text:
# What early home computers have more than one CPU, where both could be used by the programmer? [closed]

I am interested to know if any computers that are on-topic for this site had more than one CPU, where this plurality could be leveraged by a programmer.

Some cases I don't so much care about:

• The Amigas could add a faster processor, but doing so disabled the onboard CPU so that we couldn't multitask on an Amiga.
• The C128 had both a 6510 and a Z80, but these could not be run at the the same time, so a programmer had to pick one to use and leave the other.

Did any early home computers have several CPUs so that a programmer could use both?

## closed as too broad by JAL, Mark, Nick Westgate, user12, wizzwizz4♦Jan 27 '17 at 19:25

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Not exactly home computers, but e.g. the CBM 4040 disk drives had both a 6502 and a 6504 which ran at the same time. They were both free programable by the host computer. – Janka Jan 21 '17 at 13:47
• Similarly, the Apple LaserWriter had could be programmed in PostScript, and have the results returned over the serial connection. It was of comparable power to the computers it was attached to. – scruss Jan 21 '17 at 16:55
• @scruss If you want to include "intelligent peripherals" even the Commodore C64 in combination with its disk drive 1541 (that could execute custom code) would qualify. – tofro Jan 21 '17 at 17:23
• If you want to restrict the question to early home computers (which would exclude e.g. a PDP-11), it might be better to use that term instead of "retro computers" in the title. Also, a lot of the more beefier early computers/mainframes/supercomputers have several CPUs (often used to control peripherie), so it might be a good idea to make the scope of the question clearer. – dirkt Jan 21 '17 at 19:47
• Just what counts as a "CPU" for you? For example, I'd personally hesitate to apply the term to the Amiga Copper, but it is a programmable processor that runs synchronously with the main CPU. More to the point, if one ignored the topicality requirement, would the programmable GPUs found in modern PCs count? – Ilmari Karonen Jan 22 '17 at 7:42

The BBC micro family (6502 processor) had a tube interface which permitted Z80, 6502, or 80186 coprocessors (possibly others). See e.g. here.

Each of these consists of a 'cheese wedge' with a board in with its own CPU and RAM. Both processors thus can run simultaneously (though that wasn't the normal intention which is why some of the links describe the coprocessor as 'running the user's program'), and indeed do when e.g. accessing disks etc.

• Key quote from one of your links, though: “The BBC's processor no longer controls the user programme …” so the second processor relegated the main processor to IO duties only. – scruss Jan 21 '17 at 22:30
• From personal experience, that isn't really correct. You can run both processors at once (they have their own separate RAM for instance), but the OS doesn't really help you do so. – abligh Jan 21 '17 at 22:35
• The idea was also continued onto Acorn's later designs - eg en.wikipedia.org/wiki/RiscPC which featured a slot for an 80486 processor to allow you to run Windows alongside RiscOS, apparently multitasked together. – Periata Breatta Jan 23 '17 at 11:23
• The BBC was probably the earliest "Home" computer to offer two processors to the user. I beleive the enhanced version of Elite used both processors to split its workload. – Ben Hillier Jan 25 '17 at 9:50
• @BenHillier indeed it did, to achieve a higher frame rate. Thanks for the reminder. – abligh Jan 25 '17 at 10:21

## User-programmable computers

The Commodore 64 has two 6502-family CPUs. There's a 6510 on the main logic board, and there's another CPU in the 1541 disk drive. The KERNAL on the host treated the 1541 as what is now called network attached storage (NAS), sending file access commands to CBM DOS in the 1541's ROM. Because the transfer protocol in CBM DOS was so slow, many C64 users sent a "fast loader" patch to the 1541's RAM in order to make data transfer faster. The demoscene eventually figured out how to run an interrupt-driven custom DOS in the background to stream data off the disk while animation continued on the host.

## Fourth-generation game consoles

The Super Nintendo Entertainment System (called Super Famicom in East Asia) has two 6502-family CPUs. One is a second-source 3.6 MHz 65816 core licensed from WDC, combined with a DMA controller in the 5A22. The other is a 1 MHz Sony SPC700, which behaves as a 65C02 with reshuffled opcode numbers. Games run game logic on the 65816 and the sound engine on the SPC700. Some Game Paks include a third processor: either an NEC uPD77C25 DSP, a Capcom CX4 (Hitachi HG51B169) DSP, an Argonaut Super FX Graphics Support Unit (GSU), an SA1 (a 10.5 MHz 65816-based coprocessor), or even the entire Game Boy system-on-chip. (Source: "SNES Cartridges" in Fullsnes)

The Sega Mega Drive (called Genesis in North America) and Neo Geo AES have a similar setup, with a 68000 to run the game and a Z80 to run audio. The Sega CD and 32X peripherals add additional CPUs, and the game Virtua Racing contains an additional "SVP" CPU, analogous to the GSU.

• The SNES apparently had co-processors and DSPs in the carts on a fairly regular basis, custom addon hardware like that is one of the benefits of a cart based system over a disc one. The Mega Drive didn't have it utilised nearly as much, Virtua Racing being about the best example of it. – Kaithar Jan 22 '17 at 5:53
• @Kaithar Do you have a link to any articles about the snes cart co-processors. I'd be interested to read more! :) – Colin Jan 23 '17 at 14:36
• Since you could run multiple 1541 drives to one C64, how many possible 6502 CPUs could you run in parallel? Interesting. – PhasedOut Jan 23 '17 at 15:44
• Not every C64 had a 1541, particularly in countries where disk drives had high import duties like the UK. – scruss Jan 23 '17 at 16:30
• While interesting that you could treat a C64 and its connected 1541s as a primitive cluster, I don't think it counts a a single computer with several CPUs. – chepner Jan 23 '17 at 17:53

The spectacularly rare Atari Transputer Workstation from 1989 supported up to 13 parallel processing Transputer CPUs. Multiple processors communicated with one another, and allowed sharing of suitable processing tasks. Other computer manufacturers considered Transputers at the time; I know of one card developed for Macintosh.

Around the same time, linear processing array cards such as the NEC IMAP-VISION used multiple DSP-like cores to speed up tasks like edge detection in video streams. While this was not a general-purpose CPU, it did offload a large amount of computation that the CPUs of the time couldn't handle. Going too far down this route would allow maths coprocessors and DMA IO controllers to be considered as second processors, though.

I'm not sure I'd consider the BBC Micro's expansion unit as a true second processor, even if there were then-powerful CPUs such as NS 320xx and ARM 1 that used it. A little like the C128 and the CP/M card for Apple II, the expansion unit relegated the main processor to simple IO tasks.

• The original Atari 400 & 800 had a simple, programmable graphics chip, separate from the 6502, whose name escapes me at the moment. I remember you could create new graphics modes by reprogramming it. – kbelder Jan 23 '17 at 20:04
• @kbelder - AlphaNumeric Television Interface Controller (ANTIC). – Hannover Fist Jan 23 '17 at 21:27
• As discussed elsewhere, Acorn's OS decided to relegate the main processor to simple IO tasks by default but the hardware enforced no such restriction. A developer was free to do whatever they wanted with each. – Tommy Nov 17 '17 at 19:52

The Soviet PDP-11-compatible educational desktop computer UKNC, released in 1987, for example.

One unique part of the design is the usage of a peripheral processing unit (PPU). PPU unloaded the CPU, taking responsibility for the management of peripheral devices (display, audio and so on), but could run any user programs, as well as CPU.

The Commodore Super-PET had both a 6809 and a 6502.

The 6809 was used to access the Waterloo programming environment with an editor and compilers for several languages.

The 6502 accessed a non-standard variant of the PET.

Both sides had access to 64K of additional RAM. A toggle switch on the side of the unit selected which processor was active.

• Does that mean only one CPU can be active at the same time? Doesn't the question specifically mention CPUs that could be active at the same time, and so be used by the programmer? – dirkt Jan 21 '17 at 19:48
• I'm pretty sure that the 6809 used the 6502 as a sort of I/O processor handling the keyboard, screen, IEEE-488 bus etc. I don't recall if you could add custom code to the 6502 workload though. – Peter Camilleri Jan 21 '17 at 19:52

The DEC Rainbow had two processors, one for 8 bit programs and the other for 16 bit programs. I have one of these in my basement, and I think it still works!

https://en.m.wikipedia.org/wiki/Rainbow_100

• I never used more than one at a time. But the documentation says that when you were running MS-DOS or CP/M-86 in the 8086, the Z80 was used as a floppy controller. So the system software was making use of both processors. – Walter Mitty Jan 23 '17 at 12:20
• One might want to investigate the Olivetti M20 too ... but then, is it not more of a business computer, just as the DEC Rainbow? – rackandboneman Apr 14 '18 at 21:13

One more for the Amiga. If you had a bridgeboard installed (a PC-on-a-card), you could theoretically leverage an x86 processor to work alongside the 680x0. Out of the box, you more-or-less saw this straight from Commodore:

• Janus's AREAD and AWRITE commands were normally used to copy files across the PC/Amiga divide. Though fixed function, it did run code on both sides to achieve this.
• If you didn't have a VGA card installed in one of the ISA slots, the Amiga would emulate a MDA/CGA display adapter in the 680x0 environment.
• If you didn't have a hard drive installed in an ISA slot, you could have a hardfile or partition daemon running in 680x0-land to provide a hard drive to the PC.

How it did this is that there was 64KB of dual-ported RAM available to both the PC and Amiga. Documentation for how this worked was available at the time and thus I can recall at least one example where someone took advantage of this; drive a cheap ISA PC NE2000 network interface for use by the Amiga. At the time, network adapters for the Amiga were expensive so this was a cheap way of getting Ethernet on the Amiga.

So theoretically one could write a daemon of some sort on the PC side and control it from the Amiga; send a rendering file, do some automated activity, do some processing in the background, whatever. There was a mention on the newsgroups back in the day that someone was looking to build something like this but I don't recall ever seeing the result. I personally did a few experiments back in the day but never went anywhere with it.

The Atari Falcon had both a 68030 CPU and a 56001 DSP that could work and be programmed independently of each other.

If you count FPUs as some sort of "CPU": you could add an optional 68882, so it even had 3 powerful chips.

• though the DSP wasn't exactly a general purpose processor: it couldn't handle all OS tasks. – scruss Jan 21 '17 at 17:44
• The FPU doesn't count as an extra CPU; it relies entirely on instructions delegated to it by the main CPU (specifically, all those in the 0xFxxx range of opcodes, known as "F-Line"). The DSP is valid though. – Chromatix Apr 12 '18 at 20:21

Some not so vintage, but still 90's computers:

1. Apple PowerMac 6100/66 with a DOS card. It was either a 486 or Pentium CPU with 8 MB of RAM. You could switch between Macintosh & Windows by hitting Cmd+Return.
2. Apple Macintosh LC III with an Apple IIe card. The IIe expansion card provided a 65C02 CPU and 256KB of RAM, just like a real Apple IIe. You could open a program to boot into ProDOS and even hook up 5.25" floppy drives
3. Lots of add on cards to Macs (Sonnet, Newer Tech) that plugged into a slot. You would load an extension at boot and offload the heavy lifting to that CPU instead of on the mainboard.
4. The Apple Power Macintosh 9500/180MP, based on the Tsunami architecture, featured dual 180 MHz PowerPC 604e processors, 16 MB or 32 MB of RAM, a 2.0 GB hard drive, an 8X CD-ROM drive, and a 2 MB ATI video card in a very expandable tower case design with twelve RAM slots and six PCI slots.
• I wouldn't really count #3 as "more than one CPU". The presence of the expansion card disabled the CPU on the main board; they didn't work in parallel. Unless maybe you're thinking of the Radius Rocket? – Cody Gray Jan 22 '17 at 7:55

The NeXT computer (1988) had a second CPU, a Motorola 56001.

There were a number of Macintosh models with the 'AV' suffix that sported a Motorola 56000 digital signal processing second processor. Some that I recall were Quadra 660AV, Power Macintosh 6100/60AV, 7100/66AV, 8100/100AV.

These may have been used for sound processing, but aside from minor applications (software 56k modem, using the Geoport as phoneline interface) they didn't get a lot of interest.

• I wouldn't really call the NeXT a "home computer", given that it cost a third of an average person's annual salary... definitely in the realms of "professional workstation", I fear. – Periata Breatta Jan 23 '17 at 11:13
• I had a Quadra 660AV that had the DSP Chip - one feature that the DSP added was allowing windows to become transparent when you moved them. – PhasedOut Jan 23 '17 at 15:46
• Really, @phasedout? Where do you configure that "transparent windows" feature? I have a Quadra 840AV (which has a similar AT&T DSP chip), but have never seen this before nor any mention of it online, including 68k Mac enthusiast forums. – Cody Gray Jan 23 '17 at 18:20
• @CodyGray - I distinctly remember this from somewhere when i researched the 660AV I acquired at a garage sale before selling it - I will try to dig out the source, even if it means digging out my old G4 the research was done on. – PhasedOut Jan 24 '17 at 14:06
• @Cody Cray - I think this is where I learned of that concept, and after seeing that, I found something explaining that it was a function the DSP chip enabled. ( lowendmac.com/2001/treasure-your-quadra-840av ) and that led to finding some DSP 3210 chip tools here (toddp.com/classic/Software%20Install/Apple%20Source%20Code/… ) – PhasedOut Jan 24 '17 at 14:29

Ohio Scientific offered a switchable 3 CPU system in 1978, the Challenger-3 (6502, 6800, and Z80). I remember reading the description, or perhaps an advert, in a back issue of Byte(?) magazine and wondering if any of them was ever put into useful service.

On some versions, the CPU could be switched under software control, though it seems like the CPUs could not be enabled simultaneously, since the busses were shared.

I don't know if it's "vintage" enough, but

8087, 80287, 80387, 80386, and of course the 80486 had a math co-processor that you could "directly" access. You could use it for for math operations of course, but you could also access a "stack" that could be pushed, popped, and indexed of a sort.

Most applications that even used the co-processor would do so using the x87 instructions (do math on the co-processor), but you could access it directly by assembly.

Other then an academic excersize I don't know any program that would use the co-processor other then for math operations, but it could do it.

Some info can be found here http://wiretap.area.com/Gopher/Library/Techdoc/Cpu/coproc.txt

• I don't believe the x87 coprocessor exactly qualifies as a CPU. From what I recall, the x86 would stall while the x87 does its work, so not exactly parallel. While that sounds bad, keep in mind that the x87 was much faster at doing the math operations than the x86 would resulting in a net speed increase. Today, the x87 is integrated standard with the x86 CPU and has been since the original Pentium. – bjb Jan 24 '17 at 20:35
• I consider it kind of a gray area. And the x87 didn't stall the main CPU, you would send instructions starting with 11011 and the co-processor would execute them, otherwise the main processor would execute. They could actually run two instructions at the same time. Again, the x87 could do full CPU tasks, it had a stack, and could do logical operations, you could in theory send one instruction to the main cpu and one to the co-processor. But, it was rarely ever used that way. Computers and programmers just didn't think that way back then. So instead it was mainly used to process math while... – coteyr Jan 24 '17 at 20:48
• the main CPU WAITed it out. – coteyr Jan 24 '17 at 20:48
• Actually, @bjb, the original 8087 was a true co-processor, executing in parallel with the main CPU. The 8088 and 8086 did not stall while the 8087 coprocessor was doing floating-point arithmetic. But this created a serious problem if there was bus contention, and since there were no real synchronization mechanisms, programmers as a matter of habit (and indeed assemblers themselves!) inserted WAIT instructions after each FPU instruction, which is what stalled the main CPU. This design flaw/feature was removed with the 80286 and later, meaning the FPU could no longer run concurrently. – Cody Gray Jan 25 '17 at 12:15
• The relevant optimization advice from Zen of Assembly Language is here in an online copy of the book—see the last paragraph in the linked section. Scattered throughout the book is information on "cycle eaters" (Abrash's term) that make precise, predictive cycle counting essentially impossible. – Cody Gray Jan 25 '17 at 17:54


text:
Chemistry Bond Parameters-I
Click for Only Video

### Topics Covered :

● Bond Parameters
● Bond Length
● Bond Angle
● Bond Enthalpy

### Bond Parameters :

Bond parameter are given as follow :

### Bond Length :

color{red}("Definition :") Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

color{green} ✍ color{green} mathbf("KEY CONCEPT")
When atoms come closer to each other then attraction takes place between them and P.E. keeps on decreasing till at a particular distance and here potential energy is minimum. If the atoms are brought closer, the repulsion starts and therefore potential energy of the system begins to increase. Thus at equilibrium distance the atoms keep on vibrating about their mean positions.
● Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques.

● Each atom of the bonded pair contributes to the bond length (Fig. 4.1).

● In the case of a covalent bond, the contribution from each atom is called the covalent radius of that atom.

color{green}("Covalent Radii "): ● The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation.

● The covalent radius is half of the distance between two similar atoms joined by a covalent bond in the same molecule.

color{green}("van der Waal's Radii" ): ● The van der Waals radius represents the overall size of the atom which includes its valence shell in a non-bonded situation.

● The van der Waals radius is half of the distance between two similar atoms in separate molecules in a solid.

=> Covalent and van der Waals radii of chlorine are depicted in Fig.

=> Some typical average bond lengths for single, double and triple bonds are shown in Table.

=> Bond lengths for some common molecules are given in Table

=> The covalent radii of some common elements are listed in Table
•color{red}("Factors affecting bond length ") :
=> BOND LENGTH prop SIZE OF ATOMS
Eg. HI>HBr>HCl.
=> BOND LENGTH prop 1/text(MULTIPLICITY)

### Why is r_text(van der Waal) > r_text(cov.) in Covalent Compounds?

=> The van der Waals radius is greater than the covalent radius because it does not take into account overlap.

=> The overlap that exists between the two electron clouds is what causes the covalent radius to be smaller than the van der Waals radius.
• Factors affecting bond length:
=> size of the atoms:The bond length increases with increase in size of atoms. For eg. HI > HBr > HC l > H F.
=> Multiplicity of bond:The bond length decreases with multiplicity of the bond. C equiv C < C = C < C - C.

### Bond Angle :

color{red}("Definition :") It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion.

● Bond angle is expressed in degree which can be experimentally determined by spectroscopic methods.

● It gives some idea regarding the distribution of orbitals around the central atom in a molecule/complex ion and hence it helps us in determining its shape.

color{red}("Example ") H–O–H bond angle in water can be represented as shown in fig.

=> color{red}("Factors affecting bond angle ") :
•Hybridisation: Bond angle is directly proportional to % of s character.
•Lone pair: Bond angle is inversely proportional to number of lone pair.
•Central atom: Bond angle is directly proportional to electronegativity of central atom.

### Bond Enthalpy :

color{red}("Definition :") It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.

● The unit of bond enthalpy is kJ mol^(–1).

color{red}("Example :") The H – H bond enthalpy in hydrogen molecule is 435.8 kJ mol^(–1).

H_2 (g) → H (g) + H (g) ; Delta_a H^v = 435.8 kJ mol^(-1)

Similarly, the bond enthalpy for molecules containing multiple bonds, for example O_2 and N_2 will be as under :

O_2 (O = O) (g) → O(g) +O(g) ; Delta_a H^v = 498 kJ mol^(-1)

N_2 (N = N ) (g) → N (g) +N(g) ; Delta_a H^v = 946.0 kJ mol^(-1)

=> It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule.

=> For a heteronuclear diatomic molecules like HCl, we have

HCl (g) → H (g) +Cl (g) ; Delta_a H^v = 431.0 kJ mol^(-1)

=> In case of polyatomic molecules, the measurement of bond strength is more complicated.

color{red}("Example ") In case of H_2O molecule, the enthalpy needed to break the two O – H bonds is not the same.

H_2O(g) → H(g) + OH (g) ; Delta_a H_1^v = 502 kJ mol^(-1)

OH(g) → H (g) +O(g) ; Delta_a H_2^v = 427 kJ mol^(-1)

● The difference in the Δ_aH^v value shows that the second O – H bond undergoes some change because of changed chemical environment.

● This is the reason for some difference in energy of the same O – H bond in different molecules like C_2H_5OH (ethanol) and water.

● Therefore in polyatomic molecules, the term mean or average bond enthalpy is used.

● It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule,

color{green}("Average bond enthalpy") = (502+427)/2

 = 464.5 kJ mol^(-1)

• color{red}("Factors affecting bond enthalpy ") :
=> Greater the size of atoms, greater is the bond length and less is bond dissociation enthalpy.
=> For the bond between the same two atoms,greater is multiplicity of bond,greater is the bond dissociation enthalpy. Eg. H-H => Greater the number of lone pairs of electrons present on the bonded atoms, greater is the repulsion between the atoms and hence less is BDE.

format: plain


text:
## If $$4x-12 \ge x + 9,$$ which of the following must be true?

##### This topic has expert replies
Legendary Member
Posts: 2898
Joined: 07 Sep 2017
Thanked: 6 times
Followed by:5 members

### If $$4x-12 \ge x + 9,$$ which of the following must be true?

by Vincen » Wed Feb 23, 2022 3:54 am

00:00

A

B

C

D

E

## Global Stats

If $$4x-12 \ge x + 9,$$ which of the following must be true?

A. $$x > 6$$
B. $$x < 7$$
C. $$x > 7$$
D. $$x > 8$$
E. $$x < 8$$


text:
# Tex latex software

LaTeX is intended to provide a tex latex software language that accesses the power of TeX in an easier way for writers. Regular expressions support You can use regular expressions for "search and replace". LaTeX project. For other uses, see Latex disambiguation. LaTeX at Wikipedia's sister projects.

Hello! I am Ashlee. I have 22 y.o. I love dirty talks. My favorit is Busty, Fetish, Swinger

Take me now!

## Getting LaTeX

User's guide and reference. A Document Preparation System". Such macros are often collected into packages, which are available to address special formatting issues such as complicated mathematical content or graphics.

Hi! I am Merle. I have 24 y.o. I love dirty talks. My favorit is Domination, Slut sex, Upskirt

Someone has sent you a privat video :)


text:
In this paper, we shall consider - from a critical point of view - the definition of four-velocity. We consider a very precise definition of world line in space time. We emphasize that the space-time of Special Relativity must be considered as a pseudo-Riemaniann manifold, that is, a (topological) space endowed with a chart-atlas and a Riemann metric. Nevertheless, we maintain the explanation at a very elementary level, by using essential basic knowledge coming from Mathematical Analysis. Events and world-lines are elements or subsets of the space-time, therefore we cannot realize them as vectors, or subsets, of any vector space without the adoption of a reference frame. The importance of a reference frame is very well underlined and used explicitly. The definition of reference frame is explicitly emphasized. Moreover, we identify reference frames, observers and charts of the assigned atlas. In other terms, from this point of view, Differential Geometry approach asserts that we cannot interpret or visualize or study the elements (or subsets) of the space-time without the intervention of some observer. As usual, the observers, i.e. reference frames, of our differentiable manifold are smoothly correlated. In some sense, space-time is unequivocally existing but not analyzable without the intervention of some observer. World-lines are defined as smooth curves in space-time, such that, for any reference frame, they can be obtained as the graphics of convenient functions defined - by using the pre-chosen reference frame - on interval of the real line taking values on the Euclidean 3-space. We consider a very elementary example and show how to define world-lines of the space-time by using a reverse approach, starting from one fixed reference frame, hence going from the Minkovskian real (1,3)-space to the Minkovsky space time M4. Then, we explicitly define and consider the natural time parametrization of a world-line induced by a reference frame. So that we can define explicitly the natural time parameter of a particle with respect to a reference frame. Different versions of the same concept will be considered. Then, we consider the analytical definition of proper time of a moving point in the space time, starting from the natural parameter induced by a reference frame. In contrast with the usual exposition and explanation of proper time, we explicitly define the domain of the proper time and it's corresponding clear mathematical definition. We also consider the proper-time differential form, field of linear forms attached with the natural parametrization of a world-line, explicitly noting the independence by reference frame of such differential form. Finally, we define the four velocity of a particle in an unambiguous way and explicitly provide various expressions of it.

### Critical notes of Special Relativity: four-velocity

#### Abstract

In this paper, we shall consider - from a critical point of view - the definition of four-velocity. We consider a very precise definition of world line in space time. We emphasize that the space-time of Special Relativity must be considered as a pseudo-Riemaniann manifold, that is, a (topological) space endowed with a chart-atlas and a Riemann metric. Nevertheless, we maintain the explanation at a very elementary level, by using essential basic knowledge coming from Mathematical Analysis. Events and world-lines are elements or subsets of the space-time, therefore we cannot realize them as vectors, or subsets, of any vector space without the adoption of a reference frame. The importance of a reference frame is very well underlined and used explicitly. The definition of reference frame is explicitly emphasized. Moreover, we identify reference frames, observers and charts of the assigned atlas. In other terms, from this point of view, Differential Geometry approach asserts that we cannot interpret or visualize or study the elements (or subsets) of the space-time without the intervention of some observer. As usual, the observers, i.e. reference frames, of our differentiable manifold are smoothly correlated. In some sense, space-time is unequivocally existing but not analyzable without the intervention of some observer. World-lines are defined as smooth curves in space-time, such that, for any reference frame, they can be obtained as the graphics of convenient functions defined - by using the pre-chosen reference frame - on interval of the real line taking values on the Euclidean 3-space. We consider a very elementary example and show how to define world-lines of the space-time by using a reverse approach, starting from one fixed reference frame, hence going from the Minkovskian real (1,3)-space to the Minkovsky space time M4. Then, we explicitly define and consider the natural time parametrization of a world-line induced by a reference frame. So that we can define explicitly the natural time parameter of a particle with respect to a reference frame. Different versions of the same concept will be considered. Then, we consider the analytical definition of proper time of a moving point in the space time, starting from the natural parameter induced by a reference frame. In contrast with the usual exposition and explanation of proper time, we explicitly define the domain of the proper time and it's corresponding clear mathematical definition. We also consider the proper-time differential form, field of linear forms attached with the natural parametrization of a world-line, explicitly noting the independence by reference frame of such differential form. Finally, we define the four velocity of a particle in an unambiguous way and explicitly provide various expressions of it.
##### Scheda breve Scheda completa Scheda completa (DC)
File in questo prodotto:
Non ci sono file associati a questo prodotto.
##### Pubblicazioni consigliate

Caricamento pubblicazioni consigliate

I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.

Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11570/3082509`
##### Attenzione

Attenzione! I dati visualizzati non sono stati sottoposti a validazione da parte dell'ateneo

• ND
• ND
• ND


text:
# American Institute of Mathematical Sciences

November  2019, 12(7): 1879-1903. doi: 10.3934/dcdss.2019124

## Dihedral molecular configurations interacting by Lennard-Jones and Coulomb forces

 1 Department of Mathematical Sciences, University of Texas at Dallas, 800 W Campbell Road, Richardson, Texas 75080-3021, USA 2 Center for Applied Mathematics, Guangzhou University, Guangzhou 510006, China 3 University of Texas at Dallas, Department of Mathematical Sciences, 800 W Campbell Road, Richardson, Texas 75080-3021, USA 4 Department of Mathematics and Computer Science, Alcorn State University, 1000 ASU Drive, Lorman, MS, 39096, USA

* Corresponding author: Wieslaw Krawcewicz

Received  December 2017 Revised  July 2018 Published  December 2018

Fund Project: The first and second authors are supported by the Department of Mathematical Sciences University of Texas at Dallas. The third author is supported by the Center for Applied Mathematics at Guangzhou University, Guangzhou China and the Department of Mathematical Sciences University of Texas at Dallas.

In this paper, we investigate nonlinear periodic vibrations of a group of particles with a planar dihedral configuration governed by the Lennard-Jones and Coulomb forces. Using the gradient equivariant degree, we provide a full topological classification of the periodic solutions with both temporal and spatial symmetries. In the process, we provide general formulae for the spectrum of the linearized system of equations describing the above configuration, which allows us to obtain the critical frequencies of the particles' motions. The obtained frequencies represent the set of all critical periods for small amplitude periodic solutions emerging from a given stationary symmetric orbit of solutions.

Citation: Irina Berezovik, Wieslaw Krawcewicz, Qingwen Hu. Dihedral molecular configurations interacting by Lennard-Jones and Coulomb forces. Discrete & Continuous Dynamical Systems - S, 2019, 12 (7) : 1879-1903. doi: 10.3934/dcdss.2019124
##### References:

show all references

##### References:
Stationary solution to equation (7) with dihedral symmetries
Relative motions of all 6 particles with $\lambda_0^2 = \frac{l^2}{\mu}$, $l = 1$ and $\mu$ near the eigenvalue $\mu_0 = 10.10496819$ of $\nabla^2 V(u^o)$
Relative motions of all particles with $\lambda_0^2 = \frac{l^2}{\mu}$, $l = 1$ and $\mu$ near the eigenvalue $\mu = 6.442637681$ of $\nabla^2 V(u^o)$
Relative motions of all particles with $\lambda_0^2 = \frac{l^2}{\mu}$, $l = 1$ and $\mu$ near the eigenvalue $\mu = 8.469351217$ of $\nabla^2 V(u^o)$
Relative motions of all particles with Relative motions of all particles with $\lambda_0^2=\frac{l^2}{\mu}$, $l=1$ and $\mu$ near the eigenvalue $\mu=3.854423919$ of $\nabla^2 V(u^o)$
The values $\lambda_{j, l}$ in the critical set $\Lambda$
 $j$ $\mu_j$ $\lambda_{j, 1}$ $\lambda_{j, 2}$ $\lambda_{j, 3}$ $\lambda_{j, 4}$ 0 10.10496819 0.31458103 0.62916205 0.94374308 1.25832410 1 8.469351217 0.34361723 0.68723445 1.03085168 1.37446891 3 3.854423919 0.50935463 1.01870927 1.52806390 2.03741854 $2^+$ 6.442637681 0.62767390 1.25534781 1.88302171 2.51069561 $2^-$ 0.007288929 11.7130006 23.4260011 35.1390017 46.8520023
 $j$ $\mu_j$ $\lambda_{j, 1}$ $\lambda_{j, 2}$ $\lambda_{j, 3}$ $\lambda_{j, 4}$ 0 10.10496819 0.31458103 0.62916205 0.94374308 1.25832410 1 8.469351217 0.34361723 0.68723445 1.03085168 1.37446891 3 3.854423919 0.50935463 1.01870927 1.52806390 2.03741854 $2^+$ 6.442637681 0.62767390 1.25534781 1.88302171 2.51069561 $2^-$ 0.007288929 11.7130006 23.4260011 35.1390017 46.8520023
Maximal orbit types in $\mathscr W_{j,l}$
 $\mathscr W_{j,l}$, $l\ge 1$ maximal orbit types $\mathscr W_{0,l}$ $(D_6\times D_l)$ $\mathscr W_{1l}$ $({D_6}^{{\mathbb{Z}_1}}{ \times _{{D_6}}}{D_{6l}}) - ({D_2}^{{D_1}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}}) - ({D_2}^{{{\tilde D}_1}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}})$ $\mathscr W_{2,l}$ $({D_6}^{{\mathbb{Z}_2}}{ \times _{{D_3}}}{D_{3l}}) - ({D_2}^{{\mathbb{Z}_2}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}}) - ({D_2} \times {D_l})$ $\mathscr W_{3,l}$ $({D_6}^{{{\tilde D}_3}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}})$
 $\mathscr W_{j,l}$, $l\ge 1$ maximal orbit types $\mathscr W_{0,l}$ $(D_6\times D_l)$ $\mathscr W_{1l}$ $({D_6}^{{\mathbb{Z}_1}}{ \times _{{D_6}}}{D_{6l}}) - ({D_2}^{{D_1}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}}) - ({D_2}^{{{\tilde D}_1}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}})$ $\mathscr W_{2,l}$ $({D_6}^{{\mathbb{Z}_2}}{ \times _{{D_3}}}{D_{3l}}) - ({D_2}^{{\mathbb{Z}_2}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}}) - ({D_2} \times {D_l})$ $\mathscr W_{3,l}$ $({D_6}^{{{\tilde D}_3}}{ \times _{{\mathbb{Z}_2}}}{D_{2l}})$
 [1] Wenmeng Geng, Kai Tao. Large deviation theorems for dirichlet determinants of analytic quasi-periodic jacobi operators with Brjuno-Rüssmann frequency. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5305-5335. doi: 10.3934/cpaa.2020240 [2] Youshan Tao, Michael Winkler. Critical mass for infinite-time blow-up in a haptotaxis system with nonlinear zero-order interaction. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 439-454. doi: 10.3934/dcds.2020216 [3] Qingfang Wang, Hua Yang. Solutions of nonlocal problem with critical exponent. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5591-5608. doi: 10.3934/cpaa.2020253 [4] Lei Liu, Li Wu. Multiplicity of closed characteristics on $P$-symmetric compact convex hypersurfaces in $\mathbb{R}^{2n}$. Discrete & Continuous Dynamical Systems - A, 2020  doi: 10.3934/dcds.2020378 [5] Abdelghafour Atlas, Mostafa Bendahmane, Fahd Karami, Driss Meskine, Omar Oubbih. A nonlinear fractional reaction-diffusion system applied to image denoising and decomposition. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020321 [6] Manil T. Mohan. First order necessary conditions of optimality for the two dimensional tidal dynamics system. Mathematical Control & Related Fields, 2020  doi: 10.3934/mcrf.2020045 [7] Adel M. Al-Mahdi, Mohammad M. Al-Gharabli, Salim A. Messaoudi. New general decay result for a system of viscoelastic wave equations with past history. Communications on Pure & Applied Analysis, , () : -. doi: 10.3934/cpaa.2020273 [8] Sumit Arora, Manil T. Mohan, Jaydev Dabas. Approximate controllability of a Sobolev type impulsive functional evolution system in Banach spaces. Mathematical Control & Related Fields, 2020  doi: 10.3934/mcrf.2020049 [9] Yichen Zhang, Meiqiang Feng. A coupled $p$-Laplacian elliptic system: Existence, uniqueness and asymptotic behavior. Electronic Research Archive, 2020, 28 (4) : 1419-1438. doi: 10.3934/era.2020075 [10] Jianquan Li, Xin Xie, Dian Zhang, Jia Li, Xiaolin Lin. Qualitative analysis of a simple tumor-immune system with time delay of tumor action. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020341 [11] Denis Bonheure, Silvia Cingolani, Simone Secchi. Concentration phenomena for the Schrödinger-Poisson system in $\mathbb{R}^2$. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020447 [12] Xavier Carvajal, Liliana Esquivel, Raphael Santos. On local well-posedness and ill-posedness results for a coupled system of mkdv type equations. Discrete & Continuous Dynamical Systems - A, 2020  doi: 10.3934/dcds.2020382 [13] Mathew Gluck. Classification of solutions to a system of $n^{\rm th}$ order equations on $\mathbb R^n$. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5413-5436. doi: 10.3934/cpaa.2020246

2019 Impact Factor: 1.233


text:
# Description: The scenario description and requirements for Short White Paper Prepare a short... 1 answer below »

Description:The scenario description and requirements for Short White Paper

Prepare a short research paper of approximately 5 pages, double-spaced, exclusive of cover, title page, table of contents, endnotes and bibliography. Your paper must use APA formatting with the exception that tables and figures (if used) can be inserted at the appropriate location rather than added at the end. Submit the paper in your assignment folder as a Word attachment with the following file name:yourlastnameSSB.doc

For example, a submission file would be called nanceSSB.doc

Alice, a high net worth customer, banks on-line at Super Secure Bank (SSB) and has agreed to use 3DES in communicating with SSB. One day, Alice received a statement that shows a debit of $1,000,000 from her account. On inquiring, she was told that the bank manager, Bob, transferred the money out of Alice's account and into an account of his own in an offshore bank. When reached via long distance in the Cayman Islands, Bob produced a message from Alice, properly encrypted with the agreed upon 3DES keys, saying: "Thanks for your many years of fine service, Bob. Please transfer$1,000,000 from my account to yours as a token of my esteem and appreciation. Signed, Alice."

Alice filed suit against Bob, SSB and the government of the Cayman Islands, claiming that the message was a forgery, sent by Bob himself and asking for triple damages for pain and suffering. Bob has responded by claiming that all procedures were followed properly and that Alice is filing a nuisance suit. You have been employed by SSB as a cryptographic expert to assist in the investigation of this matter, and will produce a report for the SSB Board of Directors, which will assist them in determining how to proceed in this matter

.

Your report to the Board of Directors should address the following issues

:

• What can be determined from the facts as presented about whether Alice intended to make Bob a gift of \$1,000,000?

• Assuming SSB wishes to continue using only 3DES as its cryptographic system, what could SSB and Alice have done to protect against this controversy arising?

• Would this controversy have arisen if SSB had been using AES rather than 3DES?

Your report should clearly address these issues, with sufficient detail and background to allow the ?cryptographically challenged? Board of Directors to understand the issues involved and formulate plans for how to approach the immediate issue with Alice, and to continue business in the future, assuming that they want to continue using 3DES.

Amit K


text:
# Parity nonconservation in the photodisintegration of the deuteron at low energy

Abstract : The parity-nonconserving asymmetry in the deuteron photodisintegration, $\vec{\gamma}+d\to n+p$, is considered with the photon energy ranged up to 10 MeV above the threshold. The aim is to improve upon a schematic estimate assuming the absence of tensor as well as spin-orbit forces in the nucleon-nucleon interaction. The major contributions are due to the vector-meson exchanges, and the strong suppression of the pion-exchange contribution is confirmed. A simple argument, going beyond the observation of an algebraic cancellation, is presented. Contributions of meson-exchange currents are also considered, but found to be less significant.
Keywords :
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-00020467
Contributor : Lpsc Bibliotheque Connect in order to contact the contributor
Submitted on : Wednesday, March 3, 2004 - 11:31:42 AM
Last modification on : Friday, October 8, 2021 - 10:46:05 AM

### Citation

C.P. Liu, C.H. Hyun, B. Desplanques. Parity nonconservation in the photodisintegration of the deuteron at low energy. Physical Review C, American Physical Society, 2004, 69, 065502 (11 p.). ⟨10.1016/PhysRevC.69.065502⟩. ⟨in2p3-00020467⟩

Record views


text:
## anonymous 5 years ago My calculatior only displays eight digits so I need help with this problem: x+x+500,000=625,000,000.

1. anonymous

312250000

2. anonymous

What is 312,250,000+500,000?

3. anonymous

Here's how to do this without a calculator. Using scientific notation, convert 500,000 and 625,000,000 to the following:$500,000=5\times10^{5}$$625000000=625\times10^{6}$ Scientific notation means that if the exponent in the 10 is positive, you move the decimal point in the coefficient at the end of the number to the right that many times. For example 625 x 10^6 = 625. move the dot 6 places to the right filling in zeros as you go. After 3 dot moves: 625000. After 6 dot moves: 625000000. = 625 x 10^6 Now rewrite the equation as follows:$x+x+5\times10^5=625\times10^6$ 1) add the x's to get 2x$2x+5\times10^5=625\times10^6$ 2) subtract 5 x 10^5 from both sides of the equation$2x=625\times10^6-5\times10^5$ 3) On the right side of the equal sign, factor out a 10^5 leaving you with$2x=10^5(625\times10-5)=10^5(6245)$ 4) Divide both sides of the equation by 2$x=10^5(6245)/2$ 5) Divide 6245 by 2 first to get$x=10^5(3122.5)=3122.5\times10^5$ 6) Move the dot 5 places to the right After 3 moves: 3122500. After 2 more moves: 312250000. And there is your answer 312250000


text:
# Number Puzzle #6 – Solution

Two days ago, I posted this simple little number puzzle. Quite a few folks came up with the answer below.  One of the interesting questions you can ask is whether that solution is unique.

Clearly there are two lines of symmetry in the original problem so by reflection alone we come up with a total of four solutions: $\{ I, F_x, F_y, F_x \circ F_y \}$ where $F_x$ and $F_y$ represent “flips” across the lines of symmetry and $I$ represents the identity, or the solution above.  By $F_x \circ F_y$, I mean the composition of the flipping operations or just consecutive flipping.

There also exists radial symmetry at $180^{\circ}$, but this is equivalent to $F_x \circ F_y$.  So for this arrangement above, there are four solutions of the same “type”.

Are any other arrangements possible besides these four?

## One thought on “Number Puzzle #6 – Solution”

1. 1 and 8 must go on the central 2 points, as they are the only values with 6 non-neighbours.

2 must go top/bottom away from 1 as all other remaining nodes are connected to 1, and the same works for 7 (the numbers are symmetric about 4|5).

3 and 6 now similarly have only one position available each, leaving 4 and 5.

So the answer is no, there are no further arrangements.


text:
# 8.3 Phase change and latent heat  (Page 5/9)

 Page 5 / 9

Why does snow remain on mountain slopes even when daytime temperatures are higher than the freezing temperature?

Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be accumulated from the air, even if the air is above $0º\text{C}$ . The warmer the air is, the faster this heat exchange occurs and the faster the snow melts.

## Summary

• Most substances can exist either in solid, liquid, and gas forms, which are referred to as “phases.”
• Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling and freezing (or melting) points.
• During phase changes, heat absorbed or released is given by:
$Q=\text{mL,}$

where $L$ is the latent heat coefficient.

## Conceptual questions

Heat transfer can cause temperature and phase changes. What else can cause these changes?

How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below $\text{0ºC}$ , in the vicinity of large bodies of water?

What is the temperature of ice right after it is formed by freezing water?

If you place $\text{0ºC}$ ice into $\text{0ºC}$ water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?

What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?

In very humid climates where there are numerous bodies of water, such as in Florida, it is unusual for temperatures to rise above about $\text{35º}\text{C}\left(\text{95º}\text{F}\right)$ . In deserts, however, temperatures can rise far above this. Explain how the evaporation of water helps limit high temperatures in humid climates.

In winters, it is often warmer in San Francisco than in nearby Sacramento, 150 km inland. In summers, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures.

Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why.

Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result.

When still air cools by radiating at night, it is unusual for temperatures to fall below the dew point. Explain why.

In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublimate, with some crystals lingering for awhile and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from [link] .

## Problems&Exercises

How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at $0º\text{C}$ if their heat of fusion is the same as that of water?

35.9 kcal

how do I set up the problem?
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Abdullahi
hi mam
Mark
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Got questions? Join the online conversation and get instant answers!


text:
Articles
Inflation derivative - Derivative

Typically, real rate swaps also come under this bracket, such as asset swaps of inflation-indexed bonds (government-issued inflation-indexed bonds, such as the Treasury Inflation Protected Securities, UK inflation-linked gilt-edged securities (ILGs), French OATeis, Italian BTPeis, German Bundeis and Japanese JGBis are prominent examples). Inflation swaps are the linear form of these derivatives. They can take a similar form to fixed versus floating interest rate swaps (which are the derivative form for fixed rate bonds), but use a real rate coupon versus floating, but also pay a redemption pickup at maturity (i.e., the derivative form of inflation-indexed bonds).

Functional derivative - Functional derivative

the functional derivative of

Functional derivative - Functional derivative

and : as the directional derivative at point

Parametric derivative - Second derivative

The second derivative implied by a parametric equation is given by

Parametric derivative - First derivative

where the notation \dot{x}(t) denotes the derivative of x with respect to t, for example. This can be derived using the chain rule for derivatives:

Directional derivative - Normal derivative

A normal derivative is a directional derivative taken in the direction normal (that is, orthogonal) to some surface in space, or more generally along a normal vector field orthogonal to some hypersurface. See for example Neumann boundary condition. If the normal direction is denoted by \mathbf{n}, then the directional derivative of a function f is sometimes denoted as. In other notations, :

Parametric derivative - First derivative

The first derivative implied by these parametric equations is :

Functional derivative - Functional derivative

is defined to be the Radon–Nikodym derivative of this measure.

Derivative - The derivative as a function

. By the definition of the derivative function,

Covariant derivative - Derivative along curve

In particular, is a vector field along the curve \gamma itself. If vanishes then the curve is called a geodesic of the covariant derivative. If the covariant derivative is the Levi-Civita connection of a certain metric then the geodesics for the connection are precisely the geodesics of the metric that are parametrised by arc length.

Derivative - The derivative as a function

Using this idea, differentiation becomes a function of functions: The derivative is an operator whose domain is the set of all functions that have derivatives at every point of their domain and whose range is a set of functions. If we denote this operator by

Lie derivative - Nijenhuis–Lie derivative

Another generalization, due to Albert Nijenhuis, allows one to define the Lie derivative of a differential form along any section of the bundle Ω k (M, TM) of differential forms with values in the tangent bundle. If K ∈ Ω k (M, TM) and α is a differential p-form, then it is possible to define the interior product i K α of K and α. The Nijenhuis–Lie derivative is then the anticommutator of the interior product and the exterior derivative: :

Real estate derivative - Derivative efficiencies

Owning real estate assets is costly, and the transaction costs associated with purchasing commercial real estate can be prohibitive. Typical transaction costs can equal 500 - 800 basis points per transaction. Industry estimates suggest that transaction costs for commercial real estate easily surpass $10–$12 billion annually. Since the US real estate derivative market is new, the transaction costs are at this point variable. However, based on derivatives in other markets, it is anticipated that the costs will be well below the 500-800 basis points required to invest in actual real estate.

Covariant derivative - Derivative along curve

The derivative along a curve is also used to define the parallel transport along the curve.

Fractal derivative - Connection with Derivative

If for a given function f both the derivative Df and the fractal derivative D α f exists, one can find an analog to the chain rule:

Derivative - The derivative as a function

has a derivative at most, but not all, points of its domain. The function whose value at

Derivative test - First derivative test

The first derivative test examines a function's monotonic properties (where the function is increasing or decreasing) focusing on a particular point in its domain. If the function "switches" from increasing to decreasing at the point, then the function will achieve a highest value at that point. Similarly, if the function "switches" from decreasing to increasing at the point, then it will achieve a least value at that point. If the function fails to "switch", and remains increasing or remains decreasing, then no highest or least value is achieved.

Derivative - The derivative as a function

and is called the derivative function or the derivative of

Derivative - The derivative as a function

be a function that has a derivative at every point in its domain. We can then define a function that maps every point x to the value of the derivative of f at x. This function is written

Covariant derivative - Derivative along curve

Since the covariant derivative \nabla_XT of a tensor field T at a point p depends only on the value of the vector field X at p one can define the covariant derivative along a smooth curve \gamma(t) in a manifold: :Note that the tensor field T only needs to be defined on the curve \gamma(t) for this definition to make sense.


text:
## Easiest way to handle keypress with 1-bit option on memory restricted system

Started by 7 months ago43 views

I'm looking for a simple way to handle key presses as well as store an option to allow a repeating key or just a one time press and release. Through experimentation, I was able to capture the value while the key is held down and when the key is only pressed once. But that involves me adding an instruction and burning my code onto the chip again, but I don't want to do that.

Since each memory location in KEYS is 8 bits, I feel somehow I can change that 8-bit cyclic buffer that cycles the current button status in to a 7-bit cyclic buffer while using one bit exclusively for storing the option to have the key value returned only when the key is pressed once or while the key is held.

So I feel something in this fragment needs modification, but is there a way to do it with using as few clock cycles as possible?

        mov A,@R0    ;put value in our mini 8-bit cyclic buffer
rlc A
mov @R0,A

Here's my complete key scanning code. I cannot modify the first 7 lines inside the rescan loop since they load specific values from a table that makes the correct key returned.

    KEYIO equ P0
KEYS equ 0B0h ;key storage
ISPRESS equ 0Fh ;To avoid bouncing
scankey:
mov DPTR,#scans  ;to key scan table
mov R0,#KEYS+0Dh ;14 possible keys
rescan:
clr A
movc A,@A+DPTR ;load value to put on key lines to prepare matrix for that number
mov KEYIO,A    ;store in A
inc DPTR
clr A
movc A,@A+DPTR ;xor next scan value
xrl A,KEYIO    ;with captured key
inc DPTR
;If correct key is captured, then A=0
dec A        ;Here we set carry in 2 clocks because A=0
mov A,@R0    ;put value in our mini 8-bit cyclic buffer
rlc A
mov @R0,A
cjne A,#ISPRESS,npress ;See if key is literally pressed
;could reset @R0 here to allow multi-press
mov A,R0             ;it is so get pointer
subb A,#KEYS         ;and substract start of pointer
ret                  ;A=# representing key pressed
npress:
dec R0                 ;Go to next key
cjne R0,#KEYS-1,rescan ;Keep going until all keys are processed
mov A,#0FFh            ;All keys processed, nothing captured so return FFh
ret


text:
# Braces in a \@for loop

While writing macros to manipulate the items of a list with a \@for loop, I have found that braced items behave differently depending on whether they have a leading space or not:

\documentclass{article}
\usepackage[T1]{fontenc}

\begin{document}
\makeatletter

\def\uselist#1{%
\@for\temp:=#1\do{\meaning\temp\par}}

\verb|\listwithcommas{Lima,Alpha,{Delta,Oscar},Tango,{Whisky,Echo,Romeo},Xray}|

\def\listwithcommas{Lima,Alpha,{Delta,Oscar},Tango,{Whisky,Echo,Romeo},Xray}
\uselist{\listwithcommas}

\verb|\listwithcommasandspaces{ {Lima}, Alpha, {Delta,Oscar}, Tango, {Whisky,Echo,Romeo}, Xray}|

\def\listwithcommasandspaces{ {Lima}, Alpha, {Delta,Oscar}, Tango, {Whisky,Echo,Romeo}, Xray}
\uselist{\listwithcommasandspaces}

\end{document}

I would like to maintain the braces, as when there is a space before the item, after applying the macros. Is it possible to do that inside the \@for loop? The intention is that, however the list is provided (through a macro), the output list will maintain the braces. That is, both \listwithcommas and \listwithcommasandspaces should give the second output.

Another question is why the behaviour is different when there are spaces or not.

EDIT

Here is an example to clarify the intention.

Suppose I have a macro \subtractlist and that it is used two times. First I subtract Lima from the list and then {Delta,Oscar} from the resulting list. Since I have lost the braces after the first use, the second one will not work.

I can enclose all the items of the new formed list in braces. In that case, it will work with \listwithcommas but not with \listwithcommasandspaces, because there are double braces and spaces.

\documentclass{article}
\usepackage[T1]{fontenc}

\begin{document}
\makeatletter

\def\listwithcommas{{Lima},Alpha,{Delta,Oscar},Tango,{Whisky,Echo,Romeo},Xray}

\def\listwithcommasandspaces{ {Lima}, Alpha, {Delta,Oscar}, Tango, {Whisky,Echo,Romeo}, Xray}

\newif\if@isinlist
\def\subtractlist#1#2{% #1:original list, #2:remove list
\gdef\@subtractlist{}%
\@for\@tempa:=#1\do{%
\@isinlistfalse%
\@for\@tempb:=#2\do{\ifx\@tempa\@tempb\@isinlisttrue\fi}%
\if@isinlist%
\else%
\ifx\@subtractlist\empty%
\expandafter\gdef%
\expandafter\@subtractlist%
%             \expandafter{\@tempa}%
%%% extra pair of braces added
\expandafter{\expandafter{\@tempa}}%
\else%
\expandafter\@subtractlist%
%             \expandafter{\expandafter,\@tempa}%
%%% extra pair of braces added
\expandafter{\expandafter,\expandafter{\@tempa}}%
\fi%
\fi%
}%
\let\currentlist\@subtractlist}

\def\removelista{Bravo,Lima}
\def\removelistb{Bravo,{Delta,Oscar}}

\subtractlist{\listwithcommas}{\removelista}
\verb|subtract Lima:| \meaning\currentlist

\subtractlist{\currentlist}{\removelistb}
\verb|subtract {Delta,Oscar}:|\meaning\currentlist

\bigskip

\subtractlist{\listwithcommasandspaces}{\removelista}
\verb|subtract Lima:| \meaning\currentlist

\subtractlist{\currentlist}{\removelistb}
\verb|subtract {Delta,Oscar}:|\meaning\currentlist

\end{document}

Without extra braces:

With extra braces:

\@for is really minimal, so it doesn't work too hard on removing or keeping braces and spaces. The braces are removed as part of TeX's argument grabbing, so you have to add extra safety to keep them. I added an \@empty in front of every item so that TeX won't remove braces, then I expand that \@empty to remove it before passing the item to \uselistcmd.

\documentclass{article}
\usepackage[T1]{fontenc}

\begin{document}

\makeatletter
\def\q@stop{\q@stop}
\def\uselist#1{\expandafter\uselist@\expandafter\@empty#1,\q@stop,}
\def\uselist@#1,{\expandafter\uselist@@\expandafter{#1}}%
\def\uselist@@#1{%
\ifx\q@stop#1\else
\uselistcmd{#1}%
\expandafter\uselist@\expandafter\@empty
\fi}
\makeatother

\def\uselistcmd#1{\detokenize{(#1)}\par}

\verb|\listwithcommas{Lima,Alpha,{Delta,Oscar},Tango,{Whisky,Echo,Romeo},Xray}|

\def\listwithcommas{Lima,Alpha,{Delta,Oscar},Tango,{Whisky,Echo,Romeo},Xray}
\uselist{\listwithcommas}

\verb|\listwithcommasandspaces{ {Lima}, Alpha, {Delta,Oscar}, Tango, {Whisky,Echo,Romeo}, Xray}|

\def\listwithcommasandspaces{ {Lima}, Alpha, {Delta,Oscar}, Tango, {Whisky,Echo,Romeo}, Xray}
\uselist{\listwithcommasandspaces}

\end{document}

• Thank you. I have still not dared to start with expl3. It seems overwhelming. However, removing the braces is just what I do not want. I have edited the question to try to clarify it a bit. Dec 23, 2020 at 8:10
• @RaoulKessels Ah, sorry, I misunderstood. I edited with a version that doesn't change the input at all (no brace nor space removal). Also no expl3 (and you can see the amount of \expandafters that pop up :) Dec 23, 2020 at 13:13
• So, basically you have made a new \@for command. Very impressive! As per the \expandafters, don't worry. In a command for trimming spaces from a list I had to use \multiexpandafter{3} to keep the code readable. Dec 24, 2020 at 9:30
• @RaoulKessels Kind of, yes, but both \@for and \clist_map_inline:nn have a few more features than my code. The above is just a really basic mapping over a list plus not removing braces. Use with care :) Dec 24, 2020 at 11:56
• Thx, I will take care. Feliz Natal ;-) Dec 24, 2020 at 20:17

If you look closely at your output, you'll see that the output includes a space for the value of \temp which is not the case with the first set.

What happens in the first case is that TeX is parsing everything up to the comma as the argument being digested. The braces get discarded here because it's getting passed a literal argument of, e.g., {Lima} which it treats as a single argument and drops the braces, much like when you pass an argument to a macro enclosed in braces.

But in the space version, what's coming in is no longer {Lima} but  {Lima} (the space doesn't get discarded because the expansion of \@for will be inserting non-letter tokens between \@for and the argument that gets digested. Now the argument is space plus a group so it keeps the braces because they're no longer at the outer level and discardable.

It's possible to do what you want (TeX is Turing-complete, after all), but it would not be simple. Enclosing every item in the list in braces and double-bracing the items you want to keep braces would work. Or you could add the spaces and parse them out of the items that you get. There might be some fancy stuff possible with token lists.

• I already had tried too add additional braces in my macros, but it want work either. I have edited the question with an example. Dec 23, 2020 at 8:12


text:
Reusing Parts of a Proof [closed]

Hi

I have a proof for a Lemma which splits into an odd and even case.

The proof for the even case was already published by someone else in a different context and the proof for for the odd case is very similar (but not trivial) to the even case proof.

So how should I now proceed about the odd case proof? Is it ok if I make it clear that the odd case proof closely follows the idea of the even case proof, published by someone else?

-

closed as not a real question by Mark Sapir, Emil Jeřábek, Felipe Voloch, Bill Johnson, Anthony QuasFeb 1 '12 at 17:46

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Is that hard to just tell the truth? As you see it is only few lines long. –  Mark Sapir Feb 1 '12 at 16:32
@Mark Sapir I think you misunderstood me, mentioning that my proof follows the idea of the other proof is the least I would do. But I would still reuse many ideas of the original proof and I don't feel too comfortable about it. –  user695652 Feb 1 '12 at 16:45
If the odd case cannot be easily recreated (or even if it can), a few words on how to start it or modify the even case would be appreciated by future researchers. For example, "As in the even case, we partition the set, except we create A' to hold these three elements. The same machinery can now be used, except it deals with these three elements as follows:... " . If you need to reproduce the other proof (with attribution) to make the paper more self-contained, well that isn't so bad either. Gerhard "Ask Me About System Design" Paseman, 2012.02.01 –  Gerhard Paseman Feb 1 '12 at 17:07
Also, if you can summarize or exposit the other proof so that it is even more accessible (with the disclaimer that it is your interpretation of the proof), I certainly won't come banging on your door, except possibly to thank you. I suspect I am not alone in such a sentiment. Gerhard "But It's Just My Opinion" Paseman, 2012.02.01 –  Gerhard Paseman Feb 1 '12 at 17:11
If you think that your proof follows somebody else's proof, say it. There is absolutely no problems at all. I voted to close as not a real question. –  Mark Sapir Feb 1 '12 at 17:22


text:
## Specific Heat and Density of Linen

What is the spcific heat (J/kgK) and density (kg/m3) of Linen fiber? I'm testing the thermal diffusivity of nonconductive fibers and need the specific heat and density to figure out the thermal diffusivity.


text:
Home

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution will be displayed step by step (In 2 parts)
Solution
Problem: Backward, {{0,1},{1,0},{2,1},{3,10}},x=2.1 [ Calculator, Method and examples ]

Solution:
Your problem -> Backward, {{0,1},{1,0},{2,1},{3,10}},x=2.1

The value of table for X and Y

 X Y 0 1 2 3 1 0 1 10

Newton's backward differentiation method to find solution
Newton's backward differentiation table is as follows.
 X Y(X) gradY grad^2Y grad^3Y 0 1 -1 1 0 2 1 6 2 1 8 9 3 10

The value of x at you want to find the f(x) : x = 2.1

h = x_1 - x_0 = 1 - 0 = 1

q = (x - x_n) / h = (2.1 - 3) / 1 = -0.9

Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me


text:
# Fourier transform of a triangle function

## Homework Statement

Hello

I'm learning Fourier transforms via the Stanford lecture series on Youtube. In the 6th lecture, the professor claims that the FT of a triangle function is the square of the sinc function. I'm trying to derive this, but I can't get my math to work out. Could someone please help me figure out where I went wrong?

## Homework Equations

FT(f(t)) = ∫exp(i2πst)*f(t) dt

sin(t) = [exp(it) - exp(-it)] / 2i

## The Attempt at a Solution

edit: Please let me know if the images below aren't legible on your monitor. If so, I will increase the resolution.

http://imageshack.us/a/img716/9334/fourierpage1.png [Broken]

http://imageshack.us/a/img46/6585/fourierpage2.png [Broken]

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
This can be done by the convolution theorem.

$\mathcal{F} \left[ f * g\right] = \mathcal{F} \left[f \right] \mathcal{F} \left[g \right]$

So, if sinc^2(ω) corresponds to a triangle function, then a triangle function would be the convolution of the inverse Fourier transform of sinc with itself. The inverse Fourier transform of a sinc is a rectangle function. So, all you need to do is show a triangle function is the convolution of a rectangle function with itself.

Thanks for your help. Yes, I can see that it can be done via the convolution theorem, and the convolution of a rectangle with itself being a triangle is intuitively clear, but shouldn't this be a relatively simple transformation to perform directly? It's making me very frustrated that I can't get it right.

jbunniii
Homework Helper
Gold Member
The triangle function is even, so the Fourier transform takes a simpler form:
\begin{align} \int_{-1}^{1}f(t) e^{i\omega t} dt &= \int_{-1}^{1} f(t) \cos(\omega t) dt + i \int_{-1}^{1} f(t) \sin(\omega t) dt\\ &= 2 \int_{0}^{1} f(t) \cos(\omega t) dt + 0\\ &= 2 \int_{0}^{1} f(t) \cos(\omega t) dt \end{align}
This is nice because we only have to deal with one integral. Now ##f(t) = 1-t## in the interval ##[0,1]##, so the Fourier transform becomes
$$2\int_{0}^{1} \cos(\omega t) dt - 2\int_{0}^{1} t \cos(\omega t) dt$$
Try using that as a starting point. It should make your life easier.

Thanks to everyone for the help. I had to take a hiatus due to exams, but now I'm trying to figure it out again. I'm using jbunnii's tip to set it up, but I still can't get the correct answer. I've checked over my math three times and can't see what I'm doing wrong. I'm getting the ω^2 in the denominator, but I don't see how the integration can produce a sin^2 in the numerator. Could someone please help me out one more time?

http://img600.imageshack.us/img600/2717/spage001.jpg [Broken]

http://imageshack.us/a/img683/5251/spage002.jpg [Broken]

Last edited by a moderator:
jbunniii
Homework Helper
Gold Member
I think all your work is fine. To get the result into the desired form, try using this trig identity:
$$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$

jbunnii, thank you very much for your help! It's a relief that I haven't forgotten my basic calculus. I've always been horrible at trig identities; maybe this lecture series will finally bring me up to par. I have one quick question remaining:

In order to apply the identity above, I had to make the substitution x = ω/2. I ended up with FT = sinc^2(x), or equivalently FT = sinc^2(ω/2). I'm trying to understand the implications of this substitution.

ω = 2 π s, where s = k/λ. k is the integer frequency multiplier of the wave, and λ is the wavelength. For this non-periodic function, λ = ∞; therefore, s is infinitesimal. Is it valid to make the argument that since ω is 2π * an infinitesimal, x = (2π)/2 * an infinitesimal, so x = ω and therefore FT = sinc^2(ω)?

jbunniii
Homework Helper
Gold Member
ω = 2 π s, where s = k/λ. k is the integer frequency multiplier of the wave, and λ is the wavelength. For this non-periodic function, λ = ∞; therefore, s is infinitesimal. Is it valid to make the argument that since ω is 2π * an infinitesimal, x = (2π)/2 * an infinitesimal, so x = ω and therefore FT = sinc^2(ω)?
I'm not sure what you are trying to argue here.

It is true that (reasonably well behaved) periodic functions can be represented by a Fourier series, which involves terms with ##\omega = 2\pi k /\lambda## where ##k## is any integer.

For a non-periodic function, we need a "continuum" of ##\omega## values. But outside of nonstandard analysis, we don't talk about infinitesimals as they are not well defined.

If you are trying to work out whether the answer should be ##\text{sinc}^2(\omega/2)## or ##\text{sinc}^2(\omega)##, that depends on how ##\text{sinc}## is defined. This doesn't appear to be completely standardized. According to Wikipedia, there are at least two definitions in common use:
$$\text{sinc}(x) = \frac{\sin(x)}{x}$$
and
$$\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$
It would not surprise me if there are other conventions as well. Similarly, there is no universal consensus regarding the definition of the Fourier transform. Some authors move the various ##2\pi## factors around, or invert the signs on the complex exponential, etc. Best to check your textbook or instructor's notes to see what definition you should be using. Or, to be on the safe side, just leave the answer in the form ##\sin^2(\omega/2)/(\omega/2)^2##.

Thank you again. I was trying to reconcile my result with the professor's assertion that the Fourier transform will be "sinc squared". Thank you very much for your help; I really appreciate it.


text:
# [OS X TeX] Syncing in one direction only

Don Green Dragon fergdc at Shaw.ca
Mon Dec 27 00:01:18 CET 2010

Hi Alain,

On 2010-23Dec-, at 8:40 PM, Alain Schremmer wrote:

> On Dec 23, 2010, at 8:57 PM, Don Green Dragon wrote:
>
>> Now working on the suggestion made by Alain Schremmer.
>
> It should work but if not, let me know.

I've not been able to replicate your example in the case of two sources files. So I'm missing something. This is the setup:

(1) In a directory ~/Documents/Alain/ I have the three subdirectories:

Contents	Controls	WholeBook

I'm not yet at your 'ALL' stage.  ;-)

(2) In directory Contents/, I have two 'source' files '1.tex' and '2.tex'. Their content is simply that of longer chapters from which they have been culled. When typeset, each produces about three pages, mostly text, but a teeny bit of math stuff in '2.tex'. Each begins with the comment line:

%!TEX root = ../Controls/n.tex

where n is 1 or 2, as appropriate. The first non-comment line is a \chapter command.

(3) In directory Controls/, there are two 'master' or 'root' files '1.tex' and '2.tex'. Here is the essence of '2.tex'

---------------------------
\documentclass[11pt]{book}

% a very simple preamble which I'm confident is not the problem

\begin{document}
\frontmatter
\tableofcontents

\mainmatter
% encountered, then <chapter> will be incremented to 2
\include{../Contents/2}		% includes ~/Documents/Alain/Contents/2.tex
\end{document}
---------------

Structure of Controls/1.tex is essentially the same with requisite changes.

Next is my assumption about the purpose of the .tex files mentioned so far!

H1: When working on the file Contents/1.tex, one initially typesets the file Controls/1.tex and thereafter one can sync back and forth as one repairs or augments the file Contents/1.tex.

H2: Same as H1 with '1' replaced by '2'.

If H1 and H2 and not correct, then please clarify!

Assuming that H1 and H2 are okay, here is what happens:

Before continuing, I note that in the Contents/ directory, the two files '1.tex' and '2.tex' are the only files! In this directory, there is neither a '1.aux' nor a '2.aux' file.

(4) File Controls/1.tex is loaded into TeXShop, and TYPESET is requested. The main console complaint is:

--------------------------
No file ../Contents/1.aux.
) (./1.toc) [1{/usr/local/texlive/2010/texmf-var/fonts/map/pdftex/updmap/pdftex
.map}] [2]
/usr/texbin/pdflatex: Not writing to ../Contents/1.aux (openout_any = p).

./1.tex:28: I can't write on file ../Contents/1.aux'.
\@include ...\immediate \openout \@partaux #1.aux
\immediate \write \@partau...
l.28 \include{../Contents/1}
% includes ~/Documents/Schremmer/Contents/1.tex
(Press Enter to retry, or Control-D to exit; default file extension is .tex')
Please type another output file name: i /Contents/1.aux
(../Contents/1.tex
---------------------------

and --- after my input --- then the PDF version of Contents/1.tex appears and seems to be in perfect order, except for the lack of a table of contents which requires another TYPESET anyway. Notice that, by trial and error, I found that by entering

i /Contents/1.aux

after the prompt "Please type another output file name:" then the typeset would continue. However, NO file Contents/1.aux has been created!!! Consequently, if I try for the Table of Contents with another TYPESET, then the same complaint is lodged.

Same thing happens if I attempt to typeset Contents/2.tex by using Controls/2.tex; i.e., this time, if I enter

i /Contents/2.aux

Also, it is NOT necessary to enter what I've recorded above. Something like

i Strange.aux

works equally well, but I can find to trace of of 'strange.aux'.

(5) I created a file MyBook.tex in Alain/WholeBook/ --- no details given here --- and I can typeset it provided i'm willing to enter both

i /Contents/1.aux

and

i /Contents/2.aux

after the relevant prompts appear on the console window. In fact, I used 'xxxx' and 'yyyy' in one trial, and that worked as well as the i-entries above!?!?!? Perhaps, you can suggest what is missing!

> By the way, I also keep all my preambles in style sheets that I keep in a folder along with the Contents and Controls folders in one project folder.

I'm not quite there, but that is a 'next' step. I want to have a structure that works before doing what you note above. Actually, I have one, but it follows the model that Graetzer outlined.

Will look, but not today.  :-)  Thanks for the reference.

Greetings of the Season to everyone, and thanks to the many who have helped with my TeX problems as well as to those persons who continue to upgrade the fantastic TeX system.

Don Green Dragon
fergdc at Shaw.ca

-------------- next part --------------
----------- Please Consult the Following Before Posting -----------
TeX FAQ: http://www.tex.ac.uk/faq
List Reminders and Etiquette: http://email.esm.psu.edu/mac-tex/
List Archive: http://tug.org/pipermail/macostex-archives/
TeX on Mac OS X Website: http://mactex-wiki.tug.org/
List Info: http://email.esm.psu.edu/mailman/listinfo/macosx-tex


text:
# Diode current components

I have the following two questions in the derivation of current in a diode.

1) In forward bias, an assumption "abrupt junction" is made and field is equal to 0 is also made. But then we still use built in potential. How is this okay?

2) In most semiconductor classes and also in Millman and Halkias they have left reverse bias case as similar. But can anyone help me out with the actual derivation? Because i don't see how the exponential decrease in hole concentration with distance occurs if the field is non zero in the continuity equation

Any help would be appreciated. Thanks

• Could you clarify what you mean by "field is equal to 0"? I'm not familiar with that assumption for diode analysis. Basic diode analysis usually assumes an abrupt junction (a step function dopant profile) and that there are zero electrons and holes in the depletion region. I also don't understand what you mean about exponential decrease in hole concentration with distance. If you make the previous assumptions then your hole concentration will be a step function as well. 0 inside the depletion region and $N_A$ outside the depletion region. – Matt Aug 8 '18 at 13:29
• The exponential is not from pside to n-side. But rather within the n-side from the junction taken as x=0. By the field I mean the field that causes drift current . – pranav Aug 8 '18 at 13:54
• The field cant be equal to 0. A diode expierences both drift and diffusion currents and cant work if you just assume away the drift current part. You must misunderstand something. I'm still not following this exponential decrease in hole concentration. At the junction the hole concentration is zero, and since hole concentration cant be negative, I dont see what you mean. – Matt Aug 8 '18 at 14:02
• Hmn... you didnt mention where you are assuming the field is equal to zero. Its equal to zero outside the depletion region. Is this what you mean? – Matt Aug 8 '18 at 14:05
• For an abrupt junction we neglect the width of the junction and assume step function right? – pranav Aug 8 '18 at 14:06


text:
## A simple proof of Cauchy-Schwarz inequality

### September 18, 2007

[reader be aware: this is a post for getting attention from the web, you may get these information elsewhere and probably also better written]

There are a lot of proof about Cauchy-Schwarz inequality, there is also an entire book on it.

This is a proof without many advantages with respect to others, rather being very mechanic, however it is the only way someone reached my blog 🙂

Cauchy-Schwarz inequality states that:

$\langle x,y\rangle \leq \left\|x \right\| \left\|y \right\|$

Let’ s see that this inequality comes shortly from axioms of a vector space and from definition of scalar product

$\langle\frac{x}{\left\|x \right\|}-\frac{y}{\left\|y \right\|}, \frac{x}{\left\|x \right\|}-\frac{y}{\left\|y\right\|} \rangle \geq 0$

$\langle\frac{x}{\left\|x \right\|},\frac{x}{\left\|x \right\|}\rangle + \langle\frac{y}{\left\|y \right\|},\frac{y}{\left\|y \right\|}\rangle -2\langle\frac{x}{\left\|x \right\|},\frac{y}{\left\|y \right\|}\rangle \geq 0$

$1+1 -2\langle\frac{x}{\left\|x \right\|},\frac{y}{\left\|y \right\|}\rangle \geq 0$

and inequality follows.

The quality of this proof is that it comes out from the axioms of a vector space, unfortunately it does not bring an intuitive insight about this fundamental tool for analysis.


text:
PlaidCTF this year had very insane challenges. Although I have spent a lot of time working on those crypto challenges, I was only able to solve leaky block cipher.

## Challenge Summary⌗

This completely legitimate™ block cipher looks a bit like GCM, but my computer plumber keeps complaining about water residue. Can you help me spot the leak?

The challenge has a hand-crafted AEAD scheme. We are required to go through 20 rounds of challenges, and this is how each round proceeds:

1. The service sends you an IV that is used in this round.
2. You send a plaintext (with >= 7 blocks) for encryption.
3. The service encrypts your plaintext and returns the tag and 15.5 bytes of the encrypted IV ($\mathcal{C}_\text{IV}$).
4. If you are able to provide the remaining 0.5 bytes of $\mathcal{C}_\text{IV}$, you won this round.

## Solution⌗

### Part I: Understanding the scheme⌗

To begin, let's have a look on the LeakyBlockCipher class, where most of the things are happening.

class LeakyBlockCipher:
def __init__(self, key = None):
if key is None:
key = secrets.token_bytes(16)
self.key = key
self.aes = AES.new(key, AES.MODE_ECB)
self.H = self.aes.encrypt(bytes(16))
def encrypt(self, iv, data):
assert len(iv) == 16
assert len(data) % 16 == 0
ivi = int.from_bytes(iv, "big")
cip = bytes()
tag = bytes(16)
for i in range(0,len(data),16):
cntr = ((ivi + i // 16 + 1) % 2**128).to_bytes(16, byteorder="big")
block = data[i:i+16]
enced = self.aes.encrypt(xor(cntr, block))
cip += enced
tag = xor(tag, enced)
tag = gf128(tag, self.H) # multiplication under GF(2^128)
tag = xor(tag, self.aes.encrypt(iv))
return cip, tag

As mentioned in the summary, we are only given the tag for the plaintext we send. This is how the tag, $T$, for a $n$-block ciphertext $(c_1, c_2, ..., c_n)$ defined:

$T := c_1 {\mathcal{C}_0}^n + c_2 {\mathcal{C}_0}^{n-1} + ... + c_n {\mathcal{C}_0} + {\mathcal{C}_\text{IV}},$

hereby ${\mathcal{C}_0}$ is the encrypted block of null bytes and ${\mathcal{C}_\text{IV}}$ is the encrypted IV.

From above, here are two ciphertext blocks we are interested: ${\mathcal{C}_0}$ and ${\mathcal{C}_\text{IV}}$. We should pick those $c_i$'s such that $c_i \in \{{\mathcal{C}_0}, {\mathcal{C}_\text{IV}}\}$, or otherwise we are actually introducing more unknowns, which is not desired.

### Part II: Testing around and observing patterns⌗

One interesting way to set $c_i$'s is to set all of them as ${\mathcal{C}_\text{IV}}$. In that way, the tag we have is

$T := {\mathcal{C}_\text{IV}}({\mathcal{C}_0}^n + {\mathcal{C}_0}^{n-1} + ... + 1).$

It is very easy for us to exhaust ${\mathcal{C}_\text{IV}}$ since we already have 15.5 bytes of it, where there are only 16 possible cases. With that, ${\mathcal{C}_0}$ is the only unknown and its terms form an geometric series:

${\mathcal{C}_0}^n + {\mathcal{C}_0}^{n-1} + ... + 1 = \frac{T}{\mathcal{C}_\text{IV}}.$

I was expecting that the above equation has roots in ${\mathcal{C}_0}$ only if ${\mathcal{C}_\text{IV}}$ is correct. I was wrong. In average, there are ~12 out of 16 candidate ${\mathcal{C}_\text{IV}}$'s has roots for ${\mathcal{C}_0}$. Although I am able to filter a few candidates away, I am never lucky to picking one in twelve correctly for 20 times in a row.

While I was messing around with $n$, I found the number of ${\mathcal{C}_\text{IV}}$'s that the equation has roots is dropped suddenly when $n = 15$. There are only one or two candidates in most of the cases. I've executed an experiment with the below code snippet to estimate the number of possible ${\mathcal{C}_\text{IV}}$'s.

from tqdm import tqdm

R.<x> = GF(2)[]
F = GF(2^128, name='a', modulus=x^128+x^7+x^2+x+1)

counts = [0 for _ in range(17)]
for _ in tqdm(range(100)):
k0 = getrandbits(124)
k1 = getrandbits(4)

K = F.fetch_int(16*k0 + k1)
H = F.fetch_int(getrandbits(128))
T = K * sum(H^i for i in range(15+1))

c = 0
f = False

for i in range(16):
K = F.fetch_int(16*k0+i)

rs = (K * sum(x^i for i in range(15+1)) + T).roots()
for r, _ in rs: # H
if r == H: f = True
if len(rs) > 0: c += 1
assert f

counts[c] += 1

for c in range(17):
print(c, counts[c])

From the results, there are one or two ${\mathcal{C}_\text{IV}}$'s in 80% of the cases. There is a 70% chance to solve a round of the challenge. Hence, we are able to get the flag in around 1250 attempts (with winning probability 0.720). Although this is decent, I would like to investigate why is the equation $x^{15} + x^{14} + ... + 1 = K$ (where $K$ is a constant) is less likely to have roots compared with others.

### Part III: We need to go deeper⌗

…or not. It is still possible for someone to solve the challenge from the above results. I am just curious. Anyway, we will be speaking mathematics now!

Suppose that $g \in \text{GF}(2^{128})$ is a generator. Then the order of $g$ is $2^{128}-1$. Since 15 is a factor of $2^{128}-1$, $g^{15}$ has an order $(2^{128}-1)/15$. This said, there is $1/15$ chance for $x^{15} = h$ to have solutions if we pick $h \in \text{GF}(2^{128})$ randomly. On the other hand, we have an interesting fact that happens on $\text{GF}(2^{128})$:

$x^{15} + x^{14} + ... + 1 = (x + 1)^{15}.$

The probability for $x^n + x^{n-1} + ... + 1 = h$, where $h$ is randomly picked from $\text{GF}(2^{128})$, to be solvable is $1/n$ if $n \in \mathbb{N}$ satifies:

1. $x^n + x^{n-1} + ... + 1 = (x + 1)^n$, and
2. $2^{128}-1$ is divisible by $n$.

We want to find a larger $n$ to reduce the above probability, since it is correlated to the number of unwanted candidates for ${\mathcal{C}_\text{IV}}$. However $n$ should not be too large because this is the number of plaintext blocks we are sending to the server. For the first condition, it suffices to find $n$ such that $n \choose r$ are all odd for $r = 0, 1, ..., n$. This happens if and only if $n = 2^k - 1$ for some $k \in \mathbb{Z}_{\geq 0}$. In particular, $n = 255$ sounds a good pick - the probability is small enough, yet it is still reasonable to send 255 plaintext blocks for interaction. On the other hand, instead of finding the roots for $(x + 1)^{255} = K$, we can simply check if $K^{(2^{128}-1)/255}$ equals one. To summarize, this is how we solve one round of the challenge:

1. Send $m_1, m_2, ..., m_{255}$ to the server, where $m_1 = m_2 = ... = m_{255} = (\text{IV}+k) \oplus \text{IV}$. We then have $c_1 = c_2 = ... = c_k = \mathcal{C}_\text{IV}$.
2. Receive $T$ and 15.5 bytes of $\mathcal{C}_\text{IV}$.
3. For each candidate $C'$ of $\mathcal{C}_\text{IV}$ (there are only sixteen),
• Compute $A := T/C'$.
• Compute $B := A^{(2^{128}-1)/255}$.
• If $B = 1$, assume that $\mathcal{C}_\text{IV} = C'$.

After solving 20 rounds of challenge from the server, we have the flag: PCTF{you_found_the_residues!_db2f6ade22d73a90b15e8d1b06167393}.


text:
# What is the partial pressure of oxygen?

Aug 13, 2016

In the atmosphere?

#### Explanation:

At sea level, since dioxygen 21% of the the atmosphere, the partial pressure of dioxygen is $0.21 \times 1 \cdot a t m = 0.21 \cdot a t m$. What is the partial pressure of dinitrogen if dinitrogen constitutes 79% of the atmosphere?

Note that, as here and as always, the total pressure is the sum of the individual partial pressures: here $1 \cdot a t m = \left(0.21 + 0.79\right) \cdot a t m$ as required.


text:
# Contents

## Idea

In type theory, regarding propositions as types, a proof of a proposition is given by constructing a term of the corresponding type. This can hence be regarded as a program to compute a specific term (output).

## References

An exposition is in

That Martin-Löf dependent type theory can be regarded also as a functional programming language by identifying proofs as programs was stressed in

• Per Martin-Löf, Constructive mathematics and computer programming, Philosophical Transactions of the Royal Society of London (Series A) 312 (1984), no. 1522, pp. 501–518.

Revised on January 12, 2014 10:12:06 by Urs Schreiber (89.204.137.148)


text:
# Postpartum depression and its correlates: a cross-sectional study in southeast Iran | BMC Women’s Health

### Study design and participants

This cross-sectional correlation study was conducted to examine the relationship between prenatal care, depression, anxiety, stress and postpartum depression in women who have recently given birth. The research samples are women who gave birth 3 days ago (via vaginal delivery or cesarean section). The sample size included 186 women giving birth at Nik-nafs maternity ward in Rafsanjan. The inclusion criteria were: (1) mothers 18 years and older; (2) mothers with no known mental health problems or disorders; (3) Mothers without visual and auditory processing disorders. The exclusion criteria were: (1) the presence of gynecological diseases that affect the pregnancy status, the course of pregnancy and the health of mother and child, (2) termination of pregnancy due to premature premature rupture of membranes.

### Sample Size and Samples

Based on studies by Izadirad et al. [21] To determine the association between health literacy and the Prenatal Care Adequacy Index (r = 0.244) with 99% confidence and 90% power, the sample size was assumed to be 140 individuals according to the following formula. Regarding the condition of the mothers and the possibility of non-response, 200 questionnaires were distributed.

$$\upomega =\frac{1}{2}\mathit{Ln}\frac{1+r}{1-r}$$

$$n=\frac{{\left({Z}_{1-\frac{\alpha }{2}}+{Z}_{1-\beta }\right)}^{2}}{{ (\omega)}^{2}}+3$$

Finally, 186 mothers completed the questionnaires.

### Measurement

#### Demographic Information

Demographic information of participants included age, maternal body mass index (BMI), baby sex, type of delivery, previous delivery, number of deliveries, number of pregnancies, history of abortions, number of children, employment status, level education and income.

#### Edinburgh Postnatal Depression Scale (EPDS)

This 10-point self-rated measure is used to screen women for depressive symptoms during pregnancy and the postpartum period with scores from 0 to 3 (maximum EPDS score is 30). The cut-off remains at 13 or more, suggesting prenatal depressive symptoms. This questionnaire was adapted from Cox et al. [22]and has been used in various countries to study postpartum depression [23]. Montazer et al. [24] used it in Iran and the internal correlation coefficient was 0.80 [24]. In the present study, the reliability of the EPDS scale using Cronbach’s alpha coefficient was 0.79 and 0.76 at 3 days and 6 months after delivery, respectively.

#### Depression, Anxiety, Stress Scale (DASS-21)

The Depression, Anxiety, and Stress Scale (DASS-21), developed by Lovibond and Lovibond in 1995, was designed to assess the psychological constructs of depression, anxiety, and stress [25]. The scale consists of 21 items, including seven items for each of the three subscales depression (7 items), anxiety (7 items), and stress (7 items) on a four-point Likert scale (never/low/medium/high). The lowest score is zero and the highest score is three. The total score results from the sum of the scores of the associated items. The total score of the subscales should be doubled. Zakeri et al. [26] In Iran, the Iranian version of DASS-21 reported a Cronbach alpha coefficient of 0.81, 0.74, and 0.78 for depression, anxiety, and stress, respectively [26]. In the present study, the reliability of the DASS-21 scale using Cronbach’s alpha coefficient was 0.75 and 0.80, respectively, for anxiety and stress 3 days after delivery. In addition, the reliability of the DASS-21 scale 6 months after delivery was 0.93 and 0.79 for anxiety and stress, respectively. In the present study, we used two subscales of anxiety and stress.