text:
## ARDUINO BMS

#### INTRODUCTION

Now this might seem like it was a complete waste of time to try and reinvent the wheel with creating my owm custom BMS solution. You can buy them on ebay or alibaba or banggood for super cheap and it seems like a plug and play solution.

One problem I see with this, at least for me, is that I dont completely trust those cheap BMS system. I like to buy something that is reasonable as well as trusting that it will safely protect my cells from unbalancing. Another this is that I like to have flexibility with how I balance my cells or anything else.

Now this is where the custom arduino BMS came into play. I’ve designed this bms to handle a 4S battery pack so upto 16.8V fully charged and 14.8V nominal. The balance discharge current can be adjusted by simply replacing the discharge resistor with any value you wish. Just to add this does not have over current or over voltage protection..yet 🙂 I’m currently working on an add on for this so stay tuned for that future update.

Lets get started with all the cool details for this arduino bms.

#### 3D RENDERING

Here i just wanted to show you the 3D rendering images of the final board. The rendering was created using kicad and its pretty nice actually for an EE cad software.

The heart of this system is the atmega328P running the minicore arduino bootloader[1]. From here you have the 4 balancing passive resistors to the right of the board. The best part of this project is that you can replace these resistors to any value you want to customize the balance current.

The one thing that this does not have is overvoltage and overcurrent protection but that will be a separate circuit later on ;).

Now we can talk about the main features below in the schematic section.

#### SCHEMATICS

Below are images of the full schematics for this project:

The first page of the schematic files is a hierarchy block level of the system to help understand how a lot of these subcircuits are linked together. Below I’ll breakdown all the pages and what exactly they do.

1. Hierarchy Block:

• This is the overview block level that links the subcircuits on the separate pages. I normally like to design my projects this way if I have more than 2 subcircuits because its easier to see how things are working together

2. Power Supply

• This section is just the power supply section to power the arduino. This is a high differential buck converter. The input voltage can go up to 36V and output down to 1.8V with an active quiescent current of only 190uA. Now the most important part is the quiescent current because we are trying to eliminate any wasted current since this is for a battery operated system and we do not want to drain the batteries faster than needed.

3. MCU

• The MCU page has the Atmega328P and all the corresponding IO related to it. Im using A0-A3 to measure the batteries via a resistor divider
• This page also includes the external reference voltage used instead of the built in 1.1V from the atmega328p. Its a 1.25V external voltage reference that has a better tolerance at a range of temperatures.

4. Balancing Circuit

• This page has the balacing circuit which is incharge of discharging the battery cell that has a higher potentional. The heart of this is a bidirectional N-channel mosfet that allows current to flow in both directional either during discharge or during charging.

5. Connectors

• This page has the power pin connectors and the BMS pads that connectors to the lithium cells.

Now that we’ve gone through the schematic lets go into the second most important part and that is the arduino code.

#### ARDUINO CODE

Now probably the second most important part is the arduino code that will control the mosfets triggering the discharge when a preset voltage difference is hit.

/* Revision: A9
* Written by: Steven Guzman
* Date: 7/10/2018
* Description: This is a customizable BMS system for a 4
* cell lithium ion pack.
*/

#include "Wire.h"
#include "avr/wdt.h"

#define OLED_RESET 4

#define NUMFLAKES 10
#define XPOS 0
#define YPOS 1
#define DELTAY 2

//////////////////////////////////////////////
// This variable is used for checking on
// whether the OLED screen is attached or not
//////////////////////////////////////////////
int check = LOW;

//////////////////////////////////////////////
// Sampling number for analog read values.
// Will use this sample size to calculate a
// better estimation of the analog value
//////////////////////////////////////////////
int samples = 100;

//////////////////////////////////////////////
// This is the max difference that the cells
// can achieve in respect to each other
//////////////////////////////////////////////
float tol = 0.03;

//////////////////////////////////////////////
// To temporaly store analog values from the
// cell voltages to average out later with
// the number of samples taken
//////////////////////////////////////////////
float temp_cell[4];

//////////////////////////////////////////////
// Float array to store unconverted cell values
//////////////////////////////////////////////
float cell[4];

//////////////////////////////////////////////
// Variable array to store converted voltage
// values for cell measurement
//////////////////////////////////////////////
float B[4];

//////////////////////////////////////////////
// Variable array to store voltage differences
// between cells
//////////////////////////////////////////////
float diff[4];

/////////////////////////////////////////////////
// Variable used for digital output
// signals
/////////////////////////////////////////////////
int BAT[4];

/////////////////////////////////////////////////
// Variable used for storing the cell pack
/////////////////////////////////////////////////
float PACK = 0.000;

/////////////////////////////////////////////////
// Scaling factor used for converting
// the scaled down analogread values to there
// actual values from the voltage divider
/////////////////////////////////////////////////
float scale0 = 0.25; // 10k/(30k + 10k)
float scale1 = 0.125; // 10k/(69.8k + 10k)
float scale2 = 0.083; // 10k/(110k + 10K)
float scale3 = 0.0625; // 10k/(150k + 10k)

#if (SSD1306_LCDHEIGHT != 32)
#endif

void setup()
{

Serial.begin(9600);
analogReference(EXTERNAL); // Configure reference voltage for
// external 1.25V

//////////////////////////////////////////////
// After switching to external reference you
// need to read a value a couple times before
// the reference has stabalized.
//////////////////////////////////////////////
delay(20);

// Setup up watchdog timer to reset
// after 4 seconds
wdt_enable(WDTO_4S);

///////////////////////////////////////////////
// Variables to set the digital output
// pins 1-4
///////////////////////////////////////////////
BAT[0] = 5;
BAT[1] = 6;
BAT[2] = 7;
BAT[3] = 8;

///////////////////////////////////////////////
// Setting digital pins to output configuration
///////////////////////////////////////////////
pinMode(BAT[0], OUTPUT);
pinMode(BAT[1], OUTPUT);
pinMode(BAT[2], OUTPUT);
pinMode(BAT[3], OUTPUT);

///////////////////////////////////////////////
// Sets the digital outputs to an initial low
// state
///////////////////////////////////////////////

digitalWrite(BAT[0],LOW);
digitalWrite(BAT[1],LOW);
digitalWrite(BAT[2],LOW);
digitalWrite(BAT[3],LOW);

}

void loop()
{

Wire.requestFrom(0x3C, 1); // Pings OLED screen

///////////////////////////////////////////////
// If OLED pings back and check statement
// equals 1, then initiate the OLED screen
// and reset the check value to 0 so it will
// not continue to re-initiate. It needs to
// initiate another request one last time
///////////////////////////////////////////////
if (Wire.available() &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; check == HIGH )
{
display.begin(SSD1306_SWITCHCAPVCC, 0x3C);
check = 0;
Wire.requestFrom(0x3C, 1);
}
if ( Wire.available() &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; check == LOW)
{
// Leave blank
}
else if (!Wire.available())
{
check = 1;
}

///////////////////////////////////////////////
// Sets the temp variable array back to zero
///////////////////////////////////////////////
for (int x = 0; x &amp;amp;amp;amp;lt;4; x++)
{
temp_cell[x] = 0.00;
diff[x] = 0.00;
cell[x] = 0.00;
}
///////////////////////////////////////////////
// For loop used to record battery cell voltage
// information and used later to average out
///////////////////////////////////////////////
delay(100);

////////////////////////////////////////////////
// Using true RMS calculation for computing a
// better average estimate for multiple samples.
////////////////////////////////////////////////
for (int i = 0; i &amp;amp;amp;amp;lt; samples; i++)
{
temp_cell[0] = temp_cell[0] + sq((analogRead(A0)/scale0)* (1.249 / 1024));
delay(1);
temp_cell[1] = temp_cell[1] + sq((analogRead(A1)/scale1) * (1.249 / 1024));
delay(1);
temp_cell[2] = temp_cell[2] + sq((analogRead(A2)/scale2) * (1.249 / 1024));
delay(1);
temp_cell[3] = temp_cell[3] + sq((analogRead(A3)/scale3) * (1.249 / 1024));
delay(1);
}
cell[0] = sqrt(temp_cell[0] / samples);
cell[1] = sqrt(temp_cell[1] / samples);
cell[2] = sqrt(temp_cell[2] / samples);
cell[3] = sqrt(temp_cell[3] / samples);

////////////////////////////////////////////////
// Scale up the cell values and then convert to
// to voltage values
////////////////////////////////////////////////
B[0] = (cell[0]);
B[1] = ((cell[1] - cell[0]));
B[2] = ((cell[2] - cell[1] ));
B[3] = ((cell[3] - cell[2] ));
PACK = B[3] + B[2] + B[1] + B[0];

///////////////////////////////////////////////////////////////////
// Compares each cell voltage to each other and then finds the
// lowest cell value. Then it stores the difference between the
// lowest cell and the remaining cells for comparison
///////////////////////////////////////////////////////////////////
if ( (B[0] &amp;amp;amp;amp;lt;= B[1]) &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; (B[0] &amp;amp;amp;amp;lt;= B[2]) &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; (B[0] &amp;amp;amp;amp;lt;= B[3]))
{
diff[1] = B[1] - B[0];
diff[2] = B[2] - B[0];
diff[3] = B[3] - B[0];
}
else if ((B[1] &amp;amp;amp;amp;lt;= B[0]) &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; (B[1] &amp;amp;amp;amp;lt;= B[2]) &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; (B[1] &amp;amp;amp;amp;lt;= B[3]))
{
diff[0] = B[0] - B[1];
diff[2] = B[2] - B[1];
diff[3] = B[3] - B[1];
}
else if ((B[2] &amp;amp;amp;amp;lt;= B[0]) &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; (B[2] &amp;amp;amp;amp;lt;= B[1]) &amp;amp;amp;amp;amp;&amp;amp;amp;amp;amp; (B[2] &amp;amp;amp;amp;lt;= B[3]))
{
diff[0] = B[0] - B[2];
diff[1] = B[1] - B[2];
diff[3] = B[3] - B[2];
}
else
{
diff[0] = B[0] - B[3];
diff[1] = B[1] - B[3];
diff[2] = B[2] - B[3];
}

//////////////////////////////////////////////////////////////////
// It enables the balancing protocol for a given cell that has a
// greater voltage difference from the lowest cell and the set
// tolerance.
//////////////////////////////////////////////////////////////////
delay(1);
for ( int x = 0; x tol )
{
digitalWrite(BAT[x], HIGH);
delay(1);
}
else
{
digitalWrite(BAT[x], LOW);
delay(1);
}
delay(1);
}

// text display tests
display.setTextSize(1);
display.setTextColor(WHITE);
display.setCursor(0,0);
display.print("V1:");
display.print(B[0]);
display.println("V");
display.print("V2:");
display.print(B[1]);
display.println("V");
display.print("V3:");
display.print(B[2]);
display.println("V");
display.print("V4:");
display.print(B[3]);
display.println("V");

display.setCursor(70,0);
display.print("VP:");
display.print(PACK);
display.print("V");

display.setCursor(70,8);
display.print("Config:");
display.print("4S");
display.display();
delay(1);
display.clearDisplay();
wdt_reset();
}

Lets break down this code in order to understand my thought process behind this.

1. The samples variable is used to create an average sampling of the battery cells to eliminate any noise issues. The averaging technique I used in this project is a little different than most but I will explain later on
2. The next variable is the tol variable. This variable is the maximum voltage different between the lowest cell and the rest of the cells in the 4S pack. Currently its set to a maximum of 30mV difference but this can be set to a minimum of 10mV because of the 8-bit adc on the arduino.
1. Here is the scaling factor for the adc from the cell measurements. We have different values because in order to get the best accuracy measuring in reference to ground, different values were needed for measuring cell 1 which has a max of 4.2V and cell 4 which measures a max of 16.8V.
1. This portion of the code is mainly for the OLED screen. It allows for the screen to be removed and re attached without having to reset the arduino. Before you had to initialize it in the setup but with this loop it is initialized in the main loop when it detects the screen.
1. Now this section is averaging the adc values from all 4 cells to get a best measurement of the cell voltage values. Instead of using the traditional sample 25 and take the average, I’m using a true RMS calculation that will give you a slightly better voltage measurement that just taking the average. It will be at a closer measurement to your multimeter than using average sample.

#### CONCLUSION / ISSUES

Here is the final PCB:

I’m pretty excited to do more testing on this and I will post my technical review later on that includes current consumption, maybe a mini manual on how to use it or change values for your own build.

Again this does not include an over voltage or over current protection circuit but I decided to make that a separate board with its own controller in order to make the board as small as I could and isolate the two circuits as a redundancy. I’m still making some tweaks and might make another revision but all in all I’m very happy about it.

Some current issues that I know off.

1. You might have to calibrate with a known good multimeter. The arduinos ADC has some weird issues with analogPin A2 measuring higher than the rest. I added a calibration value in the code to help with that.

For the Bill of Materials, I’ve embedded the datasheets for some of the major components in case you wondered why the file is larger than it should.

## How-to: Design a Boost Converter

#### Designing a boost converter sounds complicated and intimidating, well that was always my impression when it came to this topic in school.  In reality, the design and testing of a boost converter is a lot easier than meets the eye.

Here I will walk you step by step on designing your first boost converter and how the datasheet is your best friend when designing.  For this tutorial we will be using the L6920DC IC Boost converter from skyworks.[1]

This information was referenced from TI reference report.[2]

First and foremost, download the highlighted datasheet, datasheet-l6920dc. This has all the highlighted paremeters that you will need when designing a boost converter.

Step 1:

You need to decide what are your specifications.  These are the key parameters:

• Vin(min)
• Vin(max)
• Vout
• Iout
• n = efficiency; Most boost converters average around 85 to 90% under medium load and up to 95% on heavy load.  We will use the lowest percentage to be safe.

Example:

• Vin(min): 1.8V
• Vin(max): 2.4V
• Vout: 3.3V
• Iout: 0.4A
• n = 87% or 0.87

Step 2:

With your specifications, next step is to find your DUTY CYCLE:

$\bf \Huge D= 1 - \cfrac{(V_{\text{IN}}*n)}{V_{\text{OUT}}}$

We calculated the duty cycle for both lowest input voltage and highest input voltage.

• Lowest input voltage gives you the highest switching current you will see
• Highest input voltage gives you the highest output current your converter can produce

Example:

$\bf \Huge V_{\text{IN-MIN}}$

$\Huge D = 1 - \cfrac{1.8V*0.85}{3.3V}$ = 0.52

$\bf \Huge V_{\text{IN-MAX}}$

$\Huge D = 1 - \cfrac{2.4V*0.85}{3.3V}$ = 0.36

Step 3:

Next we will estimate the switching current or CURRENT RIPPLE of the Inductor:

ΔIL = $(0.3) * I_\text{OUTmax} * \cfrac{(V_\text{OUT})}{(V_\text{IN})}$

Example:

$\bf \Huge V_{\text{IN-MIN}}$

ΔIL = $\Huge (0.3) * 0.4A * \cfrac{3.3V}{1.8V}$ = 0.22A

$\bf \Huge V_{\text{IN-MAX}}$

ΔIL = $\Huge (0.3) * 0.4A * \cfrac{3.3V}{2.4V}$ = 0.165A

Step 4:

Next we calculate the minimum INDUCTANCE we need:

$\bf \Huge L_\text{MIN} = \cfrac{(V_\text{IN})*(V_\text{OUT} - V_\text{IN})}{\Delta I_\text{L}*f_\text{S}*V_\text{OUT}}$

$\bf \Huge f_\text{S}$ – This is the switching frequency that the converter will operate at.

Example:

$\bf \Huge V_{\text{IN-MIN}}$

$\Huge L_\text{MIN} = \cfrac{(1.8V)*(3.3V - 1.8V)}{0.22A*1MHz*3.3V}$ = 3.72uH

$\bf \Huge V_{\text{IN-MAX}}$

$\Huge L_\text{MIN} = \cfrac{(2.4V)*(3.3V - 2.4V)}{0.165A*1MHz*3.3V}$ = 3.97uH

We would select the highest inductance value to meet our input voltage rage of 1.8V-2.4V

When selecting the inductor, the key parameters you need to look for is low DCR, package size, and max current the inductor can handle.

DCR – Is the resistance in the coil because at the end of the day, an inductor is still a wire. When you keep this value at a minimum, it will increase your effieciency and the ability to provide a higher output power.

In step 7 , you will calculate the maximum current the inductor will see and there you will have all the necessary parameters needed to chose the inductor.

Step 5:

Now that we have our inductor value, we can calculate the actual CURRENT RIPPLE of the Inductor:

ΔIL = $\bf \Huge \cfrac{V_{IN}*D}{f_\text{S}*L}$

Example:

$\bf \Huge V_{\text{IN-MIN}}$

ΔIL = $\Huge \cfrac{1.8V*0.525}{1MHz*3.72uH}$ = 0.19A

$\bf \Huge V_{\text{IN-MAX}}$

ΔIL = $\Huge \cfrac{2.4V*0.36}{1MHz*3.97uH}$ = 0.18A

Step 6:

Next we need to calculate the MAX OUTPUT CURRENT the boost converter can output:

$\bf \Huge I_\text{MAXOUT}$ = $\bf \Huge (I_\text{LIM} - \cfrac{\Delta I_\text{L}}{2}) * (1 - D)$

$I_\text{LIM}$ – This is the current switch limit of the boost converter.

Example:

$\bf \Huge V_{\text{IN-MIN}}$

$\Huge I_\text{MAXOUT}$ = $\bf \Huge (0.8A - \cfrac{0.19A}{2}) * (1 - 0.52)$ = 0.33A

$\bf \Huge V_{\text{IN-MAX}}$

$\Huge I_\text{MAXOUT}$ = $\bf \Huge (0.8A - \cfrac{0.18A}{2}) * (1 - 0.36)$ = 0.45A

Step 7:

Next we will calculate the MAX SWITCHING CURRENT, $I_\text{SW}$ the Inductor will see.  This value cannot exceed the ILIM value of the boost converter:

$\bf \Huge I_\text{SW-MAX} = \cfrac{\Delta I_\text{L}}{2} + \cfrac{I_\text{OUT}}{1 - D}$

Example:

$\bf \Huge V_{\text{IN-MIN}}$

$\Huge I_\text{SW-MAX} = \cfrac{0.19A}{2} + \cfrac{0.4A}{1 - 0.525}$ = 0.94A

$\bf \Huge V_{\text{IN-MAX}}$

$\Huge I_\text{SW-MIN} = \cfrac{0.18A}{2} + \cfrac{0.4A}{1 - 0.36}$ = 0.72A

Note: $\Huge I_\text{SW-MAX}$ value cannot exceed $\Huge I_\text{LIM}$ which can be found in the datasheet.  In this example we see that with a low input voltage, the switching current exceeds the limit in the datasheet.  The boost converter might still be able to output the desired current at that low input voltage because $\Huge I_\text{LIM}$ is the minimum switching current it can handle.  But better to be safe than sorry.

Here you can see the inductor will see a max of 0.94A at its lowest input voltage. Now we can chose the inductor for our design.

For this design I went with,MSS5131-472MLB, a 4.7uH inductor from coilcraft.[3]

Since I chose an inductor that has a higher value than previous calculated, the inductor current ripple and output power will be slightly lower but it will not effect your design negatively.

Step 8:

This step is only if your boost converter has an adjustable output voltage.

(This boost converter is a fixed output and does not require these resistors.  Step 8 values are dummy values but the process )

Here we will find R1 AND R2 values for the feedback network:

$\bf \Huge I_\text{R0.5} >= 100 * I_\text{FB}$

$\bf \Huge I_\text{FB}$ – This is the current that the feedback resistor draws.

$\bf \Huge R_2 = \cfrac{V_\text{FB}}{I_\text{R0.5}}$

$\bf \Huge V_\text{FB}$ – This is the feedback reference voltage

$\bf \Huge R_1 = R_2 * (\cfrac{V_\text{OUT}}{V_\text{FB}}-1)$

Example:

$\Huge I_\text{R0.5} >= 100 * 350nA$ = 35mA

$\Huge R_2 = \cfrac{1.24V}{35mA}$ = 35.4kΩ

$\Huge R_1 = 35.4k \Omega * (\cfrac{3.3V}{1.24V}-1)$ = 58.74kΩ

Step 9:

Next l, we will calculate the INPUT CAPACITOR and OUTPUT CAPACITOR needed to minimize the ripple going in and out of the system:

First, you find your input capacitor:

$\bf \Huge C_\text{IN}$: Typically this value is 4.7uF to 10uF

Next, we need to first to look at these two equations below[6]:

$\bf \Huge \Delta V_\text{OUT}= \cfrac{I_\text{OUT} *T_\text{ONmax}}{ C_\text{OUT}}$

$\bf \Huge T_\text{ONmax}$ – This is the maximum on time of the boost converter.  It is also written as

$\bf \Huge D * T_\text{S}$

Were $\bf \Huge T_\text{S} = \cfrac{1}{f_\text{S}}$

$\bf \Huge \Delta V_\text{OUTesr} = ESR* I_\text{SW-MAX}$

ESR – All capacitors are not ideal capacitors and therefore have what is known as Equivalent Series Resistance. This is an important parameter that you need to consider when choosing the right output capacitor.

Example:

Cin = 10uF

First, we need to choose a voltage ripple that we can live with. Here I chose 50mV, and if we rearrange the first equation, we get:

$\Huge \Delta C_\text{OUTmin}= \cfrac{0.4 *6.25us}{50mV}$ = 50uF

Now we have a couple of options to choose from when it comes to materials for capacitors.

Most common are ceramic and electrolytic capacitors.  Each have there own pro and con.

Ceramic capacitors offer lower ESR for lower ripple but they typically do not have the bulk capacitance.

Electrolytic capacitors have bulk capacitance but generally have a high ESR that adds to ripple.

In this case I decided to go with both, getting the benefit of bulk capacitance and low ESR.

I went with a 1206 package, 10uF ceramic capacitor and a 47uF Electrolytic in parallel. For the electrolytic, they also have an aluminum polymer that has high capacitance with the added benefit of low ESR. I went with a 47uF that has an ESR of 40mΩ.

Now we plug in the values we got back into the equations and we get:Special Note: For ceramic capacitors, you need to be careful of which class and package size you choose because you only see a certain percentage of your nominal value (ex. 1206 10uF X7R will see 73% of 10uF)[4]. Click here for more info. I generally go with 1206 or 1210 with capacitors.

$\Huge \Delta V_\text{OUT}= \cfrac{0.4A * 6.25us}{50uF}$ = 50mV

$\Huge \Delta V_\text{OUTesr} = 40m\Omega*0.94A$ = 37mV

$\Huge \Delta V_\text{OUT}$ = 87mV
Always refer to the datasheet and compare recommended value vs calculated[1]

You’ve now designed your own boost converter regulator.  See it wasnt too hard :).

I will post this project soon that has the schematic and bill of materials, it’ll be under the projects menu bar, stay tuned!!

Feel free to comment below and correct me if anything seems incorrect to you.

References:

1. L6920 Datasheet

2. TI Basic Calculations of a Boost Converter Power Stage

3. Coilcraft Inductor

4. Temperature and Voltage Variation of Ceramic Capacitors

5. Ceramic or electrolytic output capacitors
in DC/DC converters—Why not both?

6. Boost Converter Output Capacitor Selection


text:
# Math Help - Two problems with solutions...help!!

1. ## Two problems with solutions...help!!

Sergio plans to paint the 4 walls of his room with 1 coat of paint. The walls are rectangular, and, according to his measurements, each wall is 10ft by 15ft. He will not need to paint the single 3-foot-by-5-foot rectangular window in his room and the 3.5 foot-by-7-foot rectangular door. Sergio knows that each gallon of paint covers between 300 and 350 square feet. If only 1-gallon cans of paint are available, what is the minimum number of cans of paint Sergio needs to buy to paint his walls?

The book says the solution is 2 but I can't figure it out??

Also, I need to solve the following:

8^(2x + 1) = 4^(1 - x)

The solution is -1/8...not sure why though...any help would be appreciated on these.

2. There are 4 10x15 ft walls. A walls measures are 10x15 ft and its area is 150 ft². 4 walls will make 150x4 = 600 ft². As he won't paint the 3x5 ft (15 ft²) window and the 3.5x7 ft (24.5 ft²) door, we can subtract them from the total area of walls.
600 - 15 = 585
585 - 24.5 = 560.5 ft²

If 1 gallon can of paint can cover 300-350 ft², and the question asks for the best situation, we will take it as 350 ft². So he will need 2 cans to paint 560.5 ft².

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

$8^{2x + 1} = 4^{1 - x}$

$(2^3)^{2x+1} = (2^2)^{1-x}$

$2^{6x+3} = 2^{2-2x}$

$6x + 3 = 2 - 2x$

$8x = -1$

$x = - \frac{1}{8}$

3. Thanks so much wingless--can't believe I didn't think of that for both problems!

4. You're welcome. Don't worry, as many questions as you solve, as much as you think on them, you'll get better at 'seeing things' in questions

format: plain


text:
## Macroeconomics Study Guides

Keep the following two comprehensive study guides handy throughout your macroeconomics course study. They provide brief oulines for many of the major macroeconomics topics studied in this course and can help prepare you for your final economics exams.

### Principles of Macroeconomics Lecture Notes

#### Interest Rates

$i_{0,1}$ = the nominal interest rate between periods 0 and 1

(the nominal return on the asset)

$\pi^{e}_{0,1}$ = the expected inflation rate between periods 0 and 1

$r^{e}_{0,1}$ = the expected real interest rate between periods 0 and 1

Definitions

$r^e _{0,1} = i_{0,1} - \pi^e _{0,1} (or \: i_{0,1} = \pi^e _{0,1} + r^e _{0,1)}$

$r^a _{0,1} = i_{0,1} - \pi^a _{0,1} (or \: i_{0,1} = \pi^a _{0,1} + r^a _{0,1)}$

where $r^{a}$ and $\pi^{a}$ are the actual real interest rate and inflation

#### Interest Rate Notes

• The Formula given is approximate. The approximation is less accurate the higher the levels of inflation and nominal interest rates. The exact formula is $r^{e} = (1 + i) / (1 + \pi ^e) - 1$
• Central Banks are very interested in $r$ since it may affect the savings decisions of households and definitely affects the investment decisions of firms. The press talks about Central Banks setting $i$, but the Central Banks are really trying to set $r$.
• 3 easy ways of measuring expected inflation:

• Recent actual inflation.
• Survey of forecasters.
• Interest rate spread on nominal vs. inflation-indexed securities (WSJ)

#### Why We Care About Inflation?

• Note: We will have a whole lecture on this later in the course
• Inflation is Unpredictable
• Indexing Costs (even if you know the inflation rate - you have to deal with it).
• Menu Costs (have to go and re-price everything)
• Shoe-Leather Costs (you want to hold less cash - have to go to the bank more often).
• Caveat: There may be some benefits to small inflation rates - more on this later.
• An Example of how inflation can affect real returns.
• Suppose we agree that a real rate of 0.05 over the next year is fair.
• borrowing rate, salary growth rate, etc.
• Suppose we also agree that expected inflation over the next year is 0.07.
• We should then set the nominal return equal to $0.12$ $( i = r^{e} + \pi^{e} )$

Summary: $i$ = 0.12

$r^{e}$= 0.05

$\pi{e}$ = 0.07

• Suppose that actual inflation is $0.10$ $(\pi^{a} > \pi^{e} )$

In this case, $r^{a} = 0.02 (r^{a} = i - \pi^{a})$

Borrowers/Firms are better off

Lenders/Workers worse off

• Suppose that actual inflation is $0.03$ $(\pi^{a} < \pi{e} )$

In this case, $r^{a} = 0.09 \: ( r^{a} = i - \pi^{a} )$

Borrowers/Firms are worse off

Lenders/Workers better off

It has been shown that higher inflation rates are correlated with more variability.

People/Firms Don't Like the Uncertainty


text:
# Simpleton Problem

I. Mathematical Expression:

$$f(X)=\frac{\displaystyle x_1 \cdot x_2 \cdot x_3 \cdot x_4 \cdot x_5}{\displaystyle x_6 \cdot x_7 \cdot x_8 \cdot x_9 \cdot x_{10}}$$

where:

$$\bullet$$ $$1\leq x_i\leq 10$$ , $$i=1,2,\cdots,10$$

$$\bullet$$ $$f_{max}(X^*)=10^5$$

$$\bullet$$ $$x^*_i =(10,10,10,10,10,1,1,1,1,1)$$

$$\bullet$$ This benchmark function is presented in [1] as a trivial $$10-$$dimensional maximization problem. The maxima can be easily obtained by maximizing the numerator and minimizing the denominator [2].

II. References:

[1] Ariel Dolan, "A General GA Toolkit Implemented in JAVA, for Experimenting with Genetic Algorithms and Handling Optimization Problems," Jun. 1998, [Accessed August 06, 2015]. [Online]. Available: http://www.aridolan.com/ofiles/ga/gaa/gaa.aspx
[2] Ali R. Alroomi, "The Farm of Unconstrained Benchmark Functions," University of Bahrain, Electrical and Electronics Department, Bahrain, Oct. 2013. [Online]. Available: http://www.al-roomi.org/cv/publications


text:
# Electric Force/ Electrical Fields

1. Feb 6, 2009

### xswtxoj

1. The problem statement, all variables and given/known data

An electron is released a short distance above earth's surface. A 2ndelectron directly below it excerts an electrostatic force on the 1st electron just great enough to cancel the gravitational force on it How far below the 1st electron is the 2nd?

2. Relevant equations

F= Ke (q1)(q2)/r^2

3. The attempt at a solution

2. Feb 6, 2009

### DukeLuke

Since the two forces are acting in opposite directions and you want them to add to zero you can set your gravity force equation equal to the electric force equation. Then you should be able to solve for the distance "r".

3. Feb 6, 2009

### xswtxoj

Can you write it out for me? I don't quite understand what you mean!

4. Feb 6, 2009

### DukeLuke

Well the force of gravity on the 1st electron is can be found using $F=ma$ where "a" = g = acceleration due to gravity 9.8 $\frac{m}{s^2}$. You already have the equation for the force between the two electrons due to their charge. Since you want the forces to be equal(becuase they are opposing and cancel each other out) you can set both of these equations equal to each other. Then the only variable you won't know in this combined equation should be r which is the distance between the paritcles' centers. Your equation should look something like

$$mg = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2}$$

5. Feb 7, 2009

### xswtxoj

how would u solve this problem if no numbers are given? i really dont get this?

6. Feb 7, 2009

### xswtxoj

what u said so i set up the equation and to cancel is like Keqq= Frsq then qq=mg4 pi Eo rsq, then cancel all like terms, to have Ke=F=mg4 pi Eo?

7. Feb 7, 2009

8. Feb 7, 2009

### xswtxoj

considering that its an electron as 1.6 x 10^-19, and ur finding the distances, how would u apply that to the 2 equations above?

9. Feb 7, 2009

### LowlyPion

DukeLuke gave you the equation. So look up the mass.

Solve for R.

10. Feb 7, 2009

### xswtxoj

can the equation be used as q1q2/ 4 pi (8.85x 10 ^-12 ) r sq,,,, the 8.85 be used for E0?

11. Feb 7, 2009

### LowlyPion

Yes. Of course.

12. Feb 7, 2009

### xswtxoj

is i use the $$mg = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2}$$ then would it be: 9.8= 1.6 E -19 sq/ 4 (3.14)(8.85 E -12) r2, then cross multiply to get 9.8 x 4 x 3.14 x 8.85xE-12 (r sq) = 1.6x10-19 sq then get 1.089E -9 (r sq) = 2.56x -38 then divide to isolate r sq, then solve for r to get 4.84E -15


text:
# Shear And Moment Diagrams

Shear And Moment Diagrams. Shear and bending moment diagrams are analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force and bending moment at a given point of a structural element such as a beam. We will analyze bending in a beam due to two types of loads: dead loads and live loads.

Construct both shear and moment diagrams just as before. Shear and bending moment diagrams depict the variation of these quantities along the length of the member. Beam design formulas with shear and moment diagrams.

### Shear force diagram and bending moment diagrams are illustrations to describe the alterations in shear force and bending moments over the length of thebeam.

Shear and Moment Diagrams are used to help you understand what the shear and moment curve look like for different generic types of beams and loading conditions.

Shear and Moment Diagram Example 2 - Mechanics of ...

Shear Force and Bending Moment Diagrams - Wikiversity

Statics eBook: Shear and Moment Diagrams II

Shear Force and Bending Moment Diagram, Shear Force ...

Drawing Shear and Moment Diagrams Example- Mechanics of ...

Shear force and bending moment diagram practice problem #6 ...

Statics - Bending moment and shear diagram EXAMPLE ...

Bendingmomentdiagram.com is a free online calculator that generates Bending Moment Diagrams (BMD) and Shear Force Diagrams (SFD) for most simple beams. The resulting graphics are called the shear diagram and moment diagram. Construct both shear and moment diagrams just as before.


text:
## Is the Set of Linear Transformations a Group?

For the set of all linear transformations
$\left\{ T \right\}$
to be a group, thye group axioms must be satisfied.
1. The identity element must be in the set. This condition is not satisfied. Differentiation is linear. We can define a linear transformation on the set of polynomials of degree 2.
$\frac{d}{dx}(1)=0,\frac{d}{dx}(x)=1, \frac{d}{dx}(x^2)=2x$
.
We represent
$1.x.x^2$
by the vectors
$\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix}$
Hence
$\frac{d}{dx}(a+bx+cx^2)=b+2cx$
.
The columns of the matrix representing
$T$
can be found by differentiating
$1.x.x^2$
in turn and representing the results as vectors.
The matrix representing th linear transformation is
$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 &2 \end{array} \right)$
.
The polynomial
$2+3x+5x^2$
is represented by the vector
$\begin{pmatrix}2\\3\\5\end{pmatrix}$

$T(2+3x+5x^2)= \left( \begin{array}{cc} 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} \right) \begin{pmatrix}2\\3\\5\end{pmatrix}=\begin{pmatrix}3\\10\\0\end{pmatrix}$
which returns the polynomial
$3+10x$
.
The image of this transformation is the set of polynomials of degree 2, since the differentiation reduces the degree of polynomials by 1. There is thus no identity element. 2. Inverse.
$T_1 \in \left\{ T \right\} \rightarrow T_1^{-1} \in \left\{ T \right\}$
. This condition is also not satisfied since the matrix above is not square, so has no inverse. 3. Closure
$T_1 , T_2 \in \left\{ T \right\} \rightarrow T_1 T_2 \in \left\{ T \right\}$
. This condition is not satisfied either.
Take
$T_1 = \left( \begin{array}{ccc} 1 & 2 & 0 \\ 6 & 4 & 0 \\ 1 & 0 & 2 \end{array} \right) , T_2 = \left( \begin{array}{cc} 1 & 0 \\ 6 & 4 \end{array} \right)$
.
These matrices cannot be multiplied.
4. Associativity is satified for matrix multiplication
Three of the four group axioms are not satisfied so the set of linear transformations is not a group. Howver, the set of linear transformations associted with invertible square matrices do form groups for each order.


text:
# Decentralized Identity Management, Blockchain – Why Bother

This blog represents details on Decentralized Identity Management and why you should care? Given that IBM, Hyperledger has joined Blockchain Identity Consortium makes it much more important to quickly go over the concepts related with decentralized identity management. Check out a related Hyperledger project, Project Indy, on supporting independent identity on distributed ledgers.

### Traditional Centralized or Federated Identity Management System

Conventional identity management systems have always been based on centralized authorities such as corporate directory services , certificate authorities (CA) , or domain name registries. Each of these centralized identity management systems acted as a “root of trust”.

In order to have the identity management work across different systems, there is something called as federated identity management. According to Wikipedia post of Federated Identitya federated identity in IT is the means of linking a person’s electronic identity and attributes, stored across multiple distinct identity management system.

### What is Decentralized Identity Management System?

Following are some of the key points to understand decentralized identity management in better manner:

• In decentralized identity management system, there are no centralized authorities anymore. The “root of trust” lies with Distributed Ledger. The copies of ledger are maintained by multiple participants, and thus, distributed ledger. Note that Blockchain is one of the classic example of Distributed Ledger Technology.
• Entities no more identify themselves with traditional centralized authorities such as CAs, domain name registries etc. They rather identify themselves with the distributed ledger
• Each entity is assigned as Decentralized Identifiers (DIDs).
• DIDs can be used to point what is called as DID Document.
• DID Document may consists of one or more service endpoint that can be used to interact with the entity. This comes in a JSON format. Every DID document must have an associated DID. This is how a DID document would look like:
{
"@context": "https;//w3id.org/did/v1",
"id": "did:example:123456789abcdefghi",
"authorizationCapability": [{
// this entity is a delegate and may update any field in this
// DID Document using any authentication mechanism understood
// by the ledger
"permission": "UpdateDidDocument",
"entity": "did:example:zxyvwtrkpn987654321"
}],
"credentialRepositoryService": "https://vc.example.com/abcdef",
"authenticationCredential": [{
// this biometric can be used to authenticate as DID ...fghi
"id": "did:example:123456789abcdefghi/biometric/1",
"type": "PseudonymousBiometricTemplate2017",
"owner": "did:example:123456789abcdefghi",
"biometricService": "https://example.com/authenticate"
"biometricTemplateShard": "Mjk4MzQyO...5Mzg0MDI5Mwo="
}]
}

• Following are some of the key terminologies in relation with DID:
• DID path
• DID fragment
• DID normalization
• DID persistence

For details, read the W3C Community Group Draft Report on Decentralized Identifiers. Check out a related W3C Community Group draft on verifiable claims.


text:
# Uniformization theorem for 2-manifolds

A version of the uniformization theorem for the 2-manifolds that I have read about says that any connected 2-manifold is diffeomorphic to either of the 3 constant curvature model spaces modulo an action on them by a discrete subgroup of their isometries acting freely and properly discontinuously. This apparently implies that every connected 2-manifold has a complete Riemannian metric with constant Gaussian curvature.

I don't know of a reference which explains how this proof works.

I would like to understand what this really means and how does this connect to the Hopf-Rinow theorem (which I am more familiar with) which also characterizes geodesically complete spaces in general.

Especially I would like to understand why the two caveats on the group action, "freely" and "properly discontinuously" necessary. Whats the intuition behind them?

I would like to know if this is something special about 2-manifolds or are their similar analogues in higher dimensions? If not then what goes wrong?

Some geometers tell me that theorems which guarantee existence of a complete Riemannian metric on some space are in some sense "weak" since it basically says that one can make the metric "bigger" towards the edges of the spaces so that the geodesic never "hits an end".(and apparently this is not interesting) I don't quite understand these comments.

-
If the action is not properly discontinuous then the quotient space need not be a manifold! Freeness ensures that the quotient map is a covering map. The fact that a quotient by a free, properly discontinuous group action is a covering map is not special to 2-manifolds, but the existence of a uniformisation theorem is. – HJRW Aug 13 '10 at 17:45
Once you classify surfaces up to diffeomorphism it is easy to construct metrics of constant curvature by hand, e.g. 2-torus is R2 modulo the group generated by translations along coordinate axes, giving a flat metric. The case of hyperbolic surfaces isn't much harder: just consider a regular polygon in the hyperbolic disk with angle 2g and identify its sides in a standard way. The case of round sphere is trivial, and non-orientable surfaces are handled similarly. – Igor Belegradek Aug 13 '10 at 20:12
As for you other questions, I do not see why Hopf-Rinow is more relevant than Arzela-Ascoli :), and the reason "uniformization" does not work in higher dimensions is that there are many more higher-dimensional manifolds. On the other hand, geometrization does work in dimension 3 as you surely know. Finally, any manifold admits a complete Riemannian metric which is perhaps what your geometers mean, but this can hardly be compared with geometrization by nice metrics, such as of constant curvature or locally homogeneous. – Igor Belegradek Aug 13 '10 at 20:13
I messed up latex formulas in the above comment: the angle in a regular hyperbolic polygon should be $2\pi/4g$ so that the sum of angles is $2\pi$. – Igor Belegradek Aug 14 '10 at 1:44
@Igor Thanks for your insights. So what I understand is that given any 2-manifold one can put on it a metric which is complete. Can you give some explicit examples? Like what is the geodesically complete metric on the open-2-disk? (sounds surprising that there exists one!) And given some metric is there any easy to do check for completeness? Could you elaborate on the statement that "any manifold admits a complete Riemannian metric" ? (I thought one way to interprete the Hopf-Rinow theorem is that the manifold needs to be complete as a metric space to be geodesically complete) – Anirbit Aug 14 '10 at 8:41

The uniformization theorem is really about complex structures. It says that up to bi-holomorphism, there are only $3$ connected simply connected $1$-dimensional complex manifolds: the plane, the unit disc, and the Riemann sphere. This theorem is quite difficult to prove, and I do not now a reference (although a collective book on it is in preparation, initiated by Étienne Ghys).

The link with Riemannian geometry is that in real dimension $2$, a complex structure is the same than a conformal structure. So, if you consider an orientable surface, you can endow it with any riemannian metric, whose conformal class gives you a complex structure on it (rotation by an angle of $\pi/2$ plays the role of multiplication by $i$). Then its universal covering is one of the three simply connected complex surfaces above. Since each of them can be endowed with a Riemannian metric of constant curvature whose conformal class corresponds to their complex structure and the groupe of deck transformation acts by isometries, you can push this metric to your original surface.

The main reason you cannot do this in higher dimension is that you do not have complex numbers to help you. A great number of theories have been studied that are higher-dimensional analoguous, though. A few of them are: Yamabe problem (finding a constant scalar curvature metric in a conformal class), geometrization (decompose a $3$-manifold into pieces having nice, homogeneous metrics), Einstein manifolds (constant Ricci curvature metrics). If you want to know more about these topics, there are plenty of references but A panoramic view of geometry by Marcel Berger is my favorite.

-
Surely there are references to the uniformization theorem, e.g. see references at the end of math.harvard.edu/theses/senior/chan/fulldraft7.pdf, an expository thesis written under direction of Siu. One such reference is [Abikoff, William, The uniformization theorem. Amer. Math. Monthly 88 (1981), no. 8, 574--592]. – Igor Belegradek Aug 14 '10 at 13:44
Donaldson's notes on Riemann surfaces cover uniformisation: www2.imperial.ac.uk/~skdona/RSPREF.PDF I find counting degrees of freedom quite instructive. In dim $n$ the curvature tensor has $O(n^4)$ components, the metric $O(n^2)$, and a conformal factor only $1$. Trying to find a metric of prescribed curvature will give a very overdetermined system except in dim $\leq 3$. And it's only in 2 dimensions, where the curvature reduces to a scalar, that one has a hope of obtaining constant curvature by a conformal change. – Tim Perutz Aug 14 '10 at 14:06
My favorite exposition of the classical proof of the uniformization theorem is in the first chapter of John Hubbard's book on Teichmuller theory. – Andy Putman Aug 14 '10 at 15:23
Notice that this argument proves a much stronger conclusion than the "easy" gluing-of-polygons argument that Igor mentioned. The gluing-of-polygons argument proves that there exists a constant-curvature metric. The complex-analysis argument proves that there exists a constant-curvature metric <i>in the same conformal class as your favorite Riemannian metric</i>. – Dave Futer Aug 21 '10 at 12:26

There is a proof based on elliptic PDE in a paper of mine with Michael Taylor that was published in the Israel Journal of Math in 2002 Vol 130, 323-346. The emphasis is on the general open case of domains with arbitrary boundary, and the issue is to find a solution of the curvature equation which grows sufficiently strongly to produce a complete metric. --

-

There is an excellent expository article by Peter Scott titled "The Geometries of 3-Manifolds." The first section is spent talking about uniformization of surfaces (which is what you are interested in), and he also discusses 2-Oribfolds, which are what one gets when the group action is properly discontinuous but not free. The rest of the article is certainly good reading, but not really relevant to what you asked. Most importantly though, he sketches a proof of uniformazation for surfaces. The basic idea is that the only group which acts nicely on $S^2$ is $\mathbb{Z}/2\mathbb{Z}$, with quotient $P^2$. Furthermore, discrete groups of isometries of $\mathbb{R}^2$ with compact quotient are isomorphic to a group with a finite index subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$, and hence the quotient is either a torus or a Klein bottle.Then one can classify the isometries of the hyperbolic plane, $\mathbb{H}^2$, and show that $\mathbb{Z} \oplus \mathbb{Z}$ cannot be a discrete, orientation preserving subgroup of $Isom(\mathbb{H}^2)$, and hence the torus and Klein bottle cannot admit a hyperbolic structures. Furthermore, it is easy to show (with the classification of hyperbolic isometries) that other surfaces do admit hyperbolic structures.

Also, to give you an idea of why free and properly discontinuous are important, it helps to see a few examples. First, consider $\mathbb{R}^2$ and let $G$ be the group generated by rotation around the origin through the angle $2\pi/n$. This action has a fixed point at the origin. Topologically, the quotient is just $\mathbb{R}^2$, but is has a cone point with angle $2\pi/n$ at the origin, so geometrically it is different. This type of space is what is known as an orbifold.

Without proper discontinuity, things get even worse. Consider $S^1$ and the group $G$ generated by rotation through an irrational multiple of $\pi$. The resulting quotient is not even Hausdorff (the orbit of a point under $G$ is dense in $S^1$), and we generally want to avoid that sort of thing.

In dimension 3, things get a little harder, but geometrization still works. A really good reference for this dimension is William Thurston's "3 Dimensional Geometry and Topology," which is easily one of my favorite math books ever written.

In dimension 4, the same sort of approach is a lot harder. The reason is that in dimensions 2 and 3, we have a fair deal of control as to what sorts of groups show up as fundamental groups of (closed) manifolds. It can be shown (I don't recall a reference for this off the top of my head, perhaps someone else has it) that for any finitely presented group $G$, there exists a closed 4-manifold $M$ with $\pi_1(M) = G$. So, we don't have the kind of set-up like in dimension 2 where we know which fundamental groups arise, and then show that they are isometries of one of three model spaces.

Hope this helps.

-


text:
# The huygens-fresnel principle

 Page 1 / 1
The Huygens-Fresnel principle, reflection and refraction

## The huygens-fresnel principle

In order to proceed with the discussion we have to define two terms. A wave front is the surface of constant phase. In a plane wave these are planes andin a spherical wave these are spheres. A ray travels perpendicular to the fronts.

Huygens postulated that as a wave propagates through a medium each point on the advancing wavefront acts as a new point source of the wave. This iscorrect physics for the water waves but not for light waves. However the Helmholtz equation for diffraction of EM waves gives a solution identical tothat give by Huygens' principle.

Look at the figure which shows a wavefront AB coming to a surface and is reflected creating the front CD. The point A hits the surface first. The pointB hits a time $vt$ later. During that time a spherical wave is emitted from A and travels a distance $vt$ . In fact this happens for every point along the wavefront. The next figureattempts to show how a number of waves line up along the line DC and that this is perpendicular to the line AD.

From this we see that ${\mathrm{sin}}{\theta }_{i}=\frac{vt}{AC}$ and ${\mathrm{sin}}{\theta }_{r}=\frac{vt}{AC}$ so ${\theta }_{i}={\theta }_{r}$

For refraction a similar thing happens. See figure (geometric optics / Huygens refraction.vsd )

In this case the velocities are different in the two media and so one obtains: ${\mathrm{sin}}{\theta }_{i}=\frac{{v}_{i}t}{AC}$ and ${\mathrm{sin}}{\theta }_{t}=\frac{{v}_{t}t}{AC}$ which then can be rearranged $\frac{{\mathrm{sin}}{\theta }_{i}}{{v}_{i}t}=\frac{{\mathrm{sin}}{\theta }_{t}}{{v}_{t}t}$ or rearranging some more $\frac{{\mathrm{sin}}{\theta }_{i}}{{\mathrm{sin}}{\theta }_{t}}=\frac{{v}_{i}t}{{v}_{t}t}$ or $\frac{{\mathrm{sin}}{\theta }_{i}}{{\mathrm{sin}}{\theta }_{t}}=\frac{{n}_{t}}{{n}_{i}}$ finally ${n}_{t}{\mathrm{sin}}{\theta }_{t}={n}_{i}{\mathrm{sin}}{\theta }_{i}$ which is Snell's law. Now note that normally one uses rays, in which case the anglesare measured w.r.t. the normal to the surface.

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Got questions? Join the online conversation and get instant answers!


text:
# Random Matrix Theory

(Sethna, "Entropy, Order Parameters, and Complexity", ex. 1.6, developed with Piet Brouwer)

© 2016, James Sethna, all rights reserved.

This is an ipython notebook. This hints file is unusually detailed, but only covers the computational parts (a & f) of the exercise: don't neglect the analytical portions! If you are familiar with Mathematica, feel free to skip parts you don't need, or start your own notebooks.

Lectures about Python, useful both for beginners and experts, can be found at http://scipy-lectures.github.io.

I recommend installing the Anaconda distribution.

Pick your operating system (Windows, Mac OS X, Linux)

Follow instructions for Python 3.5.

To make it run much faster (probably necessary for some assignments):

Be sure not to pay for it! Get an academic subscription, using your Cornell (.edu) e-mail, from https://www.continuum.io/anaconda-academic-subscriptions-available

• Set up an account, using your Cornell e-mail
• Go to My Settings under your name in the upper right hand side of the Anaconda Cloud window, and
• Go to Add Ons, the lowest menu entry in the left-hand column
• Download the licenses, and follow the installation instructions link just above the MKL licence button (or go to http://docs.continuum.io/advanced-installation)

Open the notebook by (1) copying this file into a directory, (2) in that directory typing:

ipython notebook

and (3) selecting the notebook. In later exercises, if you do animations, you may choose to not have the plots inline.

Like Mathematica, you type in commands and then hit 'Shift Return' to execute them.

In [ ]:
a = 3.
print(a)

Python is not a complete scientific environment. Functions like log, sin, and exp are provided by external packages. The plotting package pylab supports many of these; '%pylab inline' puts the graphs into the notebook. We usually want more functionality than that, so we import a package called 'scipy' (an extension of 'numpy'). We'll also be using some of their linear algebra routines, which are in a sub-package 'scipy.linalg'.

In [ ]:
%pylab inline
from scipy import *

One of the most active and unusual applications of ensembles is random matrix theory, used to describe phenomena in nuclear physics, mesoscopic quantum mechanics, and wave phenomena. Random matrix theory was invented in a bold attempt to describe the statistics of energy level spectra in nuclei. In many cases, the statistical behavior of systems exhibiting complex wave phenomena -- almost any correlations involving eigenvalues and eigenstates -- can be quantitatively modeled using ensembles of matrices with completely random, uncorrelated entries!

The most commonly explored ensemble of matrices is the Gauusian orthogonal ensemble (GOE). Generating a member takes two steps:

• Generate an NxN matrix whose elements are independent Gaussian distributions of zero mean and standard deviation $\sigma = 1$.
• Add the matrix to its transpose to symmetrize it.

Start by finding out how Python generates random numbers. Type ?random to find out about scipy's random number generators.

Try typing 'random.random()' a few times. Try calling it with an integer argument. Use 'hist' (really pylab.hist) to make a histogram of 1000 numbers generated by random.random. Is the distribution Gaussian, uniform, or exponential? The flag 'normed' tells us to not just count the number of occurences in each bin, but to divide by the width of the bin and the number of counts (to approximate a normalized probability distribution).

In [ ]:
?random
print(random.random())
print(random.random(6))
hist(random.random(1000), normed=True);

Use random.standard_normal to generate a Gaussian distribution of one thousand random numbers. Check that it is a standard Gaussian $$\rho(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-x^2/2\sigma^2}$$ by using 'hist'.

In [ ]:
print(random.standard_normal(...))
hist(random.standard_normal(...), normed=...);
x = arange(-3,3,0.01)
plot(x,(1./sqrt(2*pi))*exp(-x**2/2),'r')
show()

One final matrix operation. To generate a symmetric matrix of the GOE form, we'll want to add a matrix to its transpose. Generate a 2x3 matrix of normally distributed random numbers, and print it and its transpose.

In [ ]:
m = random.standard_normal((2,3));
print(m)
print(transpose(m))

Now define a function GOE(N) (using 'def') which generates a GOE NxN matrix.

In [ ]:
def GOE(N):
"""
Creates an NxN element of the Gaussian Orthogonal Ensemble,
by creating an NxN matrix of Gaussian random variables
using the random array function random.standard_normal([shape])
with [shape] = (N,N).
and adding it to its transpose (applying transpose to the matrix)
#
Typing GOE(4) a few times, check it's symmetric and random
"""
m = random. ...((...,...))
m = m + transpose(m)
return m

Test it by generating a 4x4 matrix. Is it symmetric?

In [ ]:
GOE(4)

Now write a function GOE_Ensemble(num, N) generating an ensemble (a long list with 'num' elements) of GOE NxN matrices. Call GOE(N) to fill the list.

In [ ]:
def GOE_Ensemble(num, N):
"""
Creates a list "ensemble" of length num of NxN GOE matrices.
#
You can start with an empty list, and fill it with 'append'
ensemble = []
for n in range(num):
ensemble.append(GOE(N))
return ensemble
or you can use the convenient 'list comprehension'
ensemble = [GOE(N) for n in range(num)]
#
Check GOE_Ensemble(3,2) gives 3 2x2 symmetric matrices
"""
ensemble = ...
return ensemble

Here you may want to generate an ensemble, and check if the diagonal elements and the off-diagonal elements have different standard deviations, as you calculated in part (c). (This difference is part of the definition of the GOE ensemble, and is needed to make the ensemble rotation invariant.)

One of the most striking properties that large random matrices share is the distribution of level splittings.

To look at eigenvalues of our symmetric matrices, we use 'eigvalsh'. We'll want them sorted. And we'll want to find the difference between the two middle-most eigenvalues.

Notice, in python, arrays are indexed lambda = [lambda[0], ..., lambda[N-1]], so the middle two are N/2 and N/2-1. Note that in the new version of python dividing integers 4/2 = 2.0 gives a floating point number, so we need to use N//2 to get an integer. (This is a good thing: having 2/10 = 0 causes lots of problems when you write routines that don't know whether their inputs are floats or ints.)

In [ ]:
N=4
midpoint = int(N/2)
mat = GOE(N)
print(mat)
eig = sort(eigvalsh(mat))
print(eig)
print(eig[N//2]-eig[N//2-1])

In [ ]:
def CenterEigenvalueDifferences(ensemble):
"""
For each symmetric matrix in the ensemble, calculates the difference
between the two center eigenvalues of the ensemble, and returns the
differences as an array.
#
Given an ensemble of symmetric matrices,
finds their size N, using the first member ensemble[0]
len(m) gives the number of rows in m
starts with an empty list of differences
for each m in ensemble
finds the eigenvalues
eigvals(m) gives the eigenvalues of m
sorts them
sort(e) sorts a list from smallest to largest
appends eigenvalue[N/2] - eigenvalue[N/2-1] to the list
(Note: python lists go from 0 ... N-1, so for N=2 this
gives the difference between the only two eigenvalues)
#
Check
ensemble = GOE_Ensemble(3,2)
CenterEigenvalueDifferences(ensemble)
gives three positive numbers, that look like eigenvalues of the 2x2 matrices
"""
# Size of matrix
N = len(...)
diffs = []
# Note that you can 'iterate' not only over integers, but over elements of a list (for mat in ensemble: ...)
for mat in ensemble:
eigenvalues = sort(...)
diffs.append(...)
return diffs

Check that

   ensemble = GOE_Ensemble(3,2)
CenterEigenvalueDifferences(ensemble)

gives three positive numbers, that look like eigenvalue differences of 2x2 matrices, Check that

    hist(CenterEigenvalueDifferences(ensemble), bins=30, normed=True);

gives three spikes at the three eigenvalue differences.

In [ ]:
ensemble = GOE_Ensemble(...)
print(CenterEigenvalueDifferences(...))
hist(CenterEigenvalueDifferences(...), bins=50, normed=True);

Now try it with 10000 or more. Notice the smooth distribution, with a linear tail near zero.

In [ ]:
M = 10000
N = 2
ensemble = GOE_Ensemble(M, N);
diffs = CenterEigenvalueDifferences(ensemble)
hist(diffs, bins=50, normed=True);

What is this dip in the eigenvalue splitting probability near zero? It is called level repulsion.

Wigner's surmise (meaning 'guess') was that this distribution was universal, that is, independent of the ensemble of matrices. Clearly another ensemble, with all our matrix elements multiplied by two, will have eigenvalue differences twice as big. Wigner surmised that the eigenvalue differences, divided by the mean of the differences would be universal. Plot these normalized centered differences against your analytical prediction for 2x2 matrices (see text), which should be $\rho_{Wigner}(s) = (\pi s/2) \exp(-\pi s^2 / 4)$

In [ ]:
hist(diffs/mean(diffs), bins=50, normed=True);
s = arange(0.,3.,0.01)
rhoWigner = ...
plot(s, rhoWigner, 'r');
title('2x2 matrix eig diffs')

Try N=4 and N=10; do they look similar?

In [ ]:
M = 10000
N = 4
ensemble = ...
diffs = ...
hist(...);
plot(...);
title(...)

In [ ]:
M = 10000
N = 10
ensemble = ...
diffs = ...
hist(...);
plot(...);
title(...)

Note that we're typing the same commands several times. This is a bad programming practice (a 'bad smell') -- whenever writing the same code more than a couple of lines long, write it as a subroutine. That way, when you fix an error, it'll get fixed everywhere you use it.

In [ ]:
def CompareEnsembleWigner(ensemble):
diffs = ...
hist(...);
plot(...)

CompareEnsembleWigner(ensemble)

Your 2x2 formula is pretty good, but turns out to be up to 2% off for larger matrices; Wigner was wrong about that. But he was correct that the eigenvalue splitting is universal, if you consider large matrices.

We can test this by trying a different ensemble. For example, let's try random symmetric matrices filled with +-1. In the past, we've created these matrices the hard way. But let's just take the signs of the matrix elements generated by the GOE ensemble!

In [ ]:
sign(GOE_Ensemble(4,3))

Define a PM1 ensemble (Plus or Minus 1)

In [ ]:
def PM1_Ensemble(num, N):
"""
Generates a +-1 ensemble, as for GOE_Ensemble above.
Use with CompareEnsembleWigner to test that N=2 looks different
from the Wigner ensemble, but larger N looks close to the GOE ensemble.
#
This is a powerful truth: the eigenvalue differences of very general classes
of symmetric NxN matrices all have the same probability distribution
as N goes to infinity. This is what we call universality.
"""
return sign(...)

Try comparing the splittings for 2x2 matrices with the prediction.

In [ ]:
CompareEnsembleWigner(PM1_Ensemble(10000,2))

Is it well approximated by the Wigner surmise? Why not? (Hint: how many different symmetric 2x2 matrices of $\pm 1$ are there?)

Try it with 4x4 and 10x10 matrices.

In [ ]:
CompareEnsembleWigner(...)

10x10 should work better, except for a funny bump at zero.

In [ ]:
CompareEnsembleWigner(...)

The GOE ensemble has some nice statistical properties (see text).

This is our first example of an emergent symmetry. Many different ensembles of symmetric matrices, as the size $N$ goes to infinity, have eigenvalue and eigenvector distributions that are invariant under orthogonal transformations even though the original matrix ensemble did not have this symmetry. Similarly, rotational symmetry emerges in random walks on the square lattice as the number of steps $N$ goes to infinity, and also emerges on long length scales for Ising models at their critical temperatures.


text:
# Evaluation of $\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx$

Evaluate the integral $$\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx$$

My Attempt:

Let $f(x) = \frac{ax+b}{(x^3+3x+1)^2}.$ Now differentiate both side with respect to $x$, and we get

\begin{align} f'(x) &= \frac{(x^3+3x+1)^2\cdot a-2\cdot (x^3+3x+1)\cdot (3x^2+3)\cdot (ax+b)}{(x^3+3x+1)^4}\\ &= \frac{5x^3+3x-1}{(x^3+3x+1)^3}\cdot\frac{(x^3+3x+1)\cdot a-6\cdot (x^2+1)\cdot (ax+b)}{(x^3+3x+1)^3}\\ &= \frac{5x^3+3x-1}{(x^3+3x+1)^3} \frac{-5ax^3+6bx^2-3ax+(a-6b)}{(x^3+3x+1)^3}\\ &= \frac{5x^3+3x-1}{(x^3+3x+1)^3} \end{align}

for $a = -1$ and $b = 0$. Thus, by the Fundamental Theorem of Calculus,

\begin{align} \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx &= \int f'(x)\,dx\\ &= f(x)\\ &= -\frac{x}{(x^3+3x+1)^2}+\mathcal{C} \end{align}

How we can solve the above integral directly (maybe by using the substitution method)?

• the solution looks very ugly – Dr. Sonnhard Graubner Jan 12 '15 at 17:38
• You reacted very well ! I am afraid that substitution would be very difficult. – Claude Leibovici Jan 12 '15 at 17:44
• @juantheron, Why don't you start with $$\frac{ax^2+bx+c}{(x^3+3x+1)^2}$$? – lab bhattacharjee Jan 12 '15 at 17:47
• nice idea. how did you come to it? numerator has three degrees(may be four) how were able to handle with only two constants $a$ and $b?$ – abel Jan 12 '15 at 17:58

To integrate a rational function $P(x)/Q(x)$ without decomposing it into partial fractions and without finding the roots of the denominator, we can use the Ostrogradski-Hermite method, which generalizes your ansatz

$$\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx= \frac{ax+b}{(x^3+3x+1)^2}+C.$$

You can find a description of this method in section 2.1 of Gradshteyn and Ryzhik's Table of Integrals, Series, and Products, where the identity $(2)$ below is given. The formula $(1)$ appears also on the Ostrogradsky's Wikipedia page.

Assume that $\deg P(x)<\deg$ $Q(x)$. There exist polynomials $P_{1}(x)$, $P_{2}(x)$, $Q_{1}(x)$ and $Q_{2}(x)$, with $Q_{1}(x)=\gcd \left\{ Q(x), Q^{\prime }(x)\right\}$ and $Q_{2}(x)=Q(x)/Q_{1}(x)$, $\deg P_{1}(x)<\deg Q_{1}(x)$, $\deg P_{2}(x)<\deg Q_{2}(x)$, such that

$$\int \frac{P(x)}{Q(x)}dx=\frac{P_{1}(x)}{Q_{1}(x)}+\int \frac{P_{2}(x)}{ Q_{2}(x)}dx.\tag{1}$$

Then

\begin{eqnarray*} P(x) &=&\frac{P_{1}^{\prime }(x)Q_{1}(x)-P_{1}(x)Q_{1}^{\prime }(x)}{\left\{ Q_{1}(x)\right\} ^{2}}Q(x)+\frac{P_{2}(x)}{Q_{2}(x)}Q(x) \\ &=&P_{1}^{\prime }(x)\frac{Q(x)}{Q_{1}(x)}-P_{1}(x)\frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}\frac{Q(x)}{Q_{1}(x)}+P_{2}(x)\frac{Q(x)}{Q_{2}(x)} \\ &=&P_{1}^{\prime }(x)Q_{2}(x)-P_{1}(x)\left\{ \frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}Q_{2}(x)\right\} +P_{2}(x)Q_{1}(x) \end{eqnarray*}

or

$$P(x)=P_{1}^{\prime }(x)Q_{2}(x)-P_{1}(x)\left\{ T(x)-Q_{2}^{\prime }(x)\right\}+P_{2}(x)Q_{1}(x),\tag{2}$$

with $T(x)=Q^{\prime }(x)/Q_{1}(x)$, because from

\begin{equation*} Q^{\prime }(x)=\left\{ Q_{1}(x)Q_{2}(x)\right\} ^{\prime }=Q_{1}^{\prime }(x)Q_{2}(x)+Q_{1}(x)Q_{2}^{\prime }(x)=T(x)Q_{1}(x) \end{equation*}

we obtain

\begin{equation*} \frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}Q_{2}(x)+Q_{2}^{\prime }(x)=T(x). \end{equation*}

To find the coefficients of the polynomials $P_{1}(x)$ and $P_{2}(x)$ equate the coefficients of like powers of $x$.

Application to

\begin{equation*} \frac{P(x)}{Q(x)}=\frac{5x^{3}+3x-1}{\left( x^{3}+3x+1\right) ^{3}}. \end{equation*}

Since

\begin{eqnarray*} Q(x) &=&\left( x^{3}+3x+1\right) ^{3} \\ Q^{\prime }(x) &=&9\left( x^{3}+3x+1\right) ^{2}\left( x^{2}+1\right) \\ Q_{1}(x) &=&\gcd \left\{ Q(x),Q^{\prime }(x)\right\} =\left( x^{3}+3x+1\right) ^{2} \end{eqnarray*}

and \begin{equation*} Q_{2}(x)=\frac{Q(x)}{Q_{1}(x)}=\frac{\left( x^{3}+3x+1\right) ^{3}}{\left( x^{3}+3x+1\right) ^{2}}=x^{3}+3x+1, \end{equation*}

we write

$$\int \frac{5x^{3}+3x-1}{\left( x^{3}+3x+1\right) ^{3}}dx=\frac{P_{1}(x)}{ \left( x^{3}+3x+1\right) ^{2}}+\int \frac{P_{2}(x)}{x^{3}+3x+1}dx,\tag{3}$$

where

\begin{eqnarray*} P_{1}(x) &=&Ax^{5}+Bx^{4}+Cx^{3}+Dx^{2}+Ex+F \\ P_{2}(x) &=&Fx^{2}+Gx+H. \end{eqnarray*}

The identity $(2)$, with

\begin{equation*} T(x)=\frac{Q^{\prime }(x)}{Q_{1}(x)}=9\left( x^{2}+1\right), \end{equation*}

yields

\begin{eqnarray*} 5x^{3}+3x-1 &=&\left( 5Ax^{4}+4Bx^{3}+3Cx^{2}+2Dx+E\right) \left( x^{3}+3x+1\right) \\ &&-\left( Ax^{5}+Bx^{4}+Cx^{3}+Dx^{2}+Ex+F\right) \left\{ 9\left( x^{2}+1\right) -\left( 3x^{2}+3\right) \right\} \\ &&+\left( Gx^{2}+Hx+I\right) \left( x^{3}+3x+1\right) ^{2} \\ &=&Gx^{8}+\left( -A+H\right) x^{7}+\left( -2B+6G+I\right) x^{6} \\ &&+\left( 6H-3C+2G+9A\right) x^{5} \\ &&+\left( -4D+6B+9G+6I+2H+5A\right) x^{4} \\ &&+\left( 3C-5E+4B+6G+9H+2I\right) x^{3} \\ &&+\left( 3C-6F+G+6H+9I\right) x^{2} \\ &&+\left( 6I+H+2D-3E\right) x+\left( E+I-6F\right). \end{eqnarray*}

By equating coefficients we find

\begin{equation*} A=B=C=D=F=G=H=I=0,E=-1.\tag{4} \end{equation*}

Consequently,

\begin{eqnarray*} P_1(x) &=&-x \\ P_2(x) &=&0 \end{eqnarray*} and finally,

\begin{equation*} \int \frac{5x^{3}+3x-1}{\left( x^{3}+3x+1\right) ^{3}}dx=-\frac{x}{\left( x^{3}+3x+1\right) ^{2}}+C,\tag{5} \end{equation*}

as evaluated by you.

• This is so nice method which I was unaware of +1. – Paramanand Singh Apr 4 '15 at 16:34
• This is why I love this site. Something new to learn everyday. (And I who was about to put a comment to the question like: "Just do the substitution $u=-x/(x^3+3x+1)^2$", but I don't even think that'd be fun myself, after reading your nice and serious answer. – mickep Apr 4 '15 at 20:12
• Answer to a similar question here. – Américo Tavares May 1 '15 at 22:53

\begin{aligned}\int\frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3}\,\mathrm{d}x &=\int\frac{5x^3 + 3x-1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)\right)^3}\,\mathrm{d}x\\&=\int\frac{5x^3 + 3x-1}{x^{3/2}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\,\mathrm{d}x\\&=\int\frac{5x^{3/2}+3x^{-1/2}-x^{-3/2}}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\,\mathrm{d}x\\&=\int\frac{2\,\mathrm{d}(x^{5/2}+3\sqrt{x}+x^{-1/2})}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\\&=2\int\frac{\mathrm{d}\tau}{\tau^3}\\&=-\frac{1}{\tau^2}+C\\&=-\frac{1}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^2}+C\\&=\cdots\end{aligned}

Here is another solution... I also happen to think that it cannot be done with substitution.

\begin{aligned} \int \frac{5x^3+3x-1}{\left ( x^3+3x+1 \right )^3}\,dx &=\int \frac{-x^3-3x-1+6x^3+6x}{\left ( x^3+3x+1 \right )^3}\,dx \\ &= \int \frac{-x^3-3x-1+2x\left ( 3x^2+3 \right )}{\left ( x^3+x+1 \right )^3}\,dx\\ &= \int \frac{-\left ( x^3+3x+1 \right )^2+2x\left ( x^3+3x+1 \right )\left ( 3x^2+3x \right )}{\left ( x^3+3x+1 \right )^4}\,dx\\ &= \int \frac{-(x)'\left ( x^3+3x+1 \right )^2+x\left[ \left ( x^3+3x+1 \right )^2 \right]'}{\left ( x^3+3x+1 \right )^4}\,dx\\ &= \int \left [ -\frac{x}{\left ( x^3+3x+1 \right )^2} \right ]' \,dx = -\frac{x}{\left ( x^3+3x+1 \right )^2}+c, \; \; c \in \mathbb{R} \end{aligned}

• +1 I would not discover your direct algebraic transformation of the integrand. Only if I worked backwards, from the solution to the original integral. – Américo Tavares May 6 '15 at 11:40
• This is what I like doing when I work with integrals of this kind. I also liked your solution. I think I have seen the method before but I did not remember it. I guess it was just "luck" that I noticed it. If I do not , then I give up. :) – Tolaso May 6 '15 at 11:45


text:
# 1. Introduction

Algebraic fractions have properties which are the same as those for numerical fractions, the only difference being that the numerator (top) and denominator (bottom) are both algebraic expressions.

Example 1: Simplify the following fractions:

(a)$\displaystyle \frac {2b} {7b^2}$

Solution:

$$\frac{2b}{7b^2} = \frac {2 \redcancel b}{7b \redcancel b} = \frac{2}{7b}$$
(b)$\displaystyle \frac {3x+x^2} {6x^2}$

Solution:

$$\begin{eqnarray} \frac{3x+x^2}{6x^2} & = & \frac{ x \times (3+x)^\ }{x \times 6x} \\ & = & \frac{ \redcancel x \times (3+x)}{ \redcancel x \times 6x} = \frac{3+x}{6x} \end{eqnarray}$$

Note that the cancellation is allowed since $x$ is a common factor of the numerator and denominator.

Sometimes a little more work is necessary before an algebraic fraction can be reduced to a simpler form.

Example 2: Simplify the algebraic fraction:

$$\frac{x^2 - 2x + 1}{x^2 + 2x - 3}$$

Solution:

In this case, the numerator and denominator can be factored into two terms (see the module on Simplifying Expressions):

$$x^2-2x+1 = (x-1)^2 ~~ \text{and} ~~ x^2+2x-3 = (x-1)(x+3)$$

With this established the simplification proceeds as follows:

$$\begin{eqnarray} \frac{x^2-2x+1}{x^2+2x-3} & = & \frac{(x-1) \times \redcancel{(x-1)}}{(x+3) \times \redcancel{(x-1)}} \\ & = & \frac{(x-1)}{(x+3)} \end{eqnarray}$$

Exercise 1: Simplify each of the following algebraic fractions.

(a) $\displaystyle \frac{8y}{2y^3}$

Solution:

$$\begin{eqnarray} \frac{8y}{2y^3} & = & \frac{4 \times 2y}{y^2 \times 2y} \\ & = & \frac{4 \times \redcancel 2 \times \redcancel y}{y^2 \times \redcancel 2 \times \redcancel y} \\ & = & \frac{4}{y^2} \end{eqnarray}$$
(b) $\displaystyle \frac{2y}{4x}$

Solution:

$$\begin{eqnarray} \frac{2y}{4x} & = & \frac{2 \times y}{2 \times 2x} \\ & = & \frac {\redcancel 2 \times y}{\redcancel 2 \times 2x} = \frac{y}{2x} \end{eqnarray}$$
(c) $\displaystyle \frac{7a^6b^3}{14a^5b^4 }$

Solution: This time, instead of expanding the factors, it is easier to use the rule for powers (see the module on Powers):

$$\frac{a^m}{a^n} = a^{m-n}$$

Thus:

$$\begin{eqnarray} \frac{7a^6b^3}{14a^5b^4 } & = & \frac{7}{14} \times \frac{a^6}{a^5} \times \frac{b^3}{b^4} \\ & = & \frac{1}{2} \times a^{6-5} \times b^{3-4} \\ & = & \frac{1}{2} \times a^1 \times b^{-1} = \frac{a}{2b} \end{eqnarray}$$
(d) $\displaystyle \frac {(2x)^2}{4x}$

Solution:

$$\begin{eqnarray} \frac {(2x)^2}{4x} & = & \frac {2 \times x \times 2 \times x}{2 \times 2 \times x} \\ & = & \frac {\redcancel 2 \times \redcancel x \times \redcancel 2 \times x}{\redcancel 2 \times \redcancel 2 \times x} = x \end{eqnarray}$$
(e) $\displaystyle \frac{5y+2y^2}{7y}$

Solution:

$$\begin{eqnarray} \frac{5y+2y^2}{7y} & = & \frac{y \times (5+2y)}{7 \times y} \\ & = & \frac{\redcancel y \times (5+2y)}{7 \times \redcancel y} = \frac{5+2y}{7} \end{eqnarray}$$
(f) $\displaystyle \frac {5ax}{15a+10a^2}$

Solution:

$$\begin{eqnarray} \frac {5ax}{15a+10a^2} & = & \frac {5 \times a \times x}{5 \times a \times (3+2a)} \\ & = & \frac {\redcancel 5 \times \redcancel a \times x }{ \redcancel 5 \times \redcancel a \times (3 + 2a)} = \frac{x}{3+2a} \end{eqnarray}$$
(g) $\displaystyle \frac{2z^2−4z}{2z^2 −10z}$

Solution:

$$\begin{eqnarray} \frac{2z^2−4z}{2z^2 −10z} & = & \frac {2 \times z \times (z-2)}{2 \times z \times (z-5)} \\ & = & \frac {\redcancel 2 \times \redcancel z \times (z-2)}{\redcancel 2 \times \redcancel z \times (z-5)} = \frac{z-2}{z-5} \end{eqnarray}$$
(h) $\displaystyle \frac {y^2+7y+10}{y^2−25}$

Solution: In this case some initial factorisation is needed (see the module on Factorising Expressions).

$$y^2+7y+10 = (y+5)(y+2) ~~ \text{and} ~~ y^2−25=(y+5)(y−5)$$

Thus:

$$\begin{eqnarray} \frac{y^2 +7y+10}{y^2 −25} & = & \frac {\redcancel{(y+5)}(y+2)}{\redcancel{(y+5)}(y-5)} = \frac {y+2}{y-5} \\ \end{eqnarray}$$
(i) $\displaystyle \frac {w^2−5w−14}{w^2−4w−21}$

Solution: Again, in this case, some initial factorisation is needed (see the module on Factorising Expressions).

$$w^2−5w−14 = (w−7)(w+2) ~~ \text{and} ~~ w^2−4w−21 = (w−7)(w+3)$$

Thus:

$$\begin{eqnarray} \frac {w^2−5w−14}{w^2−4w−21} & = & \frac {\redcancel{(w-7)}(w+2)}{\redcancel{(w-7)}(w+3)} = \frac {w+2}{w+3} \\ \end{eqnarray}$$

Click on questions to reveal their solutions

Now try this short quiz:

Quiz 1: Which of the following is a simplified version of $\displaystyle \frac{t^2+3t−4}{t^2−3t+2}$ ?

(a) $\displaystyle \frac{t-4}{t-2}$ Incorrect - please try again!
(b) $\displaystyle \frac{t-4}{t+2}$ Incorrect - please try again!
(c) $\displaystyle \frac{t+4}{t-2}$ Correct - well done!
(d) $\displaystyle \frac{t+4}{t+2}$ Incorrect - please try again!

Explanation: The numerator and denominator respectively factorise as:

$$t^2+3t−4=(t−1)(t+4) ~~ \text{and} ~~ t^2−3t+2=(t−1)(t−2)$$

So that:

$$\begin{eqnarray} \frac{t^2+3t−4}{t^2−3t+2} & = & \frac {\redcancel{(t-1)}(t+4)}{\redcancel{(t-1)}(t-2)} = \frac {t+4}{t-2} \end{eqnarray}$$

So far, simplification has been acheived by cancelling common factors from the numerator and denominator. There are fractions which can be simplified by multiplying the numerator and denominator by an appropriate common factor, thus obtaining an equivalent, simpler expression.

Example 3: Simplify the following fractions:

(a) $\displaystyle \frac { \frac 1 4 + y}{\frac 1 2}$

Solution: In this case, multiplying both the numerator and denominator by 4 gives:

$$\frac { \frac 1 4 + y}{\frac 1 2} = \frac { 4(\frac 1 4 +y)}{4(\frac 1 2)} = \frac {1+4y}{2}$$

(b) $\displaystyle \frac { 3x + \frac 1 x}{2}$

Solution: To simplify this expression, multiply the numerator and denominator by $x$. Thus:

$$\frac { 3x + \frac 1 x}{2} = \frac { x(3x + \frac 1 x)}{2x} = \frac { 3x^2 + 1}{2x}$$

Now try this exercise on similar examples:

Exercise 2: Simplify each of the following algebraic fractions.

(a) $\displaystyle \frac {4y - \frac 3 2}{2}$

Solution: The fraction is simplified by multiplying both the numerator and denominator by $\blue 2$:

$$\frac {4y - \frac 3 2}{2} = \frac {\blue 2(4y - \frac 3 2)}{\blue 2 \times 2} = \frac {8y - 3}{4}$$
(b) $\displaystyle \frac {2x + \frac 1 2}{x + \frac 1 4}$

Solution: This fraction is simplified by multiplying both the numerator and denominator by $\blue 4$:

$$\frac {2x + \frac 1 2}{x + \frac 1 4} = \frac {\blue 4(2x + \frac 1 2)}{\blue 4(x + \frac 1 4)} = \frac {8x+2}{4x+1}$$
(c) $\displaystyle \frac {z - \frac 1 3}{z - \frac 1 2}$

Solution: In this case, since the numerator contains the fraction $\frac 1 3$ and the denominator contains the fraction $\frac 1 2$, the common factor needed is $2 \times 3 = \blue 6$. Thus:

$$\frac {z - \frac 1 3}{z - \frac 1 2} = \frac {\blue 6(z - \frac 1 3)}{\blue 6(z - \frac 1 2)} = \frac {6z - 2}{6z - 3}$$
(d) $\displaystyle \frac {2 - \frac 1 x}{2}$

Solution: For this fraction the required multiplier is $\blue x$:

$$\frac {2 - \frac 1 x}{2} = \frac {\blue x(2 - \frac 1 x)}{2\blue x} = \frac {2x - 1}{2x}$$
(e) $\displaystyle \frac {3t - \frac 2 t}{\frac 1 2}$

Solution: Here the numerator contains the fraction $\frac 2 t$ and the denominator is the fraction $\frac 1 2$, so the required multiplier is $\blue{2t}$:

$$\frac {3t - \frac 2 t}{\frac 1 2} = \frac {\blue{2t}(3t - \frac 2 t)}{\blue{2t}\frac 1 2} = \frac {6t^2 - 4}{t}$$
(f) $\displaystyle \frac {z - \frac 1 {2z}}{z - \frac 1 {3z}}$

Solution: For the last past of this exercise, since the numerator includes the fraction $\frac 1 {2z}$ and the denominator includes the fraction $\frac 1 {3z}$, the common multiplier is $\blue{6z}$:

$$\frac {z - \frac 1 {2z}}{z - \frac 1 {3z}} = \frac {\blue {6z}(z - \frac 1 {2z})}{\blue{6z}(z - \frac 1 {3z})} = \frac {6z^2 - 3}{6z^2 - 2}$$

Click on questions to reveal their solutions

For the last part of this section, try the following short quiz.

Quiz 2: Which of the following is a simplified version of $~ \displaystyle \frac {x - \frac 1 {x+1}}{x-1}$ ?

(a)$\displaystyle \frac {x^2-x+1}{x^2+x+1}$Incorrect - please try again!
(b)$\displaystyle \frac {x^2-x+1}{x^2-1}$Incorrect - please try again!
(c)$\displaystyle \frac {x^2-1}{x^2-x-1}$Incorrect - please try again!
(d)$\displaystyle \frac {x^2+x-1}{x^2-1}$Correct - well done!

Explanation: For $\displaystyle \frac {x - \frac 1 {x+1}}{x-1}$, the common multiplier is $\blue{x+1}$. Multiplying the numerator and denominator by this gives:

$$\begin{eqnarray} \frac {x - \frac 1 {x+1}}{x-1} & = & \frac {\blue{(x+1)}(x - \frac 1 {x+1})}{\blue{(x+1)}(x-1)} \\ & = & \frac {(x+1)x - (x + 1)(\frac 1 {x+1})}{(x^2-1)} = \frac {x^2+x-1}{x^2-1} \end{eqnarray}$$

# 2. Addition of Algebraic Fractions

Addition (and subtraction) of algebraic fractions proceeds in exactly the same manner as for numerical fractions.

Example 4: Write the following sum as a single fraction in is simplest form:

$$\frac 2 {x+1} + \frac 1 {x+2}$$

Solution: The least common multiple of the denominators (see the module on Fractions) is $\blue{(x+1)(x+2)}$. Thus:

$$\begin{eqnarray} \frac 2 {x+1} + \frac 1 {x+2} & = & \frac {2 \times \blue{(x+2)}}{(x+1) \times \blue{(x+2)}} + \frac {1 \times \blue{(x+1)}}{(x+2) \times \blue {(x+1)}} \\ & = & \frac {2x+4}{(x+1)(x+2)} + \frac {x+1}{(x+2)(x+1)} \\ & = & \frac {(2x+4)+(x+1)}{(x+1)(x+2)} = \frac {3x+5}{(x+1)(x+2)} \end{eqnarray}$$

Exercise 3: Evaluate each of the following fractions.

(a)$\displaystyle \frac 2 y + \frac 3 z$

Solution: The least common multiple of the denominators is $\blue{yz}$. Thus:

$$\begin{eqnarray} \frac 2 y + \frac 3 z & = & \frac {2 \times \blue{z}}{y \times \blue{z}} + \frac {3 \times \blue{y}}{z \times \blue {y}} \\ & = & \frac {2z} {yz} + \frac {3y} {yz} = \frac {3y+2z}{yz} \end{eqnarray}$$
(b)$\displaystyle \frac 1 {3y} - \frac 2 {5y}$

Solution: Here the least common multiple of the denominators is $\blue{15y}$, so

$$\begin{eqnarray} \frac 1 {3y} - \frac 2 {5y} & = & \frac {\blue{5} \times 1}{\blue{5} \times 3y} - \frac {\blue{3} \times 2}{\blue{3} \times 5y} \\ & = & \frac 5 {15y} - \frac 6 {15y} \\ & = & \frac {5-6}{15y} = - \frac 1 {15y} \end{eqnarray}$$
(c)$\displaystyle \frac {3z+1} 3 - \frac {2z+1} 2$

Solution: The least common multiple of the denominators of the two fractions in this case is $\blue{6}$. Thus:

$$\begin{eqnarray} \frac {3z+1} 3 - \frac {2z+1} 2 & = & \frac {\blue{2} \times (3z+1)}{\blue{2} \times 3} - \frac {\blue{3} \times (2z+1)}{\blue{3} \times 2} \\ & = & \frac {6z+2} 6 - \frac {6z+3} 6 \\ & = & \frac {(6z+2) - (6z+3)} 6 = - \frac 1 6 \end{eqnarray}$$

Simplification in this case has shown that the difference of these two fractions is independent of $z$.

(d)$\displaystyle \frac {3t+1} 2 + \frac 1 t$

Solution: There least common multiple of the denominators of the two fractions in this case is $\blue{2t}$. The sum simplifies as follows:

$$\begin{eqnarray} \frac {3t+1} 2 + \frac 1 t & = & \frac {\blue{t} \times (3t+1)}{\blue{t} \times 2} + \frac {\blue{2} \times 1}{\blue{2} \times t} \\ & = & \frac {3t^2+t} {2t} + \frac 2 {2t} = \frac {3t^2+t+2} {2t} \end{eqnarray}$$
(e)$\displaystyle \frac {x+1} 2 + \frac 1 {x-1}$

Solution: Here, the required least common multiple of the denominators is the factor $\blue{2(x-1)}$. Proceeding as before:

$$\begin{eqnarray} \frac {x+1} 2 + \frac 1 {x-1} & = & \frac {\blue{(x-1)} \times (x+1)}{\blue{(x-1)} \times 2} + \frac {\blue{2} \times 1}{\blue{2} \times (x-1)} \\ & = & \frac {(x^2-1)} {2(x-1)} + \frac 2 {2(x-1)} \\ & = & \frac {(x^2-1)+2} {2(x-1)} = \frac {x^2+1} {2(x-1)} \end{eqnarray}$$
(f)$\displaystyle \frac 2 {w+3} - \frac 5 {w-1}$

Solution: Here the least common multiple of the denominators is $\blue{(w+3)(w−1) = w^2+2w−3}$:

$$\begin{eqnarray} \frac 2 {w+3} - \frac 5 {w-1} & = & \frac {\blue{(w-1)} \times 2}{\blue{(w-1)} \times (w+3)} - \frac {\blue{(w+3)} \times 5}{\blue{(w+3)} \times (w-1)} \\ & = & \frac {2w-2} {(w+3)(w-1)} - \frac {5w+15} {(w+3)(w-1)} \\ & = & \frac {(2w-2)-(5w+15)} {(w+3)(w-1)} \\ & = & \frac {-3w-17} {(w+3)(w-1)} = - \left( \frac {(3w+17)} {(w+3)(w-1)} \right) \end{eqnarray}$$

Click on questions to reveal their solutions

Quiz 3: Which of the following values of $\blue{a}$ is needed if:

$$\frac {\blue{a}} {2x+1} + \frac 1 {x+2} = \frac {4x+5} {(2x+1)(x+2)} ~ \text{?}$$
(a)$\blue{a} = 3$Incorrect - please try again!
(b)$\blue{a} = -3$ Incorrect - please try again!
(c)$\blue{a} = 2$ Correct - well done!
(d)$\blue{a} = -2$ Incorrect - please try again!

Explanation: Writing all the fractions with a common denominator:

$$\begin{eqnarray} \frac {4x+5} {(2x+1)(x+2)} & = & \frac {\blue a} {2x+1} + \frac 1 {x+2} \\ & = & \frac {\blue a(x+2)} {(2x+1)(x+2)} + \frac {(2x+1)} {(2x+1)(x+2)} \\ & = & \frac {\blue ax +2 \blue a +2x+1} {(2x+1)(x+2)} \\ & = & \frac {(\blue a+2)x +(2\blue a +1)} {(2x+1)(x+2)} \end{eqnarray}$$

so that $(\blue a+2)x+(2 \blue a+1) = 4x+5$. This gives two equations

$$\begin{eqnarray} \blue a+2 & = & 4 & ~~~\text{(coefficients of } x \text{)}\\ 2\blue a+1 & = & 5 & ~~~\text{(constant coefficients)} \end{eqnarray}$$

The solution to both equations is $\blue a = 2$.

# 3. Simple Partial Fractions

The last quiz was an example of partial fractions, i.e. the technique of decomposing a fractions as a sum of simpler fractions. This section will consider the simpler forms of this technique.

Example 5: Find the partial fraction decomposition of $4 / (x^2-4)$.

Solution: The denominator factorises as $x^2-4 = (x-2)(x+2)$ (see the module on Quadratics). The partial fractions will, therefore, be of the form $a / (x-2)$ and $b / (x+2)$. Thus:

$$\begin{eqnarray} \frac a {x-2} + \frac b {x+2} & = & \frac 4 {(x-2)(x+2)} \\ \frac {a(x+2)} {(x−2)(x+2)} + \frac {b(x−2)} {(x+2)(x−2)} & = & \frac 4 {(x-2)(x+2)} \\ \frac {(a+b)x+2(a−b)} {(x−2)(x+2)} & = & \frac 4 {(x-2)(x+2)} \\ \text{so that: } (a+b)x+2(a−b) & = & 4 \end{eqnarray}$$

This enables $a$ and $b$ to be found. For the equation to be true for all values of $x$ the coefficients must match, i.e.

$$\begin{eqnarray} a+b & = & 0 & ~~~\text{(coefficients of } x \text{)}\\ 2a-2b & = & 4 & ~~~\text{(constant coefficients)} \end{eqnarray}$$

- where the first equation holds since there is no $x$ term in $4 / (x^2-4)$. This set of simultaneous equations may be solved to give $a = 1$ and $b = -1$ (see the module on Simultaneous Equations for a method of finding these equations). Thus:

$$\frac 4 {(x-2)(x+2)} = \frac 1 {x-2} - \frac 1 {x+2}$$

Exercise 4: For each of the following, find $a$ and $b$:

(a)$\displaystyle \frac a {x-2} + \frac b {x+2} = \frac {4x} {(x-2)(x+2)}$

Solution: Taking common denominators:

$$\begin{eqnarray} \frac a {x-2} + \frac b {x+2} & = & \frac {4x}{(x-2)(x+2)} \\ \frac {a(x+2)} {(x-2)(x+2)} + \frac {b(x-2)} {(x-2)(x+2)} & = & \frac {4x}{(x-2)(x+2)} \\ \frac {(a+b)x + (2a-2b)} {(x-2)(x+2)} & = & \frac {4x}{(x-2)(x+2)} \\ \text{so that } (a + b)x + (2a − 2b) & = & 4x \end{eqnarray}$$

Equating coefficients in this case gives:

$$\begin{eqnarray} a+b & = & 4 & ~~~\text{(coefficients of } x \text{)}\\ 2a-2b & = & 0 & ~~~\text{(constant coefficients)} \end{eqnarray}$$

Solving this set of equations gives $a = 2$, $b = 2$. Hence:

$$\frac 2 {x-2} + \frac 2 {x+2} = \frac {4x} {(x-2)(x+2)}$$
(b)$\displaystyle \frac a {z-3} + \frac b {z+2} = \frac {z+7} {(z-3)(z+2)}$

Solution: Taking common denominators:

$$\begin{eqnarray} \frac a {z-3} + \frac b {z+2} & = & \frac {z+7}{(z-3)(z+2)} \\ \frac {a(z+2)} {(z-3)(z+2)} + \frac {b(z-3)} {(z-3)(z+2)} & = & \frac {z+7}{(z-3)(z+2)} \\ \frac {(a+b)z + (2a-3b)} {(z-3)(z+2)} & = & \frac {z+7}{(z-3)(z+2)} \\ \text{so that } (a + b)z + (2a − 3b) & = & z+7 \end{eqnarray}$$

Equating coefficients in this case gives:

$$\begin{eqnarray} a+b & = & 1 & ~~~\text{(coefficients of } x \text{)}\\ 2a-3b & = & 7 & ~~~\text{(constant coefficients)} \end{eqnarray}$$

Solving this set of equations gives $a = 2$, $b = -1$. Hence:

$$\frac 2 {z-3} - \frac 1 {z+2} = \frac {z+7} {(z-3)(z+2)}$$
(c)$\displaystyle \frac a {w-4} + \frac b {w+1} = \frac {3w-2} {(w-4)(w+1)}$

Solution: Taking common denominators:

$$\begin{eqnarray} \frac a {w-4} + \frac b {w+1} & = & \frac {3w-2}{(w-4)(w+1)} \\ \frac {a(w+1)} {(w-4)(w+1)} + \frac {b(w-4)} {(w-4)(w+1)} & = & \frac {3w-2}{(w-4)(w+1)} \\ \frac {(a+b)w + (a-4b)} {(w-4)(w+1)} & = & \frac {3w-2}{(w-4)(w+1)} \\ \text{so that } (a + b)w + (a − 4b) & = & 3w-2 \end{eqnarray}$$

Equating coefficients in this case gives:

$$\begin{eqnarray} a+b & = & 3 & ~~~\text{(coefficients of } x \text{)}\\ a-4b & = & -2 & ~~~\text{(constant coefficients)} \end{eqnarray}$$

Solving this set of equations gives $a = 2$, $b = 1$. Hence:

$$\frac 2 {w-4} + \frac 1 {w+1} = \frac {3w-2} {(w-4)(w+1)}$$

Click on questions to reveal their solutions

Now try this short quiz:

Quiz 4: If $\displaystyle \frac a {2x-3} + \frac b {3x+4} = \frac {x+7} {(2x-3)(3x+4)}$, which of the following is the solution to the equation?

(a)$a = 1$, $b = -1$Correct - well done!
(b)$a = -1$, $b = 1$Incorrect - please try again!
(c)$a = 1$, $b = 1$Incorrect - please try again!
(d)$a = -1$, $b = -1$ Incorrect - please try again!

Explanation: Writing all the fractions with a common denominator:

$$\begin{eqnarray} \frac a {2x-3} + \frac b {3x+4} & = & \frac {x+7} {(2x-3)(3x+4)} \\ \frac {a(3x+4)} {(2x-3)(3x+4)} + \frac {b(2x-3)} {(2x-3)(3x+4)} & = & \frac {x+7} {(2x-3)(3x+4)} \\ \frac {(3a+2b)x+(4a-3b)} {(2x-3)(3x+4)} & = & \frac {x+7} {(2x-3)(3x+4)} \\ \text{so that } (3a+2b)x + (4a−3b) & = & x+7 \end{eqnarray}$$

This gives two equations:

$$\begin{eqnarray} 3a+2b & = & 1 & ~~~\text{(coefficients of } x \text{)}\\ 4a-3b & = & 7 & ~~~\text{(constant terms)} \end{eqnarray}$$

Solving these gives: $a = 1$, $b = -1$.

# 4. Quiz on Algebraic Fractions

In each of the following choose...

1. ...the simplified form of $(z^2 + 4z - 5) / (z^2 - 4z + 3)$:
(a) $(z + 5) / (z + 3)$
(b) $(z + 5) / (z - 3)$
(c) $(z - 5) / (z + 3)$
(d) $(z - 5) / (z - 3)$
2. ...the difference $[1 / (w - 2)] - [1 / (w + 7)]$:
(a) $5 / (w^2 + 5w - 14)$
(b) $5 / (w^2 - 5w + 14)$
(c) $9 / (w^2 - 5w + 14)$
(d) $9 / (w^2 + 5w - 14)$
3. ... $a$ and $b$ if $[a / (3x + 2)] + [b / (4x − 3)] = (x − 5) / [(3x + 2)(4x − 3)]$:
(a) $a = 1$, $b = 1$
(b) $a = -1$, $b = 1$
(c) $a = -1$, $b = -1$
(d) $a = 1$, $b = -1$


text:
2019
Том 71
№ 4

# Volosivets S. S.

Articles: 1
Article (English)

### Fourier cosine and sine transforms and generalized Lipschitz classes in uniform metric

Ukr. Mat. Zh. - 2012. - 64, № 5. - pp. 616-627

For functions $f \in L^1(\mathbb{R}_{+})$ with cosine (sine) Fourier transforms $\widehat{f}_c(\widehat{f}_s)$ in $L^1(\mathbb{R})$, we give necessary and sufficient conditions in terms of $\widehat{f}_c(\widehat{f}_s)$ for $f$ to belong to generalized Lipschitz classes $H^{\omega, m}$ and $h^{\omega, m}$. Conditions for the uniform convergence of the Fourier integral and for the existence of the Schwartz derivative are also obtained.