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# Modus Ponendo Ponens/Proof Rule

## Proof Rule

The Modus Ponendo Ponens is a valid deduction sequent in propositional logic.

As a proof rule it is expressed in the form:

If we can conclude $\phi \implies \psi$, and we can also conclude $\phi$, then we may infer $\psi$.

Thus it provides a means of eliminating a conditional from a sequent.

It can be written:

$\ds {\phi \qquad \phi \implies \psi \over \psi} \to_e$

### Tableau Form

Let $\phi \implies \psi$ be a propositional formula in a tableau proof whose main connective is the implication operator.

The Modus Ponendo Ponens is invoked for $\phi \implies \psi$ and $\phi$ as follows:

 Pool: The pooled assumptions of $\phi \implies \psi$ The pooled assumptions of $\phi$ Formula: $\psi$ Description: Modus Ponendo Ponens Depends on: The line containing the instance of $\phi \implies \psi$ The line containing the instance of $\phi$ Abbreviation: $\text{MPP}$ or $\implies \mathcal E$

## Also known as

Modus Ponendo Ponens is also known as:

• Modus ponens, abbreviated M.P.
• The rule of implies-elimination
• The rule of arrow-elimination
• The rule of (material) detachment
• The process of inference

## Linguistic Note

Modus Ponendo Ponens is Latin for mode that by affirming, affirms.

The shorter form Modus Ponens means mode that affirms.

## Technical Note

When invoking Modus Ponendo Ponens in a tableau proof, use the {{ModusPonens}} template:

{{ModusPonens|line|pool|statement|first|second}}

or:

{{ModusPonens|line|pool|statement|first|second|comment}}

where:

line is the number of the line on the tableau proof where Modus Ponendo Ponens is to be invoked
pool is the pool of assumptions (comma-separated list)
statement is the statement of logic that is to be displayed in the Formula column, without the $...$ delimiters
first is the first of the two lines of the tableau proof upon which this line directly depends, the one in the form $p \implies q$
second is the second of the two lines of the tableau proof upon which this line directly depends, the one in the form $p$
comment is the (optional) comment that is to be displayed in the Notes column.


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# What is the final temperature when 100 g of water at 75"^oC is mixed with 75 g of water at 80"^oC?

Dec 30, 2017

These two systems will equilibrate. The system at a higher initial temperature will be exothermic, transferring heat to the contextually endothermic system.

Hence,

-75g * (4.18J)/(g*°C) * (80°C - T_f) = 100g * (4.18J)/(g*°C) * (75°C - T_f)

therefore T_f approx 77°C

To be sure, this is but a difficult representation of the familiar scenario,

$q = m {C}_{s} \Delta T$,

where the heat supplied is merely coming from the exothermic system.

format: plain


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# Calculating prediction intervals when using cross validation

Are standard deviation estimates calculated via:

$s_N = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}.$

for prediction accuracies sampled from 10-fold cross validation? I'm concerned that the prediction accuracy calculated between each fold are dependent because of the substantial overlap between training sets (although the prediction sets are independent). Any resources that discuss this would be very helpful.

-
– chl Aug 25 '12 at 19:22
Also of possible interest: linear-regression-prediction-interval. – gung Aug 25 '12 at 20:47


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# Tag Info

0

As a general comment Aqua has function at different levels, algorithms such as VQE, QPE, HHL etc, pluggable components that were designed to be replaceable 'parts' of algorithms such as Variational Forms, Optimizers, Uncertainty models etc, and then there are circuits which can be used to build any of the above. When it comes to Phase Estimation it's a ...

0

I believe this tutorial takes you through how to find eigenvectors. They use EigsQPE, through a helper function they define themselves, called create_eigs().

0

qiskit-aqua is the github repo in the link above. Chemistry is in the qiskit-aqua repo under qiskit.chemistry so from qiskit.chemistry.components.variational_forms import UCCSD

3

The chemistry sample in the README on the qiskit-aqua repo shows how to do this now: from qiskit.chemistry import FermionicOperator from qiskit.chemistry.drivers import PySCFDriver, UnitsType from qiskit.aqua.operators import Z2Symmetries # Use PySCF, a classical computational chemistry software # package, to compute the one-body and two-body integrals in #...

1

No, it wasn't removed, it's here: https://github.com/Qiskit/qiskit-aqua/blob/master/qiskit/chemistry/components/variational_forms/uccsd.py

2

Maybe this paper can help you, that's what the implementation in Qiskit is based on. Otherwise looking at the implementation of Shor's algorithm in Qiskit itself might be insightful. The circuit for the algorithm is constructed in the method construct_circuit and can be visualized with this snippet. from qiskit.aqua.algorithms import Shor a, N = 2, 3 ...

1

It seems that docplex class also was moved to from qiskit.optimization.ising import docplex

1

You can simply control $X$ gates with qubits $q3$ and $q4$. You DO NOT have to measure them firstly and then use classical bits for controlling. The reason is that in quantum computing, controling some qubit with other qubits or with their measured results in classical register is the same. Hence, you can implement the algorithm on real quantum computer.

2

I don't think that this will be possible on real current quantum hardware. An alternative would be to run it on a simulator with a realistic noise model. This means that the circuit will be run in a non-ideal environment, and so should incur errors similar to how it would if it was executed on a real device. This tutorial teaches you how to build a noise ...

1

AFAIK, this is impossible on IBM's current hardware. See this github issue: https://github.com/Qiskit/qiskit-textbook/issues/119

3

This can be done using the statevector_simulator provided with Qiskit Aer. It will return the statevector that describes the quantum state at the end of your circuit. It can be used in the same way as the qasm_simulator, only your circuit shouldn't have measurement gates at the end. There is more information about this simulator in this tutorial.

5

The %qiskit_backend_overview magic is not deprecated and has seen updates for the most recent 0.11.0 release: https://github.com/Qiskit/qiskit-terra/blob/master/qiskit/tools/jupyter/backend_overview.py to ensure it works with new devices. The ZeroDivisionError might point to a bug in the code for the function or in your calling code, without a full traceback ...

2

A brute force solution :). You can also obtain CCH via qiskit's basic gates with help of get_controlled_circuit method. from qiskit import * from qiskit.aqua.utils.controlled_circuit import get_controlled_circuit q_reg = QuantumRegister(3, 'q') qc_h = QuantumCircuit(q_reg) qc_ch = QuantumCircuit(q_reg) qc_cch = QuantumCircuit(q_reg) qc_h.h(q_reg[0]) ...

4

Assuming you've got Toffoli and single-qubit rotations, you can implement the following: This basically works because if either of the controls is not $|1\rangle$, the Toffoli does nothing and the two single-qubit unitaries cancel each other. Whereas, if both controls are $|1\rangle$, then the net gate on the target qubit is  (\cos\frac{\pi}{8}I+i\sin\...

1

3 things I see from your implementation of inverse QFT: SWAP gates are missing prior to applying Hadamard gates and cu1 gates. The Hadamard gate should come first before cu1 gates. The angles of cu1 gates, how I understand inverse QFT, should be different. Here is inverse QFT that worked for me with not touching other parts of the code: def qft_dagger(...

0

I just copied and pasted your code above, and that is the output I get from the print() statement. What version of qiskit do you have? from qiskit import qiskit_version print(qiskit_version)

0

What does the print(qc.qasm()) yield for you? Here is what I get: OPENQASM 2.0; include "qelib1.inc"; qreg q2[2]; Controlled-Evolution^1 q2[0],q2[1];

0

It seems that this is not a problem/bug. In case the global phase gate $P$ is applied in non-controlled form, final matrix for two qubits ($P$ is acting on second qubit) is $$I \otimes P = \begin{pmatrix} P & O \\ O & P \\ \end{pmatrix} =\mathrm{e}^{i\phi}I$$ So, $P$ acts as global phase gate and the phase \$\mathrm{e}^{i\...

0

You should put measurement after all quantum gates. According to the theory, there is no difference whether gates (in this case CNOTs) are controlled by qubits or classical bits. I encountered the same problem, see solution here.

1

The error is caused by appending gates onto qubits following a measurement. On qubit 1 and qubit 0, you attach a cx gate after a measurementhas already been placed. This will compile on the simulator, but it is not something that is supported on the real hardware.

3

AFAIK, what you want can't be run on the hardware right now. See this github issue. However, you can do this in the simulators. If, for example, c[0] and c[1] make up a two-bit classical register c, you can do this: qc.x(q[0]).c_if(c,3) qc.x(q[1]).c_if(c,3)

2

This is just a warning to let you know that the API has changed for the backend configurations such as PulseBackendConfiguration. The dt (qubit drive channel timestep) and dtm (measurement drive channel timestep) parameters were previously specified in nanoseconds, but they are now specified in seconds. If you're not doing anything so advanced as to directly ...

5

Using PyCall it is possible to use arbitrary Python packages (such as Qiskit) in Julia and use the Python functions. At least, this is what PyCall advertises to be able. What you describe has been done before: here is an example on GitHub. Essentially, you would need to do this (I haven't tried this out myself): using PyCall qiskit = pyimport("qiskit") ...

Top 50 recent answers are included


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# SSH Logon takes long time?

I’ve been suffering from this on CentOS 7 for quite some time now but haven’t really have time to dig into it.

Just today, I noticed the line after a successful logon:

Last login: Fri March 27 16:03:23 2016 from gateway.

Aha, now I know where the time has been spent. The SSHd must have taken a long time to figure out the host name of my login IP.

I’ve suspected this before, but in my sshd_config file, the line “UseDNS” was commented out, so I thought it must be something else.

A simple “man sshd_config” revealed that, “UseDNS yes” is actually the default setting:

UseDNS  Specifies whether sshd(8) should look up the remote host name and check that the resolved host name for the remote IP address maps back to the very same IP address.  The default is “yes”.

So I just add “UseDNS no” in the configuration file and restarted sshd. Problem solved.


text:
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Paul's Online Notes
Home / Algebra / Common Graphs / Transformations
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### Section 4-6 : Transformations

In this section we are going to see how knowledge of some fairly simple graphs can help us graph some more complicated graphs. Collectively the methods we’re going to be looking at in this section are called transformations.

#### Vertical Shifts

The first transformation we’ll look at is a vertical shift.

Given the graph of $$f\left( x \right)$$ the graph of $$g\left( x \right) = f\left( x \right) + c$$ will be the graph of $$f\left( x \right)$$ shifted up by $$c$$ units if $$c$$ is positive and or down by $$c$$ units if $$c$$ is negative.

So, if we can graph $$f\left( x \right)$$ getting the graph of $$g\left( x \right)$$ is fairly easy. Let’s take a look at a couple of examples.

Example 1 Using transformations sketch the graph of the following functions.
1. $$g\left( x \right) = {x^2} + 3$$
2. $$f\left( x \right) = \sqrt x - 5$$
Show All Solutions Hide All Solutions
Show Discussion

The first thing to do here is graph the function without the constant which by this point should be fairly simple for you. Then shift accordingly.

a $$g\left( x \right) = {x^2} + 3$$ Show Solution

In this case we first need to graph $${x^2}$$ (the dotted line on the graph below) and then pick this up and shift it upwards by 3. Coordinate wise this will mean adding 3 onto all the $$y$$ coordinates of points on $${x^2}$$.

Here is the sketch for this one.

b $$f\left( x \right) = \sqrt x - 5$$ Show Solution

Okay, in this case we’re going to be shifting the graph of $$\sqrt x$$ (the dotted line on the graph below) down by 5. Again, from a coordinate standpoint this means that we subtract 5 from the $$y$$ coordinates of points on $$\sqrt x$$.

Here is this graph.

So, vertical shifts aren’t all that bad if we can graph the “base” function first. Note as well that if you’re not sure that you believe the graphs in the previous set of examples all you need to do is plug a couple values of $$x$$ into the function and verify that they are in fact the correct graphs.

#### Horizontal Shifts

These are fairly simple as well although there is one bit where we need to be careful.

Given the graph of $$f\left( x \right)$$ the graph of $$g\left( x \right) = f\left( {x + c} \right)$$ will be the graph of $$f\left( x \right)$$ shifted left by $$c$$ units if $$c$$ is positive and or right by $$c$$ units if $$c$$ is negative.

Now, we need to be careful here. A positive $$c$$ shifts a graph in the negative direction and a negative $$c$$ shifts a graph in the positive direction. They are exactly opposite than vertical shifts and it’s easy to flip these around and shift incorrectly if we aren’t being careful.

Example 2 Using transformations sketch the graph of the following functions.
1. $$h\left( x \right) = {\left( {x + 2} \right)^3}$$
2. $$g\left( x \right) = \sqrt {x - 4}$$
Show All Solutions Hide All Solutions
a $$h\left( x \right) = {\left( {x + 2} \right)^3}$$ Show Solution

Okay, with these we need to first identify the “base” function. That is the function that’s being shifted. In this case it looks like we are shifting $$f\left( x \right) = {x^3}$$. We can then see that,

$h\left( x \right) = {\left( {x + 2} \right)^3} = f\left( {x + 2} \right)$

In this case $$c = 2$$ and so we’re going to shift the graph of $$f\left( x \right) = {x^3}$$ (the dotted line on the graph below) and move it 2 units to the left. This will mean subtracting 2 from the $$x$$ coordinates of all the points on $$f\left( x \right) = {x^3}$$.

Here is the graph for this problem.

b $$g\left( x \right) = \sqrt {x - 4}$$ Show Solution

In this case it looks like the base function is $$\sqrt x$$ and it also looks like $$c = - 4$$ and so we will be shifting the graph of $$\sqrt x$$ (the dotted line on the graph below) to the right by 4 units. In terms of coordinates this will mean that we’re going to add 4 onto the $$x$$ coordinate of all the points on $$\sqrt x$$.

Here is the sketch for this function.

#### Vertical and Horizontal Shifts

Now we can also combine the two shifts we just got done looking at into a single problem. If we know the graph of $$f\left( x \right)$$ the graph of $$g\left( x \right) = f\left( {x + c} \right) + k$$ will be the graph of $$f\left( x \right)$$ shifted left or right by $$c$$ units depending on the sign of $$c$$ and up or down by $$k$$ units depending on the sign of $$k$$.

Let’s take a look at a couple of examples.

Example 3 Use transformation to sketch the graph of each of the following.
1. $$f\left( x \right) = {\left( {x - 2} \right)^2} + 4$$
2. $$g\left( x \right) = \left| {x + 3} \right| - 5$$
Show All Solutions Hide All Solutions
a $$f\left( x \right) = {\left( {x - 2} \right)^2} + 4$$ Show Solution

In this part it looks like the base function is $${x^2}$$ and it looks like will be shift this to the right by 2 (since $$c = - 2$$) and up by 4 (since $$k = 4$$). Here is the sketch of this function.

b $$g\left( x \right) = \left| {x + 3} \right| - 5$$ Show Solution

For this part we will be shifting $$\left| x \right|$$ to the left by 3 (since $$c = 3$$) and down 5 (since $$k = - 5$$). Here is the sketch of this function.

#### Reflections

The final set of transformations that we’re going to be looking at in this section aren’t shifts, but instead they are called reflections and there are two of them.

#### Reflection about the $$x$$-axis

Given the graph of $$f\left( x \right)$$ then the graph of $$g\left( x \right) = - f\left( x \right)$$ is the graph of $$f\left( x \right)$$ reflected about the $$x$$-axis. This means that the signs on the all the $$y$$ coordinates are changed to the opposite sign.

#### Reflection about the $$y$$-axis

Given the graph of $$f\left( x \right)$$ then the graph of $$g\left( x \right) = f\left( { - x} \right)$$ is the graph of $$f\left( x \right)$$ reflected about the $$y$$-axis. This means that the signs on the all the $$x$$ coordinates are changed to the opposite sign.

Here is an example of each.

Example 4 Using transformation sketch the graph of each of the following.
1. $$g\left( x \right) = - {x^2}$$
2. $$h\left( x \right) = \sqrt { - x}$$
Show All Solutions Hide All Solutions
a $$g\left( x \right) = - {x^2}$$ Show Solution

Based on the placement of the minus sign (i.e. it’s outside the square and NOT inside the square, or $${\left( { - x} \right)^2}$$ ) it looks like we will be reflecting $${x^2}$$ about the $$x$$-axis. So, again, the means that all we do is change the sign on all the $$y$$ coordinates.

Here is the sketch of this graph.

b $$h\left( x \right) = \sqrt { - x}$$ Show Solution

Now with this one let’s first address the minus sign under the square root in more general terms. We know that we can’t take the square roots of negative numbers, however the presence of that minus sign doesn’t necessarily cause problems. We won’t be able to plug positive values of $$x$$ into the function since that would give square roots of negative numbers. However, if $$x$$ were negative, then the negative of a negative number is positive and that is okay. For instance,

$h\left( { - 4} \right) = \sqrt { - \left( { - 4} \right)} = \sqrt 4 = 2$

So, don’t get all worried about that minus sign.

Now, let’s address the reflection here. Since the minus sign is under the square root as opposed to in front of it we are doing a reflection about the $$y$$-axis. This means that we’ll need to change all the signs of points on $$\sqrt x$$.

Note as well that this syncs up with our discussion on this minus sign at the start of this part.

Here is the graph for this function.


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# Windows CE Development Notes

English (en) 中文（台灣） (zh_TW)

## Other Interfaces

### Interfaces Development Articles

General info is here Windows_CE_Interface#Debbuging Windows CE software on the Lazarus IDE

• Sometimes GDB will crash, just reset debugger and start again.
• If you have an open call stack window it will take lots of time each time you trace or debug your program and it sometimes crashes the debugger, so use it only when needed.
• Always put breakpoints where needed; you can not-or hardly can- pause your program or even stop/reset it.
• You can strip your program with --only-keep-debug which save you around 1 mg.There are some issues with strip utility, sometimes it makes your program not executable, So I generally don't recommend using it.
• Sometimes stepping takes lots of time, just be patient ,I've experienced once a simple assignment took 1 min for debugger to execute that code.

## Alignment problems

Using ARM processors, some times you may get a EBusError exception with a message about misaligned data access. The following section explains what this is and how to fix it.

### What is misaligned data access?

Suppose a CPU has a 8bit data bus to memory and you want to read a byte. Since you have a 8bit bus, you are able to address all individual bytes, so reading is not a problem. Now imagine a CPU with a 32bit data bus and reading a byte. Data is read in chunks of 32bits (=4 bytes) and it is addressed by multiples of 4. For getting the exact byte, the CPU "shifts" the right byte.

Now we do the same for reading a 32bits integer. On an 8 bit CPU, you have to read 4 times one byte and then you have the integer. On a 32bits CPU it can be read in one go if the address is aligned to 4 bytes. If not, it has to be read in 2 halves and then combined to one integer. The x86 supports reading like this, although it is inefficient. The SPARC and some implementations of the ARM (*) don't support this and raise a bus error if you read data like this.

(*) there are extensions to the ARM core which can handle it, but this Depends on what functionality the manufacturer of a specific core has Implemented.

### How to solve this?

Use typecasts of pointers very carefully. 16-bit and 32-bit values need to be 4-byte aligned in memory. Otherwise bus error will occur. Or use newly added "unaligned" keyword before using them. You can use unaligned keyword both in left or right side of your assignments.

Examples:

var
p1: ^Longint;
l: longint;
begin
p1^ := 20;  //might cause error if p points to an unaligned memory, i.e. not 4-byte aligned.
unaligned(p1^) := 20;  //it is ok
l := pl^;  //this can also generate error
l := unaligned(pl^);  //it is ok
end.

When using Pack records, compiler automatically generates all access to the record members as unaligned access. Currently there are some small issues which sometimes with nested pack records compiler does not generate the required code. You just have to use unaligned keyword. But note that using packed records will influence speed a lot as all of its members are treated as unaligned. So use them only if needed.

## Implementation details

### ApplicationType

The Windows CE interface will try to automatically detect the kind of device which is being run if nothing is specifyed by the user. The possible devices are:

• atDefault - The LCL will detect automatically the type
• atPDA - A PDA or Smartphone with a touch screen and a small screen. Forms are usually maximized to ocupy the whole work area, and custom positioning and sizing is ignored. Windows CE in default resolution will always have a Width of 240 pixels, and the Height can change from device to device, but 320 pixels is a good guess. In some devices it is possible to specify a bigger size if the screen supports it.
• atKeyPadDevice - Simple phones and some GPS devices and Bar-code scanners. Devices of this type had are very limited and have a small screen, lacking a touch screen or other pointing device. The very limited navigation capabilities mean that the application must be simplified at a maximum. Under WindowsCE for example only 2 top level menus are possible and all forms need to be maximized.
• atDesktop - A full desktop application, for which the operating system just happens to be WinCE. Should work just like any desktop application.

Note that before Lazarus 0.9.29 atKeyPadDevice was named atSmartphone and there was an unsupported atHandheld.

How to specify a fixed device:

program MyProgram;

...

begin
// On Application.Initialize a default device would be detected if
// ApplicationType = atDefault, but now the detection is suppressed.
Application.Initialize;
...
end.

### Positioning and size of Dialogs and Forms

At run-time the position and size of forms on the WinCE interface is different then designed. Specifically, it's dependent on the kind of device being run and on the border style.

ApplicationType BorderStyle Behavior WindowState Title bar Status
atDesktop Any Not yet defined. Currently it should work just like on win32, althougth maybe the Left/Top position should always be ignored, because a Handheld may have a small screen. It's also necessary to check if all border styles are supported (how could one resize a form on a handhelp?) As on desktops As on desktops Implemented
atPDA, atKeyPadDevice bsDialog Accepts custom sizing. The X, Y position is ignored and always set to be the top-left corner, but taking into account the existence of an eventual Taskbar on the top. Not applicable Separate title bar with a "OK" close button Implemented
atPDA, atKeyPadDevice bsNone Allows full control of the size and position to the user. Not applicable Doesn't exist Implemented
atPDA, atKeyPadDevice bsSizable, bsSingle, bsToolWin, bsSizeToolWin Rejects custom size and position and is always set to fill the entire work area excluding the menu. The definition of Work Area already excludes the Taskbar Not applicable Merged into the taskbar with a "OK" close button Implemented

Obs:

• Initially atDefault was defined as atPDA, but since 0.9.27 (maybe earlier) in Application.Initialize the type will be checked automatically and the ApplicationType will be changed to the correct value.

• Code to position a form on the work area: [1]

### The Title "OK" and "X" buttons

A very common problem found in previous LCL for WinCE versions was that under Windows CE, by default, there exists no close button available in the title bar. There is, however, a "X" minimize button, which minimizes the form. Because there is a "X" written on it, and because a very similar button in desktops closes the form, the majority of Lazarus for WinCE users expected that button to close the form. To attend this expectation we have changed the default behavior for our forms and now they will all have a "OK" close button instead of a "X" minimize button.

Predicting that some people will desire to have the X button, it was provided a switch to control the title buttons policy. There are some values possible:

Value Description
tpAlwaysUseOKButton The default one and all forms will have an "OK" close button.
tpOKButtonOnlyOnDialogs Can also be choosen and all forms will have "X" minimize buttons instead of forms with bsDialog borderstyle, which will still have an "OK" close button.
tpControlWithBorderIcons If biMinimize is included in the form's BorderIcons, then a "X" minimize button will be shown. A "OK" button is shown otherwise.

Be very careful using any option which allows the "X" button, because:

• Clicking in the X in the main form of the application won't quit it. You are responsible for providing code so that when the user starts a new instance of your application it will detect the running instance, show the running instance and then quit, or else you will fill the users memory with instances of your application.

Example code to change the default policy below:

{$ifdef LCLWinCE} uses WinCEInt; {$endif}

...

begin
{$ifdef LCLWinCE} WinCEWidgetset.WinCETitlePolicy := tpOKButtonOnlyOnDialogs; {$endif}
end;

Screenshots

This is a title with "X" minimize button:

This is a title with "OK" close button:

Implementation details

The "OK" button will send a WM_COMMAND message, which is handled on wincecallback.inc, with wID=IDOK. There we respond by throwing a WM_CLOSE to the form.

Bug Reports

Showing the popup menu of a windowed control is implemented using the native behavior of the Windows CE platform for touch screens. The user holds the pointer in a given control for a certain amount of time and the popup shows. This is implemented on the WM_LBUTTONDOWN message handler by calling SHRecognizeGesture.

For more details see:

Several different problems with TCustomCheckBox descendents (like TRadioButton) and TGroupBox were found on Lazarus 0.9.25 and are being fixed.

1 - The autosize of TCustomCheckBox descendents under Windows CE produces a too small area, both inside a TRadioGroup and a alone in a form. That indicates that there was a problem with TWSWinCECustomCheckBox.GetPreferedSize. Even with the code exactly equal to the win32 interface the width was still too small, so an extra 10 pixels factor was added arbitrarely to fix that until a better solution is found - Fixed

2 - Aditionally TCustomCheckBox descendents inside a TGroupBox (but not on a form) paint a white background instead of painting the correct background color. The way to handle background color changing on a series of controls is by handling the WM_CTLCOLORSTATIC message. Even with the code exactly the same as the Win32 one that still happens, and for unknown misterious reasons if the control is on a form instead of a GroupBox it works normally.

3 - A third problem found while investigating the other ones is that the difference between the windows client are and the client area that LCL expects was being ignored under Windows CE. This would be seen when one places a control on a TGroupBox or other controls with a border that the position of the places control was wrong - Fixed

(Obs: The captions should be "Item1", "Yet another Item", "Item2", "Item3")

Not yet fixed

Bug Tracker items

### TCheckListBox

TCheckListBox presented some color problems under Windows CE which needed to be fixed. Specifically it was found that on several places where system colors were used, the constant SYS_COLOR_INDEX_FLAG wasn't added. LCLIntf.GetSysColor makes sure the constant is added, so it should always be used instead of Windows.GetSysColor. On desktop Windows this constant equals to zero, so it can be ignored.

The drawing code itself is located at TCustomListBox.LMDrawListItem and TCustomListBox.DrawItem, but what needed fixing was LCLIntf.GetSysColor.

The colors of the TCheckListBox items under Windows CE are:

Description Code Value before fix Correct value Corrected?
background of the entire item GetSysColor(COLOR_WINDOW) black white Yes
background of unselected text default background color white white Yes
background of selected text SetBkColor(GetSysColor(COLOR_HIGHLIGHT)) black blue Yes
background of selected item GetSysColor(COLOR_HIGHLIGHT) black blue Yes
color of selected text SetTextColor(GetSysColor(COLOR_HIGHLIGHTTEXT)) black white Yes
color of disabled text SetTextColor(GetSysColor(COLOR_GRAYTEXT)) black Gray Yes
color of normal text SetTextColor(GetSysColor(COLOR_WINDOWTEXT)) black black Yes

Acording to the MSDN docs a result of 0 for GetSysColor under Windows CE should identify that it isn't available. Now, 0 is also an RGB value (black), so maybe the WinCE docs are wrong and the Win32 docs should be used.

TChecklistbox before fixes

TChecklistbox after fixes

MSDN Docs

Bug tracker items

### Tab Controls (TPageControl)

While most controls work just "as-is" using normal Win32 code, Tab Controls present a big amount of "gotchas" which makes development of a tab control harder than in Win32.

A small list:

• Vertical text is not supported, so tab buttons located on the left and right won't show text. They are converted into bottom-aligned button style instead.
• Microsoft recommends using tabs with bottom style under Windows CE, so that the user won't cover the touch screen with his hand when changing tabs.
• The default style is a bit old. However, the new style seems too minimalistic. You can barely see it's a tab control at all. The call to set new style (flat stype) is on the code (TWinCEWSCustomNotebook.CreateHandle), just commented, so it will be easy to change in the future

Some docs are located here:

Background and sheet position problems

The background of each page seams not to be painted by default, and because the border around the control is very thin and incomplete (not all edges have borders), you can barely see what the client area of the sheet is.

I used a "magic" change on the sheet position given by WinCE as described here:

Here is a TPageControl before the sheet position adjustment:

And after:

Bug tracker items

### DataAware controls

For some unknown reason, data-aware controls crash the application.

Bug tracker items

### Brushes

CreateBrushIndirect doesn't exist under Windows CE, so the interface maps it to CreateSolidBrush, CreateDIBPatternBrushPt and GetStockObject.

If special flags are used they are ignored and CreateSolidBrush is used instead.

From CreateBrushIndirect on wincewinapi.inc:

  if lb.lbStyle= BS_NULL then
Result := Windows.GetStockObject(NULL_BRUSH)
else if lb.lbStyle = BS_DIBPATTERNPT then
Result := CreateDIBPatternBrushPt(pointer(lb.lbHatch), lb.lbColor)
else { lb.lbStyle = BS_SOLID }
Result := Windows.CreateSolidBrush(LB.lbColor);

### SetProp - GetProp - RemoveProp APIs

Windows CE does not have SetProp and GetProp API's which is used a lot. So I (User:Roozbeh) have created/emulated them. The current implementation is not exactly as win32 API. In win32 you can set multiple properties with different names to each window but here I ignore names so we can only have one kind of property assigned to each control. (It is easy to implement that, but I find it currently of no use). Also removeprop is not called yet! So a memory leak happens each time you exist the program. (I don't know if wince also frees memory itself or not when window handles are destroyed).Should be implemented soon.

### Modal Dialogs and the OK and X buttons

Under Windows CE some dialogs will present both the OK and the X button. It is possible to distinguish between them according to the ModalResult that they send. The OK close button will send mrOK, while the X minimize button will close the dialog and send a mrCancel result.

Here is the dialog produced by the following code under Windows CE:

MessageDlg('Shall I test a very, really, long string?', mtConfirmation, [mbYes, mbNo, mbOk, mbCancel],0);

Bug reports:

There is a flaw in Windows CE. The routine Windows.SystemParametersInfo(SPI_GETWORKAREA, 0, @WR, 0); doesn't return what we expect in some systems, and returns it on other ones. The problem is that on some systems the menu bar is considered part of the work area, and in others it isn't. So, when a menu is present we need to check if our program is overlapping the menubar, and if it is, we should make our window smaller.

Some software solve this by applying a factor of 26 to the window height, but this approach isn't that good because there is no way to determina if the device is of the kind that needs the factor.

The solution was implemented in CESetMenu, so the window will only have it's final size after the menu is set. This makes it unclear at the moment if the size of the window would be wrong in some events of the form.

Here is a screenshot of an application which is overlapping the menu:

Routine to maximize a window in Windows CE. Not very good, because it just uses a lot of ifdefs:

Workarea problem description:

Fixed in version: 0.9.27

### Standard dialogs

The native dialogs in Windows CE are too simplistic and limit the user to a small number of directories. Better dialogs will be built instead. There are some missing controls in the LCL, so those will need to be created first.

Steps to implement the standard dialogs:

1. Implement TShellTreeView
2. Implement TShellListView
3. Implement TFilterComboBox
4. Implement TOpenDialog and TSaveDialog
5. Implement other dialogs
6. Fix any bugs that appear

The implementation should be very similar to this:

In the save dialog I would remove the TFileListBox to make room for a TEdit.

A special care will need to be made to make sure that the dialog fits any resolution.

Images

State of the open and save dialogs in Lazarus 0.9.28

Bug tracker items

### Running in desktop Windows

Yes, the Windows CE interface also works for desktop Windows. I (Felipe Carvalho) implemented this feature to speed up debugging, as the desktop windows debugger is much quicker, and also the desktop compiler is much quicker. There are still bugs to be solved for this interface, but it has served well its purpose until now.

A particularly hard to implement feature are smartphone menus, which are very limited and often hard to work with. Their most visible limitation is the number of top-level menu items being at most 2, so that they can be easily accessed in devices without touchscreen.

Another important thing to notice is that PocketPC and Windows Mobile top-level menus are implemented with a TOOLBAR windows class, not with MENU class. The MENU class is destined for popup menus, those which are activated by clicking in the top-level menu, or in a popup menu.

Also, the top-level menu can only be set to a resource menu, it cannot be completely programatically created, but it can be altered later, which offers an equivalent solution. The menu is set to a resource menu with SHCreateMenuBar.

Differences between smartphone and pocketpc menus:

Some usual things don't work with smartphone menus, for example if a menu item has been set as not having a submenu, then it cannot be changed to have one, that's why we need many menu solutions in wincemenures.rc

Flow of information of menu items:

A click in a menu item is received as WM_COMMAND message, then we have a list of menu items called MenuItemsList and we search in this list for the menu "command". Note that the top-level items actually have the fixed values 1001 and 1002, so secondary forms cannot receive top-level menu clicks in atKeyPadDevice mode, because there is a clash between the menu IDs of both forms.

Revisions related to this implementation:

Bug tracker items:

Reference:

### Regressions

From time to time there can be regressions and specially in 0.9.29 a good amount of regressions were introduced in the Windows CE Widgetset. This documents the effort to find and fix those regressions.

Table about the behavior of a single issue in time:

Revision Version Compilation Obs Group Box Label PageControl ComboBox TStringGrid Window
23000 - - correct position The label inside a group box doesn't even show. Other labels show correct position. Border broken ? ? ?
23136 - - correct position The label inside a group box doesn't even show. Other labels show correct position. Border broken Works normally ? ?
23137 - - correct position The label inside a group box / pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn Works normally ? ?
23289 - - correct position The label inside a group box / pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn Works normally ? ?
23290 a 23426 - Doesn't build in Windows CE
23427 - Freeze under FPC 2.4.x caused by rev 23290
23500 - Freeze under FPC 2.4.x caused by rev 23290 correct position The label inside a group box / pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn Works normally ? ?
23618 - Fixes the Freeze caused by rev 23290 correct position The label inside a group box / pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn Works normally ? ?
23750 - - correct position The label inside a group box / pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn ? ? ?
23900 - - correct position The label inside a pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn Works normally ? ?
24084 - - correct position The label inside a pagecontrol doesn't even show. Other labels show in (0,0). Full border drawn Works normally ? ?
24085 - - ? ? ? Freezes the form ? ?
24176 - - correct position The label inside a pagecontrol doesn't show. All labels have zero X,Y position Full border drawn Freezes the form ? ?
24177 - - painted wrongly The label inside a pagecontrol doesn't show. the label inside a group box has too small x-position Border broken ? ? ?
24500 - - painted wrongly The label inside a pagecontrol doesn't show. the label inside a group box has too small x-position Border broken ? Works Maximized doesn't work
25000 0.9.29 - painted wrongly The label inside a pagecontrol doesn't show. the label inside a group box doesn't even show Border broken ? Works ?
25170 a 25178 - Doesn't build in Windows CE and rev25170 also breaks the TStringGrid scrollbar
25180 0.9.29 5 may 2010 - painted wrongly The label inside a pagecontrol doesn't show. the label inside a group box has too small x-position Border broken Freezes the form Works ?
25213 0.9.29 - painted wrongly The label inside a pagecontrol doesn't show. the label inside a group box has too small x-position Border broken ? Works ?
25214 0.9.29 - fixed The label inside a pagecontrol doesn't show fixed ? Broke in this rev ?
25300 0.9.29 - = = = = Broken ?
25483 0.9.29 - = = = = = Maximized doesn't work
25484 0.9.29 - = = = = = Maximized fixed

Table of what was broken and fixed in each revision:

Revision Version Description
23137 - Broke label in form positioning to (0, 0), fixed pagecontrol frame

The problems in rev23137 were caused by the introduction of LCLIntf.SetViewPortOrgEx which made this routine no longer be implemented, as now it was calling LCLIntf version of SetViewPortOrgEx (not implemented under WinCE) instead of the WinAPI:

function TWinCEWidgetSet.SetWindowOrgEx(DC: HDC; NewX, NewY: Integer;
OldPoint: PPoint): Boolean;
begin
{ roozbeh:does this always work?it is heavily used in labels and some other drawing controls}
Result := Boolean(SetViewPortOrgEx(DC, -NewX, -NewY, LPPoint(OldPoint)));
//  Result:=inherited SetWindowOrgEx(dc, NewX, NewY, OldPoint);
end;

rev24177 fixed that, but it didn't fix the problem completely.

23794 - Fixes label showing in Group Box
24085 - New Autosize causes freeze in TComboBox
24177 - Broke Group Box title, Group Box painting, label inside control X position, pagecontrol frame
25170 - Breaks TStringGrid Scrollbar. See http://bugs.freepascal.org/view.php?id=16576
25185, 25194, 25201, 25202, 25203, 25214 - Fixes Group Box title, Group Box painting, label painting and pagecontrol frame. Removes support to clipping to achieve that.
25213 - Breaks TStringGrid Painting
25314, 25350, 25355, 25357 - Fixes Combo Box autosize freeze.
25359 - Fixes SpinEdit autosize freeze.
25484 - Fixes Window wsMaximized
25661 - Fixes TStringGrid drawing, breaks changing a label inside a TGroupBox
25710 - Fixes TStringGrid scroll bars

Bug tracker items:

### Networking crash bug

Something in FPC introduced between 2.2.0 and 2.2.2 is causing a crash in WinCE applications.

FPC 2.2.0 was tagged in revision 8339 FPC 2.2.2 was tagged in revision 11490

The involved FPC Branch is: http://svn.freepascal.org/cgi-bin/viewvc.cgi/branches/fixes_2_2/

List of merged revisions into FPC 2.2.2:

8404, 8410 - Only linux 8411 - Compiler 8430 - RTL, Compiler 8438 - Utils

Bug tracker items:

### ShowMessage regression

Revision Observations
34300 Works correctly
34400 Doesnt compile on WinCE
34600 Doesnt compile on WinCE
34700 Works correctly
35000 Works correctly
35500 Works correctly
35700 Works correctly
35800 Works correctly
35950 Works correctly
35960 Works correctly
35961 Bug present
35962 Bug present
35963 Bug present
35965 Bug present
35975 Bug present
35980 Bug present
36000 Bug present

## List of devices with information about them

There are differences between devices, so it would be interresting to have a list with information about each of them. Please contribute.

Device Name WinCE Version ApplicationType (Auto-detection works?) Touchscreen? Width* Height* Menu Size** Observations
WM 5.0 Emulator - PocketPC 5.0 atPDA (yes) Yes 240 268 26
WM 5.0 Emulator - Smartphone 5.0 atKeyPadDevice (yes) No 176 180 0
WM 5.0 Emulator - Smartphone QVGA 5.0 atKeyPadDevice (yes) No 240 266 0
WM 2003 SE Emulator - Pocket PC 4.20 atPDA (?) Yes 240 294 0
WM 2003 SE Emulator - Pocket PC Square 4.20 atPDA (?) Yes 240 214 0
• * - Values available for a maximized program will all extras activated (start menu, menu correction, PID, etc).
• ** - The size of the correction which needs to be applyed when a menu is present. The LCL should calculate this automatically.

Abreviations:

• WM = Windows Mobile
• SE = Second Edition


text:
# What is energy

Ghetalion

Is energy simply kinetic interactions between matter BY matter? My understanding is that chemical and heat are both electromagnetic and are simply photonic flux.

Locrian
It's a scalar quantity associated with the state of a system. That's a pretty vague definition, but since we're in the QM forum, I think it's appropriate.

Ghetalion
Locrian said:
It's a scalar quantity associated with the state of a system. That's a pretty vague definition, but since we're in the QM forum, I think it's appropriate.

A state of a system

Meaning the position of matter within that system, correct?

Antiphon
Ghetalion said:
A state of a system

Meaning the position of matter within that system, correct?

There's more to it than position. But since this is a QM forum,
energy numerically defines the frequency of something.
We don't know know what that something actually is but mathematically
it looks like this:

$$\Psi = e^{i\frac{E}{\hbar}t}$$

Ghetalion
Antiphon said:
There's more to it than position. But since this is a QM forum,
energy numerically defines the frequency of something.
We don't know know what that something actually is but mathematically
it looks like this:

$$\Psi = e^{i\frac{E}{\hbar}t}$$

Frequency suggests a fluxuation of material density... doesn't it?

pmb_phy
Ghetalion said:

Is energy simply kinetic interactions between matter BY matter? My understanding is that chemical and heat are both electromagnetic and are simply photonic flux.
Nobody knows what energy is.

Pete

Homework Helper
"Energy" is a book-keeping device. It was noted pretty early on in physics that in simple collisions, two quantities mass*velocity and 1/2 * mass* velocity2 stayed the same which made them easy to use- one was called "momentum" and the other "kinetic energy". It was observed that the speed of an object (and so it's kinetic energy) increased as it fell, decreased as it went up. "Potential energy" was defined in order to keep "total energy" constant. Of course, if you added friction, you no longer had conservation of those energies but could observe that the temperature increased proportional to the lost energy. Okay, "heat energy" keeps the book keeping straight! As of the advent of relativity, because "e= mc2", we had to include mass itself as a kind of energy!

pmb_phy
HallsofIvy said:
"Energy" is a book-keeping device. ....
Nice post HallsofIvy. You are probably aware of this but that does not define energy. It only defines a property of energy. You'll note that total momentum also satisfies the property you just described.

I took a crack at this topic here

http://www.geocities.com/physics_world/mech/what_is_energy.htm

Pete

Last edited:
Joethemonkey
first of all i'd like to point out that heat is not electromagnetic, Microwave is electromagnetic.. Heat is the vibration of particles in a substance...

secondly....... its late and i am too tired to finish this post.... *laught out loud* hopefully i'll remember to add another concluding post later... Goodnight

Joethemonkey
oh by the way i mentioned microwave,. because it transfers into heat energy ... *Microwave* ovens use microwaves to heat up food... check out some articles on "howstuffworks.com" that website has some real good information on electromagnetic waves.. light and radio,. energy... a great information source for this kind of thing... Hopefully HowStuffWorks will be able to answer your question,. as i am too lazy and tired right now

DaTario
I would put it like this:

Energy is a class of mathematical combinations of quantitative aspects of one system and its enviroment which forms a way to compute the maximum work that can be done by this system on the parts of its enviroment.

Best Regards

DaTario

Masoud
In my opinion, the best definition of energy would be embedded within M-Theory.

Masoud Zargar

DaTario
Masoud said:
In my opinion, the best definition of energy would be embedded within M-Theory.

Masoud Zargar

Could you please write this best definition of M-theory ?

Best Regards

DaTario

Telos
Maybe we should ask the Department of Energy?

Zelos
im getting sick of this question. ppl ask what energy/matter/charge is. this are proepties matter and things have. its not like its particles that sourond other particles

Telos
Well, Zelos, this is a little different. There isn't a Department of Matter or a Department of Charge.

Telos
And there's a State Department. Can't tell if it obeys Pauli Exclusion, though. They all look the same.

mcah5
According to Feymann, energy is just a book-keeping device with the very useful property that energy is conserved. The conservation of energy comes from the fact that experiments are time symmetric (aka "things do not depend on the absolute time.)

Edit: I can't spell Feynman

Telos
That's very interesting, mcah5, because for me, being a human in modern civilization, if I don't have energy, I die.

What is it that I call "energy," then, since it is more than a bookkeeping device? Is it "energy transfer?"

Staff Emeritus
Gold Member
Dearly Missed
Telos said:
That's very interesting, mcah5, because for me, being a human in modern civilization, if I don't have energy, I die.

What is it that I call "energy," then, since it is more than a bookkeeping device? Is it "energy transfer?"

No; if you don't have the right chemical processes, you die. Energy, and its derivatives like free energy, are just abstractions from that. They are important to a description of what is happening, but they are part of the model, not part of the reality.

Zelos
if we can define energy we can denife matter as high concentration of matter and therefor is the matter about matter solved

inha
that makes no sense at all.

Telos
No; if you don't have the right chemical processes, you die. Energy, and its derivatives like free energy, are just abstractions from that. They are important to a description of what is happening, but they are part of the model, not part of the reality.

I understand your last statement, but your first has confused me. You seem to be saying that biochemistry is the basis for our concept of energy. Because that is from whence we abstract the notion?

How are the "right chemical processes" not also an abstraction? Why just yesterday I heard someone accuse another of having the wrong chemical processes.

mcah5
Fine, if you don't like the book-keeping definition then energy is the capacity to do work =)

I think the main problem of trying to define energy is that energy is not quantized, and so we can't imagine a chewy nugget of energy. Personally, I could care less about the "real" definition of energy; problem sets are hard enough without dealing with philosophy.

Locrian
mcah5 said:
I think the main problem of trying to define energy is that energy is not quantized,

Are you sure?

mcah5
I'm pretty sure. You could argue that E=hf means energy is quantized, but f can be any value. And I'm pretty sure there is no lower bound on the frequency of light.

pmb_phy
mcah5 said:
I'm pretty sure. You could argue that E=hf means energy is quantized, but f can be any value. And I'm pretty sure there is no lower bound on the frequency of light.
The assertion that energy is quantized is not meaningful. Before one can state that energy is quantized one must first specify what you're speaking of, i.e. energy of what??

If E = hf refers to the energy of a photon then it can't be assumed that E only comes in certain values. But if E refers to the energy state of an atom then E is quantized. In quantum mechanics there are systems which have a continuos energy spectrum and those which have a descrete energy spectrum. For example, the energy of a free particle is not quantized. The energy of a particle in a box is quantized.

Pete

pmb_phy
mcah5 said:
Fine, if you don't like the book-keeping definition then energy is the capacity to do work =)
That holds true for potential energy and kinetic energy. So that can be taken as a definition of either. But that does not give you a definition of energy, only instances of energy. What am I talking about you ask? Suppose you ask what the definition of life is. If you give examples of life then you've not actually defined life. E.g. a human being is a life form. An ant is a life form. Yet with that knowledge I still don't know what a life form is.

Pete

Telos
Once again we don't seem to have found a satisfactory answer.

Remarkably, we seem to have come to the conclusion that energy isn't real. It only exists in our models and not in reality. How strange that the entirety of physics would depend on something that isn't real? Or am I mistaken?

Enos
It exist and it doesn't. That's the simplest one can put it. It exist because it takes up space. It doesn't exist because it's value is zero.

That's what I think anyway.

pmb_phy
Telos said:
Remarkably, we seem to have come to the conclusion that energy isn't real.
WHAT??? Where did you get that impression???

Is the number 4 real? Is height real? Is life real? Don't hold that something is not real because you've been unable to define it.

Pete

Telos
Enos, that almost sounds like energy "is" space. After all, space tautologically takes up space and has an unwarpped equilibrium value we might as well call zero.


text:
# Math Help - clopen sets in Std Topology

1. ## clopen sets in Std Topology

Which sets are closed and open in R^2 with the Std topologgy?

I am confused about this, help plz!! THis is a question that was made while looking at boundaries. Can you make a general statement?

2. Originally Posted by Andreamet
Which sets are closed and open in R^2 with the Std topologgy?

I am confused about this, help plz!! THis is a question that was made while looking at boundaries. Can you make a general statement?
The sets $\emptyset, \mathbb{R}^2$ are both open and closed.
Try it out, what is the definition of being open and closed?

3. Another example can be found in a disconnected topological space.

A topological space X is disconnected iff X has a proper subset A which is both open and closed.

For instance, the subspace $X=(0,1) \cup (2,3)$ of R is a disconnected space.
Both (0,1) and (2,3) are open in X.

Definition: A set C is closed in X if X\C is open.

Now, a complement of (0,1) in X is (2,3) which is open in X. Thus, (0,1) is a closed set in X. So, we can conclude (0,1) is a clopen set in X.

You can check that a boundary of (0,1) in X is empty as well (same with (2,3)).


text:
Anual Mensual Semanal Hoy Buscar Ir al mes específico Enero Febrero Marzo Abril Mayo Junio Julio Agosto Septiembre Octubre Noviembre Diciembre 2016 2018 2019 2020 2021 2022 2023 2024 2025
SEMINARIO DE TEORÍA DE NÚMEROS

SEMINARIO DE TEORÍA DE NÚMEROS

Martes 7 de junio de 2016
12 h, Aula Gris I (ICMAT, Campus de Cantoblanco)

JEAN-MARC DESHOUILLERS
(BORDEAUX)

"Some aspects of automatic sequences
"

Resumen:
In the late 70’s, I fell by chance on “automatic sequences” through the problem of the distribution modulo 1 of the sequence ${theta^n)_n}$ where $theta$ is algebraic over $mathbb{F}_q(X)$. Some questions were then solved, but some are still opened. Some thirty years later this theme turned out to be a clue for solving a question raised by Florian Luca and Igor Shparlinski. Since then, I happened to be connected again and again to this theme. I feel this efficient tool should be somewhere in the mind of any number theorist, and I’ll advocate it. No a priori knowledge on automata theory is required.

Localización Martes 7 de junio de 2016 12 h, Aula Gris I (ICMAT, Campus de Cantoblanco)


text:
arXiv - CS - Computational Complexity Pub Date : 2020-03-18 , DOI: arxiv-2003.08331
Aviv Adler; Jeffrey Bosboom; Erik D. Demaine; Martin L. Demaine; Quanquan C. Liu; Jayson Lynch

In the Nikoli pencil-and-paper game Tatamibari, a puzzle consists of an $m \times n$ grid of cells, where each cell possibly contains a clue among +, -, |. The goal is to partition the grid into disjoint rectangles, where every rectangle contains exactly one clue, rectangles containing + are square, rectangles containing - are strictly longer horizontally than vertically, rectangles containing | are strictly longer vertically than horizontally, and no four rectangles share a corner. We prove this puzzle NP-complete, establishing a Nikoli gap of 16 years. Along the way, we introduce a gadget framework for proving hardness of similar puzzles involving area coverage, and show that it applies to an existing NP-hardness proof for Spiral Galaxies. We also present a mathematical puzzle font for Tatamibari.

down
wechat
bug


text:
## Maple 2021 Questions and Posts

##### These are Posts and Questions associated with the product, Maple 2021

After a long hiatus I have come back to the issue of null tetrads in the physics package in light of the updates to  Maple in 2021. I have uploaded a document file to illustrate. See below. My first question concerns the labelling of elements of a null tetrad. After calling the metric 27,37 from Stephani et al, and using Setup to specify a null tetrad, Maple's choice is such that the elements labelled m and bar m are not complex. Rather, both these elements are in fact real, while the elements lablled l and n are complex, with one being the negative complex conjugate of the other. While these are just labels, they don't agree with the usual conventions for the Newman-Penrose formalism, which is disorienting. What convention is Maple using to label the elments of a null tetrad?

Next, I try to specify the null tetrad used by Stephani et al., first by converting it into covariant form (which I did by hand rather than in Maple). In Maple's default null tetrad, the order in which Maple listed the elements of the null tetrad is n, m, bar m, l (as rows in the matrix display for e_[ ]), so I followed that convention (in the conventions of Stephani et al., the first and fourh element should have scalar product -1, the second and third scalar product 1, and all other scalar products zero, which is the case). After entering the matrix and using Setup to specify the null tetrad by the matrix, I get an error message saying that the components of the metric with respect to my tetrad are not just 0, 1, and -1. Yet,  executing eta_[ ]  does not confirm this warning; nor does a computation by hand.

Finally, IsTetrad asserts the tetrad is not null, contrary to the fact that it is a null tetrad.

Since I have followed the conventions implicit in Maple's default null tetrad for this metric, I am puzzled as to what has gone wrong.SKMHH27_37_2021_New.mw

On the other hand, taking into account how Maple 2019 orders the coordinates in Stephani et al 27.37 and labels the null vectors in a null ttetrad, if I translate accordingly what I have in the 2021 Maple file, Maple 2019 confirms Stephani et al.'s null tetrad is indeed a null tetrad, as one would expect. See the following file.SKMHH27_37_2019_Var.mw.

## How can the functional derivative/variation of the...

Hello Everyone,

My questions is about a straight forward as my title suggests. I know how to compute the variation by hand and of course it is in plently of books, but I would like to know how I could possibly do it in Maple as I have some terms with non-trivial contributions where I infact need that specific variation.

I have attached my file I have been working with that has more comments.

RicciAction.mw

Thank you

## How can I Simplify in the physics package a tensor...

Hi MaplePrimes,

I am interested in obtaining some gravitational field equations from an action using the FunDiff command. I have been able to write what I think is a pretty short and quick worksheet(with an arbitrary metric) and I am easily able to obtain the Einstein Field Equations. However, when I introduce some new more complicated terms into the action and apply the Simplify command maple does not appear to be able to evaluate and I end up halting the computation. When I specify a metric Maple, for example Schwarzschild, Maple will easily be able to Simplify my expression but it will use metric components during the process. Where what I am interested in is just the Tensor expression with respective indices.

I was wondering if anyone had any thoughts on how I could resolve this.

I have attached the worksheet that I am working with. I look forward to your thoughts/comments.

Thank you.

ActionFieldEquations.mw

## Copying output together vs seperately...

So I have a problem when copying multiple output from Maple to Word. I am using a simple example here.

When I copy each part of the output as an image SEPERATELY, it comes out much cleaner than if I highlight everything together and then copy. Is there a reason this is the case and is there a way to fix this problem?

Here is the Maple Code:

with(plottools);
with(plots);
ttt1 := textplot([0, 1, "3  +  5  =  __", color = "black", font = ["Arial", "bold", 120]]);
display(ttt1, size = [1000, 200], axes = none);
ttt1 := textplot([0, 1, "1  +  3  =  __", color = "black", font = ["Arial", "bold", 120]]);
display(ttt1, size = [1000, 200], axes = none);

I also uploaded the Maple Worksheet

Here is the output:

There are TWO imges in the output in this example. I tried copying both together and pasting them, and then I tried copying them individually. The output is very different in quality.

The one copied alot is much cleaner once copied and pasted in Word. Does anyone why this is happening? I don't want to have to copy images individally for obvoius reasons. I know this example is only two images but what if there was alot more, etc.

Thank you

 >
 >
 >

## Problem with Solving a PDE...

Hi Everyone,

I intend to find the invariants of a 2nd-order Tensor using the representation of the rotational group as an Infinitesimal generator. However, maple does not seem to find the correct invariants $\tr \mathbf{a}$, $\tr \mathbf{a}^2$, $\tr \mathbf{a}^3$. Maple only returns an empty set. I can't seem to figure out why that is the case.

Here is my code

 >
 >
 >
 >
 >
 (1)
 >
 (2)
 >
 (3)
 >
 >
 (4)

Any input would be appreciated!

## Make my Own Code and Call to other file and Get Re...

Hello;

Hope you are fine. I need to learn how can I call my code in the other file of maple, define new value for A and can get results. For this purpose, I just take an simple example as

questionmaple.mw

 >
 (1)

## Syrup Spice Toolbox - Netlists with arbitary volta...

Good day,

I am looking for some guidance with the Syrup toolbox

I have a netlist that was generated with the Altium designer Spice simulator (XSpice based), I was able to import it into LTSpice and verrify the results.

I want to move it into maple to make use of the modellica code generation.

I then began modifying the netlist in order to meet the netlist syntax of Syrup. One of the most prominant change I needed to make was the arbitrary input nodes for my Voltage dependant Voltage sources.

Therefore, for any line that implies arbitrary sources such as:

E_U8_E1 U8_N208620 LO
+ VALUE { IF(V(U8_N208706, 0) > 0.5, 5, -5) }

was changed to:

V1    16   17 1vdc

E_U8_E1 U8_N208620 LO 16 17
+ VALUE { IF(V(U8_N208706, 0) > 0.5, 5, -5) }

This, of course, is a non ideal workaround which brings me to question 1,

Is there a way to define a arbitrary source with the Spice parser in Syrup?

Once these changes were made, it seems the Solve function in Syrup can parse the netlist without issue, however when it gets to the Solve portion it throws the following error:

Solve("file://C:/Users/msavoy/Documents/Spice_cct/inverter_newfet_nonot.cir", 'tran');
Solve: Analyzing SPICE deck "The Apple" (ignoring this line)
Error, (in Syrup:-Solve) invalid input: rcopy uses a 1st argument, A (of type anything), which is missing

Attached is a copy of the netlist, inverter_newfet_nonot.txtI am honestly at a lost as to what could be throwing the rcopy error

Furthermore, I wanted to test out the Modellica code generation in Syurp, it appears that it does not like having subcircuits, as running the test:

V 1 0 1
L1 1 2 L
C1 2 0 C
L2 2 3 L
C2 3 0 C
L3 3 4 L
C3 4 0 C
Vshort 4 5 0
Rt 5 0 1
.SUBCKT TEST A B C Y
V1 14 15 1
EINT YINT 0 14 15
+ VALUE {IF(V(A) > 0.5 && V(B) > 0.5 && V(C) > 0.5, 1, 0)}
RINT YINT Y 1
CINT Y 0 1n
.ENDS TEST
.end

provides the error of:

test_c := ToModelica("file://C:/Users/msavoy/Documents/Spice_cct/test_cct_RCL.txt", 'probes' = ["Rt.v"], 'parameters' = {C = 2, L = 1});
Error, (in Syrup:-ToModelica) invalid input: nodeToModelica expects its 1st argument, node, to be of type nonnegint, but received Y

Is it possible to define a sub circuit in a netlist that is to be converted to Modellica, or do all subcircuits need to be defined in their own file?

Thank you

## How to compute the restricted-edge-connectivity of...

！Edits: I have found  a existing polynomial algorithm, but I still have difficulty implementing it. 2022/10/26

The edge connectivity of a connected graph G  is the minimum number of edges whose deletion from the graph G disconnects G. Below we are concerned with a particular kind of edge-cut.

• For a connected graph G=(V ,E), an edge set S ⊂ E is a restricted-edge-cut, if G−S is disconnected and every connected component of  G−S has at least 2 vertices.

Clearly, a restricted-edge-cut is an edge cut with a special requirement.

• The restricted-edge-connectivity of G, denoted by κres(G), is defined as the cardinality of a minimum restricted-edge-cut.

For example, a graph g is as follows.

Clearly, its edge-connectivity is 1 since (0,3) or (0,4) is a edge cut of g. But we can find that  if we remove the edge (0,3), then "3" is a isolated vertex. Similarly, "4" is a isolated vertex if we remove  (0,4). It is not difficult to find g has exactly the two cut-edges (0,3) and (0,4).

Based on the definition of the restricted-edge-cuts, neither {(0,3)} nor  {(0,4)} are restricted-edge-cuts A minimum restricted-edge-cut is {(0,1),(0,2)} since every connected component of G-{(0,1),(0,2)} has  (at least) 2 vertices.

So  κres(g) is 2.  My problem is:

Given a graph G, how to calculate the restricted-edge-connectivity of  G?  Moreover, how to find a minimum restricted-edge-cut?

A specific graph that I want to test is as follows. (it has 16 vertices and 56 edges.) I would like to calculate its restricted-edge-connectivity and find a minimum restricted-edge-cut.

g:=ConvertGraph("O~tIID@wL~jPbOqgLJ@p");
vertexcolor="Black",edgethickness=2])

EdgeConnectivity(g)

6

One option I came up with is to find all 6-edge subsets first. Test if they satisfy  the restricted condition (one by one). Then continue to increase to 7 or more. But this violent calculation may get stuck in the first step. That is, test all the minimum edge cut sets (Note that we will consider 32468436 edge-subsets!) I was referring to the efficiency aspect.

with(Iterator):
C:=Combination(58,6):
K:=Edges(g):
#sub:=seq( K[[seq(c[]+1)]], c=C); # do not run this code since it has 32468436 members.

Any language is acceptable.( C or C++, Python. )

PS: Some time ago, I also asked a related question (but with some differences) on mathematica stack （Find all the minimum edge cuts of a graph）. Although Bob Hanlon  gave a reply, the actual result is not good.

Edits: The following literature gives a polynomial algorithm for computing the restricted-edge-connectivity of a given  graph. The heart of it is to computing the least cardinality of some  edge-pairs's edge separator. I'm stuck here.

• Esfahanian A H, Hakimi S L. On computing a conditional edge-connectivity of a graph[J]. Information processing letters, 1988, 27(4): 195-199.

How to implement this algorithm is my current concern.

## Why the polysols does not work?...

polysols(diff(u(x), x) = u(x)^2 - 1) produces no results, while it can be verified by direct observation that u(x) = 1 is a polynomial solution.

## Error, MapleSim does not evaluate to a module...

I have maplesim installed and I get the error "Error, MapleSim` does not evaluate to a module" when I run A:=MapleSim:-LinkModel(); in a Maple worksheet.

Does anyone know how to solve this error?

## Why can't Maple solve the below code for some par...

Hello everybody.

I wrote the attached code. this code gives a good result when "H" parameter is equal to 0.5. Even in this condition if I change other parameters, I can get a true result as well. But when I change "H" parameter to 0.6,0.7,0.8 or 0.9 the code can't be executed anymore and I can't get a result.

based on the paper, for H=0.6,0.7,0.8 or 0.9. we have a result.

what's the problem?

1.mw

## Font-size of caption, labels and axes on plots sma...

Maple 2021.
I expect after exporting to pdf view of page with plot like on fig. 1.

It means with font=[TIMES, ROMAN, 12] as in maine text.
But I obtaned this view of page on pdf (fig. 2) with smaller font-size.

In previouse versions there are no the problem like this.
Dear community, how to fix this annoying inconvenience?

fig. 1.  fig. 2.

## Maple Bombs While Entering Text...

Hi.  I am using the linux version of Maple.  I seem to experience more bombs in Maple/Linux vs MathCad/Windows.  The program just froze as I was entering text.  Just text.  It is not a disaster since I make very frequent saves.  But it is very annoying.  I was a heavy user of MathCad from about 2005 to 2019.  In the early days, it bombed a lot.  But I do not recall that it ever bombed when I was entering text.  Any suggestions?  Would there be a limition on the number of text characters?  Is it Linux?  Other than this problem, it is a great program.


text:
1. ## Drug Dosage Modeling

Hopefully this is the right forum for this. I'm going over a Math Modeling book which is dealing with drug dosage modeling and i'm having trouble figuring out how they reached the following conclusion.

If we assume that C(0) is the concentration at t = 0, then we have the model:
dC / dt = -k * C
implies
C(t) = C(0) * e^(-k * t)

I've taken Calc. 3, so I feel like I should have some idea of what's going on here, but I can't remember. My idea:

dC / dt = -k * C
dC = -k * C * dt
integral(1)dC = integral(-k * C)dt
c = -k * c * t + constant
1 = -k * t + constant
-k * t = constant.
Which, is a far cry from what the book has come up with. Could someone give me an idea of what's going on here?

2. Try this

$\displaystyle \frac{dC}{dt}= -kC$

$\displaystyle \frac{dC}{C}= -k~dt$

$\displaystyle \int \frac{dC}{C}=\int -k~dt$

Can you finish him from here?

3. Originally Posted by pickslides
Try this

$\displaystyle \frac{dC}{dt}= -kC$

$\displaystyle \frac{dC}{C}= -k~dt$

$\displaystyle \int \frac{dC}{C}=\int -k~dt$

Can you finish him from here?
ln(C) = -k t
e^ln(C) = e^(-k t)
C = e^(-k t)

That makes sense to me now, thanks. I'm still not sure where the C(t) and C(0) came from though. I will have to investigate this some more.

4. Don't forget the constant of integration, lets call it K in this case

$\displaystyle \ln C= -kt+K$

$\displaystyle C= e^{-kt+K}$

$\displaystyle C= e^{-kt}\times e^{K}$

Making $\displaystyle e^{K}=C_0$

$\displaystyle C= C_0 e^{-kt}$

5. Hey, thank you. I bet that is what they did.

6. ## Re: Drug Dosage Modeling

this should need time to figure it out...


text:
Since 18 of December 2019 conferences.iaea.org uses Nucleus credentials. Visit our help pages for information on how to Register and Sign-in using Nucleus.

# 28th IAEA Fusion Energy Conference (FEC 2020)

10-15 May 2021
Virtual Event
Europe/Vienna timezone
The Conference will be held virtually from 10-15 May 2021

## Quasioptical propagation and absorption of electron cyclotron waves from both numerical and experimental point of view

11 May 2021, 14:00
4h 45m
Virtual Event

#### Virtual Event

Regular Poster Magnetic Fusion Theory and Modelling

Kota Yanagihara

### Description

The quasioptical ray tracing code named PARADE (PAraxial RAy DEscription) [a,b], which can contribute to the modeling of propagation and absorption of wave beams in both tokamak and stellarator, is newly developed based on the eXtended Geometrical Optics (XGO) theory [c]. Transmitted power profile of Electron Cyclotron Waves (ECWs) experimentally observed in Large Helical Devise (LHD) [d,e] is numerically analyzed, as one of the applications of PARADE. Qualitative consistency between numerical and experimental results shown here is one of the validations of PARADE applied on realistic magnetic confinement system, and thus, suggests the usefulness of quasioptical codes for the predictions of Electron Cyclotron Resonance Heating (ECRH) in future reactors.

ECRH is one of the essential components to achieve the nuclear fusion. Due to capabilities of the high-power density and localized heating, ECRH is expected also to play a role of the high-functional plasma control system, for example, to suppress the Neoclassical Tearing Mode. However, since ECW beam can refract, and importantly, its profile can vary along the propagation in fusion plasmas, it is important to quasioptically evaluate the propagations and absorptions of ECW from both numerical and experimental sides.

PARADE can simulate the quasioptical propagation of wave beams in inhomogeneous anisotropic media within a reasonable computational resource. The code is based on the Schrödinger-type partial differential equation, that accounts for refraction, diffraction, non-uniform dissipation across the beam, and uniquely, mode-conversion. This Schrödinger-type equation is solved along the Reference Ray (RR) trajectory given by the ray equation [a,c]. One of the advantage of PARADE is a capability of treating mode-conversion[b,c], and another one is a capability to simulate an arbitrary beam profiles with non-uniform dissipation, whereas most quasioptical codes assume the Gaussian beam profile and uniform dissipation.

An example of the PARADE results, which calculates the O-mode beam propagation with partial absorption in plasma, is shown in Fig. 1. Wave beam is absorbed non-uniformly across the beam near the cyclotron resonance. The original intensity profile is distorted, and due to the diffraction, strongly diffused propagation is obtained. Also, the beam center, which is defined as a first-moment of the beam profile and should be considered as true-ray trajectory, is diverged from the RR trajectory, which has been considered as ray trajectory in conventional approaches. Such propagation will affect the ECRH power deposition/driven current profile and the beam profile of the passed through power that can cause deteriorating effect on the first wall or sensitive diagnostics. These situations can occur also in tokamaks when the beam crosses the resonance layer in shallow angle.

As one of the applications of the PARADE code, the experimental results from the transmission power profile measurement are compared with the calculated one. A target plate facing on the ECRH antenna via plasma was set inside the LHD. By measuring the temperature increase of the target plate, transmitted intensity profile of ECW beam, which is not absorbed by the LHD plasma in a single path, is obtained. This direct measurement system gives the experimental evaluation of refraction, diffraction, and partial absorption of ECW caused by plasmas. This first quantitative comparison come to be realized due to the development of the PARADE code, because conventional multi-ray methods used in the past cannot simulate a beam profile accurately, due to an ignoring of the diffraction.

Figure 2 shows intensity profiles of ECW on the target plate, experimentally obtained by the direct measurement system and numerically simulated by PARADE. By increasing the electron density of LHD plasma, the beam positions of both numerical and experimental results are gradually shifted to clock-wise- and negative R-directions in same scale. This qualitative agreement is a good verification that the PARADE certainly traces quasioptical propagations of ECW in LHD.

As another application of PARADE, ECRH power deposition profile in a typical LHD experiment is simulated. Figure 3(a) shows the cross-section of $154$~GHz 2nd harmonics X mode ECW beam injected into the LHD, and Fig. 3(b) shows ECRH power deposition profile for same simulation with (a), where orange-line is the PARADE result, and red-line is the result from the conventional multi-ray tracing, which cannot account for the diffraction. The centers of power density of both results are very similar, but its profile of PARADE is smoother than that by multi-rays. This is because of the reproducibility of the envelope structure. The envelope with continuous and smooth amplitude profile by PARADE, which contains diffraction, gives more realistic power deposition profile than conventional multi-rays method.

Newly developed quasioptical ray tracing code PARADE'' can describe wave beams with inhomogeneous absorption across the beam, that should be treated carefully not only in LHD-type stellarator but in tokamak as well. Quasioptical propagations and absorptions of ECW in LHD, numerically performed by PARADE are in qualitative agreement with experimental results in LHD and show the superiority over the conventional multi-ray tracing, and note that, PARADE can apply to any plasma configuration such as tokamak plasmas. These confirm that PARADE is useful to predict the realistic ECRH experiments, and powerful tool for the detailed design of future tokamak and stellarator where ECRH is expected as the main heating.

[a] K. Yanagihara et al., Phys. Plasmas 26 (2019), 072111.
[b] K. Yanagihara et al., Phys. Plasmas 26 (2019), 072112.
[c] I. Y. Dodin et al., Phys. Plasmas 26 (2019), 072110.
[d] S. Kamio et al., Rev. Sci. Instrum. 85 (2014), 11E822.
[e] H. Takahashi et al., EPJ Web Conf. 87 (2015), 02019.

Country or International Organization Japan Nagoya University

### Co-authors

Shin Kubo (National Institute for Fusion Science) Dr Ilya Dodin (Princeton Plasma Physics Laboratory) the LHD experimental Group (National Institute for Fusion Science)


text:
Write the value of n for which nth terms of the A.P.s 3, 10, 17,
Question:

Write the value of n for which nth terms of the A.P.s 3, 10, 17, … and 63, 65, 67, …. are equal.

Solution:

For the first series, $a=3, d_{1}=7$

For the second series, $b=63, d_{2}=2$

Given:

$a_{n}=b_{n}$

$\Rightarrow a+(n-1) d_{1}=b+(n-1) d_{2}$

$\Rightarrow 3+(n-1) 7=63+(n-1) 2$

$\Rightarrow 3+7 n-7=63+2 n-2$

$\Rightarrow 5 n=65$

$\Rightarrow n=13$

Hence, the 13th terms of both the series are the same.


text:
## Results (1-50 of 172 matches)

Label Dim $A$ Field CM RM Traces Fricke sign $q$-expansion
$a_{2}$ $a_{3}$ $a_{5}$ $a_{7}$
405.1.h.a $2$ $0.202$ $$\Q(\sqrt{-3})$$ $$\Q(\sqrt{-15})$$ None $$-1$$ $$0$$ $$1$$ $$0$$ $$q-\zeta_{6}q^{2}-\zeta_{6}^{2}q^{5}-q^{8}-q^{10}+\zeta_{6}q^{16}+\cdots$$
405.1.h.b $2$ $0.202$ $$\Q(\sqrt{-3})$$ $$\Q(\sqrt{-15})$$ None $$1$$ $$0$$ $$-1$$ $$0$$ $$q+\zeta_{6}q^{2}+\zeta_{6}^{2}q^{5}+q^{8}-q^{10}+\zeta_{6}q^{16}+\cdots$$
405.2.a.a $1$ $3.234$ $$\Q$$ None None $$-2$$ $$0$$ $$-1$$ $$0$$ $+$ $$q-2q^{2}+2q^{4}-q^{5}+2q^{10}-5q^{11}+\cdots$$
405.2.a.b $1$ $3.234$ $$\Q$$ None None $$-1$$ $$0$$ $$1$$ $$-3$$ $+$ $$q-q^{2}-q^{4}+q^{5}-3q^{7}+3q^{8}-q^{10}+\cdots$$
405.2.a.c $1$ $3.234$ $$\Q$$ None None $$0$$ $$0$$ $$-1$$ $$2$$ $+$ $$q-2q^{4}-q^{5}+2q^{7}-3q^{11}-4q^{13}+\cdots$$
405.2.a.d $1$ $3.234$ $$\Q$$ None None $$0$$ $$0$$ $$1$$ $$2$$ $-$ $$q-2q^{4}+q^{5}+2q^{7}+3q^{11}-4q^{13}+\cdots$$
405.2.a.e $1$ $3.234$ $$\Q$$ None None $$1$$ $$0$$ $$-1$$ $$-3$$ $+$ $$q+q^{2}-q^{4}-q^{5}-3q^{7}-3q^{8}-q^{10}+\cdots$$
405.2.a.f $1$ $3.234$ $$\Q$$ None None $$2$$ $$0$$ $$1$$ $$0$$ $-$ $$q+2q^{2}+2q^{4}+q^{5}+2q^{10}+5q^{11}+\cdots$$
405.2.a.g $2$ $3.234$ $$\Q(\sqrt{3})$$ None None $$-2$$ $$0$$ $$2$$ $$-6$$ $+$ $$q+(-1+\beta )q^{2}+(2-2\beta )q^{4}+q^{5}+(-3+\cdots)q^{7}+\cdots$$
405.2.a.h $2$ $3.234$ $$\Q(\sqrt{3})$$ None None $$2$$ $$0$$ $$-2$$ $$-6$$ $-$ $$q+(1+\beta )q^{2}+(2+2\beta )q^{4}-q^{5}+(-3+\cdots)q^{7}+\cdots$$
405.2.a.i $3$ $3.234$ 3.3.564.1 None None $$-1$$ $$0$$ $$-3$$ $$5$$ $-$ $$q-\beta _{1}q^{2}+(2+\beta _{2})q^{4}-q^{5}+(2-\beta _{1}+\cdots)q^{7}+\cdots$$
405.2.a.j $3$ $3.234$ 3.3.564.1 None None $$1$$ $$0$$ $$3$$ $$5$$ $-$ $$q+\beta _{1}q^{2}+(2+\beta _{2})q^{4}+q^{5}+(2-\beta _{1}+\cdots)q^{7}+\cdots$$
405.2.b.a $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$0$$ $$0$$ $$-3$$ $$0$$ $$q-\beta _{2}q^{2}+(-2+\beta _{1}+\beta _{3})q^{4}+(-1+\cdots)q^{5}+\cdots$$
405.2.b.b $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$0$$ $$0$$ $$3$$ $$0$$ $$q-\beta _{2}q^{2}+(-2+\beta _{1}+\beta _{3})q^{4}+(1-\beta _{1}+\cdots)q^{5}+\cdots$$
405.2.b.c $4$ $3.234$ $$\Q(\sqrt{-2}, \sqrt{3})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{1}q^{2}+\beta _{2}q^{4}+(\beta _{1}+\beta _{2}-\beta _{3})q^{5}+\cdots$$
405.2.b.d $4$ $3.234$ $$\Q(\sqrt{-2}, \sqrt{3})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{1}q^{2}+\beta _{2}q^{4}+(\beta _{1}-\beta _{2}-\beta _{3})q^{5}+\cdots$$
405.2.b.e $4$ $3.234$ $$\Q(\sqrt{-2}, \sqrt{3})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{1}q^{2}+(\beta _{1}+\beta _{2})q^{5}+\beta _{3}q^{7}-2\beta _{1}q^{8}+\cdots$$
405.2.e.a $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$-2$$ $$0$$ $$-1$$ $$0$$ $$q+(-2+2\zeta_{6})q^{2}-2\zeta_{6}q^{4}-\zeta_{6}q^{5}+\cdots$$
405.2.e.b $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$-2$$ $$0$$ $$-1$$ $$3$$ $$q+(-2+2\zeta_{6})q^{2}-2\zeta_{6}q^{4}-\zeta_{6}q^{5}+\cdots$$
405.2.e.c $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$-1$$ $$0$$ $$1$$ $$0$$ $$q+(-1+\zeta_{6})q^{2}+\zeta_{6}q^{4}+\zeta_{6}q^{5}-3q^{8}+\cdots$$
405.2.e.d $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$0$$ $$0$$ $$-1$$ $$-2$$ $$q+2\zeta_{6}q^{4}-\zeta_{6}q^{5}+(-2+2\zeta_{6})q^{7}+\cdots$$
405.2.e.e $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$0$$ $$0$$ $$1$$ $$-2$$ $$q+2\zeta_{6}q^{4}+\zeta_{6}q^{5}+(-2+2\zeta_{6})q^{7}+\cdots$$
405.2.e.f $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$1$$ $$0$$ $$-1$$ $$0$$ $$q+(1-\zeta_{6})q^{2}+\zeta_{6}q^{4}-\zeta_{6}q^{5}+3q^{8}+\cdots$$
405.2.e.g $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$2$$ $$0$$ $$1$$ $$0$$ $$q+(2-2\zeta_{6})q^{2}-2\zeta_{6}q^{4}+\zeta_{6}q^{5}+2q^{10}+\cdots$$
405.2.e.h $2$ $3.234$ $$\Q(\sqrt{-3})$$ None None $$2$$ $$0$$ $$1$$ $$3$$ $$q+(2-2\zeta_{6})q^{2}-2\zeta_{6}q^{4}+\zeta_{6}q^{5}+(3+\cdots)q^{7}+\cdots$$
405.2.e.i $4$ $3.234$ $$\Q(\zeta_{12})$$ None None $$-2$$ $$0$$ $$2$$ $$6$$ $$q+(-\zeta_{12}+\zeta_{12}^{2})q^{2}+(-2+2\zeta_{12}+\cdots)q^{4}+\cdots$$
405.2.e.j $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{13})$$ None None $$-1$$ $$0$$ $$2$$ $$-2$$ $$q-\beta _{1}q^{2}+(-1+\beta _{1}+\beta _{2}+\beta _{3})q^{4}+\cdots$$
405.2.e.k $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{13})$$ None None $$1$$ $$0$$ $$-2$$ $$-2$$ $$q+\beta _{1}q^{2}+(-1+\beta _{1}+\beta _{2}+\beta _{3})q^{4}+\cdots$$
405.2.e.l $4$ $3.234$ $$\Q(\zeta_{12})$$ None None $$2$$ $$0$$ $$-2$$ $$6$$ $$q+(\zeta_{12}-\zeta_{12}^{2})q^{2}+(-2+2\zeta_{12}-2\zeta_{12}^{2}+\cdots)q^{4}+\cdots$$
405.2.f.a $16$ $3.234$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None None $$0$$ $$0$$ $$0$$ $$4$$ $$q+\beta _{3}q^{2}+(\beta _{6}-\beta _{14})q^{4}-\beta _{4}q^{5}+\beta _{10}q^{7}+\cdots$$
405.2.f.b $24$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$0$$
405.2.j.a $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$-3$$ $$0$$ $$-6$$ $$-12$$ $$q+(-1-\beta _{3})q^{2}+(2-\beta _{1}-\beta _{2}+2\beta _{3})q^{4}+\cdots$$
405.2.j.b $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$-3$$ $$0$$ $$-3$$ $$12$$ $$q+(-1-\beta _{3})q^{2}+(2-\beta _{1}-\beta _{2}+2\beta _{3})q^{4}+\cdots$$
405.2.j.c $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-5})$$ $$\Q(\sqrt{-15})$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{1}q^{2}+3\beta _{2}q^{4}+(\beta _{1}-\beta _{3})q^{5}+\beta _{3}q^{8}+\cdots$$
405.2.j.d $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$3$$ $$0$$ $$3$$ $$12$$ $$q+(1-\beta _{1})q^{2}+(1-2\beta _{1}+\beta _{2}+\beta _{3})q^{4}+\cdots$$
405.2.j.e $4$ $3.234$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$3$$ $$0$$ $$6$$ $$-12$$ $$q+(1-\beta _{1})q^{2}+(1-2\beta _{1}+\beta _{2}+\beta _{3})q^{4}+\cdots$$
405.2.j.f $8$ $3.234$ $$\Q(\zeta_{24})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\zeta_{24}^{6}q^{2}-\zeta_{24}^{4}q^{5}-\zeta_{24}q^{7}+(2\zeta_{24}^{5}+\cdots)q^{8}+\cdots$$
405.2.j.g $8$ $3.234$ $$\Q(\zeta_{24})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\zeta_{24}^{5}q^{2}+(\zeta_{24}^{2}+\zeta_{24}^{3})q^{5}+\zeta_{24}^{6}q^{7}+\cdots$$
405.2.j.h $8$ $3.234$ 8.0.12960000.1 $$\Q(\sqrt{-15})$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{1}q^{2}+(2-\beta _{3}+2\beta _{5}+\beta _{6})q^{4}+\cdots$$
405.2.k.a $30$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$0$$
405.2.k.b $42$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$0$$
405.2.m.a $8$ $3.234$ $$\Q(\zeta_{24})$$ None None $$0$$ $$0$$ $$0$$ $$8$$ $$q+\zeta_{24}^{7}q^{2}+\zeta_{24}^{2}q^{4}+(-\zeta_{24}^{3}+2\zeta_{24}^{5}+\cdots)q^{5}+\cdots$$
405.2.m.b $16$ $3.234$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None None $$0$$ $$0$$ $$0$$ $$-8$$ $$q+(\beta _{9}-\beta _{10})q^{2}+(-2\beta _{1}-\beta _{13}-\beta _{14}+\cdots)q^{4}+\cdots$$
405.2.m.c $16$ $3.234$ 16.0.$$\cdots$$.1 None None $$0$$ $$0$$ $$0$$ $$4$$ $$q-\beta _{10}q^{2}+(-\beta _{9}-\beta _{15})q^{4}+(-\beta _{7}+\cdots)q^{5}+\cdots$$
405.2.m.d $24$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$-12$$
405.2.m.e $24$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$12$$
405.2.p.a $96$ $3.234$ None None $$0$$ $$0$$ $$9$$ $$0$$
405.2.q.a $306$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$0$$
405.2.q.b $342$ $3.234$ None None $$0$$ $$0$$ $$0$$ $$0$$
405.2.r.a $192$ $3.234$ None None $$12$$ $$0$$ $$12$$ $$-12$$


text:
To input characters in alphabetic writing systems such as Russian, French and German, you don’t have to install an input method. You can just change the keyboard layout. In X system, this can be done with (Russian, for example)

setxkbmap -layout ru

But I suggest you set some shortcuts before running this command (and change layout via shortcuts). Otherwise, when you have changed to Russian, you may not be able to input Linux commands (in English), which means you cannot change back. Of course you can use mouse to change back if your DE supports it.

To set shortcuts in Openbox, add the following lines in ~/.config/openbox/rc.xml

<keybind key="A-F11">
<action name="Execute">
<execute>
setxkbmap -layout ru
</execute>
</action>
</keybind>
<keybind key="A-F12">
<action name="Execute">
<execute>
setxkbmap -layout us
</execute>
</action>
</keybind>

After reconfiguring Openbox, you’ll be able to change the keyboard layout via Alt+F11 and Alt+F12.

Finally, there’s a Russian keyboard layout


text:
Cauchy problem for the difference equation and Sobolev orthogonal functions on the finite grid, generated by discrete orthogonal functions

DOI: 10.31029/demr.7.9

We consider the system of functions ${\cal \psi}_{1,n}(x, N)$ ($n=0,1,\ldots,$ $N$), orthonormal in Sobolev sense and generated by a given orthonormal on finite grid $\Omega_N=\left\{ 0,1,\ldots,N-1 \right\}$ system of functions ${\cal\psi}_{n}(x,N)$ $( n=0,1,\ldots,N-1)$. These new functions are orthonormal with respect to the inner product of the following type: $\langle f,g\rangle = f(0)g(0)+ \sum_{j=0}^{N-1}\Delta f(j)\Delta g(j)\rho(j)$. It is shown that the finite Fourier series by the functions ${\cal\psi}_{1,n}(x)$ and their partial sums are convenient and a very effective tool for the approximate solution of the Cauchy problem for nonlinear difference equations.

Keywords: functions; functions orthogonal on the finite grid; finite grid; uniform grid; approximation of discrete functions; mixed series by the functions orthogonal on a uniform grid; iterative process for the approximate solution of difference equations.


text:
# Math Help - partial sums of sequences

1. ## partial sums of sequences

find the Sn of An = (1/n) -(1/n+2)
i got the denominator n^2 + 3n + 2 but can't find the numerator

2. $\frac{1}{n}-\frac{1}{n+2}=\left( \frac{1}{n}-\frac{1}{n+1} \right)+\left( \frac{1}{n+1}-\frac{1}{n+2} \right),$ so $\sum\limits_{n=1}^{m}{\left( \frac{1}{n}-\frac{1}{n+2} \right)}=\sum\limits_{n=1}^{m}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}+\sum\limits_{n=1}^{m}{\left( \frac{1}{n+1}-\frac{1}{n+2} \right)}.$

can you finish?

3. sorry, but no.

4. okay then, try to look at what a telescoping sum is.

5. Originally Posted by confusedgirl
find the Sn of An = (1/n) -(1/n+2)
i got the denominator n^2 + 3n + 2 but can't find the numerator
Hi confusedgirl,

You can write it this way...by placing in the values of n=1, 2, 3, 4, 5...

$A_n=\frac{1}{n}-\frac{1}{n+2}$

$S_n=A_1+A_2+A_3+A_4+A_5+....A_{n-2}+A_{n-1}+A_n$

$S_n=\left(\frac{1}{1}-\frac{1}{1+2}\right)+\left(\frac{1}{2}-\frac{1}{2+2}\right)+\left(\frac{1}{3}-\frac{1}{3+2}\right)+\left(\frac{1}{4}-\frac{1}{4+2}\right)......$

$=\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+$ $.....+\left(\frac{1}{n-2}-\frac{1}{n}\right)+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{1}{n+2}\right)$

If you examine this, you will see that virtually all of the terms cancel,
due to the same terms having + and -.

Which terms will be left standing?

Also, what process did you use to calculate your original denominator?

6. i'm sorry, but i think my question wasn't clear i need to get Sn where
Sn = Sn-1 + An
like
S1 = 1-1/3 = 2/3
S2 = 2/3 +1/2 -1/4 = 11/12
S3= 21/20
S4 = 34/30

i really didn't use any process or technique to get the denominator but just figured it out by trial and error

7. Originally Posted by Krizalid
okay then, try to look at what a telescoping sum is.
ok

8. $1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...\frac{1}{n-2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}+\frac{1}{n}-\frac{1}{n+2}$

reduces to

$1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{3}{2}-\left(\frac{1}{n+1}+\frac{1}{n+2}\right)$

since the last term to cancel is $\frac{1}{n}$

as a term can only cancel a term 3 places to the right.

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

as all the other terms cancel.

This will give you Sn for any n, without having to calculate a preceding sum.