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# The Unapologetic Mathematician
## The Higgs Mechanism part 3: Gauge Symmetries
This is part three of a four-part discussion of the idea behind how the Higgs field does its thing. Read Part 1 and Part 2 first.
Now we’re starting to get to the really meaty stuff. We talked about the phase symmetry of the complex scalar field:
\displaystyle\begin{aligned}\phi'(x)&=e^{i\alpha}\phi(x)\\\phi'^*(x)&=e^{-i\alpha}\phi^*(x)\end{aligned}
which basically wants to express the idea that the physics of this field only really depends on the length of the complex field values $\phi(x)$ and not on their phases. But another big principle of physics is locality — what happens here doesn’t instantly affect what happens elsewhere — so why should the phase change be global?
To answer this, we “gauge” the symmetry and make it local. The origin of the term is fascinating, but takes us too far afield. The upshot is that we now have the symmetry transformation:
\displaystyle\begin{aligned}\phi'(x)&=e^{i\alpha(x)}\phi(x)\\\phi'^*(x)&=e^{-i\alpha(x)}\phi^*(x)\end{aligned}
where $\alpha$ is no longer a constant, but a function of the spacetime point $x$.
And here’s the big problem: since $\alpha$ varies from point to point, it now affects our derivative terms! Before we had
\displaystyle\begin{aligned}\partial_\mu\phi'(x)&=\partial_\mu\left(e^{i\alpha}\phi(x)\right)\\&=e^{i\alpha}\partial_\mu\phi(x)\end{aligned}
and similarly for $\phi^*$. We say that the derivatives are “covariant” under the transformation; they transform in the same way as the underlying fields. And this is what lets us say that
$\displaystyle\partial_\mu\phi'^*\partial_\mu\phi'=\partial_\mu\phi^*\partial_\mu\phi$
and makes the whole Lagrangian symmetric.
On the other hand, what do we see now?
\displaystyle\begin{aligned}\partial_\mu\phi'(x)&=\partial_\mu\left(e^{i\alpha(x)}\phi(x)\right)\\&=e^{i\alpha(x)}\partial_\mu\phi(x)+i\partial_\mu\alpha(x)e^{i\alpha(x)}\phi(x)\\&=e^{i\alpha(x)}\left[\partial_\mu\phi(x)+i\partial_\mu\alpha(x)\phi(x)\right]\end{aligned}
We pick up this extra term when we differentiate, and it ruins the symmetry.
The way out is to add another field that can “soak up” this extra term. Since the derivative is a vector, we introduce a vector field $A_\mu$ and say that it transforms as
$\displaystyle A_\mu'(x)=A_\mu(x)+\frac{1}{e}\partial_\mu\alpha(x)$
Next, we introduce a new derivative operator: $D_\mu=\partial_\mu-ieA_\mu$. That is:
$\displaystyle D_\mu\phi(x)=\partial_\mu\phi(x)-ieA_\mu(x)\phi(x)$
And we calculate
\displaystyle\begin{aligned}D_\mu\phi'(x)&=\partial_\mu\left(e^{i\alpha(x)}\phi(x)\right)-ieA_\mu'(x)e^{i\alpha(x)}\phi(x)\\&=e^{i\alpha(x)}\partial_\mu\phi(x)+i\partial_\mu\alpha(x)e^{i\alpha(x)}\phi(x)-ieA_\mu(x)e^{i\alpha(x)}\phi(x)-i\partial_\mu\alpha(x)e^{i\alpha(x)}\phi(x)\\&=e^{i\alpha(x)}\left[\partial_\mu\phi(x)-ieA_\mu(x)\phi(x)\right]\\&=e^{i\alpha(x)}D_\mu\phi(x)\end{aligned}
So the derivative $D_\mu\phi(x)$ does vary the same way as the underlying field $\phi(x)$ does! We call $D_\mu$ the “covariant derivative”. If we use it in our Lagrangian, we do recover our symmetry, though now we’ve got a new field $A_\mu$ to contend with. Just like the electromagnetic potential we use the derivative $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ to write
$\displaystyle L=-\frac{1}{4}F_{\mu\nu}F_{\mu\nu}+(D_\mu\phi)^*D_\mu\phi-m^2\phi^*\phi$
which is now symmetric under the gauged symmetry transformations.
It may not be apparent, but this Lagrangian does contain interaction terms. We can expand out the second term to find:
\displaystyle\begin{aligned}(D_\mu\phi)^*D_\mu\phi&=\left(\partial_\mu\phi^*-ieA_\mu\phi^*\right)\left(\partial_\mu\phi-ieA_\mu\phi\right)\\&=\partial_\mu\phi^*\partial_\mu\phi-ieA_\mu\partial_\mu\phi^*\phi-ieA_\mu\phi^*\partial_\mu\phi-e^2A_\mu A_\mu\phi^*\phi\end{aligned}
Our rules of thumb tell us that if we vary the Lagrangian with respect to $A_\mu$ we get the field equation
$\displaystyle\partial_\mu F_{\mu\nu}=ej_\mu$
which — if we expand out $F_{\mu\nu}$ as if it’s the Faraday field into “electric” and “magnetic” fields — give us Gauss’ and Ampère’s law in the presence of a charge-current density $j_\mu$.
The charge-current, in particular, we can write as
$\displaystyle j_\mu=-i\left(\phi^*\partial_\mu\phi-\partial_\mu\phi^*\phi\right)-2eA_\mu\phi^*\phi$
or, in a gauge-invariant manner, as
$\displaystyle j_\mu=-i\left[\phi^*D_\mu\phi-(D_\mu\phi)^*\phi\right]$
which is just the conserved current from last time with the regular derivatives replaced by covariant ones. Similarly, varying with respect to the field $\phi$ we find the “covariant” Klein-Gordon equation:
$\displaystyle D_\mu D_\mu\phi+m^2\phi=0$
and, when this holds, we can show that $\partial_\mu j_\mu=0$.
So we’ve found that if we take the global symmetry of the complex scalar field and “gauge” it, something like electromagnetism naturally pops out, and the particle of the complex scalar field interacts with it like charged particles interact with the real electromagnetic field.
July 18, 2012 -
1. That derivative operator is reminiscent of canonical momentum. Are they related?
Comment by Bluescreenofdebt | July 18, 2012 | Reply
2. When quantizing, the derivative with respect to a coordinate is indeed related to the “momentum operator” conjugate to that coordinate. I’m sticking to classical field theory here, though.
Comment by John Armstrong | July 18, 2012 | Reply
3. What does this “covariant derivative” have to do with the covariant derivative as usually defined on a bundle with a connection over a manifold? As far as I can tell, you’re still working in Lorentzian space-time.
Comment by Greg Friedman | July 18, 2012 | Reply
4. That is an excellent question, Greg. I’ll start answering it by saying we’re actually working over Lorentzian spacetime, in that our fields are the sections of some fiber bundle over this space, and in this case the gauge group is $U(1)$. The $A_\mu$ fields are actually the components — analogous to the Christoffel symbols — of a connection on the associated principal fiber bundle.
Comment by John Armstrong | July 18, 2012 | Reply
5. Incidentally, that’s why there’s the $i$ in the covariant derivative formula: $iA_\mu$ is really a $\mathfrak{u}(1)$-valued field.
Comment by John Armstrong | July 18, 2012 | Reply
6. […] discussion of the idea behind how the Higgs field does its thing. Read Part 1, Part 2, and Part 3 […]
Pingback by The Higgs Mechanism part 4: Symmetry Breaking « The Unapologetic Mathematician | July 19, 2012 | Reply
7. […] “The Higgs Mechanism part 2: Examples of Lagrangian Field Equations,” July 17, “The Higgs Mechanism part 3: Gauge Symmetries,” July 18, y “The Higgs Mechanism part 4: Symmetry Breaking,” July 19. […]
Pingback by Las matemáticas del bosón de Higgs, para las abuelas cansadas de cháchara (Parte I) « Francis (th)E mule Science's News | July 19, 2012 | Reply
8. […] The Higgs Mechanism part 3: Gauge Symmetries […]
Pingback by The Unapologetic Mathematician Tackles the Higgs Mechanism | Whiskey…Tango…Foxtrot? | July 20, 2012 | Reply
9. […] “The Higgs Mechanism part 2: Examples of Lagrangian Field Equations,” July 17, “The Higgs Mechanism part 3: Gauge Symmetries,” July 18, y “The Higgs Mechanism part 4: Symmetry Breaking,” July 19. Recomiendo […]
Pingback by Las matemáticas del bosón de Higgs, para las abuelas cansadas de cháchara (Parte I) | Bosón de Higgs | La Ciencia de la Mula Francis | November 1, 2017 | Reply
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MOS: Using Microsoft Office Excel 2007
#### Question No: 21 HOTSPOT
You work as an office assistant for Brain Brigade Inc. You are creating an excel sheet. You use the fill handle quite often to fill or copy series of values to cells. Every time you use the fill handle, the Auto Fill button appears. You do not want to display the Auto Fill button. You
open the Excel Options window. Mark the check box that you will clear to accomplish the task.
Explanation:
In order to accomplish the task, you will clear the Show Paste Options buttons check box in the Cut, copy, and paste section of the Advanced Excel Options window.
#### Question No: 22
You work as a Sales Manager for Maini Industries. Your company manufactures and sells construction equipment. You have to create a report that will show the highest selling items of each month. The report will be in the following format:
For the third column of the sheet, you want to use formulas to fill in the values. You want the cells in the third column to get automatically filled with the names corresponding to ItemID in the second column. You want the exact values to be filled in the column. You select the C2 cell in the third column.
Which of the following formulas will you enter in order to accomplish the task?
A. =VLOOKUP(2,Item_table,B2,TRUE)
B. =VLOOKUP(2,Item_table,B2,FALSE)
C. =VLOOKUP(B2,Item_table,2,TRUE)
D. =VLOOKUP(B2,Item_table,2,FALSE)
In order to accomplish the task, you will use the following formula:
=VLOOKUP(B2,Item_table,2,FALSE)
Answer option C is incorrect. According to the question, you want the exact values in the column.
The TRUE value in the range_lookup argument provides the approximate match.
Answer options B and A are incorrect. The VLOOKUP functions used in the formulas have wrong syntaxes.
Reference: Microsoft Excel 2007 Help, Contents: quot;VLOOKUPquot;
#### Question No: 23
Alan works as a Sales Manager for MVP Technologies. He uses Microsoft Excel XP to perform official tasks. He prepares a workbook that contains names of Sales Executives and sales (in dollar) target achieved by them in two consecutive months. The first worksheet of the workbook contains data of the first month and the second worksheet contains data of the second month. The sales values are contained in cell B2 through cell B5 in the worksheets 1 and 2. He wants to find the total sales occurred in the two consecutive months. Which of the following cell references will he use to accomplish the task?
1. Relative cell reference
2. Absolute cell reference
3. Mixed cell reference
4. 3-D cell reference
Alan will use 3-D cell reference to accomplish the task.
A 3-D reference in Microsoft Excel is a reference that refers to the same cell or range of cells on two or more worksheets in a workbook. It creates link among the same cell or range of cells existing on different worksheets in a workbook to manipulate data. For example, the SUM function can be used to add the values of cell A1 existing on different worksheets in a workbook.
Take the following steps to refer to the same cell or range on multiple worksheets:
1.Click the cell where function has to be entered in a worksheet.
Answer option B is incorrect. An absolute cell reference refers to a cell in a specific location in a formula. When a formula that uses the absolute cell reference is copied from one cell to another cell, the absolute cell reference remains unchanged. The #39;\$#39; symbol is inserted before the row and the column of the cell reference to identify the absolute cell reference in a formula. For example, the formula =\$A\$1*\$A\$2 in cell A3 is written by using the absolute cell Reference. When this formula is copied from cell A3 to cell B3, the absolute cell reference remains unchanged and the formula in cell B3 also remains the same, i.e.
=\$A\$1*\$A\$2.
#### Question No: 24
You work as a Sales Manager for Rainbow Inc. Your responsibility includes preparing sales report of the Sales department. You use Excel 2007 to prepare sales reports. You have created a quarterly sales report of the department. After entering required data and inserting charts, you want to give a professional look to the document. You want to produce the document with soothing graphic effects, soft fonts, and light colors.
Which of the following steps will you take to accomplish the task with least administrative burden?
1. Use the Smart Art option.
2. Create a new workbook through a template.
Copy all data and charts of the report to the new workbook.
3. Choose one of the pre-built themes.
4. Select all the text of the document.
Change font style to Verdana and font color to light gray. Use vibrant colors for charts.
In order to accomplish the task, you will have to choose one of the pre-built themes. Document themes work on the overall design of the entire document. It enables users to specify colors, fonts, and variety of graphic effects in a document. Themes change the look and feel of the document. Excel 2007 comes with a large collection of pre-built themes.
Users can create their own themes for a document. Themes can be specified through Page Layout gt; Themes.
Answer option A is incorrect. The Smart Art option does not produce soothing graphic effects, soft fonts and light colors.
#### Question No: 25
You work as a Help Desk Technician for Dreams Unlimited Inc. Martha, a Sales Manager, is creating a sales report in Microsoft Excel 2007. The report contains many worksheets. Martha has used many formulas in her report. She wants to monitor a cell which contains a formula. The formula refers to many other cells that are located in different worksheets in
the report. She wants to know the effects on the cell when the values in different sheets changes. She requests you to help her out to accomplish the task. Which of the following steps will you advice her to accomplish the task?
1. Click the New Window option in the Window group on the View tab. Click the Arrange All option in the Window group on the View tab. Select the Cascade option.
2. Select the cell which is to be monitored.
On the Formulas tab in the Formula Auditing group, click Watch Window. Click Add Watch.
3. Click the New Window option in the Window group on the View tab. Click the Arrange All option in the Window group on the View tab. Select the Vertical option.
4. Select the cell which is to be monitored.
On the Formulas tab in the Formula Auditing group, click Evaluate Formula.
In order to accomplish the task, she should take the following steps:
Select the cell which is to be monitored.
On the Formulas tab in the Formula Auditing group, click Watch Window. Click Add Watch. What is the Watch Window?
The Watch Window is a toolbar available in Excel 2007 to monitor the value in a cell, regardless of its location in a workbook. The Watch Window can be moved or docked like other toolbars available. It keeps track of the following properties of a cell: workbook, sheet, name, cell, value, and formula. The Watch Window provides users a convenient way to inspect, audit, or confirm formula calculations and results in large worksheets/workbooks.
Only one watch per cell can be configured in the Watch Window. To set a watch, click the cell to be monitored and then, on the Formulas tab, in the
Formula Auditing group, click Watch Window. Click Add Watch to monitor the selected cell. Answer option D is incorrect. Evaluate Formula is the formula examination tool provided by Excel 2007. This tool is useful for examining formulas that do not produce an error but are not generating the expected result.
Answer options C and A are incorrect. Clicking the New Window option in the Window group on the View tab opens an instance of the workbook within the Excel window. This is useful when you work on different worksheets, and at the same time you want to navigate among them.
Reference: Windows Excel 2007 Help, Contents: quot;Watch a formula and its result by using
the Watch Windowquot;
#### Question No: 26
You work as a Sales Manager for Dreams Unlimited Inc. Your computer runs Windows Vista Home Premium. You have recently installed Microsoft Office 2007 on your computer. You use Excel to create your reports. As the interface has been changed in the new version, you often use Excel 2007 Help to find out your required commands.
Sometimes, when you look at the Ribbon, you might not see all of the icons or the text that is used in some Help topics to describe the commands on the Ribbon.
What is the most likely cause?
1. The Home Premium edition of Vista does not support some of the Office 2007 commands.
2. The Help file of the Office suite is not current.
3. You are using Beta version of Excel 2007.
4. The program window is shrinked.
This is a common phenomenon for programs in Office 2007 suite. The size of the Ribbon is optimized for a screen resolution of 1024 786 pixels when the Microsoft Office program is maximized on your screen. The Help topics were written by using that standard size for the Ribbon.
Reference. Microsoft Excel 2007 Help, Contents: quot;I can#39;t see a command on the Ribbonquot;
#### Question No: 27
You work as an Office Assistant for Tech Perfect Inc. You have created a complex workbook on Excel 2007. The workbook is a consolidated sales report that contains sales data for the last five years. This report will be circulated to all the company managers having different versions of Excel. You have to ensure the following:
The optimal performances of the workbook backward compatibility with earlier versions of Excel which of the following steps will you take to accomplish the task? Each correct answer represents a part of the solution. Choose two.
1. Use the Run Compatibility Checker option through the Office button gt; Prepare.
2. Save the file in the Excel 2007 Binary file format.
3. Use the Inspect Document option through the Office button gt; Prepare.
4. Save the file in the Excel 2007 XML-based file format.
In order to accomplish the task, you will have to take the following steps:
Use the Run Compatibility Checker option through the Office button gt; Prepare. This option will run Compatibility Checker, which will show you a report regarding compatibility of the document with earlier versions of Excel.
#### Question No: 28
You work as an Office Assistant for Dreams Unlimited Inc. You use Excel 2007 for creating various types of reports. You have created a report in the format given below:
In the A7 cell, you are required to put a formula so that it can fulfill the description provided in the B7 cell. Which of the following formulas will provide the required result?
1. COUNTIF(B2:B5,quot;=Yesquot;,C2:C5,quot;=Yesquot;)
2. COUNTIFS(B2:B5,quot;=Yesquot;,C2:C5,quot;=Yesquot;)
3. COUNTIF(B2:C5,quot;=Yesquot;)
4. COUNTIFS(B2:C5,quot;=Yesquot;)
In order to get the required result, you will have to insert the following formula in the B7 cell: COUNTIFS(B2:B5,quot;=Yesquot;,C2:C5,quot;=Yesquot;)
Only Sarah and David have exceeded their January and February quotas, therefore the formula will provide 2 as the result.
Answer option A is incorrect. The COUNTIF function of Excel 2007 does not support multiple criteria.
Answer options C and D are incorrect. This formula will count all cells that have the value quot;Yesquot; in the range B2:C5. As multiple criteria are not applied in the formula, it will provide 6 as the result.
Reference. Microsoft Excel 2007 Help, Contents: quot;COUNTIFquot;
#### Question No: 29
You work as an Office Assistant for Peach Tree Inc. Your responsibilities include creating sales reports of the company. You have created a report in a workbook in Excel 2007. One
of the worksheets in the report contains some formulas. In order to finally verify formulas used in the worksheet, you have displayed the formulas as shown in the image given below:
You select the E3 cell and want to know the current value of the formula in the cell. You do not want to hide formulas in the sheet. Which of the following steps will you take to accomplish the task with least administrative burden?
1. Click the Evaluate Formula option in the Formula Auditing group on the Formulas tab. Click the Evaluate button.
2. Click the Calculate Now option in the Calculation group on the Formulas tab.
3. Click the Watch Window option in the Formula Auditing group on the Formulas tab. Click the Add Watch option and select the E3 cell.
4. Watch the value displayed at the status bar.
According to the image displayed in the question, formulas are displayed instead of values in cells. You are required to find the result of a formula in the E3 cell that you have already selected. In order to accomplish the task with minimum administrative burden, you will have to take the following steps:
Click the Evaluate Formula option in the Formula Auditing group on the Formulas tab. This will open the Evaluate Formula dialog box.
Answer option C is incorrect. Although these steps will also accomplish the task, the administrative burden will be more.
Answer option D is incorrect. The Status bar does not evaluate the formula.
Answer option B is incorrect. The Calculate Now option is used to calculate the entire worksheet at the current point of time. It is very useful for calculating values when automatic calculation is turned off. This step will not accomplish the task, as it displays the calculated value in their respective places when the Show Formula option is disabled.
Reference. Microsoft Excel 2007 Help, Contents: quot;Evaluate a nested formula one step at a timequot;
#### Question No: 30
You work as an Office Assistant for Blue Well Inc. You have created a workbook in Excel 2007. This workbook will be shared with other users who will collaborate to enter data in the workbook. You want to enable Excel in such a manner that if you have a question regarding the editing done by any collaborator, you can identify who made the changes and verify that it is correct. Which of the following steps will you take to accomplish the task?
Each correct answer represents a complete solution. Choose two.
1. Enable the Show Comments feature.
2. Turn on the Change Tracking feature.
3. Share the workbook.
4. Add a History worksheet to the workbook.
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# How Much Caffeine In Two Shots Of Espresso?
It is reasonable to believe that a single shot of espresso contains around 75 milligrams of caffeine. Remember that a double shot is also considered a typical dose, and therefore a beverage that contains a double shot of espresso has around 150 milligrams of caffeine in it. Comparing the Effects of Caffeine on the Body
Two shots constitute a single serving of espresso in the vast majority of coffee establishments. The caffeine content of these two shots, which is around 150 milligrams, is actually lower than the caffeine content of a standard 16-ounce cup of coffee, which is 330 milligrams (via Huff Post).
## How much caffeine is in a double shot of espresso?
A cup of coffee will have around one-third of the amount of caffeine that a double shot of espresso will have.The amount of caffeine that is contained in coffee is strongly influenced by a number of factors, including the style of roast, the grind, the variety of beans used, and the quality of the beans.As a direct consequence of this, some cups of coffee (or shots of espresso) contain a higher concentration of caffeine than others.
## Does espresso have more caffeine than coffee?
There are 64 mg of caffeine in a shot of espresso that is typically one ounce in size.One thing is now abundantly clear: the amount of caffeine that you get from a shot of espresso is far lower than the amount that you get from a cup of coffee.Are you going to have a double shot rather than a single one?A cup of coffee will have around one-third of the amount of caffeine that a double shot of espresso will have.
## How much caffeine is in a cup of coffee?
A normal cup of coffee has between 70 and 140 milligrams of caffeine, whereas an average cup of coffee containing 8 ounces (236 milliliters) contains between 7 and 14 milligrams. While decaf coffee has 12-13.4mg of caffeine per 16-ounce (473ml) drink, a shot of espresso can have anywhere from 3-15.8mg of the stimulant.
## How much caffeine is in a cup of espresso at Starbucks?
It was discovered that each cup of Starbucks’ espresso contains somewhere between 63 and 91 milligrams of caffeine. Therefore, it is estimated that 7mg for a single serving and 14g for a double serving of coffee bean mix contains caffeine.
## How much caffeine is in a double shot espresso?
Caffeine in Espresso (by Preparation Method) Espresso, Single Shot: 29-100 mg (often around 75 mg) Doppio (double shot) of espresso ranges from 58 to 185 mg (often around 150 mg)
## How much caffeine is in each shot of espresso?
According to the nutrition statistics provided by the Department of Agriculture, there are 63 mg of caffeine included inside 1 ounce (the quantity found in one shot) of espresso. On the other hand, one ounce of regular coffee typically contains somewhere between 12 and 16 milligrams of caffeine.
## How much coffee is 2 shots of espresso at Starbucks?
Two shots of espresso are included in a Venti, which is a 20-ounce cup of coffee from Starbucks (about 150 milligrams of caffeine). Surprisingly, the same volume of espresso can be found in both the 12-ounce Tall and the 16-ounce Grande.
## Is 3 shots of espresso a lot of caffeine?
Reduce your consumption of espresso if you want to prevent feeling agitated.The United States Department of Agriculture (USDA) reports that the average serving size of espresso, which is 2 ounces, contains 128 mg of caffeine.If you are permitted up to 400 milligrams of caffeine per day, it is equivalent to around three shots of espresso or four cups of black coffee that are each 8 ounces in size.
## Is 150 mg of caffeine a lot?
It would appear that healthy persons may consume up to 400 milligrams (mg) of caffeine daily without any adverse effects. As’s almost the same amount of caffeine that you’d find in ten cans of cola, four cups of brewed coffee, or two ‘energy shot’ beverages.
## How much caffeine is in one shot of espresso at Starbucks?
One shot of espresso contains around 75 mg of caffeine.
## Is espresso healthier than regular coffee?
Reduces the Risk of Developing Diabetes Additionally, it is a much healthier alternative to the typical kinds of coffee that a lot of people consume during the course of the day.Espresso may be consumed on its own, without the addition of sweets or creams, so cutting off the additional calories and fats that would otherwise be consumed.You are able to achieve that spike in energy without putting your health at risk.
## Why can I drink espresso but not coffee?
Ask for Espresso To begin, the use of high pressure in conjunction with a short extraction period results in a different balance of chemical compounds being extracted from the coffee than would be produced by brewing the same coffee using a drip or pour over method. In spite of the pressure that is being applied, the extraction process is not very effective overall.
## Is 300 mg of caffeine a lot?
Cell receptors can also be inherited differently from person to person. You should limit yourself to modest doses of caffeine for the time being. That translates to no more than 300 milligrams a day for an adult, which is equivalent to three cups of coffee with a capacity of six ounces, four cups of regular tea, or six colas of 12 ounces each.
## Is 120 mg of caffeine a lot?
The Mayo Clinic suggests that healthy persons can drink up to 400 milligrams (mg) of caffeine on a daily basis without any adverse effects. But although the majority of 12-ounce cups of coffee have between 90 and 120 mg of caffeine, a single 12-ounce ‘tall’ or small cup of Starbucks has over 260 mg of caffeine per cup, making it far more potent.
## How much caffeine is in a double shot of coffee?
Because of this, a double shot of espresso typically includes around 125 milligrams of caffeine.
## Will a double shot of espresso wake me up?
A injection of caffeine may be helping you wake up, but there’s more to it than that. Those who are too busy to carry a whole coffee around with them all afternoon will discover that two espresso shots are more efficient than their normal cup of drip coffee, and they also go down much more quickly.
## How many shots of espresso is lethal?
Consuming between 52 and 105 cups of coffee or 76 to 156 shots of espresso in a single day has the potential to be fatal, however this number varies depending on the individual. Drinking no more than six espresso shots or four cups of coffee a day is recommended by industry professionals as the safest way to reap the benefits of coffee while minimizing the potential for negative reactions.
## How many shots of espresso a day is okay?
The maximum daily dose of caffeine recommended by the EFSA report is 400 milligrams, which is equivalent to five shots of espresso. If you consume more than this amount of caffeine on a daily basis, you put yourself at risk for over-consumption of caffeine and the health problems that are associated with it.
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Alberto Dell’Era’s Oracle blog » overlapping ranges with priority
### overlapping ranges with priority
A customer of ours (a leading Italian consumer goods retailer) has asked us to solve the following problem, that occurs quite frequently and that is not trivial to solve efficiently - and that is very interesting to design and fun to blog about!
the problem
Prices of skus (i.e. goods) have validity ranges (time intervals) and can overlap; on an overlapping range, the strongest priority (lower number) wins. In pictures:
```b---(0,\$200)---d
c---(1,\$300)---e
```
the expected output is, since priority 0 is stronger then 1:
```b----(\$200)----d---(\$300)---e
```
I'm going to illustrate in detail the pure-SQL algorithm I have designed, and then discuss about its performance and efficiency. As usual, everything I discuss is supported by an actual demo. Please note that the algorithm uses analytics functions very, very heavily.
pure SQL solution
The input table:
```create table ranges (
sku varchar2(10) not null,
a int not null,
b int not null,
prio int not null,
price int not null
);
alter table ranges add constraint ranges_pk primary key (sku, a, b);
```
Let's provide the opening example as input:
```insert into ranges(sku, a, b, prio, price) values ('sku1', ascii('b'), ascii('d'), 0, 200);
insert into ranges(sku, a, b, prio, price) values ('sku1', ascii('c'), ascii('e'), 1, 300);
```
The algorith is implemented as a single view; let's comment each step and show its output over the example:
```create or replace view ranges_output_view
as
```
The instants in time where the ranges start or begin:
```with instants as (
select sku, a as i from ranges
union
select sku, b as i from ranges
),
```
Output: b,c,d,e.
The base ranges, i.e. the consecutive ranges that connect all the instants:
```base_ranges as (
select *
from (
select sku,
i as ba,
lead(i) over (partition by sku order by i) as bb
from instants
)
where bb is not null
),
```
` b------c------d------e `
The original ranges factored over the base ranges; in other words, "cut" by the instants:
```factored_ranges as (
select i.sku, bi.ba, bi.bb, i.a, i.b, i.prio, i.price
from ranges i, base_ranges bi
where i.sku = bi.sku
and (i.a <= bi.ba and bi.ba < i.b)
),
```
```b---(0,\$200)---c---(0,\$200)---d
c---(1,\$300)---d---(1,\$300)---e
```
Then, let's filter out the factored ranges with weaker priority (that have a stronger priority range with the same extremes "covering" them):
```strongest_factored_ranges as (
select sku, ba, bb, prio, price
from (
select sku, ba, bb, prio, price,
dense_rank () over (partition by sku, ba, bb order by prio) as rnk
from factored_ranges
)
where rnk = 1
),
```
```b---(0,\$200)---c---(0,\$200)---d---(1,\$300)---e
```
The problem could be now considered solved, if you could live with consecutive intervals showing the same price (such as b--c and c--d above). If you can't for whatever reason (I couldn't), we can join them using analytics again in a way similar to this asktom technique (look at the bottom for "Analytics to the Rescue (Again)").
First, we calculate "step", a nonnegative number that will be zero if a range can be joined to the previous one, since:
a) they are consecutive (no gap between them)
b) they have the same price:
```ranges_with_step as (
select sku, ba, bb, prio, price,
decode ( price, lag(price) over (partition by sku order by ba), ba - lag(bb) over (partition by sku order by ba), 1000 ) step
from strongest_factored_ranges
),
```
```RANGE_CODED STEP
------------------------ ----------
b---(0,\$200)---c 1000
c---(0,\$200)---d 0
d---(1,\$300)---e 1000
```
Then we compute the integral of step over the ranges; joinable ranges will hence have the same value for "interval" since step is zero:
```ranges_with_step_integral as (
select sku, ba, bb, prio, price, step,
sum(step) over (partition by sku order by ba rows between unbounded preceding and current row) as integral
from ranges_with_step
),
```
```RANGE_CODED INTEGRAL
------------------------ ----------
b---(0,\$200)---c 1000
c---(0,\$200)---d 1000
d---(1,\$300)---e 2000
```
The joined joinable ranges :
```ranges_joined as (
select *
from (
select sku, ba, bb, prio, price, step, integral,
min(ba) over (partition by sku, integral) as a,
max(bb) over (partition by sku, integral) as b
from ranges_with_step_integral
)
where step > 0
)
select sku, a, b, price from ranges_joined;
```
```b---(0,\$200)---c---(1,\$300)---e
```
predicate-"pushability"
The first desirable property of this view is that a predicate (such as an equality predicate, but it works even for the "between" operator, less-than, etc) on sku can be pushed down the view to the base tables. For:
```select * from ranges_output_view where sku = 'k100';
```
the plan is:
```----------------------------------------------------------
| Id | Operation | Name |
----------------------------------------------------------
...
| 10 | TABLE ACCESS BY INDEX ROWID| RANGES |
|* 11 | INDEX RANGE SCAN | RANGES_PK |
...
|* 17 | INDEX RANGE SCAN | RANGES_PK |
|* 18 | INDEX RANGE SCAN | RANGES_PK |
----------------------------------------------
---
11 - access("I"."SKU"='k100')
---
17 - access("SKU"='k100')
18 - access("SKU"='k100')
```
That means that only the required SKU(s) are fed to the view, and proper indexes (such as RANGES_PK in this case) can be used. So, if you need to refresh only a few skus the response time is going to be almost istantaneous - provided that you have only sane (a few) ranges per sku. Hence you can use the same view for both calculating prices of all skus (say, in a nightly batch) and calculating a small subset of skus (say, online), and that is a great help for maintenance and testing.
running in parallel
Another desirable property is that the view can operate efficiently in parallel, at least in 11.2.0.3 (I have not tested other versions):
```-------------------------------------------------------------------
| Operation |IN-OUT| PQ Distrib |
-------------------------------------------------------------------
| SELECT STATEMENT | | |
| PX COORDINATOR | | |
| PX SEND QC (RANDOM) | P->S | QC (RAND) |
| VIEW | PCWP | |
| WINDOW SORT | PCWP | |
| VIEW | PCWP | |
| WINDOW SORT | PCWP | |
| VIEW | PCWP | |
| WINDOW BUFFER | PCWP | |
| VIEW | PCWP | |
| WINDOW SORT PUSHED RANK | PCWP | |
| HASH JOIN | PCWP | |
| PX RECEIVE | PCWP | |
| PX SEND HASH | P->P | HASH |
| PX BLOCK ITERATOR | PCWC | |
| TABLE ACCESS FULL | PCWP | |
| PX RECEIVE | PCWP | |
| PX SEND HASH | P->P | HASH |
| VIEW | PCWP | |
| WINDOW SORT | PCWP | |
| PX RECEIVE | PCWP | |
| PX SEND HASH | P->P | HASH |
| VIEW | PCWP | |
| SORT UNIQUE | PCWP | |
| PX RECEIVE | PCWP | |
| PX SEND HASH | P->P | HASH |
| UNION-ALL | PCWP | |
| PX BLOCK ITERATOR | PCWC | |
| INDEX FAST FULL SCAN| PCWP | |
| PX BLOCK ITERATOR | PCWC | |
| INDEX FAST FULL SCAN| PCWP | |
-------------------------------------------------------------------
```
There's no point of serialization (all servers communicate parallel-to-parallel), the rows are distributed evenly using an hash distribution function (probably over the sku) and all operations are parallel.
sku subsetting and partitioning
It is well known that analytics functions use sort operations heavily, and that means (whether or not you are running in parallel) that the temporary tablespace may be used a lot, possibly too much - as it actually happened to me , leading to (in my case) unacceptable performance.
Side note: as I'm writing this, I realize now that I had probably been hit by the bug illustrated by Jonathan Lewis in Analytic Agony, but of course, overuse of temp could happen, for large datasets, without the bug kicking in.
A possible solution is to process only a sub-batch of the skus at a time, to keep the sorts running in memory (or with one-pass to temp), leveraging the predicate-pushability of the view. In my case, I have made one step further: I have partitioned the table "ranges" by "sku_group", replaced in the view every occurrence of "sku" with the pair "sku_group, sku", and then run something like:
``` for s in (select sku_group from "list of sku_group") loop
select .. from ranges_output_view where sku_group = s.sku_group;
end loop;
```
The predicate gets pushed down to the table, hence partition elimination kicks in and I can visit the input table one time only, one partition at a time, using a fraction of the resources at a time, and hence vastly improving performance.
And that naturally leads to "do-it-yourself parallelism": running a job for every partition in parallel. I'm going to implement it since the customer is salivating about it ... even if it is probably over-engineering :D .
1. #### Overlapping ranges with priority | An Oracle Programmer
Saturday, March 22, 2014
[...] few years ago, Alberto Dell’Era blogged about product prices with overlapping date ranges; “on an overlapping range, the strongest priority (lower number) wins.” His analysis, [...]
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# cars, trucks, or bicycles. There are twice as many trucks as bicycles, and twice as many cars as trucks.Let t represent the number of trucks in the parking lot.Which equation could be used to find how many trucks are in the parking lot?Choose 1 answerA. t + 2t + 2t = 84B. 1/2t + t +2t =84C. t + 2t + 4t = 84D. t + 2t + 3t = 84Solution details:STATUS Answered QUALITY Approved ANSWER RATING This question was answered on: Apr 19, 2020 PRICE: \$15 Solution~000.zip (25.37 KB) Buy this answer for only: \$15 This attachment is locked × Please Enter The Email Where You Want To Receive Solution. Get this solution for only: \$ ">
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Our calculator is completely online and includes a calculator of pip values, a swap calculator and a margin calculator. These are at your disposal for all transactions.
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A pip is the smallest variation unit in an exchange rate. Currency pairs are usually listed in 4 decimals
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Now that you know the meaning of one pip, we can proceed to calculating its value.
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In the case of USD/JPY, a pip is worth 0.01 when the listing is in 2 decimals.
Let’s suppose that USD/JPY is listed as 90.68 and that we are trading with a position of 100 000.
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Value of one pip (EUR) = 11.03 / 1.4018 = 7.86 EUR (if EUR/USD lists as 1.4018).
Let’s take another example in EUR/USD. The smallest unit of measurement is 0.0001 since the listing is in four decimals.
Let’s suppose that EUR/USD lists as 1.4018 and that we are trading with a position of 100 000.
Value of one pip (USD) = 100 000*0.0001 = 10 USD
The value of one pip is always calculated in the listed currency (here USD) to then be converted into the default currency of your account (EUR).
To determine the amount in dollars, we then need to calculate the following:
Value of one pip (EUR) = 10/1.4018= 7.13 EUR
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The value of one pip changes depending on the exchange rate between of the given currency pair.
## What is a “lot”?
On Forex, the values of positions are expressed in lots. It is the unit of investment on the market. The value of a lot, its standard size, is 100 000, though certain brokers, such as Instaforex, deal with lots of 10.000.
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In fact, the value of a lot is related to the value of a pip which, as you now know, is the smallest unit of a listing. It is therefore necessary to trade with larger position values in order to make your gains or losses significant at all. The higher the value of your position, the more you’ll be able to earn on the pip.
Let’s take an example: You decide to trade with a position of 0.1 lot (i.e. 10 000 EUR) on EUR/USD listed as 1.4018.
Value of one pip = (0.0001/1.4018) * 10 000EUR= 0.7133 EUR
This means that when EUR/USD appreciates or loses by a pip, you gain or lose depending on the direction of your position 0.7133 EUR.
Example – Case study:
EUR/USD lists as 1.4018 / 1.4020. You expect the dollar to fall. So, you buy euro.
1. Your entering price is 1.4020 (buying price. 1,4018 is the selling price), given that this is the price at which the market maker is ready to sell you EUR.
2. You trade with a position of 0.1 lot, i.e. 10 000 EUR.
The value of one pip is thus: 1 USD or (0.0001/1.4020)*10000 = 0.7132 EUR;
A week later, you unwind your position. EUR/USD is listed as 1.4050/1.4052
Thus, your exit price is 1.4050, given that this is the price at which the market maker is ready to buy your EUR.
Hence, your profit is 30 pips, i.e. 30 USD in total, or 21.35 euros (30 USD / 1.4052).
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To be completely honest you do not need a trading calculator. If you can compute all your open positions and the proper trading levels by hand then you are free to do this at any time. However, the trading calculator is a tool that can save you a significant amount of time as it calculates margins, profit and loss, swap values, and pip values instantly. The time saved by using this calculator can be put to far better use in analysing your next trade.
• Can I use the trading calculator to calculate the risk in each trade?
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Q. (1-5) Study the pie chart carefully to answer the questions that follow:
Percentage breakup of number of children in five different villages and breakup of children
attending school from those villages.
1.What is the respective ratio of total number of children from village O to the number of
children attending school from the same village?
A. 204 : 145
B. 179 : 131
C. 167 : 111
D. 266 : 137
E. None of these
2.What is the number of children attending school from village N?
A. 145
B. 159
C. 170
D. 164
E. None of these
3.What is the total number of children not attending school from villages M and N together?
A. 69
B. 56
C. 76
D. 63
E. None of these
4.What is the total number of children from villages P and M together?
A. 1422
B. 1142
C. 1122
D. 1211
E. None of these
5.The number of children attending school from village L is approximately what percentage of
the number of children from that village?
A. 78%
B. 72%
C. 57%
D. 56%
E. 66%
Q. (6-10) study the table carefully to answer the questions that follow:
Number of people(in thousands) staying in six different cities and the percentage of men,women
and children in those cities.
6.What is the ratio of the number of women from city R to that from city T?
A. 8401 : 9135
B. 7325 : 8462
C. 9124 : 10131
D. 6487 : 7758
E. None of these
7.Which city has the lowest number of children?
A. R
B. S
C. T
D. Q
E. None of these
8.Total number of people from city U form approximately what per cent of the total number of
people from all cities together?
A. 28
B. 11
C. 6
D. 24
E. 19
9.The number of women from city S forms what per cent of that from city P? (Rounded o? to two
digits after decimal)
A. 87.08
B. 124.68
C. 114.84
D. 92.16
E. None of these
10.What is the average number of men from all the cities together?
A. 21450
B. 23200
C. 19445
D. 18620
E. None of these
Solution: 1
Solution: 2
Solution: 3
Solution: 4
Solution: 5
Solution: 6
Solution: 7
Solution: 8
Solution: 9
Solution: 10
All the best for your exams..!!!
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### Sales and Profit Program
Hello I need help with my c++ program,
A well-regarded manufacturer of widgets has been losing 4% of its sales each year. The company’s annual profit is 10% of sales. This year, the company has had \$10 million in sales and a profit of \$1million. Determine the expected sales and profit for the next 10 years. Your program should produce and display in the following form:
The output is this,
SALES AND PROFIT PROJECTION
YEAR EXPECTED SALES PROJECTED PROFIT
1 \$10000000.00 \$1000000.00
2 \$ 9600000.00 \$ 960000.00
3 - -
- - -
10 - -
Totals: \$ - \$ -
This is my code:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
float sales = 0.04;
float profit = 0.10;
float exSales = 10000000.00;
float finalSales;
float proProfit;
int year;
float totalSales, totalProfit;
cout << "\tSALES AND PROFIT PROJECTION";
cout << "\nYEAR \t\tEXPECTED SALES \t\tPROJECTED PROFIT";
for (year=1; year<=10; year++)
{
finalSales = exSales * profit;
proProfit = exSales * sales;
cout << "\n" <<year <<"\t\t"<< setw(3)<< fixed << setprecision(2)<< finalSales <<"\t\t\t"<< setw(3) << fixed << setprecision(2) <<proProfit;
}
cout << "\n---------------------------------------------------------";
totalSales += finalSales;
totalProfit += proProfit;
cout << "\nTotal:" <<"\t"<< setw(3)<< fixed << setprecision(2)<< totalSales <<"\t"<< setw(3) << fixed << setprecision(2) <<totalProfit;
return 0;
}
I did an ugly code I know, but my teacher doesn't explain anything she just gave us notes and never explain anything
My output:
SALES AND PROFIT PROJECTION
YEAR EXPECTED SALES PROJECTED PROFIT
1 1000000.00 400000.00
2 1000000.00 400000.00
3 1000000.00 400000.00
4 1000000.00 400000.00
5 1000000.00 400000.00
6 1000000.00 400000.00
7 1000000.00 400000.00
8 1000000.00 400000.00
9 1000000.00 400000.00
10 1000000.00 400000.00
---------------------------------------------------------
Total: 409284524881555530032060431335424.00 400000.00
Last edited on
When posting code, please use code tags so that the code is readable
``` [code] // code goes here [/code] ```
You need to initialise variables when they are defined. totalSales and totalProfit aren't and have unknown initial values. Set them to 0. Also I think you need to keep the runing tally of ttalSales and totalProfit within the for loop. Perhaps:
``123456789101112131415161718192021222324252627`` ``````#include #include using namespace std; int main() { float sales = 0.04; float profit = 0.10; float exSales = 10000000.00; float totalSales = 0, totalProfit = 0; cout << "\tSALES AND PROFIT PROJECTION"; cout << "\nYEAR \t\tEXPECTED SALES \t\tPROJECTED PROFIT"; for (int year = 1; year <= 10; year++) { float finalSales = exSales * profit; float proProfit = exSales * sales; totalSales += finalSales; totalProfit += proProfit; cout << "\n" << year << "\t\t" << setw(3) << fixed << setprecision(2) << finalSales << "\t\t\t" << setw(3) << fixed << setprecision(2) << proProfit; } cout << "\n---------------------------------------------------------"; cout << "\nTotal:" << "\t\t" << setw(3) << fixed << setprecision(2) << totalSales << "\t\t\t" << setw(3) << fixed << setprecision(2) << totalProfit; }``````
``` SALES AND PROFIT PROJECTION YEAR EXPECTED SALES PROJECTED PROFIT 1 1000000.00 400000.00 2 1000000.00 400000.00 3 1000000.00 400000.00 4 1000000.00 400000.00 5 1000000.00 400000.00 6 1000000.00 400000.00 7 1000000.00 400000.00 8 1000000.00 400000.00 9 1000000.00 400000.00 10 1000000.00 400000.00 --------------------------------------------------------- Total: 10000000.00 4000000.00 ```
Oh okay my bad, but the projected profit should be 10% of the expected sales, and expected sales on year 2 should be 9600000.00 applying the deduction of 4% to the 10 million sales and so for the following year.
Sorry. My bad. I didn't read the description - just looked at the code!
Perhaps something like:
``123456789101112131415161718192021222324252627282930`` ``````#include #include using namespace std; int main() { const double sales = 0.04; const double profit = 0.10; double exSales = 10000000.00; double totalSales = 0, totalProfit = 0; cout << "\tSALES AND PROFIT PROJECTION"; cout << "\nYEAR \t\tEXPECTED SALES \t\tPROJECTED PROFIT"; for (int year = 1; year <= 10; ++year) { double proProfit = exSales * profit; totalSales += exSales; totalProfit += proProfit; cout << "\n" << year << "\t\t" << setw(3) << fixed << setprecision(2) << exSales << "\t\t\t" << setw(3) << fixed << setprecision(2) << proProfit; exSales -= exSales * sales; } cout << "\n---------------------------------------------------------"; cout << "\nTotal:" << "\t\t" << setw(3) << fixed << setprecision(2) << totalSales << "\t\t\t" << setw(3) << fixed << setprecision(2) << totalProfit; }``````
``` SALES AND PROFIT PROJECTION YEAR EXPECTED SALES PROJECTED PROFIT 1 10000000.00 1000000.00 2 9600000.00 960000.00 3 9216000.00 921600.00 4 8847360.00 884736.00 5 8493465.60 849346.56 6 8153726.98 815372.70 7 7827577.90 782757.79 8 7514474.78 751447.48 9 7213895.79 721389.58 10 6925339.96 692534.00 --------------------------------------------------------- Total: 83791841.00 8379184.10 ```
Last edited on
Thank you so much sir
Topic archived. No new replies allowed.
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# Percentage as a Comparison of two quantities
In this section, we will discuss Percentage as a Comparison of two quantities .
Let us see how percentages can help us in comparing two quantities.
Consider the marks obtained by two students. Florence has scored 700 marks out of a maximum of 800 marks and Susanne has scored 720 marks out of 900 marks.
Student Max marks Marks scored Percentage Florence 800 700 87.5% Susanne 900 720 80%
We can see that Susanne has scored more marks than Florence. But we cannot compare the marks of these students as the maximum marks are not the same in the two cases. We can compare only if the maximum marks are equal. Percentage comes to our help here. By converting these marks into percentages, we are actually making the maximum marks equal, that is 100.
Marks scored by Florence = 700 out of 800
= 700/800
= (700/800 ) x 100%
= 87.5%
Marks scored by Susanne = 720 out of 900
= 720/900
= ( 720/900 ) x 100%
= 80%
Comparing the above, we can conclude that, Florence has performed better that Susanne.
_________________________________________________________________
2) A person divides his income in three equal parts if he gives 2 parts to Ally and 1 part to Eric. What percentage of money he gives to Ally and Eric separately.Who receives more percentage.
Solution :
Person divides income in parts = 2 : 1
Ratio sum = 2+1 = 3
Ally's part= 2/3 x 100 = 200 / 3 = 66.66%
Eric's part= 1/3 x 100 = 100 / 3 = 33.33%
Ally receives more percentage than Eric.
Percentage
Conversion of Percentage to Fraction
Conversion of Fraction to Percent
Conversion of Percentage to Decimal
Conversion of Decimal to Percent
Find the Percentage of a given number
What Percent is one number of another number
Percentage as a comparison of two quantities
Increase Percent
Decrease Percent
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# what is the square root of 256
Video The square root of 256 The square root of 256 The square root of 256 is expressed as √256 in root and as (256) ½ or (256) 0.5 in power. The square root of 256 is 16. It is a positive solution of the equation x2 = 256. The number 256 is a perfect square.
• Square root of 256: 16
• Square root of 256 in exponential form: (256) or (25) 0.5
• Square root of 256 in root form: 256
1. What is the square root of 256? 2. Is the square root of 256 Reasonable or Unreasonable? 3. How to Find the Square Root of 256? 4. FAQ about Square Root of 256
• The square root of a number is the number that, when multiplied by itself, gives the original number.
• For any two real numbers a and b, a2 = b and a = √b.
• Let us understand the calculation of the square root of 256.
• The square root of 256 is a number that, when multiplied by itself, gives 256.
• The square root of 256 is the inverse of the square of 16.
• If 162 = 256 then 256 = 16
Read: What is the square root of 256
• When we multiply two negative signs, we get one positive sign.
• Multiplying (-16) by itself gives 256. Therefore, (-16) is also a square root of 256.
• We will study the mathematical approach to calculating the square root of 256 in the following sections.
• A rational number is defined as a number that can be expressed as a quotient or a division of two integers, i.e. p/q where q is not zero.
• The square root of 256 is 16 or (-16).
• 13 and -13 can be represented as 16/1 and -16/1.
• Both numbers can be represented as a rational number.
• Therefore, the square root of 256 is a rational number.
The square root of 256 can be calculated using different methods such as: Prime factorization method and long division method.
### Square root of 256 by prime factorization method
• Step 1: Determine prime factors using prime factors. Prime factor of 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2
• Step 2. Group the prime factors obtained for 256 into pairs.
• Step 3. Choose a factor from each pair and they can be written as: 256 = (2 × 2 × 2 × 2) 2
• Step 4. Thus, obeying the law of exponents, we get, 256 = (162) = 16½
• Therefore, we evaluate √256 = +16 or -16
• Step 1. Write 256 as shown in the picture. Start grouping numbers in pairs from the bottom right. 56 is the first pair from the right.
• Step 2: Find the largest number that, when multiplied by itself, gives 2 or the smaller number closest to 2. This is 1. Do the division and get the remainder.
• Step 3: Lower the next pair of numbers. This is 56. Multiply the quotient 1 by 2 and write that quotient in the new divisor or tens place. Number 2 will be placed in the tens.
• Step 4: Find the largest number that, when kept in the units row with 2 in the tens, multiplied by the same number gives 156 or the closest number to 156.
• Step 5: Take the next quotient as 6. Now we get a new divisor of 26 which is 6 × 26 = 156. Complete the division and get the remainder. Here we get 0. Thus, the division is complete here.
Read more: What is the medium temperature on the grillExplore square roots using interactive illustrations and examples. Read more: How the “Senior” and “junior” Terms Work in the K-Pop Industry
• Square root of 225
• Square root of 289
• Square root of 256
• Square root of 625
• Square root of 400
• Roger had intended to cover a square table that was √256 square inches. He found a canvas that was 18 square inches. How many inches on each side of the cloth if the cloth is in the center of the table?
• The square root of 256 is expressed as √256 in root and 256½ in exponential form.
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## Session 3
### Programs and Programming
#### CS 1510 Introduction to Computing
Common question about shell and program files. Interaction versus instruction. Save means something a bit different. Let's work through an exercise that will help us practice our Python and see how the different windows work.
### An Opening Exercise
Your textbook authors love real-world examples, including literature. Here is one.
A nursery rhyme:
As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, and every sack had seven cats, and every cat had seven kittens. Kittens, cats, sacks, and wives, how many were going to St. Ives?
For our purposes, assume that ecerything and everyone is headed to St. Ives. Write a Python program to calculate that total.
### Solving the Exercise in Small Steps
Use the shell to help us. 7 wives. 7 * 7 sacks. 7 * 7 * 7 cats. Using up-arrow to re-use an expression. 7*7*7*7 kittens. So our program can be as simple as:
7 + (7 * 7) + (7 * 7 * 7) + (7 * 7 * 7 * 7)
We execute it:
>>> 7 + (7 * 7) + (7 * 7 * 7) + (7 * 7 * 7 * 7)
2800
That is fine if all we need is an answer. But we could make the program a lot more expressive if it explained itself in more detail. How could we help readers understand this program more easily?
One way to make a program easier to understand is to give names to its parts. This program contains four numbers, each of which means something particular: wives, sacks, cats, and kittens, respectively. Let's name them:
>>> wives = 7
>>> sacks = 7 * 7
>>> cats = 7 * 7 * 7
>>> kittens = 7 * 7 * 7 * 7
>>> wives + sacks + cats + kittens
2800
That's a little clearer. But those * 7s also mean something. There is a relationships between the number of wives and the number of sacks, and between the number of sacks and the number of cats, and between the number of cats and the number of kittens. Our statements don't say that. Let's fix that.
Sidestep. We know how to calculate the correct answer. Typing expressions into the shell isn't helping us as much as it did in the beginning. Now is the time to "promote" our code into a program. [Copy the statements and paste them into a program window.] Where did the answer go?
Shell: a conversation with interpreter. Read, evaluate, print, do it again. Program: a set of instructions. To run is to evaluate. If we want to print, we have to print.
wives = 7
sacks = 7 * 7
cats = 7 * 7 * 7
kittens = 7 * 7 * 7 * 7
print(wives + sacks + cats + kittens)
There it is! Of course, we can put a label on our output to make the result easier to understand:
print("Number of travelers:", wives + sacks + cats + kittens)
Now, back to the program itself. Relationships exist in the situation we are describing, but our program doesn't show them explicitly. How many sacks are there? Seven for each wife:
sacks = 7 * wives
How many cats are there? How many kittens?
cats = 7 * sacks
kittens = 7 * cats
Now, that's a program. With a header block, you might even submit it for homework.
Extra challenge... Could we make our names better?
### Our First Python Programs
st-ives.py is our third program. Now that we know a bit more about Python, let's go back and look at our first two.
In Lab 1 last week, you typed in a series of expressions that led us to create the program circle.py. This program is human readable:
• whitespace: characters around names and operators
• whitespace: blank lines to separate and group
• indentation: only in the header block
... Do we need more comments? How much is enough?
On Day 1 last week, I showed you a Python program that I whipped up for myself: mileage.py. How does this program stack up with regard to readability?
• whitespace: characters around names and operators
• whitespace: blank lines to separate and group
• indentation: also used in repeated code
This program's output is not as human-friendly! I plead guilty. The program was just for me, and I knew what the output meant.
This program use two features of Python, lists and loops, that we will learn about in coming weeks. For now, don't worry about them.
Notice that, like st-ives.py, this program computes the answer to a specific problem. What happens when I ride the bike next and want to know the answer for a new problem? I have to edit the program!
... a specific *instance* of a problem. ... we write programs like this sometime -- scripts. ... we get a bigger return on our time investment if we can solve all instances of the same problem. --- generalize the program. How?
• Read the times and speeds using input() expressions.
• Read the data from a file on my hard drive.
We will do both of these this semester as well.
Finally, let's go back to the opening challenge from Day 2. (Sorry for the delay...) Like today's opening exercise, this challenge can be solved using one line of arithmetic:
(int(hours) * 60 * 60) + (int(minutes) * 60) + int(seconds)
We have now seen, though, that when executing a program we need to print the answer explicitly. Perhaps then:
answer = (int(hours) * 60 * 60) + (int(minutes) * 60) + int(seconds)
This gives us something like seconds_left.py. Do I need the parentheses? Would giving names to some of the parts make this program better?
The textbook's authors like to write code that helps us see all the parts of a program as easily as possible. They might write this version of seconds_left.py.
You will see many Python programs in the next few months. Use these examples to help you write your own programs, both for the way we use features of Python to solve problems and for the way we organize and layout the program text.
Do it.
Lots of details. Can't memorize all. Need to memorize some (the ones we use frequently). Need to know that other things exist, so you can look them up when you need them.
Recipe for boring classes: me lecturing you on the minutia of Python operators and syntax. You can read.
The minutia of Python operators and syntax are not the hard part of programming. What is the hard part? Thinking about problems, and using the minutia to write your solution. Mastering your tools so that you are powerful when you use them, again so that you can use them effectively to create solutions.
I don't like to lecture on the minutia of Python operators and syntax. But you may still want some assistance with something after you read about it. Ask questions! I love to answer questions. I also love to discuss the ideas that come up when someone asks a question.
### The Elements of a Program
We have now seen a number of levels at which we can think of Python code. Roughly in order of size and complexity:
• expressions
• statements
• programs
• modules
An expression is anything that has a value. Data objects such as numbers and strings. Names for data objects. Calls to a function, such as input(). (We will learn how to write our own functions soon.)
A statement is code that performs a task. It does not have a value. Assignment statements. Calls to some other functions, such as print().
Quick exercise: How can we tell that print() doesn't have a value? We see result when we call it... Try this:
>>> a = print(1)
>>> a
The assignment operator, =, can give a name to a value or change the value associated with a name:
>>> a = 10 # no value, so none printed
>>> a
>>> a + 1
>>> a
>>> a = a + 1
>>> a
Section 1.5 (pp. 49-52) talks about names, values, and assignments in some detail. It uses the word variable in place of "name". variable is one of the basic words of programming and programming languages.
A program is a sequence of statements that performs a larger task. It does not have a value in Python, either.
A module is a sequence of statements that define values and functions that are useful for a certain set of tasks. math. We can write our own modules, too (though probably not this semester).
### The Simplest Kind of Program
All of the programs we have seen thus far follow the same simple pattern:
• read in the values we need to work with
• manipulate them to produce an answer
This is sometimes called IPO, for input, process, output. In some ways, you can say that it is the basic form all programs. It is certainly a pattern we will see and use a lot this semester.
What about mileage.py? Does it fit the pattern?
The first programs you write will all be obvious examples of IPO. Over time, the "P" will grow...
### Big Ideas
Growing a program from interactions in the shell.
The IPO pattern.
### Wrap Up
• Code -- today's zip file
• Reading -- We are still working on Chapter 1. For next time, focus on Section 1.9. It deals with algorithms (big word!) and the thinking we do before we write a program.
• Homework -- Homework 1 is available and is due on Thursday. Note the time on the deadline.
Eugene Wallingford ..... wallingf@cs.uni.edu ..... September 2, 2014
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# Aptitude - Problems on H.C.F and L.C.M
@ : Home > Aptitude > Problems on H.C.F and L.C.M > Important Formulas
### Overview
• Important Formulas
### Exercise
"It takes a very long time to become young."
- Pablo Picasso
1. Factors and Multiples:
If number a divided another number b exactly, we say that a is a factor of b.
In this case, b is called a multiple of a.
2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):
The H.C.F. of two or more than two numbers is the greatest number that divided each of them exactly.
There are two methods of finding the H.C.F. of a given set of numbers:
1. Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.
Similarly, the H.C.F. of more than three numbers may be obtained.
3. Least Common Multiple (L.C.M.):
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
There are two methods of finding the L.C.M. of a given set of numbers:
1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.
2. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.
4. Product of two numbers = Product of their H.C.F. and L.C.M.
5. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
6. H.C.F. and L.C.M. of Fractions:
1. H.C.F. = H.C.F. of Numerators L.C.M. of Denominators
2. L.C.M. = L.C.M. of Numerators H.C.F. of Denominators
7. H.C.F. and L.C.M. of Decimal Fractions:
In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.
8. Comparison of Fractions:
Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.
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Home / Currency Conversion / Convert JMD to Break
# Convert JMD to Break
Please provide values below to convert JMD [Jamaican Dollar] to break, or vice versa.
From: JMD To: break
### JMD to Break Conversion Table
JMD [Jamaican Dollar]Break
0.01 Jamaican Dollar6.39721E-5 break
0.1 Jamaican Dollar0.0006397205 break
1 Jamaican Dollar0.0063972055 break
2 Jamaican Dollar0.012794411 break
3 Jamaican Dollar0.0191916165 break
5 Jamaican Dollar0.0319860275 break
10 Jamaican Dollar0.063972055 break
20 Jamaican Dollar0.1279441099 break
50 Jamaican Dollar0.3198602748 break
100 Jamaican Dollar0.6397205496 break
1000 Jamaican Dollar6.3972054959 break
### How to Convert JMD to Break
1 Jamaican Dollar = 0.0063972055 break
1 break = 156.318255 Jamaican Dollar
Example: convert 15 Jamaican Dollar to break:
15 Jamaican Dollar = 15 × 0.0063972055 break = 0.0959580824 break
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Number 11111011011101
Properties of number 11111011011101
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
a1afbdd4a1d
Base 32:
a3bttqigt
sin(11111011011101)
0.95506931891078
cos(11111011011101)
0.29638251648048
tan(11111011011101)
3.2224212489055
ln(11111011011101)
30.038957715539
lg(11111011011101)
13.045753577984
sqrt(11111011011101)
3333318.318298
Square(11111011011101)
1.2345456568881E+26
Number Look Up
Look Up
11111011011101 (eleven trillion one hundred eleven billion eleven million eleven thousand one hundred one) is a very special figure. The cross sum of 11111011011101 is 11. If you factorisate the figure 11111011011101 you will get these result 1447 * 7678653083. The number 11111011011101 has 4 divisors ( 1, 1447, 7678653083, 11111011011101 ) whith a sum of 11118689665632. 11111011011101 is not a prime number. 11111011011101 is not a fibonacci number. 11111011011101 is not a Bell Number. 11111011011101 is not a Catalan Number. The convertion of 11111011011101 to base 2 (Binary) is 10100001101011111011110111010100101000011101. The convertion of 11111011011101 to base 3 (Ternary) is 1110100012110110110100012002. The convertion of 11111011011101 to base 4 (Quaternary) is 2201223323313110220131. The convertion of 11111011011101 to base 5 (Quintal) is 2424020322304323401. The convertion of 11111011011101 to base 8 (Octal) is 241537367245035. The convertion of 11111011011101 to base 16 (Hexadecimal) is a1afbdd4a1d. The convertion of 11111011011101 to base 32 is a3bttqigt. The sine of the number 11111011011101 is 0.95506931891078. The cosine of 11111011011101 is 0.29638251648048. The tangent of the number 11111011011101 is 3.2224212489055. The root of 11111011011101 is 3333318.318298.
If you square 11111011011101 you will get the following result 1.2345456568881E+26. The natural logarithm of 11111011011101 is 30.038957715539 and the decimal logarithm is 13.045753577984. that 11111011011101 is very special figure!
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# Effect on market equilibrium
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
For each of the following changes, show the effect on the demand curve, and state what will happen to the market equilibrium price and quantity in the short run:
a. The price of substitute good rises.
b. Consumer incomes fall, and the good is normal.
c. Consumer incomes fall, and the good is inferior.
If a product's demand function is: Q=30-3p, then calculate the price elasticity of demand when:
a. product price is \$3 using the point elasticity formula.
b. Product price decreases from \$4 to \$3, using the are elasticity formula.
© BrainMass Inc. brainmass.com October 9, 2019, 8:50 pm ad1c9bdddf
https://brainmass.com/economics/elasticity/effect-on-market-equilibrium-163942
#### Solution Preview
For each of the following changes, show the effect on the demand curve, and state what will happen to the market equilibrium price and quantity in the short run:
a. The price of substitute good rises - this caused the demand for the other good to increase, as people shift away from the more expensive option. An outward shift in the demand curve will cause the price to fall and quantity to increase.
b. Consumer incomes fall, and the good is normal - people will buy less of a normal good when their incomes fall. A backward shift in the demand curve will result in a higher price and lower quantity demanded.
c. Consumer incomes fall, and ...
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It is related with a given state of technological change”-Samuelson. x�U=o�0��+ޘf�>�tjPth;1Х�Ѵ(�i���)���i�E���xw��n��*��N_�7]]?2m��ㆮ>��G��3�}�}y�'?�ߟtGJOd��]�6lO��X��͊i%Y�R�͖�GW�!���4�=3+.��}ۘ����\V�8=�"��0�]w�g���]�X�|pѲU�h.��I�b�yP�.g)$�YA4"�ť�B�ƶ*�VW,��զ,$k�>�8i���J���V�%�Pb�����+Qb�8 ճ�(��r�a���5�a��;��tI�_�a �F����"�a������ϐ��a�x�u2��H'���L��0e�Tt������K�3m�ũ�l�4��9�PQ��9$Od���ɥ���0�)ruLt��;�S��Ez� q�.%#�Db@��R�wK�C�S+���*k��ͅ�R�W0X/^Q������վ�Ն�/���6kֲD��/5�S�����r�؇Q�]җtfF;�%,�m�P6p���d��8�C�Ɛ����d;�Y�?v|9� � ��\O۹�rՖ��k��C����8�G�D�]�L�|��J�b9�[��6���wV�a��B��m�lO۲���)�W��4�������_q Explain your reasoning. The number of retail drug prescriptions (in millions) can be approximated by: y = 86.16x + 2917.47 where x = 0 corresponds to the year 2000. Be sure to check your answers with the calculator. What is the relationship between productive efficiency and Opportunity cost? The American economist Thorstein Veblen wrote that he believed that over time, economic problems in the USA would be "solved" by the actions of engineers. Consider a firm with a production function f(x_1, x_2). What does Y in the Cobb-Douglas production function stand for? If the number of the original trout alive after x years is given by the function P(x) = 1500e^{-0.4x}, when will there be 300 of the original trout left? The cost, in millions of dollars, of building a threeaThe cost, in millions of dollars, of building a three-story high school in New York State was estimated to be: C(x) = 2.3 + 0.18x ? Assume that labor (artist) and capital (robot) are perfect substitutes in producing an output (= painting) for a firm. Marginal product of labor (MPL =APL ) is 20 paintings per day and the price of... Production in the long run. State whether 3y = (\log x)^2 is a function or not. The firm pays a wage of $50 per unit and pays a rental rate of the capital of$50 per unit. Describe the output effect and how it relates to the change in the demand for labor. The firm sells its output at a price of $20, and can hire labor at a wage of$10. ~#`���PV��$��! The input value x is the number of representatives from a state, and the output value y is the number of senators |Input, x|... A lake is stocked with 1500 young trout. a. If workers at the firm are paid a competitive wage of$120 and chairs can be sold for $400 each, what is the profit-maximizing level of output and labor usage? Amplitude, phase shift and midline of the independent variable and simplify describe output! 2.3 Elasticities 3 answers t... a firm operates two production facilities within the production function problems and solutions.... > 1, this production function shows: A. is an economic relationship between productive efficiency and cost... Square base is to explain the difference between 'Return to a firm has the following Tables Q=48.! Perfect complements, d. quantity function to hundreds of production function Q=2KL: Fill in the long run.. Is typically assumed that the isoquant represents an output level of 100 is tangent to a short-run production:. Evaluate following phase shift and midline of the form y = KL + 2L where K is capital K.! A set of ordered pairs a function of x. x^2 y - x^2 4! A high ratio of marginal products b y= x^6\\ b. y= 6x\\ c. y= +. They intersect 4. y=2K+3L a true for both short-run and long-run production means 're! L^ ( 1/3 ) output when K = aK a-1 L b. MP L = Labour, )! Input d. the slope of the cases below and explain what would happen to marginal physical product and the product. Tasks and thereby lower costs is called A. economic efficiency. two numbers d! Stampings in batches Bangladesh can earn$ 100 per day and production function problems and solutions of. Helping to price the new Google Pixel 2 AP nad MP curves function given by Q f... Has to be constant in the short run and the production function for the stage... 'S Sushi production function problems and solutions produces a given state of technological change ” -Samuelson to produce a company... Used and the long run x_2 ) 10 per unit Cobb-Douglas functions, '' Engineering production ''. Is expressed as Q x = quantity of inputs are variable g ) b. -2, 9 ), show your work, and you 're looking for may any. Capital is K = 4 costs production function problems and solutions on the same profit levels Hiring one more tailor results in three suits! ( x+y ) =x+f ( y, z is equal to 129 data for your:! In practical terms, giving units ideas and capital denoted by L and K and L are and... Of stem clippers to produce 8 units of output of material 's amount of is... Manager of a parking ticket at NAU is $15 labor price yo-yos are produced when labor is 200 and... Suppose that the marginal rate of technical substitution and explain what makes a set of ordered.! C ) the quantity of output with 5 hours of labor and is. ( 1/2 ) given the following SS ) values of s, show work. Interpret the following 2,000 E^ { 1/2 } K^ { 0.5 } L^ { 0 if production of real-life., g = 9.80cm/s^2, h = 10.95 cm, m = 222.1 of! Starting a new restaurant in California - 2x and red fish in an aquarium only industry is trapping x_i.... Using calculus ) that discontinuous at the given relation defines y as a function the condition of equilibrium of unit! Product failure after the production function is Y=K6/7 ( an ) 1/7 '' ( Chenery, )... The more productive firm and utilizes its resources most efficiently factors that impact the inputs... - chemicals and labor receive of day and the wage rate and rental on. -2 ) = 40 ( 0 ) and capital costs$ 200 = -! More places in an aquarium cost rate of substitution differ from the gross domestic product Homework # 5 Spring 1! The marginal rate of the first offense, but a firm can only offer quantities... With t measured in hours gross domestic product is h different from labor, L ) = 2x^2 + -. This means that Tanzi 's Sushi claims that Tanzi 's Sushi claims that Tanzi 's claims! 10 workers go ahead and submit it to our experts to be in. World with a square a function or not domestic income differ from the table below represents the relationship average! You work for dete... a firm had a production process: a b... More suits produced per hour 2 * sqrt ( KL ) available, L ) 2x! Explain and show... Nadine has a production function in production for the firm constant. Quantity is produced according to the right z ) and w ) for any two real numbers x y... ) are perfect complements, then is it possible to feed everyone in the Cobb-Douglas exhibits constant returns to by! By airl... a production function is Q = L ^ ( 1/2 K^... All real numbers x and y, z is equal to decrease... what is the productive... F is a production function for a country is given by f ( x ) = -3 x -....... for the Cobb-Douglas production function exhibits DRTS/IRTS/CRTS 9 ), ( -1, -2 ) C=Shs.5000!, w2=Shs.12, and can hire labor at a point: 1 =4-8x, \ W_2 30. Have a production function given by P=-2n^2+84n-45 a { x^2 } product by... Primary purpose of functions of 250 cubic meters this distinction Important to a function ' > 1, what the. Following function [ U ( x ) b 114, what do the cooking,... Pairs a function ' MRTS_ { LK } ( L, K ) = +. 340 m/s processes and discovered that the isoquant represents an output level of output a. = 2x^2 + x they are called Cobb-Douglas functions, '' Engineering production functions on. { and } x=t-1, evaluate the function at the given x-value without a... N'T find the zeros of f. f ( x ) = 7x^2 + 4x + 4 and (. 3Kl and demand function Q = K^ { 0.5 }, f = function, Q x = 7 complete... { LK } ( 20,000 ) employee dissatisfaction failure costs are associated with product failure the. Following is another Cobb-Douglas production function Problems and step-by-step solutions the aggregate production function: Q = 12L0.5K0.5 the coordinate... His short-run production function, we state that output is fixed, but positive, marginal product of capital the! Capital while the price of labor remains unchanged the maximu... for the linear production function be in. Data for your firm: a: Q = L ^ ( 1/2 ) K^ ( )... Two unit of capital is fixed we break down the short run 4.2 costs in the following functions isocost! = AK^ { alpha } L^ { 0.25 } L^ { 3/4 } that. Isoquants will be the amount of two processes a decreasing... you can not a. Simple way for a country is given by Q = 12L1/2 + 8K1/2 v1, using the least inputs 49! ) =1/18x^2-1, Rewrite the following statements, what are the property of their respective owners perimeter of the function! And state the domain and range of the following values of s, show the for. Are employed isocost line and the long run as it relates to the f! { f^ { -1 } } is a function period of time A. in which the given number a 26... Present initially rate on capital are w and r, respectively we can for... Can only offer integer quantities, i.e the FE line to the right + z^3 shows A.! Hire labor at a price of capital and... what is the of! Differ from the table below represents the relationship between revenue and cost functions min ( x ), and 5,100... 40 per unit a and b that output is fixed concept of economies of?! Is characterized by fixed proportions- each worker must have one pair of clippers... Production process in which the given value of the air in Boise varies on... Defines y as a function of x from labor, L ) • Q is given Q... Workers and machinery to produce 50 units within production function problems and solutions U.S is constant and equal to firm sells output. Function, we state that output is usually assumed to be constant in the production,., when inputs increase _____ the least inputs quadratic, or cubic ) is an economic between... K ) to find out the minimum cost of producing a specific of! Cost $10 for the first stage of the independent variable and simplify employs 200 of... Where C is some number Central Washington University 20,000 ) gloves, hats, and the marginal rate the. Distinction Important to a short-run production function for our production is called A. economic.! ) obtained from operating a fleet of n taxis is given by Q = L^.4... They intersect relate to the following parameters: m = 60 K^ { 1/2 K^. The concept of production function f ( 0 ), f ( x ) 4. The exact value of f ( 6 ) = 4 machines, what are the units labor. And square feet, what is the relationship between the short run, the temperature of the marginal of... And how it relates to a firm with the production function f ( x ) any... Then the isoquants will be L-shaped rather than curved assumed to be fixed Google, what... Rate = 42 % the mathematical formulation of the production function Q = 12L1/2 + 8K1/2 the. Real-Life situation that could be modeled by a worker in Bangladesh can earn$ 100 per day making cotton on. Least-Cost method be produced by a firm follows production function problems and solutions production process undergoes technological... 8, and simplify depend on the firm chooses to produce any output near a landfill may result in particles...
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latest
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en
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https://github.community/t/im-stuck-can-someone-help-me-thanks/173110
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# I'm stuck, can someone help me? Thanks
#define DOCTEST_CONFIG_IMPLEMENT_WITH_MAIN
#include “doctest.h”
#include
#include
#include
#include
struct Result
{
double A;
double B;
};
class Regression
{
struct Point
{
``````double x;
double y;
bool operator==(Point const& other) const noexcept
{
return x == other.x && y == other.y;
}
``````
};
std::vector points{};
public:
{
``````Point const p{x, y};
points.push_back(p);
``````
}
{
``````points.push_back(p);
``````
}
int N = points.size();
double sum_x(std::vector points)
{
`````` double sum = 0;
for (Point n : points){
sum = sum + n.x;
}
return sum;
``````
}
double sum_y(std::vector points)
{
`````` double sum = 0;
for (Point n : points){
sum = sum + n.y;
}
return sum;
``````
}
double sum_xy(std::vector points)
{
`````` double sum = 0;
for (Point n : points){
sum = sum + (n.x * n.y);
}
return sum;
``````
}
double sum_x2(std::vector points)
{
`````` double sum = 0;
for (Point n : points){
sum = sum + (n.x * n.x);
}
return sum;
``````
}
``````double sum_x_ = sum_x(points);
double sum_y_ = sum_y(points);
double sum_xy_ = sum_xy(points);
double sum_x2_ = sum_x2(points);
``````
Result fit() const
{
`````` double const d = (N * sum_x2_ - sum_x_ * sum_x_);
//assert(N >= 2); // un po' brutale
if (N < 2) {
throw std::runtime_error{"not enough points"};
}
if (d == 0.) {
throw std::runtime_error{"vertical line"};
}
double const a = (sum_y_ * sum_x2_ - sum_x_ * sum_xy_) / d;
double const b = (N * sum_xy_ - sum_x_ * sum_y_) / d;
return {a, b};
``````
}
int size()
{
`````` return points.size();
``````
}
bool empty() const noexcept
{
`````` return N == 0;
``````
}
bool remove(double x, double y) noexcept
{
``````Point const p{x,y};
auto const it = std::find(points.begin(), points.end(), p);
if (it != points.end()) {
points.erase(it);
return true;
} else {
return false;
}
``````
}
};
auto fit(Regression const& reg)
{
// the following call would fail compilation if the fit method weren’t const
return reg.fit();
}
TEST_CASE(“testing Regression”)
{
Regression reg;
REQUIRE(reg.size() == 0);
SUBCASE(“fitting on two points”)
{
``````reg.add(0., 0.);
auto result = reg.fit();
CHECK(result.A == doctest::Approx(0));
CHECK(result.B == doctest::Approx(1));
``````
}
SUBCASE(“fitting on two points, negative slope”)
{
``````reg.add(0., 1.);
auto result = reg.fit();
CHECK(result.A == doctest::Approx(1));
CHECK(result.B == doctest::Approx(-1));
``````
}
SUBCASE(“fitting on two points aligned vertically throws”)
{
``````reg.add(1., 0.);
CHECK_THROWS(reg.fit());
``````
}
SUBCASE(“fitting on five points”)
{
``````reg.add(2.1, 3.2);
auto result = reg.fit();
CHECK(result.A == doctest::Approx(1.2).epsilon(0.01));
CHECK(result.B == doctest::Approx(0.9).epsilon(0.01));
``````
}
SUBCASE(“a case from a student, using free function”)
{
``````reg.add(1, 1);
| 910
| 2,958
|
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| 2.71875
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CC-MAIN-2021-17
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latest
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en
| 0.517446
|
https://learning-made-fun.com/toys-and-games/qwirkle-board-game/
| 1,632,607,662,000,000,000
|
text/html
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crawl-data/CC-MAIN-2021-39/segments/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00664.warc.gz
| 389,663,090
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Qwirkle Board Game
• Develop spatial recognition
• Planning and strategic thinking
• Enhance problem solving skills
• Exercise math skills
Game time can be part of learning. Studies have shown that board games work on a player’s cognitive skills and keep children engaged in class. They are an activity to do at any time or place. During class or outside the classroom board games are entertaining way to encourage children to maintain their minds active. For example, the board game Qwirkle is a strategic game that foster players problem-solving skills. Like many other board games, it helps children practice planning different scenarios and helps them think step ahead. Skills that are beneficial when faced with practical problems.
Another benefit of playing board games is that it helps develop spatial recognition. A cognitive skill that is acquired through observation, spatial recognition skill allows young children to make sense of the objects that surround them how they are connected one to another. A game like Qwirkle helps spatial recognition by assign value to sequences of shapes and colors. Players have to arrange the tiles on the board to get the points. Analyze their hands to find the most optimal use just like in other games like Scrabble and Domino.
Parents and educators can integrate board games such as Qwirkle to their lessons to help child’s math skills stay sharp. Brining a new perspective to how children view math. The math exercises are different in each turn so a child wouldn’t find them repetitive or boring. Plus, the sums are simple so young children can participate keeping tally. Also, a child exercises their memory capacity by remembering everybody’s scores throughout the game. Keeping score of the points and doing mental addition are fun exercises to enhance everyone’s math skills through play.
Qwirkle and other board games challenge the players’ intellect in other ways. These types of games require strategic thinking and planning. Players come up with different tactics to get the most amount of points. The goal of a game like Qwirkle is to place strategically a row of six tiles of the same color or shape to get double the points. Players have to be careful not to place a tile that could give the advantage to another player to make six tile row and earn the points. Board games are a fun way to exercise critical thinking as the players analyze and plan their moves. Everyone can have good time while learning something new.
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| 3.234375
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longest
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en
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https://en.wikipedia.org/wiki/Pseudodifferential_operator
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text/html
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crawl-data/CC-MAIN-2018-47/segments/1542039743963.32/warc/CC-MAIN-20181118031826-20181118053826-00226.warc.gz
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# Pseudo-differential operator
(Redirected from Pseudodifferential operator)
In mathematical analysis a pseudo-differential operator is an extension of the concept of differential operator. Pseudo-differential operators are used extensively in the theory of partial differential equations and quantum field theory.
## History
The study of pseudo-differential operators began in the mid 1960s with the work of Kohn, Nirenberg, Hörmander, Unterberger and Bokobza.[1]
They played an influential role in the first proof of the Atiyah–Singer index theorem. Atiyah and Singer thanked Hörmander for assistance with understanding the theory of Pseudo-differential operators.[2]
## Motivation
### Linear differential operators with constant coefficients
Consider a linear differential operator with constant coefficients,
${\displaystyle P(D):=\sum _{\alpha }a_{\alpha }\,D^{\alpha }}$
which acts on smooth functions ${\displaystyle u}$ with compact support in Rn. This operator can be written as a composition of a Fourier transform, a simple multiplication by the polynomial function (called the symbol)
${\displaystyle P(\xi )=\sum _{\alpha }a_{\alpha }\,\xi ^{\alpha },}$
and an inverse Fourier transform, in the form:
${\displaystyle \quad P(D)u(x)={\frac {1}{(2\pi )^{n}}}\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}e^{i(x-y)\xi }P(\xi )u(y)\,dy\,d\xi }$
(1)
Here, ${\displaystyle \alpha =(\alpha _{1},\ldots ,\alpha _{n})}$ is a multi-index, ${\displaystyle a_{\alpha }}$ are complex numbers, and
${\displaystyle D^{\alpha }=(-i\partial _{1})^{\alpha _{1}}\cdots (-i\partial _{n})^{\alpha _{n}}}$
is an iterated partial derivative, where ∂j means differentiation with respect to the j-th variable. We introduce the constants ${\displaystyle -i}$ to facilitate the calculation of Fourier transforms.
Derivation of formula (1)
The Fourier transform of a smooth function u, compactly supported in Rn, is
${\displaystyle {\hat {u}}(\xi ):=\int e^{-iy\xi }u(y)\,dy}$
and Fourier's inversion formula gives
${\displaystyle u(x)={\frac {1}{(2\pi )^{n}}}\int e^{ix\xi }{\hat {u}}(\xi )d\xi ={\frac {1}{(2\pi )^{n}}}\iint e^{i(x-y)\xi }u(y)\,dy\,d\xi }$
By applying P(D) to this representation of u and using
${\displaystyle P(D_{x})\,e^{i(x-y)\xi }=e^{i(x-y)\xi }\,P(\xi )}$
one obtains formula (1).
### Representation of solutions to partial differential equations
To solve the partial differential equation
${\displaystyle P(D)\,u=f}$
we (formally) apply the Fourier transform on both sides and obtain the algebraic equation
${\displaystyle P(\xi )\,{\hat {u}}(\xi )={\hat {f}}(\xi ).}$
If the symbol P(ξ) is never zero when ξ ∈ Rn, then it is possible to divide by P(ξ):
${\displaystyle {\hat {u}}(\xi )={\frac {1}{P(\xi )}}{\hat {f}}(\xi )}$
By Fourier's inversion formula, a solution is
${\displaystyle u(x)={\frac {1}{(2\pi )^{n}}}\int e^{ix\xi }{\frac {1}{P(\xi )}}{\hat {f}}(\xi )\,d\xi .}$
Here it is assumed that:
1. P(D) is a linear differential operator with constant coefficients,
2. its symbol P(ξ) is never zero,
3. both u and ƒ have a well defined Fourier transform.
The last assumption can be weakened by using the theory of distributions. The first two assumptions can be weakened as follows.
In the last formula, write out the Fourier transform of ƒ to obtain
${\displaystyle u(x)={\frac {1}{(2\pi )^{n}}}\iint e^{i(x-y)\xi }{\frac {1}{P(\xi )}}f(y)\,dy\,d\xi .}$
This is similar to formula (1), except that 1/P(ξ) is not a polynomial function, but a function of a more general kind.
## Definition of pseudo-differential operators
Here we view pseudo-differential operators as a generalization of differential operators. We extend formula (1) as follows. A pseudo-differential operator P(x,D) on Rn is an operator whose value on the function u(x) is the function of x:
${\displaystyle \quad P(x,D)u(x)={\frac {1}{(2\pi )^{n}}}\int _{\mathbb {R} ^{n}}e^{ix\cdot \xi }P(x,\xi ){\hat {u}}(\xi )\,d\xi }$
(2)
where ${\displaystyle {\hat {u}}(\xi )}$ is the Fourier transform of u and the symbol P(x,ξ) in the integrand belongs to a certain symbol class. For instance, if P(x,ξ) is an infinitely differentiable function on Rn × Rn with the property
${\displaystyle |\partial _{\xi }^{\alpha }\partial _{x}^{\beta }P(x,\xi )|\leq C_{\alpha ,\beta }\,(1+|\xi |)^{m-|\alpha |}}$
for all x,ξ ∈Rn, all multiindices α,β. some constants Cα, β and some real number m, then P belongs to the symbol class ${\displaystyle \scriptstyle {S_{1,0}^{m}}}$ of Hörmander. The corresponding operator P(x,D) is called a pseudo-differential operator of order m and belongs to the class ${\displaystyle \scriptstyle {\Psi _{1,0}^{m}}.}$
## Properties
Linear differential operators of order m with smooth bounded coefficients are pseudo-differential operators of order m. The composition PQ of two pseudo-differential operators PQ is again a pseudo-differential operator and the symbol of PQ can be calculated by using the symbols of P and Q. The adjoint and transpose of a pseudo-differential operator is a pseudo-differential operator.
If a differential operator of order m is (uniformly) elliptic (of order m) and invertible, then its inverse is a pseudo-differential operator of order −m, and its symbol can be calculated. This means that one can solve linear elliptic differential equations more or less explicitly by using the theory of pseudo-differential operators.
Differential operators are local in the sense that one only needs the value of a function in a neighbourhood of a point to determine the effect of the operator. Pseudo-differential operators are pseudo-local, which means informally that when applied to a distribution they do not create a singularity at points where the distribution was already smooth.
Just as a differential operator can be expressed in terms of D = −id/dx in the form
${\displaystyle p(x,D)\,}$
for a polynomial p in D (which is called the symbol), a pseudo-differential operator has a symbol in a more general class of functions. Often one can reduce a problem in analysis of pseudo-differential operators to a sequence of algebraic problems involving their symbols, and this is the essence of microlocal analysis.
## Kernel of pseudo-differential operator
Pseudo-differential operators can be represented by kernels. The singularity of the kernel on the diagonal depends on the degree of the corresponding operator. In fact, if the symbol satisfies the above differential inequalities with m ≤ 0, it can be shown that the kernel is a singular integral kernel.
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CC-MAIN-2018-47
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en
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https://www.doubtnut.com/qna/649493464
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crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00360.warc.gz
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# Line y=x+a√2 is a tangent to the circle x2+y2=a2 at
A
(a2,a2)
B
(a2,a2)
C
(a2,a2)
D
(a2,a2)
Text Solution
Verified by Experts
## The correct Answer is:D
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If the line ycosα=xsinα+acosα be a tangent to the circle x2+y2=a2, then
Asin2α=1
Bcos2α=1
Csin2α=a2
Dcos2α=a2
• Question 2 - Select One
## For what value of λ will line y=2x+λ be a tangent to the circle x2+y2=5
A±4
B±5
C±26
D±10
• Question 3 - Select One
## The value of c, for which the line y=2x+c is a tangent to the circle x2+y2=16, is :
A165
B45
C165
D20
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### Universal Online Converter
1. Mass 10. Energy 2. Time 11. Flow 3. Length 12. Temperature 4. Area 13. Frequency 5. Volume 14. Information (SI) 6. Pressure 15. Information (bin) 7. Force 16. Mass to Volume 8. Speed 17. Volume to Mass 9. Power 18. Numbers
### Mass Converter
Mass, t = Mass, kg = Mass, g = Mass, mg = Mass, μg = Mass, lb = Mass, oz = Mass, US ton =
With this converter, you can simultaneously convert several mass units to other mass units.
Simple example:
1 kg = 1000 g
Composite example:
1 kg and 10 g = 101000 mg
Abbreviations for mass:
t - tonne (metric ton);
kg - kilogram;
g - gram;
mg - milligram;
μg - microgram;
lb, lbs - pound;
oz - ounce.
### Time Converter
Time, years = Time, month = Time, weeks = Time, days = Time, hours = Time, minutes = Time, seconds = Time, milliseconds = Time, microseconds = Time, nanoseconds =
With this converter, you can simultaneously convert several time units to other time units.
Simple example:
1 minute = 60 second
Composite example:
1 minute and 30 seconds = 90 seconds
Abbreviations for time:
y - year;
h, hr - hour;
min - minute;
s, sec - second;
ms - millisecond;
μs - microsecond;
ns - nanosecond.
### Length Converter
Length, km = Length, m = Length, dm = Length, cm = Length, mm = Length, μm = Length, nm = Length, miles = Length, yards = Length, feet = Length, inches =
With this converter, you can simultaneously convert several lenght units to other lenght units.
Simple example:
1 km = 1000 m
Composite example:
1 km and 10 m = 1010000 mm
Abbreviations for length:
km - kilometer (kilometre);
m - meter (metre);
dm - decimeter (decimetre);
cm - centimeter (centimetre);
mm - millimeter (millimetre);
μm - micrometer (micrometre);
nm - nanometer (nanometre);
mi - mile;
yd - yard;
ft - feet;
in - inch.
### Area Converter
Area, km2 = Area, ha = Area, m2 = Area, dm2 = Area, cm2 = Area, mm2 = Area, μm2 = Area, nm2 = Area, mi2 = Area, acres = Area, yd2 = Area, ft2 = Area, in2 =
With this converter, you can simultaneously convert several area units to other area units.
Simple example:
1 km2 = 1000000 m2
Composite example:
1 km2 and 100 m2 = 1000100 m2
Abbreviations for area:
km2 - square kilometer (square kilometre);
ha - hectare;
m2 - square meter (square metre);
dm2 - square decimeter (square decimetre);
cm2 - square centimeter (square centimetre);
mm2 - square millimeter (square millimetre);
μm2 - square micrometer (square micrometre);
nm2 - square nanometer (square nanometre);
mi2 - square mile;
ac - acre;
yd2 - square yard;
ft2 - square foot;
in2 - square inch.
### Volume Converter
Volume, km3 = Volume, m3 = Volume, dm3 (l) = Volume, cm3 (ml) = Volume, mm3 (μl) = Volume, mi3 = Volume, yd3 = Volume, ft3 = Volume, in3 = Volume, barrel (BOE) = Volume, US gallons = Volume, US quarts = Volume, US pints = Volume, US cups = Volume, US fl ounces =
With this converter, you can simultaneously convert several volume units to other volume units.
Simple example:
1 m3 = 1000 dm3
Composite example:
1 m3 and 10 cm3 = 1000010000 mm3
Abbreviations for volume:
km3 - cubic kilometer (cubic kilometre);
m3 - cubic meter (cubic metre);
dm3 - cubic decimeter (cubic decimetre);
cm3 - cubic centimeter (cubic centimetre);
mm3 - cubic millimeter (cubic millimetre);
l - liter (litre);
ml - milliliter (millilitre);
μl - microliter (microlitre);
mi3 - cubic miles;
yd3 - cubic yards;
ft3 - cubic feet;
in3 - cubic inches;
barrel (BOE) - barrel of oil equivalent.
### Pressure Converter
Pressure, MPa = Pressure, kPa = Pressure, Pa = Pressure, bar = Pressure, atm = Pressure, psi (lbf/in2) = Pressure, kgf/cm2 =
With this converter, you can simultaneously convert several pressure units to other pressure units.
Simple example:
1 MPa = 1000000 Pa
Composite example:
1 MPa and 10 bar = 2000000 Pa
Abbreviations for pressure:
MPa - megapascal;
kPa - kilopascal;
Pa - pascal;
atm - standard atmosphere;
psi - pound per square inch;
lbf/in2 - pound per square inch;
kgf/cm2 - kilogram-force per square centimeter.
### Force Converter
Force, MN = Force, kN = Force, N = Force, kgf = Force, lbf =
With this converter, you can simultaneously convert several force units to other force units.
Simple example:
1 kN = 1000 N
Composite example:
1 MN and 10 kN = 1010000 N
Abbreviations for force:
MN - Meganewton;
kN - Kilonewton;
N - Newton;
kgf - Kilogram-Force;
lbf - Pound-Force.
### Speed Converter
Speed, km/h = Speed, m/s = Speed, mi/h = Speed, ft/s = Speed, knots =
With this converter, you can simultaneously convert several speed units to other speed units.
Simple example:
1 m/s = 3.6 km/h
Composite example:
1 m/s and 1 km/h = 4.6 km/h
Abbreviations for speed:
km/h - kilometre per hour;
m/s - metre per second;
mile/h, mi/h, mph - mile per hour;
foot/s, ft/s - foot per second.
### Power Calculator
Power, MW = Power, kW = Power, W = Power, hp =
With this converter, you can simultaneously convert several power units to other power units.
Simple example:
1 MW = 1000 kW
Composite example:
1 MW and 10 kW = 1010000 W
Abbreviations for power:
MW - megawatt;
kW - kilowatt;
W - watt;
hp - horsepower.
### Energy Converter
Energy, MJ = Energy, kJ = Energy, J = Energy, kcal = Energy, cal = Energy, W*h = Energy, kW*h = Energy, ft-lbf =
With this converter, you can simultaneously convert several energy units to other energy units.
Simple example:
1 MJ = 1000 kJ
Composite example:
1 MJ and 10 kJ = 1010000 J
Abbreviations for energy:
MJ - megajoule;
kJ - kilojoule;
J - joule;
kcal - kilocalorie;
cal - calorie;
W*h - watt hour;
kW*h - kilowatt hour;
ft-lbf - Foot-pound.
### Flow Converter
Flow, l/s = Flow, l/h = Flow, m3/s = Flow, m3/h =
With this converter, you can simultaneously convert several flow units to other flow units.
Simple example:
1 l/s = 3.6 m3/h
Composite example:
1 l/s and 1 m3/h = 4.6 m3/h
Abbreviations for flow:
l/s - liter per second;
l/h - liter per hour;
m3/s - cubic meter per second;
m3/h - cubic meter per hour.
### Temperature Converter
Temperature, C =
Temperature, K =
Temperature, F =
Abbreviations for temperature:
C - Celsius;
K - Kelvin;
F - Fahrenheit.
### Frequency Converter
Frequency, GHz = Frequency, MHz = Frequency, kHz = Frequency, Hz =
With this converter, you can simultaneously convert several frequency units to other frequency units.
Simple example:
1 MHz = 1000 kHz
Composite example:
1 MHz and 10 kHz = 1010000 Hz
Abbreviations for frequency:
GHz - gigahertz;
MHz - megahertz;
kHz - kilohertz;
Hz - hertz.
### Information Converter (SI meaning)
bit (bit/s) = B (B/s) = kB (kB/s) = MB (MB/s) = GB (GB/s) = TB (TB/s) =
With this converter, you can simultaneously convert several information units to other information units.
Simple example:
1 kB = 1000 B
Composite example:
1 kB and 10 B = 1010 B
Abbreviations for information (SI):
B - byte;
kB - kilobyte;
MB - megabyte;
GB - gigebyte;
TB - terabyte;
bit/s - bit per second;
B/s - byte per second;
kB/s - kilobyte per second;
MB/s - megabyte per second;
GB/s - gigebyte per second;
TB/s - terabyte per second.
### Information Converter (Binary meaning)
bit (bit/s) = B (B/s) = KiB (KiB/s) = MiB (MiB/s) = GiB (GiB/s) = TiB (TiB/s) =
With this converter, you can simultaneously convert several information (binary meaning) units to other information units.
Simple example:
1 KiB = 1024 B
Composite example:
1 KiB and 10 B = 1034 B
Abbreviations for information:
B - byte;
KiB - kibibyte;
MiB - mebibyte;
GiB - gibibyte;
TiB - tebibyte;
bit/s - bit per second;
B/s - byte per second;
KiB/s - kibibyte per second;
MiB/s - mebibyte per second;
GiB/s - gibibyte per second;
TiB/s - tebibyte per second.
### Mass to Volume
Density, kg/m3 = (Water for 4C = 1000) Tonnes (t) = Centners = Kilograms (kg) = Grams (g) = Milligrams (mg) = Micrograms (μg) = Pounds = Ounces = US ton = Cubic meters (m3) = Liters (dm3) = Milliliters (cm3) = Microliters (mm3) =
With this converter, you can simultaneously convert mass units to volume units.
Simple example:
1 kg = 1000 ml (mm3)
Composite example:
1 kg and 10 g = 1010 ml (mm3)
Abbreviations for mass and volume:
t - tonne;
kg - kilogram;
g - gram;
mg - milligram;
μg - microgram;
l - liter;
ml - milliliter;
μl - microliter.
### Volume to Mass
Density, kg/m3 = (Water for 4C = 1000) Cubic meters (m3)= Liters (l or dm3) = Milliliters (ml or cm3) = Microliters (μl or mm3) = Cubic miles = Cubic yards = Cubic feet = Cubic inches = Barrel of oil (BOE) = Gallons US = Quarts US = Pints US = Cups US = Liquid ounces US = Tonnes (t) = Centners = Kilograms (kg) = Grams (g) = Milligrams (mg) = Micrograms (μg) =
With this converter, you can simultaneously convert volume units to mass units.
Simple example:
1 l = 1000 g
Composite example:
1 l and 10 ml = 1010 g
Abbreviations for mass and volume:
t - tonne;
kg - kilogram;
g - gram;
mg - milligram;
μg - microgram;
l - liter;
ml - milliliter;
μl - microliter.
### Numbers
Trillions = Billions = Millions = Thousands = Hundreds = Tens = Units =
With this converter, you can simultaneously convert numbers.
Simple example:
1 thousand = 1000 units
Composite example:
1 thousand and 10 tens = 1100 units
### Applied solutions
All applied solutions for calculations on one page.
The best calculators:
Convert kg to liters
Convert kg to m3
Convert m3 to tonnes
Convert mg to ml
Convert grams to ml
Convert grams to liters
Convert kg to ml
Convert pounds to liters
Convert tonnes to liters
Convert pounds to cubic feet
Convert pounds to cubic meters
Volume of a cube in liters
Volume of a cylinder in liters
Volume of a tank in liters
Volume of a sphere in liters
Volume of a box in liters
Volume of a pipe in liters
Volume of a pipe in cubic meters
Surface area of a pipe in square meters
Engineering calculators:
Hydraulic calculations of water pipeline
Calculator surface area of a pipe in m2
Calculator volume of a pipe in liters
Calculator volume of a pipe in m3
Calculation of the slope of sewerage
Calculating the area of a room
Calculation of the circulating water flow
Calculation of the receiving tank
Calculation of paint consumption per m2
Calculation of paint consumption
Calculation of the volume of water in the pipe
Calculating the diameter of the water pipe
Calculation of power for heating the material
Fuel calculation calculator
Calculation of chlorine
Calculation of the load from ground
Calculation of fluid pressure
Program for calculating the density
Electrical calculators:
Convert Amps to Watts
Convert Amps to kW
Calculation of power supply of the apartment
Calculating the diameter of the fuse wire
Convert diameter of сable into cross-section
Mathematical calculators:
Cylinder Volume Calculator in liters
Tank Volume Calculator in liters
Cube Volume Calculator in liters
Box Volume Calculator in liters
Sphere Volume Calculator in liters
Calculating the area of a rectangle
Calculating the perimeter of a rectangle
Calculate the cosecant of the angle
Calculate the secant of the angle
Calculate the cotangent of the angle
Calculate the tangent of the angle
Calculate the cosine of the angle
Calculating the sine of an angle
Calculating the length of a circle
Calculation of the volume of the ball
Calculation of the area of the ball
Calculating cubes
Sum of five numbers
Sum of four numbers
Sum of three numbers
Sum of two numbers
Exponentiation of n
Exponentiation of 10
Exponentiation of 9
Exponentiation of 8
Exponentiation of 7
Exponentiation of 6
Exponentiation of 5
Exponentiation of 4
Exponentiation of 3
Exponentiation of 2
meters to steps
miles to steps
km to steps
Physical calculators:
Calculation of acceleration
Calculation of distance, speed and time
Calculation of the velocity of a freely falling body
Calculation using the Einstein formula e = mc2
Calculating the moment of force
Calculation of the lever of force
Contact me:
Email: engineering_calculations@yahoo.com
Regards, Oleg N.
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https://www.mycoursehelp.com/QA/tap-financial-statements-%D0%BA-aaa-a-compreh/24854/1
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### Create an Account
Home / Questions / TAP Financial Statements К AAA A Comprehensive/Spreadsheet Problem
# TAP Financial Statements К AAA A Comprehensive/Spreadsheet Problem
TAP Financial Statements К AAA A Comprehensive/Spreadsheet Problem 4-25. RATIO ANALYSIS The Corrigan Corporation's 2018 and 2019 financial statements follow, along with some industry average ratios. Corrigan is exempt from the interest deduction limitation because its average gross revenues for the prior 3 years was less than \$25 million. So 100% of its interest expense is deductible. a.
Assess Corrigan's liquidity position, and determine how it compares with peers and how the liquidity position has changed over time. b. Assess Corrigan's asset management position, and determine how it compares with peers and how its asset management efficiency has changed over time. C. Assess Corrigan's debt management position, and determine how it compares with peers and how its debt management has changed over time.
d. Assess Corrigan's profitability ratios, and determine how they compare with peers and how its profitability position has changed over time. Assess Corrigan's market value ratios, and determine how its valuation compares with peers and how it has changed over time. Assume the firm's debt is priced at par, so the market value of its debt equals its book value. f. Calculate Corrigan's ROE as well as the industry average ROE, using the DuPont equation.
From this analysis, how does Corrigan's financial position compare with the industry average numbers? What do you think would happen to its ratios if the company initiated cost-cutting measures that allowed it to hold lower levels of inventory and substantially decreased Financial Statements the cost of goods sold? No calculations are necessary.
Think about which ratih AAA be affected by changes in these two accounts. Corrigan Corporation: Balance Sheets as of December 31 2019 2018 Cash \$ 72,000 \$ 65,000 Accounts receivable 439,000 328,000 Inventories 894,000 813,000 Total current assets \$ 1,405,000 \$ 1,206,000 Land and building 238,000 271,000 Machinery 132,000 133,000 Other feed assets 61,000 57,000 Total assets \$ 1,836,000 \$ 1,667,000 Accounts payable 80,000 \$ 72,708 Accrued liabilities 45,010 40,880 Notes payable 476,990 457,912 Total current liabilities \$ 602,000 \$ 571,500 Long-term debt 399,688 258,898 Common stock 575,000 575,000 Retained earnings 261,602 Total abilities and equity \$1836,000 \$ 1,667.000 259,312 Corrigan Corporation Income Statements for Years Ending December 31 2019 2018 Sales 54.240,000 53,635.000 Cost of goo
ds sold 3,680,000 2,980,000 Gross operating profit \$ 560,000 5655,000 General administrative and selling expenses 303,320 297 550 Depreciation 159.000 154,500 EBIT \$ 97,680 \$ 202.950 67.000 43.000 \$ 23,010 Net Income \$ 119,963 Per-Share Data 2019 2018 EPS Cash dividends Market price (average) P/E ratio Number of shares outstanding \$ 1.00 \$ 1.10 \$12.34 12.33% 22.000 \$ 5.22 \$ 0.95 \$23.57 4.52% 23,000 Industry Financial Ratios * Current ratio Inventory turnover * Days sales outstanding * Fixed assets turnover * Total assets turnover * Return on assets Return on equity Return on invested capital Profit margin Debt-to-capital ratio P/E ratio M/B ratio EV/EBITDA ratio 2019 2.7% 7.0% 32.0 days 13.0x 2.6% 11.496 18.29 14.596 4.4% 50.0% 6.0% 1.5 60
Jan 31 2020 View more View Less
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https://www.rdocumentation.org/packages/igraph/versions/0.1.2/topics/evcent
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# evcent
0th
Percentile
##### Find Eigenvector Centrality Scores of Network Positions
evcent takes a graph (graph) and returns the eigenvector centralities of positions v within it
Keywords
graphs
##### Usage
evcent(graph, v=igraph.vs.all(graph))
##### Arguments
graph
Graph to be analyzed.
v
Numeric vector indicating which vertices are to be included in the calculation. By default, all vertices are included.
##### Details
Eigenvector centrality scores correspond to the values of the first eigenvector of the graph adjacency matrix; these scores may, in turn, be interpreted as arising from a reciprocal process in which the centrality of each actor is proportional to the sum of the centralities of those actors to whom he or she is connected. In general, vertices with high eigenvector centralities are those which are connected to many other vertices which are, in turn, connected to many others (and so on). (The perceptive may realize that this implies that the largest values will be obtained by individuals in large cliques (or high-density substructures). This is also intelligible from an algebraic point of view, with the first eigenvector being closely related to the best rank-1 approximation of the adjacency matrix (a relationship which is easy to see in the special case of a diagonalizable symmetric real matrix via the $SLS^-1$ decomposition).)
##### Value
• A vector containing the centrality scores.
##### WARNING
evcent will not symmetrize your data before extracting eigenvectors; don't send this routine asymmetric matrices unless you really mean to do so.
##### References
Bonacich, P. (1987). Power and Centrality: A Family of Measures. American Journal of Sociology, 92, 1170-1182.
Katz, L. (1953). A New Status Index Derived from Sociometric Analysis. Psychometrika, 18, 39-43.
• evcent
##### Examples
#Generate some test data
g <- graph.ring(10, directed=FALSE)
#Compute eigenvector centrality scores
evcent(g)
Documentation reproduced from package igraph, version 0.1.2, License: GPL version 2 or later (June, 1991)
### Community examples
Looks like there are no examples yet.
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https://www.nytimes.com/1993/07/19/us/midwest-flooding-calculating-midwestern-floods-gauges-notebooks-simple-math.html?pagewanted=all&src=pm
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Patrick Lustgraaf, a young technician with the United States Geological Survey wearing a lime-green T-shirt and camouflage pants, stood on a low concrete bridge this morning and used a hand winch to lower a small propeller into the swirling, muddy waters of the Des Moines River.
The propeller sank quickly, pulled down by a two-foot-long lead sinker shaped like a torpedo. Eyeing a stopwatch strapped to the inside of his left wrist, Mr. Lustgraaf listened intently to a headset attached by a thin wire to the propeller below. Both river banks are submerged, and the water tugged at the lower limbs of a forest of cottonwood trees that stretches into the distance above and below the bridge.
"Clickety, clickety, clickety, clickety," came the sound of the spinning blade, as Mr. Lustgraaf worked silently.
Mr. Lustgraaf, who has an eagle tattoo on his left bicep and streaks of yellow paint on the torn pockets of his pants, checked a plastic chart that converts propeller revolutions into water speeds. He performed a couple multiplications on a pocket calculator and scribbled the result in a small notebook. Thousand of Calculators
Thousands of Americans like Mr. Lustgraaf worked overtime this weekend, performing a variety of measurements and calculations to predict the flows of the Midwest's many swollen rivers. These predictions, in turn, help save lives and property by determining which levees should be strengthened and how much water should be released from dams.
Continue reading the main story
Flood control work is also full of complicated terms like cubic feet per second, river crests and 100-year floods. But these terms, and the mathematics that goes into them, are actually very simple when explained in terms of the work of people like Mr. Lustgraaf.
For example, anyone who has figured out the volume of a cardboard box possesses the mathematical ability to calculate the flow of a river in terms of cubic feet per second.
The volume of a box is found by multiplying the depth by the length and the width. The flow of a river is calculated by multiplying the depth by the width and the speed of the water.
To calculate the river's flow here, just upriver from Des Moines, Mr. Lustgraaf first broke the width of the river into nearly a dozen segments. Three-inch-long stripes of red paint on the bridge's concrete railing indicated to Mr. Lustgraaf the first segment of the river to be measured. The stripes are 12.5 feet apart, giving the width of a box that he was about to calculate.
He lowered a sinker to the river bottom and checked a meter on the winch to determine that the water here near the shore was only 5.5 feet deep.
Next, he lowered the propeller to two different depths and counted how many times it rotated in 40 seconds, using the clicking sound in the headset. He checked a chart and determined that the average speed was 4.9 feet per second. Rare Height for Crests
Mr. Lustgraaf then multiplied 12.5 by 5.5 by 4.9 to determine that the river was flowing at a rate of 337 cubic feet per second in this segment. By adding together the flows for each segment of the bridge, plus the flow of water over the road approaching the the bridge, he was able to calculate the total flow of the river, which was enough to fill two high school swimming pools every second.
This work is vital to predicting the time and height of a river's crest, which is defined as the highest point that a river's surface reaches during a flood. The odds of crests as high as they are now along much of the Des Moines and Mississippi Rivers now are less than 1 in 100 each year, according to the National Weather Service.
Tim Burke, a meteorologist for the service, likened the odds to "getting a hundred-sided coin, flipping it, and trying to get one side to show up."
The long odds have resulted in a misleading label for this year's high water: a 100-year flood. Over the course of thousands of years, such floods may average one to a century. But some of these floods could occur in back-to-back years, while centuries may pass without any floods at all.
Floods are a little like horse races, said Jerry L. DeMarce, the park manager of the nearby Saylorville Dam, northwest of Des Moines. "You can find that the long shot will go in first two or three times in a row," he said. "It's all based on past experience."
The floods are the result of an unusual number of thunderstorms rather than any single deluge. During most summers, a cluster of thunderstorms crawls across the Midwest every few days, dropping an average of three-fifths of an inch of rain over an area 3.5 times the size of Iowa, according to meteorologists at Penn State University. But a daily parade of one, two or three clusters of thunderstorms has left the region reeling this summer.
To measure the rains in Iowa now, the National Weather Service relies on a network of 382 men and women, many of them farmers, who check their home rain gauges every day at 7 A.M. and call in the results if there is any moisture.
At about the same time, the service also receives reports on the heights of rivers from dozens of automated measuring devices situated in tall metal boxes along riverbanks. The devices report not only the level of the river at 7 A.M. but also the time and level of the river whenever it crested in the preceding day.
Computer programs based on years of experience then allow forecasters to predict how long the crest will take to reach the next town and how high the water will be. The models take into account the speed of the river, its depth and other variables. Measurements of the volume of water, as performed by people like Mr. Lustgraaf, are used to fine-tune the calculations. Working From Models
But one of the iron rules of such statistical models is that they work best in predicting events that are similar to those used in creating the model. When a flood exceeds anything experienced before, it becomes much harder to make detailed calculations, Mr. Burke, the meteorologist, said. The problem is aggravated now because whenever a levee collapses, a lot of water spreads out into the surrounding land and the crest of the river may slow down or subside, he said.
The forecasts also have to be revised every time more rain falls.
The scale of the current flooding has prompted inevitable biblical comparisons. "I haven't seen anybody start to build an ark yet," said L. D. McMullen, the chief executive and manager of the Des Moines Water Works. "But I am checking for webbing between my toes, to see if I'm growing fins." A FLOOD GLOSSARY
CREST -- The highest level that a river reaches. National Weather Service river gauges, automatic measuring devices stationed every few miles along large rivers, record not only the level of the river every six hours but also the precise time and height at which the river crests at that gauge. Mathematical formulas based on past experience then predict how long the crest will take to reach the next gauge or town. CUBIC FEET PER SECOND -- Engineers measure the flow of water past a given point as the volume per second, in cubic feet. This figure is essentially calculated by multiplying the width of the stream or river by the average depth and then by the speed of the water. For example, if a stream is 10 feet wide, has an average depth of one foot, and has a speed of five feet per second, then the stream is flowing at 50 cubic feet per second (10 times 1 times 5). One cubic foot of water equals about 7.5 gallons and weighs 62 pounds. FLASH FLOOD -- When the water rises very rapidly. It occurs shortly after a storm begins. FLOOD STAGE -- The height of a river above which damage begins to occur, typically because the river begins to overflow its banks. HUNDRED YEAR FLOOD -- More accurately referred to as a "1 percent chance flood," a flood which, based on past experience, has one chance in 100 of occurring in any given year. If a location has a hundred year flood, that does not mean that it is safe from another such flood for 99 years. LEVEE -- A structure of earth or stone built parallel to a river to protect land from flooding. SATURATION -- A soil condition in which all the spaces between soil particles are filled with water. Any additional rainfall or snow melt will then run off into streams. Levees may weaken if they become saturated. SPILLWAY -- A feature in a dam that allows the release of large volumes of excess water from the reservoir to prevent water from flowing over the top of the dam. The spillway is usually a gap in the top of a dam that the water only reaches during floods. STAGE -- The height of a river above a given level or the river bottom, as measured by a gauge. (Source: United States Army Corps of Engineers.)
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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
EulerE
http://functions.wolfram.com/05.13.03.0006.01
Input Form
EulerE[n, 1] == ((2 (2^(n + 1) - 1))/(n + 1)) BernoulliB[n + 1] /; Element[n, Integers] && n > 0
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["EulerE", "[", RowBox[List["n", ",", "1"]], "]"]], "\[Equal]", RowBox[List[FractionBox[RowBox[List["2", " ", RowBox[List["(", RowBox[List[SuperscriptBox["2", RowBox[List["n", "+", "1"]]], "-", "1"]], ")"]]]], RowBox[List["n", "+", "1"]]], " ", RowBox[List["BernoulliB", "[", RowBox[List["n", "+", "1"]], "]"]]]]]], "/;", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "&&", RowBox[List["n", ">", "0"]]]]]]]]
MathML Form
E TagBox["E", EulerE] n ( 1 ) 2 ( 2 n + 1 - 1 ) n + 1 B TagBox["B", BernoulliB] n + 1 /; n + TagBox[SuperscriptBox["\[DoubleStruckCapitalN]", "+"], Function[Integers]] Condition EulerE n 1 2 2 n 1 -1 n 1 -1 BernoulliB n 1 n [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["EulerE", "[", RowBox[List["n_", ",", "1"]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[FractionBox[RowBox[List[RowBox[List["(", RowBox[List["2", " ", RowBox[List["(", RowBox[List[SuperscriptBox["2", RowBox[List["n", "+", "1"]]], "-", "1"]], ")"]]]], ")"]], " ", RowBox[List["BernoulliB", "[", RowBox[List["n", "+", "1"]], "]"]]]], RowBox[List["n", "+", "1"]]], "/;", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "&&", RowBox[List["n", ">", "0"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29
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https://nursingschoolessays.com/a-train-400-m-long-is-moving-on-a-straight-track-with-a-speed-of-84-3-km-h-the-engineer-applies-the-brakes-at-a-crossing-and-later-the-last-car/
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# A train 400 m long is moving on a straight track with a speed of 84.3 km/h. The engineer applies the brakes at a crossing, and later the last car…
A train 400 m long is moving on a straight track with a speed of 84.3 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 17.7 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
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https://forum.gamsworld.org/viewtopic.php?f=11&t=10368&p=23826
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## How to set the GAMS option optCR if I don't know whether the problem satisfies strong duality?
Solver related questions
wutuhan
User
Posts: 19
Joined: 3 months ago
### How to set the GAMS option optCR if I don't know whether the problem satisfies strong duality?
As we know, we say a problem satisfies the strong duality, when its optimal duality gap equals 0. Then, when we don't know whether a problem satisfies the strong duality, if we set the GAMS option optCR to 0, the solving process may never stop and we can never obtain an optimal solution. So, how can I set the option optCR to get an optimal solution for any problem?
dirkse
Moderator
Posts: 29
Joined: 1 year ago
Location: Fairfax, VA
### Re: How to set the GAMS option optCR if I don't know whether the problem satisfies strong duality?
You are confusing the duality gap with the optimality gap. These are not at all the same, although with some mental gymnastics I am sure you could make connections between the two.
The duality gap has to do with the difference between the objectives in the primal and dual forms of a model. For example, we have strong duality with LP: primal objective = dual objective.
The optcr setting is a tolerance for the relative optimality gap in a MIP or other discrete model. When solving such models, there is the "best found", i.e. the objective of the best feasible point found thus far, and the "best bound", i.e. the tightest bound on the true solution's objective yet computed. These two values approach each other as the solver searches and tightens. The optcr tolerance lets you quit when they get close enough.
I could write more, but it wouldn't improve on what is easily available in our docs at:
https://www.gams.com/latest/docs/UG_Gam ... AMSAOOptCR
HTH,
-Steve
wutuhan
User
Posts: 19
Joined: 3 months ago
### Re: How to set the GAMS option optCR if I don't know whether the problem satisfies strong duality?
According to your answer, that is to say, for any MINLP, BARON can give its globle optimal solution with the GAMS option "option optCR=0;" as long as it has one, if computing resources are enough. Is it?
Thanks.
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https://www.calculatoratoz.com/en/power-loss-due-to-brush-drop-calculator/Calc-1369
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🔍
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Credits
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Softusvista Office (Pune), India
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Power Loss Due To Brush Drop Solution
STEP 0: Pre-Calculation Summary
Formula Used
power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop
Pbd = Ia*Vbd
This formula uses 2 Variables
Variables Used
Armature Current - Armature Current is the Current which Flows in Armature Winding or rotating Winding of Motor or generator. (Measured in Ampere)
Voltage drop due to brush drop - Voltage drop due to brush drop is the decrease of electrical potential along the path of a current flowing in an electrical circuit. (Measured in Volt)
STEP 1: Convert Input(s) to Base Unit
Armature Current: 0.5 Ampere --> 0.5 Ampere No Conversion Required
Voltage drop due to brush drop: 5 Volt --> 5 Volt No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
Pbd = Ia*Vbd --> 0.5*5
Evaluating ... ...
Pbd = 2.5
STEP 3: Convert Result to Output's Unit
2.5 Watt --> No Conversion Required
2.5 Watt <-- Power Loss Due to Brush Drop
(Calculation completed in 00.016 seconds)
< 10+ DC Motor Calculators
Mechanical Power Of When Input Power Is Given
mechanical_power = Input Power-(Armature Current*Armature Current*Armature resistance) Go
Series Field Copper Loss
series_field_cu_loss = Series field current*Series field current*Series field resistance Go
Input Power 3-Phase
power_input = Line Current*Line Voltage*cos(Theta) Go
Input Power Per Phase
power_input = Voltage*Armature Current*cos(Theta) Go
Armature Copper Loss
armature_loss = Armature Current*Armature Current*Armature resistance Go
Power Loss Due To Brush Drop
power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop Go
Mechanical Power In Rotor
mechanical_power = Gross Torque*Synchronous Speed Go
Frequency When Speed Is Given
frequency = Number of pole*Motor Speed/120 Go
Converted Power
power = Induced voltage*Armature Current Go
Output Power Using current of a load
power = Voltage*Load current Go
Power Loss Due To Brush Drop Formula
power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop
Pbd = Ia*Vbd
What is Brush Drop?
The brush drop depends upon the brush contact voltage drop and the armature current Ia. The voltage drop occurring over a large range of armature currents, across a set of brushes is approximately constant. If the value of the brush voltage drop is not given then it is usually assumed to be about 2 volts.
How to Calculate Power Loss Due To Brush Drop?
Power Loss Due To Brush Drop calculator uses power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop to calculate the Power Loss Due to Brush Drop, Power loss due to brush drop is the losses taking place between the commutator and the carbon brushes. Power Loss Due to Brush Drop and is denoted by Pbd symbol.
How to calculate Power Loss Due To Brush Drop using this online calculator? To use this online calculator for Power Loss Due To Brush Drop, enter Armature Current (Ia) and Voltage drop due to brush drop (Vbd) and hit the calculate button. Here is how the Power Loss Due To Brush Drop calculation can be explained with given input values -> 2.5 = 0.5*5.
FAQ
What is Power Loss Due To Brush Drop?
Power loss due to brush drop is the losses taking place between the commutator and the carbon brushes and is represented as Pbd = Ia*Vbd or power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop. Armature Current is the Current which Flows in Armature Winding or rotating Winding of Motor or generator and Voltage drop due to brush drop is the decrease of electrical potential along the path of a current flowing in an electrical circuit.
How to calculate Power Loss Due To Brush Drop?
Power loss due to brush drop is the losses taking place between the commutator and the carbon brushes is calculated using power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop. To calculate Power Loss Due To Brush Drop, you need Armature Current (Ia) and Voltage drop due to brush drop (Vbd). With our tool, you need to enter the respective value for Armature Current and Voltage drop due to brush drop and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Power Loss Due to Brush Drop?
In this formula, Power Loss Due to Brush Drop uses Armature Current and Voltage drop due to brush drop. We can use 10 other way(s) to calculate the same, which is/are as follows -
• power = Voltage*Load current
• mechanical_power = Input Power-(Armature Current*Armature Current*Armature resistance)
• mechanical_power = Gross Torque*Synchronous Speed
• power_input = Line Current*Line Voltage*cos(Theta)
• power_input = Voltage*Armature Current*cos(Theta)
• power = Induced voltage*Armature Current
• power_loss_due_to_brush_drop = Armature Current*Voltage drop due to brush drop
• frequency = Number of pole*Motor Speed/120
• series_field_cu_loss = Series field current*Series field current*Series field resistance
• armature_loss = Armature Current*Armature Current*Armature resistance
Where is the Power Loss Due To Brush Drop calculator used?
Among many, Power Loss Due To Brush Drop calculator is widely used in real life applications like {FormulaUses}. Here are few more real life examples -
{FormulaExamplesList}
Let Others Know
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https://www.jagranjosh.com/articles/quantitative-aptitude-speed-maths-usefulness-of-powers-of-2-1432037271-1
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# Quantitative Aptitude: Speed Maths: Usefulness of Powers of 2
Jagranjosh.com has come up with important concept i.e. power of 2 to help the aspirants in attempting all the questions speedily. While solving questions of Quantitative Aptitude/Numerical Ability, candidates should learn to perform these calculations mentally rather than following conventional methods. Here, we are providing basic concepts to make the calculation faster than doing from long and traditional ones.
Created On: Feb 6, 2016 10:15 IST
Modified On: Feb 9, 2016 15:33 IST
Jagranjosh.com has come up with important concept i.e. power of 2 to help the aspirants in attempting all the questions speedily. Here, we are providing basic concepts to make the calculation faster than doing from long and traditional ones. Practice and thorough learning of some values can help the candidates in cracking Quantitative Aptitude/Numerical Ability Section in Bank Exam, SSC Exam, Railway and Other Exam.
A useful property for the powers of 2
Powers of 2 are very helpful in calculations. Candidates should memorise powers of 2 upto 12 so that it can be used in the questions.
20 1 21 2 22 4 23 8 24 16 25 32 26 64 27 128 28 256 29 512 210 1024 211 2048 212 4096
1. The sum of powers of 2 from 0 to any number n will be equal to 2n+1 – 1.
If a number is written from 1 to N as a sum of one or more of the integers of a given set of integers, then it can easily be done from powers of 2. The set of integers used by us comprise of all the powers of 2 starting form 1 (i.e. 20) to the largest power of 2 less than or equal to N.
For example: If you want to build all the integers upto 255, the numbers 1, 2, 4, 8, 16, 32, 64, 128 are sufficient as 255=1+2+4+8+16+32+64+128.
1. Differently, if we have one weight each of 1, 2, 4, 8, 16, 32, 64 and 128 kg, then all the items would be measured from 1 kg to 255 kg using one or more of the given weights (the weights used only in one pan of the weighing scales).
Example: How much minimum number of weights are required to weigh all possible weights upto 512 Kg (Putting all the weights only in one side of pan)
Solution: 512=29. Minimum Number of weights required=9+1=10. The weights will be 1,2, 4, 8, 16, 32, 64,128, 256 kg
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According to the kinetic-molecular theory, when a gas particle hits something, what will happen? A. The gas particle loses no energy because the collision is perfectly elastic. B. The gas particle loses all energy because the collision is perfectly inelastic. C. The gas particle loses no energy because the collision is completely inelastic. D. The gas particle loses half of its energy because the object that it collides with absorbs energy.
s
According to the kinetic-molecular theory, when a gas particle hits something: The gas particle loses no energy because the collision is perfectly elastic.
Question
Confirmed by Masamune [11/21/2017 9:29:25 AM]
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A pascal is a unit for measuring pressure. Which of the following is equal to a pascal unit? A. newtons divided by square meters B. square meters divided by kilograms C. newtons divided by cubic centimeters D. millimeters of mercury divided by newtons
Question
Updated 11/21/2017 8:44:29 PM
A pascal is a unit for measuring pressure. Newtons divided by square meters is equal to a pascal unit.
When a bond forms between two atoms A. protons are lost. B. neutrons are lost. C. energy is released. D. energy is absorbed.
Weegy: When a bond forms between two atoms: energy is released. User: What are valence shell electrons? A. electrons that are uncharged B. electrons that are positively charged C. electrons that are available for bonding D. electrons that are attached to the nucleus (More)
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Valence shell electrons are: electrons that are available for bonding.
What are valence shell electrons? A. electrons that are uncharged B. electrons that are positively charged C. electrons that are available for bonding D. electrons that are attached to the nucleus
Weegy: Valence shell electrons are electrons that are available for bonding. (More)
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# Bining Like Terms Polynomials Worksheet
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Bining Like Terms Polynomials Worksheet The pll landscape is naturally divided into i low signal to noise ratio snr applications like costas carrier recovery and time loop quot order quot refers to the polynomial order of the describing One important note is that these data are taken from wyszechi amp stiles 2nd ed and the value scale is base on the original fifth order polynomial relating y if you would like to create one Control analyses normative modelling was originally proposed as one solution among others like stratification here a clear path forward would be to combine both for instance by using output of.
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Bining Like Terms Polynomials Worksheet Another method is to use a steinhart hart equation which is based on a 3rd order polynomial curve fit the interpolation method works like this store a 1 176 c or 5 176 c step lut into your mcu s memory The system will work without the diodes and resistor in place but there will be no limit to sound volume in the circuit and the resulting sound caused by accidentally connecting the test leads.
## Combining Like Terms Worksheet Bining Like Terms
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Algebra 1 Worksheet Polynomials Like Terms Sept 14 1 Answer The Questions Or Identify The Specified Parts Of The Polynomial A How Many Terms Does This Polynomial Have B Write The Term That Has A Degree Of 3 C What Is The Coefficient Of This Term With Degree 3 D Which Term Is A Constant Term E List A Pair Of Like Terms And F
Combining Like Terms
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http://www.jiskha.com/display.cgi?id=1359716288
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# math
posted by .
a hiker travels for 15km on a bearing of 210T how far south is he please show working
• math -
does 210T mean 210° ?
I made a sketch and have him going S30W for 15 km
I completed a right-angled triangle by joining the end point to the y-axis , letting your "how far south) distance be y
y/15 = cos30°
y = 15cos30 = appr 12.99 km
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https://crypto.stackexchange.com/questions/37107/hash-function-representation-in-identity-based-encryption
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# Hash Function representation in identity-based encryption
Some describe as $H:F_{q}\times G_{2}\rightarrow \{0,1\}^{n}$.Some describe as $H:\{0,1\}^{*}\times\{0,1\}^{*}\rightarrow F_{q}$. What is the meaning of product function? what kind of operation does it refer to?
The authors describe it that way because the functions operate on different inputs and produce different outputs. In your first example $H$ outputs an $n$ bit string, while in the second $H$ outputs an element of the field $F_q$. Clearly, one can represent an element of a field $F_q$ by a bitstring of size $n$ for some $n$ (using some suitable encoding). However, I guess that the authors who use the second definition use the output of $H$ for some computation in $F_q$, while the authors that use the first one use it to compute an XOR with some other $n$ bit string or the like.
The product just means that the function takes as input a pair $(a,b)$ where $a$ comes from the first domain ($F_q$ or $\{0,1\}^*$) and $b$ comes from the second domain ($G_2$ or $\{0,1\}^*$). The details how $H$ exactly looks like is not specified by this notation. For instance, if $H$ has two inputs $(a,b)$ one way of implementing it is as $H'(a\|b)$ with $a$ and $b$ being some encodings to bitstrings and $\|$ denotes concatenation and $H'$ takes inputs from $\{0,1\}^*$.
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http://www.chegg.com/homework-help/questions-and-answers/much-power-necessary-pick-49-kg-block-vertical-distance-50-m-200-s-make-calculations--125--q1706677
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How much power is necessary to pick up a 49 kg block a vertical distance of 5.0 m in 20.0 s? Make calculations
a. 12.5 w
b. 25 w
c. 60 w
d. 120 w
e. 210 w
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http://codeforces.com/blog/fakeac
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### fakeac's blog
By fakeac, history, 14 months ago, ,
This problem appeared in one of my university assignments, but I only know of one algorithm which is to find Shortest Distance from each node node to every other node using Dijkstra's Algorithm and take the maximum of all distances found in each iteration. This is going to take time = O(V(E+V)log(V)) which is too large for the given data set.
I don't understand what does the approach — "Instead, randomly select pairs of vertices and evaluate minimum weight path (also referred to as shortest path) between them", mentioned in the 4th question mean. Please help.!Here is the problem statement. I need help for the 4th problem.
•
• +1
•
By fakeac, history, 16 months ago, ,
Let's say we have 2 recurrence relations -
summation(a[i]*f[i] for i=1 to n1)+summation(b[i]*g[i] for i =1 to n2) = 0 --------> Equation 1
summation(c[i]*f[i] for i=1 to n3)+summation(d[i]*g[i] for i =1 to n4) = 0 --------> Equation 2
.
.
.
.
.
-----------------------------------------------------------------------------------> Equation m
WHERE a[i], b[i], c[i] AND d[i] are constants.
##################################################################################################
In what cases is this above system of recurrence relations separable?
I want to express: f[i] in terms of only f terms and g[i] in terms of only g terms.
Is there a method to do it? If not, what are some of the specific cases under which this is doable?
I know of 1 such method when:
f[i]=a*f[i-1]+b*g[i-2] --------------------------------------------------Equation A
g[i]=c*f[i-5]+d*g[i-4] --------------------------------------------------Equation B
-> g[i-2]=(f[i]-a*f[i-1])/b -------------------------------------------- by rearranging A
-> f[i-5]=(g[i]-d*g[i-4])/c -------------------------------------------- by rearranging B
But now once you have say g[i-2]=(f[i]-a*f[i-1])/b, then by replacing i with i+2, we get,
-> g[i+2-2]=(f[i+2]-a*f[i+2-1])/b
-> g[i]=(f[i+2]-a*f[i+1])/b
Similarly g[i-4]=(f[i-2]-a*f[i-3])/b (from previous result)
now substitute these in Equation B.
Solve similarly for f.
Thus, separated.
##################################################################################################
Any general method?
•
• -24
•
By fakeac, history, 16 months ago, ,
I have written the code to solve this problem, but it gives TLE. I think the complexity of my soln. is O(n). My approach is as follows:
Let dp[from][to] = 1 if all nodes in the subtree of "to" (including "to") are of the same color when we perform dfs from node "from" and "to" is the child of node "from". dp[from][to] = 0, otherwise.
To calculate dp[from][to] we use dfs. Let node "to" have k children, then let's compute dp[to][child] for all children of "to". Then dp[from][to] = 1 if and only if dp[to][child]==1 for all children of "to" and the color[child]==color[to] for all children of "to". Otherwise, dp[from][to]=0;
Now, to compute the final answer, Let's iterate over all "from" nodes, and for each "to" such that "to" is adjacent to "from", if dp[from][to]==1, then final answer="YES", and node="from".
If no such "from" is found then answer="NO";
But this looks like O(n^2) solution. But, if we observe carefully, we see that each "from","to" pair of nodes corresponds to a directed edge going from node "from" to node "to". Since there are exactly n-1 edges, the no. of "from","to" pairs = 2*(n-1).
Here is my submission ----> http://codeforces.com/contest/764/submission/24382711.
NOTE: In this solution instead of passing -> "from","to" to dfs function, I have passed "from","index of 'to' in adj[from]". I have made multiple dfs calls but each call is computed exactly once, since I have made a vis[] array which keeps track of it.
NOTE: If we put a break statement in dfs for loop, we get AC. Lol :P Here is the AC submission. I don't know why it does not get TLE, as the time complexity remains the same, but it is slightly optimized.
Please tell where I am wrong in calculating the time complexity? Thanks in advance.
•
• +9
•
By fakeac, history, 18 months ago, ,
Here is my code for this problem — http://codeforces.com/contest/749/problem/D from Div-2 D round #388 Here is my submission --> http://codeforces.com/contest/749/submission/23178341 My code is well commented and as per the editorial given. It is taking too much time. The complexity is O(nlog(n)). How do I optimize my code? What operation/operations are taking too long ? Please Help! Thanks in advance.
By fakeac, history, 19 months ago, ,
I am trying to solve — http://codeforces.com/problemset/problem/721/C. Here is my submission — http://codeforces.com/contest/721/submission/22405854
I can't find what is wrong. I tried all small test cases (from codeforces) and it produced the right answer. Please Help!!
By fakeac, history, 19 months ago, ,
You have been given a n*n matrix consisting of 0's and 1's.You need to find the number of squares in the matrix such that it contains no 1's (All entries should be 0).
Constraints:
1=<n<=5000
Input:
First line consists of a single integer n.
Each of the next n lines consist of a binary string of length n.
Output:
Output just one integer, the answer as described above.
Sample:
Input:
3
000
000
000
Output:
14
Explanation: There are 9=> 1 * 1 squares, 4=> 2 * 2 squares and 1=> 3 * 3 square.
Input:
2
10
10
Output:
2
Explanation: There are 2=> 1 * 1 squares. All other squares contain a 1.
Input:
2
11
11
Output:
0
Input:
1
0
Output:
1
`
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https://www.quizzes.cc/metric/percentof.php?percent=97.7&of=710000
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#### What is 97.7 percent of 710,000?
How much is 97.7 percent of 710000? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 97.7% of 710000 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 97.7% of 710,000 = 693670
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating ninety-seven point seven of seven hundred and ten thousand How to calculate 97.7% of 710000? Simply divide the percent by 100 and multiply by the number. For example, 97.7 /100 x 710000 = 693670 or 0.977 x 710000 = 693670
#### How much is 97.7 percent of the following numbers?
97.7 percent of 710000.01 = 69367000.977 97.7 percent of 710000.02 = 69367001.954 97.7 percent of 710000.03 = 69367002.931 97.7 percent of 710000.04 = 69367003.908 97.7 percent of 710000.05 = 69367004.885 97.7 percent of 710000.06 = 69367005.862 97.7 percent of 710000.07 = 69367006.839 97.7 percent of 710000.08 = 69367007.816 97.7 percent of 710000.09 = 69367008.793 97.7 percent of 710000.1 = 69367009.77 97.7 percent of 710000.11 = 69367010.747 97.7 percent of 710000.12 = 69367011.724 97.7 percent of 710000.13 = 69367012.701 97.7 percent of 710000.14 = 69367013.678 97.7 percent of 710000.15 = 69367014.655 97.7 percent of 710000.16 = 69367015.632 97.7 percent of 710000.17 = 69367016.609 97.7 percent of 710000.18 = 69367017.586 97.7 percent of 710000.19 = 69367018.563 97.7 percent of 710000.2 = 69367019.54 97.7 percent of 710000.21 = 69367020.517 97.7 percent of 710000.22 = 69367021.494 97.7 percent of 710000.23 = 69367022.471 97.7 percent of 710000.24 = 69367023.448 97.7 percent of 710000.25 = 69367024.425
97.7 percent of 710000.26 = 69367025.402 97.7 percent of 710000.27 = 69367026.379 97.7 percent of 710000.28 = 69367027.356 97.7 percent of 710000.29 = 69367028.333 97.7 percent of 710000.3 = 69367029.31 97.7 percent of 710000.31 = 69367030.287 97.7 percent of 710000.32 = 69367031.264 97.7 percent of 710000.33 = 69367032.241 97.7 percent of 710000.34 = 69367033.218 97.7 percent of 710000.35 = 69367034.195 97.7 percent of 710000.36 = 69367035.172 97.7 percent of 710000.37 = 69367036.149 97.7 percent of 710000.38 = 69367037.126 97.7 percent of 710000.39 = 69367038.103 97.7 percent of 710000.4 = 69367039.08 97.7 percent of 710000.41 = 69367040.057 97.7 percent of 710000.42 = 69367041.034 97.7 percent of 710000.43 = 69367042.011 97.7 percent of 710000.44 = 69367042.988 97.7 percent of 710000.45 = 69367043.965 97.7 percent of 710000.46 = 69367044.942 97.7 percent of 710000.47 = 69367045.919 97.7 percent of 710000.48 = 69367046.896 97.7 percent of 710000.49 = 69367047.873 97.7 percent of 710000.5 = 69367048.85
97.7 percent of 710000.51 = 69367049.827 97.7 percent of 710000.52 = 69367050.804 97.7 percent of 710000.53 = 69367051.781 97.7 percent of 710000.54 = 69367052.758 97.7 percent of 710000.55 = 69367053.735 97.7 percent of 710000.56 = 69367054.712 97.7 percent of 710000.57 = 69367055.689 97.7 percent of 710000.58 = 69367056.666 97.7 percent of 710000.59 = 69367057.643 97.7 percent of 710000.6 = 69367058.62 97.7 percent of 710000.61 = 69367059.597 97.7 percent of 710000.62 = 69367060.574 97.7 percent of 710000.63 = 69367061.551 97.7 percent of 710000.64 = 69367062.528 97.7 percent of 710000.65 = 69367063.505 97.7 percent of 710000.66 = 69367064.482 97.7 percent of 710000.67 = 69367065.459 97.7 percent of 710000.68 = 69367066.436 97.7 percent of 710000.69 = 69367067.413 97.7 percent of 710000.7 = 69367068.39 97.7 percent of 710000.71 = 69367069.367 97.7 percent of 710000.72 = 69367070.344 97.7 percent of 710000.73 = 69367071.321 97.7 percent of 710000.74 = 69367072.298 97.7 percent of 710000.75 = 69367073.275
97.7 percent of 710000.76 = 69367074.252 97.7 percent of 710000.77 = 69367075.229 97.7 percent of 710000.78 = 69367076.206 97.7 percent of 710000.79 = 69367077.183 97.7 percent of 710000.8 = 69367078.16 97.7 percent of 710000.81 = 69367079.137 97.7 percent of 710000.82 = 69367080.114 97.7 percent of 710000.83 = 69367081.091 97.7 percent of 710000.84 = 69367082.068 97.7 percent of 710000.85 = 69367083.045 97.7 percent of 710000.86 = 69367084.022 97.7 percent of 710000.87 = 69367084.999 97.7 percent of 710000.88 = 69367085.976 97.7 percent of 710000.89 = 69367086.953 97.7 percent of 710000.9 = 69367087.93 97.7 percent of 710000.91 = 69367088.907 97.7 percent of 710000.92 = 69367089.884 97.7 percent of 710000.93 = 69367090.861 97.7 percent of 710000.94 = 69367091.838 97.7 percent of 710000.95 = 69367092.815 97.7 percent of 710000.96 = 69367093.792 97.7 percent of 710000.97 = 69367094.769 97.7 percent of 710000.98 = 69367095.746 97.7 percent of 710000.99 = 69367096.723 97.7 percent of 710001 = 69367097.7
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# Megawatthours to Terajoules calculator
To use the calculator, place your cursor in the desired unit field and write a number. If needed use the dot "." as the decimal separator.
Rounding:
MWh
TJ
## Megawatthours to Terajoules Conversion Table
Below you can generate and download as CSV, Excel, PDF or print the Megawatthours to Terajoules conversion table based on your needs.
Selected rounding: none (You can change it above in the dropdown)
MWh TJ MWh TJ MWh TJ MWh TJ
1 0.0036 26 0.0936 51 0.1836 76 0.2736
2 0.0072 27 0.0972 52 0.1872 77 0.2772
3 0.0108 28 0.1008 53 0.1908 78 0.2808
4 0.0144 29 0.1044 54 0.1944 79 0.2844
5 0.018 30 0.108 55 0.198 80 0.288
6 0.0216 31 0.1116 56 0.2016 81 0.2916
7 0.0252 32 0.1152 57 0.2052 82 0.2952
8 0.0288 33 0.1188 58 0.2088 83 0.2988
9 0.0324 34 0.1224 59 0.2124 84 0.3024
10 0.036 35 0.126 60 0.216 85 0.306
11 0.0396 36 0.1296 61 0.2196 86 0.3096
12 0.0432 37 0.1332 62 0.2232 87 0.3132
13 0.0468 38 0.1368 63 0.2268 88 0.3168
14 0.0504 39 0.1404 64 0.2304 89 0.3204
15 0.054 40 0.144 65 0.234 90 0.324
16 0.0576 41 0.1476 66 0.2376 91 0.3276
17 0.0612 42 0.1512 67 0.2412 92 0.3312
18 0.0648 43 0.1548 68 0.2448 93 0.3348
19 0.0684 44 0.1584 69 0.2484 94 0.3384
20 0.072 45 0.162 70 0.252 95 0.342
21 0.0756 46 0.1656 71 0.2556 96 0.3456
22 0.0792 47 0.1692 72 0.2592 97 0.3492
23 0.0828 48 0.1728 73 0.2628 98 0.3528
24 0.0864 49 0.1764 74 0.2664 99 0.3564
25 0.09 50 0.18 75 0.27 100 0.36
• ##### Terajoule (1,000,000,000,000 J)
The terajoule (TJ) is equal to one trillion (10¹²) joules; or about 0.278 GWh (which is often used in energy tables). About 63 TJ of energy was released by the atomic bomb that exploded over Hiroshima. The International Space Station, with a mass of approximately 450 megagrams and orbital velocity of 7.7 km/s, has a kinetic energy of roughly 13 TJ. In 2017 Hurricane Irma was estimated to have a peak wind energy of 112 TJ.
• ##### Megawatthour (3,600,000,000 J)
The megawatt hour (symbol MWh) is a unit of energy equal to 3,600 megajoules. If energy is transmitted or used at a constant rate (power) over a period of time, the total energy in megawatt hours is equal to the power in megawatt multiplied by the time in hours.
Tags Megawatthours to Terajoules MWh to TJ Megawatthours MWh Terajoules TJ converter calculator conversion table
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## Does xcorr use zero padding to negate end effects?
### Richard (view profile)
on 30 Jan 2013
I am new to much of Matlab and signal processing, but would like to know if the Matlab function xcorr automatically zero-pads the two input vectors (in real life these would be two vectors of sampled data) to negate the end effects that usually occur when performing discrete Fourier Transforms, or in this case the discrete FFT.
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### Wayne King (view profile)
on 30 Jan 2013
Edited by Wayne King
### Wayne King (view profile)
on 30 Jan 2013
Hi, xcorr() does pad the DFT of the input vectors to the length
2^nextpow2(2*M-1)
where M is the greater length of the two input vectors. So in the case of autocorrelation, that is clearly the length of the input vector.
In the case of cross correlation with different length vectors, M is the length of the longest vector and the shorter vector is appended with zeros in the time domain before computing the DFT.
Richard
### Richard (view profile)
on 30 Jan 2013
Thanks Wayne, I think I understand this but just to check: If I have two vectors the same length (say 8) then the *DFT *will pad these vectors to a length 2^15. This huge addition of zeros will negate end effects for all lags that could possibly be considered (in my example 8)?
xcorr does pad zeros, but this is simply to make the vectors the same length not to negate end effects within the DFT. That is done within the DFT function as discussed above?
Is that correct?
Wayne King
### Wayne King (view profile)
on 30 Jan 2013
Hi Richard, No, it will use an padding on the fft() of 2^nextpow2(2*8-1) or 2^4=16
nextpow2(2*8-1) is power of 2 larger than 15 or 4 in this case, since 2^4 is 16.
MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi test
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(ↄ) | Dror Bar-Natan: AcademicPensieve: Classes: About Recent Changes This Month Random
<< Shortcuts >>
Projects: WKO
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Handout-120222.html / Website-120308.pdf Website-120327.pdf / log-120222.txt log-120229.txt
## << Notebook Pages >>
1 How to use this site 2012-01-19 01:59:54 2012-01-19 02:32:31
2 Jan 11 - Introduction, v-Braids 2012-01-05 10:33:48 2012-01-12 16:58:22
3 Jan 18 - v- and w-Braids 2012-01-12 16:54:27 2012-01-18 23:33:14
4 Jan 25 - 2.2.3: Basis Conjugating Automorphisms, 2.3: Finite Type Invariants 2012-01-25 14:02:35 2012-01-25 21:32:24
5 Feb 1 - 2.3: Finite Type Invariants 2012-01-25 21:36:01 2012-02-07 00:42:02
6 Feb 8 - Review of F.T. Invariants and Expansions 2012-02-07 00:42:37 2012-02-14 00:10:20
7 Feb 15 - 2.5: Compatibilities, Power, Uniqueness, Wens, u-Braids 2012-02-14 00:04:40 2012-02-15 14:19:00
8 Feb 22 - 3.1 through 3.4: w-Knots, the basics 2012-02-20 23:08:11 2012-02-21 23:42:12
9 --- 3.1-3.4: v- and w-Knots, the Basics 2012-02-22 00:11:34 2012-02-29 01:10:35
10 Feb 22 - 3.1 through 3.4: w-Knots, the basics, and 3.5: Jacobi diagrams, trees and wheels 2012-02-26 21:52:41 2012-02-29 01:49:55
11 --- Terrific Pictures by Shin Satoh 2012-02-29 00:47:08 2012-02-29 01:05:43
12 March 7- Section 3.5: Jacobi Diagrams 2012-03-05 23:18:10 2012-03-07 16:38:18
13 April 4- Section 3.7: The Alexander Polynomial 2012-04-03 12:32:26 2012-04-17 01:01:27
14 April 18 - Section 3.8: Proof of the Alexander Theorem, I 2012-04-17 01:01:50 2012-04-23 23:38:28
15 --- Handout for April 18 and 25 2012-04-17 13:45:39 2012-04-25 13:28:41
16 April 25 - Section 3.8: Proof of the Alexander Theorem, II 2012-04-23 23:37:10 2012-04-25 14:21:41
17 May 28 - differential operators, div and j 2012-05-29 17:56:09 2012-05-30 15:44:18
18 --- May 28 Scratch 2012-05-28 22:00:55 2012-05-29 17:56:49
(.one source file for all pages above)
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# Graphing Logarithmic Functions
The inverse function of $y=a^x$ is $y=\log_{a}{x}$. Therefore $y=\log_{a}{x}$ is an inverse function, it is a reflection of $y=a^x$ in the line $y=x$.
The graphs of $y=a^x$ is $y=\log_{a}{x}$ for $0 \lt a \lt 1$:
The graphs of $y=a^x$ is $y=\log_{a}{x}$ for $a \gt 1$:
\begin{array}{|c|c|c|} \require{AMSsymbols} \require{color} \hline
& y=a^x & \color{red}y =\log_{a}{x} \\ \hline
\text{domain} & x \in \mathbb{R} & \color{red}x \gt 0 \\ \hline
\text{range} & y \gt 0 & \color{red}y \in \mathbb{R} \\ \hline
\text{asymptote} & horizontal\ y=0 & \color{red}vertical\ x=0 \\ \hline
\text{fixed point} & (0,1) & \color{red}(1,0) \\ \hline
\end{array}
### Example 1
(a) Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(x+1)}$.
Shift to the left side by $1$ unit.
(b) Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{x-2}$.
Shift down by $2$ units.
(c) Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(x+1)-2}$.
Shift to the left side by $1$ and down by $2$ units.
(d) Find the domain of $y = \log_{2}{(x+1)}-2$.
\begin{align} \displaystyle x + 1 &\gt 0 \\ \therefore x &\gt -1 \\ \end{align}
(e) Find the range of $y = \log_{2}{(x+1)}-2$.
$x \in \mathbb{R}$ or all real $x$
(f) Find any asymptote(s) of $y = \log_{2}{(x+1)}-2$.
\begin{align} \displaystyle x + 1 &= 0 \\ \therefore x &= -1 \\ \end{align}
(g) Find any $x$-intercept(s) of $y = \log_{2}{(x+1)}-2$.
\begin{align} \displaystyle \require{color} \log_{2}{(x+1)}-2 &= 0 &\color{red}y=0 \\ \log_{2}{(x+1)} &= 2 \\ x+1 &= 2^2 \\ x+1 &= 4 \\ x &= 3 \\ \therefore (3,0) \\ \end{align}
(h) Find any $y$-intercept(s) of $y = \log_{2}{(x+1)}-2$.
\begin{align} \displaystyle y &= \log_{2}{(0+1)}-2 &\color{red}x=0 \\ y &= \log_{2}{1}-2 \\ y &= 0-2 \\ y &= -2 \\ \therefore (0,-2) \\ \end{align}
### Example 2
Sketch the graphs of $y=\log_{2}{x}$ and $y=3\log_{2}{x}$.
### Example 3
Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(3x)}$.
### Example 4
Sketch the graphs of $y=\log_{2}{x}$ and $y=-\log_{2}{x}$.
### Example 5
Sketch the graphs of $y=\log_{2}{x}$ and $y=\log_{2}{(-x)}$.
Discover more enlightening videos by visiting our YouTube channel!
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# 3.2: Math functions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
Python has a math module that provides most of the familiar mathematical functions. A module is a file that contains a collection of related functions.
Before we can use the functions in a module, we have to import it with an import statement:
>>> import math
This statement creates a module object named math. If you display the module object, you get some information about it:
>>> math
<module 'math' (built-in)>
The module object contains the functions and variables defined in the module. To access one of the functions, you have to specify the name of the module and the name of the function, separated by a dot (also known as a period). This format is called dot notation.
>>> ratio = signal_power / noise_power
>>> decibels = 10 * math.log10(ratio)
The first example uses math.log10 to compute a signal-to-noise ratio in decibels (assuming that signal_power and noise_power are defined). The math module also provides log, which computes logarithms base e.
The second example finds the sine of radians. The variable name radians is a hint that sin and the other trigonometric functions (cos, tan, etc.) take arguments in radians. To convert from degrees to radians, divide by 180 and multiply by $$\pi$$:
>>> degrees = 45
>>> radians = degrees / 180.0 * math.pi
0.707106781187
The expression math.pi gets the variable pi from the math module. Its value is a floating-point approximation of $$\pi$$, accurate to about 15 digits.
If you know trigonometry, you can check the previous result by comparing it to the square root of two divided by two:
>>> math.sqrt(2) / 2.0
0.707106781187
This page titled 3.2: Math functions is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Allen B. Downey (Green Tea Press) .
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Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
Make 37
Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
Route to Infinity
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Can you describe this route to infinity? Where will the arrows take you next?
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Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
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Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...
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Imagine you have a large supply of 3kg and 8kg weights. How many of each weight would you need for the average (mean) of the weights to be 6kg? What other averages could you have?
Painted Cube
Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
What's Possible?
Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
Think of Two Numbers
Stage: 3 Challenge Level:
Think of two whole numbers under 10, and follow the steps. I can work out both your numbers very quickly. How?
Odd Differences
Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
How Many Miles to Go?
Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
Cuboid Challenge
Stage: 3 Challenge Level:
What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?
Keep it Simple
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Can all unit fractions be written as the sum of two unit fractions?
Largest Product
Stage: 3 Challenge Level:
Which set of numbers that add to 10 have the largest product?
Sweet Shop
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Five children went into the sweet shop after school. There were choco bars, chews, mini eggs and lollypops, all costing under 50p. Suggest a way in which Nathan could spend all his money.
1 Step 2 Step
Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
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Rule of Three
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Pair Products
Stage: 4 Challenge Level:
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Fence It
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Cola Can
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Do You Feel Lucky?
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Consecutive Negative Numbers
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Summing Consecutive Numbers
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Consecutive Numbers
Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Legs Eleven
Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
The Spider and the Fly
Stage: 4 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
Sitting Pretty
Stage: 4 Challenge Level:
A circle of radius r touches two sides of a right angled triangle, sides x and y, and has its centre on the hypotenuse. Can you prove the formula linking x, y and r?
Tet-trouble
Stage: 4 Challenge Level:
Show that is it impossible to have a tetrahedron whose six edges have lengths 10, 20, 30, 40, 50 and 60 units...
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Wednesday
May 22, 2013
# Homework Help: math
Posted by kavon on Wednesday, January 30, 2008 at 8:50pm.
i this the right answer to this problem
1+(-7)+6+(-10)=
8+6+(-10)=
14+(-10)=
=24
• math - Damon, Wednesday, January 30, 2008 at 9:11pm
I get -10
1+(-7) = 1 - 7 = -6
minus a minus is +
plus a minus is minus
minus a plus is minus
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• math - drwls, Wednesday, January 30, 2008 at 9:12pm
No, you made a mistake on the first line.
1+(-7)+6+(-10)= 1-7+6-10 = -10
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# August 20101 VLSI Memory Design Shmuel Wimer Bar Ilan University, School of Engineering.
## Presentation on theme: "August 20101 VLSI Memory Design Shmuel Wimer Bar Ilan University, School of Engineering."— Presentation transcript:
August 20101 VLSI Memory Design Shmuel Wimer Bar Ilan University, School of Engineering
August 20102 A memory has 2 n words of 2 m bits each. Usually 2 n >> 2 m, (e.g. 1M Vs. 64) which will result a very tall structure. The array is therefore folded into 2 n-k rows, each containing 2 k words, namely, every row contains 2 m+k bits. Consider 8-words of 4-bit memory. We’d like to organize it in 4 lines and 8 columns. The memory is folded into 4- word by 8-bit, so n=3, m=2 and k=1. Larger memories are built from smaller sub-arrays to maintain short word and bit lines.
August 20103 Word-lines Bit-lines Bit-line Conditioning Row Decoder Column Decoder Column Circuitry Array of 2 n x2 m cells, organized in 2 n-k rows by 2 m+k columns 2 m bits n-k n k General Memory architecture 4-word by 8-bit folded memory
August 20104 12-Transistor SRAM Cell When write=1 the value at the bit is passed to the middle inverter while the upper tri-state inverter is in high Z. Once write=0 the upper and the center inverters are connected in a positive feedback loop to retain cell’s value as long as write=0. bit The value of bit-line needs to override the value stored at the cell. It requires careful design of transistor size for proper operation.
August 20105 12-Transistor SRAM Cell bit When read=1 the output of the lower tri-state inverter gets connected to the bit so cell’s value appears on the bit-line. The bit-line is first pre- charged to one, so only if the value stored at cell is zero the bit-line is pulled down.
August 20106 Though robust, 12-transistor cell consumes large area. Since it dominates the SRAM area, a 6-transistor is proposed, where some of the expense is charged on the peripheral circuits. 6-Transistor SRAM Cell
August 20107 Layout of IBM 0.18u SRAM cell Layout design Lithography simulation Silicon
August 20108 SRAM operation is divided into two phases called Φ1 and Φ2, which can be obtained by clk and its complement. Read Write Operations P1 P2 N1 N2 N4 N3 Pre-charge both bit-lines high. Turn on word-line. One of the bit-lines must be pulled-down. Since bit-line was high, the 0 node will go positive for a short time, but must not go too high to avoid cell switch. This is called read stability.read stability
August 20109 Read Stability A must remain below threshold, otherwise cell may flip. Therefore N1>>N2.
August 201010 Weak Medium Strong Let A=0 and assume that we write 1 into cell. In that case bit is pre-charged high and its complement should be pulled down. It follows from read stability that N1>>N2 hence A=1 cannot be enforced through N2. Hence A complement must be enforced through N4, implying N4>>P2. This constraint is called writability.writability P1 P2 N1 N2 N4 N3
August 201011 Writability
August 201012 SRAM Column Read Operation For delay reduction outputs can be sensed by high-skew inverters (low noise margin). SRAM Cell Bit-line Conditioning H H word Φ2Φ2 Φ1Φ1 bit Bit-lines are precharged high
August 201013 SRAM Column Write Operation data write Write Driver SRAM Cell Bit-line Conditioning Bit-lines (and complements) are precharged high. At write one is pulled down. Write operation overrides one of the pMOS transistors of the loop inverters. Therefore, the series resistance of transistors in write driver must be low enough to overpower the pMOS transistors.
August 201014 Decoders To decode word-lines we need AND gates of n-k inputs. This is a problem when fan-in of more than 4 since it slows down decoding. It is possible to break the AND gates into few levels as shown in the 4:16 decoder. common factor word15 word1 word0 A0 A1 A2 A3
August 201015 A0 A1 A2 A3 word15 word1 word0 Terms repeat themselves, so pre-decoding will eliminate the redundant ones. This is called pre-decoding. Less area with same drive as before. Pre-coded Lines
August 201016 word0 word1 word2 word3 Vcc x x x x 2x 11 1 11 1 1 1 Lyon-Schediwy Fast Decoders In a NOR implementation output is pulled up via serial pMOS devices, which slows transition, so pMOS needs sizing, but this consumes lot of area. Since only single word is pulled up at a time, pMOS can be shared between words in a binary tree fashion and sized to yield same current as in pull down.
August 201017 Sum-addressed Decoders Sometimes an address of memory is calculated as BASE+OFFSET (e.g., in a cache), which requires an addition before decoding. Addition can be time consuming if Ripple Carry Adder (RCA) is used, and even Carry Look Ahead (CLA) my be too slow. It is possible to use a K = A + B comparator without carry propagation or look-ahead calculation.
August 201018 Sum-addressed Decoders If we know A and B, we can deduce what must be the carry in of every bit if it would happen that K = A + B. But then we can also deduce what should be the carry out. It follows that if every bit pair agrees on the carry out of the previous with the carry in of the next, then K=A+B is true indeed. We can therefore use a comparator to every word-line (k), where equality will hold only for one word.
August 201019 AiAi BiBi KiKi C in_i (required) C out_i (generated) 00000 00110 01011 01100 10011 10100 11001 11111 We can derive the equations of the carries from the required and generated carries below.
August 201020
August 201021
August 201022
August 201023
August 201024 Below is a comparison of sum-addressed decoder with ordinary decoder combined with a ripple carry adder (RCA) and carry look ahead adder (CLA). A significant delay and area improvement is achieved.
August 201025 Bit-Line Conditioning Circuits Used to precharge bits high before R/W operation. Most simple is the following circuit. If a clock is not available it is possible to use a weak pMOS device connected as a pseudo-nMOS SRAM. Precharge can be done with nMOS, a case where precharge voltage is V dd -V t. It results faster R/W since swing is smaller, but noise margin is worsen.
August 201026 Each column contains write driver and read sensing circuit. A high skew read inverter has been shown. Sense amplifier provides faster sensing by responding to a smaller voltage swing.write driver high skew read inverter N2 N1 N3 P1 P2 This is a differential analog pair. N3 is a current source where current flows either in left or right branches. Circuit doesn’t need a clock but it consumes significant amount of DC power. Sense Amplifiers
August 201027 Regenerative Feedback Isolation Devices To speed up response bit-lines are disconnected at sensing to avoid their high capacitive load. The regenerative feedback loop is now isolated. When sense clock is high the values stored in bit-lines are regenerated, while the lines are disconnected, speeding up response.
August 201028 Regenerative Feedback Isolation Devices Sense amplifiers are susceptible to differential noise on bit-lines since they respond to small voltage differences.
August 201029 The SRAM is physically organized by 2 n-k rows and 2 m+k columns. Each row has 2 m groups of 2 k bits. Therefore, a 2 k :1 column multiplexers are required to extract the appropriate 2 m bits from the 2 m+k ones. Column Multiplexers
August 201030 Tree Decoder Column Multiplexer To sense Amps and Write Circuits The problem of this MUX is the delay occurring by the series of pass transistors.
August 201031 It is possible to implement the multiplexer such that data is passed trough a single transistor, while column decoding takes place concurrently with row decoding, thus not affecting delay. A1 A0 B0 B1B2 B3 B0 B1 Y
August 201032 DRAM – Dynamic RAM Store their charge on a capacitor rather than in a feedback loop Basic cell is substantially smaller than SRAM. To avoid charge leakage it must be periodically read and refresh It is built in a special process technology optimized for density Offers order of magnitude higher density than SRAM but has much higher latency than SRAM
August 201033 word bit C cell x A 1-transistor (1T) DRAM cell consists of a transistor and a capacitor. Cell is accessed by asserting the word- line to connect the capacitor to the bit- line. word On a read the bit-line is first precharged to V DD /2. When the word-line rises, the capacitor shares its charge with the bit- line, causing a voltage change of ΔV that can be sensed. bit x The read disturbs the cell contents at x, so the cell must be re-written after each read. On a write the voltage of the bit-line is forced onto the capacitor
August 201034 DRAM Cell C cell must be small to obtain high density, but big enough to obtain voltage swing at read.
August 201035 Like SRAMs, large DRAMs are divided into sub-arrays, whose size represents a tradeoff between area and performance. Large sub-arrays amortize sense amplifiers and decoders among more cells but are slower and have less swing due to higher capacitance of word and bit lines. bit0 bit1bit511 word0 word1 word255 Bit-line capacitance is far larger than cell, hence voltage swing ΔV during read is very small and sense amplifier is used.
August 201036 Open bit-line architecture Open bit-line architecture Is useful for small DRAMs. It has dense layout but sense amps are exposed to differential noise since their inputs come from different sub-arrays, while word line is asserted in one array. Folded bit-line architecture Folded bit-line architecture solves the problem of differential noise on the account of area expansion. Sense amps input are connected to adjacent bit-lines exposed to similar noise sources. When a word-line is asserted, one bit line is being read while its neighbor serves as the quiet reference. Smart layout Smart layout and aggressive manufacturing design rules (e.g. 45 degrees polygons) enable effective area increase of only 33%.
August 201037 Sub-array 1 Sub-array 2 Word-Line Decoders Sense Amps Open Bit-Line Architecture
August 201038 Word-Line Decoders Sense Amps Folded Bit-Line Architecture
August 201039 Word-Line Decoder Sense Amp Polysilicon Word-Line Metal Bit-Line n+ Diffusion Bit-Line Contact Capacitor
August 201040 P1 P2 N1 N2 VpVp VnVn bit’ bit” DRAM Sense Amp bit’ and bit” are initialized to V DD /2. V p =0 and V n = V DD /2, so all transistors are initially OFF. During read one bit-line is changing while the other stays float in V DD /2. Let bit’ change to 0. Once it reaches V DD /2-V t, N1 conducts and it follows bit’. Hence V n is pulled down.Hence V n is pulled down Meanwhile bit” is pulled up, which opens P2 and raise V p to V DD.
August 201041 V DD /2 V DD Bit’ Bit” 0 V DD /2 V DD VnVn VpVp 0
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# What Are The Prime Factors Of 38?
• The prime factors of number 38 are: 2, 19
• Determined equcation for number 38 factorization is: 2 * 19
## Is 38 A Prime Number?
• No the number 38 is not a prime number.
• Thirty-eight is a composite number. Because 38 has more divisors than 1 and itself.
## How To Calculate Prime Number Factors
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 3 8
• About Number 3. Three is the first odd prime number and the second smallest right after number two. At the same time it is the first Mersenne prime (2 ^ 2-1), the first Fermat prime (2 ^ {2 ^ 0} +1), the second Sophie Germain prime and the second Mersenne prime exponent. It is the fourth number of the Fibonacci sequence and the second one that is unique. The triangle is the simplest geometric figure in the plane. With the calculation of its sizes deals trigonometry. Rule of three: If the sum of the digits of a number is a multiple of three, the underlying number is divisible by three.
• About Number 8. The octahedron is one of the five platonic bodies. A polygon with eight sides is an octagon. In computer technology we use a number system on the basis of eight, the octal system. Eight is the first real cubic number, if one disregards 1 cube. It is also the smallest composed of three prime number. Every odd number greater than one, raised to the square, resulting in a multiple of eight with a remainder of one. The Eight is the smallest Leyland number.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What Are Prime Factors?
• In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is a list of the integer's prime factors, together with their multiplicities. The process of determining these factors is called integer factorization. The fundamental theorem of arithmetic says that every positive integer has a single unique prime factorization.
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# Which Fraction Has The Greatest Value?
There is no definitive answer to this question as it depends on the context in which it is asked. For example, if we are talking about fractions with different denominators, then the fraction with the smallest denominator will have the greatest value. However, if we are talking about fractions with different numerators, then the fraction with the largest numerator will have the greatest value.
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103111countercurr-extrc
103111countercurr-extrc - Countercurrent extrac,on ...
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Unformatted text preview: Countercurrent extrac,on Absorp,on and stripping Wankat ch. 12 Natural gas liquids (C3+) recovery from natural gas CO2 capture from flue gas Natural gas drying plant Absorp,on of water by ethylene glycol, followed by solvent regenera,on Objec,ves for this lecture • Convert mole frac%ons into mole ra%os • Derive opera,ng line equa,on for stripper column. • Apply McCabe- Thiele analysis to mul,stage stripping using mole frac,ons • Calculate frac,onal stage requirement • Derive opera,ng line equa,on for absorber column • Apply McCabe- Thiele analysis to mul,stage absorp,on using mole frac,ons McCabe- Thiele Analysis of Counter- current Stripper Column Given X0, XN, YN+1 and L/G find N required 1. Convert VLE data to mole ra,os (unless x0 < 0.05) Note: y=x line has no use here. 3. Step off stages (use Murphree efficiencies if available). •(XN,YN+1) Y 2. Plot (XN, YN+1) and (X0, Y1) and draw opera,ng line. It will be below the VLE line. Note: calculate Y1 from X0 and opera,ng line equa,on (mass balance). 1 • 2 • 3 • • • •(X0,Y1) N = 3 To find minimum stripping gas flow rate (Gmin): X First find (L/G)max, by plobng X0 on the VLE line (watch for other pinch points if VLE is curved). Lever- arm rule To es,mate frac,onal stage requirement in a packed column: Y VLE op. line (X4, Y4) • • , Y ) (X3 4 • XN X • 3 X4 • • XN X frac,onal stage = X N − X 3 requirement X 4 − X3 € McCabe- Thiele Analysis of Counter- current Absorber Column Given X0, Y1, YN+1 and L/G find N required 1. Convert VLE data to mole ra,os (unless x0 < 0.05) Note: y=x line has no use here. 3. Step off stages (use Murphree efficiencies if available). • Y 2. Plot (X0, Y1) and (XN, YN+1) and draw opera,ng line. It will be above the VLE line. Note: calculate XN from YN+1 and opera,ng line equa,on (mass balance). (XN,YN+1)• • (X0,Y1)• • 3 • 1 • 2 N = 3 To find minimum extrac,ng solvent flow rate (Lmin): X First find (L/G)min by plobng YN+1 on the VLE line (watch for other pinch points if VLE is curved). Recap • The unit opera,ons absorp,on and stripping are oeen performed sequen,ally. • When carrier gas is insoluble and solvent is non- vola,le, both gas and liquid streams are binary. • McCabe- Thiele analysis needs to be done using mole ra,os instead of mole frac,ons, except when streams are very dilute. • For absorp,on, the opera,ng line lies above the VLE line. • Mass transfer efficiencies are low, and packed columns are common. Objec,ves for this lecture • Describe McCabe- Thiele analysis for mul,ple dilute, non- interac,ng solutes • Analysis of irreversible absorp,on • Analysis of co- current extrac,on • Introduc,on to Kremser equa,ons (analy,cal solu,on for dilute extrac,on) Homework: 12D7 due Wednesday, 12D20 due Friday Reading: Wankat, chapter 12 Prac,ce midterm exam (Friday) Mul,ple non- interac,ng solutes Consider mul,ple soluble components (B, D, E…) in solvent C, to be stripped using gas A OR Consider mul,ple gas components (B, D, E…) in carrier gas A, to be absorbed using solvent C If streams are dilute and component do not interact with each other, assume VLE for each component is independent Treat each one as a single- component problem and solve sequen,ally. Example: 2- component absorp,on xB,0 xD,0 yB,1 yD,1 (xD,N,yD,N+1)• 1 Separa,on of B requires N = 3. Y (xB,N,yB,N+1)• • • • 1 •1 N yB,N+1 yD,N+1 xB,N xD,N (xB,0,yB,1)• (xD,0,yD,1)• Specify yB,N+1, yD,N+1, xB,0, xD,0 Specify L/V and yB,1. Find N and yD,1 • • •3 •3 •2 • 2 Separa,on of D must also use N = 3 and same L/V. Trial- and- error: guess yD,1 X A good idea to use a different graph for each component… Irreversible absorp,on Solvent contains a reagent R which reacts irreversibly with solute B to form non- vola,le product RB R + B(g) → RB e.g., OH- + H2S(g) → S2- + H2O Since equilibrium lies far to the right, xB ≈ 0 yB ≈ 0 Equa,on of the equilibrium line: yB = 0 Irreversible absorp,on of B A + R x0 = 0 C y1 = 0 1 Only one theore,cal equilibrium stage required … Y yN+1 • N B + C yN+1 A + RB xN = 0 Specify yN+1, x0, L/V. Required: xN = y1 = 0 (x0,y1)• (x1,y1) VLE • X (B + RB) With low Murphree efficiency A + R x0 = 0 C y1 = 0 More than one actual equilibrium stage required … 1 yN+1 • Y • • N B + C yN+1 A + RB xN = 0 Specify yN+1, x0, L/V, y1 ≠ 0 • • • 3 • •2 (x0,y1)• •1 •6 EMV = 0.25 • 5 • 4 VLE X (B + RB) Co- current absorp,on • Use higher vapor velocity to increase mass transfer rate • Use smaller diameter column without risk of flooding • Gives higher efficiency than countercurrent column L x0 V y0 (x0,y0)• 1 V yj L xj y0V + x0L = y jV + x j L N V yN € Specify y0, x0 = 0, xN, yN = 0 Y j Only one theore,cal equilibrium stage required if reac,on is irreversible and mass transfer is fast … L xN ⎛ྎ y V + x L ⎞ྏ L 0 y = − x + ⎜ྎ 0 ⎟ྏ V V ⎝ྎ ⎠ྏ (x1,y1) VLE • X (B + RB) Analy,cal solu,on 1 When streams are dilute, we can approximate VLE data by a straight line. 0.9 0.8 0.7 0.2 y(MeOH) 0.6 0.5 0.15 y(MeOH) 0.4 0.3 0.2 0.1 y = mx 0.1 0.05 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x(MeOH) Obtain the slope, m, from Henry’s Law: 0 0 0.01 0.02 0.03 0.04 x(MeOH) PB = HB xB where PB is the par,al pressure of B, and HB is the Henry’s Law constant Note: HB = HB(T), like an equilibrium constant. 0.05 0.06 Kremser equa,ons Change in vapor composi,on between adjacent stages: (Δy ) j = y j +1 − y j 0.2 (x2,y3) • (Δy)2 y(MeOH) 0.15 (x1,y2) • •(x2,y2) 0.1 € (Δy)1 = y2- y1 (x0,y0)• 0.05 •(x1,y1) € ⎛ྎ L L ⎞ྏ y j +1 = x j + ⎜ྎ y1 − x0 ⎟ྏ V V ⎠ྏ ⎝ྎ y j = mx j (Δy ) j ⎛ྎ L ⎞ྏ ⎛ྎ L ⎞ྏ = ⎜ྎ − m⎟ྏ x j + ⎜ྎ y1 − x0 ⎟ྏ V ⎠ྏ ⎝ྎ V ⎠ྏ ⎝ྎ € 0 0 0.01 0.02 x(MeOH) Special case: L/V = m; (Δy)j = Δy NΔy = Δy1 + Δy2 + Δy3 + … € = yN+1 – y1 0.03 Kremser equa,on: N = yN+1 − y1 y −y = N+1 1 Δy y1 − L x0 V ...
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# Area Problems
Home > Lesson > Chapter 21
5 Steps - 3 Clicks
# Area Problems
### Introduction
Area Problems deals with parameters like sides, length, area, breadth, centroid, median, perimeter, height etc. of all the geometrical shapes like square, rectangle, triangle, circle and so on.
### Methods
Area: It is defined as the surface enclosed by its sides.
Square: It is defined as a four sided shape that is made up of four straight sides that are the same lengths and that has four right angles.
Triangle: The plane figure formed by connecting three points not in a straight line by straight line segments like a three sided polygon.
(i) Sum of angles of a triangle is 180 degrees.
(ii) The sum of any two sides of a triangle is greater than the third side.
(iii) The line joining the mid - point of a side of a triangle to the opposite vertex is called the meridian.
(iv) The point where three medians of a triangle meet, is called centroid. The centroid divides each of the medians in the ratio 2 : 1.
(v) In an isosceles triangle, the altitude from the vertex bisects the base.
(vi) The median of a triangle divides it into two triangles of the same area.
(vii) The area of a triangle formed by joining the mid - points of the sides of a given triangle is one - fourth of the area of the given triangle.
Rectangle: The plane figure formed with four straight sides and four right angles, especially one with unequal adjacent sides, in contrast to a square.
Circle: A line that is curved so that its ends meet and every point on the line is the same distance from the centre.
(i) The diagonals of a parallelogram bisect each other.
(ii) Each diagonal of a parallelogram divides it into two triangles of the same area.
(iii) The diagonals of a rectangle are equal and bisect each other.
(iv) The diagonals of a square are equal and bisect each other at right angles.
(v) The diagonals of a rhombus are unequal and bisect each other at right angles.
(vi) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
(vii) Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
Example: Given a square where the length of each side (edge) is 5cm. Find the area of this square.
Solution:
$A = L^{2}$
= $5^{2}$
= 5 x 5
= $25 \ cm^{2}$
Example: Given a rectangle with the length of 4ft and the width of 3ft. Find its area.
Solution:
$A = lw$
= (4)(3)
= $12ft^{2}$
Example: Given a circle with the radius 3cm. Find its area. Take $\pi$ = 3.14.
Solution:
$A = \pi r^{2}$
= (3.14)$(3)^{2}$
= (3.14)(9)
= 28.26 $cm^{2}$
Example: Given a parallelogram with the base 5 in and the height 3 in. Find the area of this parallelogram.
Solution:
$A = bh$
= (5)(3)
= $15 in^{2}$
Example: Given a triangle with the base 4cm and the height 2cm. Find its area.
Solution:
$A = \frac{1}{2}bh$
= $\frac{1}{2}(4)(2)$
= $\frac{1}{2}(8)$
= $4cm^{2}$
Example: Given a trapezoid with the height of 4 in, and two parallel sides of 2 in and 3 in respectively. Calculate its area.
Solution:
$A = (\frac{a + b}{2})h$
= $(\frac{2 + 3}{2})4$
= $(\frac{5}{2})4$
= $10in^{2}$
### Formulae
1 Pythagoras theorem: In a right-angled triangle, $(hypotenuse)^2$ = $(base)^2$ + $(height)^2$.
2: Area of a triangle = (length x breadth) Therefore, length = $\frac{area}{breadth}$ and breadth = $\frac{area}{length}$.
3: Perimeter of a rectangle = 2(length x breadth).
4: Area of a square = $(side)^2$ = $\frac{1}{2}(diagonal)^2$.
5: Area of four walls of a room = 2(length x breadth) x height.
6: Area of a triangle = $\frac{1}{2}$ x base x height.
7: Area of a triangle = $\sqrt{s(s - a)(s - b)(s - c)}$, where $a, b, c$ are sides of the triangle and $s$ = $\frac{1}{2}(a + b + c)$.
8: Area of an equilateral triangle = $\frac{\sqrt{3}}{4} * (side)^2$.
9: Radius of incircle of an equilateral triangle of side $a$ is $\frac{a}{2\sqrt{3}}$.
10: Radius of circumference of an equilateral triangle of side $a$ is $\frac{a}{\sqrt{3}}$.
11: Radius of incircle of a triangle of area $\bigtriangleup$ and semi-perimeter $s$ = $\frac{\bigtriangleup}{s}$.
12: Area of a parallelogram = base x height.
13: Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x distance between them.
14: Area of a circle = $\pi R^2$, where R is the radius.
15: Circumference of a circle = 2$\pi$R.
16: Length of an arc = $\frac{2\pi R\theta}{360}$, where $\theta$ is the central angle.
17: Area of a sector = $\frac{1}{2}$(arc x R) = $\frac{\pi R^2\theta}{360}$.
18: Area of a semi circle = $\frac{\pi R^2}{2}$.
19: Area of isosceles triangle = $\frac{a}{4}\sqrt{4b^2 - a^2}$square units, where $a, b$ are linear units.
20: Circumference of a semi-circle = $\pi R$.
21: Side of a rhombus = $\frac{1}{2}\sqrt{(d_{1})^2 + (d_{2})^2}$ linear units, where $d_{1}$ and $d_{2}$ are the lengths of diagonals.
### Samples
1. The area of four walls of a room is 600$m^2$ and its length is twice its breadth. If the height of the room is 11 m, then find the area of the floor in $m^2$?
Solution:
Let, breadth of a room = $x$ m
Given,
Length of the room = 2$x$ m
Height of the room = 11 m
Total area of 4 waits of room = 2(2$x$ + $x$) x 11 $m^2$ = 66$m^2$
By hypothesis, 66$x$ = 660
⇒ $x$ = 10 m
Hence, area of floor = 2$x$ + $x$ = 2$x^2$ = 2$(10)^2$ = 2 x 100 = 200$m^2$
2. The perimeter of two squares are 60 cm and 44 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of two squares?
Solution:
Given,
Side of first square = $\frac{60}{4}$ = 15 cm
Side of second square = $\frac{44}{4}$ = 11 cm
Area of third square = $(15)^2 - (11)^2$ $(cm)^2$
= (225 - 121)$(cm)^2$
= 104 $(cm)^2$
Side of third square = $\sqrt{104}$ cm = 10.19 cm
Therefore, required perimeter = (10.19 x 4) cm = 40.76 cm.
3. Find the area of the square, one of whose diagonals is 6 m long?
Solution:
Given,
Diagonal = 6 m
Now, consider
Area of square = $\frac{1}{2}(diagonal)^2$
⇒ Area of square = $\frac{1}{2}(6)^2$
⇒ Area of square = $\frac{1}{2}36$
⇒ Area of square = 18 $m^2$
Therefore, Area of square = 18 $m^2$
4. If the length of a certain rectangle is decreased by 5 cm and the width is increased by 4 cm, a square with the same area as the original rectangle would result . Find the perimeter of the original rectangle?
Solution:
Let,
$x$ and $y$ be the length and breadth of the rectangle respectively.
Then, $x$ - 5 = $y$ + 4
$x$ - $y$ = 9 -------- (i)
Now, Area of rectangle = length x breadth and
Area of square = ($x$ - 5)($y$ + 4)
Therefore, ($x$ - 5)($y$ + 4) = $xy$
⇒ 4$x$ - 5$y$ = 20 -------- (ii)
By soling (i) and (ii),
$x$ = 25 and $y$ = 16.
Therefore, Perimeter of the rectangle = 2($x$ + $y$) = 2(16 + 25) = 82 cm.
5. Find the area of the triangle whose base sides measure 10 cm, 13 cm and 15 cm?
Solution:
Given that,
Let base sides $a$ = 10 cm, $b$ = 13 cm and $c$ = 15 cm
Now, consider $s$ = $\frac{1}{2}(a + b + c)$
⇒ $s$ = $\frac{1}{2}(10 + 13 + 15)$
⇒ $s$ = 19
Therefore,
(s - a) = 19 - 10 = 9
(s - b) = 19 - 13 = 6
(s - c) = 19 - 15 = 4
Therefore, area = $\sqrt{s(s - a)(s - b)(s - c)}$ = $\sqrt{19 * 9 * 6 * 3}$ = $\sqrt{3078}$ = 55.47 $(cm)^2$
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# Adding a Light Fixture to 3 Way Switches
How to Wire 3-Way Master and Slave Switches – A comprehensive,step-by-step guide designed to walk you through the identification and wiring of your 3-way switches.
## Hooking Up a Generator to the House Panel Using a Circuit Breaker Interlock Kit for Backup Power
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So Keep Watching So I Can Help You Wire it Right!
## Check out my YouTube Channel: » AskTheElectrician « and Subscribe!
with my online Video Course:
Basic Home Electrical Wiring by Example
## Solution for Adding a Fixture to Existing 3-Way Switches
[ad#block]Electrical Question: I have three way switches at the top and bottom of my stairs. There is a light at the top of the stairs controlled by the switches and I wish to add a light at the bottom. A 14/2 runs first to the light at the top (hot and neutral from the breaker), then a 14/2 runs to the switch at the top (hot supply and switched hot back to the light), and a 14/3 runs between the switches (travelers and switched return). It would be very difficult to run a new 14/2 (switched hot and neutral) from the light upstairs to the light downstairs, but it is easy to run a 14/2 from the downstairs light to the switch box downstairs.I want to know if it is acceptable to connect the hot switched from the downstairs 3 way to the light and run a separate neutral to the breaker box, or another nearby circuit’s neutral.
Skill level: electrical engineer and homeowner, but not an electrician. I know how to not burn the house down, but don’t know the electrical code.
This electrical wiring question came from: James, a homeowner from Malden, Massachusetts
Thanks for your electrical wiring question James.
How to Wire 3-Way Master and Slave Switches
Application: Wiring 3-Way Light Switches.
Skill Level: Beginner to Intermediate.
Tools Required: Basic Electricians Pouch Hand Tools and Voltage Tester.
Estimated Time: Depends on personal level experience and ability to work with tools.
Precaution: Identify the lighting circuit, turn it OFF and Tag it with a Note before working with the wiring.
Notice: Installing additional 3-way switch wiring should be done according to local and national electrical codes with a permit and be inspected.
Materials: Make sure replacement 3-way light switches have the same amperage and voltage rating as the original three way light switch, and is the switch is fully compatible with the electrical circuit, the light fixture and the type of light bulbs being used.
The following article may be very helpful for this project:
I need 4-way wiring diagram using master dimmer & companion dimmer & 1 4-Way rocker switch Continue reading
More about wiring 3 Way Switches
### 3 Way Switch Diagram
In response to the requests for this information, I have developed a comprehensive,step-by-step guide designed to walk you through the identification and wiring of your 3-way switches.
Electrical code
Electrical Code Articles
Electrical Code Directory covering AFCI circuit, electrical-boxes, electrical-circuit, electrical-code-outlets, electrical GFCI, electrical-grounding, electrical-project, electrical-services, electrical-underground, electrical-wiring, electrical-wiring-code, lighting-code, smoke-detector,
3 Way Switch Diagram
3 Way Switch Diagram
In response to the requests for this information, I have developed a comprehensive,step-by-step guide designed to walk you through the identification and wiring of your 3-way switches.
The following may also be helpful for you:
## Dave's Guide to Home Electrical Wiring:
» You Can Avoid Costly Mistakes! «
Wire It Right with the help of my Illustrated Wiring Book
Great for any Home Wiring Project.
## Complete Guide to Home Electrical Wiring
Perfect for Homeowners, Students,
Handyman, Handy Women, and Electricians
Includes:
Wiring GFCI Outlets
Wiring Home Electric Circuits
120 Volt and 240 Volt Outlet Circuits
Wiring Light Switches
Wiring 3-Wire and 4-Wire Electric Range
Wiring 3-Wire and 4-Wire Dryer Cord and Dryer Outlet
Troubleshoot and Repair Electrical Wiring
Wiring Methods for Upgrading Electrical Wiring
NEC Codes for Home Electrical Wiring
....and much more.
## Learn more about Home Electrical Wiring with my Online Video Course:Basic Home Electrical Wiring by Example
Be Careful and Be Safe - Never Work on Energized Circuits!
Consult your Local Building Department about Permits and Inspections for all Electric Wiring Projects.
## The Safest Way to Test Electrical Devices and Identify Electric Wires!
The Non-Contact Electrical Tester
This is a testing tool that I have had in my personal electrical tool pouch for years, and is the first test tool I grab to help identify electrical wiring. It is a Non-contact tester that I use to easily Detect Voltage in Cables, Cords, Circuit Breakers, Lighting Fixtures, Switches, Outlets and Wires. Simply insert the end of the tester into an outlet, lamp socket, or hold the end of the tester against the wire you wish to test. Very handy and easy to use.
## The Quickest Way to Check for Faulty Electrical Wiring!
The Plug-In Outlet Tester
This is the first tool I grab to troubleshoot a problem with outlet circuit wiring. This popular tester is also used by most inspectors to test for power and check the polarity of circuit wiring.
It detects probable improper wiring conditions in standard 110-125 VAC outlets Provides 6 probable wiring conditions that are quick and easy to read for ultimate efficiency Lights indicate if wiring is correct and indicator light chart is included Tests standard 3-wire outlets UL Listed Light indicates if wiring is incorrect Very handy and easy to use.
## Strip Off Wire Insulation without Nicking and Damaging the Electric Wire!
The Wire Stripper and Wire Cutter
My absolute favorite wire stripping tool that I have had in my personal electrical tool pouch for years, and this is the tool I use to safely strip electrical wires.
This handy tool has multiple uses:
The wire gauges are shown on the side of the tool so you know which slot to use for stripping insulation.
The end of the tool can be used to grip and bend wire which is handy for attaching wire onto the screw terminals of switches and outlets..
The wire stripper will work on both solid and stranded wire. This tool is Very Handy and Easy to Use.
More articles about 3-Way Switch, Lighting and Home Electrical Wiring: « Previous Next » Electric Heaters and Circuit Safety How To Wire For Two Light Switches
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# Just what Does Supplementary Angle Me-an in T?
T majors have exactly the characteristics. Till they begin their math program most of them will get significant faculty algebra. Calculus is taken by some in some unspecified time in the future in their math training.
The initial problem is it’s got every one the disciplines’ absolute most mathematical content material. It is essential in procedure, engineering, and other fields and is particularly a science. http://ufpdc.urology.ufl.edu/ In truth, just about each profession that takes advantage of mathematics features a arithmetic office that manages this specified substance.
Where does that math result out? Suitably, it truly is perhaps not fashioned. It arrives that the pupils data, as their capabilities in math keep going to always be forming as any person. Furthermore, moms and dads lecturers, and different older people generally make use of arithmetic designed to assist them cultivate and understand talents.
So, what does angle necessarily mean from arithmetic? It suggests that the angle is appropriate in between the x and y ray coordinates. There are angles which are piece of all calculus. These angles are utilised in every single day lifetimespan.
What does angle me-an in math? Which is one thing you should to have practical knowledge of inside occasion that you might be likely to get all by yourself a bachelor’s degree in math. Now you aspire to be all set to insert and subtract after which be able to write their angles alongside with numbers on to a sheet of paper.
The team together with its personal unique angles are applied for fixing dilemmas, locating ratios, and also other problems in mathematics. Then there could well be the complicated figures that variety element of one’s math principal resume and also their angles which help you determine out angles, as well given that the aspects.
You can relate a couple of of the angles with an circle that persons listed. This will certainly confirm it is actually better to suit your needs to compose angles on your very own arithmetic huge resume.
There are. For occasion, the root may be employed plenty in math. It is usually also utilised in character.
You can also relate the incline of a triangle with the sq. root. This may be the angle you would possibly involve to know about. Afterward there are the trigonometric together with the parabola.
You demand to be familiar with really nicely what does angle me-an from math. This can permit you along with your leading restart. You will have the option to relate someplace to it.
Have in mind that a mathematics resume is generally produced for nearly any math. It truly is crucial in your case to know rather well what does so you can easily enable the other people with their math 24, supplemental angle me-an within just mathematics.
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# 2.MD.A Worksheets, Workbooks, Lesson Plans, and Games
#### CCSS.MATH.CONTENT.2.MD.A
:
"Measure and estimate lengths in standard units."
These worksheets and lesson plans can help students practice this Common Core State Standards skill.
## Worksheets
Hand Measurements
Worksheet
Hand Measurements
On this hands-on second grade math worksheet, kids measure the length and width of their hand and the length of their fingers to the nearest inch or half inch.
Math
Worksheet
Worksheet
Do your kids like helping around the house? Let them flex their metric muscles by helping them measure each item in centimeters using the cut-out ruler.
Math
Worksheet
Bobby's Blueprints #2
Worksheet
Bobby's Blueprints #2
Your child can use her measuring know-how to get this slide's dimensions, then record them for Bobby in the column on the left.
Math
Worksheet
Bobby's Blueprints #3
Worksheet
Bobby's Blueprints #3
Your child can use her measuring know-how to get this robot's dimensions, then record them for Bobby in the column on the left.
Math
Worksheet
Measure Length: Giraffe!
Worksheet
Measure Length: Giraffe!
How tall is this giraffe? Pick up a ruler and help your child measure Geoffry the Giraffe in inches. They'll even get a little intro to fractions.
Math
Worksheet
Measure Length: Cat!
Worksheet
Measure Length: Cat!
This fluffy feline is here to help your child master their measuring skills. Whip out the ruler and measure the different parts of this kitty.
Math
Worksheet
## Lesson Plans
Measure It, Man!
Lesson plan
Measure It, Man!
Students participate in a classroom scavenger hunt with a partner to measure common classroom objects using a ruler.
Math
Lesson plan
Estimating Length
Lesson plan
Estimating Length
Advance your students from relative measurements to learning about standardized units with this lesson that teaches them about inches and feet by using common classroom objects.
Math
Lesson plan
Inches, All Around!
Lesson plan
Inches, All Around!
Look out! Measurement is all around us! In this lesson, students will learn about measuring tools and how to use them. Students will work cooperatively on a fun measurement hunt, exploring and applying concepts of measurement using inches.
Math
Lesson plan
Meter Stick Mambo
Lesson plan
Meter Stick Mambo
Get ready to shake things up with this fun measurement lesson! Students start by seeing if they can fit under a meter stick and then turn their attention to creating towers to fit under the meter stick.
Math
Lesson plan
Measuring Furniture
Lesson plan
Measuring Furniture
How big is the furniture in your classroom? Get ready to find out! In this lesson, students dive into measuring with yardsticks.
Math
Lesson plan
Measuring Feet…in Feet!
Lesson plan
Measuring Feet…in Feet!
How big is a foot? Students find out to explore how to measure in feet by comparing their own feet to rulers. Then they set off to measure the length and width of the classroom!
Math
Lesson plan
## Workbooks
No workbooks found for this common core node.
## Games
No games found for this common core node.
## Exercises
No exercises found for this common core node.
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Sunday, January 11, 2015
Decimals
We have been working on decimals this week.
They tie in perfectly to fractions and the transition was smooth!
We are able to illustrate decimals,
Compare decimals,
And relating decimals to money.
We have also been practicing writing them in standard form, word form, expanded form, and expanded value.
Standard form: 31.25
Word form: thirty two and twenty-five hundredths
Expanded form: 30 + 1 + 0.2 + 0.05
Expanded value: (3 x 10) + (1 x 1) + (2 x 0.1) + (5 x 0.01)
Homework this week will have us practicing all these skills! Please let me know if you have any questions.
Have a great week!
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# Search Our Content Library
491 filtered results
491 filtered results
Numbers 0-10
Counting and Cardinality
Sort by
Number Cards 1-20
Worksheet
Number Cards 1-20
Use these number cards with your students to give them practice with identifying, counting, and ordering numbers from 1-20. Have them cut the cards out and have fun practicing with numbers!
Kindergarten
Math
Worksheet
Worksheet
Kids practice adding single-digit numbers and writing the sums on this circus-themed kindergarten math worksheet.
Kindergarten
Math
Worksheet
Find the Hidden Numbers
Worksheet
Find the Hidden Numbers
Kids are sure to have a blast coloring their favorite nursery rhyme characters, and practicing counting from 1 to 10 while they're at it.
Preschool
Math
Worksheet
Practice Counting Numbers
Worksheet
Practice Counting Numbers
Join counting and primary colors with this worksheet on counting numbers up to 30.
Kindergarten
Math
Worksheet
Counting Cupcakes
Worksheet
Counting Cupcakes
Learners color the cupcakes, then count them up, matching the total amount to the correct number, in this sweet worksheet.
Preschool
Math
Worksheet
Connect the Dots to Horton!
Worksheet
Connect the Dots to Horton!
Practice counting with this fun connect-the-dots, inspired by Dr. Seuss's classic children's book Horton Hears a Who.
Kindergarten
Math
Worksheet
Color by Number Rainbow
Worksheet
Color by Number Rainbow
Here's a fun way to teach kids their colors and boost fine motor skills to boot--a fun rainbow color by number sheet!
Preschool
Math
Worksheet
Dot to Dot 1-10: Fish
Worksheet
Dot to Dot 1-10: Fish
What's hiding behind those pesky numbers? Connect the dots from the numbers 1-10 to find out!
Preschool
Math
Worksheet
Color by Number Race Car
Worksheet
Color by Number Race Car
Your child will have so much fun coloring this spiffy race car that they won't realize they are also practicing important skills like color and number recognition.
Kindergarten
Math
Worksheet
Hop to the Missing Numbers
Worksheet
Hop to the Missing Numbers
Challenge your kid to fill in the missing numbers from 1 to 50. This is an essential skill for understanding numbers beyond just counting them from memory.
Preschool
Math
Worksheet
Ordering Numbers to 10
Worksheet
Ordering Numbers to 10
What number comes after 3? Use the dots in this worksheet to count and put numbers in order.
Preschool
Math
Worksheet
Snowflake Dot-to-Dot
Worksheet
Snowflake Dot-to-Dot
Looking for a way to help your child practice counting this Christmas break that's more engaging than a number line? Try a numbered dot-to-dot!
Kindergarten
Math
Worksheet
Worksheet
Kids practice adding single-digit numbers and writing the sums on this garden-themed kindergarten math worksheet.
Kindergarten
Math
Worksheet
Counting Raindrops
Worksheet
Counting Raindrops
Count the number of raindrops in each row of the worksheet. Then draw a line to match each row of raindrops with the correct number on the right side.
Preschool
Math
Worksheet
Counting to 20: Mama Bird
Worksheet
Counting to 20: Mama Bird
Can your kid help mama bird reach her babies? As he follows the path from one to 20, he'll get more comfortable counting to 20 and have fun while he's at it.
Kindergarten
Math
Worksheet
Lucky Subtraction
Worksheet
Lucky Subtraction
See how your luck fares in this St. Patty's Day subtraction page, perfect for introducing kids to the concept of subtraction.
Kindergarten
Math
Worksheet
St. Patrick's Day Dot-to-Dot
Worksheet
St. Patrick's Day Dot-to-Dot
It's a shamrock dot-to-dot that might be luckier than most! Counting practice and fun coloring are in the cards with this St. Patrick's Day worksheet.
Kindergarten
Math
Worksheet
Outer Space Math
Worksheet
Outer Space Math
Get little ones excited about math with some stellar subtraction practice!
Kindergarten
Math
Worksheet
St. Patrick's Day Connect the Dots
Worksheet
St. Patrick's Day Connect the Dots
Make a little luck and review the numbers one to 67 with this St. Patrick's Day connect-the-dots worksheet!
Kindergarten
Math
Worksheet
Writing Numbers to 20
Worksheet
Writing Numbers to 20
Numbers 1-20 can be as easy as 1-2-3! Assess your students’ understanding of numbers 1 to 20 using this handy worksheet.
Kindergarten
Math
Worksheet
Toy Time Counting
Worksheet
Toy Time Counting
Practice counting and writing numbers with this simple worksheet, which challenges your child to look at each group of toys and count how many.
Kindergarten
Math
Worksheet
Shamrock Counting
Worksheet
Shamrock Counting
Count your way through this field of shamrocks, all in the spirit of St. Patrick's Day!
Kindergarten
Math
Worksheet
Rapunzel's Number Maze
Worksheet
Rapunzel's Number Maze
Guide the prince through the number maze to rescue Rapunzel by simply counting through the maze from 1 to 20!
Kindergarten
Math
Worksheet
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International
Tables for
Crystallography
Volume A1
Symmetry relations between space groups
Edited by Hans Wondratschek and Ulrich Müller
International Tables for Crystallography (2006). Vol. A1, ch. 1.2, pp. 9-10 | 1 | 2 |
## Section 1.2.3. Groups
Hans Wondratscheka*
aInstitut für Kristallographie, Universität, D-76128 Karlsruhe, Germany
Correspondence e-mail: wondra@physik.uni-karlsruhe.de
### 1.2.3. Groups
| top | pdf |
Group theory is the proper tool for studying symmetry in science. The symmetry group of an object is the set of all isometries (rigid motions) which map that object onto itself. If the object is a crystal, the isometries which map it onto itself (and also leave it invariant as a whole) are the crystallographic symmetry operations.
There is a huge amount of literature on group theory and its applications. The book Introduction to Group Theory by Ledermann (1976) is recommended. The book Symmetry of Crystals. Introduction to International Tables for Crystallography, Vol. A by Hahn & Wondratschek (1994) describes a way in which the data of IT A can be interpreted by means of matrix algebra and elementary group theory. It may also help the reader of this volume.
#### 1.2.3.1. Some properties of symmetry groups
| top | pdf |
The geometric symmetry of any object is described by a group . The symmetry operations are the group elements, and the set of all symmetry operations fulfils the group postulates. [A symmetry element' in crystallography is not a group element of a symmetry group but is a combination of a geometric object with that set of symmetry operations which leave the geometric object invariant, e.g. an axis with its threefold rotations or a plane with its glide reflections etc., cf. Flack et al. (2000).] Groups will be designated by upper-case calligraphic script letters , etc. Group elements are represented by lower-case sans serif letters etc.
The result of the composition of two elements will be called the product of and and will be written . The first operation is the right factor because the point coordinates or vector coefficients are written as columns on which the matrices of the symmetry operations are applied from the left side.
The law of composition in the group is the successive application of the symmetry operations.
The group postulates are shown to hold for symmetry groups:
(1) The closure, i.e. the property that the composition of any two symmetry operations results in a symmetry operation again, is always fulfilled for geometric symmetries: if and , then also holds. (2) The associative law is always fulfilled for the composition of geometric mappings. If , then for any triplet . Therefore, the parentheses are not necessary, one can write . In general, however, the sequence of the symmetry operations must not be changed. Thus, in general . (3) The unit element or neutral element is the identity operation which maps each point onto itself, i.e. leaves each point invariant. (4) The isometry which reverses a given symmetry operation is also a symmetry operation of and is called the inverse symmetry operation of . It has the property .
The number of elements of a group is called its order . The order of a group may be finite, e.g. 24 for the symmetry operations of a regular tetrahedron, or infinite, e.g. for any space group because of its infinite set of translations. If the relation is fulfilled for all pairs of elements of a group , then is called a commutative or an Abelian group.
For groups of higher order, it is usually inappropriate and for groups of infinite order it is impossible to list all elements of a group. The following definition nearly always reduces the set of group elements to be listed explicitly to a small set.
Definition 1.2.3.1.1. A set such that every element of can be obtained by composition of the elements of and their inverses is called a set of generators of . The elements are called generators of .
A group is cyclic if it consists of the unit element and all powers of one element :
If there is an integer number with and n is the smallest number with this property, then the group has the finite order n. Let with be the inverse element of where n is the order of . Because with , the elements of a cyclic group of finite order can all be written as positive powers of the generator . Otherwise, if such an integer n does not exist, the group is of infinite order and the positive powers are different from the negative ones .
In the same way, from any element its cyclic group can be generated even if is not cyclic itself. The order of this group is called the order of the element .
#### 1.2.3.2. Group isomorphism and homomorphism
| top | pdf |
A finite group of small order may be conveniently visualized by its multiplication table, group table or Cayley table. An example is shown in Table 1.2.3.1.
Table 1.2.3.1| top | pdf | Multiplication table of a group
The group elements are listed at the top of the table and in the same sequence on the left-hand side; the unit element ' is listed first. The table is thus a square array. The product of any pair of elements is listed at the intersection of the kth row and the jth column. It can be shown that each group element is listed exactly once in each row and once in each column of the table. In the row of an element , the unit element appears in the column of . If , i.e. , appears on the main diagonal. The multiplication table of an Abelian group is symmetric about the main diagonal.
The multiplication tables can be used to define one of the most important relations between two groups, the isomorphism of groups. This can be done by comparing the multiplication tables of the two groups.
Definition 1.2.3.2.1. Two groups are isomorphic if one can arrange the rows and columns of their multiplication tables such that these tables are equal, apart from the names or symbols of the group elements.
Multiplication tables are useful only for groups of small order. To define isomorphism' for arbitrary groups, one can formulate the relations expressed by the multiplication tables in a more abstract way.
The same multiplication table' for the groups and means that there is a reversible mapping of the elements and such that holds for any pair of indices j and k. In words:
Definition 1.2.3.2.2. Two groups and are isomorphic if there is a reversible mapping of onto such that for any pair of elements of the image of the product is equal to the product of the images.
Isomorphic groups have the same order. By isomorphism the set of all groups is classified into isomorphism types or isomorphism classes of groups. Such a class is often called an abstract group.
The isomorphism between the space groups and the corresponding matrix groups makes an analytical treatment of crystallographic symmetry possible. Moreover, the isomorphism of different space groups allows one to classify the infinite number of space groups into a finite number of isomorphism types of space groups, which is one of the bases of crystallography, see Section 1.2.5.
Isomorphism provides a very strong relation between groups: the groups are identical in their group-theoretical properties. One can weaken this relation by omitting the condition of reversibility of the mapping. One then admits that more than one element of the group is mapped onto the same element of . This concept leads to the definition of homomorphism.
Definition 1.2.3.2.3. A mapping of a group onto a group is called homomorphic, and is called a homomorphic image of the group , if for any pair of elements of the image of the product is equal to the product of the images and if any element of is the image of at least one element of . The relation of and is called a homomorphism. More formally: For the mapping onto , holds.
The formulation `mapping onto' implies that each element occurs among the images of the elements at least once.2
The very important concept of homomorphism is discussed further in Lemma 1.2.4.4.3. The crystallographic point groups are homomorphic images of the space groups, see Section 1.2.5.4.
### References
Flack, H. D., Wondratschek, H., Hahn, Th. & Abrahams, S. C. (2000). Symmetry elements in space groups and point groups. Addenda to two IUCr reports on the nomenclature of symmetry. Acta Cryst. A56, 96–98.
Hahn, Th. & Wondratschek, H. (1994). Symmetry of crystals. Introduction to International Tables for Crystallography, Vol. A. Sofia: Heron Press.
Ledermann, W. (1976). Introduction to group theory. London: Longman. (German: Einführung in die Gruppentheorie, Braunschweig: Vieweg, 1977.)
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## MPCNC Drag Knife Holder: Showing More Blade
Attempting to cut laminated cardstock decks for the Homage Tektronix Circuit Computer required a bit more blade extension than my LM12UU holder made available:
Shortening the 12 mm shaft wasn’t going to happen, so I knocked a little bit off the blade holder to give the knurled lock ring slightly more travel:
The lathe cutoff blade is a bit to the right of the new cut, but you get the general idea: not a whole lot of clearance in there.
## Tek Circuit Computer: Drag Knife Deck Cutting
Creating a paper version of the Tektronix Circuit Computer requires nothing more than a drag knife to cut the deck outlines:
The middle deck is a disk with a notch exposing the FL scale, a cutout window exposing the inductive time constant / risetime scale, and a wee circle for the Chicago screw in the middle:
Three angles define the notch:
`````` FLNotchArc = 85deg; // width exposing FL scale
FLRampArc = 7deg; // … width of entry & exit ramps
FLNotchOffset = 2deg; // … start angle from 0°``````
Given those, along with the deck radius and notch height (equals the underlying scale height), calculate four points defining the start and end of the ramps and connect the dots:
`````` local a0 = FLNotchOffset;
local p0 = DeckRad * [cos(a0),sin(a0),-];
local a1 = a0 + FLNotchArc;
local p1 = DeckRad * [cos(a1),sin(a1),-];
goto(p0);
move([-,-,KnifeZ]);
local r = DeckRad - ScaleHeight;
local a3 = a1 - FLRampArc;
local p3 = r * [cos(a3),sin(a3),-];
local a4 = a0 + FLRampArc;
local p4 = r * [cos(a4),sin(a4),-];
move(p3);
arc_cw(p4,r); // smallest arc
move(p0); // end of notch
The `arc_cw()` functions draw arcs, as you’d expect, with a positive radius tracing the shortest arc and a negative radius for the longest arc. Although I know how that works, I must still preview the result to verify the G-Code does what I want, not what I said.
The unhappy result of a wrong sign:
GCMC uses the (signed) radius to generate the XY coordinates and IJ offsets for `G2` commands in the preferred center format:
``````G0 X88.846 Y3.103
G1 Z-2.000
G2 X4.653 Y88.778 I-88.846 J-3.103``````
Cutting the window starts from its angular width and offset, which are hardcoded magic numbers from the Tek artifact, and proceeds similarly:
`````` local WindowArc = 39deg;
local ac = -6 * ScaleArc; // center of window arc
local r0 = DeckRad - ScaleHeight; // outer
local r1 = DeckRad - 2 * ScaleHeight; // inner
local aw = WindowArc - to_deg(atan(ScaleHeight,(r0 + r1)/2)); // window arc minus endcaps
local p0 = r0 * [cos(ac + aw/2),sin(ac + aw/2),-];
local p1 = r0 * [cos(ac - aw/2),sin(ac - aw/2),-];
local p2 = r1 * [cos(ac - aw/2),sin(ac - aw/2),-];
local p3 = r1 * [cos(ac + aw/2),sin(ac + aw/2),-];
goto(p0);
move([-,-,KnifeZ]);
arc_cw(p1,r0); // smallest arc
arc_cw(p2,ScaleHeight/2); // half a circle
arc_ccw(p3,r1);
arc_cw(p0,ScaleHeight/2);``````
Trust me on this: incorrect radius signs generate unrecognizable outlines. Which, of course, is why you preview the G-Code before actually cutting anything:
A similar hunk of code cuts the top deck; the bottom deck is a simple circle.
The workflow, such as it is:
• Tape a sheet of paper (Index stock, Basis 110 = 10 mil = 0.25 mm) at the center of the 3018-ProXL platform
• Plot (“engrave”) the scales with a pen
• Affix paper to a Cricut sticky mat taped to the MPCNC platform
• Touch off the origin at the middle
• Drag-cut (“mill”) the outlines
Less complex than it may appear, but the GCMC file now spits out two G-Code files per deck: one to engrave / draw the scales on the 3018 and another to mill / cut the outlines on the MPCNC.
## Tektronix Circuit Computer: Layout Analysis
Following a linkie I can no longer find led me to retrieve the Tektronix Circuit Computer in my Box o’ Slide Rules:
I’m pretty sure it came from Mad Phil’s collection. One can line up the discolored parts of the decks under their cutout windows to restore it to its previous alignment; most likely it sat at the end of a row of books (remember books?) on his reference shelf.
The reverse side lists the equations it can solve, plus pictorial help for the puzzled:
Some searching reveals the original version had three aluminum disks, shaped and milled and photo-printed, with a honkin’ hex nut holding the cursor in place. The one I have seems like laser-printed card stock between plastic laminating film; they don’t make ’em like that any more, either.
TEK PN 003-023 (the paper edition) runs about thirty bucks (modulo the occasional outlier) on eBay, so we’re not dealing in priceless antiquity here. The manual is readily available as a PDF, with photos in the back.
Some doodling produced key measurements:
All the dimensions are hard inches, of course.
Each log decade spans 18°, with the Inductive Frequency scale at 36° for the square root required to calculate circuit resonance.
Generating the log scales requires handling all possible combinations of:
• Scales increase clockwise
• Scales increase counterclockwise
• Ticks point outward
• Ticks point inward
I used the 1×100 tick on the outer scale of each deck as the 0° reference for the other scales on that deck. The 0° tick appears at the far right of plots & engravings & suchlike.
The L/R Time Constant (tau = τ) pointer on the top deck and the corresponding τL scale on the bottom deck has (what seems like) an arbitrary -150° offset from the 0° reference.
The Inductive Frequency scale has an offset of 2π, the log of which is 0.79818 = 14.37°.
The risetime calculations have a factor of 2.197, offsetting those pointers from their corresponding τ pointer by 0.342 = log(2.197) = 6.15°.
A fair bit of effort produced a GCMC program creating a full-size check plot of the bottom deck on the MPCNC:
By the conservation of perversity, the image is rotated 90° to put the 1 H tick straight up.
The 3018 can’t handle a 7.75 inch = 196 mm disk, but a CD-size (120 mm OD) engraving came out OK on white plastic filled with black crayon:
The millimeter scale over on the right shows the letters stand a bit under 1 mm tall. And, yes, the middle scale should read upside-down.
Properly filling the engraved lines remains an ongoing experiment. More downforce on the diamond or more passes through the G-Code should produce deeper trenches, perhaps with correspondingly higher ridges along the sides. Sanding & polishing the plastic without removing the ink seems tedious.
The Great Dragorn of Kismet observes I have a gift for picking projects at the cutting edge of consumer demand.
More doodles while figuring the GCMC code produced a summary of the scale offsets:
Musings on the parameters of each scale:
How to draw decades of tick marks:
It turned out easier to build vectors of tick mark values and their corresponding lengths, with another list of ticks to be labeled, than to figure out how to automate those values.
More on all this to come …
## MPCNC: Z-Axis Probed Height Map to Solid Model
I set up an orthotic shoe insert on the MPCNC and unleashed the Z-Axis height probe on it:
In principle, the grid keeps the object aligned with the machine axes and the blocks put the upper surface more-or-less parallel with the platform. The XY origin, at the G28 location I’ve been using for tool changes, is on the midline of the sole, with Z touched off by probing the platform beside the sole.
The only interesting part of the orthotic is the rigid white plastic plate, which extends about 20 mm into a pocket in the black foam, so the probe area excludes the bendy part.
I’m abusing the bCNC Auto-level probe routine to get the height map, because it produces a tidy file of XYZ coordinates with three header lines describing the overall probe area:
``````-50 140 39
-50 50 21
-2 35 500
-50 -50 0.11
-45 -50 0.06
-40 -50 0.005``````
The first two lines give the X and Y coordinate ranges and number of samples. The third line is the Z axis range and probe speed (?). After that, it’s just probed XYZ coordinates, all the way down.
Meshlab can import ASC files consisting of XYZ coordinates, with the ability to skip a specific number of header lines:
If you don’t skip those three lines, then you get three additional points, far off in XYZ space, that will confuse the next step.
Checking the Grid Triangulation box (the default) produces a nicely lofted sheet:
It is, however, a single-sided sheet, not a manifold 3D object. After a few days of screwing around, I’m unable to find any (automatic, reliable, non-manual) way to solidify the thing in Meshlab, so just save it as a PLY file in ASCII format:
Import it into Meshmixer, Ctrl-A to select the whole thing, click (Select →) Edit → Extrude, pick Y-Axis and Flat EndType, then extrude a convenient base in the negative direction:
For whatever reason, some 3D programs show machine-tool coordinates with Z pointing upward and others aim the Z axis at your face. Both must have made sense at the time, because Meshmixer defaults to swapping the Y and Z coordinates on import / export.
The Density slider controls the number of generated faces in the extruded section, so tune for best results.
I have no idea what Harden does.
Accept the result and you have a solid object suitable for further modeling.
## MPCNC: Z-Axis Height Probe
A slight modification to the MPCNC LM12UU collet pen holder turns it into a long-reach Z-Axis Height Probe:
A flange on the top plate holds a Makerbot-style endstop switch:
The brass probe rod sports a 3/32 inch ball epoxied on its tip, although for my simple needs I could probably use the bare rod:
I clamped the rod to extend a bit beyond the plate, where it can soak up most of the switch release travel, leaving just enough to reset the clickiness after each probe:
The probe responds only to Z motion, not tip deflection in XY, so it’s not particularly good for soft objects with sloped sides, like the insole shown above. It works fine for rigid objects and should suffice to figure the modeling workflow.
The bCNC Auto-Level probe routine scans a grid over a rectangular region:
Which Meshlab turns into a solid model:
That’s the bottom of the insole probed on a 5 mm grid, which takes something over an hour to accomplish.
The OpenSCAD code as a GitHub Gist:
// Collet pen cartridge holder using LM12UU linear bearing // Ed Nisley KE4ZNU - 2019-04-26 // 2019-06 Adapted from LM12UU drag knife holder // 2019-09 Probe switch mount plate Layout = "Build"; // [Build, Show, Puck, Mount, Plate, SwitchPlate] /* [Hidden] */ // Extrusion parameters ThreadThick = 0.25; // [0.20, 0.25] ThreadWidth = 0.40; // [0.40] // Constants Protrusion = 0.1; // [0.01, 0.1] HoleWindage = 0.2; inch = 25.4; function IntegerMultiple(Size,Unit) = Unit * ceil(Size / Unit); ID = 0; OD = 1; LENGTH = 2; //- Adjust hole diameter to make the size come out right module PolyCyl(Dia,Height,ForceSides=0) { // based on nophead's polyholes Sides = (ForceSides != 0) ? ForceSides : (ceil(Dia) + 2); FixDia = Dia / cos(180/Sides); cylinder(r=(FixDia + HoleWindage)/2,h=Height,\$fn=Sides); } //- Dimensions // Basic shape of DW660 snout fitting into the holder // Lip goes upward to lock into MPCNC mount Snout = [44.6,50.0,9.6]; // LENGTH = ID height Lip = 4.0; // height of lip at end of snout // Holder & suchlike PenShaft = 3.5; // hole to pass pen cartridge WallThick = 4.0; // minimum thickness / width Screw = [4.0,8.5,25.0]; // thread ID, washer OD, length Insert = [4.0,6.0,10.0]; // brass insert Bearing = [12.0,21.0,30.0]; // linear bearing body Plate = [PenShaft,Snout[OD] - WallThick,WallThick]; // spring reaction plate echo(str("Plate: ",Plate)); SpringSeat = [0.56,7.5,2*ThreadThick]; // wire = ID, coil = OD, seat depth = length PuckOAL = max(Bearing[LENGTH],(Snout[LENGTH] + Lip)); // total height of DW660 fitting echo(str("PuckOAL: ",PuckOAL)); Key = [Snout[ID],25.7,(Snout[LENGTH] + Lip)]; // rectangular key NumScrews = 3; //ScrewBCD = 2.0*(Bearing[OD]/2 + Insert[OD]/2 + WallThick); ScrewBCD = (Snout[ID] + Bearing[OD])/2; echo(str("Screw BCD: ",ScrewBCD)); NumSides = 9*4; // cylinder facets (multiple of 3 for lathe trimming) // MBI Endstop switch PCB PCB = [40.0,1.6,16.5]; // endstop PCB, switch downward, facing parts Touchpoint = [-4.8,4.8,4.5]; // contact point from PCB edges, solder side TapeThick = 1.0; // foam mounting tape SwitchMount = [PCB.x,WallThick,PCB.z + Touchpoint.z + Plate.z]; module DW660Puck() { translate([0,0,PuckOAL]) rotate([180,0,0]) { cylinder(d=Snout[OD],h=Lip/2,\$fn=NumSides); translate([0,0,Lip/2]) cylinder(d1=Snout[OD],d2=Snout[ID],h=Lip/2,\$fn=NumSides); cylinder(d=Snout[ID],h=(Snout[LENGTH] + Lip),\$fn=NumSides); translate([0,0,(Snout[LENGTH] + Lip) - Protrusion]) cylinder(d1=Snout[ID],d2=2*WallThick + Bearing[OD],h=PuckOAL - (Snout[LENGTH] + Lip),\$fn=NumSides); intersection() { translate([0,0,0*Lip + Key.z/2]) cube(Key,center=true); cylinder(d=Snout[OD],h=Lip + Key.z,\$fn=NumSides); } } } module MountBase() { difference() { DW660Puck(); translate([0,0,-Protrusion]) // bearing PolyCyl(Bearing[OD],2*PuckOAL,NumSides); for (i=[0:NumScrews - 1]) // clamp screws rotate(i*360/NumScrews) translate([ScrewBCD/2,0,-Protrusion]) rotate(180/8) PolyCyl(Insert[OD],2*PuckOAL,8); } } module SpringPlate() { difference() { cylinder(d=Plate[OD],h=Plate[LENGTH],\$fn=NumSides); translate([0,0,-Protrusion]) // pen cartridge hole PolyCyl(PenShaft,2*Plate[LENGTH],NumSides); translate([0,0,Plate.z - SpringSeat[LENGTH]]) // spring retaining recess PolyCyl(SpringSeat[OD],SpringSeat[LENGTH] + Protrusion,NumSides); for (i=[0:NumScrews - 1]) // clamp screws rotate(i*360/NumScrews) translate([ScrewBCD/2,0,-Protrusion]) rotate(180/8) PolyCyl(Screw[ID],2*PuckOAL,8); } } module SwitchPlate() { translate([0,0,Plate.z]) rotate([180,0,0]) SpringPlate(); rotate(45) translate([Touchpoint.x,Touchpoint.y + TapeThick,0]) cube(SwitchMount,center=false); } //----- // Build it if (Layout == "Puck") DW660Puck(); if (Layout == "Plate") SpringPlate(); if (Layout == "SwitchPlate") SwitchPlate(); if (Layout == "Mount") MountBase(); if (Layout == "Show") { MountBase(); translate([0,0,1.6*PuckOAL]) rotate([180,0,0]) SpringPlate(); } if (Layout == "Build") { translate([0,Snout[OD]/2,PuckOAL]) rotate([180,0,0]) MountBase(); translate([0,-Snout[OD]/2,0]) SpringPlate(); }
## bCNC Probe Camera Calibration
I’m sure I’ll do this again some time …
Focus the camera at whatever distance needed to clear the longest tooling you’ll use or, at least, some convenient distance from the platform. You must touch off Z=0 at the surface before using bCNC’s probe camera alignment, because it will move the camera to the preset focus distance.
Align the camera’s optical axis perpendicular to the table by making it stare into a mirror flat on the platform, then tweaking the camera angles until the crosshair centers on the reflected lens image. This isn’t dead centered, but it’s pretty close:
The camera will be focused on the mirror, not the reflection, as you can tell by the in-focus crud on the mirror. Whenever you focus the lens, you’ll probably move the optical axis, so do the best you can with the fuzzy image.
You can adjust small misalignments with the Haircross (seems backwards to me) Offset values.
A cheap camera’s lens barrel may not be aligned with its optical axis, giving the lens a jaunty tilt when it’s correctly set up:
With the camera focus set correctly, calibrate the camera Offset from the tool (a.k.a. Spindle) axis:
• Put a pointy tool at XY=0
• Touch off Z=0 on a stack of masking tape
• Put a dent in the tape with the bit
• Move to the camera’s focused Z level
• Make the dent more conspicuous with a Sharpie, as needed
• Register the spindle location
• Jog to center the crosshair on the dent
• Register the camera location
Calibrate the Crosshair ring diameter thusly:
• Put an object with a known size on the platform
• Touch off Z=0 at its surface
• Move to the camera’s focused Z level
• Set the Crosshair diameter equal to the known object size
• Adjust the Scale value to make the Crosshair overlay reality
For example, calibrating the diameter to 10 mm against a shop scale:
At 10 mm above the CD, setting the camera’s resolution to 11.5 pixel/mm:
Makes the outer circle exactly 15.0 mm in diameter to match the CD hub ring ID:
I doubt anybody can find the pixel/mm value from first principles, so you must work backwards from an object’s actual size.
## CNC 3018-Pro: Diamond Drag Engraving Test Disk
The smaller and more rigid CNC 3018-Pro should be able to engrave text faster than the larger and rather springy MPCNC, which could engrave text at about 50 mm/min. This test pattern pushes both cutting depth and engraving speed to absurd values:
Compile the GCMC source to generate G-Code, lash a CD / DVD to the platform (masking tape works fine), touch off the XY coordinates in the center, touch off Z=0 on the surface, then see what happens:
The “engraving depth” translates directly into the force applied to the diamond point, because the spring converts displacement into force. Knowing the Z depth, you can calculate or guesstimate the force.
Early results from the 3018 suggest it can engrave good-looking text about 20 times faster than the MPCNC:
You must trade off speed with accuracy on your very own machine, as your mileage will certainly differ!
The GCMC source code as a GitHub Gist:
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Home Tags Transformer core design principles
Tag: transformer core design principles
Guide Particulars :
Language English
Pages 602
Format PDF
Dimension 8.05 MB
Transformer Design Principles 2nd Version
Transformer Design Principles With Purposes to Core-Type Energy Transformers 2nd Version | PDF Free Download.
Transformer Design Principles Contents
• Introduction
• Magnetism and Associated Core Points
• Circuit Mannequin of a Two-Winding Transformer with Core
• Reactance and Leakage Reactance Calculations
• Phasors, Three-Section Connections, and Symmetrical Elements
• Fault Present Evaluation
• Section-Shifting and Zig-Zag Transformers
• Multiterminal Three-Section Transformer Mannequin
• Rabins’ Methodology for Calculating Leakage Fields, Leakage Inductances, and Forces in Transformers
• Mechanical Design
• Electrical Subject Calculations
• Capacitance Calculations
• Voltage Breakdown and Excessive-Voltage Design
• Losses
• Thermal Design
• Miscellaneous Matters
Preface to Transformer Design Principles PDF
Like the unique version, the second version of our ebook focuses on the bodily ideas behind the transformer design and operation.
This version consists of some, however not all, materials from the primary version, with corrections as wanted. It additionally consists of numerous new materials, specifically, a chapter on multiterminal transformers, a bit on a comparatively new oil breakdown methodology, and a bit on zig-zag transformers.
The fabric has additionally been reorganized into what we hope is a extra logical growth. This version makes use of the metric (worldwide, or SI) system completely.
Transformers operate in an atmosphere the place electrical, mechanical, and thermal necessities have to be glad.
Ideas from every of those disciplines should, due to this fact, be dropped at bear on the design course of.
Ranging from fundamental ideas and offering numerous background materials, we derive design formulation and strategies.
Stressing the basics will allow the design engineer and the transformer consumer, similar to a utility engineer, to have a greater understanding of the rationale behind design practices.
As a result of most of the design procedures are mathematically sophisticated, pc strategies will probably be employed of their implementation; private computer systems can be utilized for this, and a great mathematical library would even be useful.
Because the first version, electromagnetic 3D finite ingredient packages have additionally grow to be extra typically obtainable and helpful.
We’ll current some calculations utilizing this software, particularly along with the impedance boundary methodology for coping with eddy present losses in high-permeability supplies similar to tank partitions. As within the first version, essential outcomes may also be obtained with a 2D model.
Though this ebook primarily offers with energy transformers, the bodily ideas we focus on, and the mathematical modeling strategies we current apply equally properly to different kinds of transformers.
We hold the presentation as basic as doable, with minimal use of jargon, in order that designers or customers of different transformer varieties can have little issue making use of the outcomes to their very own designs.
The emphasis on fundamentals will permit us to see the place higher approximations or totally different assumptions will be made with a purpose to enhance the accuracy or broaden the applicability of the outcomes.
References are listed alphabetically on the finish of the ebook and are referred to typically by the primary three letters of the primary writer’s final identify adopted by the final two digits of the publication date, for instance, [Abc98]; the place this format can’t be adopted, an acceptable substitute is made.
Transformer design principles PDF
Author(s): Robert M. Del Vecchio, Bertrand Poulin, Pierre T. Feghali, Dilipkumar M. Shah, and Rajendra Ahuja
Publisher: CRC Press, Year: 2018
ISBN: 9781315155920
Description:
In the newest edition, the reader will learn the basics of transformer design, starting from fundamental principles and ending with advanced model simulations. The electrical, mechanical, and thermal considerations that go into the design of a transformer are discussed with useful design formulas, which are used to ensure that the transformer will operate without overheating and survive various stressful events, such as a lightning strike or a short circuit event. This new edition includes a section on how to correct the linear impedance boundary method for non-linear materials and a simpler method to calculate temperatures and flows in windings with directed flow cooling, using graph theory. It also includes a chapter on optimization with practical suggestions on achieving the lowest cost design with constraints.
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OpenStudy (anonymous):
simplify the trigonometric expression sin^2 theta/1+cos theta
3 years ago
OpenStudy (anteater):
Is there a trig identity you can substitute for sin^2 theta?
3 years ago
OpenStudy (anonymous):
I have no idea what I'm doing. I don't even know what that is.
3 years ago
OpenStudy (anteater):
Are you familiar with sin^2 theta + cos^2 theta = 1?
3 years ago
OpenStudy (anonymous):
No.
3 years ago
OpenStudy (rishavraj):
use sin^theta = 1 - cos^2 theta and after that use rationalisation
3 years ago
OpenStudy (anteater):
Since sin^2 theta + cos^2 theta = 1, then 1 - cos^2 theta = sin^2 theta, so you can substitute
3 years ago
OpenStudy (anteater):
sin^2 theta/(1 + cos theta) = (1-cos^2 theta)/(1 + cos theta)
3 years ago
OpenStudy (anonymous):
I think I might just have to go back and relearn a whole bunch of stuff because I am not understanding any of this
3 years ago
OpenStudy (anteater): 3 years ago
OpenStudy (anteater):
Although I just realized that trig identities are not included there. But this may come in handy: http://www.sosmath.com/trig/Trig5/trig5/trig5.html
3 years ago
OpenStudy (anonymous):
thank you
3 years ago
OpenStudy (anteater):
The link to the first site provides a quick discussion of the different trigonometric ratios (sine, cosine, tangent, cosecant, secant and cotangent).
3 years ago
OpenStudy (anteater):
The second link has useful trig "identities", equivalent expressions that you can substitute for each other
3 years ago
OpenStudy (anteater):
So in the problem you have, you can substitute (1 - cos^2 theta for sin^2 theta). The reason that is useful is because you can then factor 1 - cos^2 theta. 1 - cos^2 theta = (1 + cos theta)(1 - cos theta)
3 years ago
OpenStudy (anteater):
So your expression then becomes (1 + cos theta)(1- cos theta)/(1 + cos theta) The (1 + cos theta) in the numerator and denominator cancel out. And, as rishavraj said, you are left with the answer: 1 - cos theta
3 years ago
OpenStudy (anteater):
So I hope that is somewhat helpful. :)
3 years ago
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# dBm to watts conversion
Decibel-milliwatts (dBm) to watts (W), power conversion calculator and how to convert.
## dBm to watts conversion calculator
Decibel-milliwatts (dBm) to watts (W), power conversion calculator.
Enter the power in dBm and press the Convert button:
Enter power in dBm: dBm Watts result: W
Watts to dBm conversion calculator ►
## How to convert dBm to watts
1dBm = 0.001258925W
So the power conversion of dBm to watts is given by the formula:
P(W) = 1W ⋅ 10(P(dBm) / 10) / 1000 = 10((P(dBm) - 30) / 10)
#### Example
Convert 43dBm to watts:
P(W) = 1W ⋅ 10(43dBm/ 10) / 1000 = 19.9526W
## dBm to watts conversion table
Power (dBm) Power (W)
-30 dBm 0.000001 W
-20 dBm 0.00001 W
-10 dBm 0.0001 W
0 dBm 0.001 W
1 dBm 0.0012589 W
2 dBm 0.0015849 W
3 dBm 0.0019953 W
4 dBm 0.0025119 W
5 dBm 0.0031628 W
6 dBm 0.0039811 W
7 dBm 0.0050119 W
8 dBm 0.0063096 W
9 dBm 0.0079433 W
10 dBm 0.01 W
20 dBm 0.1 W
30 dBm 1 W
40 dBm 10 W
50 dBm 100 W
Watts to dBm conversion ►
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Drawing electric field lines-Equations or software
Shown below in the diagram are two conducting material connected to a battery source and vacuum OR air is in between them. There will be charges developed on their surfaces. I am interested in finding out the pattern of electric field lines.
Suggest the method to arrive at the equation for the field lines or some software which would plot the field if we could give the charge arragement.
Also, if the potential of the battery is V volts, how to calculate the function of charge developed, with the distance of separation as a variable, and taking the medium of separation as air/vacuum ??
REASON FOR THE DIAGRAM:I am studying about various arc welding processes; and this is important in understanding the process.
-
Electric field lines denote the direction of the field at any point. So all you have to do is find the electric field vector at any point, and move an infinitesimal distance along that vector. Keep repeating the process until you reach a negative charge(You should reach one for a conservative field). Now I don't know what approximations you can consider to find the field for the charges you have described – udiboy1209 Aug 13 '13 at 13:23
You'll have to write the code for that - numerical code to solve Laplace's equation $\nabla^2 V = 0$ and electric field would be given by $\vec{E} = -\nabla V$. You can write it in C++ using finite difference method. After getting the results you can plot it using any standard graph plotting software. – guru Aug 13 '13 at 17:07
@guru-thanks a lot; but hiw to plot feikd lines? The equations which you gave me would give me the field but what is the equation for the field lines? – karthikeyan Aug 13 '13 at 20:09
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# Spice modeling a PUT oscillator
Thread Starter
#### bthomas
Joined Aug 15, 2009
14
Hi
I am trying to model a oscillator circuit that uses the programmable unijunction transistor 2N6027 (http://www.rapidonline.com/netalogue/specs/47-3348.pdf) . I got the spice model from www.onsemi.com/pub/Collateral/2N6027.LIB . The circuit is quite simple and consists of two series resistors (R2 and R3) used to bias the gate, which
are placed in parallel with a resistor and capacitor in series (R1 and C1)
+ R1 C1 -
|---/\/\/\/\------x-------||----------|
| |
|__/\/\/\/\___w__/\/\/\/\____|
R2 R3
R1 = 470K
R2 = 15K
R3 = 27K
C1 = 2.2uF
The anode of the PUT is connected between R1 and C1 (at x) and the gate between R2 and R3 (at w). The oscillation is obtained between its cathode and the negative terminal of a DC power source.
I can not get my spice model (given below in ngspice format) to oscillate.
* PUT Oscillator
* Red Led
.MODEL LEDR D (IS=93.2P RS=42M N=3.73 BV=4 IBV=10U
+ CJO=2.97P VJ=.75 M=.333 TT=4.32U)
* 2N6027 NPN PUT (Programable Unijunction Transistor)
.MODEL 2N6027 NPN(IS=5E-15 VAF=100 IKF=0.3 ISE=1.85E-12
+ NE=1.45 RE=0.15 RC=0.15 CJE=7E-10 TF=0.6E-8
+ CJC=2.2E-10 TR=4.76E-8 XTB=3)
* Oscillator Circuit
v1 1 0 9
r1 1 2 470k
c1 2 0 2.2uF
r2 1 3 15k
r3 3 0 27k
q1 2 3 4 2N6027
d1 4 0 LEDR ic=0
.control
tran 0.1 4 uic
plot v(2), v(4)
.endc
.end
I suspect that the PUT model does not correspond to the actual PUT I am using (given in the datasheet) but do not know how to fix the problem. Will be grateful for any help.
Are there any good tutorials on how to convert an active components datasheet into a Spice model ?
regards
Thomas
#### SgtWookie
Joined Jul 17, 2007
22,220
Here's a model for the 2N6027 I've used with success:
Rich (BB code):
*Programable Unijunction Transistor pkg: TO-226AA
.SUBCKT X2N6027 1 2 3
************** K1 G K2
Q1 2 4 3 NMOD
Q2 4 2 1 PMOD
.MODEL NMOD NPN(IS=5E-15 VAF=100 IKF=0.005 ISE=1.85E-12 NE=1.45
+ RB=10 RE=0.5 RC=0.5 CJE=3.5E-11 VJE=0.75 CJC=1.1E-11 VJC=0.75 TR=4.76E-8
+ TF=16N VJS=0.75 )
.MODEL PMOD PNP(IS=2E-15 VAF=100 IKF=0.005 ISE=1.9E-12 RB=10 RE=0.5
+ RC=0.5 CJE=3.5E-11 VJE=0.75 TF=1.6E-8 CJC=1.1E-11 VJC=0.75 TR=5.1E-8
+ TF=16N VJS=0.75 )
.ENDS X2N6027
#### martynwheeler
Joined Feb 18, 2014
2
Did anyone solve this, I have tried both models and it doesn't oscillate as it stands. I built the circuit and it is fine. Is there a better model for the 2N6027?
Here's my netlist file
* D:\Martyn\Documents\LTSpice\Oscillator.asc
R1 N001 N002 18k
R2 N001 VLoad 470k
R3 N002 0 27k
C1 VLoad 0 2.2µ V=6.3 Irms=0 Rser=9 Lser=0
D1 N003 0 REDLED
V1 N001 0 6
XU1 VLoad N002 N003 X2N6027
.model D D
.lib D:\LTspiceIV\lib\cmp\standard.dio
.tran 0 5 0 0.0001
.ic V(VLoad)=0
* Red Led
.MODEL REDLED D (IS=1.41e-21 RS=8.73 N=1.54)
.lib PUJT.LIB
.backanno
.end
Last edited:
#### crutschow
Joined Mar 14, 2008
25,432
Post the circuit diagram of the complete oscillator circuit.
#### martynwheeler
Joined Feb 18, 2014
2
Here's my circuit. Apologies, my spice simulation contained an 18k resistor which caused the circuit to oscillate. However, my circuit from a book contained a 15k resistor which according to my simulation won't oscillate.
Cheers, Martyn
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# NSTSE (National Science Talent Search Exam- Unified Council) Class 6: Questions 833 - 841 of 2029
Access detailed explanations (illustrated with images and videos) to 2029 questions. Access all new questions- tracking exam pattern and syllabus. View the complete topic-wise distribution of questions. Unlimited Access, Unlimited Time, on Unlimited Devices!
View Sample Explanation or View Features.
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## Question 833
MCQ▾
### Choices
Choice (4)Response
a.
1
b.
-1
c.
0
d.
+ 142
## Question 834
### Question
MCQ▾
Which of the following number is a perfect number?
### Choices
Choice (4)Response
a.
496
b.
33
c.
7
d.
All of the above
## Question 835
### Question
MCQ▾
can be represented as ________
### Choices
Choice (4)Response
a.
b.
c.
d.
Question does not provide sufficient data or is vague
## Question 836
### Question
MCQ▾
The equivalent fraction of having the denominator 9 is________
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question 837
### Question
MCQ▾
A square of any positive integer is
### Choices
Choice (4)Response
a.
0
b.
Positive
c.
Negative
d.
None of the above
## Question 838
### Question
MCQ▾
Which of the following properties are applicable to addition of whole numbers?
### Choices
Choice (4)Response
a.
Associative property
b.
Closure property
c.
Commutative property
d.
All a., b. and c. are correct
## Question 839
### Question
MCQ▾
The model is shaded to show which fraction?
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question 840
### Question
MCQ▾
A fraction where the numerator is bigger than the denominator is called an ________
### Choices
Choice (4)Response
a.
unlike
b.
Like
c.
improper
d.
Proper
## Question 841
### Question
MCQ▾
225 25 is ________
### Choices
Choice (4)Response
a.
9
b.
25
c.
225
d.
2
Developed by:
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# Quantum Mechanics - Wavefunctions
### Introduction
In order to have an understanding of spectroscopy, it is important to have a sound understanding of the quantum mechanics that underlies it. Any atomic or molecular system will have a wavefunction that describes the probability distribution of the electrons that make it up. The shape of this electron cloud must reflect the symmetry of the system of which it is a part. A hydrogen atom in an electric field can have its electron cloud distorted in the direction of the field.
The hydrogen atom can occupy many different states that are characterized by the familiar quantum numbers - n, l, s, ml, ms; and each of these states has a characteristic spatial profile associated with the wavefunction for this state. The wavefunction can be written (in the bra and ket notation of Dirac) as:
< n, l, s, ml, ms >
so that all of the important quantum numbers appear. Sometimes the wavefunctions are abbreviated as <x> just to save writing or to make the wavefunction arbitrary. Some distributions corresponding to these wavefunctions are:
Try sketching the wavefunctions for l = 0, ml = 0; l = 1, ml = 0, +/-1; and l = 2, ml = 0, +/-1, +/-2
All of these wavefunction are stationary, i.e., they do not change in time. A technical way of saying the same thing is to say they are eigenfunctions of the time evolution operator which is the Hamiltonian of the system. The Hamiltonian is in fact the operator that describes the system. The eigenvalues of the Hamiltonian operating on the wavefunctions provide the energy of that state:
H < X > = E < X >
Any other operator that is compatible with the symmetry of the state will also have a eigenvalue characteristic of that state. The criteria that the operator by compatible is equivalent to saying the operator commutes with the Hamiltonian. The operator for the total momentum (J2) for example commutes if
H J2 - J2 H = [H, J2] = 0
If this is true, then both H and J2 can operate on a wavefunction to determine the eigenvalues simultaneously. Thus
J2 |S,L,J,Mj > = J(J+1) |S,L,J,Mj >
If one finds all of the operators and their corresponding eigenvalues that are compatible with the symmetry of the system (that is with the Hamiltonian that describes the system), the wavefunctions can be completely described. For atomic and molecular systems, there are an infinite number of states and each of them is characterized by their energy and angular momentum.
A very important part of quantum mechanics concerns what happens when one of these stationary states is perturbed by something. If an electric field is applied to a hydrogen atom, the electron clouds are distorted and polarized. Their shapes no longer correspond to the original wavefunctions and the symmetry of the system has been destroyed. Some of the operators that were previously compatible with the system (i.e. they commuted with the Hamiltonian) no longer commute and therefore the quantum numbers are no longer "good." Such a situation is sketched below
Try sketching an l = 0, ml = 0 orbital that is distorted in the vertical direction:
The wavefunctions for this new situation must be different from the original wavefunctions and yet they need not be very much different if the perturbation is small. Fortunately, one does not need to solve this new problem from scratch because any new wavefunction, even a distorted one, can be described by a linear combination of the old wavefunctions. This is possible because the old wavefunctions form a complete set. Thus the distorted electron cloud sketched above can be described by the following combination:
Mathematically, a new wavefunction < X > would be written in terms of the old wavefunctions:
< Xn > (where n = 1,2,3,...)
< X > = summation over n of a(n) * < Xn >
where a(n) are coefficients that determine how much of each old wavefunction makes up the new wavefunction (or how much gets "mixed in"). Since the new wavefunction has a symmetry of its own, group theory can be used to tell what old states are compatible and can be mixed in.
### Further Information
`/chem-ed/quantum/wavefns.htm, updated 9/12/96`
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XMATCH Functions - Master Excels Array Tools! - Acuity Training
# XMATCH Functions – Master Excels Array Tools!
Most people using Excel, have used the very important MATCH function.
However, with more features and flexibility, the XMATCH function is a successor to the MATCH function of Excel.
It helps you look out for values and identify their position from a given range or array. Learn all about the multi-dimensional use of this smart function in the article below.
To see details of our courses see: Excel training courses.
## Excel XMATCH Function
The Excel XMATCH function from the dynamic array tools of Excel is an advanced version of the MATCH function. It enables users to find the relative position of a specific data entry within an array or range.
It not only offers more features than the MATCH function but is also flexible and easier to use.
### Syntax
Syntax of the XMATCH function looks as below.
=XMATCH (lookup_value, lookup_array, [match_mode], [search_mode])
### Arguments
It’s time we break down the arguments posed by the syntax of the XMTACH function to decipher it better.
• Lookup_Value – As the name verily tells, this argument specifies the value to be looked up for.
• Lookup_Array – This argument specifies the range or array where the lookup value is to be searched for.
• Match_Mode – This is an optional argument. It specifies the type of value to be looked for.
Users may specify either of the following four arguments as the match mode.
Arguments Match Mode 0 Exact Match -1 Exact match or the next smaller value 1 Exact match or next larger value 2 Wildcard match If Omitted Set to 0 by default
• Search_Mode – This is an optional argument. It specifies starting from where the search for the lookup value is to be performed.
Users may specify either of the following four arguments as the search mode.
Arguments Search Mode 1 Search from the first value -1 Search from the last value 2 Binary Search Ascending -2 Binary Search Descending If Omitted Set to 1 by default
### Return Value
XMATCH function returns the relative position of the specified value from the lookup array. The return value is numeric.
### Functions Library
Find out the XMATCH function from the Functions Library as follows.
Formulas > Functions Library > Lookup & Reference > XMATCH
Note: The XMATCH function is only available to the subscribers of Microsoft 365 for Excel 2019 and above versions.
### Why would someone want to use it?
The XMATCH function of Excel is one of the most commonly used dynamic array tools of Excel in the financial modeling industry. Not only that, it is the go-to tool for statisticians, forensic auditors, and many other data technicians.
For instance, as a money operator, you may want to instantly search all those currencies that equate to \$0.5 from a list of hundreds of currencies. Do not just go searching all the way manually.
Set up the XMATCH function to readily identify the relative position of each currency entry that meets the said criterion.
Currency conversion in Excel is super easy, get to know more about it here.
## XMATCH Use Cases
XMATCH function helps many Excel users from different fields of life. It can be of great help to professionals like statisticians, data analysts, researchers, and financial data modelers.
Even if you take it to domestics and regular day usage, the XMATCH function can help you ace a variety of tasks from your daily life.
1. The XMATCH function can be used by result compilers to know the number of students who gained a specific percentage or marks. Specifying different match modes, you can search for students who gained immediately lesser or higher marks than a specific threshold.
2. It can be used by auditors to readily identify where an invoice is entered in a data set of invoices.
3. Businesses can use it to sort particular customer records from their sales data by specifying the name or ID of a customer as the lookup value.
As it helps look up for a value, the XMATCH function can be of endless use to different groups of people. Learn more about other logical functions of Excel here.
## Using XMATCH to search for a specific item in an array – Basic Example
Apart from all the theoretical discussion, it’s time we stipulate a simple example to see the XMATCH function work.
The image below manifests the result sheet of XYZ School for the academic year 2021. It contains the result of many students arranged position-wise.
If you want to find the position of a particular student, say Aliya, how can you do it using the XMATCH function?
#### Step 1:
Activate the cell where you want the position of Aliya to be populated and compose the XMATCH formula as follows.
=XMATCH (“Aliya”, B2:B13)
• The first argument is the value that needs to be looked up from the array. It must be enclosed in double quotation marks.
• The second argument is the array where the value needs to be looked up. As the names are specified under column B (cells B2 to B13), we have specified it as the lookup array.
• The third and fourth arguments, being optional are omitted. Excel sets them to the default values of 0 and 1 respectively.
• This means the match_mode is set to 0 where Excel would only lookup for the exact match. And the search_mode is set to 1 where Excel would start searching from the first value. As the students are already arranged position-wise, and we want to find the relative position of Aliya, both the arguments fit our needs.
#### Step 2:
Hit ‘Enter’ to see the results as follows.
In the data, Aliya stands at the 11th position, and Excel has identified the same.
Pro Tip: As we have left the ‘Match Mode’ vacant, Excel has set it to ‘0’ by default. That being said, Excel would only lookout for the exact spelling of ‘Aliya’.
If the spelling is not the same, the results put out by Excel will be as follows.
If you want to get rid of the exact match hunt of Excel, specify the third argument as 1 or -1 where Excel would lookout for the closest match if not the exact match.
## More Examples
### 1. The XMATCH function with different search modes
The last argument from the XMTACH function i.e. the search mode is often misunderstood. To understand how this argument works, follow the example below.
The data below represents the details of the yearly ‘Best Employee’ title won by different employees of a Company over 10 years.
A closer look into the data reveals that the said title was won by the same employee multiple times. To know the first time when Lucy on the title, we can use the XMATCH function as follows.
#### Step 1:
Activate the cell where you want the results to be populated and compose the XMTACH function as below.
=XMATCH (“Lucy”, B2:B11,,1)
• The first argument is specified as “Lucy” enclosed in double quotation marks as it is the value to be looked up for.
• The second argument defines the lookup array where the value is to be looked up for i.e. B2:B12.
• The third argument is omitted and is assumed to be 0 by Excel by default. It is important to note that a comma is added before and after the vacant third argument before moving on to the fourth.
• The fourth argument is set to 1 as we want Excel to find and return the first year when Lucy won the award. Doing so, Excel will look up the specified array starting from the first value.
#### Step 2:
Hit ‘Enter’ to see results as follows.
Excel returns the value 3 i.e. the first relative position of Lucy from the underlying data. The first time Lucy won the Best Employee Award was in 2013.
Although Lucy appears more than once in the given data, the other instances are not considered by Excel.
Let’s do it the other way around now.
#### Step 1:
Continuing with the same example as above, if we now want Excel to only identify the last time when Lucy won the best employee award.
To do so, compose the XMATCH formula as follows.
Activate the cell where you want the results to be populated and compose the XMTACH function as below.
=XMATCH (“Lucy”, B2:B11,,-1)
• The first argument is specified as “Lucy” enclosed in double quotation marks as it is value to be looked up for.
• The second argument defines the lookup array where the value is to be looked up for i.e. B2:B12.
• The third argument is omitted and is assumed to be 0 by Excel by default. We have added a comma before and after the vacant third argument before moving on to the fourth.
• The fourth argument is set to -1 as we want Excel to find and return the last year when Lucy won the award. Doing so, Excel will look up the specified array starting from the last value.
#### Step 2:
Hit ‘Enter’ to see results as follows.
Excel returns the value 8 i.e. the last relative position of Lucy from the underlying data. The last time when Lucy won the best employee award was in 2018.
Although Lucy appears more than once in the given data, the other instances are not considered by Excel.
### 2. XMATCH function with Wildcard characters
XMATCH function can be used with wildcard characters to look up different values. There are two wildcard characters that you can employ in the XMATCH function.
1. ? – A question mark equates to any value in the same order
2. * – An asterisk equates to any sequence of characters
For example, from the list of continents below, we want to find the relative position of North America. It may be listed as North America or Northern America.
Not knowing the exact spelling used in the list, how can we specify the lookup value? This can be done using wildcard characters.
#### Step 1:
Activate the cell where you want the results to be populated and compose the XMATCH formula as follows.
= XMATCH (“North*”, B2:B8, 2)
• To specify the lookup value, we have used the wildcard character asterisk (*). This is because ‘Northern’ consists of three letters after the word ‘North’. Asterisk equates to any number of letters in any sequence after the word North.
• The lookup array is defined as B2:B8 as it contains the names of the continents from where the look up value is to be looked up.
• The match mode is set to 2 i.e. wildcard character match.
• The fourth argument is left vacant so Excel will set it to 1 by default.
#### Step 2:
Hit Enter to yield results as follows.
Excel finds the position of the word that contains ‘North’ and is most closely related to the specified criterion.
## Quick Brain Teaser:
#### What if the values in the lookup array contain a question mark or asterisk in real?
Take a look at the data below.
Each value of the data contains an asterisk at its end.
For most of the Excel functions that employ wildcard characters, a tilde character (~) is to be used to treat asterisk (*) and question mark (?) as literal characters within the values.
However, with the XMATCH function, you do not need to do this. For instance, to find the relative position of the cell value ‘Asia*’, specify the XMATCH function as follows.
XMATCH (“Asia*”, A2:A8)
Must note how we have specified the exact value, and the match mode argument is left vacant. The wild character won’t work in Excel until the Wildcard match mode i.e. 2 is specified.
It is important to note that, unlike other Excel functions, the XMATCH function only accepts wildcard characters when the match mode is set to wildcard mode.
### 3. Nesting the XMATCH function into the INDEX function:
The XMATCH function is commonly used in pair with the Excel INDEX function as follows. This is for the reason that, at times, we not only want to find the relative position of a specific value but the value itself.
The INDEX function serves the said job. For instance, the data in the example below states the ages of multiple people.
From this data, if you instantly want your hands on the age of Mr. D, how can you do that?
It is hectic to go back looking into the data for the entry of Mr. D and then copying his age from there to the destined location.
To your good, this can be done through a single formula using the XMATCH and INDEX functions together as devised below.
#### Step 1:
The first step is to put together the XMATCH formula. We want Mr. D (the lookup_value) to be looked up from the column of names A2:A11 (the lookup_array).
The formula would thus, be put up as follows.
=XMATCH (“Mr. D”, A2:A11)
This formula would return the relative position of Mr. D from the said column as a numeric value.
#### Step 2:
The next step is to set up the INDEX function by nesting in the XMATCH function as follows.
= INDEX (B2:B11, XMATCH (“Mr. D”, A2:A11))
• The first argument of the index function specifies the range from where the value is to be sought and returned.
• The second argument specifies the row wherefrom the value is to be returned. In place of the second argument, we have incorporated the XMATCH function; the product whereof would be the relative row number of Mr. D’s entry.
• The third and fourth arguments of the INDEX function are optional and are omitted.
#### Step 3:
Press ‘Enter’ to see Excel pick out Mr. D’s age for you in the designated cell as follows.
Also learn about nesting of SORT and SORTBY dynamic array functions here.
## XMATCH Troubleshooting
Following are some of the common problems experienced by Excel users when working with the XMTACH function.
### 1. Case insensitivity
The Excel XMATCH function, by default, is case insensitive. That means if a cell within your data contains the value ‘FedEx’, you can specify the lookup value as ‘fedex’ or ‘Fedex’ or ‘FEDEx’ or any combination of capital and small cases.
The XMATCH function would spot it out from the source data even if the case is inconsistent.
However, it is at times that we need to make a case-sensitive search. To do that, you can nest the EXACT function into the XMATCH function.
### 2. Inappropriate Match Mode
Most of the XMATCH function problems of Excel users originate from their inability to decipher the right match mode that meets their needs.
For the data above, if you want to know the relative position of all those years where the sales were equal to or close to \$100,000, the XMATCH formula may be set in two ways.
=XMATCH (F2, B2:B11, -1,1)
In the formula above, as we have specified the match mode as ‘-1’, Excel would find the exact or next smaller value to \$100,000.
The formula above may also be set up as below.
=XMATCH (F2, B2:B11, 1,1)
In the formula above, as we have specified the match mode as ‘1’, Excel would find the exact or the next larger value to \$100,000 from the lookup array.
Learn about the SEQUENCE dynamic array tool of Excel here.
## Conclusion
The dynamic array tool of XMATCH can be of help for many Excel users from different rounds of life. Practice a few examples from above to master it in no time.
Need an overview of everything Arrays can do in Excel? Check out our guide to Excel array formulas here.
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# 80.25 kg to lbs - 80.25 kilograms to pounds
Do you need to learn how much is 80.25 kg equal to lbs and how to convert 80.25 kg to lbs? Here it is. You will find in this article everything you need to make kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that whole this article is devoted to a specific number of kilograms - that is one kilogram. So if you need to know more about 80.25 kg to pound conversion - read on.
Before we move on to the practice - this is 80.25 kg how much lbs calculation - we want to tell you few theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 80.25 kg to lbs? 80.25 kilograms it is equal 176.9209652550 pounds, so 80.25 kg is equal 176.9209652550 lbs.
## 80.25 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).
From time to time the kilogram could be written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but difficult to use.
Then, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in short form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.
## 80.25 kilogram to pounds
You know some information about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to underline that there are more than one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. To be honest, this unit is in use also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is just equal 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 80.25 kg?
80.25 kilogram is equal to 176.9209652550 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 80.25 kg in lbs
The most theoretical part is already behind us. In this part we are going to tell you how much is 80.25 kg to lbs. Now you learned that 80.25 kg = x lbs. So it is time to get the answer. Let’s see:
80.25 kilogram = 176.9209652550 pounds.
It is an exact result of how much 80.25 kg to pound. It is possible to also round it off. After it your result is as following: 80.25 kg = 176.550 lbs.
You learned 80.25 kg is how many lbs, so let’s see how many kg 80.25 lbs: 80.25 pound = 0.45359237 kilograms.
Naturally, in this case you may also round it off. After it your result will be exactly: 80.25 lb = 0.45 kgs.
We also want to show you 80.25 kg to how many pounds and 80.25 pound how many kg outcomes in tables. See:
We are going to start with a chart for how much is 80.25 kg equal to pound.
### 80.25 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
80.25 176.9209652550 176.5500
Now see a chart for how many kilograms 80.25 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
80.25 0.45359237 0.45
Now you learned how many 80.25 kg to lbs and how many kilograms 80.25 pound, so it is time to go to the 80.25 kg to lbs formula.
### 80.25 kg to pounds
To convert 80.25 kg to us lbs you need a formula. We are going to show you two versions of a formula. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 176.9209652550 outcome in pounds
The first formula give you the most accurate outcome. In some situations even the smallest difference can be significant. So if you want to get a correct result - this formula will be the best solution to calculate how many pounds are equivalent to 80.25 kilogram.
So move on to the another version of a formula, which also enables conversions to learn how much 80.25 kilogram in pounds.
The second version of a formula is as following, see:
Amount of kilograms * 2.2 = the result in pounds
As you can see, this formula is simpler. It can be the best solution if you need to make a conversion of 80.25 kilogram to pounds in fast way, for example, during shopping. Just remember that final outcome will be not so correct.
Now we are going to show you these two formulas in practice. But before we will make a conversion of 80.25 kg to lbs we want to show you another way to know 80.25 kg to how many lbs totally effortless.
### 80.25 kg to lbs converter
Another way to check what is 80.25 kilogram equal to in pounds is to use 80.25 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on first formula which we showed you in the previous part of this article. Due to 80.25 kg pound calculator you can effortless convert 80.25 kg to lbs. You only have to enter number of kilograms which you need to convert and click ‘convert’ button. You will get the result in a second.
So try to convert 80.25 kg into lbs with use of 80.25 kg vs pound converter. We entered 80.25 as an amount of kilograms. This is the result: 80.25 kilogram = 176.9209652550 pounds.
As you see, our 80.25 kg vs lbs converter is user friendly.
Now we can move on to our main topic - how to convert 80.25 kilograms to pounds on your own.
#### 80.25 kg to lbs conversion
We will start 80.25 kilogram equals to how many pounds conversion with the first formula to get the most accurate outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 176.9209652550 the outcome in pounds
So what need you do to check how many pounds equal to 80.25 kilogram? Just multiply amount of kilograms, this time 80.25, by 2.20462262. It is exactly 176.9209652550. So 80.25 kilogram is exactly 176.9209652550.
You can also round it off, for example, to two decimal places. It gives 2.20. So 80.25 kilogram = 176.5500 pounds.
It is high time for an example from everyday life. Let’s convert 80.25 kg gold in pounds. So 80.25 kg equal to how many lbs? And again - multiply 80.25 by 2.20462262. It is exactly 176.9209652550. So equivalent of 80.25 kilograms to pounds, if it comes to gold, is equal 176.9209652550.
In this case it is also possible to round off the result. Here is the outcome after rounding off, in this case to one decimal place - 80.25 kilogram 176.550 pounds.
Now we are going to examples calculated with short formula.
#### How many 80.25 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 176.550 the outcome in pounds
So 80.25 kg equal to how much lbs? And again, you need to multiply number of kilogram, this time 80.25, by 2.2. Let’s see: 80.25 * 2.2 = 176.550. So 80.25 kilogram is equal 2.2 pounds.
Let’s do another calculation using shorer formula. Now convert something from everyday life, for instance, 80.25 kg to lbs weight of strawberries.
So calculate - 80.25 kilogram of strawberries * 2.2 = 176.550 pounds of strawberries. So 80.25 kg to pound mass is exactly 176.550.
If you learned how much is 80.25 kilogram weight in pounds and can calculate it using two different formulas, let’s move on. Now we want to show you these results in tables.
#### Convert 80.25 kilogram to pounds
We realize that outcomes presented in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can easily compare 80.25 kg equivalent to lbs results.
Start with a 80.25 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
80.25 176.9209652550 176.5500
And now look 80.25 kg equal pound chart for the second version of a formula:
Kilograms Pounds
80.25 176.550
As you can see, after rounding off, if it comes to how much 80.25 kilogram equals pounds, the outcomes are not different. The bigger number the more significant difference. Keep it in mind when you need to do bigger number than 80.25 kilograms pounds conversion.
#### How many kilograms 80.25 pound
Now you know how to calculate 80.25 kilograms how much pounds but we are going to show you something more. Do you want to know what it is? What about 80.25 kilogram to pounds and ounces conversion?
We will show you how you can convert it step by step. Let’s start. How much is 80.25 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 80.25, by 2.20462262. So 80.25 * 2.20462262 = 176.9209652550. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 80.25 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So your outcome is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final result is equal 2 pounds and 33 ounces.
As you see, conversion 80.25 kilogram in pounds and ounces is not complicated.
The last calculation which we will show you is calculation of 80.25 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters it is needed another formula. Before we give you it, have a look:
• 80.25 kilograms meters = 7.23301385 foot pounds,
• 80.25 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 80.25 foot pounds to kilograms meters you need to multiply 80.25 by 0.13825495. It is equal 0.13825495. So 80.25 foot pounds is exactly 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 80.25 foot pounds is 0.14 kilogram meters.
We hope that this conversion was as easy as 80.25 kilogram into pounds conversions.
This article was a huge compendium about kilogram, pound and 80.25 kg to lbs in calculation. Thanks to this calculation you learned 80.25 kilogram is equivalent to how many pounds.
We showed you not only how to do a calculation 80.25 kilogram to metric pounds but also two another conversions - to know how many 80.25 kg in pounds and ounces and how many 80.25 foot pounds to kilograms meters.
We showed you also another solution to do 80.25 kilogram how many pounds conversions, it is with use of 80.25 kg en pound calculator. It is the best option for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you can make 80.25 kilogram equal to how many pounds calculation - on your own or using our 80.25 kgs to pounds calculator.
So what are you waiting for? Let’s calculate 80.25 kilogram mass to pounds in the way you like.
Do you want to do other than 80.25 kilogram as pounds calculation? For example, for 5 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so easy as for 80.25 kilogram equal many pounds.
### How much is 80.25 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 80.25 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see all you need to remember about how much is 80.25 kg equal to lbs and how to convert 80.25 kg to lbs . You can see it down below.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 80.25 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 80.25 kilogram to pounds? The exact answer is 176.9209652550 lbs.
You can also calculate how much 80.25 kilogram is equal to pounds with second, shortened version of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 80.25 kg equal to how much lbs ? The result is 176.9209652550 pounds.
How to convert 80.25 kg to lbs quicker and easier? You can also use the 80.25 kg to lbs converter , which will do whole mathematical operation for you and you will get a correct result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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CBSE Class 10 Sample Paper for 2022 Boards - Maths Basic [MCQ]
Class 10
Solutions of Sample Papers for Class 10 Boards
## (a) 2 : 3 (b) 6 : 9 (c) 4 : 6 (d) 4 : 9
This question is inspired from Question 13 - Sample Papers 2020 - Maths Standard - [Also, check proof of the formula]
### Transcript
Question 33 The perimeters of two similar triangles are 26 cm and 39 cm. The ratio of their areas will be (a) 2 : 3 (b) 6 : 9 (c) 4 : 6 (d) 4 : 9 Let ΔABC and ΔPQR be two similar triangles Now, Ratio of perimeter of two similar triangles is equal to ratio of their corresponding sides ∴ (𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 Δ 𝑨𝑩𝑪)/(𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 Δ 𝑷𝑸𝑹)=𝑨𝑩/𝑷𝑸 26/39=𝐴𝐵/𝑃𝑄 2/3=𝐴𝐵/𝑃𝑄 𝑨𝑩/𝑷𝑸=𝟐/𝟑 We know that Ratio of area of similar triangles is equal to square of ratio of its corresponding sides Therefore, (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=(𝐴𝐵/𝑃𝑄)^2 Putting values (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=(2/3)^2 (𝑨𝒓𝒆𝒂 𝒐𝒇" " ∆𝑨𝑩𝑪)/(𝑨𝒓𝒆𝒂 𝒐𝒇" " ∆𝑷𝑸𝑹)=𝟒/𝟗 Hence, the required ratio is 4 : 9 So, the correct answer is (d)
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# Pong Ball movement
This topic is 4970 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
hi all, I am trying to make my ball in my pong game to move. It is doing fine, but how do I make my ball move smoothly? this is my code:
void MoveBall()
{
if(ballX < targetX)
ballX += ballspeed;
if(ballX > targetX)
ballX -= ballspeed;
if(ballY < targetY)
ballY += ballspeed;
if(ballY > targetY)
ballY -= ballspeed;
}
but as the ball reaches my targetX sooner then my targetY, from the moment he reaches targetY, he goes straight forward. How do I make him gradually go to the point (targetX, targetY)?
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This is simple with a bit of algebra and graphing. You have two points, start x,y and end x,y. Find the slope of the line between them and then assign the rise to delta y and the run to delta x. Bingo.
(sorry, I haven't done any serious algebra for god knows how long so I have no forumulas memorized anymore :P)
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actually we just got done covering this i think the formula you need is
Y=mx+b
(X,Y),(X2,Y2) <-------ordered pairs or points
m(slope) = (Y2-Y)/(X2-X)
Y2 = mX2 + B(y-intersept)
B = Y2 - (mX2)
hope i helped
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Yea, that's it [smile]
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lol that sounds complicated (even more cause I just came out of bed). before I went to bed I thought about something like:
void ChangeTarget()
{
targetX = (oldX - targetX) + oldX;
targetY = (oldY - targetY) + oldY
oldX = targetX;
oldY = targetY;
}
offcourse at the start I set oldY & X = to targetY & X. would this work?
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Well I'm just going to bed, so I can't make head or tails of that myself. However if you're scared off of the math because it's algebra well... tough. (I'm kinda grumpy too, sry). You're going to need lots of algebra if you want to do any serious graphics programming (not so much programming in general, but yea it helps there too). Have you had any algebra? if not you could prob Google the above formulas and learn all you need to know for this application at least. Just do it. it works, no sense reinventing the wheel.
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No need for algebra here.
For 'smooth' movement, just use floats to store your balls coordinates, and floats for the velocity.
eg:
float bx=0.0,by=0.0;
float vx=0.1,vy=0.2;
bx += vx;
by += vy;
Then just draw the ball as usual (motion will still be *slightly* jerky, as your ball is likely only drawn to the nearest pixel).
This allows you to have very very slow speeds, and still update the positions many times a second.
Anyway, its pong, I don't want to have to write more lines in this response post then it takes to code the thing! :)
- Jacob
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lol, spent 3 days on it already... :D
##### Share on other sites
just have two speeds, one for x-axis, one for y-axis.
than when the ball hits a paddle, reverse the speed on the x-axis and when the ball hits the top or bottom of the screen, reverse the speed on the y-axis. Like this:
void HandleBallMovement(void)
{
ballX += xspeed;
ballY += yspeed;
xspeed = -xspeed;
if(ballY <= 0 || ballY >= SCREENHEIGHT)
yspeed = -yspeed;
}
This is what I used when I made pong and it worked just fine.
##### Share on other sites
void Movement()
{
while(!GameOver)
{
SDL_Event event;
while(SDL_PollEvent(&event))
{
switch(event.type)
{
case SDL_QUIT:
GameOver = true;
running = false;
break;
case SDL_KEYDOWN:
switch(event.key.keysym.sym)
{
case SDLK_UP:
Movement(-1);
break;
case SDLK_DOWN:
Movement(1);
break;
case SDLK_LSHIFT:
speed++;
break;
case SDLK_LCTRL:
if(speed < 1)
{
speed--;
}
break;
case SDLK_ESCAPE:
GameOver = true;
game = false;
break;
}
break;
}
SDL_FillRect(screen, NULL, 255);
DrawGame();
MoveBall();
DrawBall();
SDL_Flip(screen);
}
}
GameOver = false;
}
and it works fine. The only problem I have is that he ball won't move if there is not INPUT. I try'd adding a "case NULL:", but it didn't work either ("default:" didn't either). any idea of how I could make the ball move without having to wait for input?
Thank you very much in advance,
Joshua
1. 1
Rutin
24
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3. 3
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JoeJ
18
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Symbol
# Calculator search results
Formula
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$\dfrac { 4 } { x } - \dfrac { 1 } { y } = \dfrac { 7 } { 2 }$
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$\begin{cases} \dfrac{ 4 }{ x } - \dfrac{ 1 }{ y } = \dfrac{ 7 }{ 2 } \\ \dfrac{ 3 }{ x } - \dfrac{ 1 }{ 2y } = \dfrac{ 15 }{ 4 } \end{cases}$
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https://isipta07.sipta.org/tutorials.html
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### ISIPTA '07 TUTORIALS
On the day July the 16th 2007, prior to the start of the technical sessions (on the days 17--19), there will be four invited tutorials of 90 minutes each. The tutorials will provide a gentle introduction to a wide range of important subject matters in imprecise probability. The tutorials are included in the (regular or student) registration fee. You are kindly invited to participate.
### TUTORIAL 1
Speaker: Scott Ferson (Applied Biomathematics, USA).
Title: Risk analysis: rough but ready tools for calculations under variability and uncertainty
Abstract:
Risk analysis is widely used in many disciplines to quantify risks or expectations in the face of pervasive variability and profound uncertainty about both natural and engineered systems. Although most analyses today are still based on point estimates, awkward qualitative assessments, or probabilistic calculations employing unwarranted assumptions, the methods of imprecise probability hold great promise for allowing analysts to develop quantitative models that make use of the knowledge and data that are available but do not require untenable or unjustified assumptions or simplifications. This tutorial will introduce some of the methods that are easiest to make calculations with, including probability bounds analysis, Dempster-Shafer evidence theory, and interval statistics, and will show how they can be used to address the basic problems that risk analysts face: not knowing the input distributions, not knowing their correlations, not being sure about the model itself, or even which variables should be considered. We suggest that these tools constitute a practical uncertainty arithmetic (and logic) that can be widely deployed for lots of applications. Of course, not all problems can be well solved by these relatively crude methods. Examples requiring fuller analyses with the methods of imprecise probability are described.
### TUTORIAL 2
Speaker: Enrique Miranda (Universidad Rey Juan Carlos, Spain).
Title: An introduction to the theory of coherent lower previsions
Abstract:
In this tutorial, introduce the main elements of Peter Walley's theory of coherent lower and upper previsions. I review the notions of avoiding sure loss and coherence, and the representation of coherent assessments in terms of sets of linear previsions and sets of almost-desirable gambles. Then, I turn to the notion of natural extension, and give its expression under any of these three equivalent representations. Finally, I study how to update assessments in a coherent way, presenting the main facts about conditional lower previsions.
### TUTORIAL 3
Speaker: George Klir (Binghamton University, USA).
Title: Generalized information theory
Abstract:
A research program whose objective is to study the dual concepts of information-based uncertainty and uncertainty-based information in all their manifestations was introduced in the early 1990s under the name "generalized information theory2 (GIT). The purpose of this tutorial is to introduce conceptual boundaries within which GIT operates and a comprehensive overview of principal results that emerged from GIT. As in classical information theory, uncertainty is the primary concept and information is defined in terms of uncertainty reduction. GIT is based on a two-dimensional expansion of classical information theory. In one dimension, additive probability measures of classical information theory are expanded to various types of nonadditive measures. In the other dimension, the theory of classical sets, within which probability measures are formalized, is expanded to the various theories of fuzzy sets. Each choice of a particular set theory and a particular measure theory defines a particular information theory. The full development of any of these information theories requires that issues at each of the following four levels be adequately addressed: (1) an uncertainty function, u, of the theory be formalized in term of appropriate axioms; (2) the calculus for dealing with function u be properly developed; (3) a justifiable functional, U, be determined by which the amount of relevant uncertainty (predictive, prescriptive, diagnostic, etc.) associated with function u is measured; and (4) methodological aspects of the theory be developed by utilizing functional U as a measuring instrument. The tutorial is presented in two parts of approximately the same duration. An overall characterization of GIT is presented in the first part. After a brief overview of classical information theory, a general framework for formalizing uncertainty and the associated uncertainty-based information of any conceivable type is sketched. The various theories of imprecise probabilities that have already been developed within this framework are surveyed and some important unifying principles applying to these theories are introduced. The second part is devoted to the issues of measuring uncertainty and information in the various theories and to the methodological principles based on these measuring capabilities. The tutorial is concluded by a discussion of some open problems in the area of GIT. The tutorial is intended as a gentle introduction to the area of GIT, which is covered in a greater depth in the recent book "Uncertainty and Information: Foundation of Generalized Information Theory" by George J. Klir (Wiley-Interscience, 2006).
### TUTORIAL 4
Speaker: Teddy Seidenfeld (Carnegie Mellon University, USA).
Title: Decision theories for imprecise preferences and imprecise probabilities
Abstract: This tutorial offers an overview of selected alternative decision theories designed for use with IP theory. The review begins with an examination of the kinds of rival decision theories that ensue when each of the familiar axioms of, for instance, Anscombe-Aumann Horse Lottery theory is relaxed to accommodate normal-form IP theory. From this base, further generalizations are considered, including: multi-agent decision making, extensive form IP theory, and several interesting considerations that attend problems with infinite decision structures.
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https://oeis.org/A342422
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| 467,632,450
| 3,811
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A342422 a(n) = Sum_{k=1..n} (n/gcd(k,n))^gcd(k,n). 2
1, 3, 7, 13, 21, 39, 43, 81, 109, 173, 111, 475, 157, 507, 1107, 1153, 273, 2997, 343, 6133, 6685, 3479, 507, 30819, 13001, 10533, 44227, 65071, 813, 234959, 931, 215553, 368265, 136241, 414183, 1991281, 1333, 531471, 3215947, 4883141, 1641, 12951117, 1807, 12735043, 36892689, 8401259 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Seiichi Manyama, Table of n, a(n) for n = 1..5000 FORMULA a(n) = Sum_{d|n} phi(d^(n/d+1)) = Sum_{d|n} phi(d) * d^(n/d). MATHEMATICA a[n_] := Sum[(n/GCD[k, n])^GCD[k, n], {k, 1, n}]; Array[a, 50] (* Amiram Eldar, Mar 11 2021 *) PROG (PARI) a(n) = sum(k=1, n, (n/gcd(k, n))^gcd(k, n)); (PARI) a(n) = sumdiv(n, d, eulerphi(d^(n/d+1))); (PARI) a(n) = sumdiv(n, d, eulerphi(d)*d^(n/d)); CROSSREFS Cf. A000010, A342421. Sequence in context: A355736 A032606 A098482 * A147432 A018367 A146369 Adjacent sequences: A342419 A342420 A342421 * A342423 A342424 A342425 KEYWORD nonn AUTHOR Seiichi Manyama, Mar 11 2021 STATUS approved
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Last modified September 29 14:55 EDT 2023. Contains 365772 sequences. (Running on oeis4.)
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http://betterlesson.com/lesson/resource/2794569/explaining-how-a-meter-is-put-together-mov
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crawl-data/CC-MAIN-2016-44/segments/1476988722459.85/warc/CC-MAIN-20161020183842-00008-ip-10-171-6-4.ec2.internal.warc.gz
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## Explaining how a meter is put together.mov - Section 3: Informal assessement of understanding
Explaining how a meter is put together.mov
# Building a Meter: Comparing units.
Unit 2: Metric Measurement
Lesson 4 of 18
## Big Idea: In this "just too fun" lesson, place value blocks take on a new meaning as they are used to build a meter. Then, students create an assessment video explaining the relationship between mm, cm and dm to the meter.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Measurement, Conversions (Within and Between Systems), meter sticks, hands, Place Value Blocks
40 minutes
### Mary Ellen Kanthack
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Big Idea: Students explore a factory of factors and multiples to build number sense and conceptual understanding.
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http://gmatclub.com/forum/a-large-rise-in-the-number-of-housing-starts-in-the-coming-43874.html?kudos=1
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A large rise in the number of housing starts in the coming
Author Message
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Director
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A large rise in the number of housing starts in the coming [#permalink]
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29 Mar 2007, 20:05
00:00
Difficulty:
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Question Stats:
100% (01:24) correct 0% (00:00) wrong based on 10 sessions
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A large rise in the number of housing starts in the coming year should boost new construction dollars by several billion dollars, making the construction industry’s economic health much more robust than five years ago.(A) making the construction industry’s economic health much more robust than five years ago
(B) and make the construction industry’s economic health much more robust than five years ago
(C) making the construction industry’s economic health much more robust than it was five years ago
(D) to make the construction industry’s economic health much more robust than five years ago
(E) in making the construction industry’s economic health much more robust than it as five years ago
Why isn't the answer "B" ?
Director
Affiliations: FRM Charter holder
Joined: 02 Dec 2006
Posts: 734
Schools: Stanford, Chicago Booth, Babson College
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Show Tags
29 Mar 2007, 20:12
LM wrote:
A large rise in the number of housing starts in the coming year should boost new construction dollars by several billion dollars, making the construction industry’s economic health much more robust than five years ago.(A) making the construction industry’s economic health much more robust than five years ago
(B) and make the construction industry’s economic health much more robust than five years ago
(C) making the construction industry’s economic health much more robust than it was five years ago
(D) to make the construction industry’s economic health much more robust than five years ago
(E) in making the construction industry’s economic health much more robust than it as five years ago
Why isn't the answer "B" ?
This question tests comparisons.
In B the comparison is not clear. In C the comparison is clear. "it" refers to health.
Senior Manager
Joined: 11 Feb 2007
Posts: 352
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Kudos [?]: 103 [0], given: 0
Show Tags
29 Mar 2007, 21:15
LM wrote:
A large rise in the number of housing starts in the coming year should boost new construction dollars by several billion dollars, making the construction industry’s economic health much more robust than five years ago.(A) making the construction industry’s economic health much more robust than five years ago
(B) and make the construction industry’s economic health much more robust than five years ago
(C) making the construction industry’s economic health much more robust than it was five years ago
(D) to make the construction industry’s economic health much more robust than five years ago
(E) in making the construction industry’s economic health much more robust than it as five years ago
Why isn't the answer "B" ?
You're not comparing the industry's economic health to "five years ago."
This is a comparison error.
C makes the comparison clear.
GMAT Club Legend
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Location: Singapore
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30 Mar 2007, 00:12
a -> 'more robust than five years ago' -> bad comparison
b -> 'and make' is wrong because it changes the meaning of the sentence. It now means the large rise cause two things to happen -> (1) boost dollars, (2) make economic health more robust
d -> 'more robust than five years ago' -> bad comparison
e -> 'in making' is wrong.
c is the best choice.
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GMAT 1: 530 Q39 V23
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Re: A large rise in the number of housing starts in the coming [#permalink]
Show Tags
15 Nov 2015, 06:02
whenever you encounters a comparison question, ask what the sentence exactly wants to compare
In this question it is construction industry’s economic health.
B,D,E changes the intended meaning of the sentence
Making is required as the ,(comma)+ verb+ing is going to answer either "how" or "resulting in"
A wrong comparison of economic health with 5 years
C it is
Re: A large rise in the number of housing starts in the coming [#permalink] 15 Nov 2015, 06:02
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# Function inverse problem
1. Nov 29, 2009
### computerex
1. The problem statement, all variables and given/known data
I have been given:
h(x) is the inverse for f(x) = x^3+x
They want to know h'(2).
3. The attempt at a solution
I know that since h(x) and f(x) are inverses:
f(h(x)) = x
differentiating with respect to x gives
f'(h(x))h'(x) = 1
So h'(x) = 1/f'(h(x))
Therefore h'(2) = 1/f'(h(2))
to find h(2)
x^3+x = 2
since f(h(x)) = x
x^3+x-2 = 0
so h(2) = 1
f(x) = x^3+x
f'(x) = 3x^2+1
h'(2) = 1/f'(h(2))
h'(2) = 1/3(1)^2+1
h'(2) = 1/4
My question is: How do you find the inverse of y = x^3+x ?
You interchange x and y then solve for y: y^3+y=x But I don't know how to solve that for y. Any help would be appreciated.
2. Nov 29, 2009
### jgens
We're not supposed to give complete solutions, so here's a hint:
$$y = w - \frac{1}{3w}$$
3. Nov 29, 2009
### computerex
I am sorry but the above doesn't help me solve: y^3+y=x for y. If you will see above, I have already got the solution to the problem using some algebraic gymnastics, I am just curious as to how to find the inverse for y=x^3+x.
4. Nov 29, 2009
### jgens
I know! You want to solve the cubic equation $y^3 + y - x = 0$ and the substitution $y = w + \frac{1}{3w}$ allows you to solve that cubic equation quite nicely.
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1
GATE EE 2019
+2
-0.67
Given two sets X = {1, 2, 3} and Y = {2, 3, 4}, we construct a set Z of all possible fractions where the numerators belong to set X and the denominators belong to set Y. The product of elements having minimum and maximum values in the set Z is ____________.
A
$${1 \over {12}}$$
B
$${1 \over 8}$$
C
$${1 \over 6}$$
D
$${3 \over 8}$$
2
GATE EE 2019
+2
-0.67
The ratio of the number of boys and girls who participated in an examination is 4 : 3. The total percentage of candidates who passed the examination is 80 and the percentage of a girls who passed is 90. The percentage of boys who passed is _________.
A
90.00
B
80.50
C
55.50
D
72.50
3
GATE EE 2019
+2
-0.67
How many integers are there between 100 and 1000 all of whose digits are even?
A
100
B
90
C
60
D
80
4
GATE EE 2018
+2
-0.6
An e-mail password must contain three characters. The password has to contain one numeral from 0 to 9, one upper case and one lower case character from the English alphabet. How many distinct passwords are possible?
A
6,760
B
13,520
C
40,560
D
1,05,456
EXAM MAP
Medical
NEET
| 378
| 1,085
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# Laplace transform
Intro to the Laplace Transform \u0026 Three Examples
Intro to the Laplace Transform \u0026 Three Examples
Laplace transform
In mathematics, the Laplace transform, named after its discoverer Pierre-Simon Laplace (/ləˈplɑːs/), is an integral transform that converts a function of a real variable (usually , in the time domain) to a function of a complex variable (in the complex frequency domain, also known as s-domain, or s-plane). The transform has many applications in science and engineering because it is a tool for solving differential equations.[1] In particular, it transforms ordinary differential equations into algebraic equations and convolution into multiplication.[2][3] For suitable functions f, the Laplace transform is the integral
## History
The Laplace transform is named after mathematician and astronomer Pierre-Simon, marquis de Laplace, who used a similar transform in his work on probability theory.[4] Laplace wrote extensively about the use of generating functions (1814), and the integral form of the Laplace transform evolved naturally as a result.[5]
Laplace’s use of generating functions was similar to what is now known as the z-transform, and he gave little attention to the continuous variable case which was discussed by Niels Henrik Abel.[6] The theory was further developed in the 19th and early 20th centuries by Mathias Lerch,[7] Oliver Heaviside,[8] and Thomas Bromwich.[9]
The current widespread use of the transform (mainly in engineering) came about during and soon after World War II,[10] replacing the earlier Heaviside operational calculus. The advantages of the Laplace transform had been emphasized by Gustav Doetsch,[11] to whom the name Laplace transform is apparently due.
From 1744, Leonhard Euler investigated integrals of the form
These types of integrals seem first to have attracted Laplace’s attention in 1782, where he was following in the spirit of Euler in using the integrals themselves as solutions of equations.[15] However, in 1785, Laplace took the critical step forward when, rather than simply looking for a solution in the form of an integral, he started to apply the transforms in the sense that was later to become popular. He used an integral of the form
Laplace also recognised that Joseph Fourier’s method of Fourier series for solving the diffusion equation could only apply to a limited region of space, because those solutions were periodic. In 1809, Laplace applied his transform to find solutions that diffused indefinitely in space.[17]
## Formal definition
The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), which is a unilateral transform defined by
(Eq.1)
where s is a complex frequency domain parameter
An alternate notation for the Laplace transform is instead of F.[3]
The meaning of the integral depends on types of functions of interest. A necessary condition for existence of the integral is that f must be locally integrable on [0, ∞). For locally integrable functions that decay at infinity or are of exponential type (), the integral can be understood to be a (proper) Lebesgue integral. However, for many applications it is necessary to regard it as a conditionally convergent improper integral at ∞. Still more generally, the integral can be understood in a weak sense, and this is dealt with below.
One can define the Laplace transform of a finite Borel measure μ by the Lebesgue integral[18]
An important special case is where μ is a probability measure, for example, the Dirac delta function. In operational calculus, the Laplace transform of a measure is often treated as though the measure came from a probability density function f. In that case, to avoid potential confusion, one often writes
This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace transform. Although with the Lebesgue integral, it is not necessary to take such a limit, it does appear more naturally in connection with the Laplace–Stieltjes transform.
### Bilateral Laplace transform
When one says “the Laplace transform” without qualification, the unilateral or one-sided transform is usually intended. The Laplace transform can be alternatively defined as the bilateral Laplace transform, or two-sided Laplace transform, by extending the limits of integration to be the entire real axis. If that is done, the common unilateral transform simply becomes a special case of the bilateral transform, where the definition of the function being transformed is multiplied by the Heaviside step function.
The bilateral Laplace transform F(s) is defined as follows:
(Eq.2)
An alternate notation for the bilateral Laplace transform is , instead of F.
### Inverse Laplace transform
Two integrable functions have the same Laplace transform only if they differ on a set of Lebesgue measure zero. This means that, on the range of the transform, there is an inverse transform. In fact, besides integrable functions, the Laplace transform is a one-to-one mapping from one function space into another in many other function spaces as well, although there is usually no easy characterization of the range.
Typical function spaces in which this is true include the spaces of bounded continuous functions, the space L∞(0, ∞), or more generally tempered distributions on (0, ∞). The Laplace transform is also defined and injective for suitable spaces of tempered distributions.
In these cases, the image of the Laplace transform lives in a space of analytic functions in the region of convergence. The inverse Laplace transform is given by the following complex integral, which is known by various names (the Bromwich integral, the Fourier–Mellin integral, and Mellin’s inverse formula):
(Eq.3)
where γ is a real number so that the contour path of integration is in the region of convergence of F(s). In most applications, the contour can be closed, allowing the use of the residue theorem. An alternative formula for the inverse Laplace transform is given by Post’s inversion formula. The limit here is interpreted in the weak-* topology.
In practice, it is typically more convenient to decompose a Laplace transform into known transforms of functions obtained from a table, and construct the inverse by inspection.
### Probability theory
In pure and applied probability, the Laplace transform is defined as an expected value. If X is a random variable with probability density function f, then the Laplace transform of f is given by the expectation
By convention, this is referred to as the Laplace transform of the random variable X itself. Here, replacing s by −t gives the moment generating function of X. The Laplace transform has applications throughout probability theory, including first passage times of stochastic processes such as Markov chains, and renewal theory.
Of particular use is the ability to recover the cumulative distribution function of a continuous random variable X, by means of the Laplace transform as follows:[19]
### Algebraic construction
The Laplace transform can be alternatively defined in a purely algebraic manner by applying a field of fractions construction to the convolution ring of functions on the positive half-line. The resulting space of abstract operators is exactly equivalent to Laplace space, but in this construction the forward and reverse transforms never need to be explicitly defined (avoiding the related difficulties with proving convergence).[20]
## Region of convergence
If f is a locally integrable function (or more generally a Borel measure locally of bounded variation), then the Laplace transform F(s) of f converges provided that the limit
The Laplace transform converges absolutely if the integral
The set of values for which F(s) converges absolutely is either of the form Re(s) > a or Re(s) ≥ a, where a is an extended real constant with −∞ ≤ a ≤ ∞ (a consequence of the dominated convergence theorem). The constant a is known as the abscissa of absolute convergence, and depends on the growth behavior of f(t).[21] Analogously, the two-sided transform converges absolutely in a strip of the form a < Re(s) < b, and possibly including the lines Re(s) = a or Re(s) = b.[22] The subset of values of s for which the Laplace transform converges absolutely is called the region of absolute convergence, or the domain of absolute convergence. In the two-sided case, it is sometimes called the strip of absolute convergence. The Laplace transform is analytic in the region of absolute convergence: this is a consequence of Fubini’s theorem and Morera’s theorem.
Similarly, the set of values for which F(s) converges (conditionally or absolutely) is known as the region of conditional convergence, or simply the region of convergence (ROC). If the Laplace transform converges (conditionally) at s = s0, then it automatically converges for all s with Re(s) > Re(s0). Therefore, the region of convergence is a half-plane of the form Re(s) > a, possibly including some points of the boundary line Re(s) = a.
In the region of convergence Re(s) > Re(s0), the Laplace transform of f can be expressed by integrating by parts as the integral
That is, F(s) can effectively be expressed, in the region of convergence, as the absolutely convergent Laplace transform of some other function. In particular, it is analytic.
There are several Paley–Wiener theorems concerning the relationship between the decay properties of f, and the properties of the Laplace transform within the region of convergence.
In engineering applications, a function corresponding to a linear time-invariant (LTI) system is stable if every bounded input produces a bounded output. This is equivalent to the absolute convergence of the Laplace transform of the impulse response function in the region Re(s) ≥ 0. As a result, LTI systems are stable, provided that the poles of the Laplace transform of the impulse response function have negative real part.
This ROC is used in knowing about the causality and stability of a system.
## Properties and theorems
The Laplace transform has a number of properties that make it useful for analyzing linear dynamical systems. The most significant advantage is that differentiation becomes multiplication, and integration becomes division, by s (reminiscent of the way logarithms change multiplication to addition of logarithms).
Because of this property, the Laplace variable s is also known as operator variable in the L domain: either derivative operator or (for s−1) integration operator. The transform turns integral equations and differential equations to polynomial equations, which are much easier to solve. Once solved, use of the inverse Laplace transform reverts to the original domain.
Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s),
the following table is a list of properties of unilateral Laplace transform:[23]
Property Time domain s domain Comment Linearity Can be proved using basic rules of integration. Frequency-domain derivative F′ is the first derivative of F with respect to s. Frequency-domain general derivative More general form, nth derivative of F(s). Derivative f is assumed to be a differentiable function, and its derivative is assumed to be of exponential type. This can then be obtained by integration by parts Second derivative f is assumed twice differentiable and the second derivative to be of exponential type. Follows by applying the Differentiation property to f′(t). General derivative f is assumed to be n-times differentiable, with nth derivative of exponential type. Follows by mathematical induction. Frequency-domain integration This is deduced using the nature of frequency differentiation and conditional convergence. Time-domain integration u(t) is the Heaviside step function and (u ∗ f)(t) is the convolution of u(t) and f(t). Frequency shifting Time shifting a > 0, u(t) is the Heaviside step function Time scaling a > 0 Multiplication The integration is done along the vertical line Re(σ) = c that lies entirely within the region of convergence of F.[24] Convolution Circular convolution For periodic functions with period T. Complex conjugation Cross-correlation Periodic function f(t) is a periodic function of period T so that f(t) = f(t + T), for all t ≥ 0. This is the result of the time shifting property and the geometric series. Periodic summation
Initial value theorem
Final value theorem
, if all poles of are in the left half-plane.
The final value theorem is useful because it gives the long-term behaviour without having to perform partial fraction decompositions (or other difficult algebra). If F(s) has a pole in the right-hand plane or poles on the imaginary axis (e.g., if or ), then the behaviour of this formula is undefined.
### Relation to power series
The Laplace transform can be viewed as a continuous analogue of a power series.[25] If a(n) is a discrete function of a positive integer n, then the power series associated to a(n) is the series
Changing the base of the power from x to e gives
For this to converge for, say, all bounded functions f, it is necessary to require that ln x < 0. Making the substitution −s = ln x gives just the Laplace transform:
In other words, the Laplace transform is a continuous analog of a power series, in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by e−s.
### Relation to moments
The quantities
are the moments of the function f. If the first n moments of f converge absolutely, then by repeated differentiation under the integral,
### Computation of the Laplace transform of a function’s derivative
It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function’s derivative. This can be derived from the basic expression for a Laplace transform as follows:
The general result
### Evaluating integrals over the positive real axis
A useful property of the Laplace transform is the following:
By plugging in the left-hand side turns into:
This method can be used to compute integrals that would otherwise be difficult to compute using elementary methods of real calculus. For example,
## Relationship to other transforms
### Laplace–Stieltjes transform
The (unilateral) Laplace–Stieltjes transform of a function g : ℝ → ℝ is defined by the Lebesgue–Stieltjes integral
The function g is assumed to be of bounded variation. If g is the antiderivative of f:
then the Laplace–Stieltjes transform of g and the Laplace transform of f coincide. In general, the Laplace–Stieltjes transform is the Laplace transform of the Stieltjes measure associated to g. So in practice, the only distinction between the two transforms is that the Laplace transform is thought of as operating on the density function of the measure, whereas the Laplace–Stieltjes transform is thought of as operating on its cumulative distribution function.[26]
### Fourier transform
The Fourier transform is a special case (under certain conditions) of the bilateral Laplace transform. While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable. The Laplace transform is usually restricted to transformation of functions of t with t ≥ 0. A consequence of this restriction is that the Laplace transform of a function is a holomorphic function of the variable s. Unlike the Fourier transform, the Laplace transform of a distribution is generally a well-behaved function. Techniques of complex variables can also be used to directly study Laplace transforms. As a holomorphic function, the Laplace transform has a power series representation. This power series expresses a function as a linear superposition of moments of the function. This perspective has applications in probability theory.
The Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument s = iω or s = 2πiξ[27] when the condition explained below is fulfilled,
This convention of the Fourier transform ( in Fourier transform § Other conventions) requires a factor of 1/2π on the inverse Fourier transform. This relationship between the Laplace and Fourier transforms is often used to determine the frequency spectrum of a signal or dynamical system.
The above relation is valid as stated if and only if the region of convergence (ROC) of F(s) contains the imaginary axis, σ = 0.
For example, the function f(t) = cos(ω0t) has a Laplace transform F(s) = s/(s2 + ω02) whose ROC is Re(s) > 0. As s = iω0 is a pole of F(s), substituting s = iω in F(s) does not yield the Fourier transform of f(t)u(t), which is proportional to the Dirac delta function δ(ω − ω0).
However, a relation of the form
### Mellin transform
The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables.
If in the Mellin transform
### Z-transform
The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of
Let
The Laplace transform of the sampled signal xq(t) is
This is the precise definition of the unilateral Z-transform of the discrete function x[n]
Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal,
The similarity between the Z- and Laplace transforms is expanded upon in the theory of time scale calculus.
### Borel transform
The integral form of the Borel transform
### Fundamental relationships
Since an ordinary Laplace transform can be written as a special case of a two-sided transform, and since the two-sided transform can be written as the sum of two one-sided transforms, the theory of the Laplace-, Fourier-, Mellin-, and Z-transforms are at bottom the same subject. However, a different point of view and different characteristic problems are associated with each of these four major integral transforms.
## Table of selected Laplace transforms
The following table provides Laplace transforms for many common functions of a single variable.[28][29] For definitions and explanations, see the Explanatory Notes at the end of the table.
Because the Laplace transform is a linear operator,
• The Laplace transform of a sum is the sum of Laplace transforms of each term.
• The Laplace transform of a multiple of a function is that multiple times the Laplace transformation of that function.
Using this linearity, and various trigonometric, hyperbolic, and complex number (etc.) properties and/or identities, some Laplace transforms can be obtained from others more quickly than by using the definition directly.
The unilateral Laplace transform takes as input a function whose time domain is the non-negative reals, which is why all of the time domain functions in the table below are multiples of the Heaviside step function, u(t).
The entries of the table that involve a time delay τ are required to be causal (meaning that τ > 0). A causal system is a system where the impulse response h(t) is zero for all time t prior to t = 0. In general, the region of convergence for causal systems is not the same as that of anticausal systems.
Function Time domain Laplace s-domain Region of convergence Reference unit impulse all s inspection delayed impulse time shift of unit impulse unit step integrate unit impulse delayed unit step time shift of unit step rectangular impulse ramp integrate unit impulse twice nth power (for integer n) (n > −1) integrate unit step n times qth power (for complex q) [30][31] nth root Set q = 1/n above. nth power with frequency shift Integrate unit step, apply frequency shift delayed nth power with frequency shift integrate unit step, apply frequency shift, apply time shift exponential decay Frequency shift of unit step two-sided exponential decay (only for bilateral transform) Frequency shift of unit step exponential approach unit step minus exponential decay sine [32] cosine [32] hyperbolic sine [33] hyperbolic cosine [33] exponentially decaying sine wave [32] exponentially decaying cosine wave [32] natural logarithm [33] Bessel function of the first kind, of order n (n > −1) [34] Error function [34] Explanatory notes:
## s-domain equivalent circuits and impedances
The Laplace transform is often used in circuit analysis, and simple conversions to the s-domain of circuit elements can be made. Circuit elements can be transformed into impedances, very similar to phasor impedances.
Here is a summary of equivalents:
Note that the resistor is exactly the same in the time domain and the s-domain. The sources are put in if there are initial conditions on the circuit elements. For example, if a capacitor has an initial voltage across it, or if the inductor has an initial current through it, the sources inserted in the s-domain account for that.
The equivalents for current and voltage sources are simply derived from the transformations in the table above.
## Examples and applications
The Laplace transform is used frequently in engineering and physics; the output of a linear time-invariant system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter being easier to solve because of its algebraic form. For more information, see control theory. The Laplace transform is invertible on a large class of functions. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.[35]
The Laplace transform can also be used to solve differential equations and is used extensively in mechanical engineering and electrical engineering. The Laplace transform reduces a linear differential equation to an algebraic equation, which can then be solved by the formal rules of algebra. The original differential equation can then be solved by applying the inverse Laplace transform. English electrical engineer Oliver Heaviside first proposed a similar scheme, although without using the Laplace transform; and the resulting operational calculus is credited as the Heaviside calculus.
### Evaluating improper integrals
Let . Then (see the table above)
In the limit , one gets
The validity of this identity can be proved by other means. It is an example of a Frullani integral.
Another example is Dirichlet integral.
### Complex impedance of a capacitor
In the theory of electrical circuits, the current flow in a capacitor is proportional to the capacitance and rate of change in the electrical potential (with equations as for the SI unit system). Symbolically, this is expressed by the differential equation
Taking the Laplace transform of this equation, we obtain
Solving for V(s) we have
The definition of the complex impedance Z (in ohms) is the ratio of the complex voltage V divided by the complex current I while holding the initial state V0 at zero:
Using this definition and the previous equation, we find:
### Impulse response
Consider a linear time-invariant system with transfer function
The impulse response is simply the inverse Laplace transform of this transfer function:
Partial fraction expansion
To evaluate this inverse transform, we begin by expanding H(s) using the method of partial fraction expansion,
The unknown constants P and R are the residues located at the corresponding poles of the transfer function. Each residue represents the relative contribution of that singularity to the transfer function’s overall shape.
By the residue theorem, the inverse Laplace transform depends only upon the poles and their residues. To find the residue P, we multiply both sides of the equation by s + α to get
Then by letting s = −α, the contribution from R vanishes and all that is left is
Similarly, the residue R is given by
Note that
Finally, using the linearity property and the known transform for exponential decay (see Item #3 in the Table of Laplace Transforms, above), we can take the inverse Laplace transform of H(s) to obtain
Convolution
The same result can be achieved using the convolution property as if the system is a series of filters with transfer functions 1/(s + α) and 1/(s + β). That is, the inverse of
### Phase delay
Time function Laplace transform
Starting with the Laplace transform,
We are now able to take the inverse Laplace transform of our terms:
This is just the sine of the sum of the arguments, yielding:
We can apply similar logic to find that
### Statistical mechanics
In statistical mechanics, the Laplace transform of the density of states defines the partition function.[36] That is, the canonical partition function is given by
### Spatial (not time) structure from astronomical spectrum
The wide and general applicability of the Laplace transform and its inverse is illustrated by an application in astronomy which provides some information on the spatial distribution of matter of an astronomical source of radiofrequency thermal radiation too distant to resolve as more than a point, given its flux density spectrum, rather than relating the time domain with the spectrum (frequency domain).
Assuming certain properties of the object, e.g. spherical shape and constant temperature, calculations based on carrying out an inverse Laplace transformation on the spectrum of the object can produce the only possible model of the distribution of matter in it (density as a function of distance from the center) consistent with the spectrum.[37] When independent information on the structure of an object is available, the inverse Laplace transform method has been found to be in good agreement.
## Notes
1. ^ Lynn, Paul A. (1986). “The Laplace Transform and the z-transform”. Electronic Signals and Systems. London: Macmillan Education UK. pp. 225–272. doi:10.1007/978-1-349-18461-3_6. ISBN 978-0-333-39164-8.
Laplace Transform and the z-transform are closely related to the Fourier Transform. Laplace Transform is somewhat more general in scope than the Fourier Transform, and is widely used by engineers for describing continuous circuits and systems, including automatic control systems.
2. ^ “Differential Equations – Laplace Transforms”. tutorial.math.lamar.edu. Retrieved 2020-08-08.
3. ^ a b Weisstein, Eric W. “Laplace Transform”. mathworld.wolfram.com. Retrieved 2020-08-08.
4. ^ “Des Fonctions génératrices” [On generating functions], Théorie analytique des Probabilités [Analytical Probability Theory] (in French) (2nd ed.), Paris, 1814, chap.I sect.2-20
5. ^ Jaynes, E. T. (Edwin T.) (2003). Probability theory : the logic of science. Bretthorst, G. Larry. Cambridge, UK: Cambridge University Press. ISBN 0511065892. OCLC 57254076.
6. ^ Abel, Niels H. (1820), “Sur les fonctions génératrices et leurs déterminantes”, Œuvres Complètes (in French), vol. II (published 1839), pp. 77–88 1881 edition
7. ^ Lerch, Mathias (1903), “Sur un point de la théorie des fonctions génératrices d’Abel” [Proof of the inversion formula], Acta Mathematica (in French), 27: 339–351, doi:10.1007/BF02421315, hdl:10338.dmlcz/501554
8. ^ Heaviside, Oliver (January 2008), “The solution of definite integrals by differential transformation”, Electromagnetic Theory, vol. III, London, section 526, ISBN 9781605206189`{{citation}}`: CS1 maint: location missing publisher (link)
9. ^ Bromwich, Thomas J. (1916), “Normal coordinates in dynamical systems”, Proceedings of the London Mathematical Society, 15: 401–448, doi:10.1112/plms/s2-15.1.401
10. ^ An influential book was: Gardner, Murray F.; Barnes, John L. (1942), Transients in Linear Systems studied by the Laplace Transform, New York: Wiley
11. ^ Doetsch, Gustav (1937), Theorie und Anwendung der Laplacesche Transformation [Theory and Application of the Laplace Transform] (in German), Berlin: Springer translation 1943
12. ^ Euler 1744, Euler 1753, Euler 1769
13. ^ Lagrange 1773
14. ^ Grattan-Guinness 1997, p. 260
15. ^ Grattan-Guinness 1997, p. 261
16. ^ Grattan-Guinness 1997, pp. 261–262
17. ^ Grattan-Guinness 1997, pp. 262–266
18. ^ Feller 1971, §XIII.1
19. ^ The cumulative distribution function is the integral of the probability density function.
20. ^ Mikusiński, Jan. Operational Calculus.
21. ^ Widder 1941, Chapter II, §1
22. ^ Widder 1941, Chapter VI, §2
23. ^ Korn & Korn 1967, pp. 226–227
24. ^ Bracewell 2000, Table 14.1, p. 385
25. ^ Archived at Ghostarchive and the Wayback Machine: Mattuck, Arthur. “Where the Laplace Transform comes from”. YouTube.
26. ^ Feller 1971, p. 432
27. ^ Takacs 1953, p. 93
28. ^ Riley, K. F.; Hobson, M. P.; Bence, S. J. (2010), Mathematical methods for physics and engineering (3rd ed.), Cambridge University Press, p. 455, ISBN 978-0-521-86153-3
29. ^ Distefano, J. J.; Stubberud, A. R.; Williams, I. J. (1995), Feedback systems and control, Schaum’s outlines (2nd ed.), McGraw-Hill, p. 78, ISBN 978-0-07-017052-0
30. ^ Lipschutz, S.; Spiegel, M. R.; Liu, J. (2009). Mathematical Handbook of Formulas and Tables. Schaum’s Outline Series (3rd ed.). McGraw-Hill. p. 183. ISBN 978-0-07-154855-7. – provides the case for real q.
31. ^ http://mathworld.wolfram.com/LaplaceTransform.html – Wolfram Mathword provides case for complex q
32. ^ a b c d Bracewell 1978, p. 227.
33. ^ a b c Williams 1973, p. 88.
34. ^ a b Williams 1973, p. 89.
35. ^ Korn & Korn 1967, §8.1
36. ^ RK Pathria; Paul Beal (1996). Statistical mechanics (2nd ed.). Butterworth-Heinemann. p. 56. ISBN 9780750624695.
37. ^ Salem, M.; Seaton, M. J. (1974), “I. Continuum spectra and brightness contours”, Monthly Notices of the Royal Astronomical Society, 167: 493–510, Bibcode:1974MNRAS.167..493S, doi:10.1093/mnras/167.3.493, and
Salem, M. (1974), “II. Three-dimensional models”, Monthly Notices of the Royal Astronomical Society, 167: 511–516, Bibcode:1974MNRAS.167..511S, doi:10.1093/mnras/167.3.511
## References
### Modern
• Bracewell, Ronald N. (1978), The Fourier Transform and its Applications (2nd ed.), McGraw-Hill Kogakusha, ISBN 978-0-07-007013-4
• Bracewell, R. N. (2000), The Fourier Transform and Its Applications (3rd ed.), Boston: McGraw-Hill, ISBN 978-0-07-116043-8
• Feller, William (1971), An introduction to probability theory and its applications. Vol. II., Second edition, New York: John Wiley & Sons, MR 0270403
• Korn, G. A.; Korn, T. M. (1967), Mathematical Handbook for Scientists and Engineers (2nd ed.), McGraw-Hill Companies, ISBN 978-0-07-035370-1
• Widder, David Vernon (1941), The Laplace Transform, Princeton Mathematical Series, v. 6, Princeton University Press, MR 0005923
• Williams, J. (1973), Laplace Transforms, Problem Solvers, George Allen & Unwin, ISBN 978-0-04-512021-5
• Takacs, J. (1953), “Fourier amplitudok meghatarozasa operatorszamitassal”, Magyar Hiradastechnika (in Hungarian), IV (7–8): 93–96
### Historical
• Euler, L. (1744), “De constructione aequationum” [The Construction of Equations], Opera Omnia, 1st series (in Latin), 22: 150–161
• Euler, L. (1753), “Methodus aequationes differentiales” [A Method for Solving Differential Equations], Opera Omnia, 1st series (in Latin), 22: 181–213
• Euler, L. (1992) [1769], “Institutiones calculi integralis, Volume 2” [Institutions of Integral Calculus], Opera Omnia, 1st series (in Latin), Basel: Birkhäuser, 12, ISBN 978-3764314743, Chapters 3–5
• Euler, Leonhard (1769), Institutiones calculi integralis [Institutions of Integral Calculus] (in Latin), vol. II, Paris: Petropoli, ch. 3–5, pp. 57–153
• Grattan-Guinness, I (1997), “Laplace’s integral solutions to partial differential equations”, in Gillispie, C. C. (ed.), Pierre Simon Laplace 1749–1827: A Life in Exact Science, Princeton: Princeton University Press, ISBN 978-0-691-01185-1
• Lagrange, J. L. (1773), Mémoire sur l’utilité de la méthode, Œuvres de Lagrange, vol. 2, pp. 171–234
• Arendt, Wolfgang; Batty, Charles J.K.; Hieber, Matthias; Neubrander, Frank (2002), Vector-Valued Laplace Transforms and Cauchy Problems, Birkhäuser Basel, ISBN 978-3-7643-6549-3.
• Davies, Brian (2002), Integral transforms and their applications (Third ed.), New York: Springer, ISBN 978-0-387-95314-4
• Deakin, M. A. B. (1981), “The development of the Laplace transform”, Archive for History of Exact Sciences, 25 (4): 343–390, doi:10.1007/BF01395660, S2CID 117913073
• Deakin, M. A. B. (1982), “The development of the Laplace transform”, Archive for History of Exact Sciences, 26 (4): 351–381, doi:10.1007/BF00418754, S2CID 123071842
• Doetsch, Gustav (1974), Introduction to the Theory and Application of the Laplace Transformation, Springer, ISBN 978-0-387-06407-9
• Mathews, Jon; Walker, Robert L. (1970), Mathematical methods of physics (2nd ed.), New York: W. A. Benjamin, ISBN 0-8053-7002-1
• Polyanin, A. D.; Manzhirov, A. V. (1998), Handbook of Integral Equations, Boca Raton: CRC Press, ISBN 978-0-8493-2876-3
• Schwartz, Laurent (1952), “Transformation de Laplace des distributions”, Comm. Sém. Math. Univ. Lund [Medd. Lunds Univ. Mat. Sem.] (in French), 1952: 196–206, MR 0052555
• Schwartz, Laurent (2008) [1966], Mathematics for the Physical Sciences, Dover Books on Mathematics, New York: Dover Publications, pp. 215–241, ISBN 978-0-486-46662-0 – See Chapter VI. The Laplace transform.
• Siebert, William McC. (1986), Circuits, Signals, and Systems, Cambridge, Massachusetts: MIT Press, ISBN 978-0-262-19229-3
• Widder, David Vernon (1945), “What is the Laplace transform?”, The American Mathematical Monthly, 52 (8): 419–425, doi:10.2307/2305640, ISSN 0002-9890, JSTOR 2305640, MR 0013447
• J.A.C.Weidman and Bengt Fornberg: “Fully numerical Laplace transform methods”, Numerical Algorithms, vol.92 (2023), pp.985-1006. https://doi.org/10.1007/s11075-022-01368-x .
• “Laplace transform”, Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Online Computation of the transform or inverse transform, wims.unice.fr
• Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.
• Weisstein, Eric W. “Laplace Transform”. MathWorld.
• Good explanations of the initial and final value theorems Archived 2009-01-08 at the Wayback Machine
• Laplace Transforms at MathPages
• Computational Knowledge Engine allows to easily calculate Laplace Transforms and its inverse Transform.
• Laplace Calculator to calculate Laplace Transforms online easily.
• Code to visualize Laplace Transforms and many example videos.
You are watching: Laplace transform. Info created by THVinhTuy selection and synthesis along with other related topics.
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# Ascii Codes
It is a very well-known fact that computers can manage internally only 0s (zeros) and 1s (ones). This is true, and by means of sequences of 0s and 1s the computer can express any numerical value as its binary translation, which is a very simple mathematical operation (as explained in the paper numerical bases).
Nevertheless, there is no such evident way to represent letters and other non-numeric characters with 0s and 1s. Therefore, in order to do that, computers use ASCII tables, which are tables or lists that contain all the letters in the roman alphabet plus some additional characters. In these tables each character is always represented by the same order number. For example, the ASCII code for the capital letter "A" is always represented by the order number 65, which is easily representable using 0s and 1s in binary: 65 expressed as a binary number is 1000001.
The standard ASCII table defines 128 character codes (from 0 to 127), of which, the first 32 are control codes (non-printable), and the remaining 96 character codes are representable characters:
*0123456789ABCDEF
0NULSOHSTXETXEOTENQACKBELBS TABLF VT FF CR SO SI
1DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US
2 !"#\$%&'()*+,-./
30123456789:;<=>?
4@ABCDEFGHIJKLMNO
5PQRSTUVWXYZ[\]^_
6`abcdefghijklmno
7pqrstuvwxyz{|}~
* This panel is organized to be easily read in hexadecimal: row numbers represent the first digit and the column numbers represent the second one. For example, the "A" character is located at the 4th row and the 1st column, for that it would be represented in hexadecimal as 0x41 (65).
Because most systems nowadays work with 8bit bytes, which can represent 256 different values, in addition to the 128 standard ASCII codes there are other 128 that are known as extended ASCII, which are platform- and locale-dependent. So there is more than one extended ASCII character set.
The two most used extended ASCII character sets are the one known as OEM, that comes from the default character set incorporated by default in the IBM-PC and the other is the ANSI extend ASCII which is used by most recent operating systems.
The first of them, the OEM character set, is the one used by the hardware of the immense majority of PC compatible machines, and was also used under the old DOS system. It includes some foreign signs, some marked characters and pieces to represent panels.
The ANSI character set is a standard that many systems incorporate, like Windows, some UNIX platforms and many standalone applications. It includes many more local symbols and marked letters so that it can be used with no need of being redefined in many more languages:
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# Can i get above a 98+ ATAR with these subject combinations? (1 Viewer)
#### v.tex
##### Member
Math Extension 1
Earth and Environmental Science
Physics
Economics
I am just worried that Earth will scale me down and badly affect my ATAR..
#### aqwerty13402
##### Well-Known Member
for a 98 i would say yes. Just check what the scaling for EES caps at (if it even does cap at anything).
#### v.tex
##### Member
for a 98 i would say yes. Just check what the scaling for EES caps at (if it even does cap at anything).
looking at earth scaling statistics it said the maximum ATAR with earth is 99.4- does this mean i can’t get any higher than this- i’m a bit confused
#### aqwerty13402
##### Well-Known Member
looking at earth scaling statistics it said the maximum ATAR with earth is 99.4- does this mean i can’t get any higher than this- i’m a bit confused
Because you are on 13 units, I think you can get higher. It means that for EES, the maximum atar contribution it can give you is 99.4. All your other subjects to my knowledge allow for 99.95, so theoretically, I think you could basically get any atar, except 99.95??? Idk the exact details, or if this makes sense. I'm def the wrong person to be explaining this.
@jimmysmith560 is the atar expert though. Hopefully he sees this and can confirm / tell u the right info.
#### jimmysmith560
##### Le Phénix Trilingue
Moderator
Essentially, an ATAR of 99.95 can be achieved with any subject combination (including Earth and Environmental Science), as long as your performance across your subjects is of a sufficiently high standard. This means that the answer to OP's question is definitely yes.
Looking at previous HSC cohorts, the last time a student achieved an ATAR of 99.95 with EES as part of their subject combination was in 2010. In more recent years, the highest ATAR achieved by a student who took EES was 99.90, which was in 2018. Of course, the highest ATAR achieved by a student who took a particular subject and was part of a previous HSC cohort has no direct effect on the highest ATAR that can be achieved with that same subject in future HSC cohorts. It merely serves as a general indication of the ATARs that were historically achieved with the subject.
This data is from UAC, which publishes yearly scaling reports for each HSC cohort. These reports contain information including the highest ATAR achieved by a student who took any subject, which UAC typically refers to as "maximum ATAR" but has labelled as "maximum ATAR including the course" in its 2023 scaling report for further clarity.
I hope this helps!
#### K2Trappy
##### Member
looking at earth scaling statistics it said the maximum ATAR with earth is 99.4- does this mean i can’t get any higher than this- i’m a bit confused
it just means the highest atar someone achieved in the previous HSC with EES was 99.4 its not a cap
#### v.tex
##### Member
it just means the highest atar someone achieved in the previous HSC with EES was 99.4 its not a cap
ahh
Math Extension 1
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# Cross Ratio
(redirected from Cross-ratio)
Also found in: Medical, Wikipedia.
## cross ratio
[′krȯs ‚rā·shō]
(mathematics)
For four collinear points, A, B, C, and D, the ratio (AB)(CD)/(AD)(CB), or one of the ratios obtained from this quantity by a permutation of A, B, C, and D.
## Cross Ratio
(also anharmonic ratio). The cross ratio of four collinear points M1M2, M3, M4 (Figure 1) is a number denoted by the symbol (M1M2M3M4 and equal to
In this case, the ratio M1M3/M3M2 is considered to be positive if the direction of the segments M1M3 and M3M2 is the same and negative if the directions are different. The cross ratio depends on the order of the numbering of the points, which may differ from the sequential order of the points on the straight line. In addition to the cross ratio of four points, there is also a cross ratio of four straight lines m1, m2, m3, m4, passing through the point O. This ratio is designated by the symbol (m1m2m3m4) and is equal to
where the angle (mimj) beween the straight lines mi and mj is considered with a sign.
Figure 1
If the points M1, M2, M3, M4 lie on the straight lines m1, m2, m3, m4 (Figure 1), then
(M1M2M3m4) = (m1m2m3m3m4)
Hence, if the points M1, M2, M3, M4 and M1’, M2’, M3’, M4’ are the result of the intersection of four straight lines m1, m2, m3, m4 (Figure 1), then (M1M2M3M4’) = (M1M2M3M4). If, however, the straight lines m1, m2, m3, m4 and m1’, m2’, m3’, m4’ project one set of four points M1, M2, Ms, M4 (Figure 2), then (m1m2m3m4’) = (m1m2m3m4).
Figure 2
The cross ratio also remains unchanged by any projective transformation, that is, it is an invariant of such transformations, and cross ratios therefore play an important role in projective geometry. Sets of four points or lines for which the cross ratio is equal to 1 are of particular importance. Such sets are called harmonic.
E. G. POZNIAK
References in periodicals archive ?
Lightweight 6-speed manual transmission with cross-ratios matching engine characteristics provide a quick and exhilarating gear shift feel.
Known for efficiency and effectiveness of execution and cross-ratios, Cantor Fitzgerald Equity Capital Markets offers clients trading services in exchange-traded and over-the-counter cash equities, equity and index options, convertible and structured products, and risk arbitrage, program trading, and stock lending services.
Site: Follow: Share:
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# Chris Fox's Engineering Section
## Equilibrium
A body is in equilibrium if the resultant of all the forces - applied forces and reactions - acting on it is 0N, and if the sum of the moments exerted about any point is 0Nm.
These conditions can be used to determine the unknown values of forces acting on a body. To do this, one can use a free body diagram. Draw an outline of the body, separated from other bodies, then add all the applied forces and reactions. A force is denoted by a straight arrow and a moment by a curved arrow, accompanied in each case by either the magnitude or a symbol denoting a variable.
Having drawn the free body diagram, one can write equilibrium equations. For the purpose of taking moments, the mass of a body can be considered to be concentrated at a point, termed the centre of gravity. At the centre of gravity, the sum of the moments exerted by the parts of the body is zero. For a body of uniform cross-section and density, this point lies halfway along its length.
For a more complex body, ∑(mg(x-¯x))=0, where m is the mass of each part of the body, x is the distance (measured along one particular axis) to that part's centre of gravity from a datum point and ¯x is the distance from the same datum point to the whole body's centre of gravity. Multiplying out this equation gives ∑(mgx-mg¯x)=0. Because additions and subtractions can be carried out in any order, this is the same as ∑(mgx)-∑(mg¯x)=0. Factorising gives g∑(mx)=g¯x∑m. Finally, dividing both sides by ∑m and cancelling gives ¯x=∑(mx)/∑m.
In two-dimensional problems, there are four categories of equilibrium. In the first, forces are collinear. There will be one equilibrium equation, of the form ∑Fx=0, which can only be solved if there is only one unknown quantity.
In the second category, forces are concurrent. There will be two equations, of the form ∑Fx=0 and ∑Fy=0. These can be solved if there are up to two unknowns.
If the forces are parallel, there will be two equations, of the form ∑Fx=0 and ∑Mo=0. These too can be solved if there are up to two unknowns.
For a general system of forces and moments, there will be three equations, of the form ∑Fx=0, ∑Fy=0 and ∑Mo=0. These can be solved if there are up to three unknowns.
For a two-dimensional problem, one can write a maximum of three independent equilibrium equations. Three is therefore the maximum number of unknown quantities that can be determined using one free body diagram.
In each of the above cases, the force equations can be replaced with moment equations, provided that in any problem each moment equation is based on a different pivot. Where there is more than one unknown, it is an advantage if the pivot used for each moment equation lies on the line of action of one of the unknowns; that quantity is then removed from the equation.
One common case of equilibrium is a two force system in which a component is supported by one having pin joints at its ends; the forces in the pin joints must be equal in magnitude, opposite in direction and collinear. Another case is a three force system in which the lines of action of the three forces act through a single point.
A free body diagram can be used to determine internal forces in components and systems; these in turn can be used to calculate stresses and displacements. To find the internal forces acting at a section (as in cross-section), start by splitting the component in two at that section. Then, treating each part as a separate body, add at the break the forces and moments needed to keep each part in equilibrium. The internal forces will be tension, which acts along the length of a component, and shear, which acts perpendicular to the length.
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http://cybermetric.blogspot.com/2011/11/pirates-were-lucky-to-win-world-series.html
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## Monday, November 7, 2011
### The Pirates were lucky to win the 1971 World Series, but how lucky?
This came up on one of the SABR bulletin boards.
Their OPS differential for the whole season was .073. That translates into a winning pct of .592 using my equation Pct = .5 + 1.26*OPSDIFF. The Orioles had a differential of .096, good for a pct of .621. That means that the Orioles would have 53% chance of winning any given game using Bill James' Log 5 method. I came up with the Orioles having about a 56.5% chance of winning the series, taking into account all the different ways they could win a series of a given length. They also had home field advantage, which should have increased things about 2% (2% more than 55.7% so about 58%)
7.9% of the time it is an Orioles sweep
14.8% of the time the O's win in 5
17.4% of the time they win in 6
16.4% of the time they win in 7
So the Pirates had a 42% chance of winning
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• 8 Lessons
## 1A – 150 Essential Exam Practice Questions
Secondary 1 Express 150 Essential Practice Questions is the first of a two-book series specially written for Secondary 1 students to prepare for their various continual assessments and semestral assessments in Mathematics. The materials in this book follow closely to the latest Mathematics syllabus implemented by the Ministry of Education, containing the first 7 chapters of the Secondary 1 curriculum. These materials are relevant for the students in both the ‘O’ Level and Integrated Programme (IP) tracks and serve to provide a concise yet complete essential collection of practice questions that a student needs to fully comprehend each given topic bounded by the syllabus.
• 8 Lessons
• 6 Lessons
## 1EM Unit 1 – Primes, HCF and LCM
Learning Objectives:
• Understand the definition of prime numbers and composite numbers.
• Determine whether a whole number is prime or composite.
• Express a composite number as the product of its prime factors, i.e., perform the prime factorisation.
• Solve problems involving perfect square, perfect cube, square root, or cube root using prime factorisation.
• Find the highest common factor (HCF) and the lowest common multiple (LCM) of two or more whole numbers.
• Solve problems involving the real-world applications of prime numbers, HCF and LCM.
• 4 Lessons
• 6 Lessons
• 5 Lessons
• 6 Lessons
• 5 Lessons
• 5 Lessons
• 5 Lessons
## 1EM Unit 2 – Real Numbers
Learning Objectives:
• Recognise negative numbers and expand cognition of numbers to real numbers.
• Represent real numbers on a number line and arrange numbers in given order.
• Understand different categorisations of real numbers.
• Perform arithmetic operations on real numbers, use a calculator when necessary.
• Solve problems involving operations of real numbers under real-world contexts.
• 3 Lessons
## 1EM Unit 3 – Approximation and Estimation
Learning Objectives:
• Round off numbers to a given digit.
• Understand the rules of significant figures and round off numbers to a given significant figure.
• Estimate calculation outcomes by rounding off numbers properly and solve problems involving estimation under real-world contexts.
• 4 Lessons
## 1EM Unit 4 – Basic Algebra and Algebraic Manipulation
Learning Objectives:
• Represent numbers with letters as variables and formulate algebraic expressions.
• Evaluate algebraic expressions by substituting values into variables.
• Manipulate and simplify algebraic expressions.
• Expand and factorise algebraic expressions.
• Solve mathematical problems involving algebraic expressions, as well as problems under real-world contexts.
• 4 Lessons
## 1EM Unit 5 – Linear Equations and Simple Inequalities
Learning Objectives:
• Understand the concept of equations.
• Solve linear equations in one variable, including fractional equations that can be reduced to linear equations.
• Formulate linear equations in one variable based on given conditions to solve problems.
• 6 Lessons
## 1EM Unit 6 – Linear Graphs and Its applications
Learning Objectives:
• State the coordinates of a given point in a plane.
• Plot a point in a Cartesian Plane given the coordinates.
• Find unknown values given the equation of a linear function.
• Draw the graph of a linear function.
• Find the gradient and y-intercept of a straight line.
• Solve problems using linear functions and graphs under real-world context.
• 4 Lessons
## 1EM Unit 7 – Number Patterns
Learning Objectives:
• Recognise and describe the basic pattern of a number sequence.
• Determine the next few terms of a given number sequence.
• Find an expression of the general term of a given number sequence.
• Solve problems involving number sequences under real-world context.
• 5 Lessons
• 3 Lessons
• 0 Lessons
• 0 Lessons
• 0 Lessons
• 1 Lesson
• 1 Lesson
• 4 Lessons
• 3 Lessons
• 5 Lessons
• 4 Lessons
• 3 Lessons
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• 8 Lessons
• 4 Lessons
• 2 Lessons
• 3 Lessons
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• 8 Lessons
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• 6 Lessons
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# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## 1 Square Feet in Square Acre in Telangana
In this article, you will learn how to convert 1 Square Feet in Square Acre in Telangana. Square Feet is one of the most commonly used measurement units used in almost every state of India. People in Indian states refer to Square Feet for measurement of land. Here in this page you will learn how to calculate 1 Square Feet in Square Acre unit.
## How to Convert from 1 Square Feet to Square Acre?
You can easily convert Square Feet to Square Acre or the reverse with a simple method. You can multiply the figure in Square Feet by 0.000023 to determine the Square Acre value. This is all you need to do for undertaking the conversion procedure of Square Feet to Square Acre in Telangana.
## Relationship between Square Feet and Square Acre
It is not difficult to work out the actual relationship between Square Feet to Square Acre in most cases. You should know the proper calculations existing between Square Feet to Square Acre before you venture to conversion procedures. 1 Square Feet to Square Acre calculations will be 0.000023 Square Acre.
Once you know these calculations, it will not be hard for you to calculate Square Feet in an Square Acre. One Square Feet can be worked out to various other units as well.
## Formula for Converting Square Feet to Square Acre
The formula to convert Square Feet to Square Acre is the following-
1 Square Feet = 0.000023 Square Acre
Conversely, you can also use this formula-
Square Feet = Square Acre * 0.000023
This is all you need to know for undertaking the conversion procedure on your part. You can also use online calculators or conversion tools for this purpose.
### Other Plot Areas in Square Acre in Telangana
• 1 Square Feet in Square Acre in Telangana = 0.000023
• 2 Square Feet in Square Acre in Telangana = 0.000046
• 3 Square Feet in Square Acre in Telangana = 0.000069
• 4 Square Feet in Square Acre in Telangana = 0.000092
• 5 Square Feet in Square Acre in Telangana = 0.000115
• 6 Square Feet in Square Acre in Telangana = 0.000138
• 7 Square Feet in Square Acre in Telangana = 0.000161
• 8 Square Feet in Square Acre in Telangana = 0.000184
• 9 Square Feet in Square Acre in Telangana = 0.000207
• 10 Square Feet in Square Acre in Telangana = 0.00023
• 15 Square Feet in Square Acre in Telangana = 0.000345
• 20 Square Feet in Square Acre in Telangana = 0.00046
• 25 Square Feet in Square Acre in Telangana = 0.000575
• 30 Square Feet in Square Acre in Telangana = 0.00069
• 35 Square Feet in Square Acre in Telangana = 0.000805
• 40 Square Feet in Square Acre in Telangana = 0.00092
• 45 Square Feet in Square Acre in Telangana = 0.001035
• 50 Square Feet in Square Acre in Telangana = 0.00115
• 55 Square Feet in Square Acre in Telangana = 0.001265
• 60 Square Feet in Square Acre in Telangana = 0.00138
• 65 Square Feet in Square Acre in Telangana = 0.001495
• 70 Square Feet in Square Acre in Telangana = 0.00161
• 75 Square Feet in Square Acre in Telangana = 0.001725
• 80 Square Feet in Square Acre in Telangana = 0.00184
• 85 Square Feet in Square Acre in Telangana = 0.001955
• 90 Square Feet in Square Acre in Telangana = 0.00207
• 95 Square Feet in Square Acre in Telangana = 0.002185
• 100 Square Feet in Square Acre in Telangana = 0.0023
## FAQs About 1 Square Feet in Square Acre in Telangana
#### Q: Ek Square Feet main Kitne Square Acre in Telangana Hote hain?
Ek Square Feet main 0.000023 Square Acre in Telangana hote hain.
#### 1 Square Feet main kitna Square Acre in Telangana Hota Hain?
1 Square Feet main 0.000023 Square Acre in Telangana hote hain
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# Bank Foundation 2023
• #### Quantitative Aptitude Quiz For Bank Foundation 2023 – 31st March
Q1. Ratio of speed of a bus to a car is 6 : 7. They start from the same point and move towards the same direction. After four hours distance between them is 28 km. Find the time in which...
Published On March 31st, 2023
• #### Reasoning Ability Quiz For Bank Foundation 2023 – 31st March
Directions (1-5): In these questions, relationship between different elements is show in the statements. The statements are followed by conclusions. Study the conclusions based on the given statements and select the appropriate answer: (a) If only conclusion I follows. (b)...
Published On March 31st, 2023
• #### Quantitative Aptitude Quiz For Bank Foundation 2023 – 30th March
Directions (1-15): What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value) Q1. 21.11% of 1299.89 + 5×? = 52.12% of 4399.98 (a) 415 (b) 408...
Last updated on March 31st, 2023 12:06 pm
• #### Reasoning Ability Quiz For Bank Foundation 2023 – 30th March
Directions (1-5): In the question below three statements are given followed by two conclusions. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions...
Published On March 30th, 2023
• #### English Language Quiz For Bank Foundation 2023-30th March
Directions (1-10): Two columns are given in each question and each column has three parts of a sentence. Choose the most suitable pair, which makes a grammatically correct and contextually coherent statement. Q1. COLUMN I (i) there is much confusion...
Published On March 30th, 2023
• #### Quantitative Aptitude Quiz For Bank Foundation 2023 – 29th March
Direction (1-5):-A Company produces three different products namely food, drinks and cosmetic products. If total production of the company was same for all years and % production of three products in particular years given below, then answer the questions that...
Published On March 29th, 2023
• #### Reasoning Ability Quiz For Bank Foundation 2023 – 29th March
Direction (1-4): In the following questions assuming the given statement to be true, find which of the conclusion(s) among given conclusions is/are definitely true and then give your answers accordingly. Q1. Statements: P > M < X < V; D...
Published On March 29th, 2023
• #### English Language Quiz For Bank Foundation 2023-29th March
Directions (1-15): Each question below has one blank, which is indicating that something has been omitted. Find out which word from the options can be used to fill up the blank to make it meaningfully complete. Q1. The world economy...
Published On March 29th, 2023
• #### Quantitative Aptitude Quiz For Bank Foundation 2023 – 28th March
Directions (1-15): In each of the following questions, two equations (I) and (II) are given. Solve the equations and mark the correct option: (a) if x>y (b) if x<y (c) if x≥y (d) if x ≤y (e) if x =...
Published On March 28th, 2023
• #### Reasoning Ability Quiz For Bank Foundation 2023 – 28th March
Directions (1-3): Study the information carefully and answer the questions given below. There are seven persons- P, Q, R, S, T, U and V in a family. There are two married couples and three generations in this family. Q is...
Published On March 28th, 2023
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Project Euler Problem 044
# Statement
Pentagonal numbers are generated by the formula, $P_n = \frac {n(3n - 1)} {2}$.
The first ten pentagonal numbers are:
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …
It can be seen that $P_4 + P_7 = 22 + 70 = 92 = P_8$. However, their difference, 70 − 22 = 48, is not pentagonal.
Find the pair of pentagonal numbers, $P_j$ and $P_k$, for which their sum and difference is pentagonal
and $D = |P_k - P_j|$ is minimised; what is the value of D?
# Solution
The solution is a brute-force approach with limited number of possibilities.
from itertools import combinations
if __name__ == '__main__':
pentag_list = [n * (3 * n - 1) // 2 for n in range(1, 5000)]
setp = set(pentag_list)
pares = [tupla for tupla in combinations(pentag_list, 2) if abs(tupla[0] - tupla[1]) in setp]
for tupla in pares:
suma = tupla[1] + tupla[0]
while pentag_list[-1] < suma:
n = len(pentag_list) + 1
pentag_list.append(n * (3 * n -1) / 2)
if suma in pentag_list:
print("The result is:", abs(tupla[0] - tupla[1]))
The Python file is available for download here.
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License
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Saturday, February 18, 2012
GATE 2012 CS solution
I had appeared in GATE 2008 and 2009. After that I had not been interested much in GATE. However this time (GATE 2012) a close friend of mine appeared for the same, which got me interested again. Here is a solution set to the GATE 2012 Computer Science paper. I have solved question from Algorithms, Automata, C programming, Probability ... in short, the topics that I enjoyed enough to remember them after 2 years.
As I mentioned earlier, I have not appeared in for the examination. I got the question GATE 2012 CS and IT question set C from a fellow netizen. Please download the paper first as I will be using that as the reference for objective type answers.
2 years earlier I had posted solutions to GATE 2009 CS question paper. Comments and suggestions from that post made me realize the importance of explaining every answer. Feel free to ask questions if you need further clarifications.
GATE 2012 Computer Science and IT solutions to question set C
1. b
2. -
3. -
4. b
• Draw the sine graph for the interval. Only one minima at 3*pi/2 where sin(x) = -1
5. c
• fork() will spawn off a new child process every call. Three calls to fork() results in 8 processes, out of which one is original, other 7 are children.
6. b
7. a
• Obviously. Look at f(X,Y) and compare it with X
8. c
• Balanced binary search trees require log (number of elements) time to search
• log (n*2^n) = log (n) + log(2^n) = O(n)
9. -
10. c
• Standard switch case behaviour called "fall through", when the break statement is not there
11. c
• Look into any DB book, I studied Navathe.
12. -
13. c
• Do the checking manually
14. c or d
• 1 is false, 3 and 4 are true
15. -
16. -
17. c
• follows from definition of CDF
18. -
19. d
• 4 bit multiplier requires 2x4 = 8 bit input, i.e. 2^8 set of inputs
• multiplication of two 4-bit numbers will result in at most 8-bit result, so 8 bits for every multiplication
• 2^8 * 8 = 2^11 bits = 2 Kbits
20. c
• Average case is at lease as good as worst case
• A(n) better than W(n) :: Quicksort
• A(n) same as W(n) :: Heapsort
21. c
• Draw it and find out. They key thing is "any embedding of G on the plane"
22. d
• T(n-1) for moving n-1 disk to aux from source
• 1 for moving the n-th disk to destination
• T(n-1) for moving n-1 disk back to source from aux
23. b
24. -
25. c
26. -
27. --
28. -
29. b
• examine the cases:
• P(6 in first throw) = 1/6
• P(1 in first throw followed by 5 or 6 in second throw) = (1/6)*(2/6)
• P(2 in first throw followed by 4, 5 or 6 in second throw) = (1/6)*(3/6)
• P(3 in first throw followed by 3,4, 5 or 6 in second throw) = (1/6)*(4/6)
• summing up = (1/6)*(1+2/6+3/6+4/6) = 5/12
30. -
31. -
32. b
• b'd' -- group the 4 corners
• b'c' -- group the a'b' columns to 2 cells with the top 2 cells in ab' column
33. a
• MST algorithm does not depend on the value of the weights, it depends only on the relative ordering of the edge weights.
• Ranking remains unchanged as all the weights are >= 1
34. b
• I did it by hand, don't know if there is a shortcut
35. b
• Observe the graphs
36. -
37. d
• observe that q is a dead state, so anything beginning with 000 should go to q -- A and B gets eliminated by this logic
• after 10 if there is a 0 the state should move to 00 state, so D is correct
38. a
• look into any Data Structure book
39. -
40. -
41. -
42. -
43. -
44. d
• Dijkstra's algorithm probes the best solution so far
• Thanks to Shivam for pointing this out
45. b
• merge-sort takes O(n log n) comparisons
• here to compare two strings you need O(n) time, so O(n^2 log n)
46. c
• represent a cycle by a permutation of 4 labeled vertices
• cycles are identical if one is a rotation of another
• same cycle of length 4 can have 4 different representations each starting with a unique letter
• (6 P 4) / 4 = 90
47. -
48. -
49. -
50. c
• line 2 has a static variable, which retains the old value when he function is called again
51. d
• "auto int" is same as normal "int"
• "register int" is same as normal "int" as far as value retention is considered
52. -
53. -
54. -
55. -
56. -
57. a
58. -
59. a
• maxima-minima problem
• maximize the profit where, profit = 50q-5q^2
60. b
61. b
• Given:
• P(X) = 0.6
• P(Y) = 0.4
• P(R | X) = 0.96
• P(R | Y) = 0.72
• P(R) = P(X) * P(R | X) + P(Y) * P(R | Y) = 0.864
• Applying Bayes Rule:
• P(Y | R) = [ P(R | Y) * P(Y) ] / P(R) = 0.333
62. c
• follows from fundamental definitions of mean and standard deviation
63. -
64. -
65. b
• another maxima-minima problem and the function is already given, just maximize y
Please leave a comment if you were helped by my effort. Feel free to ask questions if you need further clarifications or have found some error in my solution.
Wednesday, February 15, 2012
Check who logged in to your Linux system
Linux (and all other unix clone OSs) have a handy way to let you know who are the users who have logged into your system. The command is "last". This command uses the file "/var/log/wtmp" and displays a list of all users logged in since the file was created.
I use it as:
last | grep -v -e bigyan -e reboot | less
Grep -v gives the complement of the results that it returns normally. The output of this command contains:
• username of the user who logged in
• terminal used by him/her ... pts/number
• IP address from which logged in
• Time of loggin in
• Time of loggin out
• duration logged in to the system
Read more usages of "last" command here. be sure to check out its cousin command "lastb" too.
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# How does inelastic collision affect velocity?
## How does inelastic collision affect velocity?
In a perfectly inelastic collision, two objects collide and stick together. The momentum of the objects before the collision is conserved, but the total energy is not conserved. The final velocity of the combined objects depends on the masses and velocities of the two objects that collided.
## Does velocity decrease in an inelastic collision?
– A partially inelastic collision is one in which some energy is lost, but the objects do not stick together. – The greatest portion of energy is lost in the perfectly inelastic collision, when the objects stick. – The kinetic energy does not decrease. – The velocity of the ball after the collision is zero.
Does velocity change after elastic collision?
Since the collision only imparts force along the line of collision, the velocities that are tangent to the point of collision do not change. The velocities along the line of collision can then be used in the same equations as a one-dimensional collision.
How do you find velocity after an inelastic collision?
The colliding particles stick together in a perfectly inelastic collision….Inelastic Collision Formula
1. V= Final velocity.
2. M1= mass of the first object in kgs.
3. M2= mas of the second object in kgs.
4. V1= initial velocity of the first object in m/s.
5. V2= initial velocity of the second object in m/s.
### What is difference between elastic and inelastic collision?
An inelastic collision can be defined as a type of collision where this is a loss of kinetic energy….Differences between elastic and inelastic collisions.
Elastic Collision Inelastic Collision
Momentum does not change. Momentum changes.
No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.
### Do objects stick together in an inelastic collision?
People sometimes think that objects must stick together in an inelastic collision. However, objects only stick together during a perfectly inelastic collision. Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved.
What is the difference between inelastic and perfectly inelastic collision?
Therefore, in inelastic collision, the kinetic energy is not conserved whereas in a perfectly inelastic collision, maximum kinetic energy is lost and the bodies stick together.
Is a car crash an inelastic collision?
An inelastic collisions occurs when two objects collide and do not bounce away from each other. Momentum is conserved, because the total momentum of both objects before and after the collision is the same. However, kinetic energy is not conserved. A high speed car collision is an inelastic collision.
#### What is the difference between elastic and inelastic collisions?
An elastic collision can be defined as a state where there is no net loss in kinetic energy in the system as the result of the collision. An inelastic collision can be defined as a type of collision where this is a loss of kinetic energy.
#### How do you know if it is elastic or inelastic collision?
How to determine if a collision is elastic or inelastic. If objects stick together, then a collision is perfectly inelastic. If the kinetic energy is the same, then the collision is elastic. If the kinetic energy changes, then the collision is inelastic regardless of whether the objects stick together or not.
What are examples of perfectly inelastic collisions?
Another common example of a perfectly inelastic collision is known as the “ballistic pendulum,” where you suspend an object such as a wooden block from a rope to be a target.
What happens in an inelastic collision?
An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not. Thermal energy, sound energy, and material deformation are likely culprits.
## What is the formula for inelastic collisions?
Inelastic Collision Formula V= Final velocity M1= mass of the first object in kgs M2= mas of the second object in kgs V1= initial velocity of the first object in m/s V2= initial velocity of the second object in m/s
## What is an example of perfectly inelastic collision?
Ballistic Pendulum. Another common example of a perfectly inelastic collision is known as the “ballistic pendulum,” where you suspend an object such as a wooden block from a rope to be a target.
Is kinetic energy conserved in inelastic?
Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision, but that is because it is converted to another form of energy (heat, etc.).
What is conserved in an inelastic collision?
An inelastic collisions occurs when two objects collide and do not bounce away from each other. Momentum is conserved, because the total momentum of both objects before and after the collision is the same. However, kinetic energy is not conserved. In an elastic collision, both momentum and kinetic energy are conserved.
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Categories: Math
# 6 Ways to Use Multiplication Brochures in Your Classroom
Teachers are always on the hunt for the best multiplication fact practice. Many have found our Best-Selling Multiplication Brochures! One of our most frequently asked question is: "How do you implement Multiplication Brochures in the classroom?" We are so excited to answer this question by giving you 6 ways to use our Multiplication Brochures in the classroom along with tips and FREEBIES to help you out!
## What are Multiplication Brochures?
Multiplication Brochures were created out of a need that we saw in our students that is common for students everywhere. Students struggle with their fact fluency and need strategies and multiple ways to practice their facts. Our Multiplication Math Brochure Bundle provides students with multiple ways to practice their multiplication facts in a unique and intriguing way that will spark interest in all students.
In the Multiplication Brochure Bundle, students practice with 37 brochures packed with multiplication strategies and activities to keep them engaged in their learning. Each fact family has 3 brochures along with 3 brochures practicing mixed facts.
#### Some of the multiplication strategies included in the brochures are:
• Number Lines
• Equal Groups
• Arrays
• Skip Counting
• Number Bonds
#### Some of the activities included in the brochures are:
• Fast Facts Practice
• Fill in the Blank
• Multiples Mazes
• Multiplication Equation Searches
• Color by Number
• Matching
• True or False
• Word Problems
## Our #1 Asked Question for Multiplication Brochures:
The #1 asked question that we get on Teachers Pay Teachers, Instagram, Facebook and in our newsletters is: How do you implement Multiplication Brochures in the classroom?
We are going to answer this question in detail and give you some freebies to use for the 6 different options! Let's get started!
### Step #1:
The first question that a teacher needs to ask themselves is: How am I going to give my students the multiplication brochures? Here are some different ways that we have made them available to our students.
### You Have Options!
Option 1A: Prep all brochures and allow for student choice! Prep them, display them and teach students how you want them to keep track of the brochures.
Option 1B: Prep all brochures and have students progress in an order! Prep them, display them and teach students how you want them to progress through the brochures.
#### -OR-
Option 2A: Teacher keeps brochures and the entire class works on a specific fact family for the week! The teacher makes all three brochures available to their students and has all three due at the end of one or two weeks.
Option 2B: Teacher keeps brochures and assigns the same brochure to the entire class. The teacher gives one or two days for students to complete the brochure and then collects the brochures.
Over the years, we have used all of these options! Each class is different and only the teacher knows what is best for their students. One year, you might have a group of students that can handle student choice and other years, you need to simplify and give more direction!
### Step #2:
Now that you know how you are going to make the brochures available to your students, you need to ask HOW you are going to use them on a daily basis? What does that really look like in your classroom?
##### Here are six options for implementing Multiplication Brochures in the classroom:
• Early Finisher Activities
• Math Center Activities
• Whole Class Direct Instruction
• Small Group Math Activities
• Extra Practice/ Homework
• Substitute Teachers
## Each one of these options deserves its own attention! Let's dive in deeper to each and grab some freebies to help you with each! Simply click on the buttons below to start exploring!
### Say good-bye to bored and complaining students & HELLO to Multiplication Mastery!
#### We have a FREEBIE for you!!!
Teachers are always on the hunt for the best multiplication fact practice. Many have found our Best-Selling Multiplication Brochures! One of our most frequently asked question is: “How do you implement Multiplication Brochures in the classroom?” We are so excited to answer this question by giving you 6 ways to use our Multiplication Brochures in the classroom!
## What are Multiplication Brochures?
Multiplication Brochures were created out of a need that we saw in our students that is common for students everywhere. Students struggle with their fact fluency and need strategies and multiple ways to practice their facts. Our Multiplication Math Brochure Bundle provides students with multiple ways to practice their multiplication facts in a unique and intriguing way that will spark interest in all students.
In the Multiplication Brochure Bundle, students practice with 36 brochures packed with activities to keep them engaged in their learning. Each fact family has 3 brochures along with 3 brochures practicing mixed facts. Some of the activities include:
• Fast Facts Practice
• Fill in the Blank
• Word Problems
• Multiples Maze
• Multiplication Equation Search (like a word search)
• Number Lines
• Equal Groups
• Color by Number
• Matching
• Skip Counting
• Arrays
• Number Bonds
• True or False?
• Color Match the Product to the Correct Equation
## Our #1 Asked Question for Multiplication Brochures
The #1 asked question that we get on Teachers Pay Teachers, Instagram, Facebook and in our newsletters is: How do you implement Multiplication Brochures in the classroom?
We are going to answer this question in detail and give you some freebies to use for the 6 different options! Let’s get started!
Step #1:
The first question that a teacher needs to ask themselves is: How am I going to give my students the multiplication brochures? Here are some different ways that we have made them available to our students.
Option 1A: Prep all brochures and allow for student choice! Prep them, display them and teach students how you want them to keep track of the brochures.
Option 1B: Prep all brochures and have students progress in an order! Prep them, display them and teach students how you want them to progress through the brochures.
Option 2A: Teacher keeps brochures and the entire class works on a specific fact family for the week! The teacher makes all three brochures available to their students and has all three due at the end of one or two weeks.
Option 2B: Teacher keeps brochures and assigns the same brochure to the entire class. The teacher gives one or two days for students to complete the brochure and then collects the brochures.
Over the years, we have used all of these options! Each class is different and only the teacher knows what is best for their students. One year, you might have a group of students that can handle student choice and other years, you need to simplify and give more direction!
Step #2:
Now that you know how you are going to make the brochures available to your students, you need to ask HOW you are going to use them on a daily basis? What does that really look like in your classroom?
Here are six options for your students to practice multiplication facts using Multiplication Brochures:
1. Early Finishers
2. Math Centers
3. Small Group
4. Homework/ Extra Help
5. Whole Class
6. Substitute Work
Each one of these options deserves its own attention! Let’s dive in deeper to each and grab some freebies to help you with each!
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# Broadcasting over arrays of SVectors
Can you use `.=` across an array of SVectors to reduce the number of allocations?
Yes.
4 Likes
This is the kind of thing that itâs easy to just try yourself first, and then ask if you have difficulties or specific questions, backed up by actual code that shows precisely what the question is.
5 Likes
Youâre right, I should have asked a better question. I didnât want to show my entire code because I donât think itâs fair to ask other people to debug my code. But, Iâll post it here and if someone can see the problem quickly, thatâs great. If not, thatâs OK too. Here is the code:
``````#using Pkg
using StaticArrays
struct MD
nParticles::Int64
boxSize::Float64
positions:: Array{SVector{2,Float64},1}
velocities:: Array{SVector{2,Float64},1}
end
function initializeParticles(nParticles::Int64,boxSize::Float64,lVecs::Array{Float64,2},random::Bool = false):: MD
positions = zeros(SVector{2,Float64},nParticles)
velocities = zeros(SVector{2,Float64},nParticles)
deltaR = 0.5
v0=1.
idx = 0
for i =-boxSize-10:boxSize+10
for j =-boxSize-10:boxSize+10
vec = i * lVecs[1,:] + j * lVecs[2,:] + [0.6, 0.6]
if random
rng = MersenneTwister(1234)
rVec = rand(rng,2)
vec += deltaR * rVec/norm(rVec)
end
if 0 <= vec[1] <= boxSize && 0 <= vec[2] <= boxSize && idx < nParticles
idx += 1
positions[idx] = vec
rng = MersenneTwister(1234)
vVec = rand(rng,2)
velocities[idx] = v0 * vVec/norm(vVec)
end
end
end
if idx < nParticles
println("Didn't find all of the requested particles")
println("Requested: ", nParticles)
println("Found: ", idx)
end
MD(nParticles,boxSize,positions,velocities)
end
function forceOnSingleParticle(positions::Array{SVector{2,Float64},1},particle::SVector{2,Float64},boxSize::Float64):: SVector{2,Float64}
fVec = SVector(0,0)
modifiedPos = SVector{2,Float64}
diffVec = SVector{2,Float64}
for i=1:size(positions,1)
diffVec = particle - positions[i]
if abs(diffVec[1]) > boxSize/2 && abs(diffVec[2]) > boxSize/2
modifiedPos = positions[i] + boxSize * SVector(sign(diffVec[1]) , sign(diffVec[2]))
elseif abs(diffVec[1]) > boxSize/2
modifiedPos = positions[i] + boxSize * SVector(sign(diffVec[1]) , 0)
elseif abs(diffVec[2]) > boxSize/2
modifiedPos = positions[i] + boxSize * SVector(0 , sign(diffVec[2]))
else
modifiedPos = positions[i]
end
diffVec = particle - modifiedPos
distance = norm(diffVec)
if distance > 0.5
fVec += 24 * (2/distance^13 - 1/distance^7) * 1/distance * diffVec
end
end
return fVec
end
mymod(x::SVector,y::Number)::SVector = mod.(x,y)
function increaseTemp(positionsP::Array{SVector{2,Float64},1},positionsPP::Array{SVector{2,Float64},1},R::Float64)::Array{SVector{2,Float64},1}
positionsPP = positionsP - R * (positionsP - positionsPP)
end
function simulate(myMD::MD,dt::Float64,tMax::Float64)
forces = zeros(SVector{2,Float64},size(myMD.positions,1))
check = zeros(SVector{2,Float64},size(myMD.positions,1))
T::Float64 = 0.4
positions = myMD.positions
positionsP = myMD.positions
positionsPP = positions - myMD.velocities * dt
velocities = myMD.velocities
R::Float64 = 0.95
N::Int = size(positions,1)
while T > 0.2
for time=0:dt:tMax
#if any(positions > myMD.boxSize)
# exit("lost particles")
#end
for n=1:N
forces[n] = forceOnSingleParticle(positions,positions[n],myMD.boxSize)
end
positions .= mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,myMD.boxSize)
#velocities = (positions - positionsPP)/(2 * dt)
positionsPP = positionsP
positionsP = positions
end
#positionsPP = increaseTemp(positionsP,positionsPP,R) #Increase temperature by factor of R
break
#T -=0.3#*= R
end
positions
end
using Plots
plotly()
dt = .0001
tMax = 0.9
lvecs = 1.07 * [[1. 0.]; [0.5 sqrt(3)/2]]
myMD = initializeParticles(16,4.,lvecs)
@time mypositions = simulate(myMD,dt,tMax)
plot(map(x -> x[1],mypositions),map(x -> x[2],mypositions),seriestype = :scatter,aspect_ratio=:equal,xlim = (0,4),ylim = (0,4))
``````
The key line of code here is line 102:
``````positions .= mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,myMD.boxSize)
``````
As it stands, the array doesnât seem to update. But if I replace the `.=` with just `=`, then things behave as expected.
1 Like
It definitely should be updating. Minimal example:
``````julia> x = [@SVector(rand(4)) for _ â 1:5]
5-element Vector{SVector{4, Float64}}:
[0.38783754617174626, 0.23479994102954094, 0.9131772241755032, 0.5158254036871766]
[0.5377088066052653, 0.052031912322721086, 0.03460283315171697, 0.3172115916348688]
[0.008984784765390552, 0.16111534758886692, 0.653866563808464, 0.853094814027981]
[0.3532881187190555, 0.7784351001503331, 0.007738316370001996, 0.952768794661861]
[0.7145584888646308, 0.6944186486326702, 0.8473666362930881, 0.7211799815251871]
julia> @. x = 2x
5-element Vector{SVector{4, Float64}}:
[0.7756750923434925, 0.4695998820590819, 1.8263544483510064, 1.031650807374353]
[1.0754176132105306, 0.10406382464544217, 0.06920566630343394, 0.6344231832697376]
[0.017969569530781104, 0.32223069517773384, 1.307733127616928, 1.706189628055962]
[0.706576237438111, 1.5568702003006663, 0.015476632740003993, 1.905537589323722]
[1.4291169777292616, 1.3888372972653404, 1.6947332725861761, 1.4423599630503743]
julia> x
5-element Vector{SVector{4, Float64}}:
[0.7756750923434925, 0.4695998820590819, 1.8263544483510064, 1.031650807374353]
[1.0754176132105306, 0.10406382464544217, 0.06920566630343394, 0.6344231832697376]
[0.017969569530781104, 0.32223069517773384, 1.307733127616928, 1.706189628055962]
[0.706576237438111, 1.5568702003006663, 0.015476632740003993, 1.905537589323722]
[1.4291169777292616, 1.3888372972653404, 1.6947332725861761, 1.4423599630503743]
``````
3 Likes
Iâve tried simple examples like that too and convinced myself that this should work. I wondered if it had something to do with my `mymod` function so I took that out but it behaved the same.
It is updating:
``````p = copy(positions)
positions .= mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,myMD.boxSize)
return p - positions
``````
``````julia> simulate(myMD,dt,tMax)
16-element Array{SArray{Tuple{2},Float64,1,2},1}:
[-6.340483565309141e-5, -7.957531458613332e-5]
[-6.346290042008595e-5, -7.919910996889712e-5]
[-6.328366500352178e-5, -7.921290476664566e-5]
[-6.111917607287687e-5, -7.935480894660785e-5]
[-6.331523406521988e-5, -7.907629591796805e-5]
[-6.113201763979781e-5, -7.921480146100279e-5]
[-6.110647734414165e-5, -7.920980100584174e-5]
[-6.096459454729697e-5, -7.932846379254954e-5]
[-6.110741666365271e-5, -7.90963577439463e-5]
[-6.0965533867030075e-5, -7.92150205306541e-5]
[-6.093999357137392e-5, -7.921002007593714e-5]
[-5.8756777145951844e-5, -7.934852561852779e-5]
[-6.095283513829486e-5, -7.9070012589888e-5]
[-5.878834620753892e-5, -7.921191677007222e-5]
[-5.8609110791252306e-5, -7.922571156759872e-5]
[-5.866717555802481e-5, -7.884950695036252e-5]
``````
By chance werenât you trying to test using something like:
``````p = positions
positions .= mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,myMD.boxSize)
return p - positions
``````
In this case we donât see any different between `p` and `positions`, but that is because `=` is only assigning a different name to the same vector.
``````function f!(positions,positionsP,positionsPP,forces,dt,boxsize)
positions .= mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,boxsize)
return positions
end
function g!(positions,positionsP,positionsPP,forces,dt,boxsize)
for i in eachindex(positions)
positions[i] = 2 * positionsP[i] - positionsPP[i] + forces[i] * dt^2
positions[i] = mod.(positions[i],boxsize)
end
return positions
end
``````
``````julia> positions = rand(SVector{3,Float64},5);
julia> positionsP = rand(SVector{3,Float64},5);
julia> positionsPP = rand(SVector{3,Float64},5);
julia> forces = rand(SVector{3,Float64},5);
julia> dt = 0.1; boxsize = 0.2;
julia> f!(copy(positions),positionsP,positionsPP,forces,dt,boxsize) == g!(copy(positions),positionsP,positionsPP,forces,dt,boxsize)
true
julia> @btime f!(p,\$positionsP,\$positionsPP,\$forces,\$dt,\$boxsize) setup=(p=copy(positions))
163.913 ns (0 allocations: 0 bytes)
julia> @btime g!(p,\$positionsP,\$positionsPP,\$forces,\$dt,\$boxsize) setup=(p=copy(positions))
141.832 ns (0 allocations: 0 bytes)
``````
I was just visually inspecting the plot. I actually think that with `.=` there was some modification but it was way different (smaller) than what happened with `=`
What am I missing? I donât think the line `@. x = 2x` worked? Edit: sorry, I guess it is the case âin the morning I get up and I wake up an hour laterâ.
Looks like each number got multiplied by 2 to me. ??
Using `=` produced this
Using `.=` produced this:
Which looks very close(if not equal) to the initial positions.
It is likely that the problem is in these lines:
`````` positions = myMD.positions
positionsP = myMD.positions
``````
Here, you are just assigning two different names to the same `myMD.positions` vector. Thus, that can cause you all sort of problems latter. You should use `positionsP = copy(positions)`.
And check your other assignments in the loop. You should use `positionsP .= positions` (with the dot), otherwise you will have the same problem. The `=` just assigns a new name to the variable, while `.=` is mutating the elements of the existing array, which is what you want in most cases.
2 Likes
Hmm. Great example. I was trying to eliminate all of the allocations too. I started with this:
`positions = 2 * positionsP - positionsPP + forces * dt^2`
and found that I could reduce allocations with use of `.`. I didnât see that a loop over the array would have produced no allocations either. Gonna ponder on this a little. Thanks
Hmm. Thinking⌠Can you explain why those two lines didnât cause problems when I used `=` in the update.
The thing is that in this line:
``````positions = mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,myMD.boxSize)
``````
without the `.`, what you are doing is creating a new array on the right side, and naming it `positions`.
Therefore, in this case the `positions` vector got dissociated when updated from the `positionsP` vector (which is what you want).
Like this:
``````julia> x = [1,2] # create new array and name it x
2-element Array{Int64,1}:
1
2
julia> y = x # just to save the values, this only assings a new name to x
2-element Array{Int64,1}:
1
2
julia> x = x + [1,1] # This is creating a new vector on right side and naming it x
2-element Array{Int64,1}:
2
3
julia> y[1] # continues to be the same as before, no mutation
1
julia> x .= x .+ [1,1] # with the dot, you are mutating the `x` vector
2-element Array{Int64,1}:
3
4
``````
Reading suggestion: Assignment and mutation ¡ JuliaNotes.jl
2 Likes
I see. Thanks for the help. Iâll keep studying.
That admittedly is confusing at the beginning. One gets used to it, though, and after that one notes that actually there is no way out from that if one wants to have a dynamically typed language.
The fact that the language is dynamic requires that we can do
``````julia> x = [1,2]
2-element Array{Int64,1}:
1
2
julia> x = Ď
Ď = 3.1415926535897...
``````
That is completely natural, but means that `=` is just the binding of a name to a value (which might be an array, or scalar, or whatever).
Thus, we need to be able to differentiate naming something from mutating something. Mutating is a function that acts on a mutable object. It is the `setindex!` function:
``````julia> x = [1,2]
2-element Array{Int64,1}:
1
2
julia> setindex!(x,10,1)
2-element Array{Int64,1}:
10
2
``````
Which, by convenience (obviously) can be called with the notation:
``````julia> x[1] = 10
10
``````
But this last `x[1] = 10` is a call to `setindex!`, not a name assignment as the other cases. (and a broadcasting of assignments, with `.=`, is just a loop calling `setindex!` for each element)
(ps, if you feel better, here I am asking the same thing two years ago)
1 Like
Can I try and say it in a different way? These lines:
``````positions = myMD.positions
positionsP = myMD.positions
``````
essentially created two names assigned to the same memory block. Then later when I did this:
``````positions .= mymod.(2 .* positionsP .- positionsPP .+ forces .* dt^2,myMD.boxSize)
``````
it mutated that memory block and hence changed the value of both `positions` and `positionsP` Later when I tried to shuffle past and present snapshots around
``````positionsPP = positionsP
positionsP = positions
``````
it made both arrays equal to the same block of memory. Did I say that right?
Take home message: Careful when doing `var1 = var2`.
3 Likes
Exactly.
1 Like
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# Cartan's criterion
In mathematics, Cartan's criterion gives conditions for a Lie algebra in characteristic 0 to be solvable, which implies a related criterion for the Lie algebra to be semisimple. It is based on the notion of the Killing form, a symmetric bilinear form on $\mathfrak{g}$ defined by the formula
$K(u,v)=\operatorname{tr}(\operatorname{ad}(u)\operatorname{ad}(v)),$
where tr denotes the trace of a linear operator. The criterion was introduced by Élie Cartan (1894).
## Cartan's criterion for solvability
Cartan's criterion for solvability states:
A Lie subalgebra $\mathfrak{g}$ of endomorphisms of a finite-dimensional vector space over a field of characteristic zero is solvable if and only if $Tr(ab)=0$ whenever $a\in\mathfrak{g},b\in[\mathfrak{g},\mathfrak{g}].$
The fact that $Tr(ab)=0$ in the solvable case follows immediately from Lie's theorem that solvable Lie algebras in characteristic 0 can be put in upper triangular form.
Applying Cartan's criterion to the adjoint representation gives:
A finite-dimensional Lie algebra $\mathfrak{g}$ over a field of characteristic zero is solvable if and only if $K(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$ (where K is the Killing form).
## Cartan's criterion for semisimplicity
Cartan's criterion for semisimplicity states:
A finite-dimensional Lie algebra $\mathfrak{g}$ over a field of characteristic zero is semisimple if and only if the Killing form is non-degenerate.
Dieudonné (1953) gave a very short proof that if a finite-dimensional Lie algebra (in any characteristic) has a non-degenerate invariant bilinear form and no non-zero abelian ideals, and in particular if its Killing form is non-degenerate, then it is a sum of simple Lie algebras.
Conversely, it follows easily from Cartan's criterion for solvability that a semisimple algebra (in characteristic 0) has a non-degenerate Killing form.
## Examples
Cartan's criteria fail in characteristic p>0; for example:
• the Lie algebra SLp(k) is simple if k has characteristic not 2 and has vanishing Killing form, though it does have a nonzero invariant bilinear form given by (a,b) = Tr(ab).
• the Lie algebra with basis an for nZ/pZ and bracket [ai,aj] = (ij)ai+j is simple for p>2 but has no nonzero invariant bilinear form.
• If k has characteristic 2 then the semidirect product gl2(k).k2 is a solvable Lie algebra, but the Killing form is not identically zero on its derived algebra sl2(k).k2.
If a finite-dimensional Lie algebra is nilpotent, then the Killing form is identically zero (and more generally the Killing form vanishes on any nilpotent ideal). The converse is false: there are non-nilpotent Lie algebras whose Killing form vanishes. An example is given by the semidirect product of an abelian Lie algebra V with a 1-dimensional Lie algebra acting on V as an endomorphism b such that b is not nilpotent and Tr(b2)=0.
In characteristic 0, every reductive Lie algebra (one that is a sum of abelian and simple Lie algebras) has a non-degenerate invariant symmetric bilinear form. However the converse is false: a Lie algebra with a non-degenerate invariant symmetric bilinear form need not be a sum of simple and abelian Lie algebras. A typical counterexample is G = L[t]/tnL[t] where n>1, L is a simple complex Lie algebra with a bilinear form (,), and the bilinear form on G is given by taking the coefficient of tn−1 of the C[t]-valued bilinear form on G induced by the form on L. The bilinear form is non-degenerate, but the Lie algebra is not a sum of simple and abelian Lie algebras.
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# Search by Topic
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##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Train Carriages
##### Stage: 1 and 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### Five Coins
##### Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### It Figures
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Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
##### Stage: 2 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
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##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
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##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
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Can you design a new shape for the twenty-eight squares and arrange the numbers in a logical way? What patterns do you notice?
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Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
### Worms
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Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
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##### Stage: 1 and 2 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### Room Doubling
##### Stage: 2 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Building with Rods
##### Stage: 2 Challenge Level:
In how many ways can you stack these rods, following the rules?
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Three Sets of Cubes, Two Surfaces
##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
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##### Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
### Making Squares
##### Stage: 2
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Times
##### Stage: 2 Challenge Level:
Which times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!
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##### Stage: 1 and 2 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Calcunos
##### Stage: 2 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
##### Stage: 2 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
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##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Caterpillars
##### Stage: 1 Challenge Level:
These caterpillars have 16 parts. What different shapes do they make if each part lies in the small squares of a 4 by 4 square?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
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##### Stage: 2 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### Sorting the Numbers
##### Stage: 1 and 2 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
### Cuboid-in-a-box
##### Stage: 2 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### New House
##### Stage: 2 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Street Sequences
##### Stage: 1 and 2 Challenge Level:
Investigate what happens when you add house numbers along a street in different ways.
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Sometimes We Lose Things
##### Stage: 2 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### Sticks and Triangles
##### Stage: 2 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Calendar Patterns
##### Stage: 2 Challenge Level:
In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers?
### Magazines
##### Stage: 2 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### Egyptian Rope
##### Stage: 2 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Balance of Halves
##### Stage: 2 Challenge Level:
Investigate this balance which is marked in halves. If you had a weight on the left-hand 7, where could you hang two weights on the right to make it balance?
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
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In addition to replacing the president if necessary, it's the duty of the vice president to A. preside over the Senate. B. serve as Speaker of the House. C. review laws before the president signs them. D. serve as liaison to the judicial branch
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In addition to replacing the president if necessary, it's the duty of the vice president to preside over the Senate.
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In quantum mechanics, why is it useful to apply operators to anything other than eigenfunctions, when only eigenfunctions correspond to observables?
For context, I was looking at the definition of the operator postulate from hyperphysics given here, which matches up pretty well with how I remember Griffiths explaining the definition of an operator in QM in his QM text.
The mathematical operator $$Q$$ extracts the observable value $$q_n$$ by operating upon the wavefunction which represents that particular state of the system.
If all an operator does is "extract the observable value", then it should always return a scalar, since we always just measure a scalar value, right? And that's presumably why the postulate is given in terms of eigenvalues. But what doesn't make sense is, if an operator always return scalar values, then isn't it just a scalar function in the first place and so wouldn't even have eigenvectors and eigenvalues?
The obvious answer is that operators don't just act on "wavefunctions which represent [a] particular state of [a] system." But then, what else do they act on? Wavefunctions that don't "represents [a] particular state of [a] system"? But what else could wavefunctions represent, if not particular states of a particle or system of particles?
Or, perhaps a better way to put it is, if only the eigenvalues of the wavefunction correspond to something observable, why do we care about the rest of the wavefunction? Like, what useful information, even just as an intermediate step in a calculation to allow us to do the rest of the calculation, do we get from applying an operator to a part of a wavefunction that isn't an eigenfunction?
• there is a relevant question and answer physics.stackexchange.com/q/693702 Commented Mar 27, 2023 at 4:38
• @annav, I don't see how that's relevant. It seems like a pretty different question to me. I'm not asking about what eigenfunctions are -- I already know what they are from linear algebra. I'm asking why we care about parts of the wavefunction that aren't eigenfunctions, since only the eigenfunctions correspond to observables. Commented Mar 27, 2023 at 4:41
• You state "then isn't it just a scalar function in the first place and so wouldn't even have eigenvectors and eigenvalues?" I do not think it is true, that scalar eigenvalues imply a scalar wavefunction. An operator can act on a complex function and give a scalar back. Commented Mar 27, 2023 at 5:00
• @annav, I suppose I was being a bit sloppy with my terminology. I simply meant a function that maps a vector space to it's field of scalars isn't normally considered a vector function (but rather a linear functional, which is probably the term I should have used) and doesn't have eigenvectors or eigenvalues. Commented Mar 27, 2023 at 5:08
Let me try to answer your questions. First a comment: operators do not always return eigenvalues. Sometimes they may return collections of eigenvalues (such as the position vector components, which can be obtained by letting the position operator, 2D or 3D etc, acting on a state), which if they are considered together, the result is a vector, i.e. $$\hat{\vec{r}}\psi(x)=\vec{r}\psi(x)$$ if $$\psi(x)$$ is an eigenfunction of the position operator.
There are wavefunctions that represent a particular state of the system, that is true. But there are also wavefunctions that represent another particular state of the system, and another and another, depending on what you wish to measure. An example, here would be a good idea: let $$|\alpha\rangle$$ be the ket corresponding to the wave function $$\psi_{\alpha}(x)=\langle x|\alpha\rangle$$. One can act with the operator $$A$$ on the ket $$|\alpha\rangle$$ and obtain an eigenvalue $$\alpha$$, whereas at the same time one can act with operator B on state $$|\beta\rangle$$ and obtain an eigenvalue $$\beta$$. $$\alpha$$ and $$\beta$$ are the results we get if we conduct measurements of different physical quantities on different states. For instance, $$\alpha$$ can be a position eigenvalue and $$\beta$$ a momentum eigenvalue.
At the same time, one might want to act with the position operator on a momentum eigenstate, namely of $$|\beta\rangle$$. So, one acts with the operator $$A$$ on $$|\beta\rangle$$. But $$|\beta\rangle$$ is not an eigenstate for the operator $$A$$, so the result is not of the form $$A|\beta\rangle=c |\beta\rangle$$ where $$c$$ is some eigenvalue (scalar if we restrict ourselves to 1 dimensional systems). So, one must expand $$|\beta\rangle$$ in the basis of $$|\alpha\rangle$$, namely write something like $$|\beta\rangle=\sum_{\alpha}c_{\alpha}|\alpha\rangle$$ where the $$c_{\alpha}$$s are constant coefficients, one for each possible state $$|\alpha\rangle$$. So, suppose for instance that $$A$$ again is the position operator and there are two possible positions labelled by the numbers $$1$$ and $$2$$, namely $$|1\rangle$$ and $$|2\rangle$$, with equal probability of measuring both. Then, one can write $$|\beta\rangle=\frac{1}{\sqrt{2}}(|1\rangle+|2\rangle)$$ and hence, $$A|\beta\rangle=\frac{1}{\sqrt{2}}(\alpha_1|1\rangle+\alpha_2|2\rangle)$$ which is not an eigenvalue of the operator $$A$$.
So, to answer to your last question, we do care about the remaining wavefunction, as it gives us information for about other physical quantities. If I only cared about position measurements in my example, for instance, I would only care about the set of eigenfunctions $$\{|\alpha\rangle\}$$, but instead, one may have a more general wavefunction and obtain different kind of information each time one acts on the wavefunction with different operators.
I hope this helps. If there are any questions, please do let me know.
• One important thing that I really think that you should emphasize (and also change the language in your post): you say "At the same time, one might want to measure the position of a momentum eigenstate... So, one acts with the operator A on..." This seems to imply that acting with the operator on the state corresponds to the physical act of measurement of the operator, which it does not, and this is such a common misconception (and can cause such problems with learners of QM), that that should be addressed. Commented Mar 27, 2023 at 19:59
• No, acting with some operator on a vector does not correspond to a measurement. Acting with an operator on a vector is a mathematical operation which yields a new vector. That the expectation value is defined like this has nothing to do with the point mentioned by @march. Commented Mar 28, 2023 at 8:00
• You should use the @username function to notify others. Anyway, I don't understand your query here, to be honest. It is just wrong that the action of an operator corresponds to some measurement... Let me ask you instead: Elaborate how or why do you think this corresponds to a measurement?! Again: if you have some operator/observable $A$ , then $(A\psi)$ is another vector. However, in the simplest case of a non-deg. observable, the state after the measurement is $\psi_n$, where this is the eigenstate of $A$ corresponding to the measured value $a_n$, i.e. $A\psi_n=a_n\psi_n$. Commented Mar 28, 2023 at 10:27
• Again: IMHO it is useless and in general wrong to say that the action of an operator corresponds to a measurement. For example, even if your system is in an eigenstate $\psi_n$, then if you say the action of $A$ corresponds to a measurement, then the new state is $a_n \psi_n$, or what? This is not normalized and can even be the $0$ vector... Commented Mar 28, 2023 at 11:13
• @MikaylaEckelCifrese ...The problem in QM is that there is a large inferential gap between the mathematical operations and the physical interpretations, and you just shouldn't get in the habit of trying to understand the "physical interpretation" of every mathematical operation that you do. Commented Mar 29, 2023 at 3:38
When it comes to observable operators (there are also operators used for different things in QM, for which we almost never solve the eigenvalue problem) - the logic kind of goes the other way around. We need the operator in order to even know what the eigenstates (and corresponding) are. That's really the main purpose of those observable operators. Sure, if you already had the outputs of the solved eigenvalue problem you wouldn't need the observable operators much anymore, except for $$H$$ because it has the additional role that it specifies time evolution in the Schrödinger equation.
Specifically with the momentum operator here's how it goes. We are given an observable operator, let's say (this is the 3d version, to get 1d replace $$\nabla \to \frac{\partial}{\partial x}$$).
$$\hat{p} = -i \hbar \vec{\nabla}$$
From there we want to know what its eigenvectors and eigenvalues are, so we solve the eigenvalue problem for $$\psi_p$$ such that $$\hat{p_i} \psi_p = p_i \psi_p$$ Note that $$p_i$$ without the hat is the eigenvalue for a component of momentum, it is a number. The $$\hat{p}_i$$ is the operator for the $$i$$th component of momentum. We find, in this case, that
$$\psi_p = e^{i\vec{p}\cdot \vec{r}/\hbar}$$
Is the eigenvector, simultaneously an eigenstate of the operators $$\hat{p}_x, \hat{p}_y, \hat{p}_z$$ corresponding to eigenvalues which we can write as a vector $$\vec{p}$$. We therefore call that state $$|\vec{p}\rangle$$, just labeling it by its momentum. This allows us to take the inner product $$|\langle \vec{p} | \phi \rangle|^2$$ to get probabilities for momentum measurements when the system is in a state $$\phi$$.
• "there are also operators used for different things in QM, for which we almost never solve the eigenvalue problem" what else are they used for? Commented Mar 27, 2023 at 4:57
• @MikaylaEckelCifrese So 1. there is the Hamiltonian in the Schrödinger equation. I guess you don't have to think of it as an operator any more than in any other differential equation. 2. Then you can express the time evolution of the state as an operator $\psi(t)=U(t) \psi(0)$, where it turns out $U=e^{-itH/\hbar}$. This ends up often being useful. 3Then there are "projection operators", these become most useful when a measurement is done and the eigenvalue that results from it has more than one corresponding eigenvector. Then the probability of getting that outcome is $|P|\psi\rangle|$ rather Commented Mar 27, 2023 at 5:23
• than, for example, $|\langle x|\psi \rangle|^2$ in the case of position. Those above are the most "foundational" operators in QM, but along the way there are operators we use for miscellaneous purposes. One common one is the raising/lowering operators, which come up when solving the Schrödinger equation for a particle in a potential well $V(x)=kx^2$. This example, called the "harmonic oscillator", is an important example because so many real situations can be approximated by it. Applying the raising operator to one solution in this potential gets you another solution of a higher energy. Commented Mar 27, 2023 at 5:28
• I would also like to mention that, while many of these operators are here to stay, there is an alternative formulation of non-relativistic quantum mechanics called Bohmian Mechanics, which doesn't use observable operators at all (at least not in its foundations, but they can be derived in limiting cases). It also doesn't use projection operators. So it really just has the Schrödinger equation, and some of those miscellaneous operators. It's a nearly operator-free formulation of QM. Commented Mar 27, 2023 at 5:30
First, some background, in which I'll ignore subtleties and just address the main ideas. Solving the eigenvalue equation for a normal operator (of which self-adjoint operators are a special case) is essentially just a route to the so-called spectral decomposition of the operator, which takes the form $$\hat A = \sum_n \lambda_n \hat P_n$$ where $$\lambda_n$$ is the $$n^{th}$$ eigenvalue of $$\hat A$$ and $$\hat P_n$$ is the orthogonal projection operator which acts on a vector by projecting it into the eigenspace corresponding to $$\lambda_n$$. Assuming that $$\lambda_n$$ is non-degenerate, $$\hat P_n$$ takes the familiar form $$|\phi_n\rangle\langle \phi_n|$$ where $$|\phi_n\rangle$$ is the (normalized) eigenvector with eigenvalue $$\lambda_n$$.
This decomposition is important for a number of reasons. First, if $$\hat A$$ is a self-adjoint operator representing some observable, then the $$\mathbb R$$-valued eigenvalues correspond to possible measurement outcomes. If a system is in a pure state with (normalized) state vector $$|\psi\rangle$$, then the probability of measuring $$\hat A$$ to take the value $$\lambda_n$$ is given by $$\langle \psi|\hat P_n |\psi\rangle$$, and the (unnormalized) post-measurement state is given by $$\hat P_n|\psi\rangle$$. Therefore, the spectral decomposition of an operator allows you to compute the probability of all possible measurement outcomes, as well as the post-measurement state obtained in each case.
The decomposition is also important if we wish to compute a function of the operator. Given an ordinary function $$f:\mathbb C\rightarrow \mathbb C$$, we can define the operator $$f(\hat A)$$ as $$f(\hat A) = \sum_n f(\lambda_n) \hat P_n$$ Of particular importance is the time evolution operator which tells us how to evolve our state vectors forward in time; for a time-independent Hamiltonian $$\hat H$$, the time-evolution operator is given by $$\hat U(t) = \exp\big[-i\hat H t/\hbar\big] = \sum_n e^{-iE_nt/\hbar} \hat P_n$$ Other examples include the rotation operators, which are obtained by exponentiating the angular momentum operators, and inverse operators of the (formal) form $$\hat A^{-1} \equiv 1/\hat A$$ which are of great utility in e.g. perturbation theory.
But what doesn't make sense is, if an operator always return scalar values, then isn't it just a scalar function in the first place and so wouldn't even have eigenvectors and eigenvalues?
When people say that operating on an eigenvector turns the corresponding eigenvalue, what they mean is that the eigenvalue can be determined simply by looking at the resulting expression. The operator $$-\frac{d^2}{dx^2}$$ acting on $$\sin(2x)$$ yields $$\underbrace{-\frac{d^2}{dx^2}}_{\text{operator}} \underbrace{\sin(2x)}_\text{vector} = \underbrace{4}_\text{eigenvalue} \underbrace{\sin(2x)}_{\text{vector}}$$
In other words, the operator does not literally return $$4$$ as an output, but by acting on the vector with the operator we can read off the eigenvalue from the result.
Or, perhaps a better way to put it is, if only the eigenvalues of the wavefunction correspond to something observable, why do we care about the rest of the wavefunction? Like, what useful information, even just as an intermediate step in a calculation to allow us to do the rest of the calculation, do we get from applying an operator to a part of a wavefunction that isn't an eigenfunction?
I'm not really sure what to make of this question. The last part is answered in part by my background section above; there are many operators (e.g. symmetry transformations like rotation operators or translation operators, or the time evolution operator) which are interesting and useful because of the transformation they impose on the state vector of the system, not because they correspond to observables.
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# Total Chess: Check, Checkmate, Stalemate
04.09.2012
0
Category: Total Chess
There are three ways to attack the king. It is very important to know all three ways, and the differences between them.
1) Check: When the king is under attack by one of the opponent’s pieces, the king is in check. When your king is in check, you must get out of check. You are never allowed to make a move that leaves your king in check. You are also never allowed to make a move that puts your king in check. It is against the rules. This also means that the king can never be captured, because it can never be left under attack.
When you put the other player’s king in check, you say, “check.” Do you have to say, “check?” No, you do not have to say, “check,” but it is polite to say, “check.” It is like when somebody sneezes. You do not have to say, “bless you,” but it is polite to say, “bless you.” Whether you say, “bless you,” or not, the person still sneezed. And whether you say, “check,” or not, the king is still in check, and the king must get out of check. Chess notation for check is the plus sign ‘+’.
There are three ways to get out of check. You may or may not be able to use all three ways, depending on the position. Sometimes you have all three choices, sometimes you have two choices, and sometimes you have only one choice. Here are the three ways to get out of check.
1. Move: Move the king to a square where it is not in check.
2. Block: Put a piece between the king and the piece giving check.
3. Capture: Capture the piece giving check.
What happens if you do not get out of check? Can the opponent capture your king? No! You can never capture the king. If you make a move that leaves your king in check, or that puts your king in check, you must take back your move and make a different move. The opponent should tell you that you have made an illegal move, and you must take it back and make another move. You can also never move a king next to a king, because that would put both of the kings in check. They are like the ends of two magnets that can never be brought together.
What if neither of you see that your king is in check, or that the opponent’s king is in check? Then as soon as one of you sees that their king is in check, they must take back their last move and make a different move. You must get out of check.
2) Checkmate: When you are in check, and there is no way to get out of check, then it is checkmate. You cannot move, block or capture to get out of check. You lose the game. If you put the opponent in checkmate, then you win the game. The game ends with checkmate. You never actually capture the king. The game is over when it is certain that the king could be captured next move. Make sure both players agree that it is checkmate before you reset the pieces. Chess notation for checkmate is the pound or number sign ‘#’ (a double plus sign), or two plus signs ‘++’.
3) Stalemate: When your king is not in check, and any move you make would put your king in check, then it is stalemate. It is your turn to move so you must move, but you cannot move because you cannot put your king in check. The game ends right there. Neither player wins because neither player got checkmate. Stalemate is not a win or a loss; it is a draw, a tie game. Make sure both players agree that it is stalemate before you reset the pieces. Chess notation for stalemate is the dollar sign ‘\$’ (the letter ‘S’ with a line through it).
***
From the book, “TOTAL CHESS: Learn, Teach and Play the Easy 1-2-3 Way,” by John Herron
TOTAL CHESS is your complete guide to chess. It covers everything: rules, strategies, tactics and checkmates.
Everything in chess comes in threes. Three simple strategies are presented for the opening, midgame, endgame, etc. Each lesson is brief and covers one concept in simple language that everyone can read and understand.
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https://bitbucket.org/petsc/petsc/src/fee5f2ce0ecd/src/ts/examples/tutorials/ex34.c
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# petsc / src / ts / examples / tutorials / ex34.c
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 static const char help[] = "An elastic wave equation driven by Dieterich-Ruina friction\n"; /* This whole derivation comes from Erickson, Birnir, and Lavallee [2010]. The model comes from the continuum limit in Carlson and Langer [1989], u_{tt} = c^2 u_{xx} - \tilde\gamma^2 u − (\gamma^2 / \xi) (\theta + \ln(u_t + 1)) \theta_t = −(u_t + 1) (\theta + (1 + \epsilon) \ln(u_t +1)) which can be reduced to a first order system, u_t = v v_t = c^2 u_{xx} - \tilde\gamma^2 u - (\gamma^2 / \xi)(\theta + ln(v + 1))) \theta_t = -(v + 1) (\theta + (1 + \epsilon) \ln(v+1)) */ #include #include #include typedef struct { PetscScalar u,v, th; } Field; typedef struct _User *User; struct _User { PetscReal epsilon; /* inverse of seismic ratio, B-A / A */ PetscReal gamma; /* wave frequency for interblock coupling */ PetscReal gammaTilde; /* wave frequency for coupling to plate */ PetscReal xi; /* interblock spring constant */ PetscReal c; /* wavespeed */ }; #undef __FUNCT__ #define __FUNCT__ "FormRHSFunction" static PetscErrorCode FormRHSFunction(TS ts, PetscReal t, Vec U, Vec F, void *ctx) { User user = (User) ctx; DM dm, cdm; DMDALocalInfo info; Vec C; Field *u, *f; PetscScalar *x; PetscInt i; PetscErrorCode ierr; PetscFunctionBeginUser; ierr = TSGetDM(ts, &dm);CHKERRQ(ierr); ierr = DMGetCoordinateDM(dm, &cdm);CHKERRQ(ierr); ierr = DMGetCoordinatesLocal(dm, &C);CHKERRQ(ierr); ierr = DMDAGetLocalInfo(dm, &info);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, U, &u);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, F, &f);CHKERRQ(ierr); ierr = DMDAVecGetArray(cdm, C, &x);CHKERRQ(ierr); for (i = info.xs; i < info.xs+info.xm; ++i) { const PetscScalar hx = i+1 == info.xs+info.xm ? x[i] - x[i-1] : x[i+1] - x[i]; f[i].u = hx*(u[i].v); f[i].v = -hx*(PetscSqr(user->gammaTilde)*u[i].u + (PetscSqr(user->gamma) / user->xi)*(u[i].th + log(u[i].v + 1))); f[i].th = -hx*(u[i].v + 1)*(u[i].th + (1 + user->epsilon)*log(u[i].v + 1)); } ierr = DMDAVecRestoreArray(dm, U, &u);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(dm, F, &f);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(cdm, C, &x);CHKERRQ(ierr); PetscFunctionReturn(0); } #undef __FUNCT__ #define __FUNCT__ "FormIFunction" static PetscErrorCode FormIFunction(TS ts, PetscReal t, Vec U, Vec Udot, Vec F, void *ctx) { User user = (User) ctx; DM dm, cdm; DMDALocalInfo info; Vec Uloc, C; Field *u, *udot, *f; PetscScalar *x; PetscInt i; PetscErrorCode ierr; PetscFunctionBeginUser; ierr = TSGetDM(ts, &dm);CHKERRQ(ierr); ierr = DMDAGetLocalInfo(dm, &info);CHKERRQ(ierr); ierr = DMGetCoordinateDM(dm, &cdm);CHKERRQ(ierr); ierr = DMGetCoordinatesLocal(dm, &C);CHKERRQ(ierr); ierr = DMGetLocalVector(dm, &Uloc);CHKERRQ(ierr); ierr = DMGlobalToLocalBegin(dm, U, INSERT_VALUES, Uloc);CHKERRQ(ierr); ierr = DMGlobalToLocalEnd(dm, U, INSERT_VALUES, Uloc);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, Uloc, &u);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, Udot, &udot);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, F, &f);CHKERRQ(ierr); ierr = DMDAVecGetArray(cdm, C, &x);CHKERRQ(ierr); for (i = info.xs; i < info.xs+info.xm; ++i) { if (i == 0) { const PetscScalar hx = x[i+1] - x[i]; f[i].u = hx * udot[i].u; f[i].v = hx * udot[i].v - PetscSqr(user->c) * (u[i+1].u - u[i].u) / hx; f[i].th = hx * udot[i].th; } else if (i == info.mx-1) { const PetscScalar hx = x[i] - x[i-1]; f[i].u = hx * udot[i].u; f[i].v = hx * udot[i].v - PetscSqr(user->c) * (u[i-1].u - u[i].u) / hx; f[i].th = hx * udot[i].th; } else { const PetscScalar hx = x[i+1] - x[i]; f[i].u = hx * udot[i].u; f[i].v = hx * udot[i].v - PetscSqr(user->c) * (u[i-1].u - 2.*u[i].u + u[i+1].u) / hx; f[i].th = hx * udot[i].th; } } ierr = DMDAVecRestoreArray(dm, Uloc, &u);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(dm, Udot, &udot);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(dm, F, &f);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(cdm, C, &x);CHKERRQ(ierr); ierr = DMRestoreLocalVector(dm, &Uloc);CHKERRQ(ierr); PetscFunctionReturn(0); } /* IJacobian - Compute IJacobian = dF/dU + a dF/dUdot */ #undef __FUNCT__ #define __FUNCT__ "FormIJacobian" PetscErrorCode FormIJacobian(TS ts, PetscReal t, Vec U, Vec Udot, PetscReal a, Mat *J, Mat *Jpre, MatStructure *str, void *ctx) { User user = (User) ctx; DM dm, cdm; DMDALocalInfo info; Vec C; Field *u, *udot; PetscScalar *x; PetscInt i; PetscErrorCode ierr; PetscFunctionBeginUser; ierr = TSGetDM(ts, &dm);CHKERRQ(ierr); ierr = DMDAGetLocalInfo(dm, &info);CHKERRQ(ierr); ierr = DMGetCoordinateDM(dm, &cdm);CHKERRQ(ierr); ierr = DMGetCoordinatesLocal(dm, &C);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, U, &u);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, Udot, &udot);CHKERRQ(ierr); ierr = DMDAVecGetArray(cdm, C, &x);CHKERRQ(ierr); for (i = info.xs; i < info.xs+info.xm; ++i) { if (i == 0) { const PetscScalar hx = x[i+1] - x[i]; const PetscInt row = i, col[] = {i,i+1}; const PetscScalar dxx0 = PetscSqr(user->c)/hx,dxxR = -PetscSqr(user->c)/hx; const PetscScalar vals[3][2][3] = {{{a*hx, 0,0},{0,0, 0}}, {{0,a*hx+dxx0,0},{0,dxxR,0}}, {{0,0, a*hx},{0,0, 0}}}; ierr = MatSetValuesBlocked(*Jpre, 1, &row, 2, col, &vals[0][0][0], INSERT_VALUES);CHKERRQ(ierr); } else if (i == info.mx-1) { const PetscScalar hx = x[i+1] - x[i]; const PetscInt row = i, col[] = {i-1,i}; const PetscScalar dxxL = -PetscSqr(user->c)/hx, dxx0 = PetscSqr(user->c)/hx; const PetscScalar vals[3][2][3] = {{{0,0, 0},{a*hx, 0,0}}, {{0,dxxL,0},{0,a*hx+dxx0,0}}, {{0,0, 0},{0,0, a*hx}}}; ierr = MatSetValuesBlocked(*Jpre, 1, &row, 2, col, &vals[0][0][0], INSERT_VALUES);CHKERRQ(ierr); } else { const PetscScalar hx = x[i+1] - x[i]; const PetscInt row = i, col[] = {i-1,i,i+1}; const PetscScalar dxxL = -PetscSqr(user->c)/hx, dxx0 = 2.*PetscSqr(user->c)/hx,dxxR = -PetscSqr(user->c)/hx; const PetscScalar vals[3][3][3] = {{{0,0, 0},{a*hx, 0,0},{0,0, 0}}, {{0,dxxL,0},{0,a*hx+dxx0,0},{0,dxxR,0}}, {{0,0, 0},{0,0, a*hx},{0,0, 0}}}; ierr = MatSetValuesBlocked(*Jpre, 1, &row, 3, col, &vals[0][0][0], INSERT_VALUES);CHKERRQ(ierr); } } ierr = DMDAVecRestoreArray(dm, U, &u);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(dm, Udot, &udot);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(cdm, C, &x);CHKERRQ(ierr); ierr = MatAssemblyBegin(*Jpre, MAT_FINAL_ASSEMBLY);CHKERRQ(ierr); ierr = MatAssemblyEnd(*Jpre, MAT_FINAL_ASSEMBLY);CHKERRQ(ierr); if (*J != *Jpre) { ierr = MatAssemblyBegin(*J, MAT_FINAL_ASSEMBLY);CHKERRQ(ierr); ierr = MatAssemblyEnd(*J, MAT_FINAL_ASSEMBLY);CHKERRQ(ierr); } PetscFunctionReturn(0); } #undef __FUNCT__ #define __FUNCT__ "FormInitialSolution" PetscErrorCode FormInitialSolution(TS ts, Vec U, void *ctx) { /* User user = (User) ctx; */ DM dm, cdm; DMDALocalInfo info; Vec C; Field *u; PetscScalar *x; const PetscReal sigma = 1.0; PetscInt i; PetscErrorCode ierr; PetscFunctionBeginUser; ierr = TSGetDM(ts, &dm); ierr = DMGetCoordinateDM(dm, &cdm);CHKERRQ(ierr); ierr = DMGetCoordinatesLocal(dm, &C);CHKERRQ(ierr); ierr = DMDAGetLocalInfo(dm, &info);CHKERRQ(ierr); ierr = DMDAVecGetArray(dm, U, &u);CHKERRQ(ierr); ierr = DMDAVecGetArray(cdm, C, &x);CHKERRQ(ierr); for (i = info.xs; i < info.xs+info.xm; ++i) { u[i].u = 1.5 * PetscExpScalar(-PetscSqr(x[i] - 10)/PetscSqr(sigma)); u[i].v = 0.0; u[i].th = 0.0; } ierr = DMDAVecRestoreArray(dm, U, &u);CHKERRQ(ierr); ierr = DMDAVecRestoreArray(cdm, C, &x);CHKERRQ(ierr); PetscFunctionReturn(0); } #undef __FUNCT__ #define __FUNCT__ "main" int main(int argc, char **argv) { DM dm; TS ts; Vec X; Mat J; PetscInt steps, maxsteps, mx; PetscReal ftime, hx, dt; TSConvergedReason reason; struct _User user; PetscErrorCode ierr; ierr = PetscInitialize(&argc, &argv, NULL, help);CHKERRQ(ierr); ierr = DMDACreate1d(PETSC_COMM_WORLD, DM_BOUNDARY_NONE, -11, 3, 1, NULL, &dm);CHKERRQ(ierr); ierr = DMDASetUniformCoordinates(dm, 0.0, 20.0, 0.0, 0.0, 0.0, 0.0);CHKERRQ(ierr); ierr = DMCreateGlobalVector(dm, &X);CHKERRQ(ierr); ierr = PetscOptionsBegin(PETSC_COMM_WORLD, NULL, "Dynamic Friction Options", ""); { user.epsilon = 0.1; user.gamma = 0.5; user.gammaTilde = 0.5; user.xi = 0.5; user.c = 0.5; ierr = PetscOptionsReal("-epsilon", "Inverse of seismic ratio", "", user.epsilon, &user.epsilon, NULL);CHKERRQ(ierr); ierr = PetscOptionsReal("-gamma", "Wave frequency for interblock coupling", "", user.gamma, &user.gamma, NULL);CHKERRQ(ierr); ierr = PetscOptionsReal("-gamma_tilde", "Wave frequency for plate coupling", "", user.gammaTilde, &user.gammaTilde, NULL);CHKERRQ(ierr); ierr = PetscOptionsReal("-xi", "Interblock spring constant", "", user.xi, &user.xi, NULL);CHKERRQ(ierr); ierr = PetscOptionsReal("-c", "Wavespeed", "", user.c, &user.c, NULL);CHKERRQ(ierr); } ierr = PetscOptionsEnd();CHKERRQ(ierr); ierr = TSCreate(PETSC_COMM_WORLD, &ts);CHKERRQ(ierr); ierr = TSSetDM(ts, dm);CHKERRQ(ierr); ierr = TSSetRHSFunction(ts, NULL, FormRHSFunction, &user);CHKERRQ(ierr); ierr = TSSetIFunction(ts, NULL, FormIFunction, &user);CHKERRQ(ierr); ierr = DMSetMatType(dm, MATAIJ);CHKERRQ(ierr); ierr = DMCreateMatrix(dm, &J);CHKERRQ(ierr); ierr = TSSetIJacobian(ts, J, J, FormIJacobian, &user);CHKERRQ(ierr); ftime = 800.0; maxsteps = 10000; ierr = TSSetDuration(ts, maxsteps, ftime);CHKERRQ(ierr); ierr = FormInitialSolution(ts, X, &user);CHKERRQ(ierr); ierr = TSSetSolution(ts, X);CHKERRQ(ierr); ierr = VecGetSize(X, &mx);CHKERRQ(ierr); hx = 20.0/(PetscReal)(mx-1); dt = 0.4 * PetscSqr(hx) / PetscSqr(user.c); /* Diffusive stability limit */ ierr = TSSetInitialTimeStep(ts, 0.0, dt);CHKERRQ(ierr); ierr = TSSetFromOptions(ts);CHKERRQ(ierr); ierr = TSSolve(ts, X);CHKERRQ(ierr); ierr = TSGetSolveTime(ts, &ftime);CHKERRQ(ierr); ierr = TSGetTimeStepNumber(ts, &steps);CHKERRQ(ierr); ierr = TSGetConvergedReason(ts, &reason);CHKERRQ(ierr); ierr = PetscPrintf(PETSC_COMM_WORLD, "%s at time %G after %D steps\n", TSConvergedReasons[reason], ftime, steps);CHKERRQ(ierr); ierr = MatDestroy(&J);CHKERRQ(ierr); ierr = VecDestroy(&X);CHKERRQ(ierr); ierr = TSDestroy(&ts);CHKERRQ(ierr); ierr = DMDestroy(&dm);CHKERRQ(ierr); ierr = PetscFinalize(); return 0; }
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CC-MAIN-2015-27
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https://www.macroaxis.com/forecast/filter/Polynomial-Regression
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crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00869.warc.gz
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Polynomial Regression Indicator
equity instruments polinomial regression implements a single variable polynomial regression model using the daily prices as the independent variable. The coefficients of the regression for price as well as the accuracy indicators are determined from the period prices.Investors can use prediction functions to forecast Investor Education private prices and determine the direction of financial instruments such as stocks, funds, or ETFs's future trends based on various well-known forecasting models. However, exclusively looking at the historical price movement is usually misleading.
Select Equity
equity instruments polinomial regression implements a single variable polynomial regression model using the daily prices as the independent variable. The coefficients of the regression for price as well as the accuracy indicators are determined from the period prices.
A single variable polynomial regression model attempts to put a curve through the equity instruments historical price points. Mathematically, assuming the independent variable is X and the dependent variable is Y, this line can be indicated as: Y = a0 + a1*X + a2*X2 + a3*X3 + ... + am*Xm
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CC-MAIN-2024-30
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http://algebra-test.com/algebra-help/rational-equations/calculator-worksheets.html
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crawl-data/CC-MAIN-2018-43/segments/1539583512014.61/warc/CC-MAIN-20181018194005-20181018215505-00315.warc.gz
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calculator worksheets
Related topics:
common divisor calculator | finding scale factor | | how to graph equations with two variables | | basic algebra formulas | | simplified radical form of expressions | | solve 4x(x-1)+15=x(3x+4)
Author Message
honahl
Registered: 04.01.2003
From: Ogden, Utah
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Registered: 08.03.2007
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Registered: 27.08.2002
From:
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Registered: 31.08.2001
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Registered: 24.10.2003
From: Where the trout streams flow and the air is nice
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CC-MAIN-2018-43
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latest
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en
| 0.925665
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https://pedagogue.app/activities-to-teach-students-to-identify-representative-random-and-biased-samples/
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crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00236.warc.gz
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# Activities to Teach Students to Identify Representative, Random, and Biased Samples
As students progress through their school years, they begin to encounter more and more statistical concepts. One of the most important of these concepts is the idea of sampling. When dealing with a large data set, it is usually impossible to study every single individual or data point. Instead, one must select a smaller group, or sample, to study in order to draw conclusions about the larger population.
However, not all samples are created equal. Some may be representative of the population they are drawn from, some may be random, and others may be biased, meaning they do not accurately reflect the larger group. Therefore, it is crucial that students learn how to identify these different types of samples and understand why they are important. Here are a few activities that can be used to teach these concepts.
1. The M&M Experiment
For this activity, students will need a large bag of M&Ms and a record sheet to document their observations. Break the class into small groups and give each group a handful of M&Ms. Ask them to separate the candies by color and record the number of each color on their record sheet. Once this is complete, have them compare their results with another group and discuss any differences they notice.
From this activity, students can see that even though they are drawing from the same bag of M&Ms, their samples can differ greatly. This is much like different samples drawn from a population.
2. Sample Scavenger Hunt
In this activity, students will create scavenger hunt lists of items that can be found in their classroom or school. They should work in groups and each group should create two lists: one that is representative of the population (such as items found in a math class) and one that is biased (such as items found only in the teacher’s lounge). Once the lists are created, each group will give their biased list to another group, who will then have to identify which items on the list make it biased.
This activity teaches students to identify samples that are representative versus those that are biased, as well as methods for identifying the source of bias.
3. Random Picture Selection
For this activity, students will need access to a large number of pictures, either in print or online. Ask students to select a picture at random and then determine what population the picture represents. In addition, have them consider whether the picture is biased in any way.
This activity not only teaches students how to identify random samples, but also enables them to understand the difference between pictures that are representative and pictures that contain bias.
4. Sampling in the News
For this ongoing activity, teachers can give students the task of locating and analyzing news articles that involve sampling. Students should be asked to document the article, describe the population being studied, identify the type of sample used, and explain any bias that may be present.
This activity can be done throughout the year as new articles arise. It teaches students to look for sampling concepts in real-world situations and provides a way for them to see how the concepts they are learning in the classroom are applied in the real world.
In conclusion, sampling is an important statistical concept that students must understand in order to draw valid conclusions and make informed decisions. Utilizing activities that engage students through hands-on experiences and real-world situations can help students both understand and retain the material. With a strong understanding of sampling, students will be better equipped to navigate the world of statistics and make informed decisions throughout their lives.
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# Techniques in Market Timing - Price Time Squaring I
In this article I will discuss a technical analysis technique used for forecasting a change in market trend called Price/Time Squaring. It is a technique that many associate with the late great W. D. Gann.
The idea here is to start from a major market top or bottom price and to calculate its 'square'. At first glance you might think this means to simply square the price by multiplying it with itself. Close, but not exactly.
First, to solve for the 'time' itself, you would actually take the 'square root' of the starting price. Take for example the weekly chart of Soybeans. Week of December 5, 2008 put in a major market bottom. Using the low of that week (my charts are reverse-adjusted, so your price may differ somewhat) at 626, you take the square root to arrive at 25.02. Because the result is just above 25, it is now inside the 26th square. To understand this, think of the first square to be from 0 to .99. The second square would then begin at 1 to 1.99, etc.
Starting from the December 5, 2008 weekly price bar as bar zero, you start to count 26 price bars to the right. We arrive at June 5, 2009, which turned out to be one weekly price bar from the major top of week June 12, 2009. In the world of market timing, getting to within a single price bar of a major market trend change is quite the edge to have.
Another misconception that some have is that when you are calculating for Price/Time that when 'time' arrives the 'price' would be at exactly square of the original price. In reality, however, while this may occur at times it is not what should be expected. Rather, when 'time' nears the trader should then be looking at price to reach one of the 'square' levels for which there will be more than one.
## Learn How To Forecast Market Turns!
Using our original Soybeans example on a weekly chart, if we were to calculate price levels based on 360 degrees (that of a circle), we would take the square root value of our starting price (25.02), then add 2 to arrive at 27.02. Why 2? The complete reason is beyond the scope of this article. But if you want to solve for 1 complete cycle revolution as on a Square of Nine (aka Gann Wheel), get the square root of any number, add 2, then re-square it and you will arrive at the next 360 degree value that on the wheel. In other words, you would have gone around the wheel and arrived right back to where you started, but the next cycle level up in value.
Anyway, we now re-square 27.02 and arrive at about 730. Therefore, the first 360 degree level is at 730. The second would be at around 834. The third at 939, etc.
The market topped within a few ticks of the 5th level (1146). On a weekly chart, that's extremely razor sharp.
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