can always turn a radical into a rational (fractional) exponent; here is an example: $${{\left( {\sqrt[3]{x}} \right)}^{4}}=\sqrt[3]{{{{x}^{4}}}}={{x}^{{\frac{4}{3}}}}$$.
And don’t forget when we end up with a quadratic equation, put everything on one side (set to 0) and factor, or use the quadratic formula.
Here are some of the problems we solved previously and a few more (since we know how to FOIL now!):
Radical Equation Notes $$\displaystyle \sqrt{x}=-3$$ $$\displaystyle \begin{array}{c}{{\left( {\sqrt{x}} \right)}^{2}}={{\left( {-3} \right)}^{2}}\\x=9\end{array}$$ Doesn’t work, so no solution! After we square each side and solve for $$x$$, we get $$x=9$$. Since we squared each side, we need to check our answer to see if it works. When we take the square root of a number, we only take the positive value, so our answer doesn’t work! You can also see early on in the problem that you can’t get a negative number from a square root, so the answer is no solution, or $$\emptyset$$. $$4\sqrt{{x-1}}=\sqrt{{x+1}}$$ \begin{align}{{\left( {4\sqrt{{x-1}}} \right)}^{2}}&={{\left( {\sqrt{{x+1}}} \right)}^{2}}\\\,{{4}^{2}}\left( {x-1} \right)&=\left( {x+1} \right)\\16x-16&=x+1\\15x&=17;\,\,\,\,\,x=\frac{{17}}{{15}}\end{align} Since we have square roots on both sides, we can simply square both sides to get rid of them. We have to make sure we square the 4 too, since it is on the outside of the radical. Then we solve for $$x$$ to get $$\displaystyle \frac{{17}}{{15}}$$. We have to make sure our answers don’t produce any negative numbers under the square root; this looks good. Also, since we squared both sides, let’s check our answer: $$\displaystyle 4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\frac{{32}}{{15}}}}?\,\,\,\,\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\left( {16} \right)\left( 2 \right)\frac{1}{{15}}}}\,\,?\,\,\,\,\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=4\sqrt{{\frac{2}{{15}}}}\,\,\,\,\surd$$ $$\sqrt[4]{{2-x}}=\sqrt[4]{{x-4}}$$ \begin{align}{{\left( {\sqrt[4]{{2-x}}} \right)}^{4}}&={{\left( {\sqrt[4]{{x-4}}} \right)}^{4}}\\2-x&=x-4\\2x&=6\\x&=3\end{align} Doesn’t work, so no solution! Here’s one where we have fourth roots instead of square roots. Since we have the 4th root on each side, we can just raise each side to the 4th power and solve. We correctly solved the equation, but notice that when we plug in 3 in the first radical (and the second one too!), we get a negative number ($$2-3=-1$$). We can’t take even roots of negative numbers. We have to “throw away” our answer and the correct answer is “no solution” or $$\emptyset$$. $$\displaystyle {{\left( {x-1} \right)}^{{\frac{1}{2}}}}=x-3$$ $$\require{cancel} \displaystyle \begin{array}{c}{{\left( {{{{\left( {x-1} \right)}}^{{\frac{1}{2}}}}} \right)}^{2}}={{\left( {x-3} \right)}^{2}}\\x-1={{x}^{2}}-6x+9\\{{x}^{2}}-7x+10=0\\\left( {x-5} \right)\left( {x-2} \right)=0\\\text{x=5,}\,\,\,\cancel{{x=2}}\end{array}$$ 2 doesn’t work! First of all, remember that $$\displaystyle {{x}^{{\frac{1}{2}}}}=\sqrt{x}$$, so we still solve by squaring both sides (see that the exponents “cancel out” on the left?). When we square each side to get rid of the radical sign, we find we have to FOIL, or multiply two binomials on the right side. Now we’re left with a quadratic equation, so we have to get everything to one side and set to 0. When we factor, we get both 5 and 2 for answers. 5 works, but when we put 2 in the original problem, we get a negative number on the right-hand side. Since the square root of something can never be a negative number, we have to eliminate this solution (so 2 is an extraneous solution). The answer is $$x=5$$. $$\displaystyle \begin{array}{c}{{\left( {\sqrt[3]{{x+2}}} \right)}^{4}}+2=18\\{{\left( {x+2} \right)}^{{\frac{4}{3}}}}=16\end{array}$$ \displaystyle \begin{align}{{\left( {{{{\left( {x+2} \right)}}^{{\frac{4}{3}}}}} \right)}^{{\frac{3}{4}}}}&={{16}^{{\frac{3}{4}}}}\\x+2&=\pm {{2}^{3}}\\x&=\pm {{2}^{3}}-2\\x&=8-2=6\,\,\,\,\,\text{and}\\x&=-8-2=-10\end{align} With both a root and an exponent, turn the radical into a rational (fractional) exponent, and raise each side to the reciprocal of the exponent. If the even number is on the top of the fraction, you have to take the positive and negative solutions. We also have to make sure the number on the right that we’re raising to an exponent is positive, or there would be no answers. Let’s check our answers: $$\displaystyle \begin{array}{c}{{\left( {\sqrt[3]{{6+2}}} \right)}^{4}}+2={{\left( {\sqrt[3]{8}} \right)}^{4}}+2={{2}^{4}}+2=18\,\,\,\,\,\,\surd \\{{\left( {\sqrt[3]{{-10+2}}} \right)}^{4}}+2={{\left( {\sqrt[3]{{-8}}} \right)}^{4}}+2={{\left( {-2} \right)}^{4}}+2=18\,\,\,\,\,\,\surd \end{array}$$
Here are a few where we have to square both sides two times to get rid of the radicals. Note the second problem has a radical inside of a radical.
Radical Equation Notes $$\displaystyle 5+\sqrt{{3x-8}}-\sqrt{{4x}}=3$$ $$\displaystyle \begin{array}{c}\sqrt{{3x-8}}=\sqrt{{4x}}-2\\{{\left( {\sqrt{{3x-8}}} \right)}^{2}}={{\left( {\sqrt{{4x}}-2} \right)}^{2}}\\\,3x-8=4x-4\sqrt{{4x}}+4\\\,4\sqrt{{4x}}=x+12\\{{\left( {4\sqrt{{4x}}} \right)}^{2}}={{\left( {x+12} \right)}^{2}}\\\,16\left( {4x} \right)={{x}^{2}}+24x+144\\{{x}^{2}}-40x+144=0\\\left( {x-4} \right)\left( {x-36} \right)=0\\x=4,\,\,\,x=36\,\end{array}$$ Both work! Since we have two radicals, let’s put one on each side to make the squaring easier. Also, combine any like terms before squaring each side. After squaring each side, we see that we still have a radical, so combine like terms again, and square each side once more. Then we have a quadratic, so we need to put everything to one side and set to 0. Factor, set each factor to 0, and then get answers. We have to make sure the answers work under the radicals (causing no negative numbers under the radicals) – this seems fine. Also, since we squared both sides, we have to check our answers in the original problem – both work. $$\displaystyle \surd$$ $$\displaystyle \sqrt{{{{x}^{2}}-2+2\sqrt{{3x+3}}}}=x+1$$ $$\displaystyle \begin{array}{c}{{x}^{2}}-2+2\sqrt{{3x+3}}={{\left( {x+1} \right)}^{2}}\\\cancel{{{{x}^{2}}}}-2+2\sqrt{{3x+3}}=\cancel{{{{x}^{2}}}}+2x+1\\\,2\sqrt{{3x+3}}=2x+3\\\,{{\left( {2\sqrt{{3x+3}}} \right)}^{2}}={{\left( {2x+3} \right)}^{2}}\\4\left( {3x+3} \right)=4{{x}^{2}}+12x+9\\\,\,\cancel{{12x}}+12=4{{x}^{2}}+\cancel{{12x}}+9\\\,4{{x}^{2}}=3\\\,{{x}^{2}}=\frac{3}{4}\end{array}$$ $$\displaystyle x=\,\frac{{\sqrt{3}}}{2}\,\,\,\,\,\,\,\,\left( {-\frac{{\sqrt{3}}}{2}\text{ doesn }\!\!’\!\!\text{ t work}} \right)$$ Square both sides first to get rid of the outside square root. Then simplify, get the radical by itself on one side, and square both sides again. Check both answers to make sure we have no negatives under radicals. Note that $$\displaystyle -\frac{{\sqrt{3}}}{2}$$ makes the outside radical negative, so it is extraneous. The positive answer works in the original problem! $$\displaystyle \surd$$
And are a couple of examples with odd-indexed radicals, where we can sit back and relax and just solve – everything we get should work!
Radical Equation Notes $$2\sqrt[3]{{x+2}}=\sqrt[3]{{x+8}}$$ $$\begin{array}{c}{{\left( {2\sqrt[3]{{x+2}}} \right)}^{3}}={{\left( {\sqrt[3]{{x+8}}} \right)}^{3}}\\8\left( {x+2} \right)=x+8\\8x+16=x+8\\\,7x=-8\\\,x=-\frac{8}{7}\end{array}$$ We cube both sides, not forgetting to cube the 2 since it’s on the outside of the radical sign. We then solve for $$x$$, not worrying about the sign! Much easier! $$\displaystyle 2\sqrt[5]{x}-\sqrt[5]{{{{x}^{2}}-4x-8}}=0$$ $$\displaystyle \begin{array}{c}2\sqrt[5]{x}=\sqrt[5]{{{{x}^{2}}-4x-8}}\\{{\left( {\,2\sqrt[5]{x}} \right)}^{5}}={{\left( {\sqrt[5]{{{{x}^{2}}-4x-8}}} \right)}^{5}}\\32x={{x}^{2}}-4x-8\\{{x}^{2}}-36x-8=0\end{array}$$ $$\displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{36\pm \sqrt{{1328}}}}{2}$$ $$\displaystyle =\frac{{36\pm 4\sqrt{{83}}}}{2}=18\pm 2\sqrt{{83}}$$ First, move one of the radicals to the other side so we can raise each side to the 5th power. We end up with a quadratic to solve. Sorry – I snuck in one where we couldn’t factor – so we have to use the quadratic formula to solve the quadratic. You can actually check this answer (or any of these) by storing the solution(s) in $$x$$ in your calculator. For example, to check in the positive value, type $$18+2\sqrt{{83}}$$, and then hit STO> X,T,,t. Then type in $$2x\hat{\ }(1/5)-(x\hat{\ }2-4x-8)\hat{\ }(1/5)$$, using thebutton for $$x$$. You should get 0.
We can graph radical functions either with a t-chart or in the graphing calculator. Later, we’ll learn how to transform functions more easily in the Parent Graphs and Transformations section.
Since we’re so good with the graphing calculator (yeah!), let’s solve a radical function equation using the calculator:
Radical Equation and Graph Calculator Screens $$\sqrt{{5x-16}}=\sqrt{{2x-4}}$$ Notes: Push Y = and enter the two equations in Y1 = and Y2 =, respectively. Push GRAPH. You may need to hit “ZOOM 6” (ZoomStandard) and/or “ZOOM 0” (ZoomFit), and maybe “ZOOM 3” (Zoom Out) ENTER to make sure you see the lines crossing in the graph. Hit TRACE and then use the round right arrow to move the cursor up to one of the graphs above the intersection (otherwise, the Intersect may not work with these square root functions). You may have to do this again for the other curve after the first ENTER (Second Curve?) below. To get the point of intersection, push “2nd TRACE” (CALC), and then push 5 (for intersect) (First Curve?), ENTER (Second Curve?), ENTER (Guess?), ENTER. You should see the point of intersection on the bottom! Now we are solving for the $$x$$ in these types of problems, so our answer is 4.
Note that we saw some of these same examples in the Exponents and Radicals in Algebra section.
Remember these rules for solving inequalities algebraically:
• When solving inequalities, we need to be careful with multiplying and dividing by anything negative, where we have to change the direction of the inequality sign.
• What’s under an even radical has to be positive (domain restriction); we have to create another inequality and set what’s under the even radical to $$\boldsymbol{{\ge 0}}$$. We then solve for $$x$$, and take the intersection of both solutions. The reason we take the intersection of the two solutions is because both must work.
• For more advanced solving, we’ll want to use a sign chart to show the intervals that work and don’t work; when we solve for $$x$$ in these situations, we get the critical values for the sign chart. We then have to check each interval to see if the inequality is true or false.
• For radicals, when there is a variable not under a square root and we square both sides, we have to be careful since don’t know if the side without the square is positive or negative (and thus if we should switch the sign). In these cases, check the interval test values in the original inequality, and use T or F (or Y or N) to indicate whether or not they work.
• If we get something like $$\sqrt{n}<0$$ (or a negative number), there is no solution, and something like $$\sqrt{n}\ge 0$$ (or a negative number), we get all real numbers, except for the domain restriction ($$n\ge 0$$).
• You can check these inequalities in your graphing calculator to make sure they are correct. For example, for the first one use $${{Y}_{1}}=\sqrt{{{{x}^{2}}-2x-8}}$$ and $${{Y}_{2}}=x+2$$.
Here are some examples; note that we just raise each side to the root to get rid of it. We can do that in these examples, since we know the sign of the values on both sides. It gets trickier when we don’t know the sign of one of the sides.
Radical Inequality Notes $$\sqrt{{x+2}}\le 4$$ $$\begin{array}{c}\sqrt{{x+2}}\le 4\text{ }\,\text{ }\,\text{and }x+2\ge 0\\{{\left( {\sqrt{{x+2}}} \right)}^{2}}\le {{4}^{2}}\text{ }\,\,\text{and }x+2\ge 0\text{ }\\x+2\le 16\text{ }\,\text{and }x\ge 0-2\text{ }\\x\le 14\text{ }\,\text{and }x\ge -2\\\\\{x:-2\le x\le 14\}\text{ or }\left[ {-2,14} \right]\end{array}$$ We have to solve two inequalities, since our $$x$$ must work in the original, but also work so anything under the even radical is positive (domain restriction): $$\displaystyle x+2\ge 0$$. To get rid of the square roots, we square each side, and we can leave the inequality signs the same since we’re multiplying by positive numbers. Then we just solve for $$x$$, just like we would for an equation. We need to check our answer by trying random numbers in our solution (like $$x=2$$) in the original inequality (which works). We also need to try numbers outside our solution (like $$x=-6$$ and $$x=20$$) and see that they don’t work. $$\displaystyle \sqrt{{5x-16}}<\sqrt{{2x-4}}$$ $$\displaystyle \begin{array}{c}{{\left( {\sqrt{{5x-16}}} \right)}^{2}}<{{\left( {\sqrt{{2x-4}}} \right)}^{2}}\\5x-16<2x-4\\3x<12\\x<4\end{array}$$ Also: $$\displaystyle \begin{array}{c}5x-16 \,\ge 0\text{ and 2}x-4 \,\ge 0\\x\ge \frac{{16}}{5}\text{ and }x\ge 2\\x<4\,\,\,\cap \,\,\,x\ge \frac{{16}}{5}\,\,\,\cap \,\,\,x\ge 2\end{array}$$ $$\displaystyle \{x:\,\,\frac{{16}}{5}\le x<4\}\text{ or }\left[ {\frac{{16}}{5},\,\,4} \right)$$ Now we have to solve three inequalities: one for the main problem, and one each for the even root radicands, that have to be $$\ge 0$$. Since both sides are positive (even roots), we can safely take the square of each side. We need to take the intersection (all must work) of the inequalities: $$\displaystyle x<4$$ and $$\displaystyle x\ge \frac{{16}}{5}$$ and $$\displaystyle x\ge 2$$. This will give us $$\displaystyle \frac{{16}}{5}\le \,\,x<4$$. (Try it yourself on a number line). Watch out for the hard and soft brackets. We need to check our answer by trying $$x=3.5$$ in the original inequality (which works) and $$x=3$$ or $$x=5$$ (which don’t work). $$\sqrt{{x+6}}\le -2$$ $$\{\}\text{ or }\emptyset$$ Before we even need to get started with this inequality, we can notice that the square root of anything can never $$\displaystyle \le -2$$ (or $$\boldsymbol{<0}$$), by definition. We know right away that the answer is no solution, or $$\{\}\text{ or }\emptyset$$. $$\sqrt[3]{{x-3}}>4$$ $$\begin{array}{c}{{\left( {\sqrt[3]{{x-3}}} \right)}^{3}}>{{4}^{3}}\,\,\,\,\\x-3>64\\x>67\end{array}$$ With odd roots, we don’t have to worry about checking underneath the radical sign, since we could have positive or negative numbers as a radicand.
Here are more complicated problems where we need to use a sign chart to solve radical inequalities.
Note that we have to be careful when there is a variable on a side and it’s not under a square root; when we square both sides, we’re not really sure if we’d have to switch the inequality sign. This is because we don’t know if the side without the square is positive or negative. Since this is the case, it’s best to check the intervals in the original inequality, and use T or F (or Y or N) when checking.
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Radical Inequality Notes $$\sqrt{{{{x}^{2}}-2x-8}}>x+2$$ $$\displaystyle \begin{array}{c}\sqrt{{{{x}^{2}}-2x-8}}\,\,>x+2\,\,\text{and }{{x}^{2}}-2x-8\ge 0\\{{\left( {\sqrt{{{{x}^{2}}-2x-8}}} \right)}^{2}}>{{\left( {x+2} \right)}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{and}\,\,\left( {x+2} \right)\left( {x-4} \right)\ge 0\text{ }\\\cancel{{{{x}^{2}}}}-2x-8>\cancel{{{{x}^{2}}}}+4x+4\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{and }\left( {x+2} \right)\left( {x-4} \right)\ge 0\text{ }\\\,-6x>12\text{ and }\left( {x+2} \right)\left( {x-4} \right)\ge 0\\\,\,\,\,x<-2\text{ }\,\text{and }\left( {x+2} \right)\left( {x-4} \right)\ge 0\end{array}$$ Here is the sign chart for the right-hand inequality, including the restriction for the left-hand: We have to solve two inequalities, since our $$x$$ must work in the original, but also work so anything under the even radical is positive (domain restriction): $$\displaystyle {{x}^{2}}-2x-8\ge 0$$. When we solve, we’ll get the critical values, or critical points. In this case, we have –2, –2 again, and 4 (use closed circles for $$\ge$$, $$\le$$ and open circles for just $$>,<$$). Since $$x$$ has to be less than –2, we can exclude the other intervals. But, using the sign chart, it’s a good idea to check each interval with random points to see if the original equation works (T or F). The answer is $$\left( {-\infty ,-2} \right)$$. $$\displaystyle \sqrt{{2-x}},<$$). At this point, we need to check the original equation, since we don’t know if the right-hand side of the original inequality is positive or negative, which would create a sign change when solving. So, for the sign chart, we check each interval with random points to see if the original equation works (T or F). Don’t forget that we can’t have a negative number under the even radical sign and the square root will always be positive. The answer is $$\left( {1,2} \right]$$. Try it in the graphing calculator – it works! √ $$3\sqrt{{{{x}^{2}}-9}}+2<14$$ $$\begin{array}{c}3\sqrt{{{{x}^{2}}-9}}+2<14\,\,\,\text{and}\,\,\,{{x}^{2}}-9\ge 0\\\sqrt{{{{x}^{2}}-9}}<4\,\,\,\,\,\text{and}\,\,\,\left( {x-3} \right)\left( {x+3} \right)\ge 0\\{{\left( {\sqrt{{{{x}^{2}}-9}}} \right)}^{2}}<{{4}^{2}}\,\,\,\text{and}\,\,\,\left( {x-3} \right)\left( {x+3} \right)\ge 0\\{{x}^{2}}<25\,\,\,\,\,\text{and}\,\,\,\left( {x-3} \right)\left( {x+3} \right)\ge 0\\\left( {x-5} \right)\left( {x+5} \right)<0\,\,\,\,\,\text{and}\,\,\,\left( {x-3} \right)\left( {x+3} \right)\ge 0\end{array}$$ We have to solve two inequalities, since our $$x$$ must work in the original, but also work so anything under the even radical is positive (domain restriction): $$\displaystyle {{x}^{2}}-9\ge 0$$, or $$\left( {x-3} \right)\left( {x+3} \right)\ge 0$$. We also see that $${{x}^{2}}<25$$, or $$\left( {x-5} \right)\left( {x+5} \right)<0$$. We’ll use a sign chart with all four critical values and check the intervals by checking original (simplified) inequality: $$\sqrt{{{{x}^{2}}-9}}<4$$. Note the open and close circles, used for inclusion ($$<$$)or exclusion ($$\ge$$). The answer is $$\left( {-5,3} \right]\cup \left[ {3,5} \right)$$.
Here are a few more problems with radical inequalities:
Radical Inequality Notes $$\sqrt[4]{x}>\sqrt[3]{x}$$ $$\displaystyle \begin{array}{c}{{x}^{{\frac{1}{4}}}}>{{x}^{{\frac{1}{3}}}}\text{ }\,\text{and}\text{ }\,\,x\ge 0\\{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{{12}}}>{{\left( {{{x}^{{\frac{1}{3}}}}} \right)}^{{12}}}\text{ and}\text{ }\,x\ge 0\\{{x}^{3}}>{{x}^{4}}\text{ }\,\text{and}\,\text{ }\,x\ge 0\\{{x}^{4}}-{{x}^{3}}<0\text{ }\,\text{and}\,\text{ }\,x\ge 0\\{{x}^{3}}\left( {x-1} \right)<0\text{ }\,\text{and }x\ge 0\text{ }\end{array}$$ For this inequality, let’s first turn the roots into exponents so we can more easily get rid of them. We also have to remember to set $$x\ge 0$$ since $$x$$ is under an even radical (4th root). Raise each side to the 12th power, since that will cancel out both of the fractional exponents (neat trick – like a common denominator for the exponents!) Note that we can safely do this, since $$x$$ has to be positive (because of even root), so both sides will be positive. When we solve, we’ll get the critical values, or critical points. In this case, we have 0 and 1. We need an open circle on 0 because of the primary inequality. For the sign chart, we check each interval with random points to see if the original equation works (T or F). Don’t forget that we can’t have a negative number under the even radical sign. The answer is $$\left( {0,\,\,1} \right)$$. $$\sqrt{x}\ge 3\sqrt[4]{x}\,$$ $$\displaystyle \begin{array}{c}\sqrt{x}\,\,\le \,\,3\,\sqrt[4]{x}\,\text{ and }\,\,x\ge 0\\{{\left( {\sqrt{x}} \right)}^{4}}\ge \,\,{{\left( {3\,\sqrt[4]{x}} \right)}^{4}}\,\text{ and }\,x\ge 0\\{{x}^{2}}\ge 81x\text{ }\,\,\text{and }\,\,x\ge 0\\{{x}^{2}}-81x\ge 0\,\,\text{ and }\,\,x\ge 0\\x\left( {x-81} \right)\ge 0\,\text{ and }\,x\ge 0\end{array}$$ Raise each side to the 4th power, since that will cancel out both of the fractional exponents (neat trick – like a common denominator for the exponents!) Note that we can safely do this since both sides (even roots) have to be positive. We see that the critical values are 0 and 81. Use close circles because of the $$\ge$$. For the sign chart, we check each interval with random points to see if the original equation works (T or F). Don’t forget that we can’t have a negative number under the even radical sign. The answer is $$\left[ {81,\,\,\infty } \right)$$.
Let’s first just graph a simple radical inequality to show what the shading looks like; you may have to make some graphs like this. We saw earlier what the radical function looks like, and we can use the “rain up” (for $$>$$) and “rain down” (for $$<$$) shading like we did here in the Quadratics Inequalities section. Remember that with “$$<$$” and “$$>$$” inequalities, we draw a dashed (or dotted) line to indicate that we’re not really including that line (but everything up to it), whereas with “$$\le$$” and “$$\ge$$”, we draw a regular line, to indicate that we are including it in the solution.
Note that we also had to check so that anything under the even radical is positive; this is why the graph is shaded for $$x\ge 5$$. We still have to keep this vertical line dotted, since we take the intersection (both have to work) of the two inequalities, and in this example, we have $$<$$.
Note that we can put this in the graphing calculator, too. We had to move the cursor way to the left of “$${{\text{Y}}_{1}}=$$” and change to an inequality (play around with it; it’s different with the color calculator!) and then graph:
Radical Inequality and Graph Calculator Screens $$y<\sqrt{{x-5}}$$
We can also solve radical inequalities graphically. To get the intervals for $$x$$ for these graphs, you have to look and see whichever graph is on the bottom or below the other one (has the smaller $$y$$ for that interval) if it is a “less than” problem. For a “greater than” problem, you find the interval of the graph that is on the top or above the other one (has a larger $$y$$ for that interval).
Sometimes (like in the third example below), there are no values that make the inequality true.
Radical Inequalities and Graphs $$\sqrt{{x+2}}\le 4$$ $$\{x:-2\le x\le 14\}\text{ or }\left[ {-2,14} \right]$$ $$\sqrt{{5x-16}}\,\,<\,\,\sqrt{{2x-4}}$$ $$\displaystyle \,\{x:\frac{{16}}{5}\le x<4\}\text{ or }\left[ {\frac{{16}}{5},4} \right)$$ $$\,\sqrt{{x+6}}\le -2$$ $$\,\{\}\text{ or }\emptyset$$ $$\sqrt[3]{{x-3}}>4$$ $$\displaystyle \{x:x>67\}\text{ or }\left( {67,\,\,\,\infty } \right)$$
NOTE: We could solve use graphing calculator as we did for the equalities above, but it’s really difficult to get the point of intersection where the graphs hit the $$x$$-axis, since the graphs are just starting there (you might use the table). Also, when you get the other points of intersection, it’s easiest if you have use TRACE to move the cursor above the intersection of the functions, if there is an intersection. When you get the point of intersection, use the $$x$$value, since we’re solving for $$x$$.
Learn these rules, and practice, practice, practice!
For Practice: Use the Mathway widget below to try an Inequality problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve for x to see the answer.
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You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.
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On to Solving Absolute Value Equations and Inequalities – you’re ready!
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References of "Southworth, J" in Complete repository Arts & humanities Archaeology Art & art history Classical & oriental studies History Languages & linguistics Literature Performing arts Philosophy & ethics Religion & theology Multidisciplinary, general & others Business & economic sciences Accounting & auditing Production, distribution & supply chain management Finance General management & organizational theory Human resources management Management information systems Marketing Strategy & innovation Quantitative methods in economics & management General economics & history of economic thought International economics Macroeconomics & monetary economics Microeconomics Economic systems & public economics Social economics Special economic topics (health, labor, transportation…) Multidisciplinary, general & others Engineering, computing & technology Aerospace & aeronautics engineering Architecture Chemical engineering Civil engineering Computer science Electrical & electronics engineering Energy Geological, petroleum & mining engineering Materials science & engineering Mechanical engineering Multidisciplinary, general & others Human health sciences Alternative medicine Anesthesia & intensive care Cardiovascular & respiratory systems Dentistry & oral medicine Dermatology Endocrinology, metabolism & nutrition Forensic medicine Gastroenterology & hepatology General & internal medicine Geriatrics Hematology Immunology & infectious disease Laboratory medicine & medical technology Neurology Oncology Ophthalmology Orthopedics, rehabilitation & sports medicine Otolaryngology Pediatrics Pharmacy, pharmacology & toxicology Psychiatry Public health, health care sciences & services Radiology, nuclear medicine & imaging Reproductive medicine (gynecology, andrology, obstetrics) Rheumatology Surgery Urology & nephrology Multidisciplinary, general & others Law, criminology & political science Civil law Criminal law & procedure Criminology Economic & commercial law European & international law Judicial law Metalaw, Roman law, history of law & comparative law Political science, public administration & international relations Public law Social law Tax law Multidisciplinary, general & others Life sciences Agriculture & agronomy Anatomy (cytology, histology, embryology...) & physiology Animal production & animal husbandry Aquatic sciences & oceanology Biochemistry, biophysics & molecular biology Biotechnology Entomology & pest control Environmental sciences & ecology Food science Genetics & genetic processes Microbiology Phytobiology (plant sciences, forestry, mycology...) Veterinary medicine & animal health Zoology Multidisciplinary, general & others Physical, chemical, mathematical & earth Sciences Chemistry Earth sciences & physical geography Mathematics Physics Space science, astronomy & astrophysics Multidisciplinary, general & others Social & behavioral sciences, psychology Animal psychology, ethology & psychobiology Anthropology Communication & mass media Education & instruction Human geography & demography Library & information sciences Neurosciences & behavior Regional & inter-regional studies Social work & social policy Sociology & social sciences Social, industrial & organizational psychology Theoretical & cognitive psychology Treatment & clinical psychology Multidisciplinary, general & others Showing results 1 to 20 of 88 1 2 3 4 5 WASP-South transiting exoplanets: WASP-130b, WASP-131b, WASP-132b, WASP-139b, WASP-140b, WASP-141b & WASP-142bHellier, Coel; Anderson, D. R.; Collier Cameron, A. et alin Monthly Notices of the Royal Astronomical Society (2017), 465We describe seven new exoplanets transiting stars of V = 10.1 to 12.4. WASP-130b is a "warm Jupiter" having an orbital period of 11.6 d, the longest yet found by WASP. It transits a V = 11.1, G6 star with ... [more ▼]We describe seven new exoplanets transiting stars of V = 10.1 to 12.4. WASP-130b is a "warm Jupiter" having an orbital period of 11.6 d, the longest yet found by WASP. It transits a V = 11.1, G6 star with [Fe/H] = +0.26. Warm Jupiters tend to have smaller radii than hot Jupiters, and WASP-130b is in line with this trend (1.23 Mjup; 0.89 Rjup). WASP-131b is a bloated Saturn-mass planet (0.27 Mjup; 1.22 Rjup). Its large scale height coupled with the V = 10.1 brightness of its host star make the planet a good target for atmospheric characterisation. WASP-132b is among the least irradiated and coolest of WASP planets, being in a 7.1-d orbit around a K4 star. It has a low mass and a modest radius (0.41 Mjup; 0.87 Rjup). The V = 12.4, [Fe/H] = +0.22 star shows a possible rotational modulation at 33 d. WASP-139b is the lowest-mass planet yet found by WASP, at 0.12 Mjup and 0.80 Rjup. It is a "super-Neptune" akin to HATS-7b and HATS-8b. It orbits a V = 12.4, [Fe/H] = +0.20, K0 star. The star appears to be anomalously dense, akin to HAT-P-11. WASP-140b is a 2.4-Mjup planet in a 2.2-d orbit that is both eccentric (e = 0.047) and with a grazing transit (b = 0.93) The timescale for tidal circularisation is likely to be the lowest of all known eccentric hot Jupiters. The planet's radius is large (1.4 Rjup), but uncertain owing to the grazing transit. The host star is a V = 11.1, [Fe/H] = +0.12, K0 dwarf showing a prominent 10.4-d rotational modulation. The dynamics of this system are worthy of further investigation. WASP-141b is a typical hot Jupiter, being a 2.7 Mjup, 1.2 Rjup planet in a 3.3-d orbit around a V = 12.4, [Fe/H] = +0.29, F9 star. WASP-142b is a typical bloated hot Jupiter (0.84 Mjup, 1.53 Rjup) in a 2.1-d orbit around a V = 12.3, [Fe/H] = +0.26, F8 star. [less ▲]Detailed reference viewed: 129 (6 ULiège) Orbital alignment and star-spot properties in the WASP-52 planetary systemMancini, L.; Southworth, J.; Raia, G. et alin Monthly Notices of the Royal Astronomical Society (2017), 465We report 13 high-precision light curves of eight transits of the exoplanet WASP-52 b, obtained by using four medium-class telescopes, through different filters, and adopting the defocussing technique ... [more ▼]We report 13 high-precision light curves of eight transits of the exoplanet WASP-52 b, obtained by using four medium-class telescopes, through different filters, and adopting the defocussing technique. One transit was recorded simultaneously from two different observatories and another one from the same site but with two different instruments, including a multiband camera. Anomalies were clearly detected in five light curves and modelled as star-spots occulted by the planet during the transit events. We fitted the clean light curves with the JKTEBOP code, and those with the anomalies with the PRISM+GEMC codes in order to simultaneously model the photometric parameters of the transits and the position, size and contrast of each star-spot. We used these new light curves and some from the literature to revise the physical properties of the WASP-52 system. Star-spots with similar characteristics were detected in four transits over a period of 43 d. In the hypothesis that we are dealing with the same star-spot, periodically occulted by the transiting planet, we estimated the projected orbital obliquity of WASP-52 b to be λ = 3.8° ± 8.4°. We also determined the true orbital obliquity, ψ = 20° ± 50°, which is, although very uncertain, the first measurement of ψ purely from star-spot crossings. We finally assembled an optical transmission spectrum of the planet and searched for variations of its radius as a function of wavelength. Our analysis suggests a flat transmission spectrum within the experimental uncertainties. [less ▲]Detailed reference viewed: 9 (0 ULiège) MiNDSTEp differential photometry of the gravitationally lensed quasars WFI 2033-4723 and HE 0047-1756: Microlensing and a new time delayGiannini, E.; Schmidt, R. W.; Wambsganss, J. et alin Astronomy and Astrophysics (2017), 597Aims. We present V and R photometry of the gravitationally lensed quasars WFI 2033-4723 and HE 0047-1756. The data were taken by the MiNDSTEp collaboration with the 1.54 m Danish telescope at the ESO La ... [more ▼]Aims. We present V and R photometry of the gravitationally lensed quasars WFI 2033-4723 and HE 0047-1756. The data were taken by the MiNDSTEp collaboration with the 1.54 m Danish telescope at the ESO La Silla observatory from 2008 to 2012. Methods. Differential photometry has been carried out using the image subtraction method as implemented in the HOTPAnTS package, additionally using GALFIT for quasar photometry. Results. The quasar WFI 2033-4723 showed brightness variations of order 0.5 mag in V and R during the campaign. The two lensed components of quasar HE 0047-1756 varied by 0.2-0.3 mag within five years. We provide, for the first time, an estimate of the time delay of component B with respect to A of Δt = (7.6 ± 1.8) days for this object. We also find evidence for a secular evolution of the magnitude difference between components A and B in both filters, which we explain as due to a long-duration microlensing event. Finally we find that both quasars WFI 2033-4723 and HE 0047-1756 become bluer when brighter, which is consistent with previous studies. © ESO, 2016. [less ▲]Detailed reference viewed: 13 (0 ULiège) From Dense Hot Jupiter to Low Density Neptune: The Discovery of WASP-127b, WASP-136b and WASP-138bLam, K. W. F.; Faedi, F.; Brown, D. J. A. et alin Astronomy and Astrophysics (2017), 599We report three newly discovered exoplanets from the SuperWASP survey. WASP-127b is a heavily inflated super-Neptune of mass 0.18Mj and radius 1.35Rj. This is one of the least massive planets discovered ... [more ▼]We report three newly discovered exoplanets from the SuperWASP survey. WASP-127b is a heavily inflated super-Neptune of mass 0.18Mj and radius 1.35Rj. This is one of the least massive planets discovered by the WASP project. It orbits a bright host star (V = 10.16) of spectral type G5 with a period of 4.17 days.WASP-127b is a low density planet which has an extended atmosphere with a scale height of 2500+/-400 km, making it an ideal candidate for transmission spectroscopy. WASP-136b and WASP-138b are both hot Jupiters with mass and radii of 1.51 Mj and 1.38 Rj, and 1.22 Mj and 1.09 Rj, respectively. WASP-136b is in a 5.22-day orbit around an F9 subgiant star with a mass of 1.41 Msun and a radius of 2.21 Rsun. The discovery of WASP-136b could help constraint the characteristics of the giant planet population around evolved stars. WASP-138b orbits an F7 star with a period of 3.63 days. Its radius agrees with theoretical values from standard models, suggesting the presence of a heavy element core with a mass of 10 Mearth. The discovery of these new planets helps in exploring the diverse compositional range of short-period planets, and will aid our understanding of the physical characteristics of both gas giants and low density planets. [less ▲]Detailed reference viewed: 35 (7 ULiège) WASP-92b, WASP-93b and WASP-118b: Three new transiting close-in giant planetsHay, K. L.; Collier-Cameron, A.; Doyle, A. P. et alin Monthly Notices of the Royal Astronomical Society (2016), 463We present the discovery of three new transiting giant planets, first detected with the WASP telescopes, and establish their planetary nature with follow up spectroscopy and ground-based photometric ... [more ▼]We present the discovery of three new transiting giant planets, first detected with the WASP telescopes, and establish their planetary nature with follow up spectroscopy and ground-based photometric lightcurves. WASP-92 is an F7 star, with a moderately inflated planet orbiting with a period of 2.17 days, which has R[SUB]p[/SUB] = 1.461 ± 0.077R[SUB]J[/SUB] and M[SUB]p[/SUB] = 0.805 ± 0.068M[SUB]J[/SUB]. WASP-93b orbits its F4 host star every 2.73 days and has R[SUB]p[/SUB] = 1.597 ± 0.077R[SUB]J[/SUB] and M[SUB]p[/SUB] = 1.47 ± 0.029M[SUB]J[/SUB]. WASP-118b also has a hot host star (F6) and is moderately inflated, where R[SUB]p[/SUB] = 1.440 ± 0.036R[SUB]J[/SUB] and M[SUB]p[/SUB] = 0.514 ± 0.020M[SUB]J[/SUB] and the planet has an orbital period of 4.05 days. They are bright targets (V = 13.18, 10.97 and 11.07 respectively) ideal for further characterisation work, particularly WASP-118b, which is being observed by K2 as part of campaign 8. The WASP-93 system has sufficient angular momentum to be tidally migrating outwards if the system is near spin-orbit alignment, which is divergent from the tidal behaviour of the majority of hot Jupiters discovered. [less ▲]Detailed reference viewed: 47 (6 ULiège) WASP-157b, a Transiting Hot Jupiter Observed with K2Močnik, T.; Anderson, D. R.; Brown, D. J. A. et alin Publications of the Astronomical Society of the Pacific (2016), 970We announce the discovery of the transiting hot Jupiter WASP-157b in a 3.95-d orbit around a V = 12.9 G2 main-sequence star. This moderately inflated planet has a Saturn-like density with a mass of $0.57 ... [more ▼]We announce the discovery of the transiting hot Jupiter WASP-157b in a 3.95-d orbit around a V = 12.9 G2 main-sequence star. This moderately inflated planet has a Saturn-like density with a mass of$0.57 \pm 0.10$M$_{\rm Jup}$and a radius of$1.06 \pm 0.05$R$_{\rm Jup}\$. We do not detect any rotational or phase-curve modulations, nor the secondary eclipse, with conservative semi-amplitude upper limits of 250 and 20 ppm, respectively. [less ▲]Detailed reference viewed: 46 (2 ULiège) Five transiting hot Jupiters discovered using WASP-South, Euler, and TRAPPIST: WASP-119 b, WASP-124 b, WASP-126 b, WASP-129 b, and WASP-133 bMaxted, P. F. L.; Anderson, D. R.; Collier Cameron, A. et alin Astronomy and Astrophysics (2016), 591We have used photometry from the WASP-South instrument to identify 5 stars showing planet-like transits in their light curves. The planetary nature of the companions to these stars has been confirmed ... [more ▼]We have used photometry from the WASP-South instrument to identify 5 stars showing planet-like transits in their light curves. The planetary nature of the companions to these stars has been confirmed using photometry from the EulerCam instrument on the Swiss Euler 1.2-m telescope and the TRAPPIST telescope, and spectroscopy obtained with the CORALIE spectrograph. The planets discovered are hot Jupiter systems with orbital periods in the range 2.17 to 5.75 days, masses from 0.3 M[SUB]Jup[/SUB] to 1.2 M[SUB]Jup[/SUB] and with radii from 1 R[SUB]Jup[/SUB] to 1.5 R[SUB]Jup[/SUB]. These planets orbit bright stars (V = 11-13) with spectral types in the range F9 to G4. WASP-126 is the brightest planetary system in this sample and hosts a low-mass planet with a large radius (0.3 M[SUB]Jup[/SUB],0.95 R[SUB]Jup[/SUB]), making it a good target for transmission spectroscopy. The high density of WASP-129 A suggests that it is a helium-rich star similar to HAT-P-11 A. WASP-133 A has an enhanced surface lithium abundance compared to other old G-type stars, particularly other planet host stars. These planetary systems are good targets for follow-up observations with ground-based and space-based facilities to study their atmospheric and dynamical properties. Full Tables 2 and 3 are only available at the CDS via anonymous ftp to http://cdsarc.u-strasbg.fr (http://130.79.128.5) or via http://cdsarc.u-strasbg.fr/viz-bin/qcat?J/A+A/591/A55 [less ▲]Detailed reference viewed: 18 (1 ULiège) Three irradiated and bloated hot Jupiters:. WASP-76b, WASP-82b, and WASP-90bWest, R. G.; Hellier, C.; Almenara, J.-M. et alin Astronomy and Astrophysics (2016), 585We report on three new transiting hot Jupiter planets, discovered from the WASP surveys, which we combine with radial velocities from OHP/SOPHIE and Euler/CORALIE and photometry from Euler and TRAPPIST ... [more ▼]We report on three new transiting hot Jupiter planets, discovered from the WASP surveys, which we combine with radial velocities from OHP/SOPHIE and Euler/CORALIE and photometry from Euler and TRAPPIST. The planets WASP-76b, WASP-82b, and WASP-90b are all inflated, with radii of 1.7-1.8 R[SUB]Jup[/SUB]. All three orbit hot stars, of type F5-F7, with orbits of 1.8-3.9 d, and all three stars have evolved, post-main-sequence radii (1.7-2.2 R[SUB]⊙[/SUB]). Thus the three planets fit a known trend of hot Jupiters that receive high levels of irradiation being highly inflated. We caution, though, about the presence of a selection effect, in that non-inflated planets around ~2 R[SUB]⊙[/SUB] post-MS stars can often produce transits too shallow to be detected by the ground-based surveys that have found the majority of transiting hot Jupiters. Tables of the photometry and radial velocity are only available at the CDS via anonymous ftp to http://cdsarc.u-strasbg.fr (ftp://130.79.128.5) or via http://cdsarc.u-strasbg.fr/viz-bin/qcat?J/A+A/585/A126 [less ▲]Detailed reference viewed: 33 (2 ULiège) Erratum to A detailed census of variable stars in the globular cluster NGC 6333 (M9) from CCD differential photometryArellano Ferro, A.; Bramich, D. M.; Figuera Jaimes, R. et alin Monthly Notices of the Royal Astronomical Society (2016), 458(2), 1188-1189Detailed reference viewed: 18 (0 ULiège) High-precision photometry by telescope defocussing - VIII.WASP-22, WASP-41,WASP-42 andWASP-55Southworth, J.; Tregloan-Reed, J.; Andersen, M. I. et alin Monthly Notices of the Royal Astronomical Society (2016), 457(4), 4205-4217We present 13 high-precision and four additional light curves of four bright southernhemisphere transiting planetary systems: WASP-22, WASP-41, WASP-42 and WASP-55. In the cases of WASP-42 and WASP-55 ... [more ▼]We present 13 high-precision and four additional light curves of four bright southernhemisphere transiting planetary systems: WASP-22, WASP-41, WASP-42 and WASP-55. In the cases of WASP-42 and WASP-55, these are the first follow-up observations since their discovery papers. We present refined measurements of the physical properties and orbital ephemerides of all four systems. No indications of transit timing variations were seen. All four planets have radii inflated above those expected from theoretical models of gas-giant planets; WASP-55 b is the most discrepant with a mass of 0.63MJup and a radius of 1.34 RJup. WASP-41 shows brightness anomalies during transit due to the planet occulting spots on the stellar surface. Two anomalies observed 3.1 d apart are very likely due to the same spot. We measure its change in position and determine a rotation period for the host star of 18.6 ± 1.5 d, in good agreement with a published measurement from spot-induced brightness modulation, and a sky-projected orbital obliquity of λ = 6 ± 11°. We conclude with a compilation of obliquity measurements from spot-tracking analyses and a discussion of this technique in the study of the orbital configurations of hot Jupiters. © 2016 The Authors. Published by Oxford University Press on behalf of the Royal Astronomical Society. [less ▲]Detailed reference viewed: 21 (0 ULiège) SPITZER PARALLAX of OGLE-2015-BLG-0966: A COLD NEPTUNE in the GALACTIC DISKStreet, R. A.; Udalski, A.; Novati, S. C. et alin Astrophysical Journal (2016), 819(2), We report the detection of a cold Neptune mplanet = 21 ± 2 M⊕ orbiting a 0.38 Mo M dwarf lying 2.5-3.3 kpc toward the Galactic center as part of a campaign combining ground-based and Spitzer observations ... [more ▼]We report the detection of a cold Neptune mplanet = 21 ± 2 M⊕ orbiting a 0.38 Mo M dwarf lying 2.5-3.3 kpc toward the Galactic center as part of a campaign combining ground-based and Spitzer observations to measure the Galactic distribution of planets. This is the first time that the complex real-time protocols described by Yee et al., which aim to maximize planet sensitivity while maintaining sample integrity, have been carried out in practice. Multiple survey and follow up teams successfully combined their efforts within the framework of these protocols to detect this planet. This is the second planet in the Spitzer Galactic distribution sample. Both are in the near to mid-disk and are clearly not in the Galactic bulge. © 2016. The American Astronomical Society. All rights reserved. [less ▲]Detailed reference viewed: 24 (0 ULiège) SPITZER OBSERVATIONS of OGLE-2015-BLG-1212 REVEAL A NEW PATH TOWARD BREAKING STRONG MICROLENS DEGENERACIESBozza, V.; Shvartzvald, Y.; Udalski, A. et alin Astrophysical Journal (2016), 820(1), Spitzer microlensing parallax observations of OGLE-2015-BLG-1212 decisively break a degeneracy between planetary and binary solutions that is somewhat ambiguous when only ground-based data are considered ... [more ▼]Spitzer microlensing parallax observations of OGLE-2015-BLG-1212 decisively break a degeneracy between planetary and binary solutions that is somewhat ambiguous when only ground-based data are considered. Only eight viable models survive out of an initial set of 32 local minima in the parameter space. These models clearly indicate that the lens is a stellar binary system possibly located within the bulge of our Galaxy, ruling out the planetary alternative. We argue that several types of discrete degeneracies can be broken via such space-based parallax observations. © 2016. The American Astronomical Society. All rights reserved.. [less ▲]Detailed reference viewed: 26 (0 ULiège) Erratum: Estimating the parameters of globular cluster M 30 (NGC 7099) from time-series photometry (Astronomy and Astrophysics (2013) 555 (A36))Kains, N.; Bramich, D. M.; Arellano Ferro, A. et alin Astronomy and Astrophysics (2016), 588[No abstract available]Detailed reference viewed: 22 (0 ULiège) Physical properties of the planetary systems WASP-45 and WASP-46 from simultaneous multiband photometryCiceri, S.; Mancini, L.; Southworth, J. et alin Monthly Notices of the Royal Astronomical Society (2016), 456(1), 990-1002Accurate measurements of the physical characteristics of a large number of exoplanets are useful to strongly constrain theoretical models of planet formation and evolution, which lead to the large variety ... [more ▼]Accurate measurements of the physical characteristics of a large number of exoplanets are useful to strongly constrain theoretical models of planet formation and evolution, which lead to the large variety of exoplanets and planetary-system configurations that have been observed. We present a study of the planetary systemsWASP-45 andWASP-46, both composed of a mainsequence star and a close-in hot Jupiter, based on 29 new high-quality light curves of transits events. In particular, one transit of WASP-45 b and four of WASP-46 b were simultaneously observed in four optical filters, while one transit of WASP-46 b was observed with the NTT obtaining a precision of 0.30 mmag with a cadence of roughly 3 min. We also obtained five new spectra of WASP-45 with the FEROS spectrograph. We improved by a factor of 4 the measurement of the radius of the planet WASP-45 b, and found that WASP-46 b is slightly less massive and smaller than previously reported. Both planets now have a more accurate measurement of the density (0.959 ± 0.077 ρJup instead of 0.64 ± 0.30 ρJup for WASP-45 b, and 1.103 ± 0.052 ρJup instead of 0.94 ± 0.11 ρJup for WASP-46 b). We tentatively detected radius variations with wavelength for both planets, in particular in the case of WASP-45 b we found a slightly larger absorption in the redder bands than in the bluer ones. No hints for the presence of an additional planetary companion in the two systems were found either from the photometric or radial velocity measurements. © 2015 The Authors. [less ▲]Detailed reference viewed: 12 (0 ULiège) Exploring the crowded central region of ten Galactic globular clusters using EMCCDs: Variable star searches and new discoveriesFiguera Jaimes, R.; Bramich, D. M.; Skottfelt, J. et alin Astronomy and Astrophysics (2016), 588Aims. We aim to obtain time-series photometry of the very crowded central regions of Galactic globular clusters; to obtain better angular resolution thanhas been previously achieved with conventional CCDs ... [more ▼]Aims. We aim to obtain time-series photometry of the very crowded central regions of Galactic globular clusters; to obtain better angular resolution thanhas been previously achieved with conventional CCDs on ground-based telescopes; and to complete, or improve, the census of the variable star population in those stellar systems. Methods. Images were taken using the Danish 1.54-m Telescope at the ESO observatory at La Silla in Chile. The telescope was equipped with an electron-multiplying CCD, and the short-exposure-time images obtained (ten images per second) were stacked using the shift-and-add technique to produce the normal-exposure-time images (minutes). Photometry was performed via difference image analysis. Automatic detection of variable stars in the field was attempted. Results. The light curves of 12 541 stars in the cores of ten globular clusters were statistically analysed to automatically extract the variable stars. We obtained light curves for 31 previously known variable stars (3 long-period irregular, 2 semi-regular, 20 RR Lyrae, 1 SX Phoenicis, 3 cataclysmic variables, 1 W Ursae Majoris-type and 1 unclassified) and we discovered 30 new variables (16 long-period irregular, 7 semi-regular, 4 RR Lyrae, 1 SX Phoenicis and 2 unclassified). Fluxes and photometric measurements for these stars are available in electronic form through the Strasbourg astronomical Data Center. © ESO, 2016. [less ▲]Detailed reference viewed: 18 (0 ULiège) High-resolution Imaging of Transiting Extrasolar Planetary systems (HITEP): I. Lucky imaging observations of 101 systems in the southern hemisphereEvans, D. F.; Southworth, J.; Maxted, P. F. L. et alin Astronomy and Astrophysics (2016), 589Context. Wide binaries are a potential pathway for the formation of hot Jupiters. The binary fraction among host stars is an important discriminator between competing formation theories, but has not been ... [more ▼]Context. Wide binaries are a potential pathway for the formation of hot Jupiters. The binary fraction among host stars is an important discriminator between competing formation theories, but has not been well characterised. Additionally, contaminating light from unresolved stars can significantly affect the accuracy of photometric and spectroscopic measurements in studies of transiting exoplanets. Aims. We observed 101 transiting exoplanet host systems in the Southern hemisphere in order to create a homogeneous catalogue of both bound companion stars and contaminating background stars, in an area of the sky where transiting exoplanetary systems have not been systematically searched for stellar companions. We investigate the binary fraction among the host stars in order to test theories for the formation of hot Jupiters. Methods. Lucky imaging observations from the Two Colour Instrument on the Danish 1.54 m telescope at La Silla were used to search for previously unresolved stars at small angular separations. The separations and relative magnitudes of all detected stars were measured. For 12 candidate companions to 10 host stars, previous astrometric measurements were used to evaluate how likely the companions are to be physically associated. Results. We provide measurements of 499 candidate companions within 20 arcsec of our sample of 101 planet host stars. 51 candidates are located within 5 arcsec of a host star, and we provide the first published measurements for 27 of these. Calibrations for the plate scale and colour performance of the Two Colour Instrument are presented. Conclusions. We find that the overall multiplicity rate of the host stars is 38-13 +17%, consistent with the rate among solar-type stars in our sensitivity range, suggesting that planet formation does not preferentially occur in long period binaries compared to a random sample of field stars. Long period stellar companions (P > 10 yr) appear to occur independently of short period companions, and so the population of close-in stellar companions is unconstrained by our study. © ESO, 2016. [less ▲]Detailed reference viewed: 14 (0 ULiège) THE SPITZER MICROLENSING PROGRAM AS A PROBE for GLOBULAR CLUSTER PLANETS: ANALYSIS of OGLE-2015-BLG-0448Poleski, R.; Zhu, W.; Christie, G. W. et alin Astrophysical Journal (2016), 823(1), The microlensing event OGLE-2015-BLG-0448 was observed by Spitzer and lay within the tidal radius of the globular cluster NGC 6558. The event had moderate magnification and was intensively observed, hence ... [more ▼]The microlensing event OGLE-2015-BLG-0448 was observed by Spitzer and lay within the tidal radius of the globular cluster NGC 6558. The event had moderate magnification and was intensively observed, hence it had the potential to probe the distribution of planets in globular clusters. We measure the proper motion of NGC 6558 ((μcl (N, E) = +0.36 ± 0.10, +1.42 ± 0.10 mas yr-1) as well as the source and show that the lens is not a cluster member. Even though this particular event does not probe the distribution of planets in globular clusters, other potential cluster lens events can be verified using our methodology. Additionally, we find that microlens parallax measured using Optical Gravitational Lens Experiment (OGLE) photometry is consistent with the value found based on the light curve displacement between the Earth and Spitzer. © 2016. The American Astronomical Society. All rights reserved. [less ▲]Detailed reference viewed: 14 (0 ULiège) MASS MEASUREMENTS of ISOLATED OBJECTS from SPACE-BASED MICROLENSINGZhu, W.; Calchi Novati, S.; Gould, A. et alin Astrophysical Journal (2016), 825(1), We report on the mass and distance measurements of two single-lens events from the 2015 Spitzer microlensing campaign. With both finite-source effect and microlens parallax measurements, we find that the ... [more ▼]We report on the mass and distance measurements of two single-lens events from the 2015 Spitzer microlensing campaign. With both finite-source effect and microlens parallax measurements, we find that the lens of OGLE-2015-BLG-1268 is very likely a brown dwarf (BD). Assuming that the source star lies behind the same amount of dust as the Bulge red clump, we find the lens is a 45 ±7 BD at 5.9 ±1.0 kpc. The lens of of the second event, OGLE-2015-BLG-0763, is a 0.50 ±0.04 star at 6.9 ±1.0 kpc. We show that the probability to definitively measure the mass of isolated microlenses is dramatically increased once simultaneous ground- and space-based observations are conducted. © 2016. The American Astronomical Society. All rights reserved. [less ▲]Detailed reference viewed: 8 (0 ULiège) Many new variable stars discovered in the core of the globular cluster NGC 6715 (M 54) with EMCCD observationsFiguera Jaimes, R.; Bramich, D. M.; Kains, N. et alin Astronomy and Astrophysics (2016), 592Context. We show the benefits of using electron-multiplying CCDs and the shift-and-add technique as a tool to minimise the effects of atmospheric turbulence, such as blending between stars in crowded ... [more ▼]Context. We show the benefits of using electron-multiplying CCDs and the shift-and-add technique as a tool to minimise the effects of atmospheric turbulence, such as blending between stars in crowded fields, and to avoid saturated stars in the fields observed. We intend to complete, or improve on, the census of the variable star population in globular cluster NGC 6715. Aims. Our aim is to obtain high-precision time-series photometry of the very crowded central region of this stellar system via the collection of better angular resolution images than has been previously achieved with conventional CCDs on ground-based telescopes. Methods. Observations were carried out using the Danish 1.54-m telescope at the ESO La Silla observatory in Chile. The telescope is equipped with an electron-multiplying CCD that enables short-exposure-time images to be obtained (ten images per second) that were stacked using the shift-and-add technique to produce the normal-exposure-time images (minutes). The high precision photometry was performed via difference image analysis employing the DanDIA pipeline. We attempted automatic detection of variable stars in the field. Results. We statistically analysed the light curves of 1405 stars in the crowded central region of NGC 6715 to automatically identify the variable stars present in this cluster. We found light curves for 17 previously known variable stars near the edges of our reference image (16 RR Lyrae and 1 semi-regular) and we discovered 67 new variables (30 RR Lyrae, 21 irregular (long-period type), 3 semi-regular, 1 W Virginis, 1 eclipsing binary, and 11 unclassified). Photometric measurements for these stars are available in electronic form through the Strasbourg Astronomical Data Centre. © 2016 ESO. [less ▲]Detailed reference viewed: 8 (0 ULiège) OGLE-2015-BLG-0479LA,B: BINARY GRAVITATIONAL MICROLENS CHARACTERIZED by SIMULTANEOUS GROUND-BASED and SPACE-BASED OBSERVATIONSHan, C.; Udalski, A.; Gould, A. et alin Astrophysical Journal (2016), 828(1), We present a combined analysis of the observations of the gravitational microlensing event OGLE-2015-BLG-0479 taken both from the ground and by the Spitzer Space Telescope. The light curves seen from the ... [more ▼]We present a combined analysis of the observations of the gravitational microlensing event OGLE-2015-BLG-0479 taken both from the ground and by the Spitzer Space Telescope. The light curves seen from the ground and from space exhibit a time offset of ∼13 days between the caustic spikes, indicating that the relative lens-source positions seen from the two places are displaced by parallax effects. From modeling the light curves, we measure the space-based microlens parallax. Combined with the angular Einstein radius measured by analyzing the caustic crossings, we determine the mass and distance of the lens. We find that the lens is a binary composed of two G-type stars with masses of ∼1.0 M⊙ and ∼0.9 M⊙ located at a distance of ∼3 kpc. In addition, we are able to constrain the complete orbital parameters of the lens thanks to the precise measurement of the microlens parallax derived from the joint analysis. In contrast to the binary event OGLE-2014-BLG-1050, which was also observed by Spitzer, we find that the interpretation of OGLE-2015-BLG-0479 does not suffer from the degeneracy between (±, ±) and (±, ∓) solutions, confirming that the four-fold parallax degeneracy in single-lens events collapses into the two-fold degeneracy for the general case of binary-lens events. The location of the blend in the color-magnitude diagram is consistent with the lens properties, suggesting that the blend is the lens itself. The blend is bright enough for spectroscopy and thus this possibility can be checked from future follow-up observations. © 2016. The American Astronomical Society. All rights reserved. [less ▲]Detailed reference viewed: 10 (0 ULiège) 1 2 3 4 5
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## Description
Title: Studies of trap generation in silicon-dioxide and electromigration in aluminum thin films Author(s): Chen, Ann J. Doctoral Committee Chair(s): Sah, C.T. Department / Program: Electrical and Computer Engineering Discipline: Electrical Engineering Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Engineering, Electronics and Electrical Abstract: The effects of HCl on trap generation in SiO$\sb2$ film and at Si/SiO$\sb2$ interface at high electric fields are studied by avalanche electron injection. A comparison between a 9%-HCl-oxide MOS capacitor and a dry-oxide MOS capacitor shows that HCl decreases the hydrogenation rate of the boron acceptor in the silicon surface layer and increases the density of the peaked interface trap at 0.3 eV above the silicon midgap. A new chlorine-related, positively-charged electron trap in the oxide is observed and isolated from the chlorine-independent, negatively-charged oxide hole trap. Chlorine also reduces the density of the smaller cross-section oxide electron trap, which gives the turn-around phenomenon.Electromigration in pure aluminum thin film is studied at high-current density and low-test temperature. The resistance of the aluminum line is measured during current stress. The following points are concluded from this study: (1) Due to the thermal resistances of the substrate and the package, the temperature at the thin film surface is higher than that at the ambient. This additional temperature rise has to be corrected in order to compare the data. (2) A damage relaxation is observed when the current is turned off; therefore, the lifetime of pulse operation cannot be obtained by dividing a duty factor from the lifetime of direct current operation. (3) The anode is damaged more seriously than the cathode in our samples. Issue Date: 1989 Type: Text Language: English URI: http://hdl.handle.net/2142/19715 Rights Information: Copyright 1989 Chen, Ann J. Date Available in IDEALS: 2011-05-07 Identifier in Online Catalog: AAI8916225 OCLC Identifier: (UMI)AAI8916225
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# Troubleshooting¶
If you’re having any trouble running your Pyquil programs locally or on the QPU, please check the following things before sending a support request. It will save you time and make it easier for us to help!
1. Ensure that your pyQuil version is up to date. If you’re using pip, you can do this with pip freeze. Within your script, you can use __version__:
import pyquil
print(pyquil.__version__)
You can update pyQuil with pip using pip install pyquil --upgrade. You can find the latest version available at our releases page or on PyPi.
2. If the error appears to be authentication-related, or makes any mention of your user_auth_token, then please update your token following the directions at https://qcs.rigetti.com/auth/token.
3. Run your script with debug logging enabled. If you’re running a script, you can enable that using an environment variable:
LOG_LEVEL=DEBUG pyquil my_script.py
import logging
from pyquil.api._logger import logger
logger.setLevel(logging.DEBUG)
If the problem still isn’t clear, then we can help! Please send your debug log to us, along with the contents of your ~/.qcs_config and ~/.forest_config files, at our support page. Thanks for using pyQuil!
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# zbMATH — the first resource for mathematics
Empirical validation of a queueing approach to uninterrupted traffic flows. (English) Zbl 1124.60079
The paper reports an empirical validation of the validity of queueing models for modeling uninterrupted highway traffic flows. A dataset collected by the ministry of transportation of the Flemish Government (Belgium) was used in the analysis. Since traffic flows change during the day, different queueing models were used for different periods of the day. The utilized approach was to find a queueing model with a small difference between the observed speed and the speed predicted by a queueing model. The selected measure of the difference was the Theil coefficient. The conclusion of the study is that during non-congested hours the best model is $$M/G/1$$. During the congested hours, the state dependent $$GI/G/z$$ models are more realistic. For $$GI/G/z$$ models the authors use approximate formulae.
##### MSC:
60K25 Queueing theory (aspects of probability theory) 90B20 Traffic problems in operations research 90B22 Queues and service in operations research 60K30 Applications of queueing theory (congestion, allocation, storage, traffic, etc.) 68M20 Performance evaluation, queueing, and scheduling in the context of computer systems
##### Keywords:
traffic flows; queueing theory; empirical validation
AIMSUN
Full Text:
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Many influences are cited as antecedents to the modern guitar. Although the development of the earliest "guitars" is lost in the history of medieval Spain, two instruments are commonly cited as their most influential predecessors, the European lute and its cousin, the four-string oud; the latter was brought to Iberia by the Moors in the 8th century.[7]
The loud, amplified sound and sonic power of the electric guitar played through a guitar amp has played a key role in the development of blues and rock music, both as an accompaniment instrument (playing riffs and chords) and performing guitar solos, and in many rock subgenres, notably heavy metal music and punk rock. The electric guitar has had a major influence on popular culture. The guitar is used in a wide variety of musical genres worldwide. It is recognized as a primary instrument in genres such as blues, bluegrass, country, flamenco, folk, jazz, jota, mariachi, metal, punk, reggae, rock, soul, and many forms of pop.
YellowBrickCinema’s Sleep Music is the perfect relaxing music to help you go to sleep, and enjoy deep sleep. Our music for sleeping is the best music for stress relief, to reduce insomnia, and encourage dreaming. Our calm music for sleeping uses Delta Waves and soft instrumental music to help you achieve deep relaxation, and fall asleep. Our relaxing sleep music can be used as background music, meditation music, relaxation music, peaceful music and sleep music. Let our soothing music and calming music help you enjoy relaxing deep sleep.
The ratio of the spacing of two consecutive frets is {\displaystyle {\sqrt[{12}]{2}}} (twelfth root of two). In practice, luthiers determine fret positions using the constant 17.817—an approximation to 1/(1-1/ {\displaystyle {\sqrt[{12}]{2}}} ). If the nth fret is a distance x from the bridge, then the distance from the (n+1)th fret to the bridge is x-(x/17.817).[15] Frets are available in several different gauges and can be fitted according to player preference. Among these are "jumbo" frets, which have much thicker gauge, allowing for use of a slight vibrato technique from pushing the string down harder and softer. "Scalloped" fretboards, where the wood of the fretboard itself is "scooped out" between the frets, allow a dramatic vibrato effect. Fine frets, much flatter, allow a very low string-action, but require that other conditions, such as curvature of the neck, be well-maintained to prevent buzz.
The playing of (3-5 string) guitar chords is simplified by the class of alternative tunings called regular tunings, in which the musical intervals are the same for each pair of consecutive strings. Regular tunings include major-thirds tuning, all-fourths, and all-fifths tunings. For each regular tuning, chord patterns may be diagonally shifted down the fretboard, a property that simplifies beginners' learning of chords and that simplifies advanced players' improvisation. On the other hand, in regular tunings 6-string chords (in the keys of C, G, and D) are more difficult to play.
Open tuning refers to a guitar tuned so that strumming the open strings produces a chord, typically a major chord. The base chord consists of at least 3 notes and may include all the strings or a subset. The tuning is named for the open chord, Open D, open G, and open A are popular tunings. All similar chords in the chromatic scale can then be played by barring a single fret.[16] Open tunings are common in blues and folk music,[17] and they are used in the playing of slide and bottleneck guitars.[16][18] Many musicians use open tunings when playing slide guitar.[17]
Learning guitar is a lot of fun, and with the right lessons anyone can become a great guitar player. However, to be successful it's important to pick the right learning method and stay focused. We designed our Core Learning System to be a step-by-step system that keeps beginners on-track and having fun. Give it a try today by becoming a Full Access member.
Electric guitars and bass guitars have to be used with a guitar amplifier and loudspeaker or a bass amplifier and speaker, respectively, in order to make enough sound to be heard by the performer and audience. Electric guitars and bass guitars almost always use magnetic pickups, which generate an electric signal when the musician plucks, strums or otherwise plays the instrument. The amplifier and speaker strengthen this signal using a power amplifier and a loudspeaker. Acoustic guitars that are equipped with a piezoelectric pickup or microphone can also be plugged into an instrument amplifier, acoustic guitar amp or PA system to make them louder. With electric guitar and bass, the amplifier and speaker are not just used to make the instrument louder; by adjusting the equalizer controls, the preamplifier, and any onboard effects units (reverb, distortion/overdrive, etc.) the player can also modify the tone (aka timbre or "colour") and sound of the instrument. Acoustic guitar players can also use the amp to change the sound of their instrument, but in general, acoustic guitar amps are used to make the natural acoustic sound of the instrument louder without changing its sound that much.
Anyone playing and/or teaching guitar needs staff paper, blank tab, guitar chord charts, guitar scale charts, and fretboard diagrams to chart their guitar lessons and musical ideas. You can find books with some combination of these blank charts and grids, but you can’t find one with all of them organized in a practical way. That’s why we chose to design our own.
##### Justin is an instructor with that rare combination that encompasses great playing in conjunction with a thoughtful, likable personality. Justin's instruction is extremely intelligent because he's smart enough to know the 'basics' don't have to be served 'raw' - Justin keenly serves the information covered in chocolate. Justin's site is like a free pass in a candy store!
A few years back, I dusted off the ol' Takamine I got in high school to try some 'music therapy' with my disabled son, who was recovering from a massive at-birth stroke. This reignited my long dormant passion to transform myself from a beach strummer to a 'real' musician; however, as a single mom, taking in-person lessons was financially difficult. Then I found Justinguitar! Flash forward to today; my son is almost fully recovered (YAY!), my guitar collection has grown significantly, and I'm starting to play gigs. None of this would have been possible without your guidance and generosity, Justin. Thank you for being part of the journey!
When all is said and done, the only way to know for sure what a certain set of strings will sound like on your guitar is to take them for a spin. For that reason, it's not a bad idea to try out a few different ones that interest you to see which ones you like best - then you can stock up! However you like to approach your string shopping, the amazing selection you'll find here is sure to satisfy.
## Our Study Music for concentration uses powerful Alpha Waves and Binaural Beats to boost concentration and brain power and is ideal relaxing music for stress relief. This Study Music and Focus Music is relaxing instrumental music that will help you study, focus and learn for that big test or exam and naturally allow your mind to reach a state of focus, perfect for work and study.
Adding a minor seventh to a major triad creates a dominant seventh (denoted V7). In music theory, the "dominant seventh" described here is called a major-minor seventh, emphasizing the chord's construction rather than its usual function.[27] Dominant sevenths are often the dominant chords in three-chord progressions,[18] in which they increase the tension with the tonic "already inherent in the dominant triad".[28]
F major. This is fairly similar to the C, but a little more difficult to play. Press the fourth string down at the third fret with your ring finger, the third string down at the second fret with your middle finger, and the first and second strings down at the first fret with your index. You just flatten your index finger down across the two strings; lower your thumb if you struggle. You don't play the fifth or sixth strings in this chord.
Learn the C chord. The first chord we will cover is a C chord—one of the most basic chords in music. Before we do, let's break down just what that means. A proper chord, whether played on a piano, a guitar, or sung by well-trained mice, is simply three or more notes sounded together. (Two notes is called a "diad," and while musically useful, is not a chord.) Chords can also contain far more than three notes, but that's well beyond the scope of this article. This is what a C chord looks like on the guitar:
In Mexico, the popular mariachi band includes a range of guitars, from the small requinto to the guitarrón, a guitar larger than a cello, which is tuned in the bass register. In Colombia, the traditional quartet includes a range of instruments too, from the small bandola (sometimes known as the Deleuze-Guattari, for use when traveling or in confined rooms or spaces), to the slightly larger tiple, to the full-sized classical guitar. The requinto also appears in other Latin-American countries as a complementary member of the guitar family, with its smaller size and scale, permitting more projection for the playing of single-lined melodies. Modern dimensions of the classical instrument were established by the Spaniard Antonio de Torres Jurado (1817–1892).[12]
Spiral bound guitar book arrived on time as promised. As reference book for guitar chords, it's quite convenient to use for all levels of guitar expertise. It also provides alternatives to play a certain chord. It's easy to follow and to use. Using the tabs near the edge of the page, chords are arranged from A to G & "other chords". Obviously, the guitar greenhorn needs to learn a few basic chords first, and this book builds on those skills. Although the first edition was published in 2006, guitar chords don't really change, unlike other fields of study, so it's relevant today as it was years ago. I deducted 1 star because the back cover arrived crumpled, and I like to keep my books pristine. This book is supposed to be brand new. The person who packed the box was not careful. I still recommend this guitar book as a quick reference. It's faster to use this than look up chords individually on the web.
The headstock is located at the end of the guitar neck farthest from the body. It is fitted with machine heads that adjust the tension of the strings, which in turn affects the pitch. The traditional tuner layout is "3+3", in which each side of the headstock has three tuners (such as on Gibson Les Pauls). In this layout, the headstocks are commonly symmetrical. Many guitars feature other layouts, including six-in-line tuners (featured on Fender Stratocasters) or even "4+2" (e.g. Ernie Ball Music Man). Some guitars (such as Steinbergers) do not have headstocks at all, in which case the tuning machines are located elsewhere, either on the body or the bridge.
With the C major chord, put that shape on the guitar for thirty seconds, take it off, shake it out, and repeat the process a few times. As you’re making the shape, remember to come right behind the frets on the tips of your fingers. When you’re starting out, you may have to place each finger down one at a time, but that’s natural. You’ll get better with time and eventually be able to go right to the chord.
The guitar is a fretted musical instrument that usually has six strings.[1] It is typically played with both hands by strumming or plucking the strings with either a guitar pick or the finger(s)/fingernails of one hand, while simultaneously fretting (pressing the strings against the frets) with the fingers of the other hand. The sound of the vibrating strings is projected either acoustically, by means of the hollow chamber of the guitar (for an acoustic guitar), or through an electrical amplifier and a speaker.
Play the A Major. This is another "big chord," sonically. There are several ways to play this. You can use one finger across the 2nd fret of the B, G, and D strings (playing C#, A, and E, respectively), or any other combination of fingers. For this example, we'll use the 4th finger on the B string, 3rd finger on the G string, and 2nd finger on the D string.
YellowBrickCinema’s Sleep Music is the perfect relaxing music to help you go to sleep, and enjoy deep sleep. Our music for sleeping is the best music for stress relief, to reduce insomnia, and encourage dreaming. Our calm music for sleeping uses Delta Waves and soft instrumental music to help you achieve deep relaxation, and fall asleep. Our relaxing sleep music can be used as background music, meditation music, relaxation music, peaceful music and sleep music. Let our soothing music and calming music help you enjoy relaxing deep sleep.
Learn a D major. This chord only requires the bottom four strings. Place your index finger on the 3rd string, 2nd fret. Your ring finger then goes on the 2nd string, 3rd fret, and your middle finger is the 1st string, second fret. You'll form a little triangle shape. Only strum these three strings and the 4th string -- the open D -- to sound out the chord.
Open tunings improve the intonation of major chords by reducing the error of third intervals in equal temperaments. For example, in the open-G overtones tuning G-G-D-G-B-D, the (G,B) interval is a major third, and of course each successive pair of notes on the G- and B-strings is also a major third; similarly, the open-string minor-third (B,D) induces minor thirds among all the frets of the B-D strings. The thirds of equal temperament have audible deviations from the thirds of just intonation: Equal temperaments is used in modern music because it facilitates music in all keys, while (on a piano and other instruments) just intonation provided better-sounding major-third intervals for only a subset of keys.[65] "Sonny Landreth, Keith Richards and other open-G masters often lower the second string slightly so the major third is in tune with the overtone series. This adjustment dials out the dissonance, and makes those big one-finger major-chords come alive."[66]
Guitar chords are dramatically simplified by the class of alternative tunings called regular tunings. In each regular tuning, the musical intervals are the same for each pair of consecutive strings. Regular tunings include major-thirds (M3), all-fourths, augmented-fourths, and all-fifths tunings. For each regular tuning, chord patterns may be diagonally shifted down the fretboard, a property that simplifies beginners' learning of chords and that simplifies advanced players' improvisation.[70][71][72] The diagonal shifting of a C major chord in M3 tuning appears in a diagram.
With the advent of YouTube tutorials for just about everything, it's only fitting that there are a lot of videos out there claiming to be able to teach you how to play guitar. While you might think this is a quick and easy way to become a pro, you'll want to make sure you have all of the information before diving in headfirst. Online Tutorials When you're surfing the Internet, you'll come across many different websites with prerecorded tutorials to help you learn guitar online. But the keywo
The three notes of a major triad have been introduced as an ordered triplet, namely (root, third, fifth), where the major third is four semitones above the root and where the perfect fifth is seven semitones above the root. This type of triad is in closed position. Triads are quite commonly played in open position: For example, the C-major triad is often played with the third (E) and fifth (G) an octave higher, respectively sixteen and nineteen semitones above the root. Another variation of the major triad changes the order of the notes: For example, the C-major triad is often played as (C,G,E), where (C,G) is a perfect fifth and E is raised an octave above the perfect third (C,E). Alternative orderings of the notes in a triad are discussed below (in the discussions of chord inversions and drop-2 chords).
Stacking the C-major scale with thirds creates a chord progression Play (help·info), traditionally enumerated with the Roman numerals I, ii, iii, IV, V, vi, viio. Its major-key sub-progression C-F-G (I-IV-V) is conventional in popular music. In this progression, the minor triads ii-iii-vi appear in the relative minor key (Am)'s corresponding chord progression.
Of course, there are a few ways to narrow down the string options. For starters, since guitars come in different scales, you need a set that's the right length for your instrument. You also need to match the type of guitar: electric strings for an electric guitar, acoustic strings for acoustic. If you play an acoustic-electric, you'll usually be looking for acoustic strings since those instruments use non-magnetic pickups. For classical and Latin guitar types directly descended from ancient gut-stringed instruments, the right strings are generally going to be nylon.
Getting to grips with how chords are formed gives you a basic introduction to music theory and helps you understand the ways you can alter them to create more interesting sounds. All chords are built from certain notes in scales. The C major scale is the easiest, because it just runs C, D, E, F, G, A and B. These notes are numbered (usually using Roman numerals) in that order, from one (I) to seven (VII).
You need to place one finger on whatever fret you want to bar and hold it there over all of the strings on that fret. The rest of your fingers will act as the next finger down the line (second finger barring, so third finger will be your main finger, and so on). You can also buy a capo, so that you don't have to deal with the pain of the guitar's strings going against your fingers. The capo bars the frets for you. This also works with a ukulele.
## The fingerboard, also called the fretboard, is a piece of wood embedded with metal frets that comprises the top of the neck. It is flat on classical guitars and slightly curved crosswise on acoustic and electric guitars. The curvature of the fretboard is measured by the fretboard radius, which is the radius of a hypothetical circle of which the fretboard's surface constitutes a segment. The smaller the fretboard radius, the more noticeably curved the fretboard is. Most modern guitars feature a 12" neck radius, while older guitars from the 1960s and 1970s usually feature a 6-8" neck radius. Pinching a string against a fret on fretboard effectively shortens the vibrating length of the string, producing a higher pitch.
My personal opinion on the topic, as one musician to another, is that the best thing you can possibly do when trying to figure out which string to go with is to try out as many different brands and types of guitar strings as you can. Strings are cheap enough that most people are going to be able to afford to experiment, and the truth of the matter is that you’re probably not going to really know what works best for you until you have hands on experience.
Our intermediate and advanced guitar lessons are tailored to build on the skills that students have developed through their previous beginner lessons and programs. These programs pair private guitar lessons with full-band group rehearsals and live performances, allowing student guitarists to showcase their skills by playing advanced songs from famous musicians. Our world-class guitar instructors and teaching system are proven to help students play and perform at a higher level.
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# Multiplicity one conjecture
I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one.
Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder 1) for which lattices this conjecture had been conjectured? 2) what is the motivation for this conjecture? 3) to whom this conjecture is due? 4) is it published somewhere? 5) is it proven in some cases?
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In Sarnak's article http://web.math.princeton.edu/sarnak/baltimore.pdf, he recalls one famous conjecture (Conjecture I, due to himself) that $\Gamma \backslash \mathbb{H}$ should have very few Maass forms for "most" Fuchsian lattices $\Gamma$.
Btw, he attributes the simplicity conjecture (Conjecture 3) for $SL_2(\mathbb{Z})$ to Cartier.
Wolpert (Theorem I) has shown that Sarnak's conjecture would follow if the simplicity conjecture holds for the congruence subgroup $\Gamma(2)$.
Also GH last conjecture that the multiplicity is uniformly bounded in $N$ would suffice for Sarnak's conjecture, but current knowledge is that the multiplicity of an eigenvalue of magnitude $T$ is at most $\ll_N \sqrt{T}/ \log T$ and not $\ll_N 1$.
In fact, Sarnak conjectures that there exist $\Gamma$ with only finitely many Maass wave forms, but this does not follow from the simplicity conjecture for $\Gamma(2)$.
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This conjecture is usually stated for $\mathrm{SL}_2(\mathbb{Z})$, and it is widely open. I think it is folklore, and is stated in several papers, e.g. in Luo: Nonvanishing of $L$-values and the Weyl law (before (3)).
The motivation, I think, is similar as with the conjecture for the multiplicity of the Riemann zeta zeros. The belief is that there is no "accidental" algebraic independence among the eigenvalues of the Laplacian or the zeros of an automorphic $L$-function. For example, the Laplacian eigenvalue $1/4$ is expected to "come from" an even Galois representation, while the zero $1/2$ is expected to "come from" rational points of infinite order on an abelian variety. For $\mathrm{SL}_2(\mathbb{Z})$ or $\zeta(s)$ we don't know of any object that would "impose" any algebraic independence on the data, hence we believe that in those cases the data is entirely transcendental.
For congruence subgroups $\Gamma_0(N)$ the "multiplicity one conjecture" is false, because the eigenvalue $1/4$ is known to occur with multiplicity for some $N$'s. The known examples come from even Galois representations. I think it is safe to believe that the multiplicities are bounded for any $N$.
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Is it also know for newforms of $\Gamma_0(N)$? – Marc Palm Jul 19 '12 at 14:29
I think the eigenvalue 1/4 occurs with multiplicity even among newforms of some level. All we need is two even Galois representations with the same Artin conductor. – GH from MO Jul 19 '12 at 19:16
Sorry for bumping this old question, but is the multiplicity bounded uniformly with respect to $N$? – user31814 May 15 at 18:07
@user31814: I think the multiplicity is bounded (conjecturally) for a fixed $N$, but it can get arbitrary large (provably) for varying $N$. – GH from MO May 15 at 18:11
@GHfromMO, thanks! Is there a reference for what you said? – user31814 May 16 at 1:36
Multiplicity one refers to something else, related but much weaker.
For the analogue question for lattices, there are trivial counter examples: Induction by steps for example suggests on the level of Lie groups $$Ind_{\Gamma(N)} ^{PSL_2(\mathbb{R})} 1 \cong Ind_{PSL_2(\mathbb{Z})} ^{PSL_2(\mathbb{R})} Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$ and e.g. by the Peter-Weyl theorem, we know that $$Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$ the multiplicity of an irreducible representation equals its dimension.
Note that GH's example is less trivial, since $$Ind_{\Gamma_0(N)}^{PSL_2(\mathbb{Z})} 1$$ decomposes with multiplicity one.
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Could you elaborate to what it does refer, please? – Ruedi Meier Aug 15 '12 at 17:49
For every prime, you get also an eigenvalue. Multiplicity one theorem states that their exists only one eigenfunction with all the same eigenvalues. More rigorously put, you find this here: en.wikipedia.org/wiki/Multiplicity-one_theorem – Marc Palm Sep 4 '12 at 9:33
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Viewpoint
Towards an Atomtronic Diode
Physics 8, 72
Rubidium atoms in an optical trap have been made to exhibit negative differential conductance, a phenomenon normally found in semiconductor diodes.
Atomtronics—an area in which highly controlled atomic gases mimic the properties of electrical circuits [1]—has reached a series of milestones as a result of novel experimental techniques. Being able to tailor the microscopic behavior of these gases opens opportunities to explore fundamental aspects of electronic transport, while also contributing new tools for applications, such as quantum devices for measurement and sensing. Now in a new study, Herwig Ott, from the University of Kaiserslautern in Germany, and colleagues [2] have used techniques in which an atomic gas is prepared and probed with a focused electron beam to realize a system exhibiting a phenomenon known as negative differential conductance. In this phenomenon, the current across a tunnel junction in a circuit decreases as the voltage, or chemical potential difference, is increased. Their experiment demonstrates a building block for complex atomic circuits and opens new avenues for investigating how many-particle transport processes equilibrate to a steady state.
In atomtronics, the circuit components are determined via laser-field or magnetic traps for atoms, which can be designed flexibly. The atoms can be fermionic or bosonic, and their interactions can also be controlled using magnetic fields. This allows for the realization of phenomena beyond classical electronics, such as analogs to spintronics and mesoscopic transport systems. Several such phenomena were recently demonstrated, including the quantized conductance of atoms [3] and thermoelectric transport [4] in quantum wires made with light. Quantized flow of atoms without dissipation of energy has also been observed in a toroidal optical trap with a superfluid gas of bosonic atoms, together with a phenomenon known as hysteresis—which is common in electrical circuits—when a weak barrier is added in the trap [5, 6]. By using laser light to couple the motion of atoms to transitions between different internal atomic states, the spin-Hall effect [7] has also been realized.
Ott and colleagues exploited a major recent experimental development in this field—the possibility of manipulating and probing single microtraps of atoms that are separated by distances of less than a micrometer. This advance has been achieved using optical devices called quantum gas microscopes [8] and also using a focused electron beam [9]. On such short length scales, particles can tunnel quantum mechanically through the barriers between the microtraps formed by laser light on time scales of the order of milliseconds. Ott and co-workers’ system is a one-dimensional periodic arrangement of light called an optical lattice, with each lattice site corresponding to a microtrap that is initially occupied by around 700 bosonic rubidium atoms (Fig. 1, top). The authors used an electron beam to empty one of the lattice’s microtraps. Interactions between atoms lead to a chemical potential difference between the emptied site and the neighboring filled sites. This difference corresponds to a voltage across the barrier (tunnel junction) between the sites. Using the same electron beam, they then probed the atom number as atoms tunnel to repopulate the emptied site (Fig. 1, bottom). These data allowed them to calculate the relationship between the chemical potential difference and the current of atoms flowing across the junction.
The researchers observed the normal increase in current with increasing potential difference when each of these quantities is small—that is, they detected a positive differential conductance. However, when the chemical potential difference was large, they instead detected a negative differential conductance. Such a characteristic is well known in fast semiconductor diodes known as tunnel diodes. In such diodes, negative differential conductance typically occurs as a result of the interplay between the chemical potential on one side of the diode’s junction and the availability of electron holes on the other side of the junction. In Ott and colleagues’ atomic system, however, this behavior occurs because the tunneling rate between sites depends on the atom-number difference. There is a marked decrease in the tunneling rate as the number imbalance becomes large, leading to a decrease in the current as the chemical potential difference is increased.
One of the most interesting aspects of this work is the role of thermalization in the form of “collisional dephasing,” a process that induces a thermal steady state for the current. Ott and co-workers’ setup—a tunnel junction for bosonic atoms—might be expected to exhibit Josephson oscillations [10], which would lead to fluctuations in the atomic density back and forth between the two microtraps. The authors instead observed a steady flow of atoms across the junction until the number of atoms on each site was essentially equal, and the chemical potential difference returned to zero. This can be attributed directly to collisional dephasing: that is, the collisions between atoms induce a thermal steady-state current. This arises because many atoms are involved in the tunneling process, and for each atom that tunnels, there are many available motional modes in the direction transverse to the tunnel junction, which can be populated as a result of collisions between atoms. The population of many modes leads to rapid equilibrium in the system, so that the current is always in a steady state. The high availability of such excited modes allows Ott and colleagues to build a simple model for this mechanism and obtain excellent agreement with their experimental measurements.
How quantum many-particle systems relax to a thermal steady state is a key fundamental question, both in general and, more specifically, in the case of transport systems. This work paves the way for using atomtronic systems to study thermalization under a range of conditions. In particular, it is possible to engineer the processes that cause thermalization or dissipation of energy in these systems, controlling the times on which thermalization occurs. For the present setup, the collisional dephasing could be reduced by working with a smaller initial atom-number imbalance. Future experiments could build on Ott and colleagues’ work by exploring the crossover from Josephson oscillations to the dynamics observed here or by generalizing these studies to other regimes, including quantized conductance. On a more practical level, the demonstration of negative differential conductance can form the basis for building complex, interaction-controlled atomtronic circuits. These could include components such as diodes or amplifiers, which, combined with the controllability and precise measurement techniques available for ultracold atoms, point towards their potential uses in quantum measurement devices.
This research is published in Physical Review Letters.
References
1. B. T. Seaman, M. Krämer, D. Z. Anderson, and M. J. Holland, “Atomtronics: Ultracold-Atom Analogs of Electronic Devices,” Phys. Rev. A 75, 023615 (2007).
2. R. Labouvie, B. Santra, S. Heun, S. Wimberger, and H. Ott, “Negative Differential Conductivity in an Interacting Quantum Gas,” Phys. Rev. Lett. 115, 050601 (2015).
3. S. Krinner, D. Stadler, D. Husmann, J.-P. Brantut, and T. Esslinger, “Observation of Quantized Conductance in Neutral Matter,” Nature 517, 64 (2015).
4. J.-P. Brantut, C. Grenier, J. Meineke, D. Stadler, S. Krinner, C. Kollath, T. Esslinger, and A. Georges, “A Thermoelectric Heat Engine with Ultracold Atoms,” Science 342, 713 (2013).
5. A. Ramanathan, K. C. Wright, S. R. Muniz, M. Zelan, W. T. Hill III, C. J. Lobb, K. Helmerson, W. D. Phillips, and G. K. Campbell, “Superflow in a Toroidal Bose-Einstein Condensate: An Atom Circuit with a Tunable Weak Link,” Phys. Rev. Lett. 106, 130401 (2011).
6. S. Eckel, J. G. Lee, F. Jendrzejewski, N. Murray, C. W. Clark, C. J. Lobb, W. D. Phillips, M. Edwards, and G. K. Campbell, “Hysteresis in a Quantized Superfluid ‘Atomtronic’ Circuit,” Nature 506, 200 (2014).
7. M. C. Beeler, R. A. Williams, K. Jiménez-García, L. J. LeBlanc, A. R. Perry, and A. R. Perry, “The Spin Hall Effect in a Quantum Gas,” Nature 498, 201 (2013).
8. W. S. Bakr, J. I. Gillen, A. Peng, S. Fölling, and M. Greiner, “A Quantum Gas Microscope for Detecting Single Atoms in a Hubbard-Regime Optical Lattice,” Nature 462, 74 (2009); J. F. Sherson, C. Weitenberg, M. Endres, M. Cheneau, I. Bloch, and S. Kuhr, “Single-Atom-Resolved Fluorescence Imaging of an Atomic Mott Insulator,” 467, 68 (2010).
9. T. Gericke, P. Würtz, D. Reitz, T. Langen, and H. Ott, “High-Resolution Scanning Electron Microscopy of an Ultracold Quantum Gas,” Nature Phys. 4, 949 (2008).
10. M. Albiez, R. Gati, J. Fölling, S. Hunsmann, M. Cristiani, and M. K. Oberthaler, “Direct Observation of Tunneling and Nonlinear Self-Trapping in a Single Bosonic Josephson Junction,” Phys. Rev. Lett. 95, 010402 (2005).
Andrew J. Daley is Professor of Theoretical Quantum Optics in the Department of Physics at the University of Strathclyde in Glasgow, Scotland. He completed his doctoral studies at the University of Innsbruck, Austria, in 2005. He was a senior scientist at Innsbruck and then a faculty member at the University of Pittsburgh before moving to Scotland in 2013. His research centers on the interface between quantum optics and many-body physics, especially on the exploration of new possibilities to study out-of-equilibrium dynamics with strongly interacting gases of atoms and molecules in optical potentials (http://qoqms.phys.strath.ac.uk).
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Hellenica World
# .
In geometry, the parallel postulate, also called Euclid's fifth postulate because it is the fifth postulate in Euclid's Elements, is a distinctive axiom in Euclidean geometry. It states that, in two-dimensional geometry:
If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
Euclidean geometry is the study of geometry that satisfies all of Euclid's axioms, including the parallel postulate. A geometry where the parallel postulate does not hold is known as a non-Euclidean geometry. Geometry that is independent of Euclid's fifth postulate (i.e., only assumes the first four postulates) is known as absolute geometry (or, in other places known as neutral geometry).
Converse of Euclid's parallel postulate
If the sum of the two interior angles equals 180°, the lines are parallel and will never intersect.
Euclid did not postulate the converse of his fifth postulate, which is one way to distinguish Euclidean geometry from elliptic geometry. The Elements contains the proof of an equivalent statement (Book I, Proposition 27): If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another. As De Morgan[1] pointed out, this is logically equivalent to (Book I, Proposition 16). These results do not depend upon the fifth postulate, but they do require the second postulate[2] which is violated in elliptic geometry.
Logically equivalent properties
Probably the best known equivalent of Euclid's parallel postulate is Playfair's axiom, named after the Scottish mathematician John Playfair, which states:
At most one line can be drawn through any point not on a given line parallel to the given line in a plane.[3]
Many other statements equivalent to the parallel postulate have been suggested, some of them appearing at first to be unrelated to parallelism, and some seeming so self-evident that they were unconsciously assumed by people who claimed to have proven the parallel postulate from Euclid's other postulates. This is a summary
There is at most one line that can be drawn parallel to another given one by an external point. (Playfair's axiom)
The sum of the angles in every triangle is 180° (triangle postulate).
There exists a triangle whose angles add up to 180°.
The sum of the angles is the same for every triangle.
There exists a pair of similar, but not congruent, triangles.
Every triangle can be circumscribed.
If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.
There exists a quadrilateral of which all angles are right angles.
There exists a pair of straight lines that are at constant distance from each other.
Two lines that are parallel to the same line are also parallel to each other.
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides (Pythagoras' Theorem).[4][5]
There is no upper limit to the area of a triangle. (Wallis axiom)[6]
The summit angles of the Saccheri quadrilateral are 90°.
If a line intersects one of two parallel lines, both of which are coplanar with the original line, then it also intersects the other. (Proclus' axiom)[7]
However, the alternatives which employ the word "parallel" cease appearing so simple when one is obliged to explain which of the three common definitions of "parallel" is meant – constant separation, never meeting, or same angles where crossed by a third line – since the equivalence of these three is itself one of the unconsciously obvious assumptions equivalent to Euclid's fifth postulate. For example, if the word "parallel" in Playfair's axiom is taken to mean 'constant separation', then it is no longer equivalent to Euclid's fifth postulate, and is provable from the first four (the axiom says 'There is at most one line...', which is consistent with there being no such lines). However, if the definition is taken so that parallel lines are lines that do not intersect, Playfair's axiom is equivalent to Euclid's fifth postulate and is thus independent of the first four postulates.
History
For two thousand years, many attempts were made to prove the parallel postulate using Euclid's first four postulates. The main reason that such a proof was so highly sought after was that, unlike the first four postulates, the parallel postulate isn't self-evident. If the order the postulates were listed in the Elements is significant, it indicates that Euclid included this postulate only when he realised he could not prove it or proceed without it.[8] Many attempts were made to prove the fifth postulate from the other four, many of them being accepted as proofs for long periods of time until the mistake was found. Invariably the mistake was assuming some 'obvious' property which turned out to be equivalent to the fifth postulate (Playfair's axiom). Although known from the time of Proclus, this became known as Playfair's Axiom after John Playfair wrote a famous commentary on Euclid in 1795 in which he proposed replacing Euclid's fifth postulate by his own axiom.
Proclus (410-485) wrote a commentary on The Elements where he comments on attempted proofs to deduce the fifth postulate from the other four, in particular he notes that Ptolemy had produced a false 'proof'. Proclus then goes on to give a false proof of his own. However he did give a postulate which is equivalent to the fifth postulate.
Ibn al-Haytham (Alhazen) (965-1039), an Arab mathematician, made the first attempt at proving the parallel postulate using a proof by contradiction,[9][not in citation given] where he introduced the concept of motion and transformation into geometry.[10][not in citation given] He formulated the Lambert quadrilateral, which Boris Abramovich Rozenfeld names the "Ibn al-Haytham–Lambert quadrilateral",[11] and his attempted proof also shows similarities to Playfair's axiom.[12]
Omar Khayyám (1050–1123), a Persian, made the first attempt at formulating a non-Euclidean postulate as an alternative to the parallel postulate,[13] and he was the first to consider the cases of elliptical geometry and hyperbolic geometry, though he excluded the latter.[14] The Khayyam-Saccheri quadrilateral was also first considered by Omar Khayyam in the late 11th century in Book I of Explanations of the Difficulties in the Postulates of Euclid.[11] Unlike many commentators on Euclid before and after him (including Giovanni Girolamo Saccheri), Khayyam was not trying to prove the parallel postulate as such but to derive it from an equivalent postulate: "Two convergent straight lines intersect and it is impossible for two convergent straight lines to diverge in the direction in which they converge."[15] He recognized that three possibilities arose from omitting Euclid's Fifth; if two perpendiculars to one line cross another line, judicious choice of the last can make the internal angles where it meets the two perpendiculars equal (it is then parallel to the first line). If those equal internal angles are right angles, we get Euclid's Fifth; otherwise, they must be either acute or obtuse. He persuaded himself that the acute and obtuse cases lead to contradiction, but had made a tacit assumption equivalent to the fifth to get there.[citation needed]
Nasir al-Din al-Tusi (1201–1274), in his Al-risala al-shafiya'an al-shakk fi'l-khutut al-mutawaziya (Discussion Which Removes Doubt about Parallel Lines) (1250), wrote detailed critiques of the parallel postulate and on Khayyám's attempted proof a century earlier. Nasir al-Din attempted to derive a proof by contradiction of the parallel postulate.[16] He was also one of the first to consider the cases of elliptical geometry and hyperbolic geometry, though he ruled out both of them.[14]
Euclidean, elliptical and hyperbolic geometry. The Parallel Postulate is satisfied only for models of Euclidean geometry.
Nasir al-Din's son, Sadr al-Din (sometimes known as "Pseudo-Tusi"), wrote a book on the subject in 1298, based on his father's later thoughts, which presented one of the earliest arguments for a non-Euclidean hypothesis equivalent to the parallel postulate. "He essentially revised both the Euclidean system of axioms and postulates and the proofs of many propositions from the Elements."[16][17] His work was published in Rome in 1594 and was studied by European geometers. This work marked the starting point for Saccheri's work on the subject [16] wherein he criticised this work as well as the work of Wallis.[18]
Giordano Vitale (1633-1711), in his book Euclide restituo (1680, 1686), used the Khayyam-Saccheri quadrilateral to prove that if three points are equidistant on the base AB and the summit CD, then AB and CD are everywhere equidistant. Girolamo Saccheri (1667-1733) pursued the same line of reasoning more thoroughly, correctly obtaining absurdity from the obtuse case (proceeding, like Euclid, from the implicit assumption that lines can be extended indefinitely and have infinite length), but failing to debunk the acute case (although he managed to wrongly persuade himself that he had).
Where Khayyám and Saccheri had attempted to prove Euclid's fifth by disproving the only possible alternatives, the nineteenth century finally saw mathematicians exploring those alternatives and discovering the logically consistent geometries which result. In 1829, Nikolai Ivanovich Lobachevsky published an account of acute geometry in an obscure Russian journal (later re-published in 1840 in German). In 1831, János Bolyai included, in a book by his father, an appendix describing acute geometry, which, doubtlessly, he had developed independently of Lobachevsky. Carl Friedrich Gauss had also studied the problem, but he did not publish any of his results. Upon hearing of Bolyai's results in a letter from Bolyai's father, Farkas Bolyai, he stated:
"If I commenced by saying that I am unable to praise this work, you would certainly be surprised for a moment. But I cannot say otherwise. To praise it would be to praise myself. Indeed the whole contents of the work, the path taken by your son, the results to which he is led, coincide almost entirely with my meditations, which have occupied my mind partly for the last thirty or thirty-five years."[19]
The resulting geometries were later developed by Lobachevsky, Riemann and Poincaré into hyperbolic geometry (the acute case) and spherical geometry (the obtuse case). The independence of the parallel postulate from Euclid's other axioms was finally demonstrated by Eugenio Beltrami in 1868.
Criticism
Attempts to logically prove this postulate, rather than the eighth axiom, were criticized by Schopenhauer, as described in Schopenhauer's criticism of the proofs of the Parallel Postulate. However, the argument used by Schopenhauer was that the postulate is evident by perception, not that it was not a logical consequence of the other axioms.
Notes and references
^ Heath, T.L., The thirteen books of Euclid's Elements, Vol.1, Dover, 1956, pg.309.
^ Coxeter, H.S.M., Non-Euclidean Geometry, 6th Ed., MAA 1998, pg.3
^ Euclid's Parallel Postulate and Playfair's Axiom
^ Eric W. Weisstein (2003), CRC concise encyclopedia of mathematics (2nd ed.), p. 2147, ISBN 1-58488-347-2, "The parallel postulate is equivalent to the Equidistance postulate, Playfair axiom, Proclus axiom, the Triangle postulate and the Pythagorean theorem."
^ Alexander R. Pruss (2006), The principle of sufficient reason: a reassessment, Cambridge University Press, p. 11, ISBN 0-521-85959-X, "We could include...the parallel postulate and derive the Pythagorean theorem. Or we could instead make the Pythagorean theorem among the other axioms and derive the parallel postulate."
^ Bogomolny, Alexander. "Euclid's Fifth Postulate". Cut The Knot. Retrieved 30 September 2011.
^ Weisstein, Eric W.. "Proclus' Axiom – MathWorld". Retrieved 2009-09-05.
^ Florence P. Lewis (Jan 1920), "History of the Parallel Postulate", The American Mathematical Monthly (The American Mathematical Monthly, Vol. 27, No. 1) 27 (1): 16–23, doi:10.2307/2973238, JSTOR 2973238.
^ Eder (2000)
^ (Katz 1998, p. 269):
In effect, this method characterized parallel lines as lines always equidistant from one another and also introduced the concept of motion into geometry.
^ a b (Rozenfeld 1988, p. 65)
^ (Smith 1992)
^ Victor J. Katz (1998), History of Mathematics: An Introduction, p. 270, Addison-Wesley, ISBN 0-321-01618-1:
"In some sense, his treatment was better than ibn al-Haytham's because he explicitly formulated a new postulate to replace Euclid's rather than have the latter hidden in a new definition."
^ a b Boris A. Rosenfeld and Adolf P. Youschkevitch (1996), "Geometry", in Roshdi Rashed, ed., Encyclopedia of the History of Arabic Science, Vol. 2, p. 447-494 [469], Routledge, London and New York:
"Khayyam's postulate had excluded the case of the hyperbolic geometry whereas al-Tusi's postulate ruled out both the hyperbolic and elliptic geometries."
^ Boris A Rosenfeld and Adolf P Youschkevitch (1996), Geometry, p.467 in Roshdi Rashed, Régis Morelon (1996), Encyclopedia of the history of Arabic science, Routledge, ISBN 0-415-12411-5.
^ a b c Victor J. Katz (1998), History of Mathematics: An Introduction, p. 270-271, Addison-Wesley, ISBN 0-321-01618-1:
"But in a manuscript probably written by his son Sadr al-Din in 1298, based on Nasir al-Din's later thoughts on the subject, there is a new argument based on another hypothesis, also equivalent to Euclid's, [...] The importance of this latter work is that it was published in Rome in 1594 and was studied by European geometers. In particular, it became the starting point for the work of Saccheri and ultimately for the discovery of non-Euclidean geometry."
^ Boris A. Rosenfeld and Adolf P. Youschkevitch (1996), "Geometry", in Roshdi Rashed, ed., Encyclopedia of the History of Arabic Science, Vol. 2, p. 447-494 [469], Routledge, London and New York:
"In Pseudo-Tusi's Exposition of Euclid, [...] another statement is used instead of a postulate. It was independent of the Euclidean postulate V and easy to prove. [...] He essentially revised both the Euclidean system of axioms and postulates and the proofs of many propositions from the Elements."
^ MacTutor's Giovanni Girolamo Saccheri
^ On Gauss' Mountains. http://www.mathpages.com/rr/s8-06/8-06.htm
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Dynamics of Singular Traveling Wave Solutions of a Short Capillary-Gravity Wave Equation Leta Temesgen Desta,Liu Wenjun,El Achab Abdelfattah Keywords:Singular parabola, singular periodic wave solutions, multi-peaked periodic wave solutions, two-peaked solitary wave solutions. Abstract: In this paper, dynamical behaviour of traveling wave solutions to a short capillary-gravity is analyzed by using the method of bifurcation. When the phase orbits intersects the singular parabola $y^2=\frac{2}{\lambda} \phi$ on the phase plane, then the trajectory creates a more weaker wave fronts than the regular traveling wave solutions. By using proper Euler transformations, we reformulate the model as a singular chaotic problem, which can then be analyzed using the singularity study. We prove the existence of three types of physically realistic traveling wave solutions to the case of small diffusion for the first time, two-peaked solitary waves, three-peaked and multi-peaked periodic wave solutions.
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# What major problem did the Roman Empire face in the mid to late Fourth Century
###### Question:
What major problem did the Roman Empire face in the mid to late Fourth Century?
A. Poor infrastructure
B. Lack of education within the empire
D. Immigration and invasion
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# EXERCISE 24.3
QUESTION 1
Find the equation of the circle, the end points of whose diameter are $(2,-3)$ and $(-2,4)$ . Find its center and radius.
Sol :
$(2,-3)$ and $(-2,4)$ are the ends points of the diameter of a circle. The equation of this circle is $(x-2)(x+2)+(y+3)(y-4)=0$
$\Rightarrow x^{2}-4+y^{2}-4 y+3 y-12=0$
$\Rightarrow x^{2}+y^{2}-y-16=0 \quad \ldots(1)$
Equation (1) can be rewritten as
$\Rightarrow x^{2}+\left(y-\frac{1}{2}\right)^{2}-\frac{1}{4}-16=0$
$\Rightarrow x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{65}{4}$
$\therefore$ Center is $\left(0, \frac{1}{2}\right)$ and radius is $\frac{\sqrt{65}}{2}$
QUESTION 2
Find the equation of the circle the end points of whose diameter are the centers of the circles $x^{2}+y^{2}+6 x-14 y-1=0$ and $x^{2}+y^{2}-4 x+10 y-2=0$
Sol :
Given :
$x^{2}+y^{2}+6 x-14 y-1=0 \ldots(1)$
$x^{2}+y^{2}-4 x+10 y-2=0 \ldots (2)$
Equations $(1)$ and $(2)$ can be rewritten as follows:
$(x+3)^{2}+(y-7)^{2}=59$
$(x-2)^{2}+(y+5)^{2}=31$
Thus, the centers of the circles are $(-3,7)$ and $(2,-5)$
Hence, the equation of the circle, the end points of whose diameter are the centers of the given
circles, is $(x+3)(x-2)+(y-7)(y+5)=0,$ i.e. $x^{2}+y^{2}+x-2 y-41=0$
QUESTION 3
The sides of a square are $x=6, x=9, y=3$ and $y=6 .$ Find the equation of a circle drawn on the diagonal of the square as its diameter.
Sol :
According to the question:
Sides of the square are $x=6$ $x=9$ $y=3$ and $y=6$
The vertices of the square are $(6,6)$ $(9,6)$ $(9,3)$ and $(6,3) .$
And, the vertices of two diagonals are $(6,6)$ $(9,3)$ and $(9,6)$ $(6,3)$
Hence, the equation of the circle is $(x-6)(x-9)+(y-6)(y-3)$ or $x^{2}+y^{2}-15 x-9 y+72=0$
QUESTION 4
Find the equation of the circle circumscribing the rectangle whose sides are $x-3 y=4$ $3x+y=22$ $x-3 y=14$ and $3 x+y=62$
Sol :
Sides of the rectangle:
$x-3 y=4 \quad \ldots$ (1)
$3 x+y=22 \quad \ldots(2)$
$x-3 y=14 \quad \ldots(3)$
$3 x+y=62 \quad \ldots(4)$
The intersection of (1) and (2) is $(7,1)$
The intersection of (2) and (3) is $(8,-2)$
The intersection of (3) and (4) is $(20,2)$ .
The intersection of (1) and (4) is $(19,5)$ .
Hence, the vertices of the rectangle are $(7,-1),(8,-2),(20,2)$ and $(19,5)$
The vertices of the diagonals are $(7,-1),(20,2)$ and $(19,5),(8,-2)$
Thus, the required equation of the circle is $(x-7)(x-20)+(y-1)(y-2)=0$ or $x^{2}+y^{2}-27 x-3 y+142=0$
QUESTION 5
Find the equation of the circle passing through the origin and the points where the line $3x+4y=12$ meets the axes of coordinates.
Sol :
Putting $x=0$ in $3 x+4 y=12$ :
$y=3$
Putting $y=0$ in $3 x+4 y=12 :$
$x=4$
Thus, the line $3 x+4 y=12$ meets the axes of coordinates at points $A(0,3)$ and $B(4,0)$
The equation of the circle with AB as the diameter is $(x-0)(x-4)+(y-3)(y-0)=0$ or $x^{2}-4 x+y^{2}-3 y=0$
Hence, the required equation is $x^{2}-4 x+y^{2}-3 y=0$
QUESTION 6
Find the equation of the circle which passes through the origin and cuts off intercepts a and b
respectively from x and y-axes.
Sol :
CASE-1
If the required circle passes through the origin and $(a, b),$ then the end points of the diameter of the circle will be $(0,0)$ and $(a, b)$ and $(a, b)$
Required equation of circle:
$(x-0)(x-a)+(y-0)(y-b)$
$x^{2}+y^{2}-a x-b y=0$
CASE-2
If the required circle passes through the origin and $(-a,-b),$ then the end points of the diameter of the circle will be $(0,0)$ and $(-a,-b)$
Required equation of circle :
$(x-0)(x+a)+(y-0)(y+b)$
$x^{2}+y^{2}+a x+b y=0$
Hence, the equation of the required circle is $x^{2}+y^{2} \pm a x \pm b y=0$
QUESTION 7
Find the equation of the circle whose diameter is the line segment joining $(-4,3)$ and $(12,-1)$ Find also the intercept made by it on y-axis.
Sol :
It is given that the end points of the diameter of the circle are $(-4,3)$ and $(12,-1)$
Required equation of circle :
$(x+4)(x-12)+(y-3)(y+1)$
$x^{2}+y^{2}-8 x-2 y-51=0 \quad \ldots(1)$
Putting $x=0$ in (1) :
$y^{2}-2 y-51=0$
$\Rightarrow y^{2}-2 y-51=0$
$\Rightarrow y=1 \pm 2 \sqrt{13}$
Hence, the intercepts made by it on the $y$ -axis is $1+2 \sqrt{13}-1+2 \sqrt{13}=4 \sqrt{13}$
QUESTION 8
The abscissae of the two points A and B are the roots of the equation $x^{2}+2 a x-b^{2}=0$ and their ordinates are the roots of the equation $x^{2}+2 p x-q^{2}=0$ Find the equation of the circle with AB as diameter. Also, find its radius.
Sol :
Roots of equation $x^{2}+2 a x-b^{2}=0$ are
$x=\frac{-2 a \pm \sqrt{4 a^{2}+4 b^{2}}}{2}$ $=-a \pm \sqrt{a^{2}+b^{2}}$
Roots of equation $x^{2}+2 p x-q^{2}=0$ are
$y=\frac{-2 p \pm \sqrt{4 p^{2}+4 q^{2}}}{2}$ $=-P \pm \sqrt{p^{2}+q^{2}}$
Co-ordinates of A and B respectively :
$A=\left(-a+\sqrt{a^{2}+b^{2}},-p \sqrt{p^{2}+q^{2}}\right)$
$B=\left(-a-\sqrt{a^{2}+b^{2}},-P-\sqrt{p^{2}+q^{2}}\right)$
Hence, equation of circle is
$\left(x+a-\sqrt{a^{2}+b^{2}}\right)\left(x+a+\sqrt{a^{2}+b^{2}}\right)+\left(y+p-\sqrt{p^{2}+q^{2}}\right)\left(y+p+\sqrt{p^{2}+q^{2}}\right)=0$
$\Rightarrow(x+a)^{2}-a^{2}-b^{2}+(y+p)^{2}-p^{2}-q^{2}=0$
$\Rightarrow x^{2}+y^{2}+2 a x+2 y p-p^{2}-q^{2}=0$
Also, radius of circle is $\sqrt{a^{2}+b^{2}+p^{2}+q^{2}}$
QUESTION 9
ABCD is a square whose side is a taking AB and AD as axes, prove that the equation of the circle circumscribing the square is $x^{2}+y^{2}-a(x+y)=0$
Sol :
ABCD is a square with side a units.
Let AB and AD represent the x-axis and the y-axis, respectively.
Thus, the coordinates of B and D are $(a, 0)$ and $(0, a),$ respectively.
The end points of the diameter of the circle circumscribing the square are B and D
Thus, equation of the circle circumscribing the square is
$(x-a)(x-0)+(y-0)(y-a)=0$ or $x^{2}+y^{2}-a(x+y)=0$
QUESTION 10
The line $2 x-y+6=0$ meets the circle $x^{2}+y^{2}-2 y-9=0$ at A and B .Find the equation of the circle on AB as diameter.
Sol :
The equation of the line can be rewritten as $x=\frac{y-6}{2}$
Substituting the value of $x$ in the equation of the circle, we get:
$\Rightarrow \left(\frac{y-6}{2}\right)^{2}+y^{2}-2 y-9=0$
$\Rightarrow(y-6)^{2}+4 y^{2}-8 y-36=0$
$\Rightarrow y^{2}+36-12 y+4 y^{2}-8 y-36=0$
$\Rightarrow 5 y^{2}-20 y=0$
$\Rightarrow y^{2}-4 y=0$
$\Rightarrow y(y-4)=0$
$\Rightarrow y=0,4$
At $y=0, x=-3$
At $y=4, x=-1$
Therefore, the coordinates of A and B are $(-1,4)$ and $(-3,0)$
$\therefore$ Equation of the circle with AB as its diameter:
$(x+1)(x+3)+(y-4)(y-0)=0$
$\Rightarrow x^{2}+4 x+y^{2}-4 y+3=0$
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## Fundamenta Mathematicae
2002 | 171 | 3 | 213-222
Tytuł artykułu
### Borsuk-Sieklucki theorem in cohomological dimension theory
Autorzy
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Języki publikacji
EN
Abstrakty
EN
The Borsuk-Sieklucki theorem says that for every uncountable family ${X_{α}}_{α∈A}$ of n-dimensional closed subsets of an n-dimensional ANR-compactum, there exist α ≠ β such that $dim (X_{α} ∩ X_{β}) = n$. In this paper we show a cohomological version of that theorem:
Theorem. Suppose a compactum X is $clc^{n+1}_{ℤ}$, where n ≥ 1, and G is an Abelian group. Let ${X_{α}}_{α∈J}$ be an uncountable family of closed subsets of X. If $dim_{G}X = dim_{G}X_{α} = n$ for all α ∈ J, then $dim_{G}(X_{α}∩ X_{β}) = n$ for some α ≠ β.
For G being a countable principal ideal domain the above result was proved by Choi and Kozlowski [C-K]. Independently, Dydak and Koyama [D-K] proved it for G being an arbitrary principal ideal domain and posed the question of validity of the Theorem for quasicyclic groups (see Problem 1 in [D-K]).
As applications of the Theorem we investigate equality of cohomological dimension and strong cohomological dimension, and give a characterization of cohomological dimension in terms of a special base.
Słowa kluczowe
Kategorie tematyczne
Czasopismo
Rocznik
Tom
Numer
Strony
213-222
Opis fizyczny
Daty
wydano
2002
Twórcy
autor
• Instituto de Matemáticas, UNAM, Av. Universidad S/N, Col. Lomas de Chamilpa, 62210 Cuernavaca, Morelos, México
autor
• Department of Mathematics, University of Tennessee, Knoxville, TN 37996, U.S.A.
autor
• Instituto de Matemáticas, UNAM, Av. Universidad S/N, Col. Lomas de Chamilpa, 62210 Cuernavaca, Morelos, México
autor
• Division of Mathematical Sciences, Osaka Kyoiku University, Kashiwara, Osaka 582-8582, Japan
autor
• Steklov Institute of Mathematics, Gubkina 8, 117966 Moscow GSP-1, Russia
Bibliografia
Typ dokumentu
Bibliografia
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# Manifold for Schwarzschild and Bertotti-Robinson
In short: what is the manifold in discussion for Schwarzschild metric $$ds^2 = -(1-\frac {2M}r)dt^2 + \frac1{1-\frac{2M}r} dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2)$$ and Bertotti-Robinson metric (cf. Thorne and Blandford, "Applications of Classical Physics" http://www.pma.caltech.edu/Courses/ph136/yr2011/1025.1.K.pdf, Exercise 24.2) $$ds^2 = Q^2(-dt^2 + \sin^2 t dz^2 + d\theta^2 + \sin ^2 \theta d\phi^2)$$
(where $Q=$const, $0\leq t \leq \pi, -\infty < z < +\infty, 0\leq \theta \leq \pi, , 0\leq \phi \leq 2\pi$)?
More description:
It seems that often, in general relativity, spacetimes are discussed purely by describing their metric tensors but not the manifold in question. This is very troublesome for me, since as far as I know, a manifold must be defined before describing metric tensor on it. For example, if one is to talk about metric tensor of a sphere $ds^2 = r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$ before defining what a sphere is, then it would be ambiguous to see what exactly $\theta$ and $\phi$ means (unless one is judging from the context, which somewhat works in this case because $\theta,\phi$ usually means angle)
So when I saw Schwarzschild metric and Bertotti-Robinson metric suddenly showing up without giving any description of which manifold we are talking about, I became very confused (and still am, having been unable to find any fruitful answer to such question of mine).
My guess for Schwarzschild is that, judging from the context, $r$ is "radius", $\theta,\phi$ are "angles" so it is "obviously true" that Schwarzschild metric is defined on the manifold $\mathbb R^4$ endowed with spherical polar coordinates for the spatial part. For Bertotti-Robinson, I'm guessing that $z,\theta,\phi$ together describe a somewhat alternative spherical polar coordinate where $z$ is height and so on, and thus that Bertotti-Robinson is also defined on $\mathbb R^4$.
• Locally, any manifold is $\mathbb{R}^4$, and GR almost always quietly supposes these local patches are large enough that we don't have to switch between them when doing physics. – ACuriousMind Dec 7 '14 at 14:01
• So does that mean that, in theoretical settings, manifold in discussion is generally $\mathbb R^4$? – progressiveforest Dec 7 '14 at 14:05
• Also, I'd like to know the following important additional detail: I found a claim that Bertotti-Robinson spacetime is spherically symmetric. How can one see this? The only way that I can think of is via showing that any element in $SO(3)$ induces an isometry on Bertotti-Robinson. However, I failed to find any general, explicit formula for rotation in terms of spherical coordinates. Therefore I am clueless in trying to demonstrate it. – progressiveforest Dec 7 '14 at 14:08
• There is some details on the matter on these notes arxiv.org/abs/1403.2371, I suggest you take a look. – user1620696 Mar 16 '18 at 17:44
There are many manifolds that admit the Schwarzschild metric, but the standard manifold to discuss it is the maximally extended Schwarzschild solution, which is $\mathbb R^2 \times S^2$. The easiest way to see this is to pick the Kruskal coordinates, which cover the whole manifold
$$ds^{2} = \frac{32M^3}{r}e^{-r/2M}(-dT^2 + dX^2) + r^2 d\Omega^2$$
with coordinates $(T,X,\theta,\varphi)$, with the angular coordinates describing a 2-sphere while $(T,X)$ indeed describe a region homeomorphic to $\mathbb R^2$. Since we have
$$X \in \mathbb R,\ T^2 - X^2 \in (-\infty, 1)$$
and
$$\frac {r}{2GM}=1+W_{0}\left({\frac {X^{2}-T^{2}}{e}}\right)$$
with $W_0$ the Lambert function, this means that the argument of the Lambert function is in the range $(-e^{-1}, \infty)$, so that $r \in (0, \infty)$, the metric never runs into any singularity (the values $T^2 - X^2 = 1$ corresponding to the actual singularities of the metric).
Once you have the maximally extended Schwarzschild solution, you can define a lot of other manifolds from it, either subsets or quotients. The exterior Schwarzschild solution still has the same topology : it's just a restriction to a quadrant of $(T,X)$, and in Schwarzschild coordinates you can easily see that it is just $\mathbb R^4$ with a sphere removed from every Cauchy surface, which is also equal to $\mathbb R^2 \times S^2$. The same goes for the interior Schwarzschild solution, or the Schwarzschild-Eddington-Finkelstein coordinate patch.
There are weirder topologies for the Schwarzschild solution, such as the elliptic Schwarzschild solution, which corresponds to the quotient by identification of the black hole and white hole solution (which is not causal or time-orientable), which I think has topology $\mathbb R \times S \times S^2$
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### C-73.2, r. 4 - Regulation respecting records, books and registers, trust accounting and inspection of brokers and agencies
38. Each year, before 31 March, and each time the Organization so requests, a licence holder must send the Organization
(1) a summary of the deposits into and withdrawals from the holder’s general trust account and all special trust accounts including
(a) identification of the period covered;
(b) for the general trust account:
i. the balance of the general trust account according to the accounting register at the start of the period;
ii. the total of the sums deposited during the period;
iii. the sum of the amounts referred to in subparagraphs i and ii;
iv. the total of the sums withdrawn during the period;
v. the balance of the account according to the accounting register at the end of the period, established by subtracting the sum mentioned in subparagraph iv from the sum obtained in subparagraph iii;
(c) for all the special trust accounts:
i. the balance of the special trust accounts according to the accounting registers at the start of the period;
ii. the total of the sums deposited during the period;
iii. the interest deposited during the period;
iv. the sum of the amounts mentioned in subparagraphs i, ii and iii;
v. the total of the sums withdrawn during the period;
vi. the total of the interest withdrawn during the period;
vii. the sum of the amounts mentioned in subparagraphs v and vi;
viii. the balance for all the special accounts, according to the accounting register at the end of the period, established by subtracting the sum mentioned in subparagraph vii from the sum obtained in subparagraph iv;
(d) the total of the balances, according to the accounting registers at the end of the period, established by adding the balances in subparagraph v of subparagraph b and subparagraph viii of subparagraph c;
(2) a copy of the bank reconciliation statement, established at the end of the calendar year or for the period for which the Organization requests it, for the general trust account and each of the special trust accounts including
(a) the date on which ends the period covered;
(b) for the general trust account:
i. the name and address of the financial institution and the number of the general trust account;
ii. the balance of the general trust account, according to the financial institution’s statement;
iii. the total of the sums not yet deposited;
iv. the sum of the amounts mentioned in subparagraphs ii and iii;
v. the total of the outstanding cheques, bills of exchange and transfer slips;
vi. the balance of the general trust account after reconciliation, established by subtracting the sum mentioned in subparagraph v from the sum mentioned in subparagraph iv;
vii. the balance mentioned in subparagraph v of subparagraph b of subparagraph 1;
viii. the difference between what is mentioned in subparagraphs vi and vii;
(c) for all the special trust accounts:
i. the name and address of the financial institution and the number of each special trust account;
ii. the balance of each special trust account, according to the financial institution’s statement;
iii. the total of the sums not yet deposited in each special trust account;
iv. the sum of the amounts mentioned in subparagraphs ii and iii;
v. the total of the outstanding cheques, bills of exchange and transfer slips for each special trust account;
vi. the balance of each special trust account after reconciliation, established by subtracting the sum mentioned in subparagraph v from the sum mentioned in subparagraph iv;
vii. the total of all the special trust accounts;
viii. the balance mentioned in subparagraph viii of subparagraph c of subparagraph 1;
ix. the difference between what is mentioned in subparagraphs vii and viii;
(d) the total of the balances according to the accounting registers at the end of the period, established by adding the balances provided for in subparagraph vi of subparagraph b and in subparagraph vii of subparagraph c;
(3) the detailed list of the sums held in the licence holder’s general trust account and special trust accounts at the end of the calendar year or the period for which the Organization requests the list. The list must contain
(a) the date of the end of the period covered;
(b) for the general trust account:
i. the number attributed to each transaction by the licence holder;
ii. the sum held with respect to each transaction;
iii. the total of the sums held in the general account;
(c) for the special trust accounts:
i. the number attributed to each transaction by the licence holder;
ii. the sum held in each special trust account;
iii. the number of the special trust account;
iv. the total of the sums held in the special trust accounts;
(d) the total of the balances according to the detailed list of the sums held at the end of the period, established by adding the balances provided for in subparagraph iii of subparagraph b and in subparagraph iv of subparagraph c.
The total amounts appearing under the headings mentioned in subparagraph d of subparagraph 1, subparagraph d of subparagraph 2 and subparagraph d of subparagraph 3 must coincide.
Each document required under subparagraphs 1, 2 and 3 of the first paragraph must contain the name of the licence holder, be signed by a person authorized by the licence holder and bear the date of the signature.
O.C. 296-2010, s. 38.
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# Are there other practical means of 'slowing' down an aircraft, other than parachutes? [closed]
Whenever an airliner crashes, inevitably questions arise as to how can the planes be made safe and ultimately, how can we avoid loss of lives.
Though statistically, modern airliners are safest ever, but due to the nature and outcome of even a single crash, it affects people psychologically worldwide.
Questions like parachutes for planes or even passengers are numerous on this site. For example, I posted this questions after reading this question.
The way I understand, if an aircraft can be slowed down, by means of a parachute, the resultant damgage to human lives would be a lot less. This led me to wonder, what can be other means of slowing down an aircraft, which would crash otherwise.
## closed as too broad by CGCampbell, mins, Ralph J, vasin1987, Federico♦May 25 '15 at 15:18
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• what can be other means of slowing down an aircraft, which would crash otherwise. If it's going to crash, what makes you think that a parachute will stop the crash? Are you able to reference any crash that might have been prevented by a parachute? Remember, the big problem is that if you slow it down more than about 140kts, which is going to be fatal, it will fall out of the sky because you slowed it down. – Simon May 24 '15 at 18:50
• Voting to close as Too Broad because this kind of question doesn't fit in the Stack Exchange paradigm. SE is designed so that every question should have one answer that is more 'right' or 'acceptable' and therefore chosen as accepted. This question could very easily have two (or more) equally valid and acceptable answers, therefore must be considered as Too Broad. OP, please rewrite to be less so. Perhaps come up with an alternate solution idea yourself and present it as a question. – CGCampbell May 25 '15 at 12:18
• Do you mean a runway overshoot? Tail hooks and steel cables are used on aircraft carriers. – user7241 May 25 '15 at 19:27
For an aircraft to fly, it needs forward speed. If you slow down an aircraft too much, it will stall and fall out of the sky. Airplanes are designed to not slow down. It's not "if a Boeing 737 is slowed down to 80 knots, more people will survive". A Boeing 737 has to fly at 130 knots minimum. If you want a machine that can, it's a helicopter.
Most crashes happen suddenly, e.g. approaching a runway while not paying attention to airspeed. It's not like "Hmmm, I think it's going to crash in 3 minutes, let's arm this system to minimize injury". It's like, "we're stable, runway ahead, everything's good......(terrain! terrain! PULL UP!)......oh sh!t ! Max power!!!". If the pilot can foresee the accident, most of the time he can avoid it altogether.
If you know well in advance you ARE going to crash, chances are the aircraft is completely out of control, spinning around and falling. In this case I think a large enough parachute will be effective.
• I would think a plane completely out of control has most likely suffered loss of some pretty important control systems, who is to say that the parachute will deploy? And of course, a tumbling aircraft is probably more likely to cut and destroy the parachute. – JustSid May 25 '15 at 13:51
There are a few things you could do but a full frame chute would be the most effective.
Speed brakes and flaps (really anything that creates drag) can act to slow a plane down even in a glide (engine out) situation.
In the realm of more unlikely things some sort of rocket style short burn booster can be deployed and discharged apposing the direction of movement thats how we landed on the moon. But these would add weight to the aircraft and could (if deployed incorrectly) put the aircraft in a worse situation. This does NOT include deploying thrust reversers that can be dangerous.
Slipping an aircraft (rudder opposite aileron deflection) can aid in slowing an airplane down. All student pilots in the US are required to know how to do this. Its strange the first time but quite interesting to execute.
• Please explain how a chute could prevent an airliner crash, – Simon May 24 '15 at 22:12
• @Simon There's other questions about that. – Zizouz212 May 25 '15 at 0:28
• In the realm of "totally impossible" we also get the Kulbit, which is quite effective for slowing the aircraft down but rather tricky for an airliner to pull off. – raptortech97 May 25 '15 at 3:24
• @raptortech97 - Isn't that a post-stall maneuver, like Pugachev's Cobra, that requires vectored thrust to accomplish? – KeithS May 27 '15 at 0:49
• @KeithS yes. As I said, rather tricky for an airliner to pull off. – raptortech97 May 27 '15 at 0:50
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# Solving Equation with Second Derivative
1. ### Swimmingly!
45
1. The problem statement, all variables and given/known data
Solve the following equation for x(y). (use no differential functions)
$image=http://latex.codecogs.com/gif.latex?-x^{-2}=d^{2}x/dy^{2}&hash=0b6db8781e593b1854b7258796e704bf$
x(0)' and x(0) are known.
2. Relevant equations
$image=http://latex.codecogs.com/gif.latex?-x^{-2}=d^{2}x/dy^{2}&hash=0b6db8781e593b1854b7258796e704bf$
3. The attempt at a solution
I'm a bit unsure as to what to do next but I can easily make a messy formula up to approximate the result.
$image=http://latex.codecogs.com/gif.latex?\left\{\begin{matrix}%20x(0)=k_{1}%20\\%20x(0)%27=k_{2}%20\\%20x(y)%27%27=-x^{-2}%20\\%20x(y+E)\approx%20x(y)+x(y)%27\times%20E+x(y)''\times%20E^{2}/2%20\\%20x(y+E)%27\approx%20x(y)%27+%20x(y)%27%27\times%20E%20\end{matrix}\right.&hash=ade0b5b60f18ff4b7daf16c3aac1ab3f$
k1 and k2 are a constant value.
The bigger E is, the bigger the error. For an infinite recursive use of this formula with an infinitesimal E the right result would be achieved. I don't know what it converges to though.
Any help for this problem or/and similar problems would be great. Thank you.
Last edited: Jan 4, 2012
2. ### CompuChip
4,296
What does "use no differential functions" mean?
[strike]Why don't you just integrate the whole thing twice, or separate variables?[/strike]
It was a bit more complex than I thought - never mind the second line there[/edit]
Last edited: Jan 4, 2012
3. ### quZz
125
Hi,
the equation does not contain explicit dependency on y, so try multiplying both sides on dx/dy. You'll get a sort of "conservation of energy" for the problem:
(dx/dy)^2/2 + 1/x = const
const depends on the initial conditions. This way you are left with solving 1st order diff equation
4. ### Swimmingly!
45
CompuChip:
It means that I want something with which I can calculate values directly. I do not want Integrals or Derivatives in my final formula and I think this is very much possible.
quZz:
I'm not sure what you mean by explicit but for a given k1, k2 and y there's a single well defined value of x.
Also I'm sorry I'm not very good at handling differentials, I don't even have much practice with calculus. I don't understand how you got that from multiplying the first equation by dx/dy. (assuming that was what you did?)
Last edited: Jan 4, 2012
5. ### quZz
125
so you just need an answer, right?
6. ### Swimmingly!
45
I've had my time for solving it alone for fun. But it doesn't look so simple and I am extremely curious as for the answer.
I'm trying to find an answer but with the proof too. Otherwise I'll just forget it because I won't understand it.
If you could do that, it'd be really nice. Thanks.
7. ### quZz
125
try wolframalpha.com...
http://www.wolframalpha.com/input/?_=1325716343605&i=d^2x%2fdy^2%3d-1%2fx^2&fp=1&incTime=true
8. ### HallsofIvy
41,265
Staff Emeritus
$$\frac{d^2x}{dy^2}= x^{-2}$$
As quZz said, y does not appear explictly so you can use "quadrature". let v= dx/dy. Then
$$\frac{d^2x}{dy^2}= \frac{dv}{dy}= \frac{dv}{dx}\frac{dx}{dy}= v\frac{dv}{dx}= x^{-2}$$
That is now a separable first order equation:
$$v dv= x^{-2}dx$$
$$\frac{1}{2}v^2= -x^{-1}+ C$$
$$v^2= 2(C- x^{-1})$$
$$v= \frac{dx}{dy}= 2\sqrt{C- x^{-1}}$$
which is also a separable first order equation:
$$\frac{dx}{\sqrt{C- x^{-1}}}= 2dy$$
9. ### Swimmingly!
45
Thanks!
I didn't know how to write second derivatives in Wolfram Alpha and I thought it wouldn't give a proof for such a complex problem.
Anyway thanks a lot. Problem solved.
Edit: There's a slight mistake on the your result by the way, v^2=2*(C-x^-1). It's times 2, not times "2 squared". It was much easier to follow then wolfram though. Thanks.
Last edited: Jan 4, 2012
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# Tag Info
1
Let´s go: [1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition. We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations: From a cross-section of longitudinal (scalar) ...
2
1) Note that the non-relativistic gluon model of glueballs has even less justification than the non-relativistic quark model of baryons and mesons. This is because the model messes up the spin assignments: Massless gluons have only two spin states, but non-relativistic gluons have three. 2) The color quantum numbers are trivial: The product [8]\times ...
4
Experimentally the charge distribution of protons and neutrons has been measured as a function of the radius. So the different charge content of the two nucleons does affect the distributions. As the other answer states this is the regime where only quantum chromodynamics models can attempt to describe the wavefunctions of the quarks within the ...
0
This is not a meaningful question. No, really, it isn't. Quarks don't exist as free charged objects on which we could take the classical limit and consider "forces" on them. They are confined, and occur only as constituents of bound states. In quantum mechanics, it doesn't make sense to ask whether the constituents of a bound state "repel" or "attract" each ...
Top 50 recent answers are included
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# F.c. analysis
1. Mar 13, 2010
### Rajini
Dear all,
Any quantum mechanical software (e.g., Gaussian03) can compute the fundamental frequencies (3N-6) and force constant for each frequency..
I don't understand exactly about force constant..??
For a particular mode of vibration what does this force constant mean?
For a molecular with 2 atoms we can tell force constant is between two atoms..but for polyatomic molecules is there any good way to understand??
thanks
2. Mar 13, 2010
### SpectraCat
The vibrational motions of polyatomic molecules are more complicated than those of diatomic molecules, but the idea of the force constant is the same. I have written up a description below, which uses the concept of a force constant in two distinct ways: to describe a single chemical bond connecting two atoms, and to describe a harmonic normal mode of the entire molecule. Both definitions are consistent, and are correlated to the associated harmonic motion through the angular frequency, which is defined as:
$$\omega=\sqrt{\frac{k}{\mu}}$$
where k is the force constant and mu is the reduced mass.
For any system of N atoms, you can keep track of the motion of the molecule by plotting the displacements of each of the atoms along the x,y and z Cartesian axes, describing a system with 3N overall degrees of freedom (DoF's). Now, if there are chemical bonds connecting those atoms in some pattern to form a molecule, there will be correlations between the motions of the atoms. You can think of each bond as a localized spring connecting two atoms, with an associated force constant that is proportional to the strength of the bond. If you were to average all the motions of the molecule for a long time, you would find the there is some "average" configuration, for which you can define a center of mass (CoM) in the usual way.
You can then describe the translation of the system through space in terms of the translation of its center of mass, using up 3 DoF's. You could also describe the overall rotation of the system around its COM, using up 3 DoF's (or just 2 if the molecule has a linear structure). The remaining 3N-6 DoF's (or 3N-5 for a linear molecule) then describe the vibrational motions of the molecule around it's averge configuration. The key point here is that this set of 3N-6 DoF's is *complete*, that it, you can use it to represent *any* motion of the molecule, therefore it represents a complete basis for the motions of the molecule. This means that you can set up a system of 3N-6 linear equations in terms of the masses of the atoms and force constants of the individual bonds, and solve it to obtain the orthogonal basis of harmonic normal modes for the molecule, which also form a complete set. These normal modes are more "natural" for describing the molecular motions, because they are orthogonal ... if you excite a vibration of one of them (say by absorption of an infrared photon of the appropriate frequency), then to a good approximation, the vibrational energy stays in that mode, and does not get transferred to the other modes. The motions described by local vibrational of chemical bonds are *not* orthogonal in general .. ff you were to excite the motion of a single bond between two atoms, the energy would rapidly (over a few vibrational periods) start to leak out into the other bonds in the molecule. This is why vibrational spectroscopy (infrared and Raman) measures the fundamental frequencies of the normal modes of the molecules.
Ok ... that got a little long-winded .. hope it's helpful.
3. Mar 13, 2010
### Rajini
Hi,
So force constant calculated for a normal mode by such quantum calculations can be said as...force constant of that particular normal mode instead of just saying it as simply force constant between bonds??
or
average force constant?
Can Raman or IR measure overtones??
Also can you also elaborate reduced mass for polyatomic molecules??
thanks
4. Mar 13, 2010
### SpectraCat
Yes
In principle yes, there is always some anharmonicity of the vibrational potential, which allows for non-zero intensity of overtones. However in practice, this intensity is often too small to be measured.
This is very tricky to do ... the way it is handled in normal mode analysis is to define a system of mass-weighted coordinates ... these automatically include the contributions of the different masses to the system. The reduced masses can then be calculated once the normal modes are known, but there is no easy way to write them in a closed form. If you are really interested in the details of all of this, I suggest you have a look at http://www.gaussian.com/g_whitepap/vib.htm" [Broken] from the Gaussian website.
Last edited by a moderator: May 4, 2017
5. Mar 14, 2010
### Rajini
Hi spectracat,
i already saw that link..for reduced mass and freq. calculations..
actually why i ask this question?
I have a phonon spectrum..which displays only the vibrations of iron nucleus..from that spectrum i can get the force constant of iron with neighboring atoms/ligands..
and
i have another spectrum of a molecular which is calculated using gaussian03..i just want to compare the force constants of experimentally obtained ones with calculated ones..
By using visualizing software i can roughly predict the pure iron vibrations..so i guess if everything is correct then the f.c should be same for that vibration..??
if not i can assume that the calculated ones may deviate from experimentally obtained f.c due to non pure iron vibrations..
so
is there any other suggestions??
another small question..normally overtones should appear roughly at two times the fundamental frequency?? is that correct
thanks
6. Mar 14, 2010
### SpectraCat
Ok, I am not really that familiar with phonon spectra ... I assume that you can extract force constants from them *if* you know the reduced mass of the phonon, which seems to be to be a highly non-trivial quantity to obtain.
This is a good question ... and I have to say that I don't really know. I would be very cautious drawing parallels between force constants calculated for molecules to those measured for solid-state phonons. The reason for this is that phonons are global normal mode excitations of a lattice, whereas molecular normal modes have much more localized character. So, the force constants for a vibration between a group of atoms in a molecular calculation may not be directly comparable to the force constants for phonons in a solid state lattice.
Having said that, I should think that the molecular force constants could be used as inputs into a calculation to derive the phonon frequency spectrum. This is not my field, so I can't really help you much ... perhaps ZapperZ or another condensed phase specialist could jump in here?
Yes ... but not necessarily for that reason .. see above
Yes, that is correct. Since overtones arise from anharmonic effects, there will be a small red shift of the overtone .. that is, it should appear at a frequency that is slightly less than twice the fundamental.
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User ashin - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:12:57Z http://mathoverflow.net/feeds/user/18036 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83452/convergence-of-eigenvalues Convergence of Eigenvalues Ashin 2011-12-14T18:42:43Z 2012-01-16T20:22:12Z <p>Suppose we have a matrix $A_n = \frac{1}{n}\sum_{i=1}^nX_i X_i^T$, where $X_i$ is a $p$-dimensional random-vector. We also know that $E(XX^T) = \Sigma_{p \times p}$. Let us denote the $j$-th largest eigenvalue of a symmetric positive-definite matrix $M$ by $\lambda_j(M)$. Then can we say anything about convergence of $\lambda_j(A_n) \rightarrow \lambda_j(\Sigma)$ as $n \rightarrow \infty$, that is, whether it converges in probability or in distribution and if so can we characterize the rate of convergence.</p> <p>Thanks a lot, any help is much appreciated.</p> <p>Best Ashin</p> http://mathoverflow.net/questions/77057/partial-derivatives-of-eigen-value-decomposition-or-singular-value-decomposition partial Derivatives of Eigen value decomposition or Singular value decomposition Ashin 2011-10-03T16:51:27Z 2011-10-04T06:11:05Z <p>Hi All,</p> <p>Suppose I've a symmetric matrix $A_{N \times N} = (A_{ij})$ which has a eigen value decomposition $A = UDU'$. I would like to know under what conditions $\frac{\partial U}{\partial A_{ij}}$ exists for all $i,j = 1,2, \ldots, N$. I found the following paper which talks about estimating the Jacobian of the SVD transformation</p> <p><a href="http://www.ics.forth.gr/cvrl/publications/conferences/2000_eccv_SVD_jacobian.pdf" rel="nofollow">http://www.ics.forth.gr/cvrl/publications/conferences/2000_eccv_SVD_jacobian.pdf</a></p> <p>But its not very clear regarding the conditions that the matrix $A$ would need to satisfy. Any help is much appreciated.</p> <p>Thanks Ashin</p> http://mathoverflow.net/questions/76245/measure-of-subspace-of-matrices-with-repeated-singular-values/76260#76260 Answer by Ashin for Measure of Subspace of Matrices with repeated Singular Values Ashin 2011-09-24T03:33:41Z 2011-09-24T03:33:41Z <p>Hi Chris,</p> <p>Thanks a lot for your quick response. But I'm not that familiar with the results that you are using to arrive at the conclusion. Would you kindly be a bit more elaborate. Or maybe you can guide me to the resources where I could find the results that you are using here.</p> <p>Thanks a lot for your help. Ashin</p> http://mathoverflow.net/questions/83452/convergence-of-eigenvalues Comment by Ashin Ashin 2011-12-14T19:59:48Z 2011-12-14T19:59:48Z @ Robert : Yes I do mean $A_n = \frac{1}{n}\sum_{i = 1}^{n}X_i X_i^T$. Sorry about the typo. http://mathoverflow.net/questions/77057/partial-derivatives-of-eigen-value-decomposition-or-singular-value-decomposition/77108#77108 Comment by Ashin Ashin 2011-11-07T03:22:14Z 2011-11-07T03:22:14Z But my question is more about the singular vectors. Are they also going to admit partial derivatives w.r.t. the entries of the matrix almost everywhere. Best
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# Group Testing with Two and Three Defectives
@article{Chang1989GroupTW,
title={Group Testing with Two and Three Defectives},
author={X. M. Chang and F. Hwang and J. Weng},
journal={Annals of the New York Academy of Sciences},
year={1989},
volume={576}
}
• Published 1989
• Computer Science
• Annals of the New York Academy of Sciences
Consider a population of n items consisting of d defectives and n d good ones. The problem is to find the d defectives by means of a sequence of group tests. We will call a set of items contaminated if it contains a t least one defective, pure if it contains no defective, and free if there is no information on it. Then a group test is a test on a given set with two possible outcomes: the set is identified either as a contaminated set or as a pure set. Let M,(d, n ) be the maximum number of… Expand
Quaternary splitting algorithm in group testing
• Computer Science, Mathematics
• J. Comb. Optim.
• 2021
This paper focuses on estimating M ( d, n ) and obtaining a better result than known ones in various cases of d and n and denotes the minimum number of tests in the worst case situation. Expand
An Efficient Algorithm for Combinatorial Group Testing
• Andreas Allemann
• Mathematics, Computer Science
• Information Theory, Combinatorics, and Search Theory
• 2013
A new algorithm which in the worst case needs less than $0.255d+\frac{1}{2}\log d+5.5$ tests more than the information lower bound for n/d≥2 and the behaviour for large n and d of the difference is optimal for $\frac{n}{d}\leq4$. Expand
Competitive Group Testing
• Computer Science, Mathematics
• Discret. Appl. Math.
• 1993
This paper presents some competitive group testing algorithms and defines M(n,d) = minA MA(n), the maximum number of group tests for a group testing algorithm A to identify d defectives from a set of n items when d is known. Expand
Group testing problem with two defectives
• Mathematics, Computer Science
• Probl. Inf. Transm.
• 2013
A new adaptive algorithm is proposed such that for N = 2 + 1 - t - t 2 - t 4, the problem of finding two defectives among N elements can be solved in t tests. Expand
Updating a Tale of Two Coins
The criterion used to evaluate a procedure A for a model M is the worst case number of tests required to detect the two irregulars among n items, denoted by T,(n). Expand
An NP-Completeness Result of Edge Search in Graphs
• T. Gerzen
• Mathematics, Computer Science
• Graphs Comb.
• 2014
The present paper shows that the computation of cp(G), which contains n vertices two of which are defective and adjacent, is an NP-complete problem, and establishes some results on it for random graphs. Expand
A group testing problem for graphs with several defective edges
• Petra Johann
• Computer Science, Mathematics
• Discret. Appl. Math.
• 2002
This paper proves the conjecture that finding all defective edges by testing whether an induced subgraph contains a defective edge or not can be done by using at most d log 2 m d +c tests for some constant c. Expand
#### References
SHOWING 1-4 OF 4 REFERENCES
An optimal search procedure
Abstract Consider the following problem. There are exactly two defective (unknown) elements in the set X ={ x 1 , x 2 ,…, x n }, all possibilities occuring with equal probabilities. We want toExpand
A Group Testing Problem on Two Disjoint Sets
• Mathematics
• 1981
Recently the following group testing problem has been studied. We have two disjoint sets of items with cardinalities m and n respectively, where each set is known to contain exactly one defectiveExpand
Group testing with two defectives
• Computer Science, Mathematics
• Discret. Appl. Math.
• 1982
This paper gives a partial solution to the problem of determining the minimax number of group tests for finding two defectives separately contained in two disjoint sets. Expand
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My Math Forum Need Help Finding the lengths.
Trigonometry Trigonometry Math Forum
May 12th, 2015, 02:53 AM #1
Newbie
Joined: May 2015
From: england
Posts: 1
Thanks: 0
Need Help Finding the lengths.
Hi all,
I'm so stuck on this question, can you help me find the lengths and include the workings out?
Attached Images
Fullscreen capture 12052015 092346.bmp.jpg (56.8 KB, 13 views)
May 12th, 2015, 01:42 PM #2 Math Team Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 You haven't shown what you have tried or told us what you do know about this, so I don't know if this method is reasonable for you, but here is how I would do it: Set up a coordinate system with point F at the origin, the x-axis along GE and the y-axis along FB. Then A has coordinates (-13, 3.2), E has coordinate (13, 0), and B has coordinates (0, 10). The distance from $\displaystyle (x_0, y_0)$ to $\displaystyle (x_1, y_1)$ is $\displaystyle \sqrt{(x_1- x_0)^2+ (y_1- y_0)^2}$. Once you have found the length of AE and BE, as asked, do the same thing to find the length of AB so that you know all three sides of triangle ABE. The angle AEB can be gotten from the "cosine law": $\displaystyle AB^2= AE^2+ BE^2- 2(AE)(BC)\cos(AEB)$. Last edited by skipjack; May 12th, 2015 at 11:49 PM.
May 13th, 2015, 08:03 PM #3 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,688 Thanks: 701 Calculate BE from right triangle BEF (EF = 13, BF = 10). Calculate AE from right triangle AEG (GE = 26, AG = 3.2) Calculate AB from right triangle ABM (AM = 13, BM = 10-3.2 = 6.8) Use Cosine Law to calculate angle AEB.
Tags finding, lengths, triganometry
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Processed foods do not contain water
By R Mutt (Mon Mar 20, 2006 at 01:22:07 AM EST) MLP (all tags)
Key:
[MeFi] = Stolen from Metafilter
[/.] = Stolen from Slashdot
[M] = Stolen from Memepool
[BX] = Stolen from Blogdex
[X.] = Stolen from Christdot
[)] = Stolen from Monkeyfilter
[B] = Stolen from B3ta
[GG] = Stolen from Green Gabbro
[BFB] = Stolen from Big Fat Blog
[DD] = Stolen from Daily Depravity
[S2MM] = Stolen from Stuffisendtomymates
[BB] = Stolen from BoingBoing.net
[[:)] = Needs sound
[:(] = Serious
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[;)] = Ironic
[:o] = Strange
[*] = Flash
[#] = Free registration required
[$] = Possible corporate shill [NSFW] = Not Safe For Work [NSFWFUP] = Not Safe For Work For Ultra-Prudish [(UK)] = UK-centric [LL] = Late or repeated link Processed foods do not contain water | 35 comments (35 topical, 0 hidden) | Trackback Brain Gym by gazbo (2.00 / 0) #1 Mon Mar 20, 2006 at 01:36:23 AM EST I am so glad to have some references. A friend of mine (housemate's girlfriend, as it happens) is a primary school teacher and she's fully bought into this. I've delicately tried to imply that it's a load of rubbish, but the thing is she did her dissertation on it, and as such considers herself quite the expert (at parroting the materials supplied by the quacks). So, me telling her that she's talking rubbish and pedalling snake-oil in the classroom is really quite a personal attack; it's implying that not only has she been wasting her time for the past year in the classroom, but that her entire dissertation may as well have been about pixies and fairies. Perhaps a couple of links passed on without further comment may enlighten her by gradual realisation. Though I shall not hold my breath. I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme. I didn't read the article, by ambrosen (2.00 / 0) #2 Mon Mar 20, 2006 at 01:42:29 AM EST but I disagree about not drinking water when thinking. I've normally found myself drinking 4 or 5 litres of water a day when I do intellectually demanding work. Mind you, half of that's psychological. Also, good energy levels are key to brainwork. It's more efficient than your average computer, but it still uses 25 watts average. I have no figures for the no load/100% load energy consumption. [ Parent ] You would agree with the article by gpig (2.00 / 0) #6 Mon Mar 20, 2006 at 01:59:33 AM EST Dr. Goldacre wasn't arguing with the idea of drinking water and doing some gentle exercise or movement during lessons. His criticism was directed at the fact that bollocks science is used to justify it. There is probably room in the market for something which has much the same practices, but doesn't teach kids that they have 'brain buttons'. On the other hand, Gillian McKeith exists. --- (, ,') -- eep [ Parent ] Thanks. by ambrosen (2.00 / 0) #7 Mon Mar 20, 2006 at 02:14:08 AM EST I always find those kind of articles to annoying for me with their preaching to the choirness. Still, now I know the word "brain buttons". That was worth it. Now, off to email my brother to tell him why there was no need to put the freshwater fish on Noah's Ark. (Apparently, pockets of fresh water may stay stable in seawater for a period well in excess of forty days and forty nights) [ Parent ] Ben Goldacre does have an awful lot of choir by gpig (2.00 / 0) #10 Mon Mar 20, 2006 at 02:31:19 AM EST Even the troll (with account name "ANGRY BRAIN") just turned out to be a scientist having a laugh. It was funny though. --- (, ,') -- eep [ Parent ] Hmm by gazbo (2.00 / 0) #3 Mon Mar 20, 2006 at 01:42:46 AM EST Although although those articles are entertaining for a critical thinker, they will do nothing to convince someone who has bought into the crap. Sneering at the concept of "brain buttons" is fine and dandy, but only if you already know what a load of shit it is. Maybe I should email the guy to get him more information from my unwitting source. I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme. [ Parent ] Suspect it's a lost cause by R Mutt (4.00 / 1) #4 Mon Mar 20, 2006 at 01:43:34 AM EST ...Before doing these tasks, children are required to take a swig of water and hold it in their mouths for a few seconds until the teacher tells them they can swallow. When I asked why, the teacher, who had been sent on a Brain Gym course by the school, informed me that the water was partially absorbed through the roof of the children's mouths and was absorbed by the brain, improving learning. If you'll swallow that, I think you're pretty much lost to reason. [ Parent ] yikes by Merekat (4.00 / 3) #9 Mon Mar 20, 2006 at 02:18:43 AM EST I think I'd be looking for a meeting with the head teacher if any child I might have came home from school telling me that. [ Parent ] I dunno by gazbo (2.00 / 0) #13 Mon Mar 20, 2006 at 02:42:06 AM EST I've seen that quote before (from the randi.org forums when I first searched for references to Brain Gym), and I keep meaning to ask her if that's what she's heard. One of the Brain Gym apologists practitioners on the forum claimed they'd never been taught that reason at all, and that the person who claimed it was simply confused. A not unbelievable possibility. But I really hope they do teach that, because that's such an easy door in to make people see how wrong the practice is. And once I've convinced her that the roof of the mouth is not made of wick material, maybe the seeds of doubt will be sown...? Or maybe I'm being optimistic. I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme. [ Parent ] I suspect... by bob6 (2.00 / 0) #35 Mon Mar 20, 2006 at 07:51:08 AM EST ... they're teaching authority rather than anything else. What schoolroom will hold water in their mouth and not a single one spits it on the next geek? This exercise should end in water fight. Btw Brain buttons lie directly over and stimulate the carotid arteries. The actual carotid artery lays under your skin on the right side of your neck, where cops in the movies try to feel the pulse of dead people. Well, if you stimulate it by rubbing it slowly and long enough... you pass out, plain an simple. Cheers. [ Parent ] Dissertation for what? by komet (4.00 / 1) #5 Mon Mar 20, 2006 at 01:55:34 AM EST What degree is she getting? This surely can't be much good for the reputation of the university in question. -- <ni> komet: You are functionally illiterate as regards trashy erotica. [ Parent ] I don't know, I didn't ask. by gazbo (2.00 / 0) #11 Mon Mar 20, 2006 at 02:36:52 AM EST PGCE maybe? Some form of teaching qualification anyway. One day I may pluck up the courage to ask to see it - I'll offer to swap a copy of mine for hers to even things out. I consider my dissertation to be a fairly poor example of a write-up: interesting and successful project ("Sound synthesis using genetic programming") but the write-up tells the truth about how much time I put into it (48 hours frantic final write-up). Weighing in at around 60 pages of text, with only a handful of references, and then a couple of dozen pages of graphs and numbers from runs of the system. I suspect - and I base this on nothing other than hunch - that her dissertation will be a fraction the size, consist of no references other than Brain Gym publications, and contain zero new research. Well, if the subject comes up again I'll ask her for a copy. But I'll not introduce the subject myself. I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme. [ Parent ] Brain Gym Stage #2: by Rogerborg (4.00 / 1) #8 Mon Mar 20, 2006 at 02:15:33 AM EST Submit to the E-Meter. Google page rank dropped? Simple solution: sue Google and watch it soar again. Sometimes I worry that I'm not cynical enough. - Metus amatores matrum compescit, non clementia. brain gym stage #3 by martingale (4.00 / 1) #25 Mon Mar 20, 2006 at 04:27:07 AM EST Convince someone else to exercise your brain for you. I must admit my first thought when I read the headline was handbag -> Margaret Thatcher -> Dennis. Zombies on the brain... --$E(X_t|F_s) = X_s,\quad t > s\$
[ Parent ]
What is it with Germans and severed body parts? by Rogerborg (2.00 / 0) #26 Mon Mar 20, 2006 at 04:44:18 AM EST
Hang on, a German Turk? Well, now I don't know which racist stereotype to pillory.
-
Metus amatores matrum compescit, non clementia.
[ Parent ]
Google sue by TurboThy (2.00 / 0) #12 Mon Mar 20, 2006 at 02:41:33 AM EST
The complaint accuses Google, as the dominant provider of Web searches, of violating KinderStart's constitutional right to free speech by blocking search engine results showing Web site content and other communications.
I'd like to start a donation drive to sic a huge bloke with a cricket bat on these people's arses.
__
Sommerhus til salg, første række til Kattegat.
Wow! by ReallyEvilCanine (2.00 / 0) #14 Mon Mar 20, 2006 at 02:48:05 AM EST
ASCII art from 18 years before ASCII was developed. I guess those Popular Mechanics articles about time machines were right on the money.
# include WheresMyFlyingCarRant.h;
the internet: amplifier of stupidity -- discordia
Far be it by yicky yacky (4.00 / 1) #15 Mon Mar 20, 2006 at 03:04:40 AM EST
from me to be pugnacious, contradictory and obstreperous (forgive the florid vocabulary; I've just done my early-afternoon 'Brain Buttons' ...), but who's to say that this 'Brain Gym' doesn't actually work on some practical meta-level?
I recently had a debate with a family member over homeopathic medicine. Things were proceeding nicely when, in the middle of the chat and having well and truly "got my debunk on", it was pointed-out that, given that the "placebo effect" is real / demonstrable / scientifically established etc., can one not simply defend homeopathy on the grounds of being an effective, powerful, folk-wisdom-backed, pseudo-organic, peer-pressurized placebo delivery mechanism?
That's all very easy to say, certainly, but might the 'Brain Gym' be valid inasmuch as it gets kids drinking water, shutting-up, behaving and working instead of being juiced-up on Sunny D, giving their vocal chords a thrashing and smearing chocolate / snot / whatever up the walls?
----
Vacuity abhors a vacuum.
It's tempting to argue along those lines, but by DesiredUsername (4.00 / 1) #16 Mon Mar 20, 2006 at 03:16:45 AM EST
A medical system based on homeopathy will appear to work at some low level of effectiveness but will fail anybody who doesn't (or can't--the young, unconscious, mentally disabled, etc) believe in it. A medical system based on science will work at a higher level (real medicine plus the placebo effect) and will help even unbelievers.
---
Now accepting suggestions for a new sigline
[ Parent ]
caveat by Merekat (2.00 / 0) #20 Mon Mar 20, 2006 at 03:31:02 AM EST
A medical system based on science will work at a higher level (real medicine plus the placebo effect) and will help even unbelievers...
...when properly applied and not just when operated on a quasi mystical faith in the effectiveness of antibiotics which causes the practitioner to completely fail to listen to any of the patient's symptoms.
[ Parent ]
Ah, but by yicky yacky (2.00 / 0) #21 Mon Mar 20, 2006 at 03:32:24 AM EST
Homeopathy self-selects for believers. Non-believers simply don't bother with it, which gets rid of as much as half your point ;)
----
Vacuity abhors a vacuum.
[ Parent ]
Oh, very possibly by gazbo (4.00 / 1) #18 Mon Mar 20, 2006 at 03:18:53 AM EST
I have no doubt that it may work on a placebo level. And maybe more directly on a calm-them-down, shut-them-up, get-their-attention level. In fact, the same person teaches at a Catholic school (she's not religious herself) and if they are being rowdy she'll announce "In the name of the father..." because that gets them to suddenly sit still and shut up.
But does the end justify the means? Does the potential for a tiny increase in the effectiveness of a lesson justify a new wave of 20 year olds in the future who have no critical thinking skills, because they've been indoctrinated with pseudo-science since age 5? How about when they have children who slowly starve to death because the parents willingly believe that centering the child's chi by massaging the meridians* is more effective and less harmful that taking modern medical advice?
Now I know you're at least partially playing devil's advocate here, so this isn't a rant directed at you, but when I read cases of patients dying because they were informed that modern medicine was making them worse and that the true way to treat brain tumours was with coffee enemas (I didn't just make that up), or of people being fleeced for huge amounts of money by mediums preying on the lonely and vulnerable, or even wars started over religion, I can't help but see all such made-up nonsense as being potentially dangerous.
*Because that's so different to increasing brain activity by massaging the brain buttons.
I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme.
[ Parent ]
I don't disagree by yicky yacky (2.00 / 0) #22 Mon Mar 20, 2006 at 03:48:52 AM EST
particularly with anything you say and, yes, was playing devil's advocate (perhaps even partially trolling ...) to an extent in the original post.
However: I think there is a fundamental contradiction in there somewhere. Many of the most critical thinkers I know were raised on no small measure of unfalsifiable creed (Santa Claus, the Tooth Fairy, the whole God thing, misunderstood science as delivered by parents), and didn't really develop genuine (as opposed to 'aped') critical thinking skills until their mid or late teens.
What I'm saying is that, given what Goldacre calls "the predisposition of children to learn from adults", how different is blind acceptance of established scientific fact A from blind acceptance of unfalsifiable gibberish B, given that both are assented-to equally? The well-intentioned answer is that fact A is correct whereas B is not (at which point we could go into a long and not-particularly-productive discourse involving Wittgenstein and French deconstructionism) but that would ignore the fact that people capable of critical thinking will figure it out for themselves anyway, regardless of whether they were taught A or B initially (as both assertions are simply indoctrination at that stage).
My point is just that people like Dr. Goldacre (and many of us here, including myself) have a predisposition towards the importance of teaching correct facts (to whatever extent that is a tautology) when it is just as likely that it is the teaching of utter gibberish which motivates critical thinking.
----
Vacuity abhors a vacuum.
[ Parent ]
I only went to a Comprehensive by R Mutt (4.00 / 1) #24 Mon Mar 20, 2006 at 04:01:29 AM EST
But when I went to school, we were generally taught the evidence and reasoning behind the stuff we were taught.
For History, we learned what Primary and Secondary sources were, and how we could use them to find out what probably happened.
For Science, we'd learn about the experiments that led to certain theories, and when possible we repeated the experiments themselves to confirm them.
Maybe things have changed so much that kids are taught scientifc theories without experiments and history without sources, but I doubt it.
The problem with Brain Gym isn't that they're being taught incorrect facts, but that they're being taught blind acceptance of supposed facts.
[ Parent ]
I think you're being disingenuous by yicky yacky (2.00 / 0) #28 Mon Mar 20, 2006 at 04:56:01 AM EST
I went to a comprehensive, too, and had a nigh-on identical experience to you. So tell me:
How much of the calculus did you understand, aged eleven? When the time came (early GCSE, IIRC) to discuss Newtonian motion in more detail, did you query where those equations came from, or did they spring, silken and fully-formed, from the forehead of El Guru? What thoughts did Einstein's relativistic effects and equations conjure in your thirteen-year-old brain? How did your twelve-year-old self cope with Maxwell and grad, div and curl?
See: You didn't and you weren't thinking critically. You took it on faith that the person at the front of the class was telling the truth when it came time to discuss gravity and electricity. The fact that your teacher probably didn't understand them properly didn't occur to you. I've already gone too far for illustratory purposes, though. You were assenting to facts you had no way of proving, some of which were such gross simplifications (I'm thinking basic chemistry) as to be outright lies, all the way through school. How does this differ from the 'Brain Gym'? It would be possible, if advocating diabolically, to assert that, at the very least, the 'Brain Gym' encourages good practice (drinking water, a relaxed and focussed state of mind etc.), even if it's for spurious reasons.
I presume you've read this. All I'm saying is that these issues are far more complex, especially with regard to people who are incapable of understanding the fundamental proofs and salient ideas (be they child or adult) than these kinds of knee-jerk, banner-of-wisdom discussion imply.
Two further points: If 'Brain Gym' encouraged exactly the same practices, but dressed it in less scientifically-inaccurate verbiage, what would the consensus be? (That it was no great problem, I'm guessing). Secondly: In individuals perfectly capable of understanding the salient points, I completely agree with yourself (and gazbo / DU etc.). A counterargument to that last point is that children are, but that takes us back to my original comment: I don't think they are - not at the meta-level of efficacy. Bear in mind that we're not talking about GCSE students here; we're talking about primary school kids.
----
Vacuity abhors a vacuum.
[ Parent ]
Well by R Mutt (4.00 / 1) #30 Mon Mar 20, 2006 at 05:11:00 AM EST
With Maths, I didn't have any problem understanding the proofs until we got into A-Level further maths. With Physics, regardless of where the equations come from, the experiments are understandable, and even worked in lab sometimes. But more important than the details was the principle that these things have to be tested and proved before they can be relied on.
But in the case of science, I'd say that's less important; as apart from a few oddities like evolution no-one's trying to deliberately mislead the kids into believing stuff that's not true. When it comes to things like Brain Gym, people are consistently, systematically targetted with misinformation; encouraging them to believe falsehoods for the sake of selling spurious cures and improvements to them. It's not just advertising: newspapers, magazines and TV documentaries are full of this crap; and their parents probably believe it too. If the schools are full of the same bullshit, then I don't see anywhere left where the kids can find out that the bullshit is bullshit.
[ Parent ]
Like I said, by yicky yacky (2.00 / 0) #34 Mon Mar 20, 2006 at 07:16:04 AM EST
I don't particularly disagree with the general principle (and most of the time fervently agree), and certainly don't care enough to flog this to an eternal death, but I would argue that they'll learn the same place we all learn. I stopped believing the bull about chewing-gum-wrapping-round-the-heart the instant I saw an X-Ray / autopsy / Gray's anatomy. By the time they've done secondary school biology for a couple of terms they'll realize the brain gym stuff was a crock, but they also might wonder why, if it was so patently a crock, whether it had any other point (which it does). We learn. That's pretty much all we do for the first couple of decades.
----
Vacuity abhors a vacuum.
[ Parent ]
Me. by Rogerborg (2.00 / 0) #27 Mon Mar 20, 2006 at 04:51:50 AM EST
I'm to say it.
If homeoepathy works, so does saying "Pull yourself together."
If Brain Gym works, so does giving them a Nintendo and a bottle of Evian.
-
Metus amatores matrum compescit, non clementia.
[ Parent ]
In all seriousness by yicky yacky (4.00 / 1) #29 Mon Mar 20, 2006 at 05:00:31 AM EST
I'd argue that well-designed games (Zelda etc.) encourage logic and rational thought in the young as well as any other method; game logic, after all, being essentially "Hypothesis => Test, Hypothesis => Test, Hypothesis => Test ..."
At least; that was my excuse ...
----
Vacuity abhors a vacuum.
[ Parent ]
Who's to say that I wasn't being serious? by Rogerborg (4.00 / 1) #31 Mon Mar 20, 2006 at 05:12:17 AM EST
Who? Go on, tell me.
-
Metus amatores matrum compescit, non clementia.
[ Parent ]
the real question is by 256 (4.00 / 1) #32 Mon Mar 20, 2006 at 05:56:57 AM EST
which is the purview of elementary schools:
1. making sure that kids drink enough water
2. filling their heads with information that is, at the very least, not willful nonsense
---
I don't think anyone's ever really died from smoking. --ni
[ Parent ]
Oddly enough by ucblockhead (4.00 / 1) #33 Mon Mar 20, 2006 at 06:58:03 AM EST
The simple sugars in a juice drink like "Sunny D" may well have a measurable effect on short-term mental performance.
Though true nerds know that the demon caffiene is the real brain drug.
---
[ Parent ]
More Brain Gym stuff... by squigs (2.00 / 0) #17 Mon Mar 20, 2006 at 03:17:58 AM EST
On Ben Goldacre's blog.
Seems to be a general mix of agreement, and people complaining because he fails to consider the benefits of excercising and keeping hydrated.
Brain Gym by priestess (2.00 / 0) #19 Mon Mar 20, 2006 at 03:26:08 AM EST
I think I'd have prefered it to actuall Gym, sounds a little better than running around in the mud and the cold chasing after a funny shaped ball, certainly.
Strange what some folks will believe though.
Pre...........
---------
Chat to the virtual me...
re: modernism by lm (2.00 / 0) #23 Mon Mar 20, 2006 at 03:57:39 AM EST
The chief problem with modernism is that it sees the sad plight of people with religion over the past 2000 years and assumes that their plight will be better without religion, ignoring the possibility that the problem with humanity stem from being human.
Or as Leo Strauss put it, ...no bloody or unbloody change of society can eradicate the evil in man: as long as there will be men, there will be malice, envy and hatred...''
Aside from which, I am somehow not surprised that an analysis of modernism in architecture that starts with the German built fortresses along the Atlantic wall should find that modernism is all about death.
There is no more degenerate kind of state than that in which the richest are supposed to be the best.
Cicero, The Republic
Processed foods do not contain water | 35 comments (35 topical, 0 hidden) | Trackback
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# Math in Focus Grade 6 Chapter 1 Lesson 1.4 Answer Key Squares and Square Roots
Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 1 Lesson 1.4 Squares and Square Roots to finish your assignments.
## Math in Focus Grade 6 Course 1 A Chapter 1 Lesson 1.4 Answer Key Squares and Square Roots
### Math in Focus Grade 6 Chapter 1 Lesson 1.4 Guided Practice Answer Key
Find the square of each number.
Question 1.
2
2 x 2 = 4
Explanation:
The square of 2 is 4.
Question 2.
6
6 x 6 = 36
Explanation:
The square of 6 is 36.
Question 3.
9
9 x 9 = 81
Explanation:
The square of 9 is 81.
Question 4.
11
11 x 11 = 121
Explanation:
The square of 11 is 121.
Find a square root of a perfect square.
a) A square has an area of 9 square inches. Find the length of each side of the square.
You know that
Area of square = length × length.
To find the length of a side of the square, you need to find the number whose square is 9.
Recalling the multiplication facts of 3, you know that 3 × 3 = 9.
So, the length of each side of the square is 3 inches.
3 is called a square root of 9. This can be written as $$\sqrt{9}$$ = 3.
You read this as “the square root of 9 equals 3”.
b) Find the square root of 100.
I can relate this to finding the length of the side of a square, given that it has an area of 1oo units2.
Method 1
Recalling the multiplication facts of 10, you know that
10 × 10 = 100
So, $$\sqrt{100}$$ = 10.
Method 2
By prime factorization,
Find the square root of each number.
Question 5.
25
5
Explanation:
Recalling the multiplication facts of 25, you know that
25 = 5 x 5
= 5²
So, $$\sqrt{25}$$ = 5
Question 6.
64
8
Explanation:
Recalling the multiplication facts of 64, you know that
64 = 8 x 8
= 8²
So, $$\sqrt{64}$$ = 8
Question 7.
144
12
Explanation:
Recalling the multiplication facts of 144, you know that
144 = 12 x 12
= 12²
So, $$\sqrt{144}$$ = 12
Question 8.
196
13
Explanation:
Recalling the multiplication facts of 169, you know that
169 = 13 x 13
= 13²
So, $$\sqrt{169}$$ = 13
### Math in Focus Course 1A Practice 1.4 Answer Key
Find the square of each number.
Question 1.
3
3 x 3 = 9
Explanation:
The square of 3 is 9.
Question 2.
7
7 x 7 = 49
Explanation:
The square of 7 is 49.
Question 3.
12
144
Explanation:
The square of 12 is 144.
Question 4.
10
100
Explanation:
The square of 10 is 100.
Find the square root of each number.
Question 5.
36
6
Explanation:
Recalling the multiplication facts of 36, you know that
36 = 6 x 6
= 6²
So, $$\sqrt{36}$$ = 6
Question 6.
81
9
Explanation:
Recalling the multiplication facts of 81, you know that
81 = 9 x 9
= 9²
So, $$\sqrt{81}$$ = 9
Question 7.
121
11
Explanation:
Recalling the multiplication facts of 121, you know that
121 = 11 x 11
= 11²
So, $$\sqrt{121}$$ = 11
Question 8.
49
7
Explanation:
Recalling the multiplication facts of 49, you know that
49 = 7 x 7
= 7²
So, $$\sqrt{49}$$ = 7
Solve.
Question 9.
List the perfect squares that are between 25 and 100.
Find the value of each of the following.
Question 10.
352
35 x 35 = 1225
Explanation:
The square of 35 is 1225.
Question 11.
562
56 x 56 = 3136
Explanation:
The square of 56 is 3136.
Question 12.
642
64 x 64 = 4096
Explanation:
The square of 64 is 4096.
Question 13.
$$\sqrt{289}$$
17
Explanation:
Recalling the multiplication facts of 289, you know that
289 = 17 x 17
= 17²
So, $$\sqrt{289}$$ = 17
Question 14.
$$\sqrt{400}$$
20
Explanation:
Recalling the multiplication facts of 400, you know that
400 = 20 x 20
= 20²
So, $$\sqrt{400}$$ = 20
Question 15.
$$\sqrt{484}$$
22
Explanation:
Recalling the multiplication facts of 484, you know that
484 = 22 x 22
= 22²
So, $$\sqrt{484}$$ = 22
Solve.
Question 16.
Given that 412 = 1,681, find the square of 410.
168100
Explanation:
If 412 = 1,681 then 4102 = 168100.
Question 17.
Given that 512 = 2,601, find the square root of 260,100.
510
Explanation:
If 512 = 2601 then the square root of 260100 will be 510.
Question 18.
Given that $$\sqrt{676}$$ = 26, evaluate $$\sqrt{2,704}$$.
52
Explanation:
Recalling the multiplication facts of 400, you know that
2704 = 52 x 52
= 52²
So, $$\sqrt{2704}$$ = 52
Question 19.
Given that $$\sqrt{1,521}$$ = 39, evaluate 3902.
152100
Explanation:
If $$\sqrt{1,521}$$ = 39 then 390² = 152100.
Question 20.
Heather wants to make a giant square quilt with sides of length 28 feet. She uses square patches of fabric that have sides of length 4 feet. How many patches of fabric will Heather need to make the giant square quilt?
7 patches of fabric Heather will need to make the giant square quilt.
Explanation:
Heather wants to make a giant square quilt with sides of length 28 feet.
She uses square patches of fabric that have sides of length 4 feet.
4 x 7 = 28
So, 7 patches of fabric Heather will need to make the giant square quilt.
Question 21.
This week, customers at a carpet store pay $3 for a square foot of carpet. Next week the store will be having a sale. During the sale, each square foot of carpet will cost only$2. Neil wants to carpet two square rooms in his house. The floor in one room is 10 feet by 10 feet. The floor in the other room is 14 feet by 14 feet. How much money will Neil save if he waits to buy carpet during the sale?
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# generating function of Pn(x)
• April 16th 2012, 12:04 AM
Jesssa
generating function of Pn(x)
hey guys,
my lecturer skipped the proof to show that $\frac{1}{\sqrt{1+u^2 -2xu}}$ is a generating function of the polynomials,
he told us that we should do it as an exercise by first finding the binomial series of
$\frac{1}{\sqrt{1-s}}$ then insert s = -u2 + 2xu
he then said to expand out sn and group together all the un terms,
this is what i've been doing for a while and i haven't been able to see the end,
$\frac{1}{\sqrt{1-x}}=\sum\limits_{n=0}^{\infty }{\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)....\left( \frac{1}{2}-n \right)}{n!}}{{(x)}^{n}}\,\,\,\,\,let\,\,x=2xu-{{u}^{2}}$
${{(u(-u+2x))}^{n}}={{(-1)}^{n}}{{u}^{n}}\left[ {{u}^{n}}-2x\frac{n!}{(n-1)!}{{u}^{n-1}}+4{{x}^{2}}\frac{n!}{(n-2)!}{{u}^{n-2}}+...+\frac{n!}{r!(n-r)!}{{(-1)}^{-r}}{{u}^{n-r}}{{(2x)}^{n}}+...+{{(-1)}^{-n}}{{(2x)}^{n}} \right]$
is my approach incorrect?
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