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# What constitutes measuring in the double slit experiment? [duplicate]

In the double slit experiment attempting to measure which slit the particle passed through causes the wave function to collapse.

According to the question: What is the quantum mechanical definition of a measurement?

Until we have an accepted solution of the Measurement Problem there is no definitive definition of quantum measurement, since we don't know exactly what happens at measurement.

And:

The many-worlds interpretation defines measurement as any physical procedure in which the observer gets entangled with a quantum system.

To me, the most obviously arising avenue of investigation would be to narrow down on precisely what does or doesn't cause the wave function to collapse.

Have physicists extensively experimented with what conditions cause it to collapse? Do you have to be taking a measurement to make it collapse or will any interaction cause it to collapse? For example, what if you measure it with an apparatus that then destroys the data gathered without allowing the data to exit a faraday cage, so that it is impossible for any information to ever be accessible to the outside universe? Then what ever the answer, one might invent even more obscure hypothetical circumstance to test...

Has this avenue been explored?

## marked as duplicate by John Rennie, user191954, Jon Custer, Ben Crowell, Kyle KanosOct 26 '18 at 10:05

• If we destroy the information, then how can we possibly know which slit the electron passed through? – Mechanic7 Oct 21 '18 at 9:57
• We can't. But can you tell whether or not the wave function has collapsed without knowing which slit it passed through? – Lorry Laurence mcLarry Oct 21 '18 at 10:02
• – John Rennie Oct 21 '18 at 10:04
• @Mechanic7: In quantum mechanics, it is impossible to destroy information. You can scramble it so that it is utterly inaccessible to any experimenters, but it's still there. – Peter Shor Oct 21 '18 at 15:02

What causes interference pattern to vanish? Suppose we send a particle through two slits. Then you will see an interference pattern.

Now, suppose there is some asymmetry in the two slots, so that somewhere in the universe (not necessarily accessible to the experimenter) there is some information telling which of the two slits the particle went through. Then the interference pattern vanishes.

There might be incomplete information that only gives us a probabilistic guess as to which slit the particle went through. In this case, the interference pattern is dimmer, but it doesn't go away completely.

Finally, to make things even more complicated, there is a very clever experiment that shows that, after we make a measurement $$M_1$$ that gives us some information (we keep the results of this measurement in quantum superposition) telling us that which slit the particles went through, we can make a second measurement $$M_2$$ on this information we kept which destroys it. What happens here is if we just look at the results of $$M_1$$, the interference pattern is gone. But, conditioned on the result of the measurement $$M_2$$, the interference pattern reappears. This is called the quantum eraser experiment.

So does the wave function collapse after measurement $$M_1$$? No, it can't because we can restore the interference pattern with measurement $$M_2$$. But it's not the experimenter looking at the results of $$M_1$$ that collapses the wave function, because we can just throw the results of measurement $$M_1$$ away, rather than remeasuring the outcome with measurement $$M_2$$. If we do this, the interference pattern is gone forever, without the experimenter ever learning anything about the measurement.

I would stop worrying about what makes the wave function collapse, and start worrying about what causes the interference pattern to go away. Nobody knows when the wave function collapses. And some of the people who believe in the many-worlds interpretation of QM claim that the wave function never collapses; what happens instead is that the experimenter gets entangled with the outcome of the measurement.

What makes the wave function collapse is not the fact that anyone sees the result of the observation, or that the information is accessible to an observer, it is the interaction between the macroscopic apparatus and the microscopic quantum system. I don’t understand how one could think that physics would work in such a exoteric way such as the one proposed, where the state of a system would only be determined if someone checked the results of an experiment, or only if these results were somehow available. It is clear that what changes the state of the quantum system being measured is the interaction with apparatus. You can make a machine that interacts with the passing particles in the double-slit experiment that works just like the one where you try to measure which slit it went through, but this machine does not record any data, and in this case the interference pattern would be destroyed just like in the other case.

• But you could also make a machine that interacts with the passing particles in the double-slit experiment, but that interacts with both of them in the same way, so that it can tell whether a particle went by, but not which slit the particle went through. And this machine won't destroy the interference. – Peter Shor Oct 21 '18 at 11:00
• I don’t think that is exactly true, having an interference pattern is not binary, it’s not like you have it or you don’t. What happens is that the more energetic the interaction, the less of the pattern you see, which means the more you try to localize the particle in space, the more you spread it over momentum space, changing its wavelength, so breaking the coherence to have interference. So you can use a machine that interacts minimally just to see whether a particle passed by, but that still affects the pattern, even if it’s not visible. – Hugo V Oct 21 '18 at 11:31
• Anyway, that was not what the question was about, it was about whether the accessibility of the information in the measurement is what caused the collapse, and that’s not true, even thought it is true that measurements that look for different information affect the state differently, but that’s just because of how much energy in the interaction each kind of measurement will need – Hugo V Oct 21 '18 at 11:32
• I never said that having an interference pattern was binary. And you can interact with a system without spreading it all across momentum space; you just don't learn anything about its position if you use these kinds of interaction. – Peter Shor Oct 21 '18 at 15:04
• I never said anything like “spread all across momentum space”, but there is no way to restrict the position of a previously delocalized state without spreading it in momentum space. If you have a definite momentum to get a good interference pattern, then if you try to localize the particle, even minimally, the spread in momentum space will occur, even if also minimally, and that will affect the interference pattern. But this has nothing to do with my answer. The fact is that it does not matter whether you record this data, the effect on the interference pattern comes from the interaction – Hugo V Oct 21 '18 at 15:27

In the specific double slit situation, measuring which slit a particle goes through changes the boundary conditions of the quantum mechanical solution, which is what gives the interference pattern as an accumulated probability distribution. A different wave function is generated by the change in the boundary conditions.

Here is how the probability distribution predicted by the wave function in the experiment "electron scattering off double slits of specific width, distance and medium (vacuum)"

The way that the "which way" detection was attempted in this particular experiment shows the boundary condition changes clearly:

With a filter over the right slit, electrons are more likely to undergo inelastic scattering and act like a spherical wave. Electrons passing through an uncovered slit are more likely to undergo elastic scattering and act like a cylindrical wave. The two different waves do not have a phase correlation and so, even if an electron passed through both slits, it could not create an interference pattern.

Italics mine.

There is no longer the same "medium" in each slit, which changes the wavefunctions. It is worth reading the link.

Although the electrons (which were shot one by one) could still pass through the filtered slit, the filter caused more of the electrons to undergo inelastic scattering rather than elastic scattering. As the physicists explained, an electron undergoing inelastic scattering is localized at the covered slit, and acts like a spherical wave after passing through the slit. In contrast, an electron passing through the unfiltered slit is more likely to undergo elastic scattering, and act like a cylindrical wave after passing through that slit. The spherical wave and cylindrical wave do not have any phase correlation, and so even if an electron has a probability to pass through both slits, the two different waves that come out cannot create an interference pattern on the wall behind them.

Italics mine, it is a different wave function.

This should be a clear demonstration that the quantum mechanical probability interference pattern depends on the boundary conditions, and any detector/interaction at the slits will change them and thus give a different incoherent wave function and so the interference pattern in the accumulation will be lost.


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# zbMATH — the first resource for mathematics

A local analysis of congruences in the $$(p,p)$$ case. I. (English) Zbl 1072.11506
Summary: Given an irreducible, modular, mod $$p$$ representation $$p$$, we analyse the local components at $$p$$ of newforms $$f$$ which give rise to it.

##### MSC:
 11F33 Congruences for modular and $$p$$-adic modular forms 11F32 Modular correspondences, etc. 11F80 Galois representations 11F85 $$p$$-adic theory, local fields
Full Text:

format: plain


text:
Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: English - Português - 日本語 - Русский

See the recommended documentation of this function

Aide de Scilab >> Graphiques > text > text properties

text properties

description of the Text entity properties

Description

The Text entity is a leaf of the graphics entities hierarchy. This entity defines the parameters for string drawing

parent:

This property contains the handle of the parent. The parent of the text entity should be of the type "Axes" or "Compound".

children:

This property contains a vector with the children of the handle. However, text handles currently do not have any children.

visible:

This field contains the visible property value for the entity . It should be "on" or "off" .By default, the text is visible, the value's property is "on". If "off" the text is not displayed on the screen.

text:

the matrix containing the strings of the object. The rows of the matrix are displayed horizontally and the columns vertically.

Starting from Scilab 5.2, it is possible to write LaTeX or MathML expression.

alignment:

Specify how the strings are aligned in their columns. The value must be 'left','center'or 'right'.

data:

This field is the vector [x,y,[z]] of the origin of the text in the data units of the axes.

box:

This field takes the values "on" or "off". If "on" a box is draw around the text with a line on its edge and a background.

line_mode:

This boolean property allows to draw or not a line around the box when the box property is "on". If line_mode is "off", the line of the box is not drawn.

fill_mode:

This boolean property allows to draw or not the background of the box when the box property is "on". If fill_mode is "off", the background of the box is not transparent.

text_box:

A two dimensional vector specifying the size of a rectangle in user coordinates. The rectangle is used when the text_box_mode property is set to 'centered' or 'filled'.

text_box_mode:

May have three different value : 'off', 'centered' or 'filled'. If 'off', the strings are displayed using the given font and the data field specifies the position of the lower-left point of the text. If 'centered', the text is displayed in the middle of the rectangle whose size is given by text_box. If 'filled' the font size of the strings will be expanded to fill the rectangle.

When using 'off' or 'centered' modes, text size remains constant upon zooming. They are the best modes to create annotations in a graph. On the contrary, when using the 'filled' mode, the text size follow the graphic scale. It is then possible to zoom upon text objects.

auto_dimensionning:

auto_dimensionning set to "off" is equivalent to text_box_mode set to "filled". If text_box_mode is "off" or "centered" then it implies that auto_dimensionning is "on".

font_foreground:

This field contains the color used to display the characters of the text. Its value should be a color index (relative to the current colormap).

foreground:

This field contains the color used to display the line on the edge of the box. Its value should be a color index (relative to the current colormap).

background:

This field contains the color used to fill the box around of the text. Its value should be a color index (relative to the current colormap).

font_size:

It is a scalar specifying the displayed characters size. If fractional_font property is "off" only the integer part of the value is used. For more information see graphics_fonts.

font_style:

Specifies the font used to display the character strings. This is a positive integer referecing one of the loaded fonts. Its value must be between 0, referecing the first font, and the number of loaded fonts minus one, referencing the last font. For more information see graphics_fonts.

fractional_font:

This property specify whether text is displayed using fractional font sizes. Its value must be either "on" or "off". If "on" the floating point value of font_size is used for display and the font is anti-aliased. If "off" only the integer part is used and the font is not smoothed.

font_angle:

This property determines the orientation of the text string. Specify value of rotation in degrees.

clip_state:

This field contains the clip_state property value for the text. Its value should be :

• "off" this means that the text is not clipped.

• "clipgrf" this means that the text is clipped outside the Axes box.

• "on" this means that the text is clipped outside the rectangle given by the property clip_box.

clip_box:

This field contains the clip_box property. Its value should be an empty matrix if clip_state is "off" or the vector [x,y,w,h] (upper-left point width height).

user_data:

This field can be use to store any scilab variable in the text data structure, and to retreive it.

Examples

a=get("current_axes");
a.data_bounds=[0,0;1,1];
a.axes_visible = 'on' ;

xstring(0.5,0.5,"Scilab is not esilaB",0,0)

t=get("hdl")   //get the handle of the newly created object

t.font_foreground=6; // change font properties
t.font_size=5;
t.font_style=5;

t.text=["SCILAB","is";"not","esilaB"] ; // change the text
t.font_angle=90 ; // turn the strings
t.text_box = [0,0] ;
t.text_box_mode = 'centered' ; // the text is now centered on [0.5,0.5].
t.alignment = 'center' ;
t.box = 'on' ; // draw a box around the text

// Only valid from Scilab 5.2
mathml="<mrow>      <mfrac>        <mrow>          <mi>d</mi>          <mi>y</mi>        </mrow>        <mrow>          <mi>d</mi>          <mi>x</mi>        </mrow>      </mfrac>      <mo>=</mo>      <mfrac>        <mn>1</mn>        <msup>          <mi>y</mi>          <mn>2</mn>        </msup>      </mfrac>    </mrow>";
t.text=["SCILAB","can write LaTeX :","$\frac{abc}{xyz}$";"or","MathML :",mathml] ;

• set — set a property value of a graphic entity object or of a User Interface object.
• get — Retrieve a property value from a graphics entity or an User Interface object.
• delete — delete a graphic entity and its children.
• xtitle — add titles on a graphics window
• graphics_entities — description of the graphics entities data structures


text:
## Distance traveled

Imagine we are traveling by car from Santurce to Bilbao. Before leaving we reset the odometer to zero. As we move forward on our way the odometer reading will increase its value until we arrive to Bilbao, where the odometer will show the distance traveled between the two cities (for example 22 km). The length of road, measured over the trajectory, is called distance traveled, or simply distance. This measurement is a scalar, and like any length, its unit in the International System (S.I.) is the meter (m).

The distance traveled,  $∆s$  is the length of the trajectory  followed between the initial and final position of the body in motion.

Distance Traveled

The distance traveled by a body, between two positions, is the length of the trajectory between those two positions, or what is the same, it is the difference of the distance traveled up to the last position (S2) minus the distance traveled in the previous stretch (S1)

$\Delta S={S}_{2}-{S}_{1}$

In general, the calculation of the distance traveled is equivalent to the calculation of the length of the trajectory curve. This calculation is not immediate and requires mathematical tools which exceed the level in which we find ourselves. However, it is important to remember that, in the case of circular trajectories, the calculation of the distance traveled is equivalent to the calculation of the length of the circumference arc and that is given by:

$L=\theta \cdot r$

Where:

• : Arc length
• $\theta$ : Angle of the circumference arc in radians
• : Radius of the arc

Finally, remember that the distance traveled is always positive, since the length of the curve described by a moving body will always be a scalar greater than or equal to zero.


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What does HLSL's tex2D return at (0,0)?

I'm trying to understand the mapping of texels to pixels, especially in the context of pixel shaders. I already found out the following:

• When I define a standard quad with vertices at integer positions (texture coordinates ranging from 0 to 1), a pixel shader for the upper left pixel will be called with texture coordinates (0,0)

• When I subtract 0.5 from the vertices' coordinates (as it's recommended when drawing sprites), the pixel shader for the upper left pixel will be called with texture coordinates (1/(2*quadWidth), 1/(2*quadHeight))

My question is about the behaviour of the tex2D function in HLSL:

I would have assumed that it just returns the color of the upper left pixel of the texture when called with texture coordinates of (0,0). But obviously this is not the case, since the second of the above-mentioned variants actually leads to the desired 1:1 mapping.

So is it correct that tex2D returns the original pixel values of a texture when given the following texture coordinates?

• Note if you use a clamp address mode tex2D @ (0,0) will be the same as tex2D @(0+0.5PX,0+0.5PX) – bobobobo Aug 30 '13 at 1:43
• @bobobobo: OK, but I guess this special case is only true for (0,0) because there are no adjacent pixels to the left/above for interpolation. So tex2D @ (1,1) will not be the same as tex2D @ (1+0.5PX,1+0.5PX), right? Otherwise the clamp mode would be a much easier way to draw sprites with 1:1 pixel mapping than subtracting 0.5 from every vertex. – ph4nt0m Aug 30 '13 at 8:04

This applies to bilinear filtering, clamp-to-edge, etc. So if you call tex2D at (0, 0), or any other position that corresponds to an integer number of texels, with bilinear filtering you'll get a blend of four adjacent texels. If called at a position corresponding to an integer-plus-half number of texels, you'll get one texel exactly.


text:
# How reassuring is CRC-then-XTS as authenticated encryption?

It is known that the "xz" compression format embeds a checksum in the compressed file to verify integrity during decompression or standalone testing. The algorithm of the checksum can be one of: CRC32, CRC64, and SHA256, and the default is CRC64.

Now, we put an "xz" file into a XTS-encrypted disk image. The XTS mode is a pure confidentiality mode, and have no authenticity or integrity feature at all. But the "xz" file inside it has integrity feature. So the question is:

When combining the CRC64 integrity of "xz" with the confidentiality of XTS mode of operation, do we get a secure deterministic authenticated encryption (DAE, a.k.a. key-wrap) scheme?

How do you break it? First, find a pair of messages $$m = m_1 \mathbin\| m_2$$ and $$m' = m'_1 \mathbin\| m'_2$$ so that $$m_i \ne m'_i$$ and $$\operatorname{CRC}(m) = \operatorname{CRC}(m')$$. (Finding collisions in CRCs is not hard.) Then:
1. Query the oracle for $$m_1 \mathbin\| m_2$$ to find the ciphertext $$c_1 \mathbin\| c_2 \mathbin\| c_3$$; assume the CRC is in the plaintext of block $$c_3$$.
2. Query the oracle for $$m'_1 \mathbin\| m_2$$ to find the ciphertext $$c'_1 \mathbin\| c_2 \mathbin\| c_{\mathit{garbage}}$$.
3. Query the oracle for $$m_1 \mathbin\| m'_2$$ to find the ciphertext $$c_1 \mathbin\| c'_2 \mathbin\| c_{\mathit{garbage}}$$.
Finally, yield $$c'_1 \mathbin\| c'_2 \mathbin\| c_3$$ as a forgery, for the message $$m'_1 \mathbin\| m'_2$$ not previously sent to the oracle.


text:
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# Questions about a finite element modeling situation ?

## Questions about a finite element modeling situation ?

(OP)
Hello everyone,

I am currently modelling a part with finite elements in Patran, and I have encountered a small problem. Basically I have a part that is made out of several hollow beams and solid plates, and I have modeled everything with shell elements. The problem occurs at 4 beams that run through the whole part, these beams have a square hollow variable section (it starts at 14 mm edge length and ends at 10 mm). All the beams in the part were modeled with shell elements constructed at the mid-surface of the section, except for these 4 beams for which the elements where constructed on the exterior surface (I had to use the exterior surface because of the way the part was designed, it was easier to model this way).

Now I have 2 questions regarding my FE model:
1. As you can see on the picture I have attached, when I input the shell properties I use a shell offset for the thickness, this generates a material overlap which basically increases my overall mass. Because the thickness is quite high (3mm) and there are 4 beams I get a difference between the mass of my FE model and the mass of the CAD model of 170 grams (CAD Model: 2800 g and FEM Model: 2970), an error of 6%. How can I reduce this difference ? (without changing the FEM model)

2. Is this material overlap increasing the stiffness of the model ? will this have a very big impact on the overall strength and give wrong results ?

Thank you very much,
Cosmin

PS: Sorry for my English, I am not a native.

### RE: Questions about a finite element modeling situation ?

1) why do you Have to use the external surface for some shells ? You Can always measure to the OML, then calc where the mid-surface is; and these are flat shells (ie 4 points per surface)

2) you are modelling all these shells, ie as you show in the overlap. why not delete one set within the overlap ? how is the box beam ass'y made ? are pieces welded together ? (then you can model that way)

### RE: Questions about a finite element modeling situation ?

(OP)
1) Because of the design of the part I had to use exterior surface on those beams to generate the shell elements, it was easier for me to construct the model because you have other beams and plates intersecting with this one. I don t really understand what OML means?

2) The shells can't be deleted because they will intersect other beams or plates. This part is quite complex, so complex that it can't be realized by traditional manufacturing, we will 3D Print the part using AlSi10Mg metallic powder.

I talked with some fellow stress engineers and they have suggested that I could locally modify the density of the material in the overlapping zones. Is this a good practice regarding FE modelling ?

### RE: Questions about a finite element modeling situation ?

1) OML = outer mould line (the outer face) I think you're using the external face of the part for your model because it's easier for you. It doesn't look to be a lot of work to extract the midplane. This will affect the weight of the model, if that's important to you.

2) As I understand it you have overlapping shells. These overlapping areas either have the same mesh, or could be modified to have the same mesh. I can't see why you can't remove one of the duplicates.

changing density would improve the weight estimate of the model (if that is that important to you).

a thought ... why not TET mesh (use 3D TET10 elements) the CAD model part ? Then you model would be exactly the same as your part.

### RE: Questions about a finite element modeling situation ?

(OP)
1) Yes, the shell elements are constructed at the outer surface of the beam. It would be easy to construct the mid-surface but it would take very long because there are a lot of things (connection with other beams, plates, etc.) to be modified/resolved, and I don't have the time because the project is ending soon

2) No, the shells are not overlapping. What I was trying to say is that when I input the thickness with offset to the plate, I will basically have "double" the material in the overlapping zone because the shells are at the outer surface and the offset is directed inwards.

Originally the plan was to use solid elements (because of the complexity of the part) but we are running a random vibration analysis and it wont run with more than 20,000 solid elements (at least the license that we have doesn't allow this). Right now, by using a coarser mesh, the model has approximately 50,000 shell elements so it's impossible to go below 20k tet elements.

### RE: Questions about a finite element modeling situation ?

Hi Ionut,

Depending on what results you're after, the stiffness (and stress) may be very different if you use exterior surface instead of the mid-surface. Let's say your thickness is 3mm, let's also say at a particular cross section of your beam, the mid-surface model width is 12mm and the exterior surface width model is 15mm. The tensile area per side is respectively 36mm^2 and 45mm^2 (25% difference). The area moment of inertia difference between the two cases would be much larger.

I agree with rb1957 where a solid model is the way to go if the shell approximation doesn't fit your requirements. You could try CMS to get around the element count limitation.

Best regards,
Sze Kwan (Jason) Cheah

### RE: Questions about a finite element modeling situation ?

1) the project ends when the work's done !

2) I don't understand. your pic shows an overlap. the talk has been about shell elements. So I assume the pic shows shell elements overlapping.

3) maybe you can get a one-time license ?

hi jason !

### RE: Questions about a finite element modeling situation ?

Ionut Cosmin

1)Your seniors rightly pointed to change the density of the model for the extra material you are putting. This will take care of mass part of your vibration analysis. But see if you could be able to find verified solution w.r.t. any similar past project or analytical solution if available. Then you can compare the result and get even.

2)The stiffness part is also important. The extra material you put will surely increase the stiffness and hence the stress and frequency results will differ. How much wrong? Compare the analytical/past project/reference paper results if available.

One way I see is to model solid and shell together where solid for complex part and shell for simple part assuming you have overall limitation of 20k elements on solids only.

### RE: Questions about a finite element modeling situation ?

Assuming your beams effectively react as 'beams' (normal stress + bending stress), then you need to divide by 2, in the corners, not only the density, but also the elastic modulus of the material. Otherwise, using 12 as the side and 3 as the thickness, you would get an increase of 33% in axial stiffness and of nearly 50% in bending stiffness.
I think this is an acceptable practice, the results will for sure be much more correct than with no density and modulus reductions.

prex
http://www.xcalcs.com : Online engineering calculations
https://www.megamag.it : Magnetic brakes and launchers for fun rides

### RE: Questions about a finite element modeling situation ?

(OP)
Jason

I have already used the exterior surface for these beams, and it will be time consuming to modify everything. I can't use CMS because it is forbidden by the FEM Requirements of the project.

rb1957

1) well this part will need to be manufactured and send to space and we have a strict deadline so as to not miss the launch date of the rocket :))

2) No, the shells are constructed on the exterior surface of the beams and when I plot the thickness of the shell you will get an overlap in the corners (I"ve attached a picture to better understand, although in the picture the shell are at mid-surface)

3) not enough time right now

NRP99

Unfortunately I don't have any projects results for a case like this. Initially I have made a model with solid elements and shell elements with RSSCON connections but the design was changed pretty much and I went only with shell elements because it was simpler to do.

prex

This is what I am going to do, it seems to be the best solution at the time.

### RE: Questions about a finite element modeling situation ?

sorry, I thought the overlap was on the face of the elements, not in the thickness.

I've added a little red square to show the overlap, and a little blue square for missing material, right?

So you have elements finely meshed in the corners ? (like t/2, to capture the overlap)

### RE: Questions about a finite element modeling situation ?

(OP)
rb1957

Yes the red is the zone with the overlap. Yes the elements are finely meshed on the corners

### RE: Questions about a finite element modeling situation ?

prex

#### Quote (prex)

Assuming your beams effectively react as 'beams' (normal stress + bending stress), then you need to divide by 2, in the corners, not only the density, but also the elastic modulus of the material.

I get the point of decreasing density to decrease the mass and also reducing the modulus for reducing the the stiffness for the extra material considered. But why divide it by 2? Why not some random number? Is there any guideline for situations like this?

### RE: Questions about a finite element modeling situation ?

(OP)
Okay guys I have run some simulations on a single beam modeled with solid elements, shell elements on outer layer and shell elements on mid-plane. The models can be seen here. The beams are blocked in translation at one end and a mass (CONM2 property) was tied using RBE2 at the other end.

I have run a modal analysis (SOL103) on the elements with the default properties and then other 3 simulations, on the shell_out model, modifying the density and elasticity modulus on the elements from the corner, as seen here

The results are centralized here. Results show that the outer model adds some stiffness to the model and even while decreasing E by half, the stiffness remains high. It seems that only by decreasing E a lot (something like 80%), I would reach the same results as the solid and shell_mid model. Also the density had to be reduced by 40% on the corner zones to reach the same mass between the models.

### RE: Questions about a finite element modeling situation ?

Ionut Cosmin

At corners you don't have material. So ideally the stiffness should be zero which is why you get stresses close when you reduce the E to 80% compared to solid model. Check whether you get results as close as possible by reducing E to 80-99%.

Density reduction will depend on the amount of mass to be reduced. We generally calculate it by taking ratio of mass vs density.

I am waiting for the reply of the my earlier question. Anybody have pointers/ideas?

### RE: Questions about a finite element modeling situation ?

(OP)
NRP99

Yes, if I reduce my E by 84% I will get the same results in frequency. Is it correct to reduce it so much ?

### RE: Questions about a finite element modeling situation ?

I guess so. Solid model will reflect your actual condition. All other models will be approximate than solid one(even solid also approximation to real but comparatively close to real than other approaches).

### RE: Questions about a finite element modeling situation ?

Sorry Ionut, I am not sure to understand how the shell offset feature really works.
However this shouldn't have any effect on mass calculation, so I don't understand why the mass is not correct: as in the corners, for the outer shell, you have twice the area, so reducing the density by 50% should give exactly the same figure as for the solid model (and the mid-shell model).
Concerning the modulus reduction and its influence on the bending stiffness, that's less simple. The reduction of E by 50% is what one expects, at first sight, to keep the same EJ product for the whole section. However this is not entirely correct, as for the 'vertical' walls of the beam a stronger reduction of E is needed to keep the same EJ: by recalculating the whole section with 12 as the outer edge and 3 for the thickness, I get a reduction factor of E for the corners of 42%, instead of 50%.
Another factor that may affect this matter is that everything is based on the basic assumption of bending theory, that the sections remain plane after deformation: now this is likely strongly incorrect in the model with the corners having a lower E, as the 'flanges' of the section (the 'horizontal' walls) certainly deviate from that assumption (you could easily check by how much).
I'm afraid that not much more can be said or done on this subject, the solid model is the only one that may solve all the issues (but, as expected, you get a better approximation with the density and E reductions).

prex
http://www.xcalcs.com : Online engineering calculations
https://www.megamag.it : Magnetic brakes and launchers for fun rides

### RE: Questions about a finite element modeling situation ?

Ionut,

Another dimension to keep in mind before moving forward is the shaft is tapered. Any knockdown factor that may be somewhat valid for one cross section but not valid for the entire length.

Instead of modifying the corners, an unverified suggestion is to modify the entire cross section properties (thickness, E & density) to be more like the Shell_Mid model. These values would change depending on the tapered cross section edge lengths. Perhaps you or others could comment if this approach is valid.

Step 1. Adjust each shell thickness of outer-surface(OS) such that it's area is the same as the original mid-surface(MS). The thicknessOS is a function of length of the tapered shaft.

#### CODE --> math

widthMS*thicknessMS = widthOS*thicknessOS
widthOS = widthMS+thicknessMS
thicknessOS = widthMS*thicknessMS/(widthMS+thicknessMS)
Step 2. Determine the modulus of elasticity scale factor for the cross section that is proportional to inverse of width3 where the width is varying along the length of the shaft. This is done via trial-and-error/optimization. I assume a simple relationship of inverse cubic but could be wrong. A more complicated cubic or quadratic function may fit better.
Step 3. Adjust the densities to match original mass.

I hope this spark some ideas.

Best regards,
Sze Kwan (Jason) Cheah

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PINNup: Robust neural network wavefield solutions using frequency upscaling and neuron splitting

Highlights
1. We propose a novel physics-informed neural network framework using frequency upscaling and neuron splitting, yielding efficient, highly accurate, high-frequency wavefield solutions.
2. We propose an empirical formula that relates neuron splitting to frequency upscaling that allows for high accuracy and fast convergence.
3. It offers a compressed representation of wavefield solution, allowing for easy storage, and instant access of wavefield on dynamic grids.

Abstract

Seismic wave-equation based methods, for example, full waveform inversion, are currently used to illuminate the interior of Earth. Solving for the frequency-domain scattered wavefield via physics-informed neural network (PINN) has great potential in increasing the flexibility and reducing the computational cost of seismic modeling and inversion. However, when dealing with high-frequency wavefields using PINN, its accuracy and training cost limit its application. Thus, we propose a novel implementation of PINN using frequency upscaling and neuron splitting, which allows the neural network model to grow in size as we increase the frequency while leveraging the information from the pre-trained model for lower-frequency wavefields, resulting in fast convergence to highly accurate wavefield solutions. Numerical results show that, compared to the commonly used PINN with random initialization, the proposed PINN exhibits notable superiority in terms of convergence and accuracy and can achieve neuron based high-frequency wavefield solutions with a shallow model.

Theory
Similar to FWI, our NN is first trained for low-frequency wavefields, and gradually optimized for high-frequency wavefields using the lower frequency NN parameters to initialize the model. The information contained in the NN for low-frequency wavefields is beneficial for higher-frequency wavefield training, leading to faster convergence and better prediction than training from scratch. From the prospective of deep learning, our source domain (low-frequency wavefield) and target domain (a higher-frequency wavefield) are inherently related through the kinematic properties for a given velocity model and source location. However, higher frequency wavefields are dynamically more complex, and thus, will require larger neural network models that can represent the complex features. The workflow of the proposed method is illustrated in Figure 1. After we train the neural network to predict low-frequency wavefields using a small network, we increase the size of the network through neural splitting and then use it to learn higher frequency representation. Thus, we can benefit from the NN training experience at low frequency to help us converge faster at high frequencies.

As mentioned before, we want to leverage the information from trained models, while increasing the convergence speed of larger NNs needed for higher frequency representations. The concept of neuron splitting allows us to increase the network size without effecting its output. The splitting process of the neurons in a hidden layer involves duplicating the weights coming into the neuron to all of its off springs, while dividing the weights connecting the neuron with the next layer by the number of offsprings. In the case of splitting all the neurons, the weights and biases are given by the following formulas: $$\label{equ:inputl} \mathbf{W}^{(1)}_{split} = \begin{bmatrix} \mathbf{W}^{(1)} \dots \mathbf{W}^{(1)} \end{bmatrix}^T, \mathbf{b}^{(1)}_{split} = \begin{bmatrix} \mathbf{b}^{(1)}\dots\mathbf{b}^{(1)}\end{bmatrix}^T,$$ $$\label{equ:hidl} \mathbf{W}^{(\ell)}_{split} = \frac{1}{n}\begin{bmatrix} \mathbf{W}^{(\ell)} \dots \mathbf{W}^{(\ell)} \\ \vdots \\ \mathbf{W}^{(\ell)} \dots \mathbf{W}^{(\ell)} \end{bmatrix}, \mathbf{b}^{(\ell)}_{split} = \begin{bmatrix} \mathbf{b}^{(\ell)}\dots\mathbf{b}^{(\ell)}\end{bmatrix}^T,$$ $$\label{equ:outl} \mathbf{W}^{(L)}_{split} = \frac{1}{n}\begin{bmatrix} \mathbf{W}^{(L)} \dots \mathbf{W}^{(L)} \end{bmatrix}, \mathbf{b}^{(L)}_{split} = \mathbf{b}^{(L)}.$$ where the size of the vector and the number of columns in the matrix are equal to $n$.

Results

Even though the representations of wavefield via NN at every frequency may include some errors, those errors, as we saw, are generally small, and we managed to capture the most important features of the wavefield. To further highlight the importance of our approach for PINN, we refer the reader to Sitzmann et al. (2020) in which they use a sine activation function to predict a 3.2 Hz wavefield corresponding to the Helmholtz equation, and to do so they needed a network of 5 layers with 256 neurons per layer, and that prediction was for a single source, not a Green's function.

From the experiments, we observe that with our proposed method, we can easily predict highly accurate multi-frequency wavefield solutions at any location in the domain of interest corresponding to any source location on the surface (because the input of our network includes the coordinates of source on the surface and arbitrary space coordinates). In other words, no interpolation is needed. Such a continuous wavefield representation is stored in a much more compressed form even for the 32 Hz representation.

Experiments on the Overthrust model demonstrate that the proposed method outperforms the vanilla PINN in accuracy and efficiency.

Citation

If you found the paper useful, please cite it via:

Huang, X., & Alkhalifah, T. (2022). PINNup: Robust neural network wavefield solutions using frequency upscaling and neuron splitting. Journal of Geophysical Research: Solid Earth, 127, e2021JB023703. https://doi.org/10.1029/2021JB023703

@article{Huang2022PINNup,
doi = {10.1029/2021jb023703},
url = {https://doi.org/10.1029%2F2021jb023703},
year = 2022,
month = {jun},
volume={127},
issue={6},
publisher = {American Geophysical Union ({AGU})},
author = {Xinquan Huang and Tariq Alkhalifah},
title = {{PINNup}: Robust neural network wavefield solutions using frequency upscaling and neuron splitting},
journal = {Journal of Geophysical Research: Solid Earth} }


text:
User tdnoe - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:23:15Z http://mathoverflow.net/feeds/user/2360 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62444/are-there-ever-three-perfect-powers-between-consecutive-squares Are there ever three perfect powers between consecutive squares? tdnoe 2011-04-20T18:20:41Z 2012-12-11T05:04:00Z
I have not seen this question treated in the literature. Does anyone have more information? There are several OEIS sequences (A097056, A117896, A117934) dealing with this question, but no answers.
 http://mathoverflow.net/questions/26032/intervals-with-large-numbers-of-primes/26052#26052 Answer by tdnoe for Intervals with large numbers of primes tdnoe 2010-05-26T19:10:28Z 2011-11-16T16:50:37Z
OEIS sequence A120934 gives the least prime $p$ such that the interval $[p,p+\log(p)]$ contains $n$ primes.
 http://mathoverflow.net/questions/47738/what-is-the-name-fora2-b2-c2-a-b-c What is the name for(a^2 + b^2 + c^2 +...)/(a + b + c +...)? tdnoe 2010-11-30T00:24:55Z 2010-11-30T11:08:41Z
That is, the sum of squares of some numbers divided by the sum of the numbers. The term "anti-harmonic mean" has been coined for this quantity. I'm hoping there is a better name.
 http://mathoverflow.net/questions/44865/are-the-largely-composite-numbers-the-same-as-the-fully-composite-numbers Are the largely composite numbers the same as the fully composite numbers? tdnoe 2010-11-04T19:37:26Z 2010-11-07T18:22:23Z
Let $d(n)$ be the number of divisors of $n$. Let $p(n)$ be the product of the divisors of $n$. Ramanujan called a number $n$ largely composite if $d(n) \ge d(m)$ for $m < n$. Let's call $n$ fully composite if $p(n) \ge p(m)$ for $m < n$. It is conjectured that largely composite numbers (LCN) are the same as fully composite numbers (FCN). It is easy to show that an LCN is an FCN. Is every FCN an LCN? In the OEIS, these are sequences http://oeis.org/classic/A067128 and http://oeis.org/classic/A034287. These two sequences are the same for the 105834 terms less than 10^150.

This question is interesting because it connects the number of divisors to the product of divisors.
 http://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers/44894#44894 Answer by tdnoe for What is the lower bound for highly composite numbers? tdnoe 2010-11-04T23:27:02Z 2010-11-05T16:57:20Z
It would be nice to have an inequality $n \ge f(x)$. If the poser wants numerical results, here are two:

The least number having exactly x divisors is given by OEIS sequence http://www.oeis.org/classic/A005179. It is a pretty wild function. The nice paper by Grost is recommended.

The least number having x (or more) divisors is given by the OEIS sequence http://www.oeis.org/classic/A061799.
 http://mathoverflow.net/questions/41725/factoring-blocks-of-numbers/41831#41831 Answer by tdnoe for Factoring blocks of numbers tdnoe 2010-10-11T20:30:51Z 2010-10-11T20:30:51Z
Using a sieve of Eratosthenes approach, it is easy to create a list $S$ of the smallest prime factor of every number less than $N$. Then to factor any $n < N$, we just recursively look up the factors: let $n_1=n$, $f_i = S(n_i)$, and $n_{i+1} = n_i/f_i$. With 4 GB of memory, numbers less than ${10}^9$ are quickly factored. Although this algorithm takes more memory, it is much faster than trial division.
 http://mathoverflow.net/questions/41802/a-hierarchy-of-k-highly-composite-numbers A hierarchy of k-highly composite numbers tdnoe 2010-10-11T16:13:49Z 2010-10-11T19:05:17Z
Let $\sigma_k(n)$ denote the sum of the k-th powers of the divisors of n. For any real value of k, we can find a sequence of numbers $s_k$ that has increasing values of n at which $\sigma_k(n)$ attains a new maximum. For k=0, that sequence is called the highly composite numbers, which is A002182 in the OEIS database. From numerical experimentation, it appears that if i > j >= 0, then $s_j$ is a subsequence of $s_i$. Is this a known result? If not, any ideas on how to prove it?
 http://mathoverflow.net/questions/32771/does-there-exist-a-pure-recurrence-formula-with-polynomial-coefficients-for-fibon/32829#32829 Answer by tdnoe for Does there exist a pure recurrence formula with polynomial coefficients for Fibonacci(2^n)? tdnoe 2010-07-21T18:31:24Z 2010-07-21T18:31:24Z
Interestingly, the recursion $u_{n+1} = (u_n + 5/u_n)/2$, with $u_0=1$, gives the fractions $Lucas(2^n)/Fibonacci(2^n)$.
 http://mathoverflow.net/questions/19146/do-six-consecutive-numbers-always-contain-an-abundant-or-perfect-number Do six consecutive numbers always contain an abundant or perfect number? tdnoe 2010-03-23T21:15:34Z 2010-03-24T00:10:24Z
Let sigma(n) be the sum of the divisors of n. Take six consecutive numbers. It appears that at least one of the six has sigma(n) >= 2n. Has this been proved?
 http://mathoverflow.net/questions/7969/irreducible-polynomials-with-constrained-coefficients/8086#8086 Answer by tdnoe for Irreducible polynomials with constrained coefficients tdnoe 2009-12-07T07:34:52Z 2009-12-07T07:34:52Z
Sequence http://www.research.att.com/~njas/sequences/A087481 lists the number of such irreducible polynomials up to degree 18.
 http://mathoverflow.net/questions/47738/what-is-the-name-fora2-b2-c2-a-b-c Comment by tdnoe tdnoe 2010-11-30T00:46:02Z 2010-11-30T00:46:02Z Yes, I do. At least it has some history. Thanks! http://mathoverflow.net/questions/46068/how-small-can-a-sum-of-a-few-roots-of-unity-be/46069#46069 Comment by tdnoe tdnoe 2010-11-16T18:35:48Z 2010-11-16T18:35:48Z The OEIS sequence oeis.org/A108380 gives the least number of distinct n-th roots of unity summing to the smallest possible nonzero magnitude. There is also a plot of the least magnitude for n up to 81. http://mathoverflow.net/questions/44865/are-the-largely-composite-numbers-the-same-as-the-fully-composite-numbers Comment by tdnoe tdnoe 2010-11-04T21:53:49Z 2010-11-04T21:53:49Z It appears to be something like c (log x)^2. http://mathoverflow.net/questions/44844/galois-groups-of-a-family-of-polynomials Comment by tdnoe tdnoe 2010-11-04T20:28:48Z 2010-11-04T20:28:48Z For 241, the discriminant is 2^240 11^478 241^238 -- clearly a square. http://mathoverflow.net/questions/44865/are-the-largely-composite-numbers-the-same-as-the-fully-composite-numbers Comment by tdnoe tdnoe 2010-11-04T20:10:27Z 2010-11-04T20:10:27Z Actually, it can be proved that $p(n) \ne p(m)$ for $n \ne m$. Using the identity $p(n) = n^{(d(n)/2)}$, it is easy to show that $d(n) \ge d(m)$ implies $p(n) \ge p(m)$. http://mathoverflow.net/questions/44844/galois-groups-of-a-family-of-polynomials Comment by tdnoe tdnoe 2010-11-04T19:50:15Z 2010-11-04T19:50:15Z Your three exceptions 7, 17, 97 are the first three primes of the form 2*q^2-1 where q is prime, which is sequence A092057. Is the next exception 241? http://mathoverflow.net/questions/42897/how-many-solutions-are-there-to-abcd-mod-p-where-p-is-a-prime-and-1a-b-c-dp Comment by tdnoe tdnoe 2010-10-20T22:05:52Z 2010-10-20T22:05:52Z See OEIS sequence oeis.org/A020478 for the general case of p being a positive integer http://mathoverflow.net/questions/41802/a-hierarchy-of-k-highly-composite-numbers Comment by tdnoe tdnoe 2010-10-19T15:56:59Z 2010-10-19T15:56:59Z There are ten counterexamples in the new sequence oeis.org/classic/A181309 http://mathoverflow.net/questions/41802/a-hierarchy-of-k-highly-composite-numbers Comment by tdnoe tdnoe 2010-10-11T17:32:18Z 2010-10-11T17:32:18Z It was just pointed out to me that 1084045767585249647898720000 provides a counterexample: it is in $s_0$ but not in $s_1$. http://mathoverflow.net/questions/39373/does-nm-t-have-infinitely-many-solutions-besides-trivial-ones/39374#39374 Comment by tdnoe tdnoe 2010-09-20T19:54:26Z 2010-09-20T19:54:26Z In the 1975 Erdos paper mentioned in OEIS sequence A003135, this is considered a trivial solution. See pages 27-28 of that paper. http://mathoverflow.net/questions/39373/does-nm-t-have-infinitely-many-solutions-besides-trivial-ones Comment by tdnoe tdnoe 2010-09-20T18:21:10Z 2010-09-20T18:21:10Z The OEIS sequence oeis.org/classic/A003135 is related to this question. http://mathoverflow.net/questions/34669/is-there-any-progress-toward-solving-gilbreaths-conjecture/34691#34691 Comment by tdnoe tdnoe 2010-08-06T05:22:44Z 2010-08-06T05:22:44Z The question of how many increasing sequences of integers have the Gilbreath property is answered in oeis.org/classic/A080839. It doesn't make the primes seem that special. http://mathoverflow.net/questions/34474/conjecture-about-prime-gaps Comment by tdnoe tdnoe 2010-08-05T16:07:12Z 2010-08-05T16:07:12Z The terms of -h[2n] are in sequence oeis.org/classic/A121573, prime-gap race; difference of the cumulative sums of gaps above and below prime(2n). http://mathoverflow.net/questions/33411/a-generalisation-of-the-equation-n-ab-ac-bc Comment by tdnoe tdnoe 2010-07-26T22:32:25Z 2010-07-26T22:32:25Z If we allow equality, the $n_0$ have been conjectured in oeis.org/classic/A027565, which appears to be growing exponentially. http://mathoverflow.net/questions/32771/does-there-exist-a-pure-recurrence-formula-with-polynomial-coefficients-for-fibon/32776#32776 Comment by tdnoe tdnoe 2010-07-22T15:58:43Z 2010-07-22T15:58:43Z It there a typo here? Starting with $x_0=1$ and $x_1=1$, the recursion produces $x_2=3$ (correct), $x_3=21$ (correct), and $x_4=2016$ (should be 987).


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# OpenGL OpenGL ignores glTexCoordPointer

This topic is 3617 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

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This suddenly started to happen and I have no clue what's going on. I've got a VBO with texture coordinates ( mixed with others ). Using it for rendering and depth testing works. Now I want to use it for some other shadow caster and render it using depth ( something which is proven to work! ). Still OpenGL insists on delivering for each and every vertex (0,0) as the texture coordinates. I'm using the exact same code as I did for the other place:
glClientActiveTextureARB( GL_TEXTURE0_ARB );
glEnableClientState( GL_TEXTURE_COORD_ARRAY );
glBindBufferARB( GL_ARRAY_BUFFER_ARB, vboModel );
glTexCoordPointer( 2, GL_FLOAT, VBO_PNTC_STRIDE, ( const GLvoid * )VBO_PNTC_TEXCOORD );
Still though OpenGL plainly ignores this. How can this be? Exists there some utterly strange and unholy OpenGL State that can magically force it to ignore texture coordinates? I'm really out of ideas since I can see no error, not even the slightest, in this code. EDIT: VBO_PNTC_ are just the stride and offset variables. Nothing fancy here and it worked at the other place so this is not the problem

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The last parameter for glTexCoordPointer() should be NULL or 0 since the glBindBuffer() already took care of telling OpenGL where the data is. You'll still need the offset in glTexCoordPointer() since it sounds like you're using an interleaved array.

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@V-Man: Nice try but I'm developing on Linux.

Looks like I found the problem. The object in question had been a wire fence with the appropriate alpha texture ( mask style, not full alpha ). As it looks like the sun shadow texture with 1024 ( on the lower spec machine ) caused troubles. The texture got indeed sampled ( why I got (0,0) texture coordinates is beyond me ) but simply got skipped hence I got no shadow at all. Why though it worked when I used the vertex coordinates as texture coordinates ( sort of pseudo-planar projection for testing ) is also beyond me. Bumping up the texture resoution to 2048 or 4096 allows the shadow to be seen ( although, well, it's rather noisy ). Looks like PCF performs very bad in this situation ( I should not be astonished since I've seen nothing good from PCF so far... except being slow :/ ) too. Maybe I find a better way to smooth the shadows, let's see.

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# Vector bundles trivial on complement of a hypersurface section

Assume we have a smooth projective variety $$X$$ over a field (finite field if matters). Consider a smooth hypersurface section $$H$$ of $$X$$. Let $$Spec(A)=U=X \setminus H$$ be an affine variety. I want to know whether there are any nice description of vector bundles on $$X$$ such that they are trivial on $$U$$ and $$H$$? My suspicion was that they are all direct sums of line bundles of the form $$O(nH)$$ is that correct?

Edit: Let's assume that we are talking about rank $$m$$ vector bundles that are subsheaves of the constant sheaf given by the vector space $$F(X)^{\oplus m}$$ which $$F(X)$$ is the function field or equivalently field of fractions of $$A$$. Does that change anything?

• This is not true. Try the projective plane and $H$ a hyeprplane. – Mohan Apr 10 '19 at 1:33
• @Mohan Is this true if we assume the variety is 1 dimensional? And assume the $H$ is just 1 point. – user127776 Apr 10 '19 at 3:29
• This is definitely false. Take a smooth projective curve $X$ of genus $\geq 2$. There are of course many nontrivial vector bundles on $X$ with trivial determinant. Their restriction to the complement of a point is trivial, because a projective module over a Dedekind ring with trivial determinant is free. – abx Apr 10 '19 at 5:24
• I made an edit, I'd really appreciate your opinions on it. – user127776 Apr 10 '19 at 6:55
• All rank $m$ vector bundles embed into $F(X)^{\oplus m}$ (by restriction to the generic fiber). – abx Apr 10 '19 at 7:07

In many cases, $$\mathcal{O}_X(n \cdot H)$$ itself will not be trivial on both $$H$$ and $$U$$, because it will be nontrivial on $$H$$. For example, if $$H \subset \mathbb{P}^n$$ is a degree $$d$$ hypersurface, then $$\mathcal{O}_{\mathbb{P}^n}(H) \simeq \mathcal{O}_{\mathbb{P}^n}(d)$$ and so $$\mathcal{O}_{\mathbb{P}^n}(H)|_H \simeq \mathcal{O}_H(d)$$ which is non-trivial as long as $$n> 1$$, for instance because it is ample.
EDIT: I realize what this shows is: not all sums of line bundles of the form $$\cal{O}_X(n \cdot H)$$ are trivial on both $$U$$ and $$H$$. All the same it seemed relevant.


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# Fourier inversion theorem

Fourier inversion theorem In mathematics, the Fourier inversion theorem says that for many types of functions it is possible to recover a function from its Fourier transform. Intuitively it may be viewed as the statement that if we know all frequency and phase information about a wave then we may reconstruct the original wave precisely.

The theorem says that if we have a function {displaystyle f:mathbb {R} to mathbb {C} } satisfying certain conditions, and we use the convention for the Fourier transform that {displaystyle ({mathcal {F}}f)(xi ):=int _{mathbb {R} }e^{-2pi iycdot xi },f(y),dy,} then {displaystyle f(x)=int _{mathbb {R} }e^{2pi ixcdot xi },({mathcal {F}}f)(xi ),dxi .} In other words, the theorem says that {displaystyle f(x)=iint _{mathbb {R} ^{2}}e^{2pi i(x-y)cdot xi },f(y),dy,dxi .} This last equation is called the Fourier integral theorem.

Another way to state the theorem is that if {displaystyle R} is the flip operator i.e. {displaystyle (Rf)(x):=f(-x)} , then {displaystyle {mathcal {F}}^{-1}={mathcal {F}}R=R{mathcal {F}}.} The theorem holds if both {displaystyle f} and its Fourier transform are absolutely integrable (in the Lebesgue sense) and {displaystyle f} is continuous at the point {displaystyle x} . However, even under more general conditions versions of the Fourier inversion theorem hold. In these cases the integrals above may not converge in an ordinary sense.

Contents 1 Statement 1.1 Inverse Fourier transform as an integral 1.2 Fourier integral theorem 1.3 Inverse transform in terms of flip operator 1.4 Two-sided inverse 2 Conditions on the function 2.1 Schwartz functions 2.2 Integrable functions with integrable Fourier transform 2.3 Integrable functions in one dimension 2.4 Square integrable functions 2.5 Tempered distributions 3 Relation to Fourier series 4 Applications 5 Properties of inverse transform 6 Proof 7 Notes 8 References Statement In this section we assume that {displaystyle f} is an integrable continuous function. Use the convention for the Fourier transform that {displaystyle ({mathcal {F}}f)(xi ):=int _{mathbb {R} ^{n}}e^{-2pi iycdot xi },f(y),dy.} Furthermore, we assume that the Fourier transform is also integrable.

Inverse Fourier transform as an integral The most common statement of the Fourier inversion theorem is to state the inverse transform as an integral. For any integrable function {displaystyle g} and all {displaystyle xin mathbb {R} ^{n}} set {displaystyle {mathcal {F}}^{-1}g(x):=int _{mathbb {R} ^{n}}e^{2pi ixcdot xi },g(xi ),dxi .} Then for all {displaystyle xin mathbb {R} ^{n}} we have {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)=f(x).} Fourier integral theorem The theorem can be restated as {displaystyle f(x)=int _{mathbb {R} ^{n}}int _{mathbb {R} ^{n}}e^{2pi i(x-y)cdot xi },f(y),dy,dxi .} If f is real valued then by taking the real part of each side of the above we obtain {displaystyle f(x)=int _{mathbb {R} ^{n}}int _{mathbb {R} ^{n}}cos(2pi (x-y)cdot xi ),f(y),dy,dxi .} Inverse transform in terms of flip operator For any function {displaystyle g} define the flip operator[note 1] {displaystyle R} by {displaystyle Rg(x):=g(-x).} Then we may instead define {displaystyle {mathcal {F}}^{-1}f:=R{mathcal {F}}f={mathcal {F}}Rf.} It is immediate from the definition of the Fourier transform and the flip operator that both {displaystyle R{mathcal {F}}f} and {displaystyle {mathcal {F}}Rf} match the integral definition of {displaystyle {mathcal {F}}^{-1}f} , and in particular are equal to each other and satisfy {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)=f(x)} .

Since {displaystyle Rf=R{mathcal {F}}^{-1}{mathcal {F}}f=RR{mathcal {FF}}f} we have {displaystyle R={mathcal {F}}^{2}} and {displaystyle {mathcal {F}}^{-1}={mathcal {F}}^{3}.} Two-sided inverse The form of the Fourier inversion theorem stated above, as is common, is that {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)=f(x).} In other words, {displaystyle {mathcal {F}}^{-1}} is a left inverse for the Fourier transform. However it is also a right inverse for the Fourier transform i.e.

{displaystyle {mathcal {F}}({mathcal {F}}^{-1}f)(xi )=f(xi ).} Since {displaystyle {mathcal {F}}^{-1}} is so similar to {displaystyle {mathcal {F}}} , this follows very easily from the Fourier inversion theorem (changing variables {displaystyle zeta :=-zeta } ): {displaystyle {begin{aligned}f&={mathcal {F}}^{-1}({mathcal {F}}f)(x)\[6pt]&=int _{mathbb {R} ^{n}}int _{mathbb {R} ^{n}}e^{2pi ixcdot xi },e^{-2pi iycdot xi },f(y),dy,dxi \[6pt]&=int _{mathbb {R} ^{n}}int _{mathbb {R} ^{n}}e^{-2pi ixcdot zeta },e^{2pi iycdot zeta },f(y),dy,dzeta \[6pt]&={mathcal {F}}({mathcal {F}}^{-1}f)(x).end{aligned}}} Alternatively, this can be seen from the relation between {displaystyle {mathcal {F}}^{-1}f} and the flip operator and the associativity of function composition, since {displaystyle f={mathcal {F}}^{-1}({mathcal {F}}f)={mathcal {F}}R{mathcal {F}}f={mathcal {F}}({mathcal {F}}^{-1}f).} Conditions on the function When used in physics and engineering, the Fourier inversion theorem is often used under the assumption that everything "behaves nicely". In mathematics such heuristic arguments are not permitted, and the Fourier inversion theorem includes an explicit specification of what class of functions is being allowed. However, there is no "best" class of functions to consider so several variants of the Fourier inversion theorem exist, albeit with compatible conclusions.

Schwartz functions The Fourier inversion theorem holds for all Schwartz functions (roughly speaking, smooth functions that decay quickly and whose derivatives all decay quickly). This condition has the benefit that it is an elementary direct statement about the function (as opposed to imposing a condition on its Fourier transform), and the integral that defines the Fourier transform and its inverse are absolutely integrable. This version of the theorem is used in the proof of the Fourier inversion theorem for tempered distributions (see below).

Integrable functions with integrable Fourier transform The Fourier inversion theorem holds for all continuous functions that are absolutely integrable (i.e. {displaystyle L^{1}(mathbb {R} ^{n})} ) with absolutely integrable Fourier transform. This includes all Schwartz functions, so is a strictly stronger form of the theorem than the previous one mentioned. This condition is the one used above in the statement section.

A slight variant is to drop the condition that the function {displaystyle f} be continuous but still require that it and its Fourier transform be absolutely integrable. Then {displaystyle f=g} almost everywhere where g is a continuous function, and {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)=g(x)} for every {displaystyle xin mathbb {R} ^{n}} .

Integrable functions in one dimension Piecewise smooth; one dimension If the function is absolutely integrable in one dimension (i.e. {displaystyle fin L^{1}(mathbb {R} )} ) and is piecewise smooth then a version of the Fourier inversion theorem holds. In this case we define {displaystyle {mathcal {F}}^{-1}g(x):=lim _{Rto infty }int _{-R}^{R}e^{2pi ixxi },g(xi ),dxi .} Then for all {displaystyle xin mathbb {R} } {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)={frac {1}{2}}(f(x_{-})+f(x_{+})),} i.e. {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)} equals the average of the left and right limits of {displaystyle f} at {displaystyle x} . At points where {displaystyle f} is continuous this simply equals {displaystyle f(x)} .

A higher-dimensional analogue of this form of the theorem also holds, but according to Folland (1992) is "rather delicate and not terribly useful".

Piecewise continuous; one dimension If the function is absolutely integrable in one dimension (i.e. {displaystyle fin L^{1}(mathbb {R} )} ) but merely piecewise continuous then a version of the Fourier inversion theorem still holds. In this case the integral in the inverse Fourier transform is defined with the aid of a smooth rather than a sharp cut off function; specifically we define {displaystyle {mathcal {F}}^{-1}g(x):=lim _{Rto infty }int _{mathbb {R} }varphi (xi /R),e^{2pi ixxi },g(xi ),dxi ,qquad varphi (xi ):=e^{-xi ^{2}}.} The conclusion of the theorem is then the same as for the piecewise smooth case discussed above.

Continuous; any number of dimensions If {displaystyle f} is continuous and absolutely integrable on {displaystyle mathbb {R} ^{n}} then the Fourier inversion theorem still holds so long as we again define the inverse transform with a smooth cut off function i.e.

{displaystyle {mathcal {F}}^{-1}g(x):=lim _{Rto infty }int _{mathbb {R} ^{n}}varphi (xi /R),e^{2pi ixcdot xi },g(xi ),dxi ,qquad varphi (xi ):=e^{-vert xi vert ^{2}}.} The conclusion is now simply that for all {displaystyle xin mathbb {R} ^{n}} {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)=f(x).} No regularity condition; any number of dimensions If we drop all assumptions about the (piecewise) continuity of {displaystyle f} and assume merely that it is absolutely integrable, then a version of the theorem still holds. The inverse transform is again defined with the smooth cut off, but with the conclusion that {displaystyle {mathcal {F}}^{-1}({mathcal {F}}f)(x)=f(x)} for almost every {displaystyle xin mathbb {R} ^{n}.} [1] Square integrable functions In this case the Fourier transform cannot be defined directly as an integral since it may not be absolutely convergent, so it is instead defined by a density argument (see the Fourier transform article). For example, putting {displaystyle g_{k}(xi ):=int _{{yin mathbb {R} ^{n}:leftvert yrightvert leq k}}e^{-2pi iycdot xi },f(y),dy,qquad kin mathbb {N} ,} we can set {displaystyle textstyle {mathcal {F}}f:=lim _{kto infty }g_{k}} where the limit is taken in the {displaystyle L^{2}} -norm. The inverse transform may be defined by density in the same way or by defining it in terms of the Fourier transform and the flip operator. We then have {displaystyle f(x)={mathcal {F}}({mathcal {F}}^{-1}f)(x)={mathcal {F}}^{-1}({mathcal {F}}f)(x)} in the mean squared norm. In one dimension (and one dimension only), it can also be shown that it converges for almost every x∈ℝ- this is Carleson's theorem, but is much harder to prove than convergence in the mean squared norm.

Tempered distributions The Fourier transform may be defined on the space of tempered distributions {displaystyle {mathcal {S}}'(mathbb {R} ^{n})} by duality of the Fourier transform on the space of Schwartz functions. Specifically for {displaystyle fin {mathcal {S}}'(mathbb {R} ^{n})} and for all test functions {displaystyle varphi in {mathcal {S}}(mathbb {R} ^{n})} we set {displaystyle langle {mathcal {F}}f,varphi rangle :=langle f,{mathcal {F}}varphi rangle ,} where {displaystyle {mathcal {F}}varphi } is defined using the integral formula. If {displaystyle fin L^{1}(mathbb {R} ^{n})cap L^{2}(mathbb {R} ^{n})} then this agrees with the usual definition. We may define the inverse transform {displaystyle {mathcal {F}}^{-1}colon {mathcal {S}}'(mathbb {R} ^{n})to {mathcal {S}}'(mathbb {R} ^{n})} , either by duality from the inverse transform on Schwartz functions in the same way, or by defining it in terms of the flip operator (where the flip operator is defined by duality). We then have {displaystyle {mathcal {F}}{mathcal {F}}^{-1}={mathcal {F}}^{-1}{mathcal {F}}=operatorname {Id} _{{mathcal {S}}'(mathbb {R} ^{n})}.} Relation to Fourier series When considering the Fourier series of a function it is conventional to rescale it so that it acts on {displaystyle [0,2pi ]} (or is {displaystyle 2pi } -periodic). In this section we instead use the somewhat unusual convention taking {displaystyle f} to act on {displaystyle [0,1]} , since that matches the convention of the Fourier transform used here.

The Fourier inversion theorem is analogous to the convergence of Fourier series. In the Fourier transform case we have {displaystyle fcolon mathbb {R} ^{n}to mathbb {C} ,quad {hat {f}}colon mathbb {R} ^{n}to mathbb {C} ,} {displaystyle {hat {f}}(xi ):=int _{mathbb {R} ^{n}}e^{-2pi iycdot xi },f(y),dy,} {displaystyle f(x)=int _{mathbb {R} ^{n}}e^{2pi ixcdot xi },{hat {f}}(xi ),dxi .} In the Fourier series case we instead have {displaystyle fcolon [0,1]^{n}to mathbb {C} ,quad {hat {f}}colon mathbb {Z} ^{n}to mathbb {C} ,} {displaystyle {hat {f}}(k):=int _{[0,1]^{n}}e^{-2pi iycdot k},f(y),dy,} {displaystyle f(x)=sum _{kin mathbb {Z} ^{n}}e^{2pi ixcdot k},{hat {f}}(k).} In particular, in one dimension {displaystyle kin mathbb {Z} } and the sum runs from {displaystyle -infty } to {displaystyle infty } .

Applications Some problems, such as certain differential equations, become easier to solve when the Fourier transform is applied. In that case the solution to the original problem is recovered using the inverse Fourier transform.

In applications of the Fourier transform the Fourier inversion theorem often plays a critical role. In many situations the basic strategy is to apply the Fourier transform, perform some operation or simplification, and then apply the inverse Fourier transform.

More abstractly, the Fourier inversion theorem is a statement about the Fourier transform as an operator (see Fourier transform on function spaces). For example, the Fourier inversion theorem on {displaystyle fin L^{2}(mathbb {R} ^{n})} shows that the Fourier transform is a unitary operator on {displaystyle L^{2}(mathbb {R} ^{n})} .

Properties of inverse transform The inverse Fourier transform is extremely similar to the original Fourier transform: as discussed above, it differs only in the application of a flip operator. For this reason the properties of the Fourier transform hold for the inverse Fourier transform, such as the Convolution theorem and the Riemann–Lebesgue lemma.

Tables of Fourier transforms may easily be used for the inverse Fourier transform by composing the looked-up function with the flip operator. For example, looking up the Fourier transform of the rect function we see that {displaystyle f(x)=operatorname {rect} (ax)quad Rightarrow quad ({mathcal {F}}f)(xi )={frac {1}{|a|}}operatorname {sinc} left({frac {xi }{a}}right),} so the corresponding fact for the inverse transform is {displaystyle g(xi )=operatorname {rect} (axi )quad Rightarrow quad ({mathcal {F}}^{-1}g)(x)={frac {1}{|a|}}operatorname {sinc} left(-{frac {x}{a}}right).} Proof The proof uses a few facts, given {displaystyle f(y)} and {displaystyle {mathcal {F}}f(xi )=int _{mathbb {R} ^{n}}e^{-2pi iycdot xi }f(y),dy} .

If {displaystyle xin mathbb {R} ^{n}} and {displaystyle g(xi )=e^{2pi mathrm {i} xcdot xi }psi (xi )} , then {displaystyle ({mathcal {F}}g)(y)=({mathcal {F}}psi )(y-x)} . If {displaystyle varepsilon in mathbb {R} } and {displaystyle psi (xi )=varphi (varepsilon xi )} , then {displaystyle ({mathcal {F}}psi )(y)=({mathcal {F}}varphi )(y/varepsilon )/|varepsilon |} . For {displaystyle f,gin L^{1}(mathbb {R} ^{n})} , Fubini's theorem implies that {displaystyle textstyle int g(xi )cdot ({mathcal {F}}f)(xi ),dxi =int ({mathcal {F}}g)(y)cdot f(y),dy} . Define {displaystyle varphi (xi )=e^{-pi vert xi vert ^{2}}} ; then {displaystyle ({mathcal {F}}varphi )(y)=varphi (y)} . Define {displaystyle varphi _{varepsilon }(y)=varphi (y/varepsilon )/varepsilon ^{n}} . Then with {displaystyle ast } denoting convolution, {displaystyle varphi _{varepsilon }} is an approximation to the identity: for any continuous {displaystyle fin L^{1}(mathbb {R} ^{n})} and point {displaystyle xin mathbb {R} ^{n}} , {displaystyle lim _{varepsilon to 0}(varphi _{varepsilon }ast f)(x)=f(x)} (where the convergence is pointwise).

Si quieres conocer otros artículos parecidos a Fourier inversion theorem puedes visitar la categoría Generalized functions.

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# Kalman filter

This post details the Kalman filter equations.

## Predict

State prediction:

$\hat{x}_{i}^p = F_i.\hat{x}_{i-1} + B_i.u_i$

Where:

• $\hat{x}_i^p$ is the predicted state at time step $i$.
• $\hat{x}_{i}$ is the estimate of state at time step $i$.
• $F_i$ is the transition matrix. It describes how the state will change according to the previous state (prediction).
• $B_i$ is a matrix that translates control input at time step $i$ into a predicted change in state. In another words, it maps an input vector $u_i$ into the state space.
• $u_i$ is the system input at time step $i$.

Uncertainty (or covariance) prediction:

$P_i^p = F_i.P_{i-1}.F_i^\top + Q_i$

Where:

• $P_i^p$ is the error covariance matrix predicted at time step $i$.
• $P_i$ is the estimated error covariance matrix associated with the estimated state $\hat{x}_{i}$.
• $Q_i$ is the system noise covariance matrix.

## Update

Innovation or measurement residual:

$\widetilde{y}_i = z_i - H_i.\hat{x}_i^p$

Where:

• $\widetilde{y}_i$ is a measurement error : this is the difference between the measurement $z_i$ and the estimate measurement from state $\hat{x}_i^p$.
• $z_i$ is an observation (or measurement) from the true state $x_i$.
• $H_i$ is a transition matrix which maps the state space into the observed space.

Innovation (or residual) covariance:

$S_i=H_i.P_i^p.H_i^\top+R_i$

Where:

• $S_i$ is the covariance matrix associated to the measurement error $\widetilde{y}_i$.
• $R_i$ is the covariance matrix for the measurement noise.

Optimal Kalman gain

$K_i=P_i^p.H_i^\top.S_i^{-1}$

Where

• $K_i$ is the Kalman gain, this matrix contains the balance between prediction and observations. This matrix will weight the merging between predicted state and observations.

Updated state estimate:

$\hat{x}_i=\hat{x}_i^p + K_i.\widetilde{y}_i$

Updated estimate covariance

$P_i=(I-K_i.H_i).P_i^p$

Where:

• $I$ is the identity matrix.


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Dot Product Involving Path of a Curve

1. Feb 13, 2013

Karnage1993

1. The problem statement, all variables and given/known data
Let $\gamma(t)$ be a path describing a level curve of $f : \mathbb{R}^2 \to \mathbb{R}$. Show, for all $t$, that $( \nabla f ) (\gamma(t))$ is orthogonal to $\gamma ' (t)$

2. Relevant equations
$\gamma(t) = ((x(t), y(t))$
$\gamma ' (t) = F(\gamma(t))$
$F = \nabla f = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}\right)$ [this is called a gradient field]

None of these were given for the question at hand but I think they might be useful.

3. The attempt at a solution
If $x(t)$ and $y(t)$ are the parameters for $\gamma(t)$, then $( \nabla f ) (\gamma(t)) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$

But both $\nabla f (\gamma(t))$ and $\gamma ' (t)$ are equal to $F(\gamma(t))$, so the dot product is

$F(\gamma(t)) \cdot F(\gamma(t)) = ||F(\gamma(t))||^2$

At this point, I'm stuck. I don't think $||F(\gamma(t))||^2$ would be 0 for any $t$ let alone for any function $f : \mathbb{R}^2 \to \mathbb{R}$. Did I make a bad assumption/simplification somewhere?

2. Feb 13, 2013

HallsofIvy

Staff Emeritus
Note that $f(\lambda(t))$ is a constant for all t and use the chain rule.

3. Feb 13, 2013

Karnage1993

EDIT: I see now, thank you for the help!

Last edited: Feb 13, 2013


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Proof of Correctness for Sparse Tiling of Gauss-Seidel
Gauss-Seidel is an iterative computation used for solving sets of simulataneous linear equations, $Au=f$. When these unknowns are associated with nodes in an irregular mesh, then the Gauss-Seidel computation structure is related to the mesh structure. We use this structure to subdivide the computation at runtime using a technique called {\em sparse tiling}. The rescheduled computation exhibits better data locality and therefore improved performance. This paper gives a complete proof that a serial schedule based on sparse tiling generates results equivalent to those that a standard Gauss-Seidel computation produces.


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## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 3 years ago Let f(x) = x + 7 and g(x) = x – 2. What is f(x) • g(x)?

• This Question is Closed
1. anonymous

$f(x)\circ g(x)$versus$f(x)\cdot g(x)$

2. anonymous

I dont think thats an answer

3. anonymous

I think he means which is the problem.... is it a circle or dot?

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# Change border color of a \boxed equation?

I recently asked this question. The solution worked for me but I have a problem. cropped images are as small as possible. I would like to have a gif that bounds to borders of box of the equation. so I add \boxed to all my equations. Now sizes are good but I don't know how do I remove black borders around equation. Probably changing its color to white will help. [?]

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You could redefine the \boxed command. For example:

\renewcommand{\boxed}[1]{\textcolor{\boxedcolor}{%
\fbox{\normalcolor\m@th$\displaystyle#1$}}}

It's originally defined in amsmath.sty:

\newcommand{\boxed}[1]{\fbox{\m@th$\displaystyle#1$}}

Here's a complete minimal example:

\documentclass{article}
\usepackage{amsmath}
\usepackage{xcolor}
\newcommand*{\boxedcolor}{red}
\makeatletter
\renewcommand{\boxed}[1]{\textcolor{\boxedcolor}{%
\fbox{\normalcolor\m@th$\displaystyle#1$}}}
\makeatother
\begin{document}
text
$$\boxed{1+1=2}$$
text
\end{document}

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Is it possible to add a background??? –  71GA Oct 3 '12 at 8:31

Another idea of Herbert in mathmode.pdf for coloured boxes is to use package empheq (now in the mh bundle)

\documentclass{article}
\usepackage{amsmath,empheq}
\usepackage{xcolor}

\begin{document}

$f(x)=\int_1^{\infty}\frac{1}{x^2}\,\mathrm{d}x=1$
\begin{empheq}[box={\color{white}\fboxsep=10pt\fbox}]{align}
\color{red}
f(x)=\int_1^{\infty}\frac{1}{x^2}\,\mathrm{d}x=1
\end{empheq}

\begin{empheq}[box={\fboxsep=10pt\fbox}]{align}
\color{red}
f(x)=\int_1^{\infty}\frac{1}{x^2}\,\mathrm{d}x=1
\end{empheq}
\end{document}

-

Set a border for the preview.

\documentclass{scrartcl}
\usepackage{amsmath}
\usepackage[active,displaymath,tightpage]{preview}
\setlength\PreviewBorder{10pt}
\begin{document}

$y=x$
\begin{align} g(x)=x^2 \end{align}

\end{document}

-


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# Element $a$ of a C*-algebra such that $f(a^*a) = 0$ but $f(aa^*) \neq 0$ for some positive linear functional $f$

Could somebody please give me an example, assuming one exists, of an element $a$ of a C*-algebra $A$ such that $f(a^*a) = 0$ but $f(aa^*) \neq 0$ for some positive linear functional $f:A \to \mathbb{C}$? I tried to prove there was no example, but that didn't seem to work out so now I'm guessing it's false. Thanks in advance.

-

Let $e_0=(1,0,0,0,0,\ldots)\in\ell^2$, let $f:B(\ell^2)\to\mathbb C$ be defined by $f(T)=\langle Te_0,e_0\rangle$, and let $S\in B(\ell^2)$ be the "backward shift" defined by $S(a_0,a_1,a_2,a_3,\ldots)=(a_1,a_2,a_3,a_4,\ldots)$. Then $f(SS^*)=1$ but $f(S^*S)=0$.
Similar but simpler: $A=M_2(\mathbb C)$, $f(T)=\left\langle T\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}1\\0\end{bmatrix} \right\rangle$, $a=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.
Aghhh I was trying to make that exact example $A=M_2(\mathbb{C})$, $a=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ work, but the only positive linear functional I could think of was the trace, which didn't work. – Mike F Dec 5 '11 at 5:36
So, generally, if $A = B(H)$ for some Hilbert space, then $T \mapsto \langle Tx,x \rangle$ is always an example of a positive linear functional when $x \in H$? I didn't realize they were so easy to conjure up. By the way, are those all of the positive linear functionals when $A = B(H)$? – Mike F Dec 5 '11 at 5:39
@Mike: $T\mapsto\langle Tx,x\rangle$ is sometimes called a positive vector functional, and a vector state if $\|x\|=1$. No, not all positive linear functional on $B(H)$ have this form. Perhaps the simplest way to see this is to note that a sum of two positive linear functionals is a positive linear functional, but not typically a vector functional itself. The trace on $M_2(\mathbb C)$ is a sum of two vector functionals, but is not a vector functional. The positive linear functionals on $B(H)$ that are finite sums of positive vector functionals are precisely those that are continuous... – Jonas Meyer Dec 5 '11 at 5:57
...with respect to the weak operator topology on $B(H)$. Allowing infinite sums gives the functionals continuous with respect to the ultraweak operator topology (which coincides with the weak* topology), also known as positive normal functionals. There are also nonnormal functionals when $H$ is infinite dimensional. E.g., there are nonzero positive linear functionals on $B(H)$ that vanish on the ultraweakly dense subalgebra of compact operators (as can be obtained using Hahn-Banach). – Jonas Meyer Dec 5 '11 at 6:00
Righhht the trace on $M_2(\mathbb{C})$ is not of of that form. I had momentarily bamboozled myself into thinking $x=(1,1)$ worked. Also I had some kind of vague idea along the lines of 1) take a positive linear functional $f$ on $A$, 2) carry out the GNS construction to get a representation $\pi$ of $A$ on an $H$ that is a quotient of $A$ 3) OK now $f(a) = \langle \pi (a) x,x \rangle$ where $x$ is the class of the unit (assuming such exists) in $H$, so $f$ is "sort of" a coming from a vector, but the Hilbert space might change... – Mike F Dec 5 '11 at 6:19


text:
Article

Full entry | PDF   (0.2 MB)
Keywords:
$p$-limit; $p$-compact; almost $p$-compact; quasi $M$-compact; countably compact
Summary:
For $\emptyset \neq M \subseteq \omega^*$, we say that $X$ is quasi $M$-compact, if for every $f: \omega \rightarrow X$ there is $p \in M$ such that $\overline{f}(p) \in X$, where $\overline{f}$ is the Stone-Čech extension of $f$. In this context, a space $X$ is countably compact iff $X$ is quasi $\omega^*$-compact. If $X$ is quasi $M$-compact and $M$ is either finite or countable discrete in $\omega^*$, then all powers of $X$ are countably compact. Assuming $CH$, we give an example of a countable subset $M \subseteq \omega^*$ and a quasi $M$-compact space $X$ whose square is not countably compact, and show that in a model of A. Blass and S. Shelah every quasi $M$-compact space is $p$-compact (= quasi $\{p\}$-compact) for some $p \in \omega^*$, whenever $M \in [\omega^*]^{< {\frak c}}$. We prove that if $\emptyset \notin \{ T_\xi :\, \xi < 2^{{\frak c}} \} \subseteq [\omega^*]^{< 2^{{\frak c}}}$, then there is a countably compact space $X$ that is not quasi $T_\xi$-compact for every $\xi < 2^{{\frak c}}$; hence, if $2^{{\frak c}}$ is regular, then there is a countably compact space $X$ such that $X$ is not quasi $M$-compact for any $M \in [\omega^*]^{< 2^{{\frak c}}}$. We list some open problems.
References:
[Be] Bernstein A.R.: A new kind of compactness for topological spaces. Fund. Math. 66 (1970), 185-193. MR 0251697 | Zbl 0198.55401
[Bl] Blass A.: Near coherence of filters I: Cofinal equivalence of models of arithmetic. Notre Dame J. Formal Logic 27 (1986), 579-591. MR 0867002 | Zbl 0622.03040
[BL] Blass A., Laflamme C.: Consistency results about filters and the number of inequivalent growth types. J. Symbolic Logic 54 (1989), 50-56. MR 0987321 | Zbl 0673.03038
[BS1] Blass A., Shelah S.: Near coherence of filters III: A simplified consistency proof. to appear. MR 1036674 | Zbl 0702.03030
[BS2] Blass A., Shelah S.: There may be simple $P_{\aleph_1}$ and $P_{\aleph_2}$ points and the Rudin-Keisler ordering may be downward directed. Ann. Pure Appl. Logic 33 (1987), 213-243. MR 0879489
[Co] Comfort W.W.: Ultrafilters: some old and some new results. Bull. Amer. Math. Soc. 83 (1977), 417-455. MR 0454893
[CN] Comfort W., Negrepontis S.: The Theory of Ultrafilters. Springer-Verlag, Berlin, 1974. MR 0396267 | Zbl 0298.02004
[G] Garcia-Ferreira S.: Quasi $M$-compact spaces. Czechoslovak Math. J. 46 (1996), 161-177. MR 1371698 | Zbl 0914.54019
[GJ] Gillman L., Jerison M.: Rings of continuous functions. Lectures Notes in Mathematics No. 27, Springer-Verlag, 1976. MR 0407579 | Zbl 0327.46040
[GS] Ginsburg J., Saks V.: Some applications of ultrafilters in topology. Pacific J. Math. 57 (1975), 403-418. MR 0380736 | Zbl 0288.54020
[Ku] Kunen K.: Weak $P$-points in $ømega^*$. Colloquia Math. Soc. János Bolyai 23 (1978), North-Holland, Amsterdam, pp.741-749. MR 0588822
[vM] van Mill J.: An introduction to $\beta(ømega)$. in K. Kunen and J.E. Vaughan, eds., Handbook of Set-Theoretic Topology, North-Holland, Amsterdam, 1984, pp.503-567. MR 0776630 | Zbl 0555.54004

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## Precalculus (6th Edition)

$\color{blue}{(a-b)(a-b+2)}$
Factor out the GCF $a-b$ to obtain: $=(a-b)[(a-b)+2] \\=\color{blue}{(a-b)(a-b+2)}$


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## Intermediate Algebra (6th Edition)

We are given that $x-2\lt6$ or $3x+1\gt1$. Therefore, a solution only needs to satisfy one of the inequalities. We can first solve $x-2\lt6$ for x. Add 2 to both sides. $x\lt8$ Next, can first solve $3x+1\gt1$ for x. Subtract 1 from both sides. $3x\gt0$ Divide both sides by 3. $x\gt0$ Therefore, the solution set consists of all numbers that are either greater than 0 or less than 8, which includes all real numbers.