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# Description

I want to use a 50 Watt LED like this one (42mm*40mm)

The space to integrate this component is limited. I have only 55mm*55mm*30mm in total. So the LED + Heatsink size must not be higher than that.

I found one that is just with these dimensions: 55mm*55mm*13mm

However the description says Fit High Power Led 5 watt 10 watt.

The second option is to use a passive 40mm*40mm*20mm heatsink like this

The LED will be put in a tight spot: plastic under and metal (don't know which one) on the sides.

2cm away from the back (from the heatsink) would be some wires.

I can leave the top cover of my device open in order for the heat to go away, no problem for that.

# Questions

Is one of these two heatsink appropriate in my case? Which one would be better? I have no idea how much heat produces an LED like this

What is the risk of using such heatsink with a 50 Watt LED? Could the LED blow up? Burn?

Is it possible the sheath of the wires on the back melt?

• What is the specified thermal resistance of the heatsink in the (no doubt) excellent product data sheets? A 50 watt LED will probably produce somewhere between 49 watts and 51 watts of heat but, in order to verify this you should provide a link to its data sheet AND NOT the bang good page that you are buying the heatsink from. – Andy aka Jan 27 at 12:33
• @Andyaka I just updated my question to add a 2nd option with another passive heatsink. However I can't find any data sheet for both products and no mention of thermal resistance in their description – Jérôme MEVEL Jan 27 at 12:37
• Without that data it is guesswork. – Andy aka Jan 27 at 12:46
• Is it possible the sheath of the wires on the back melt? Sure, the wire isolation could melt. But even if it didn't your LED will not last long. LEDs suffer from heat, it limits their operating life. If you do not keep the LED at a low enough temperature (like 70 - 80 degrees C) it will develop "black spots" that will go open circuit, then more current will flow through the other LED chips so they will fail sooner. I suggest to use temperature control and never let the LED become hotter than a certain temperature. You will get less light but that's the price for bad cooling. – Bimpelrekkie Jan 27 at 13:05

If ambient temperature is 30°C and you want to keep the LED under 80°C, ignoring thermal resistance from junction to heat sink, you'll need a bit less than 1°C/W heat sink. Passive heat sinks with this thermal resistance exist but they're going to be huge and heavy (about half a kilo, and 200x70x50mm).

Using Mouser search engine, passive heat sinks with dimensions similar to your 40x40x20mm example score about 8-12°C/W so that won't work.

So you have to use a fan, like a CPU cooler.

The example heat sink with fan you link from banggood has no specs but it is way too small anyway.

This one could work but it is too big.

I have no idea how much heat produces an LED like this

About 50W, and the LED should be kept reasonably not too hot (like 85°C maximum).

What is the risk of using such heatsink with a 50 Watt LED? Could the LED blow up? Burn?

With the heat sinks given as examples it'll fry in less than a minute.

Is it possible the sheath of the wires on the back melt?

The LED will probably fry first.

You have to do the thermal design first because that will set quite a lot of constraints on your whole design, like the shape of the enclosure, air flow, etc.

If you don't do the thermal design first then you end up painting yourself in a corner.

Your options are either to pick a cheap cooling solution and then design around that, or use something a lot more expensive that will fit in your current design, probably something like a 1U server cpu cooler with heat pipes and a flat turbine fan that will be very loud.

If you need flat, but can tolerate length, then a GPU cooler from an obsolete GPU could work too.

You could put the fan somewhere else and duct the air through the fins of a passive cooler similar to the one you posted, but you'll have to check its thermal resistance versus airspeed.

It's always a good idea to add a temperature control, or at least a failsafe which turns the LED off if it is too hot, in case dust or cat hair clogs the vents.

Remember the air has to flow, so the enclosure needs enough holes to let the cool air in and the hot air out... and it needs to prevent the hot exhaust air from being sucked back into the fan and recycled... I'm not sure how you intend to build this but if the LED has its back against a wall, then you're going to have to blow the air from one side of the heat sink for example.

• Ok I see. Given that I want to go cheap, those Chinese products on BangGood don't have any data sheet available so I can't do the thermal design. I could actually cut a small hole in the plastic at the bottom and put a 50*50mm computer fan on the top of the heatsink. The LED + heatsink are limited in a 50*50mm space but I could add up another extra fan over that to suck air up I think. I will try that and at worst the LED will fry... Seems acceptable to me because it's very cheap. I just requires time. Thanks for your answer – Jérôme MEVEL Jan 28 at 7:59
• You can try and measure the temperature of the heatsink... – bobflux Jan 28 at 9:18

I'm concerned that you have packed the LED in so tightly that you won't get any airflow around it. Leaving just the top open won't be enough to establish good convection...you need a way for cool air to get in at the bottom. You might be better off using the metal sides as the radiator, and thermally connect the LED to the sides with thick metal.

Is one of these two heatsink appropriate in my case? Which one would be better? I have no idea how much heat produces an LED like this

If you had the datasheet you could estimate its efficiency, but probably somewhere around 10-15% of the power becomes light and 85-90% becomes heat for a cheap COB.

The space to integrate this component is limited. I have only 55mm*55mm*30mm in total. So the LED + Heatsink size must not be higher than that.

You've picked the wrong COB for your application. Using a high efficiency COB will more than halve the input power requirements as compared to low end models, while further reducing the fraction of your input power that becomes heat. With a more appropriately chosen COB, you could probably get your thermal load closer to 15W at the same optical power radiated as a cheap generic unit. This will be much more manageable in a small space.


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Microprocessor Design/Microprocessors

Microprocessors

Microprocessors are the devices in a computer which make things happen. Microprocessors are capable of performing basic arithmetic operations, moving data from place to place, and making basic decisions based on the quantity of certain values.

The components of a PC computer. Part number 3 is the CPU.

Types of Processors

The vast majority of microprocessors can be found in embedded microcontrollers. The second most common type of processors are common desktop processors, such as Intel's Pentium or AMD's Athlon. Less common are the extremely powerful processors used in high-end servers, such as Sun's SPARC, IBM's Power, or Intel's Itanium.

Historically, microprocessors and microcontrollers have come in "standard sizes" of 8 bits, 16 bits, 32 bits, and 64 bits. These sizes are common, but that does not mean that other sizes are not available. Some microcontrollers (usually specially designed embedded chips) can come in other "non-standard" sizes such as 4 bits, 12 bits, 18 bits, or 24 bits. The number of bits represent how much physical memory can be directly addressed by the CPU. It also represents the amount of bits that can be read by one read/write operation. In some circumstances, these are different; for instance, many 8 bit microprocessors have an 8 bit data bus and a 16 bit address bus.

• 8 bit processors can read/write 1 byte at a time and can directly address 256 bytes
• 16 bit processors can read/write 2 bytes at a time, and can address 65,536 bytes (64 Kilobytes)
• 32 bit processors can read/write 4 bytes at a time, and can address 4,294,967,295 bytes (4 Gigabytes)
• 64 bit processors can read/write 8 bytes at a time, and can address 18,446,744,073,709,551,616 bytes (16 Exabytes)

General Purpose Versus Specific Use

Microprocessors that are capable of performing a wide range of tasks are called general purpose microprocessors. General purpose microprocessors are typically the kind of CPUs found in desktop computer systems. These chips typically are capable of a wide range of tasks (integer and floating point arithmetic, external memory interface, general I/O, etc). We will discuss some of the other types of processor units available:

General Purpose
A general purpose processing unit, typically referred to as a "microprocessor" is a chip that is designed to be integrated into a larger system with peripherals and external RAM. These chips can typically be used with a very wide array of software.
DSP
A Digital Signal Processor, or DSP for short, is a chip that is specifically designed for fast arithmetic operations, especially addition and multiplication. These chips are designed with processing speed in mind, and don't typically have the same flexibility as general purpose microprocessors. DSPs also have special address generation units that can manage circular buffers, perform bit-reversed addressing, and simultaneously access multiple memory spaces with little to no overhead. They also support zero-overhead looping, and a single-cycle multiply-accumulate instruction. They are not typically more powerful than general purpose microprocessors, but can perform signal processing tasks using far less power (as in watts).
Embedded Controller
Embedded controllers, or "microcontrollers" are microprocessors with additional hardware integrated into a single chip. Many microcontrollers have RAM, ROM, A/D and D/A converters, interrupt controllers, timers, and even oscillators built into the chip itself. These controllers are designed to be used in situations where a whole computer system isn't available, and only a small amount of simple processing needs to be performed.
Programmable State Machines
The most simplistic of processors, programmable state machines are a minimalist microprocessor that is designed for very small and simple operations. PSMs typically have very small amount of program ROM available, limited scratch-pad RAM, and they are also typically limited in the type and number of instructions that they can perform. PSMs can either be used stand-alone, or (more frequently) they are embedded directly into the design of a larger chip.
Graphics Processing Units
Computer graphics are so complicated that functions to process the visuals of video and game applications have been offloaded to a special type of processor known as a GPU. GPUs typically require specialized hardware to implement matrix multiplications and vector arithmetic. GPUs are typically also highly parallelized, performing shading calculations on multiple pixels and surfaces simultaneously.

Types of Use

Microcontrollers and Microprocessors are used for a number of different types of applications. People may be the most familiar with the desktop PC, but the fact is that desktop PCs make up only a small fraction of all microprocessors in use today. We will list here some of the basic uses for microprocessors:

Signal Processing
Signal processing is an area that demands high performance from microcontroller chips to perform complex mathematical tasks. Signal processing systems typically need to have low latency, and are very deadline driven. An example of a signal processing application is the decoding of digital television and radio signals.
Real Time Applications
Some tasks need to be performed so quickly that even the slightest delay or inefficiency can be detrimental. These applications are known as "real time systems", and timing is of the utmost importance. An example of a real-time system is the anti-lock braking system (ABS) controller in modern automobiles.
Throughput and Routing
Throughput and routing is the use of a processor where data is moved from one particular input to an output, without necessarily requiring any processing. An example is an internet router, that reads in data packets and sends them out on a different port.
Sensor monitoring
Many processors, especially small embedded processors are used to monitor sensors. The microprocessor will either digitize and filter the sensor signals, or it will read the signals and produce status outputs (the sensor is good, the sensor is bad). An example of a sensor monitoring processor is the processor inside an antilock brake system: This processor reads the brake sensor to determine when the brakes have locked up, and then outputs a control signal to activate the rest of the system.
General Computing
A general purpose processor is like the kind of processor that is typically found inside a desktop PC. Names such as Intel and AMD are typically associated with this type of processor, and this is also the kind of processor that the public is most familiar with.
Graphics
Processing of digital graphics is an area where specialized processor units are frequently employed. With the advent of digital television, graphics processors are becoming more common. Graphics processors need to be able to perform multiple simultaneous operations. In digital video, for instance, a million pixels or more will need to be processed for every single frame, and a particular signal may have 60 frames per second! To the benefit of graphics processors, the color value of a pixel is typically not dependent on the values of surrounding pixels, and therefore many pixels can typically be computed in parallel.
   \mbox{Clock Time} = \frac{1}{\mbox{Clock Rate}}

Abstraction Layers

Computer systems are developed in layers known as layers of abstraction. Layers of abstraction allow people to develop computer components (hardware and software) without having to worry about the internal design of the other layers in the system. At the highest level are the user-interface programs that people use on their computers. At the lowest level are the transistor layouts of the individual computer components. Some of the layers in a computer system are (listed from highest to lowest):

1. Application
2. Operating System
3. Firmware
4. Instruction Set Architecture
5. Microprocessor Control Logic
6. Physical Circuit Layout

This book will be mostly concerned with the Instruction Set Architecture (ISA), and the Microprocessor Control Logic but we will also describe the Operating System (OS) in brief. Topics above these are typically the realm of computer programmers. The bottom layer, the Physical Circuit Layout is the job of hardware and VLSI engineers.

Operating System

Operating System is a program which acts as an interface between the system user and the computer hardware and controls the execution of application programs. It is the program running at all times on the computer, usually called the Kernel.

ISA

The Instruction Set Architecture is a long name for the assembly language of a particular machine, and the associated machine code for that assembly language. We will discuss this below.

Assembly Language

An assembly language is a small language that contains a short word or "mnemonic" for each individual command that a microcontroller can follow. Each command gets a single mnemonic, and each mnemonic corresponds to a single machine command. Assembly language gets converted (by a program called an "assembler") into the binary machine code. The machine code is specific to each different type of machine.

Common ISAs

Wikibooks contains books about programming in multiple different types of assembly language. For more information about Assembly language, or for books on a particular ISA, see Assembly Language.

Some of the most common ISAs, listed in order of popularity (most popular first) are:

• ARM
• IA-32 (Intel x86)
• MIPS
• Motorola 68K
• PowerPC
• Hitachi SH
• SPARC

Moore's Law

A common law that governs the world of microprocessors is Moore's Law. Moore's Law, originally by Dr. Carver Mead at Caltech, and summarized famously by Intel Founder Gordon Moore. Moore's Law states that the number of transistors on a single chip at the same price will double every 18 to 24 months. This law has held without fail since it was originally stated in 1965. Current microprocessor chips contain millions of transistors and the number is growing rapidly. Here is Moore's summarization of the law from Electronics Magazine in 1965:

The complexity for minimum component costs has increased at a rate of roughly a factor of two per year...Certainly over the short term this rate can be expected to continue, if not to increase. Over the longer term, the rate of increase is a bit more uncertain, although there is no reason to believe it will not remain nearly constant for at least 10 years. That means by 1975, the number of components per integrated circuit for minimum cost will be 65,000. I believe that such a large circuit can be built on a single wafer.
—Gordon Moore

Moore's Law has been used incorrectly to calculate the speed of an integrated circuit, or even to calculate its power consumption, but neither of these interpretations are true. Also, Moore's law is talking about the number of transistors on a chip for a "minimum component cost", which means that the number of transistors on a chip, for the same price, will double. This goes to show that chips for less price can have fewer transistors, and that chips at a higher price can have more transistors. On an economic note, a consequence of Moore's Law is that companies need to continue to innovate and integrate more transistors onto a single chip, without being able to increase prices.

Moore's Law does not require that the speed of the chip increase along with the number of transistors on the chip. However, the two measurements are typically related. Some points to keep in mind about transistors and Moore's Law are:

1. Smaller Transistors typically switch faster then larger transistors.
2. To get more transistors on a single chip, the chip needs to be made larger, or the transistors need to be made smaller. Typically, the transistors get smaller.
3. Transistors tend to leak electrical current as they get smaller. This means that smaller transistors require more power to operate, and they generate more heat.
4. Transistors tend to generate heat as a function of frequencies. Higher clock rates tend to generate more heat.

Moore's law is occasionally misinterpreted to mean that the speed of processors, in hertz will double every 18 months. This is not strictly true, although the speed of processors does tend to increase as transistors are made smaller and more compact. With the advent of multi-core processors, some people have used Moore's law to mean that processor throughput increases with time, which is not strictly the case either (although it is a likely side effect of Moore's law).

Clock Rates

Microprocessors are typically discussed in terms of their clock speed. The clock speed is measured in hertz (or megahertz, or gigahertz). A hertz is a "cycle per second". Each cycle, a microprocessor will perform certain tasks, although the amount of work performed in a single cycle will be different for different types of processors. The amount of work that a processor can complete in a single cycle is measured in "cycles per instruction". For some systems, such as MIPS, there is 1 cycle per instruction. For other systems, such as modern x86 chips, there are typically very many cycles per instruction.

The clock rate is equated as such:

$\mbox{Clock Time} = \frac{1}{\mbox{Clock Rate}}$

This means that the amount of time for a cycle is inversely proportional to the clock rate. A computer with a 1MHz clock rate will have a clock time of 1 microsecond. A modern desktop computer with a 3.2 GHz processor will have a clock time of approximately 3× 10-10 seconds, or 300 picoseconds. 300 picoseconds is an incredibly small amount of time, and there is a lot that needs to happen inside the processor in each clock cycle.

Basic Elements of a Computer

There are a few basic elements that are common to all computers. These elements are:

• CPU
• Memory
• Input Devices
• Output Devices

Depending on the particular computer architecture, these elements may be available in various sizes, and they may be accompanied by additional elements.

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Server Not Reachable.

This may be due to your internet connection or the nubtrek server is offline.

Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.
mathsTrigonometryBasics of Trigonometry

### Importance of Right Angled Triangle

In this page, the importance of right angled triangles in application is explained. This justifies that problems of any polygon is simplified into right-triangles.

click on the content to continue..

Before going into details of right angled triangles, students need to understand one application scenario for the same.

A person walks 3 meter north and continues in the same direction for another 4 meter. The distance he has to walk to return back to the starting position is 3+4 = 7 meter.

Two directional quantities add up in magnitude if they are in same direction.

A person walks 3 meter north and continues towards east for another 4 meter. The distance he has to walk to return back to the starting position is not 3 + 4

Two directional quantities do not add up in magnitude if they are not in same direction. The result magnitude will be smaller to the magnitudes of the quantities being added.

A person walks 1 unit at angle 24^@ and then continues to walk 2.3 unit at angle 43^@.
The problem is illustrated in figure. To find, the distance he has to walk to return back to the starting position, Principles of right angled triangles can help.

The first walk of 1 unit at 24^@ is considered as hypotenuse of right angle triangle. The triangle is specified by three parameters 1. right angle 2. given angle 3. length of one side.

The same for the 2.3 unit at angle 43^@, consider that as a right angled triangle. This two can be used to find the sides for the result bar(OP).

To find the bar(OP) the following steps are followed

Combine x & y axes components using Pythagoras theorem

The distances covered at different angles is a good example of directional quantities. Principles and properties of Right angled triangles are useful to solve problems like that.

Where "sides of right angled triangles" are used in practice?

•  Area of a quadrilateral is sum of area of two triangles. Any polygon can be handled as sum of triangles.

•  Any triangle can be considered as combination of two right angled triangles.

•  Directional quantities are best represented in orthogonal components. The orthogonal components are the legs of the right angled triangles.

•  Some Directional quantities have their basis in rotational elements. An example is the induced electric current in a wire that rotates in a magnetic field. The angular speed of rotation is constant, and the directional quantity, derived out of it at a given angle theta, is a leg of the right angled triangle defined by theta.

•  Length of a leg of a right angled triangle, as a function of angle, serves as a mathematical model for analysis.

Students may skip these if these are not easy to understand now. When you use the trigonometry in applications, you may return to refer and understand the above.

Right angled triangles play a significant role in orthogonal components of quantities with directions and orthogonal components of circular rotation.

Application of Right-Angled-Triangle: Application can be classified into

•  orthogonal components of quantities having direction.

•  orthogonal components of quantities with circular rotation.

•  mathematical model for analysis.

switch to interactive version


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# sudoku algorithm explanation formula

I'm implementing a sudoku solver using human way algorithm. Which have 3 constraint, different number ini row, cell and box.

I googled and I got http://www.emanueleferonato.com/2008/12/09/sudoku-creatorsolver-with-php/. But I cannot understand how this guy get floor($cell / 9) for return_row function or floor(return_row($cell) / 3) * 3 + floor(return_col($cell) / 3) for return_block. I try to figure it out by write down the data in excel and I know there is some pattern like this : [cell] [column] 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 0  But how did he figure it out that the formula was $cell % 9 ?

I want to know, if I don't know the answer for the formula, how can I calculate that ? How can I determine that formula ? What method should I use?

Thanks

--- UPDATE ---

Based on @peterwhy answer now I understand about the row, col and block. I know about is_possible_col and is_possible_row , which show that index = 9*row + col so from here we get row = floor(index/9) and col = index % 9.

Now I confused about is_possible_block function. @peterwhy said :

floor($block/3)*27 + 9*floor($x/3) is 9 times the number of rows before the cell, $x%3 + 3*($block%3) is the number of columns before the cell.

Which make me think like this:

9row+col
9(floor($block/3)*3 + floor($x/3)) + ($x%3 + 3*($block%3))

row = 3row_block+row_x_block

9(3(floor($block/3) + floor($x/3))) + ($x%3 + 3*($block%3))
floor($block/3) = row_block floor($x/3) = row_x_block

col = 3col_block+col_x_block

How col = 3col_block+col_x_block , because what I know, the formula should be like this : col = 3row_block+col.

I know col_x_block mean column position on 0-8 block. And row_block mean row position on 0-2 block.

Please see this to see the detail calculation.

How do you explain this?

## migrated from scicomp.stackexchange.comDec 25 '14 at 15:46

This question came from our site for scientists using computers to solve scientific problems.

• Hi zae and welcome to scicomp! I really don't think this is the best site for your question, since it deals with the algorithm to solve a game. It might do well in the game developers SE or the math SE site. – Paul Dec 25 '14 at 8:21
• @Paul oh, oakay I will ask there. I'm sorry, i'm so confused abaut this question. I already ask in stackoverflow but someone said this is math question. – john Dec 25 '14 at 8:23
• If the question is "how do I come up with these formulas", it's just logic/mathematization and I'd ask on math.se. That site is less research-oriented and more suited to explaining basic maths; one can get really insightful answers on these mental processes there. – Federico Poloni Dec 25 '14 at 15:31
• The author: It’s an array (line 193) with 81 values representing positions from 0 to 80. The value at n-th position is the one you will find on the table at row floor(n/9) and column n%9. – peterwhy Dec 25 '14 at 15:53

floor($cell/9) and $cell % 9 are just how (integer) quotient and remainder are calculated in programming language. The $cell indices are arranged row by row. The row number can be obtained as a quotient and the column number can be obtained as a remainder. Take the cell numbered$42$as an example. By division, $$42 = 9\cdot4 + 6$$ hence the cell is located on the 4th row (0-based) and the 6th column (also 0-based). The block number is just sightly more complicated. The first part, floor(return_row($cell) / 3) * 3, first calculates the quotient of the row number divided by $3$. So rows $0$ to $2$ are in the first third, and $6$ to $8$ are in the last third. Then this "group of 3 rows" index is multiplied by $3$, because there are 3 blocks for each group of 3 rows.
The second part, floor(return_col($cell) / 3), again find the quotient of the column number divided by$3$. Taking the same example of cell$42$, which is on row 4 and column 6. The row number $$4 = 3\cdot\color{red}1 + 1;\;\left\lfloor\frac43\right\rfloor = \color{red}1$$ the quotient$\color{red}1$means the cell is on the 1st "group of 3 rows" (0-based). And the column number $$6 = 3\cdot\color{blue}2+0;\;\left\lfloor\frac63\right\rfloor = \color{blue}2$$ the quotient$\color{blue}2$means the cell is on the 2nd "group of 3 columns" (also 0-based). Combining this two information, the block where cell$42$is at is assigned a block index $$\color{red}1\cdot 3 + \color{blue}2 = 5$$ Again, take$42$as an example: $$\begin{array}{l|l|ccc|ccc|ccc|} \text{Row}&\text{Cells}&0&1&2&3&4&5&6&7&8\\ \hline 0&0-8&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc\\ 1&9-17&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc\\ 2&18-26&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc\\ \hline 3&27-35&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc\\ 4&36-44&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\color{red}\bigcirc&\cdots\\ \end{array}$$ Notice the red$\color{red}\bigcirc$is the 43rd$\bigcirc$. Elementary division says$\lfloor42/9\rfloor=4$gives how many complete groups of$9$the$42$black$\bigcirc$s can form: $$\begin{array}{l|l|ccc|ccc|ccc|} \text{Row}&\text{Cells}&0&1&2&3&4&5&6&7&8\\ \hline \color{blue}0&0-8&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc\\ \color{green}1&9-17&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc\\ \color{blue}2&18-26&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc&\color{blue}\bigcirc\\ \hline \color{green}3&27-35&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc&\color{green}\bigcirc\\ 4&36-44&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\bigcirc&\color{red}\bigcirc&\cdots\\ \end{array}$$ And the remaining$42\bmod 9=6$black$\bigcirc$s count how many columns are before the red$\color{red}\bigcirc$. Since the row number and column number are$0$-based, the quotient$4$and the remainder$6$also denote the row and column numbers respectively. • I know how they work. But I don't know to make the formula. For example if I have another table data, for another case, I want to make formula, how can I determine the formula for that pattern. I tried with arhitmatic and geometry but it fails. Also I tried regression, but cam with non integer formula.. – john Dec 26 '14 at 3:04 • What are some examples of your table data and formula? – peterwhy Dec 26 '14 at 3:06 • This is for sudoku case : docs.google.com/spreadsheets/d/… . The number marked with red is the pattern. It's like f(x) function. x is the black number, and f(x) is the red one. I just confuse how to derive the formula like f(x) = x mod 9 or something else. Thanks – john Dec 26 '14 at 4:11 • Which is what I wrote in my answer, find the quotient and the remainder when the cell number is divided by$9$. Focus on the$9\times 9$grid in your spreadsheet and the cell$42$. The row number, or how many complete rows are before$42$, is$\lfloor 42/9\rfloor$. The remainder of the division,$42 \bmod 9$, gives how many complete columns are before$42$. – peterwhy Dec 26 '14 at 4:12 • I know how the function work, I just don't know how to explain that from the table, how can I get the function. I mean, how to get the formula if I dont have the answer (I don't know floor($cell/9) or $cell % 9). I just have the table. How can I explain that from the table I can conclude that the pattern was$cell%9, is that only by guessing or there is a math method ? It also to explain the block formula. – john Dec 26 '14 at 4:18


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# Mathematical functions

## ABS

Syntax:ABS( number )

Returns the absolute value of number.

## ACOS

Syntax:ACOS( number )

Returns the arccosine of number in radians.

## ASIN

Syntax:ASIN( number )

Returns the arcsine of number in radians.

## ATAN

Syntax:ATAN( number )

Returns the arctangent of number in radians.

## ATAN2

Syntax:ATAN2( x, y )

Returns the arctangent in radians for the specified coordinates x and y.

## CEILING

Syntax:CEILING( number )

Rounds the value up to the nearest integer.

## COMPARE

Syntax:COMPARE( left, right, epsilon )

Returns:

• 0 if left and right differs by not more than epsilon.
• -1 if left is less than right by more than epsilon.
• 1 if left is greater than right by more than epsilon.

## COS

Syntax:COS( number )

Returns the cosine of number in radians.

## COT

Syntax:COT( number )

Returns the cotangent of number in radians.

## DEGREES

Syntax:DEGREES( radians )

Converts radians to degrees.

## DIV

Syntax:DIV( number_1, number_2 )

Divides number_1 by number_2. The result is rounded down to the nearest integer.

## EXP

Syntax:EXP( number )

Returns the result of raising the number 'e' to the power of number.

## FLOOR

Syntax:FLOOR( number )

Rounds the value down to the nearest integer.

## GREATEST

Syntax:GREATEST( value_1, value_2, value_3 [ , ... ] )

Returns the greatest value.

See also LEAST.

Depending on the specified data type, it returns:

• The greatest number.
• The last string in alphabetical order.
• The latest date.
• TRUE when selecting between TRUE and FALSE for Boolean type.

## LEAST

Syntax:LEAST( value_1, value_2, value_3 [ , ... ] )

Returns the smallest value.

See also GREATEST.

Depending on the specified data type, it returns:

• The smallest number.
• The first string in alphabetical order.
• The earliest date.
• FALSE when selecting between TRUE and FALSE for Boolean type.

## LN

Syntax:LN( number )

Returns the natural logarithm of the number number. Returns 'NULL' if the number is less than or equal to 0.

## LOG

Syntax:LOG( value, base )

Returns the logarithm of value to base base. Returns 'NULL' if the number value is less than or equal to 0.

## LOG10

Syntax:LOG10( number )

Returns the logarithm of the number number to base 10. Returns 'NULL' if the number is less than or equal to 0.

## PI

Syntax:PI()

Returns PI. The accuracy depends on the data source.

## POWER

Syntax:POWER( base, power )

Raises base to the power of power.

## RADIANS

Syntax:RADIANS( degrees )

Converts degrees degrees to radians.

## ROUND

Syntax:ROUND( number [ , precision ] )

Rounds the number number to the number of decimal digits specified in precision.
If the number precision is omitted, number is rounded to the nearest integer.

## SIGN

Syntax:SIGN( number )

Returns the sign of the number number:

• -1 if the number is negative.
0 if the number is zero.
• 1 if the number is positive.

## SIN

Syntax:SIN( number )

Returns the sine of number in radians.

## SQRT

Syntax:SQRT( number )

Returns the square root of the specified number.

## SQUARE

Syntax:SQUARE( number )

Returns the number number raised to the power of 2.

## TAN

Syntax:TAN( number )

Returns the tangent of number in radians.


text:
Spatiotemporal-Aware Augmented Reality: Redefining HCI in Image-Guided Therapy

Suboptimal interaction with patient data and challenges in mastering 3D anatomy based on ill-posed 2D interventional images are essential concerns in image-guided therapies. Augmented reality (AR) has been introduced in the operating rooms in the last decade; however, in image-guided interventions, it has often only been considered as a visualization device improving traditional workflows... As a consequence, the technology is gaining minimum maturity that it requires to redefine new procedures, user interfaces, and interactions. The main contribution of this paper is to reveal how exemplary workflows are redefined by taking full advantage of head-mounted displays when entirely co-registered with the imaging system at all times. The proposed AR landscape is enabled by co-localizing the users and the imaging devices via the operating room environment and exploiting all involved frustums to move spatial information between different bodies. The awareness of the system from the geometric and physical characteristics of X-ray imaging allows the redefinition of different human-machine interfaces. We demonstrate that this AR paradigm is generic, and can benefit a wide variety of procedures. Our system achieved an error of $4.76\pm2.91$ mm for placing K-wire in a fracture management procedure, and yielded errors of $1.57\pm1.16^\circ$ and $1.46\pm1.00^\circ$ in the abduction and anteversion angles, respectively, for total hip arthroplasty. We hope that our holistic approach towards improving the interface of surgery not only augments the surgeon's capabilities but also augments the surgical team's experience in carrying out an effective intervention with reduced complications and provide novel approaches of documenting procedures for training purposes. read more

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text:
# A variation of the Banach fixed-point theorem

Let $(X, d)$ be complete metric space, $q \in [0, 1)$ be a real number, and $f$ be a map that satisfies $$d(f(x), f(y)) \leq q \cdot d(x, y)$$ for all $x, y \in X$. Then, Banach fixed-point theorem says that there is a unique fixed-point of $f$ in $X$.

I am interested in figuring out what happens if we relax the condition on $f$ to require, for all $x, y \in X$:

$$d(f^n(x), f^m(y)) \leq q \cdot d(x, y)\tag{1}$$ for some $n, m \geq 0.$

In general, there need not be a fixed-point when the constraint on $f$ is weakened to the above form. For example, points on a periodic orbit satisfy this weakened constraint, yet they never converge to a fixed-point. This means that, even though we conserve the notion of pairwise convergence, it is not enough to guarantee the existence of a fixed-point.

What can we say about maps that exhibit a fixed point under $(1)$? In other words, what properties do such maps have?

-
What exactly is the question? The way it is stated, the answer seems to be "maps satisfying (1) and having a fixed point". Just (1) guarantees, for example, that $f^{m+n}$ has a fixed point. –  Alex Degtyarev Mar 6 at 0:02
@AlexDegtyarev: First of all, numbers $m$ and $n$ are not fixed. Each $x,y \in X$ might have a different $m$ and $n$. Also, I don't understand what you mean by "maps satisfying (1) and having a fixed point". It seems like you are answering the question circularly. –  Mehmet Ozan Kabak Mar 6 at 0:12
That's exactly what I mean when I ask "what's the question". I answered the question as it was stated. (You can take it as a joke as well.) Speaking about $m$, $n$, then you should change the order of the quantifiers in (1), as otherwise it's misleading. –  Alex Degtyarev Mar 6 at 0:23
I will rephrase (1) to make it more clear. Anyway, I am interested in finding out the properties of maps that exhibit a fixed point whenever (1) holds. –  Mehmet Ozan Kabak Mar 6 at 0:27
If $m=n$, then it works. –  Suvrit Mar 6 at 2:15

If you really mean $m=m(x,y)$ and $n=n(x,y)$ (as per your comments), your assumption is equivalent to say that every pair of orbits have minimum distance $0$: for any $x\in X$ and $y\in X$ $$\inf_{ n\in\mathbb{N}\atop m\in\mathbb{N}}d\big((f^n(x),f^m(y)\big)=0\, .$$
A map $f$ satisfying this condition may be discontinuous. Even assuming continuity, it may have no fixed point: for instance, any irrational rotation on the unit circle (one can then fix $m=0$).


text:
Skip to main content

27.4D: Development of the Cardiovascular System

The circulatory system develops initially via vasculogenesis, with the arterial and venous systems developing from distinct embryonic areas.

Learning Objectives

• Outline the development of the cardiovascular system

Key Points

• The aortic arches are a series of six, paired, embryological vascular structures that give rise to several major arteries. The first and second arches disappear early. The third arch becomes the carotid artery.
• The fourth right arch forms the right subclavian artery, while the fourth left arch forms the arch of the aorta. The fifth arch disappears on both sides.The proximal part of the sixth right arch persists as the proximal right pulmonary artery. The sixth left arch gives off the left pulmonary artery.
• Approximately 30 posterolateral branches arise off the dorsal aortae and will form the intercostal arteries, the upper and lower extremity arteries, the lumbar arteries, and the lateral sacral arteries. The lateral branches of the aorta form the definitive renal, suprarenal, and gonadal arteries.
• The ventral branches consist of the vitelline and umbilical arteries. The vitelline arteries form the celiac, and superior and inferior mesenteric arteries of the gastrointestinal tract. After birth, the umbilical arteries will form the internal iliac arteries.
• The venous system develops from the vitelline veins, umbillical veins, and the cardinal veins, all of which empty into the sinus venosus.

Key Terms

• sinus venosus: A large quadrangular cavity that precedes the atrium on the venous side of the chordate heart. In humans, it exists distinctly only in the embryonic heart, where it is found between the two venae cavae.
• aortic arches: Also known as pharyngeal arch arteries, this is a series of six, paired, embryological vascular structures that give rise to several major arteries. They are ventral to the dorsal aorta.
• cardinal vein: The precardinal veins or anterior cardinal veins contribute to the formation of the internal jugular veins and, together with the common cardinal vein, form the superior vena cava. In an anastomosis by anterior cardinal veins, the left brachiocephalic vein is produced.

Vasculogenesis

The human arterial system originates from the aortic arches and from the dorsal aortae starting from week 4 of embryonic life.
The development of the circulatory system initially occurs by the process of vasculogenesis, the formation of new blood vessels when there are no preexisting ones.

Embryonic cardiovascular system: A profile view of a human embryo estimated at twenty or twenty-one days old.

Vasculogenesis is when endothelial precursor cells (angioblasts) migrate and differentiate in response to local cues (such as growth factors and extracellular matrix) to form new blood vessels. The human arterial and venous systems develop from different embryonic areas.

Aortic Arches

The aortic arches—or pharyngeal arch arteries—are a series of six, paired, embryological vascular structures that give rise to several major arteries. They are ventral to the dorsal aorta and arise from the aortic sac.

Arches 1 and 2

Aortic arches: A schematic of the aortic arches and their arterial destinations.

The first and second arches disappear early, but the dorsal end of the second gives origin to the stapedial artery, a vessel that atrophies in humans, but persists in some mammals. It passes through the ring of the stapes and divides into supraorbital, infraorbital, and mandibular branches that follow the three divisions of the trigeminal nerve.

The infraorbital and mandibular branches arise from a common stem, the terminal part of which anastomoses with the external carotid. On the obliteration of the stapedial artery, this anastomosis enlarges and forms the internal maxillary artery; the branches of the stapedial artery are now branches of this vessel.

The common stem of the infraorbital and mandibular branches passes between the two roots of the auriculotemporal nerve and becomes the middle meningeal artery. The original supraorbital branch of the stapedial artery is represented by the orbital branches of the middle meningeal artery.

Arches 3 and 4

The third aortic arch constitutes the commencement of the internal carotid artery, and is named the carotid arch. The fourth right arch forms the right subclavian artery as far as the origin of its internal mammary branch. The fourth left arch constitutes the arch of the aorta between the origin of the left carotid artery and the termination of the ductus arteriosus.

Arches 5 and 6

The fifth arch disappears on both sides.The proximal part of the sixth right arch persists as the proximal part of the right pulmonary artery, while the distal section degenerates. The sixth left arch gives off the left pulmonary artery and forms the ductus arteriosus.

This duct remains during fetal life, but closes within the first few days after birth due to increased O2 concentration. This causes the production of bradykinin which causes the ductus to constrict, occluding all flow. Within one to three months, the ductus is obliterated and becomes the ligamentum arteriosum.

Aortic Branches

The dorsal aortae are initially bilateral and then fuse to form the definitive dorsal aorta. Approximately 30 posterolateral branches arise off the aorta and will form the intercostal arteries, upper and lower extremity arteries, lumbar arteries, and the lateral sacral arteries.

The lateral branches of the aorta form the definitive renal, suprarenal, and gonadal arteries. Finally, the ventral branches of the aorta consist of the vitelline arteries and umbilical arteries.

The vitelline arteries form the celiac, and superior and inferior mesenteric arteries of the gastrointestinal tract. After birth, the umbilical arteries will form the internal iliac arteries.

The human venous system develops mainly from the vitelline, umbilical, and cardinal veins, all of which empty into the sinus venosus. The venous system arises during the fourth to eighth weeks of human development.

CLINICAL EXAMPLE

Most defects of the great arteries arise as a result of the persistence of aortic arches that normally should regress or due to the regression of arches that normally should not.

A double aortic arch occurs with the development of an abnormal right aortic arch, in addition to the left aortic arch, forming a vascular ring around the trachea and esophagus, which usually causes difficulty breathing and swallowing.

Occasionally, the entire right dorsal aorta abnormally persists and the left dorsal aorta regresses. In this case, the right aorta will have to arch across from the esophagus, causing difficulty breathing or swallowing.

27.4D: Development of the Cardiovascular System is shared under a CC BY-SA license and was authored, remixed, and/or curated by LibreTexts.

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text:
# Measurements of groomed-jet substructure of charm jets tagged by ${\rm D}^0$ mesons in proton-proton collisions at $\sqrt{s}$ = 13 TeV

Understanding the role of parton mass and Casimir colour factors in the quantum chromodynamics parton shower represents an important step in characterising the emission properties of heavy quarks. Recent experimental advances in jet substructure techniques have provided the opportunity to isolate and characterise gluon emissions from heavy quarks. In this work, the first direct experimental constraint on the charm-quark splitting function is presented, obtained via the measurement of the groomed shared momentum fraction of the first splitting in charm jets, tagged by a reconstructed ${\rm D}^0$ meson. The measurement is made in proton--proton collisions at $\sqrt{s}$ = 13 TeV, in the low jet transverse-momentum interval of $15 \leq p_{\rm T}^{\rm jet~ch} <~ 30$ GeV/$c$ where the emission properties are sensitive to parton mass effects. In addition, the opening angle of the first perturbative emission of the charm quark, as well as the number of perturbative emissions it undergoes, are reported. Comparisons to measurements of an inclusive-jet sample show a steeper splitting function for charm quarks compared to gluons and light quarks. Charm quarks also undergo fewer perturbative emissions in the parton shower, with a reduced probability of large-angle emissions.

Submitted to: PRL
e-Print: arXiv:2208.04857 | PDF | inSPIRE
CERN-EP-2022-171


text:
# Find $\lim\limits_{x \rightarrow \infty} x e^{x^2} \int_x^{\infty}e^{-u^2}du$

$$\lim\limits_{x \rightarrow \infty} x e^{x^2} \int_x^{\infty}e^{-u^2}du$$

That $x$ on the lower bound of the Gaussian integral is really confusing.

Can we solve it without the fact $\int_{-\infty}^{\infty}e^{-u^2}du = \sqrt\pi \ge \int_x^{\infty}e^{-u^2}du$ ?

(original question have been edited to fix mistakes pointed out in the comments by anon)

• Do you mean $\int_x^\infty e^{-u^2}du$? You can't have $x$ be the integral's dummy variable and have $x$ appear outside of the integral. You just can't. – anon Jul 22 '15 at 5:30
• no but i'll edit to make it clearer. sorry this is exactly how i got it on the exam .... ? why its not possible ? – Boris Morozov Jul 22 '15 at 5:30
• Then I'll fix your exam for you - edited. Do you think $\lim\limits_{x\to\infty}x\cdot x^{-1}$ is $\infty$ because $x\to\infty$ but $x^{-1}$ is bounded for $x>1$? Notice how $x^{-1}\to0$ though! – anon Jul 22 '15 at 5:37
• i wish you were at the exam to fix it :D your right – Boris Morozov Jul 22 '15 at 5:38

Why not rewrite $$x e^{x^2} \int_x^{\infty}e^{-u^2}du=\frac{\int_x^{\infty}e^{-u^2}du}{\frac 1xe^{-x^2}} =\frac u v$$ Use the fundamental theorem of calculus and L'Hospital $$u'=-e^{-x^2}$$ (this comes from the fundamental theorem of calculus) $$v'=-2 e^{-x^2}-\frac{e^{-x^2}}{x^2}$$ $$\frac{u'}{v'}= \frac{-e^{-x^2}}{-2 e^{-x^2}-\frac{e^{-x^2}}{x^2}}=\frac{1}{2-\frac 1 {x^2} }$$
This was the short way. Now, the long long way : assuming that you know the error function $$\int_x^{\infty}e^{-u^2}du=\frac{1}{2} \sqrt{\pi } \,\text{erfc}(x)$$ So $$A=x e^{x^2} \int_x^{\infty}e^{-u^2}du=\frac{1}{2} \sqrt{\pi } e^{x^2} x \, \text{erfc}(x)$$ For large values of the argument, you could find the expansion $$\text{erfc}(x)=e^{-x^2} \left(\frac{1}{\sqrt{\pi } x}-\frac{1}{2 \sqrt{\pi } x^3}+O\left(\left(\frac{1}{x}\right)^4\right)\right)$$ which makes $$A=\frac{1}{2}-\frac{1}{4 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$


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# Kernel of the Extension of a Bounded Linear Operator

Suppose $T\colon E\to F$ is a bounded linear operator between Banach spaces. Moreover let $i\colon E\to E’$ be a dense, compact inclusion of $E$ into some other Banach space $E’$. Finally assume that there exists a positive constant $c\in\mathbb{R}$ such that $\|T(x)\|_F\leq c\|x\|_{E’}$ for all $x\in E'$.

By uniform continuity there exists a unique bounded linear operator $T’\colon E’\to F$ such that $T=T’\circ i$. How is $\operatorname{ker}T$ related to $\operatorname{ker}T’$? My guess is that $i(\operatorname{ker}T)$ is in general not dense in $\operatorname{ker}T’$. Can someone give an example where it is not dense? Or a proof that it is dense? Does the situation change if the spaces are assumed to be Hilbert spaces?

As you suspected $i($ker $T)$ needn't be dense in ker $T'$. Take $E=\ell^2$ and $E'=\lbrace (x_n)_{n\in\mathbb N}: (x_n/n)\in \ell^2\rbrace$ endowed with the obvious Hilbert norm. The inclusion $E\hookrightarrow E'$ is dense and compact. Define $T':E'\to E'$, $(x_n)_{n\in\mathbb N} \mapsto (x_n-x_{n+1})_{n\in\mathbb N}$. This is a continuous linear operator with $(1,1,1,\ldots)\in$ ker $T'$ and the restriction $T=T'|_{\ell^2}$ is injective.


text:
Question

# The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickl

Other
The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing - I am sorry that I am not able to provide the drawing at this time as my scanner is broken). Suppose that your loft the ball with an initial speed of 15.0 m/s, at an angle of 50.0 (degrees) above the horizontal. At this instant your opponent is 10.0 m away from

2021-02-26

The components of the velocity are $$\displaystyle{v}_{{{o}{y}}}={v}_{{o}}{\sin{{50}}}={15}{\sin{{50}}}={11.5}\ \frac{{m}}{{s}}$$
$$\displaystyle{v}_{\otimes}={v}_{{o}}{\cos{{50}}}={9.64}\ \frac{{m}}{{s}}$$ The time taken by the ball to reach the point P ' is
$$\displaystyle{S}={v}_{{o}}{y}{t}-{\left({\frac{{{1}}}{{{2}}}}\right)}{>}^{{2}}$$
$$\displaystyle{0}{r}{2.10}={11.5}{t}-{\left({\frac{{{1}}}{{{2}}}}\right)}{\left({9.8}\right)}{t}^{{2}}$$
$$\displaystyle{4.9}{t}^{{2}}-{11.5}{t}+{2.1}={0}$$
This Quadractic equation in t, is solved for the values of t
$$\displaystyle{t}={2.15}\ {s}$$ and 0.2 s
as the opponent begins to move only after 0.3s later, we taket = 2.15s
Therefore, he catches the ball at time t'=(2.15-0.3)s
The total horizontal distance
$$\displaystyle{x}={v}_{\otimes}\cdot{t}$$
$$\displaystyle{\left({10}+{x}\right)}={v}_{\otimes}\cdot{2.15}{s}$$
$$\displaystyle={9.64}\cdot{2.15}'$$
The value of ' x ' is calculated.
Thus the required speed of the opponent $$\displaystyle{v}={\frac{{{x}}}{{{t}'}}}$$


text:
# Exponential operators

1. Dec 31, 2015

### Happiness

Why is (5.302) an approximation instead of an equality?

Let $T$ be the operator $\frac{p_x^2}{2m}$.

By the law of indices, we should have $e^{-\frac{i}{\hbar}(T+V)\Delta t}=e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}$ exactly. Is it because $T$ and $V$ do not commute? So the correct equation should be $e^{-\frac{i}{\hbar}(T+V)\Delta t}=\frac{1}{2}[e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}+e^{-\frac{i}{\hbar}V\Delta t} e^{-\frac{i}{\hbar}T\Delta t}]$?

But if this is so, shouldn't (5.302) be correct up to terms of order $\Delta t$ instead of order $(\Delta t)^2$ as claimed by the book?

Last edited: Dec 31, 2015
2. Dec 31, 2015

### atyy

3. Dec 31, 2015

### strangerep

Yes.

Not in general. Check out the Baker-Campbell-Hausdorff formula(s).

4. Dec 31, 2015

### Happiness

When I work out the term of order $(\Delta t)^2$, the left hand side gives $\frac{1}{2}(-\frac{i}{\hbar})^2(T^2+TV+VT+V^2)(\Delta t)^2$ but the right hand side gives $\frac{1}{2}(-\frac{i}{\hbar})^2(T^2+2TV+V^2)(\Delta t)^2$. They are not equal. So is the book wrong?

5. Jan 1, 2016

### strangerep

Which book is it?

6. Jan 1, 2016

### Happiness

It's Quantum Mechanics 2nd ed. by Bransden & Joachain, page 243.

7. Jan 1, 2016

### samalkhaiat

To 2nd order, the BCH identity reads
$$e^{\lambda (T+V)/N} = e^{\lambda T/N}\ e^{\lambda V/N} - \frac{1}{2}(\frac{\lambda}{N})^{2} [ T , V ]$$
However, one can prove that, as $N \to \infty$,
$$\left( e^{\lambda (T+V)/N} \right)^{N} = \left( e^{\lambda T/N}\ e^{\lambda V/N} \right)^{N}$$
This helps you to rewrite the Green's function as
$$\langle x | e^{\lambda H} | y \rangle = \langle x | ( e^{\lambda H /N} )^{N}| y \rangle = \langle x | ( e^{\lambda T /N} e^{\lambda V /N} )^{N}| y \rangle ,$$
which leads to the Path Integral when you insert the identity operators
$$\int dx_{j} |x_{j}\rangle \langle x_{j}| = I, \ \ \ \int dp |p \rangle \langle p| = I.$$

8. Jan 1, 2016

### atyy

I think you are right and the book is wrong.

9. Jan 1, 2016

### Happiness

But this equation can be simplified to one independent of $N$:

$$e^{\lambda (T+V)} = e^{\lambda T}\ e^{\lambda V}$$

So if it is true as $N \to \infty$, then shouldn't it be true for all values of $N$? But this is not true, so the equation should not be true for any value of $N$?

10. Jan 1, 2016

### blue_leaf77

If $T$ and $V$ do not commute, you can't interchange $e^{\lambda T}$ and $\ e^{\lambda V}$.

11. Jan 1, 2016

### Happiness

So that means
$$\left( e^{\lambda (T+V)/N} \right)^{N} = \left( e^{\lambda T/N}\ e^{\lambda V/N} \right)^{N}$$
is false even as $N \to \infty$?

I think the correct statement should instead be as $N \to \infty$,

$$e^{\lambda (T+V)/N} = e^{\lambda T/N}\ e^{\lambda V/N}$$

Is this correct?

Last edited: Jan 1, 2016
12. Jan 1, 2016

### blue_leaf77

What I implied is that, I don't think you can reduce the RHS of $\left( e^{\lambda (T+V)/N} \right)^{N} = \left( e^{\lambda T/N}\ e^{\lambda V/N} \right)^{N}$ to $\left( e^{\lambda T}\ e^{\lambda V} \right)$ because in doing so, you must have brought $N$ in the outer most to inside, which means you are interchanging some pairs of the two exponential operators in the process.

13. Jan 1, 2016

### Happiness

Since (https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula)

then to the second order, $\ e^Xe^Y=e^{X+Y+\frac{1}{2}[ X , Y ]}$.

So shouldn't it be

$e^{\lambda (T+V)/N} = e^{\lambda T/N}\ e^{\lambda V/N}\ e^{\frac{1}{2}(\frac{\lambda}{N})^{2} [ T , V ]}$?

14. Jan 1, 2016

### strangerep

@Happiness,

Sam has (almost) given you the crucial missing piece in post #7. I'm reasonably sure Sam meant "in the limit as $N\to\infty$". In general, some properties might be true only in the sense of limits, though untrue for finite values of $N$.

I should have also given you the Wiki link to the Lie-Trotter product formula. I.e., $$e^{A+B} ~=~ \lim_{N\to\infty} \Big( e^{A/N} e^{B/N} \Big)^N ~.$$The way your textbook expresses it seems a bit misleading, imho, but the proper Lie-Trotter formula can be applied more successfully in the path integral material.

15. Jan 2, 2016

### samalkhaiat

No. You can write
$$e^{H} = (e^{H/N})^{N} ,$$
But
$$(e^{T/N}e^{V/N})^{N} = (e^{T/N}e^{V/N}) \cdots N \mbox{-factors} \cdots (e^{T/N}e^{V/N}) .$$

16. Jan 2, 2016

### samalkhaiat

No. It is true and has a name. It is called the Lie-Trotter product formula. You should not jump to a wrong conclusion just because you can not prove it. Do you think I lied to you when I said “we can prove that”?

Yes, it is correct because as I said
$$e^{T/N}e^{V/N} - e^{(T+V)/N} = \frac{1}{2N^{2}} [ T , V ] \sim \mathcal{O}(\frac{1}{N^{2}}) . \ \ \ \ (1)$$
So,
$$\lim_{N \to \infty} \left( e^{T/N}e^{V/N} - e^{(T+V)/N} \right) = 0 .$$
But this does not help you in setting up the Path integral representation of the propagator. In fact, we can use this result to prove the Lie-Trotter formula which is essential in deriving the path integral. Let me show you how. Let $A$ and $B$ be two operators, and consider the difference $A^{N} - B^{N}$. By adding and subtracting equal terms, I can write the following identity
$$\begin{equation*} \begin{split} A^{N} - B^{N} =& (AB^{N-1} - B^{N}) + (A^{2}B^{N-2} - AB^{N-1}) \\ & + (A^{3}B^{N-3} - A^{2}B^{N-2}) + (A^{4}B^{N-4} - A^{3}B^{N-3}) \\ & + \cdots + (A^{N} – A^{N-1}B) . \end{split} \end{equation*}$$
This can be rewritten as
$$\begin{equation*} \begin{split} A^{N} - B^{N} =& (A - B ) B^{N-1} + A (A - B ) B^{N-2} \\ & + A^{2} (A - B ) B^{N-3} + A^{3} (A - B ) B^{N-4} \\ & + \cdots + A^{N-1} ( A - B ) . \end{split} \end{equation*}$$
Okay, now take
$$A = e^{T/N}e^{V/N} , \ \ B = e^{(T+V)/N} .$$
But, we know that
$$A - B = \frac{1}{2N^{2}} [ T , V ] \sim \mathcal{O}(N^{-2}) .$$
So, $(A-B) \to 0$ as $N \to \infty$. Now, the above identity consists of $N$ terms each of which has the factor $(A-B)$ which is of order $(1/N^{2})$. Hence, in the limit $N \to \infty$, the difference $A^{N} - B^{N}$ is zero. Thus, we obtain the Lie-Trotter formula
$$\lim_{N \to \infty} \left( e^{T/N}e^{V/N} \right)^{N} = \lim_{N \to \infty} \left( e^{(T+V)/N} \right)^{N} = e^{(T+V)} .$$

17. Jan 2, 2016

### samalkhaiat

This equation is correct to all orders provided that
$$[ X , [ X , Y ] ] = [ Y , [ X , Y ] ] = 0 .$$
We speak of “orders” when we expand the exponentials.

No. Again, if $[ T , V ]$ commutes with both $T$ and $V$, you can write
$$e^{\lambda (T + V)/N } = e^{\lambda T/N} e^{\lambda V/N} e^{- \frac{1}{2}(\frac{\lambda}{N})^{2} [ T , V ] } .$$
Notice the minus sign on the right hand side.
But, when I wrote
$$e^{X+Y} = e^{X}e^{Y} - \frac{1}{2} [ X , Y ], \ \ \ \ (2)$$
I did not assume that $[ X , Y ]$ commutes with both $X$ and $Y$. Equation (2) is an identity up to quadratic terms, i.e., when you expand and keep only the quadratic terms $X^{2},Y^{2},XY$ and $YX$.

18. Jan 2, 2016

### Happiness

Yes, I realized my mistake when @blue_leaf77 pointed it out earlier in post #12, sorry!


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# What is a Dotted Word™?

This is based on the What is a Word/Phrase™ series of Phrase™ and Word™ puzzles, started by JLee.

If a word follows a certain rule, then I call it a Dotted Word™.

Use the example word lists below to find the rule. Each word can be tested for whether it is a Dotted Word™ without depending on other words in these lists.

Dotted Word™, Not Dotted Word™
ONLINE, DIGITAL
HOWDY, HEY
URGES, INSTINCTS
NASAL, BARITONE
POLAND, UKRAINE
WATER, LUBRICATE
FOLIO, DOCUMENT
COBRAS, PYTHONS
REBIND, UNSEAL
PRATTLER, BABBLER
LUAU, FIESTA
RECEPT, RECEIVING
KNIFE, SPOON
FRAMES, BOXES

Hint:

The number of possible Dotted Words™ is relatively small.

Hint 2:

This is a very elementary puzzle. You don't need more than one hint.

• Perhaps the fact that "recept" is included in the list indicates that Dotted Words are relatively hard to find and/or that they were found using a very wide net. – GoldenGremlin Jan 6 '17 at 3:31
• @tilper, the fact that the word is very uncommon may suggest that Mike Q had trouble finding words with this property, and so needed to resort to an uncommon one to fill out his list. This suggests to me that the words might have been found with CPU aid, using a large lexicon. – GoldenGremlin Jan 6 '17 at 16:12
• @MikeQ, it's not a criticism at all! I'm trying to help others by reverse engineering your path to finding these buggers! – GoldenGremlin Jan 6 '17 at 16:39
• I liked the old format. T'was a tradition :/. – Karan Atree Jan 6 '17 at 16:50
• @Joe Sometimes when I'm stuck on a puzzle, I find that taking a break and stretching may help. – MikeQ Jan 12 '17 at 13:39

The last hint and also Gareth's answer make it clear that ...

... all Dotted Words can be split into symbols for chemical elements. But this is not a sufficient condition, because some of the non-dotted words, but not all, can also be "elementarily split".

The distinguishing trait is ...

... that all elements used in the dotted words must be of the same period, i.e. they must occur on the same row in the periodic table:

O·N·Li·Ne: Oxygen (8), Nitrogen (7), Lithium (3), Neon (10): period 2
Ho·W·Dy: Holmium (67), Tungsten (74), Dysprosium (66): period 6
U·Rg·Es: Uranium (92), Roentgenium (111), Einsteinium (99): period 7
Na·S·Al: Sodium (11), Sulfur (16), Aluminium (13): period 3
Fr·Og: Francium (87), Oganesson (118): period 7
Po·La·Nd: Polonium (84), Lanthanum (57), Neodymium (60): period 6
W·At·Er: Tungsten (74), Astatine (85), Erbium (68): period 6
F·O·Li·O: Fluorine (9), Oxygen (8), Lithium (3), Oxygen (8): period 2
Co·Br·As: Cobalt (27), Bromine (35), Arsenic (33): period 4
Re·Bi·Nd: Rhenium (75), Bismuth (83), Neodymium (60): period 6
Pr·At·Tl·Er: Praseodymium (59), Astatine (85), Thallium (81), Erbium (68): period 6
Lu·Au: Lutetium (71), Gold (79): period 6
Re·Ce·Pt: Rhenium (75), Cerium (58), Platinum (78): period 6
K·Ni·Fe: Potassium (19), Nickel (28), Iron (26): period 4
Fr·Am·Es: Francium (87), Americium (95), Einsteinium (99): period 7
La·Nd·La·Dy: Lanthanum (57), Neodymium (60), Lanthanum (57), Dysprosium (66): period 6

(There are several possible divisions for some words, but only one that makes it count as a Dotted Word.)

Why are they called Dotted Words?

No, it has nothing to do with Lewis dots, which would describe the electron configuration in an atom and therefore would refer to the columns in the periodic table.

They are called Dotted words, because some people call a dot a period.

• I think you missed the link to the Lewis Dot. – Sid Jan 12 '17 at 18:10
• There is no link to Lewis ... oh, I see. Fixed, thanks! – M Oehm Jan 12 '17 at 18:14
• Ohhh! Well done. – Gareth McCaughan Jan 12 '17 at 18:18

I thought, after looking at just a few of the words, that a Dotted Word might be one

made out of chemical element abbreviations.

Thus, e.g.,

ONLINE = O/N/Li/Ne and FRAMES = Fr/Am/Es.

Hence the hint saying that the puzzle

is "elementary".

However, this is demonstrably wrong; see below.

ONLINE = O/N/Li/Ne; DIGITAL fails because neither D nor Di exists.
HOWDY = H/O/W/Dy; HEY =He/Y.
URGES = U/Re/Ge/S; INSTINCTS fails because none of {Ct,T,Ts} exists.
NASAL = Na/S/Al; BARITONE fails because none of {It,T,To} exists.
FROG = Fr/Og; TOAD fails because neither T nor To exists.
POLAND = Po/La/Nd; UKRAINE = U/K/Ra/I/Ne.
WATER = W/At/Er; LUBRICATE = Lu/Br/I/Ca/Te.
FOLIO = F/O/Li/O; DOCUMENT fails because neither D nor Do exists.
COBRAS = C/O/Br/As; PYTHONS =P/Y/Th/O/N/S.
REBIND = Re/B/I/Nd; UNSEAL = U/N/Se/Al.
PRATTLER = Pr/At/Tl/Er, BABBLER fails because none of {Bl,L,Le} exists.
LUAU = Lu/Au; FIESTA =F/I/Es/Ta.
RECEPT = Re/Ce/Pt; RECEIVING fails because neither Ng nor G exists.
KNIFE = K/N/I/Fe; SPOON =S/Po/O/N.
FRAMES = Fr/Am/Es; BOXES =B/O/Xe/S.
LANDLADY = La/Nd/La/Dy; DUCHESS fails because neither D nor Du exists.

There are several counterexamples. But this seems to be close to working, and it's hard to believe it's coincidence. I suspect the intended answer is a variation on this theme and perhaps the variation explains why "Dotted Word" is the name.

It's very noticeable that

all the counterexamples are non-DWs that have decompositions, suggesting a rule of the form "a Dotted Word is one that has an 'elementary' decomposition such that ...". (This would also explain how it's possible that relatively few Dotted Words exist, even though a substantial fraction of all words have 'elementary' decompositions.)

A substantial fraction of the counterexamples

are non-DWs whose only decompositions involve noble gases, which (approximately) never form actual chemical compounds. But not all, and O/N/Li/Ne depends on Ne, so this is probably coincidence and in any case can be at most part of the answer.

A related conjecture would be

something along the lines of "a DW is a word with an 'elementary' decomposition that would be a possible chemical compound" or "... where each consecutive pair of elements can actually combine together somehow", but (1) that seems rather complicated and (2) S/Po/O/N seems like it fits any criterion of that sort at least as well as, say, O/N/Li/Ne does.

I notice that

all the Dotted Words have decompositions into at most four elements, and some of the Non-DWs with decompositions require more. (But note e.g. that Fr/Og uses only two elements; and that B/O/Xe/S requires only four but isn't a DW.) This is kinda suggestive of IP addresses, which is particularly interesting given Mike Q's comment to Glorfindel.

An obvious guess if it had worked in every case would be that "Dotted Word"

is just meant to hint at abbreviations, which are often indicated by trailing dots.

Or, more specifically,

conventionally one puts a dot at the end of an abbreviation exactly when its last letter is not the last letter of the thing abbreviated, so perhaps we could allow "Ra" but forbid "Rn" since it comes from Radon and wouldn't be dotted if we followed that convention for element abbreviations.

But clearly we need something smarter...

• What about BOXeS, HeY, and LuBrICaTe? You should check all of the words first. – MikeQ Jan 12 '17 at 16:03
• I already put a comment on the original post - FIESTA can be comprised of chemical element symbols – Joe Jan 12 '17 at 16:04
• Yup, I found some counterexamples once testing. But it's ... surprising ... how large a fraction of words seem to be classified correctly by this criterion. Perhaps it's a deliberate red herring, though. – Gareth McCaughan Jan 12 '17 at 16:05
• (I hadn't seen Joe's comment, but now I have. Sorry, Joe!) – Gareth McCaughan Jan 12 '17 at 16:05
• @Silenus Coincidence. "Dotted" is intentionally vague, but designing a rule that involves solving chemistry problems seems too complicated for this type of puzzle. You should expand your mind, otherwise you'll miss the big picture. – MikeQ Jan 12 '17 at 16:18


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# Variable tension in rope connected to mass

Problem 3.9 from Kleppner and Kolenkow's text An Introduction to Mechanics involves a uniform rope of length $L$ and mass $m$ that is connected at one end (its "bottom" end) to a block of mass $M$ and is being pulled with a force $F$ at its other end (its "top" end). The problem asks to find the tension $T(x)$ in the rope a distance $x$ away from the end being pulled, while neglecting gravity. I am very unsure of the answer I came up with.

First, I assumed that the acceleration of the block and each segment of rope would be equal and that the force acting the block would be upwards with magnitude $T(L) = F$. Each small segment of rope (length $\Delta x$, mass $\Delta m = \frac{m}{L} \Delta x$), undergoes an upward force $T(x)$ and a downward force $T(x+\Delta x)$ such that

$$\frac{-F}{M} = \frac{T(x+\Delta x) - T(x)}{\Delta m}$$

Isolating the variables related to the variable tension and taking the limit $\Delta x \rightarrow 0$ yields

$$\frac{dT}{dx} = \frac{m}{ML}F$$

After integrating using the assumed boundary condition $T(L) = F$, I arrived at the answer

$$T(x) = (1 - \frac{m}{M})F + \frac{mF}{ML}x$$

Is there anything wrong with my thought process? I thought that the net force on any segment of rope would always be zero because of Newton's Third Law, yet my use of $T(x + \Delta x) - T(x)$ (which should be zero) provided an answer that seems reasonable. Does my answer contradict Newton's law, or is my understanding of it flawed?


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# What is Proverbs 8:23 asserting about the origin of wisdom?

Young's Literal Translation has this:

YLT Proverbs 8:23 From the age I was anointed, from the first, From former states of the earth.

So it has:

• from the age (I was anointed)
• from the first
• from the former states

Is this giving the sense of the underlying Hebrew?

When are we to understand this to be?

Here is the Hebrew:

8:23 Hebrew OT: Westminster Leningrad Codex מֵ֭עֹולָם נִסַּ֥כְתִּי מֵרֹ֗אשׁ מִקַּדְמֵי־אָֽרֶץ׃

This is the LXX:

LES before the present age he founded me in the beginning, before making the earth

  23 πρὸ τοῦ αἰῶνος ἐθεμελίωσέν με ἐν ἀρχῇ,
πρὸ τοῦ τὴν γῆν ποιῆσαι

Swete, H. B. (1909). The Old Testament in Greek: According to the Septuagint (Pr 8:23). Cambridge, UK: Cambridge University Press.

Also, why does YLT have "anointed"?

The schemes, plans, and circumstances, relative to creation, government, providence, and to all being, material, animal, and intellectual, were conceived in the Divine mind, by the Divine wisdom, from eternity, or ever the earth was. There was no fortuitous creation, no jumbling concourse of original atoms, that entered into the composition of created beings; all was the effect of the plans before conceived, laid down, and at last acted upon by God's eternal wisdom.

I think the question of it's origin might be already tackled in Proverbs 8:22. Where Septuagint uses the word έκτισεν, which is generally translated as to create, form, shape, make.

We may also note this word in the Greek translation of Genesis 1:1, by Aquila of Sinope, where it stands, as generally accepted, for creation.

According to 1 Corinthians 1:24:

But to them who are called, both Jews and Greeks, Christ is the power of God and the wisdom of God.

From letter to Colossians:

who is the image of the invisible God, the firstborn of all creation. And He is before all things, and by Him all things consist. ... that He may be pre-eminent in all things.

Here all-things I think refer also to his position and glory from the past, as well into future (John 17:5; Revelation 3:14).

God's personal wisdom was always with God. But when God began creation, first he produced his own Image of himself, his Son in form of his word by which all other things were made (John 1:1-3). He, as his image, reflects wisdom of his God. We can also name Christ as wisdom of God. Who is his mediator to us people. (1 Timothy 2:5)

Septuagint / LXX

22 "κυριοςεκτισενμε αρχην οδων αυτου εις εργα αυτου"
23 "προ του αιωνος εθεμελιωσεν με εν αρχη"

LXX 2012:

   22 "The Lord made me the beginning of his ways for his works."
23 "He established me before time [was] in the beginning, before he made the earth:"

Mishneh Torah and JPS Tanakh:

   22 "The LORD made me as the beginning of His way, the first of His works of old."
23 "I was set up from everlasting, from the beginning, or ever the earth was."

Aramaic Targum from Adam Clarke's Commentary:

   22 "God created me in the beginning of his creatures."

• That's what Paul's epistles are saying, but where do you see it in the words of Proverbs? – b a Aug 26 '18 at 13:26
• As a combination of Old Testament prophecies from those sources (i.e. Isaiah 9, 11 with Proverbs, Psalms etc). Also as God's word, Memra. As to Messianic expectations, of which Paul and others were part of. – ChristGaveTruth Aug 28 '18 at 10:13