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# Introduction: generic algorithms with AbstractArrays

Somewhat unusually, this blog post is future-looking: it mostly focuses on things that don’t yet exist. Its purpose is to lay out the background for community discussion about possible changes to the core API for AbstractArrays, and serves as background reading and reference material for a more focused “julep” (a julia enhancement proposal). Here, often I’ll use the shorthand “array” to mean AbstractArray, and use Array if I explicitly mean julia’s concrete Array type.

As the reader is likely aware, in julia it’s possible to write algorithms for which one or more inputs are only assumed to be AbstractArrays. This is “generic” code, meaning it should work (i.e., produce a correct result) on any specific concrete array type. In an ideal world—which julia approaches rather well in many cases—generality of code should not have a negative impact on its performance: a generic implementation should be approximately as fast as one restricted to specific array type(s). This implies that generic algorithms should be written using lower-level operations that give good performance across a wide variety of array types.

Providing efficient low-level operations is a different kind of design challenge than one experiences with programming languages that “vectorize” everything. When successful, it promotes much greater reuse of code, because efficient, generic low-level parts allow you to write a wide variety of efficient, generic higher-level functions.

Naturally, as the diversity of array types grows, the more careful we have to be about our abstractions for these low-level operations.

# Examples of arrays

In discussing general operations on arrays, it’s useful to have a diverse collection of concrete arrays in mind.

In core julia, some types we support fairly well are:

• Array: the prototype for all arrays

• Ranges: a good example of what I often consider a “computed” array, where essentially none of the values are stored in memory. Since there is no storage, these are immutable containers: you can’t set values in individual slots.

• BitArrays: arrays that can only store 0 or 1 (false or true), and for which the internal storage is packed so that each entry requires only one bit.

• SubArrays: the problems this type introduced, and the resolution we adopted, probably serves as the best model for the generalizations considered here. Therefore, this case is discussed in greater detail below.

Another important class of array types in Base are sparse arrays: SparseMatrixCSC and SparseVector, as well as other sparse representations like Diagonal, Bidiagonal, and Tridiagonal. These are good examples of array types where access patterns deserve careful thought. Notably, despite many commonalities in “strategy” among the 5 or so sparse parametrizations we have, implementations of core algorithms (e.g., matrix multiplication) are specialized for each sparse-like type—in other words, these mimic the “high level vectorized functions” strategy common to other languages. What we lack is a “sparse iteration API” that lets you write the main algorithms of sparse linear algebra efficiently in a generic way. Our current model is probably fine for SparseLike*Dense operations, but gets to be harder to manage if you want to efficiently compute, e.g., Bidiagonal*SparseMatrixCSC: the number of possible combinations you have to support grows rapidly with more sparse types, and thus represents a powerful incentive for developing efficient, generic low-level operations.

Outside of Base, there are some other mind-stretching examples of arrays, including:

• DataFrames: indexing arrays with symbols rather than integers. Other related types include NamedArrays, AxisArrays.

• Interpolations: indexing arrays with non-integer floating-point numbers

• DistributedArrays: another great example of a case in which you need to think through access patterns carefully

# SubArrays: a case study

For arrays of fixed dimension, one can write algorithms that index arrays as A[i,j,k,...] (good examples can be found in our linear algebra code, where everything is a vector or matrix). For algorithms that have to support arbitrary dimensionality, for a long time our fallback was linear indexing, A[i] for integer i. However, in general SubArrays cannot be efficiently accessed by a linear index because it results in call(s) to div, and div is slow. This is a CPU problem, not a Julia-specific problem. The slowness of div is still true despite the recent addition of infrastructure to make it much faster—now one can make it merely “really bad” rather than “Terrible, Horrible, No Good, and Very Bad”.

The way we (largely) resolved this problem was to make it possible to do cartesian indexing, A[i,j,k,...], for arrays of arbitrary dimensionality (the CartesianIndex type). To leverage this in practical code, we also had to extend our iterators with the for I in eachindex(A) construct. This allows one to select an iterator that optimizes the efficiency of access to elements of A. In generic algorithms, the performance gains were not small, sometimes on the scale of ten- to fifty-fold. These types were described in a previous blog post.

To my knowledge, this approach has given Julia one of the most flexible yet efficient “array view” types in any programming language. Many languages base views on array strides, meaning situations in which the memory offset is regular along each dimension. Among other things, this requires that the underlying array is dense. In contrast, in Julia we can easily handle non-strided arrays (e.g., sampling at [1,3,17,428,...] along one dimension, or creating a view of a SparseMatrixCSC). We can also handle arrays for which there is no underlying storage (e.g., Ranges). Being able to do this with a common infrastructure is part of what makes different optimized array types useful in generic programming.

It’s also worth pointing out some problems:

• Most importantly, it requires that one adopt a slightly different programming style. Despite being well into another release cycle, this transition is still not complete, even in Base.

• For algorithms that involve two or more arrays, there’s a possibility that their “best” iterators will be of different types. In principle, this is a big problem. Consider matrix-vector multiplication, A[i,j]*v[j], where j needs to be in-sync for both A and v, yet you’d also like all of these accesses to be maximally-efficient. In practice, right now this isn’t a burning problem: even if our arrays don’t all have efficient linear indexing, to my knowledge all of our (dense) array types have efficient cartesian indexing. Since indexing by N integers (where N is equal to the dimensionality of the array) is always performant, this serves as a reliable default for generic code. (It’s worth noting that this isn’t true for sparse arrays, and the lack of a corresponding generic solution is probably the main reason we lack a generic API for writing sparse algorithms.)

Unfortunately, I suspect that if we want to add support for certain new operations or types (specific examples below), it will force us to set the latter problem on fire.

# Challenging examples

Some possible new AbstractArray types pose novel challenges.

## ReshapedArrays (#15449)

These are the front-and-center motivation for this post. These are motivated by a desire to ensure that reshape(A, dims) always returns a “view” of A rather than allocating a copy of A. (Much of the urgency of this julep goes away if we decide to abandon this goal, in which case for consistency we should always return a copy of A.) It’s worth noting that besides an explicit reshape, we have some mechanisms for reshaping that currently cause a copy to be created, notably A[:] or A[:, :] applied to a 3D array.

Similar to SubArrays, the main challenge for ReshapedArrays is getting good performance. If A is a 3D array, and you reshape it to a 2D array B, then B[i,j] must be expanded to A[k,l,m]. The problem is that computing the correct k,l,m might result in a call to div. So ReshapedArrays violate a crutch of our current ecosystem, in that indexing with N integers might not be the fastest way to access elements of B. From a performance perspective, this problem is substantial (see #15449, about five- to ten-fold).

In simple cases, there’s an easy way to circumvent this performance problem: define a new iterator type that (internally) iterates over the parent A’s indexes directly. In other words, create an iterator so that B[I] immediately expands to A[I'], and so that the latter has “ideal” performance.

Unfortunately, this strategy runs into a lot of trouble when you need to keep two arrays in sync: if you want to adopt this strategy, you simply can’t write B[i,j]*v[j] for matrix-vector multiplication anymore. A potential way around this problem is to define a new class of iterators that operate on specific dimensions of an array (#15459), writing B[ii,jj]*v[j]. jj (whatever that is) and j need to be in-sync, but they don’t necessarily need to both be integers. Using this kind of strategy, matrix-vector multiplication

for j = 1:size(B, 2)
vj = v[j]
for i = 1:size(B, 1)
dest[i] += B[i,j] * vj
end
end

might be written in a more performant manner like this:

for (jj, vj) in zip(eachindex(B, Dimension{2}), v)
for (i, ii) in zip(eachindex(dest), eachindex(B, (:, jj)))
dest[i] += B[ii,jj]*vj
end
end

It’s not too hard to figure out what eachindex(B, Dimension{2}) and eachindex(B, (:, jj)) should do: ii, for example, could be a CartesianInnerIndex (a type that does not yet exist) that for a particular column of B iterates from A[3,7,4] to A[5,8,4], where the dth index component wraps around at size(A, d). The big performance advantage of this strategy is that you only have to compute a div to set the bounds of the iterator on each column; the inner loop doesn’t require a div on each element access. No doubt, given suitable definition of jj one could be even more clever and avoid calculating div altogether. To the author, this strategy seems promising as a way to resolve the majority of the performance concerns about ReshapedArrays—only if you needed “random access” would you require slow (integer-based) operations.

However, a big problem is that compared to the “naive” implementation, this is rather ugly.

## Row-major matrices, PermutedDimensionArrays, and “taking transposes seriously”

Julia’s Array type stores its entries in column-major order, meaning that A[i,j] and A[i+1,j] are in adjacent memory locations. For certain applications—or for interfacing with certain external code bases—it might be convenient to support row-major arrays, where instead A[i,j] and A[i,j+1] are in adjacent memory locations. More fundamentally, this is partially related to one of the most commented-on issues in all of julia’s development history, known as “taking transposes seriously” aka #4774. There have been at least two attempts at implementation, #6837 and the mb/transpose branch, and for the latter a summary of benefits and challenges was posted.

One of the biggest challenges mentioned was the huge explosion of methods that one would need to support. Can generic code come to the rescue here? There are two related concerns. The first is linear indexing: oftentimes this is conflated with “storage order,” i.e., given two linear indexes i and j for the same array, the offset in memory is proportional to i-j. For row-major arrays, this notion is not viable, because otherwise a loop

function copy!(dest, src)
for i = 1:length(src)
dest[i] = src[i]  # trouble if i means "memory offset"
end
dest
end

would end up taking a transpose if src and dest don’t use the same storage order. Consequently, a linear index has to be defined in terms of the corresponding cartesian (full-dimensionality) index. This isn’t much of a real problem, because it’s one we know how to solve: use ind2sub (which is slow) when you have to, but for efficiency make row major arrays belong to the category (LinearSlow) of arrays that defaults to iteration with cartesian indexes. Doing so will ensure that if one uses generic constructs like eachindex(src) rather than 1:length(src), then the loop above can be fast.

The far more challenging problem concerns cache-efficiency: it’s much slower to access elements of an array in anything other than storage-order. Some reasonably fast ways to write matrix-vector multiplication are

for j = 1:size(B, 2)
vj = v[j]
for i = 1:size(B, 1)
dest[i] += B[i,j] * vj
end
end

for a column-major matrix B, and

for i = 1:size(B, 1)
for j = 1:size(B, 2)
dest[i] += B[i,j] * v[j]
end
end

for a row-major matrix. (One can do even better than this by using a scalar temporary accumulator, but let’s not worry about that here.) The key point to note is that the order of the loops has been switched.

One could generalize this by defining a RowMajorRange iterator that’s a lot like our CartesianRange iterator, but traverses the array in row-major order. eachindex claims to return an “efficient iterator,” and without a doubt the RowMajorRange is a (much) more efficient iterator than a CartesianRange iterator for row-major arrays. So let’s imagine that eachindex does what it says, and returns a RowMajorRange iterator. Using this strategy, the two algorithms above can be combined into a single generic implementation:

for I in eachindex(B)
dest[I[1]] += B[I]*v[I[2]]
end

Yay! Score one for efficient generic implementations.

But our triumph is short-lived. Let’s return to the example of copy! above, and realize that dest and src might be two different array types, and therefore might be most-efficiently indexed with different iterator types. We’re tempted to write this as

function copy!(dest, src)
for (idest, isrc) in zip(eachindex(dest), eachindex(src))
dest[idest] = src[isrc]
end
dest
end

Up until we introduced our RowMajorRange return-type for eachindex, this implementation would have been fine. But we just broke it, because now this will incorrectly take a transpose in certain situations.

In other words, without careful design the goals of “maximally-efficient iteration” and “keeping accesses in-sync” are in conflict.

## OffsetArrays and the meaning of AbstractArray

Julia’s arrays are indexed starting at 1, whereas some other languages start numbering at 0. If you take comments on various blog posts at face value, there are vast armies of programmers out there eagerly poised to adopt julia, but who won’t even try it because of this difference in indexing. Since recruiting those armies will lead to world domination, this is clearly a problem of the utmost urgency.

More seriously, there are algorithms which simplify if you can index outside of the range from 1:size(A,d). In my own lab’s internal code, we’ve long been using a CenterIndexedArray type, in which such arrays (all of which have odd sizes) are indexed over the range -n:n and for which 0 refers to the “center” element. One package which generalizes this notion is OffsetArrays. Unfortunately, in practice both of these array types produce segfaults (due to built-in assumptions about when @inbounds is appropriate) for many of julia’s core functions; over time my lab has had to write implementations specialized for CenterIndexedArrays for quite a few julia functions.

OffsetArrays illustrates another conceptual challenge, which can easily be demonstrated by copy!. When dest is a 1-dimensional OffsetArray and src is a standard Vector, what should copy! do? In particular, where does src[1] go? Does it go in the first element of dest, or does it get stored in dest[1] (which may not be the first element).

Such examples force us to think a little more deeply about what an array really is. There seem to be two potential conceptions. One is that arrays are lists, and multidimensional arrays are lists-of-lists-of-lists-of… In such a world view, the right thing to do is to put src[1] into the first slot of dest, because 1 is just a synonym for first. However, this world view doesn’t really endow any kind of “meaning” to the index-tuple of an array, and in that sense doesn’t even include the distinction conveyed by an OffsetArray. In other words, in this world an OffsetArray is simply nonsensical, and shouldn’t exist.

If instead one thinks OffsetArrays should exist, this essentially forces one to adopt a different world view: arrays are effectively associative containers, where each index-tuple is the “key” by which one retrieves a value. With this mode of thinking, src[1] should be stored in dest[1].

# Formalizing AbstractArray

These examples suggest a formalization of AbstractArray:

• AbstractArrays are specialized associative containers, in that the allowable “keys” may be restricted by more than just their julia type. Specifically, the allowable keys must be representable as a cartesian product of one-dimensional lists of values. The allowed keys may depend not just on the array type but also the specific array (e.g., its size). Attempted access by keys that cannot be converted to one of the allowed keys, for that specific array, result in BoundsErrors.

• For any given array, one must be able to generate a finite-dimensional parametrization of the full domain of valid keys from the array itself. This might only require knowledge of the array size, or the keys might depend on some internal storage (think DataFrames and OffsetArrays). In some cases, just the array type might be sufficient (e.g., FixedSizeArrays). By this definition, note that a Dict{ASCII5,Int}, where ASCII5 is a type that means an ASCII string with 5 characters, would qualify as a 5-dimensional (sparse) array, but that a Dict{ASCIIString,Int} would not (because there is no length limit to an ASCIIString, and hence no finite dimensionality).

• An array may be indexed by more than one key type (i.e., keys may have multiple parametrizations). Different key parametrizations are equivalent when they refer to the same element of a given array. Linear indexes and cartesian indexes are simple examples of interconvertable representations, but specialized iterators can produce other key types as well.

• Arrays may support multiple iterators that produce non-equivalent key sequences. In other words, a row-major matrix may support both CartesianRange and RowMajorRange iterators that access elements in different orders.

# Finding a way forward

Resolving these conflicting demands is not easy. One approach might be to decree that some of these array types simply can’t be supported with generic code. It is possible that this is the right strategy. Alternatively, one can attept to devise an array API that handles all of these types (and hopefully more).

In GitHub issue #15648, we are discussing APIs that may resolve these challenges. Readers are encouraged to contribute to this discussion.


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# The effect of wavefunction collapse on spacetime?

#### asimov42

Hi folks,

I'm not sure if it's best to ask this question here, or in the Special & General Relativity section - it's probably more appropriate for this forum.

I've been wondering about the following question: what effect does wavefunction collapse have on space-time? For example, if we measure, say, the position of an electron at some point in time, its wavefunction should collapse instantaneously. The collapse should localize the 'peak' of the wavefunction to a smaller region of space-time, and this should happen instantaneously.

Shouldn't this collapse have an effect on the space-time background? I.e. because the position of the electron is now localized in a smaller region of space-time, shouldn't the shape of space-time also change (due to gravity), albeit by a very small amount? If the collapse occurs instantaneously, does this imply an instantaneous change in space-time also? And if so, wouldn't this make space-time discontinuous?

Thanks all. As a final remark, perhaps the questions about can't really be answered until there is a workable theory of quantum gravity?

J.

Related Quantum Physics News on Phys.org

#### Dmitry67

In the most commonly accepted version of CI, neither wavefunction nor collapse are REAL. It is our 'knowledge' about the system, and it is observer dependent (for different observers there are different collapses - check Wigner's friend experiment)

After the discovery of the Quantum Decoherence (wiki it) there is no longer need in the mysterious 'collapse' as measurements can be explained using just pure QM without any additional assumptions about special role of measurements

For that reason MWI is more and more accepted and CI is no longer #1.

#### lalbatros

But then I doubt that it would be observer-dependent.

Before decoherence was a widely used word, the interaction of the microscopic system with the macroscopic world (measuring device) was also used for the interpretation. Long ago, this was the pov of Landau who probably did not consider the "measurement problem" was anything of interrest.

Since I have learned physics mainly in Landau book, this pov is my basic hypothesis, and I really prefer not discussing this question unless there is some prospect for real physics instead of lounge philosophy or even new mysticism.

It seems that today this has really become a laboratory topic and this make me very happy. I don't expect decoherence to have any relation with our understanding of spacetime.
But your question is not very clear in this respect.

#### Fra

Another personal opinon to add to your calculation.

I've been wondering about the following question: what effect does wavefunction collapse have on space-time? For example, if we measure, say, the position of an electron at some point in time, its wavefunction should collapse instantaneously. The collapse should localize the 'peak' of the wavefunction to a smaller region of space-time, and this should happen instantaneously.

Shouldn't this collapse have an effect on the space-time background? I.e. because the position of the electron is now localized in a smaller region of space-time, shouldn't the shape of space-time also change (due to gravity), albeit by a very small amount?
For me the best way of seeing it is that the wavefunction represents the observers information about it's own environment (spacetime is part of the structure of that information). But this information is local, to the observer. Thus the collapse is not non-local if you look at where the information is encoded. The collapse is merely a local information update. No mystery at all.

In this view, the question is then (IMHO) rather different: how an spacetime emerges in the sense that the different observer views are related exactly by the observer-observer transformations of SR and GR. At first glance this would suggest that each observer would see a totally different spacetime, unrelated to it's neighbouring observers. But this would be the case only if they didn't interact, but of course the do.

Ie. how can the disitributed observer images, by interaction synchronize as to be consistent with some global, or quasi-global(some local neighbourhoods) symmetries.

I think this is what the result of observers evolving together in ineraction results in. And how can we understand, how the interaction process itself, produces distributed images, but that are related by a symmetry. (IE. the logic of emergent symmetry.) But the details on an exact model is still missing.

Thanks all. As a final remark, perhaps the questions about can't really be answered until there is a workable theory of quantum gravity?
I think the the last questions I posed, won't be answered until there is a theory of QG.

/Fredrik

#### Ilja

In the most commonly accepted version of CI, neither wavefunction nor collapse are REAL. It is our 'knowledge' about the system, and it is observer dependent (for different observers there are different collapses - check Wigner's friend experiment)

After the discovery of the Quantum Decoherence (wiki it) there is no longer need in the mysterious 'collapse' as measurements can be explained using just pure QM without any additional assumptions about special role of measurements

For that reason MWI is more and more accepted and CI is no longer #1.
Decoherence needs some undefined "decomposition of the world into systems" (some tensor product structure).

Moreover, it is irrelevant for the measurement problem. If you have a superposition of a living and a dead cat, decoherence does not help you. Instead, it gives you only a superposition of a living cat + environment and a dead cat + environment.

The problem remains the same.

#### Dmitry67

Moreover, it is irrelevant for the measurement problem. If you have a superposition of a living and a dead cat, decoherence does not help you. Instead, it gives you only a superposition of a living cat + environment and a dead cat + environment.

The problem remains the same.
Cat exchanges photons with the environment, and the enviroment get also decoherenced. So we get a superposition of:

living cat + happy observer

but not
superposition of both cats + confused observer

--> exactly what we observe.

#### Ilja

Cat exchanges photons with the environment, and the enviroment get also decoherenced. So we get a superposition of:

living cat + happy observer

but not
superposition of both cats + confused observer

--> exactly what we observe.
No. What we observe, is something describable by the wave function living cat + happy observer. Or, in other cases, by a wave function of dead cat + sad observer. And we have learned, in the beginners course of quantum mechanics,
that we should never never interpret a state psi1 + psi2 as being or in psi1 or in psi2. The superposition of them is something very different.

But my main objection against MWI is the missing of precise definitions. No definition of the decomposition into systems. No precise definition of the branches, nor of their evolution, nor of their splitting. Many words instead of many worlds.

#### Dmitry67

And we have learned, in the beginners course of quantum mechanics,
that we should never never interpret a state psi1 + psi2 as being or in psi1 or in psi2. The superposition of them is something very different.

But my main objection against MWI is the missing of precise definitions. No definition of the decomposition into systems. No precise definition of the branches, nor of their evolution, nor of their splitting. Many words instead of many worlds.
"beginners course" is based on CI for purely historical reasons. So many people start to assume that CI = QM and they even never learn about other interpretations.

MWI = pure (interpretation-less) QM + Quantum Decoherence (QD).
And MWI does not need any new axioms or assumptions, it is just a consequence of QD

Regarding "definitions, evolutions and spliiting" - there are all there:
http://en.wikipedia.org/wiki/Quantum_decoherence

#### Ilja

MWI = pure (interpretation-less) QM + Quantum Decoherence (QD).
And MWI does not need any new axioms or assumptions, it is just a consequence of QD

Regarding "definitions, evolutions and spliiting" - there are all there:
http://en.wikipedia.org/wiki/Quantum_decoherence
Your formula for MWI is wrong. Interpretationless QM gives a single state vector, and no containment relation between this vector and other vectors, which may appear in some decompositions.

Quantum decoherence needs a decomposition into systems to start. This decomposition is not provided by standard QM. Thus, you cannot even start with decoherence if you have only interpretationless QM.

I have not asked about definitions of decoherence, this is quite clear, but for the many worlds interpretation: branches, their evolution laws, any justification for the strange containment relation between the state and the branches, and the decomposition into systems you need to start decoherence.

#### Dmitry67

Your formula for MWI is wrong. Interpretationless QM gives a single state vector, and no containment relation between this vector and other vectors, which may appear in some decompositions.
However, decoherence by itself may not give a complete solution of the measurement problem, since all components of the wave function still exist in a global superposition, which is explicitly acknowledged in the many-worlds interpretation. All decoherence explains, in this view, is why these coherences are no longer available for inspection by local observers. To present a solution to the measurement problem in most interpretations of quantum mechanics, decoherence must be supplied with some nontrivial interpretational considerations (as for example Wojciech Zurek tends to do in his Existential interpretation). However, according to Everett and DeWitt the many-worlds interpretation can be derived from the formalism alone, in which case no extra interpretational layer is required.
Do you agree that QD explains measurements? Before QD was discovered MWI required some additional assumptions, like other iterpretations. QD made CI total BS: CI had invented the whole thing of 'collapse' just to explain measurements and the macroscopic behavior, and then oooops, we dont need collapse at all, we have QD! I wonder what CI guys do with both things.

On the contrary, MWI does not need any additional axioms. So CI is effectively cut by the Occams razor.

#### ueit

MWI = pure (interpretation-less) QM + Quantum Decoherence (QD).
And MWI does not need any new axioms or assumptions, it is just a consequence of QD
1. How does MWI account for the Born postulate?

2. How does MWI account for the observed reality (3d space, point particles)?

#### Dmitry67

1. How does MWI account for the Born postulate?

2. How does MWI account for the observed reality (3d space, point particles)?
1. http://en.wikipedia.org/wiki/Quantum_decoherence#Loss_of_interference_and_the_transition_from_quantum_to_classical

2.
As a consequence, the system behaves as a classical statistical ensemble of the different elements rather than as a single coherent quantum superposition of them. From the perspective of each ensemble member's measuring device, the system appears to have irreversibly collapsed onto a state with a precise value for the measured attributes, relative to that element.

#### ueit

The link does not address my questions. It does not explain how MWI (which accepts only the reality of the wave-function) is supposed to account for our observations consisting more or less of point particles in a 3D space. Only after this question is addressed we can speak about decoherence.

#### Ilja

Do you agree that QD explains measurements?

No. It needs a decomposition of the universe into systems. This decomposition is not defined.

See my paper arXiv:0901.3262

#### Dmitry67

The link does not address my questions. It does not explain how MWI (which accepts only the reality of the wave-function) is supposed to account for our observations consisting more or less of point particles in a 3D space. Only after this question is addressed we can speak about decoherence.
On the contrary, QD explains why you observe 'particles'
Say, you register 1 photon using your photocamera. Photon (wave) hits the matrix, and it splits/decoherenced into 4000000 branches (for 4Megpixel camera). Then you say: look,photon is particle, it is almost a point, it hit just one pixel!

#### ueit

On the contrary, QD explains why you observe 'particles'
Say, you register 1 photon using your photocamera. Photon (wave) hits the matrix, and it splits/decoherenced into 4000000 branches (for 4Megpixel camera). Then you say: look,photon is particle, it is almost a point, it hit just one pixel!
Yeah, but you have already assumed the 3D classical world with ready-made 3d objects (the 4Megpixel camera for example). First, you need to explain how these objects are supposed to appear from your universal wavefunction. Then you have to explain how to ascribe probabilities for different branches and so on.

#### Dmitry67

QM explain the existence of complex systems: atoms, moleculs, crystalls from the first principles without any additional assumptions. There is nothing new.

#### lalbatros

What about entanglement as the fabric of space-time?

#### bg032

Regarding "definitions, evolutions and spliiting" - there are all there:
http://en.wikipedia.org/wiki/Quantum_decoherence
In the above link I have not found an answer to the following simple question: let $$\Psi(t)$$ be the wave function of the universe at a suitable time t. According to MWI+decoherence, it is split into n branches or worlds:

$$\Psi(t)=\Psi_1 + \ldots + \Psi_n$$

What is the criterion defining this splitting? Can some expert in MWI+decoherence answer this?

Note 1: if the criterion requires to split the universe into subsystems, the criterion for this splitting is also required.

Note 2: of course the criterion must be mathematical; thus a criterion of the type

$$|\hbox{cat alive}\rangle + |\hbox{cat dead}\rangle$$

is not acceptable.

#### Ilja

QM explain the existence of complex systems: atoms, moleculs, crystalls from the first principles without any additional assumptions. There is nothing new.
QM explains it, but not from first principles. Instead, one needs some classical background, or something else which replaces this classical background. Say, the configuration of pilot wave theory.

#### Dmitry67

Wait, lets take for example hydrogen atom (lets assume that proton is elementary)
Hydrogen atom can be calculated from the first principles using QFT

Or are you talking about some phylosophical issues like 'yes, we have a result, some numbers, but what we have actually calculated'?

#### Dmitry67

In the above link I have not found an answer to the following simple question: let $$\Psi(t)$$ be the wave function of the universe at a suitable time t. According to MWI+decoherence, it is split into n branches or worlds:

$$\Psi(t)=\Psi_1 + \ldots + \Psi_n$$

What is the criterion defining this splitting? Can some expert in MWI+decoherence answer this?

Note 1: if the criterion requires to split the universe into subsystems, the criterion for this splitting is also required.

Note 2: of course the criterion must be mathematical; thus a criterion of the type

$$|\hbox{cat alive}\rangle + |\hbox{cat dead}\rangle$$

is not acceptable.
Whats about density matrix approach in the same article?

#### LaserMind

Hi folks,

I'm not sure if it's best to ask this question here, or in the Special & General Relativity section - it's probably more appropriate for this forum.

I've been wondering about the following question: what effect does wavefunction collapse have on space-time? For example, if we measure, say, the position of an electron at some point in time, its wavefunction should collapse instantaneously. The collapse should localize the 'peak' of the wavefunction to a smaller region of space-time, and this should happen instantaneously.

Shouldn't this collapse have an effect on the space-time background? I.e. because the position of the electron is now localized in a smaller region of space-time, shouldn't the shape of space-time also change (due to gravity), albeit by a very small amount? If the collapse occurs instantaneously, does this imply an instantaneous change in space-time also? And if so, wouldn't this make space-time discontinuous?

Thanks all. As a final remark, perhaps the questions about can't really be answered until there is a workable theory of quantum gravity?

J.
My view is that the electron does not collapse into a tiny heap when observed but the observation gets a snapshot at one time only. I don't know why we call it 'collapse' come to think of it - what collapses?

#### bg032

Whats about density matrix approach in the same article?
This approach requires to take the partial trace of a density matrix, which requires the decomposition into subsystems, and this decomposition is not mathematically defined.

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text:
Integrals of complex functions and Integration by parts

In this class, we will deal only with integrals of complex functions of a real variable integrated with respected to the real variable. This is identical to integration of real functions of real variables. The $j$ is simply treated as a constant in all these cases. In this class, we are typically interested in time $t$ or frequency $\omega$ being the independent variable. Hence, the integrals will be with respect to $t$ or $\omega$.

For example,

$\int_0^{\pi/4} e^{j 2 t} \ dt = \left[\frac{e^{j 2 t}}{2 j}\right]_{0}^{\pi/4} = \frac{j-1}{2j}$

Sometimes we will have to use Integration by parts to evaluate integrals. The main result to recall is

(1)
\begin{align} \displaystyle{\int_a^b u(t) \ dv(t) \ dt = \left[ u(t) \ v(t) \right]_a^b - \int_a^b v(t) \ du(t) \ dt} \end{align}

Example:

Evaluate $\displaystyle{\int_0^1 t e^{-j \omega t} dt}$, where $\omega$ is any complex number
We choose

(2)
\begin{eqnarray} \nonumber u(t) = t &,& dv(t) = e^{-j \omega t} \ dt \\ \nonumber \Longrightarrow du(t)=dt &,& v(t) = \frac{e^{-j \omega t}}{-j \omega} \end{eqnarray}
(3)
\begin{eqnarray} \nonumber \int_0^1 t e^{-j \omega t} dt & = & \displaystyle{\left[\frac{t e^{-j \omega t}}{-j \omega}\right]_0^1 - \int_0^1 \frac{e^{-j \omega t}}{-j \omega} \ dt}\\ \nonumber & = & - e^{-j \omega} - \displaystyle{\left[\frac{e^{-j \omega t}}{-\omega^2} \right]_0^1} = - e^{-j \omega} + \frac{e^{-j \omega}-1}{\omega^2} \end{eqnarray}


text:
# Introduction¶

Supervised learning:

• Classification (binary/not)
• regression: continuous numeric labels
• ranking

Batch assumption: iid

• Distribution of things and measurements defines some unknown $$P(\mathbf{x}, y)$$ or $$D(\mathbf{x}, y)$$ over domain-label pairs
• We want to find a hypothesis $$f(\mathbf{x})$$ that is close to the truth
• so we need a loss fcn $$L(y, t)$$ where y = prediction, t = truth
• We want to minimize $$\int P(\mathbf{x},y) L(y, f(\mathbf{x}))$$

Overfitting and Underfitting

Overfitting is when your hypothesis is too complex for the data, and

Underfitting is when your hypothesis is too simple.

## Main Steps¶

### Feature Extraction¶

In feature extraction, you need to extract some useful data dimension from your sample. It’s very important to extract good features from your data - there’s a whole field around it (feature engineering).

Most commonly, an instance is represented as a vector of features $$\mathbf{x}$$, along with a ground-truth label $$y$$. This can be represented as $$(\mathbf{x}, y)$$.

They can be categorical/nominal, ordinal, or numeric.

Example

For example, in a binary spam/ham email classifier, some features you might extract might be:

• are there words in all caps?
• is the email long?
• this can be represented as a binary attribute (after a certain length), or a numeric one.
• how many ! are in the email?

### Training¶

In the training step, you choose a hypothesis space/hypothesis class and learn a hypothesis: a function that maps an instance to a class/label.

The hypothesis $$g$$ is a learned model defined by parameters. We use ML when $$g(\mathbf{x})$$ is unknown to us and we can’t think about how to implement it algorithmically.

### Testing¶

Now, given a new instance, we use our classifier to predict a label. Did it predict it correctly?

$y' = g(\mathbf{x'})$

It’s important for classifiers to be able to generalize: to predict correctly when encountering a situation that has never been seen before.

The power for a classifier to generalize often depends just as much on how you’re representing instances as it does the classification algorithm.

Generally, test data and training data should be drawn from the same population.

Note

Independent, Identically Distributed (iid) assumption for same population:

We assume the distribution of instances and labels defines some unknown, but fixed, $$P(\mathbf{x}, y)$$.

We also assume that all training and all test instances/labels are independent and identically distributed.

That is - a collection of random variables is iid if they all have the same probability distribution, and are all mutually independent.

## Supervised Learning¶

Supervised learning is the primary type of ML. In this approach, training instances come with a ground-truth label.

### Classification¶

In a classification problem, labels are nominal (an unordered set). The labels can be either binary, or multiclass (for example, a spam/ham classifier or a MNIST digit classifier).

### Regression¶

In a regression problem, labels are numeric (and continuous). For example, predicting the price of a house might be a regression problem.

### Ranking¶

In a ranking problem, the model is asked to order a set of objects. Usually, in this case, the input to these models are keywords and a prior (e.g. prior data gathered on you), and the output is the objects’ ranking.

### More¶

There’s a lot more examples, too!

• Disease diagnosis
• x: patient properties
• y: disease/recommended therapy
• Part-of-speech tagging
• x: an english sentence
• y: the part of speech of a word in there
• Face recognition
• x: bitmap of a person’s face
• y: identity of a person
• Reinforcement Learning
• output: a sequence of actions (policy). Individual actions aren’t important, but the overall policy is.
• no supervised output, but delayed rewards
• e.g. game playing, robot in a maze
• Online Learning
• train on one instance at a time, as opposed to batch learning
• e.g. perceptron

## Other Forms¶

• unsupervised learning: no labels provided during train time
• clustering
• e.g. image compression, bioinformatics
• semi-supervised learning: use partially labeled data as well as unlabeled data

## Training¶

But what does training mean?

The hypothesis, $$g(\mathbf{x})$$, is defined by parameters. How do we optimize these parameters?

We need to use a loss function.

### Loss Function¶

To learn, we want to minimize the loss function $$L(y, y')$$. This function measures the error of the prediction $$y'$$ on the train set, and tells you how good $$g(\mathbf{x})$$ is at this point.

Choosing the right loss function is important, and different functions will want different functions.

Note

For example, a simple loss function for binary classification is as follows:

def L(y, y_pred):
if y_pred == y:
return 1
else:
return 0

Minimization is done using an optimization algorithm like gradient descent.

aka SGD

In a sense, a function that, at an arbitrary point, finds the steepest slope that moves you in the direction of the local minimum.

\begin{align}\begin{aligned}\begin{split}\text{Given:} \\ t = [1..] \\ \text{samples } (\mathbf{x}_t, y_t) \\ \text{current model } f_t \\ \text{learning rate } \eta_t \\ \\\end{split}\\f_{t+1} = f_t - \eta_t \cdot \nabla_{f_t} L(f(\mathbf{x}_t), y_t)\end{aligned}\end{align}

### Choices¶

Model

So really, our goal is to learn a generalizable $$g(x)$$ that is a good approximation of the truth. You should choose a model that is capable of approximating the truth so.

Hypothesis Space

If we choose a hypothesis space that is too simple, there are fewer parameters to learn, but the model is less powerful.

The model will have less variance - fewer changes with changing training data - but more bias (making more assumptions).

Buuut, choosing a hypothesis space that is too complex may cause high variance. This is called the bias-variance tradeoff.

Variance

Having super high variance means you can perfectly represent the training data - but this is overfitting!

Different training sets may produce wildly different hypotheses - this is what high variance is.

Bias

Similarly, we say a model has high inductive bias when the model makes a lot of assumptions about the data.

For example, when we choose to use a linear regression, that model has a high bias towards a linear relationship between the features and the labels.

Overall, variance represents estimation error (limits due to data) and bias represents approximation error (limits due to model family).

These concepts are closely linked to the concept of overfitting/underfitting.

## Evaluation¶

There are lots of ways to measure predictive performance. The most popular are:

• Accuracy and Error Rate
• Precision Recall and F-measure

### Accuracy¶

Useful in X vs. Y problems, where any classes are equally important.

$accuracy = \frac{\text{# correct predictions}}{\text{# test instances}}$
$error = 1 - accuracy = \frac{\text{# incorrect predictions}}{\text{# test instances}}$

### Confusion Matrix¶

An extension of accuracy - given a matrix of positive and negative instances, it is possible to make a matrix

         Predicted
Y   N
+-------
Actual Y | TP  FN       Where TP = True Positive, FN = False Negative,
N | FP  TN       FP = False Positive, TN = True Negative

Example: Given the result table

         Predicted
Y    N
+--------
Actual Y | 100  5
N | 10   50

Total corpus size: 165
Total predicted yes: 110
Total predicted no: 55
Actual yes: 105
Actual no: 60

Accuracy = (100 + 50) / 160 = 0.91
Error = (5 + 10) / 160 = 0.09

### Precision, Recall, F-measure¶

What about “X vs. non-X” types of problems: spam v. not spam, relevant v. not relevant?

$precision = \frac{\text{# relevant records retrieved}}{\text{total # of records retrieved}}$
$recall = \frac{\text{# relevant records received}}{\text{total # of relevant records}}$
$F measure = \frac{2 * P * R}{P + R}$

So, going back to our example, but now we’re predicting whether or not something is relevant…

         Predicted
Y    N
+--------
Actual Y | 100  5
N | 10   50

Accuracy = (100 + 50) / 160 = 0.91

Precision = TP / (TP + FP) = 100 / 110 = 0.91
Recall = TP / (TP + FN) = 100 / 105 = 0.95
F1 score = (2 * 0.91 * 0.95) / (0.91 + 0.95) = 0.93

TLDR:

         Predicted
Y   N
+-------
Actual Y | TP  FN
N | FP  TN

$accuracy = \frac{TP + TN}{P + N}$
$precision = \frac{TP}{TP + FP}$
$recall = \frac{TP}{TP + FN}$

### Reporting Performance¶

• separate training and test data
• never see test data in training
• usually 80-20 split
• k-fold cross validation
• why just split once?
• run k iterations, in each iteration hold out a different test set
• improves robustness of reported set


text:
# Confidence interval for mean of a normal population

Consider a normal population with unknown mean $$\mu$$ and variance $$\sigma^2=9$$. To test $$H_{0}:\mu=0$$,against $$H_{1}:\mu \ne 0$$. A random sample of size 100 is taken. Based on this sample, the test of the form $$|\bar{X_n}| > K$$ rejects the null hypothesis at 5% level of significance. Then, which of the following is a possible 95% confidence interval for $$\mu$$ ?

(A) (-0.488, 0.688) (B) (-1.96, 1.96) (C) (0.422, 1.598) (D) (0.588, 1.96)

Now, if we construct the test statistic $$\tau=\frac{\bar{X_n}-\mu}{0.3} \sim N(0,1)$$ Therefore our test ,actually based on the observed value of $$\tau=\tau_{0}$$ is given by $$|\tau_{0}|> \tau_{\frac{\alpha}{2}}$$ where $$\alpha=0.05$$ Now,we get $$K=\frac{3}{10}\tau_{0.025}=0.588$$ Thus the 95% confidence interval for $$\mu$$ is coming to be $$(\bar{X_n}-0.588 ,\bar{X_n}+0.588)$$So, looking at the options it seems that both (A) , (C) are correct but only (C) is given to be correct. Please help!

• Consider the width of the CI. – GNUSupporter 8964民主女神 地下教會 Dec 2 '18 at 14:51
• sorry,i edited one of the options – Legend Killer Dec 2 '18 at 14:52
• @GNUSupporter8964民主女神地下教會they have the same width,I mean options A,C – Legend Killer Dec 2 '18 at 14:55


text:
# Solving for the length of velocity of a particle in space (Calc III, due Tomorrow!)

1. Sep 28, 2008

### JoeSabs

1. The problem statement, all variables and given/known data
r(t) is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of t. Write the particle's velocity at that time as the product of its speed and direction.

r(t)=(2cost)i+(3sint)j+4tk; t=pi/2

2. Relevant equations
sin^2+cos^2=1
length of v= sqrt(vi+vj+vk) if vi is the first coefficient of velocity equation, vj 2nd, vk 3rd.

3. The attempt at a solution
v(t)= (-2sint)i+(3cost)j+4k

|v(t)| (length)=2sqrt(5), but I don't know how to get this.

sqrt(((-2sin(t))^2)+((3cos(t))^2)+(4^2))
sqrt((4sin^2(t))+(9cos^2(t))+(16))
?
Are you supposed to somehow remove the coefficients to use cos^2+sin^2? I tried this to no avail. I also tried using the sin^2(x)= (1-cos2x)/2, and the cos one too, but that didn't work either.

2. Sep 28, 2008

### Pacopag

Re: Solving for the length of velocity of a particle in space (Calc III, due Tomorrow

It looks to me like you have the correct answer. Just put t=Pi/2 into your last expression. You get sqrt(20) which is the same as 2sqrt(5).

3. Sep 28, 2008

### JoeSabs

Re: Solving for the length of velocity of a particle in space (Calc III, due Tomorrow

Ah, I see! Plug in before trying to simplify. That's very easy, just out of the normal order of operations I'm used to. Thanks a lot!


text:
# Finding a basis such that the $\mathcal{B}$-matrix is diagonal for orthogonal projection and reflection

Find a basis of the transformations such that the $\mathcal{B}$-matrix is diagonal.

1. Orthogonal projection $T$ onto the plane $3x_1+x_2+2x_3=0$ in $\mathbb{R^3}$.

2. Reflection $T$ about the plane $x_1+2x_2+x_3=0$ in $\mathbb{R^3}$.

In both cases, the $\mathcal{B}$-matrix will be diagonal iff $T(v_i)=kv_i$ for all vectors $v_i$ in the basis.

For number 1, we know the plane is defined by $\left( \begin{array}{ccc}3 \\ 1\\2 \end{array} \right)$, and we choose our basis to be three vectors $v1 = \left( \begin{array}{ccc}3\\1\\2 \end{array} \right), v_2 = \left( \begin{array}{ccc}2\\0\\-3 \end{array} \right), v_3 = \left( \begin{array}{ccc}1\\-3\\0 \end{array} \right)$ so that $v_2$ and $v_3$ are perpendicular to $v_1$. Then we have $T(v_1) = v_1$, but I'm not sure how to calculate $T(v_2)$ and $T(v_3)$.

For number 2, we know the plane is defined by $\left( \begin{array}{ccc}1\\2\\1 \end{array} \right)$ and we again consider three vectors $v1 = \left( \begin{array}{ccc}1\\2\\1 \end{array} \right), v_2 = \left( \begin{array}{ccc}1\\0\\-1 \end{array} \right), v_3 = \left( \begin{array}{ccc}2\\-1\\0 \end{array} \right)$ so that $v_1$ is parallel to the plane and $v_2,v_3$ are perpendicular to the plane. In this case, I know $T(v_1)=v_1$, but again I'm not sure how to calculate $T(v_2)$ and $T(v_3)$.

In either case, is my initial work a correct procedure? Also, how do we calculate $T(v_2)$ and $T(v_3)$ in either case for $\mathbb{R^3}$. It doesn't seem so clear to see this geometrically as it is in $\mathbb{R^2}$.

For 1: $v_2$ and $v_3$ are constructed to be in the plane under question. Thus, the projection leaves these vectors as they are, $T(v_2)=v_2$ and $T(v_3)=v_3$. (Expanded: These vectors are already in the plane on which $T$ projects. The projection onto a plane takes vectors outside the plane and throws them onto the plane. If such a vector is already on a plane, then the projection does nothing.)

Moreover, $v_1$ is the direction normal to the plane, hence $T(v_1)=0$.

For 2: With similar considerations, one finds $T(v_1)=-v_1$, $T(v_2)=v_2$ and $T(v_3)=v_3$.

(Expanded: Reflection changes the direction of incoming rays. If the ray is perpendicular to the plane (or the mirror) it gets thrown back and its direction changes: $T(v_1)=-v_1$.)

• Can you please elaborate? I'm trying to imagine these geometrically and I'm failing to see what you describe. For the first one, I understand why $T(v_1)=0$, but I don't understand why $T(v_2)=v_2$ and $T(v_3)=v_3$. Also for the second one, why is $T(v_1)=-v_1$? Why is it not $T(v_1)=v_1$ if we're just reflecting the plane on itself (thus keeping it the same)? And I honestly don't understand the $T(v_2)$ nor $T(v_3)$ here either. Oct 14 '14 at 10:14
• Actually, I'm not sure I understand why $T(v_1)=0$. Isn't the orthogonal of something onto itself just that something? Oct 14 '14 at 10:29
• en.wikipedia.org/wiki/Projection_%28linear_algebra%29
– daw
Oct 14 '14 at 11:10
• en.wikipedia.org/wiki/Reflection_%28mathematics%29
– daw
Oct 14 '14 at 11:11
• Can you explain why $v_1$ is perpendicular to the plane in both cases? I thought $v_1$ was parallel to the planes. Similarly, why are $v_2,v_3$ parallel to the plane? Oct 15 '14 at 8:15


text:
# The single-slit diffraction pattern shown in the Figure was produced with red light of wavelength lambda...

###### Question:

The single-slit diffraction pattern shown in the Figure was produced with red light of wavelength lambda = 633 nm. The screen on which the pattern was projected was located a distance D = 2.0m from the slit. The slit has a width of a = 0.30mm. What is the width w of the central maximum? (The width is equal to the distance between the two first diffraction minima located on either side of the center.)

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##### [z Points]DETAILSMY NOTESWhen the eigenvalues ofA =4 = 3 and0, what are the possible values of J and d? (Select all that apply:)and-J and dandand d =andSubminansyer
[z Points] DETAILS MY NOTES When the eigenvalues of A = 4 = 3 and 0, what are the possible values of J and d? (Select all that apply:) and -J and d and and d = and Subminansyer...
##### $\text { Find the indicated partial sums for the sequence.}$ $1,2,3,4,5,6,7, \ldots ; S_{3} \text { and } S_{7}$
$\text { Find the indicated partial sums for the sequence.}$ $1,2,3,4,5,6,7, \ldots ; S_{3} \text { and } S_{7}$...
##### INSTRUCTOR: BYUNGIK KAHNG Answer the questions regarding the given graph y = (sin â‚¬) e(2cosz+1) 0 < I < 2t.definite integral. points) Express the area of the shaded region through Calculate the anti-derivative (indefinite integral). (10 points) and (b) to find the precise arca of the shaded region. (5 points) Use
INSTRUCTOR: BYUNGIK KAHNG Answer the questions regarding the given graph y = (sin â‚¬) e(2cosz+1) 0 < I < 2t. definite integral. points) Express the area of the shaded region through Calculate the anti-derivative (indefinite integral). (10 points) and (b) to find the precise arca of the sh...
##### An agreement between two parties to exchange specified cash flows at specified intervals in the future...
An agreement between two parties to exchange specified cash flows at specified intervals in the future is called a/an Derivative. Futures Forward O Option. Swap Оо ООО...
##### IIn Exercises 16-24. identify the initial value and the rate of change. and explain their meanings in practical terms using unitsIS. After & rain storm. the water in & trough begins t0 evap- orate . The amount in gallons remaining alter ! days is given by V = 50 - 1.2.36. PSEG Long Island charges its residential customers monthly service charge plus an energy charge based on the amount of electricity used: The monthly cost of electricity is approximated by the funetion: â‚¬ = f(h) 37.60
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##### What are the eritical points olf" b. On what open intervals is inereasing or decreasing? c At what points, if any, does f assume local maximum and m mum values? 1. f'() = A6 - I) 2. f"(x) = (r - I)(x + 2) 3. f'(x) = ( - I)6 + 2) 4. f"'(x) = (x - I)(x + 2)2 5. f'() = ( - Ie * 6. f'(x) = (r _ 7) + I)x + 5) r6 - 1) 7. f'(x) 1 + 2 17-2 8. 2)6 + 4) f'(x) (x + I)G - 3) x-1,3 9. f'(x) =1-4- Xf0 10. f'(x) = 3 - X f0 Vx 11. f'(x) = x-W6x +
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##### Problem 1: Four identical charges, each with charge Q and mass m, are placed at four corners of a square with side They are initially held at rest and then released_ The charges then start moving apart due to electrostatic repulsion. What is the magnitude of the velocity, V, of each particle, at the instant when they form square with side 2a? (Hint: use energy conservation:)Problem 2: Consider the circuit shown in Figure 1_ All four resistors are iden- tical with resistances 3k02 each: Find the
Problem 1: Four identical charges, each with charge Q and mass m, are placed at four corners of a square with side They are initially held at rest and then released_ The charges then start moving apart due to electrostatic repulsion. What is the magnitude of the velocity, V, of each particle, at the...
##### Let the "Tribonacci sequence" be defined by T; Tn-3 for n > 4. Prove that Tn 2n for alln > 1.TzT31 and Tn Tn-1 +Tn-2 +
Let the "Tribonacci sequence" be defined by T; Tn-3 for n > 4. Prove that Tn 2n for alln > 1. Tz T3 1 and Tn Tn-1 +Tn-2 +...
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##### Saying yes Frequency 30 0 56 2 73 3 41
Saying yes Frequency 30 0 56 2 73 3 41...


text:
# Given a sequence with Fourier transform X ( e j ω ) = 1 /( 1 + 0.8 − j ω ) ,

1. Given a sequence with Fourier transform X ( e j ω ) = 1 /( 1 + 0.8 j ω ) , determine the

Fourier transform of the following signals:

(a) x 1 [ n ]= e j π n / 2 x [ n + 2],

(b) x 2 [ n ]= x [ n ] cos ( 0.4 π n ) ,

(c) x 3 [ n ]= x [ n ]∗ x [− n ],

(d) x 4 [ n ]= x [2 n ],

1. x 5 [ n ]= x [ n ], n = even and x 5 [ n ]= 0, n = odd.

## Plagiarism Checker

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text:
## Abstract Algebra: Groups and Subgroups

The problem says: Suppose that * is an associative binary operation on a set S.
Let H= {a ε S l a * x = x * a for all x ε s}. Show that H is closed under *. ( We think of H as consisting of all elements of S that commute with every element in S)

My teacher is horrible so I am pretty lost in the class. I am aware of what the associative property is, but I'm not sure how to go about solving this question when it comes to the binary operation. This is going to be on my exam so I need to know how to solve it.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 A binary operation is nothing but a two-variable function. It takes two elements of the set G and gives a new element. In other words, a binary operation is a function *: G$\times$G → G and G is a set. Since * is a function on two-variables and its domain is a cartesian product of two sets, the elements in * are ordered pairs like (x,y), we define (x,y) := x*y. If x*y is in G, we say G is closed under the operation *. Are those definitions clear? Now, your problem asks us to show that H is closed under *. To do that you should take two arbitrary elements of H, like h and h' and show that h*h' is also in H. You don't need to use associativity to solve this problem, the associativity is later needed when you want to show that H is a subgroup of G.
 That helps a lot! Thank you!

Recognitions:

## Abstract Algebra: Groups and Subgroups

 Quote by AdrianZ A binary operation is nothing but a two-variable function. It takes two elements of the set G and gives a new element. In other words, a binary operation is a function *: G$\times$G → G and G is a set. Since * is a function on two-variables and its domain is a cartesian product of two sets, the elements in * are ordered pairs like (x,y), we define (x,y) := x*y. If x*y is in G, we say G is closed under the operation *. Are those definitions clear? Now, your problem asks us to show that H is closed under *. To do that you should take two arbitrary elements of H, like h and h' and show that h*h' is also in H. You don't need to use associativity to solve this problem, the associativity is later needed when you want to show that H is a subgroup of G.
(emphasis mine)

yes, you DO. suppose we want to show that h*h' is in H whenever h,h' are. by definition, this means we want to show that:

(h*h')*x = x*(h*h'), for all x in S.

to actually DO this, we might proceed like this:

(h*h')*x = h*(h'*x) <---this is where we need associativity

= h*(x*h') (by definition of H, since h' is in H)

= (h*x)*h' <---associativity used AGAIN

= (x*h)*h' (since h is in H)

= x*(h*h') <---associativity used for a THIRD time.

 Quote by Deveno (emphasis mine) yes, you DO. suppose we want to show that h*h' is in H whenever h,h' are. by definition, this means we want to show that: (h*h')*x = x*(h*h'), for all x in S. to actually DO this, we might proceed like this: (h*h')*x = h*(h'*x) <---this is where we need associativity = h*(x*h') (by definition of H, since h' is in H) = (h*x)*h' <---associativity used AGAIN = (x*h)*h' (since h is in H) = x*(h*h') <---associativity used for a THIRD time.
indeed. you're right, I didn't do all the steps because the problem looked so straight and simple but you're right.

 Thank you so much


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# scaling laws for density and temperature of high-energy explosions

I'm wondering if there are heuristic ways to derive how the peak density and temperature of nuclear explosions scale with the amount of fissile/fusible material.

Does it matter what the explosion energy is made of? what if the source of the energy is matter antimatter annihilation? how would that scale? any intuitive way to derive those as well?

-
Certainly for nuclear explosives the design of the device matters--there are bomb designs that get larger (and generally cleaner) or smaller (and generally dirtier) explosions from similar amounts of fissile material. There are even rumors that the nuclear powers were investigating "dial-a-yield" devices where the explosive power could be set (over some range) in the field. –  dmckee Sep 5 '12 at 14:58
@dmckee, interesting. So maybe what is relevant for this question would be a maximum achievable bound on the yield? –  lurscher Sep 5 '12 at 16:41
The existence of dial-a-yield technology was declassified (item 8.c) in the 90s, but only some UK weapons were confirmed to have selectable yield. –  mmc Sep 6 '12 at 1:10
@mmc What a fascinating document. All the instances of "the mere fact" are interesting as are the degree to which many of these facts were widely known long before that time. On the other had I recall watching a episode of Mutual of Omaha'a Wild Kingdom on finding whales at sea one evening after which my father (a retired naval officer) said something like "I think they just covered an entire classified book called 'How to find a submarine'". –  dmckee Sep 6 '12 at 21:29


text:
# zbMATH — the first resource for mathematics

Blowing-up construction of maximal smoothings of real plane curve singularities. (English) Zbl 0861.14022
Broglia, Fabrizio (ed.) et al., Real analytic and algebraic geometry. Proceedings of the international conference, Trento, Italy, September 21-25, 1992. Berlin: Walter de Gruyter. 169-188 (1995).
Let $$f(x,y)=0$$ be the germ at $$O\in \mathbb{R}^2$$ of a (reduced) real analytic singularity. We identify it with its analytic representative in a suitable neighborhood of $$O$$ and we look at local real deformations $$f_\varepsilon$$ of it into a smooth curve (in other words, at local level sets of real morsifications of the singularity); we call such curves smoothings of $$f$$. The first natural question, which goes back to V. I. Arnol’d [in: Singularities, Summer Inst., Arcata 1981, Proc. Symp. Pure Math. 40, Part I, 57-69 (1983; Zbl 0519.58019)] is: What topological types of smoothings are possible for a given singularity? We address the problem of construction of smoothings of real plane curve singularities with the possible maximum of connected components. The technique of construction is by using blowing-ups, and lower bounds on the maximal number of connected components are given in the general case (i.e., in the case of a non irreducible curve). The case of a curve with branches having distinct tangents is especially emphasized.
For the entire collection see [Zbl 0812.00016].
##### MSC:
 14H20 Singularities of curves, local rings 14P05 Real algebraic sets 14N10 Enumerative problems (combinatorial problems) in algebraic geometry
##### Keywords:
real analytic singularity; real deformations; smoothings


text:
Exercise26

Use the distributive property to write an equivalent expression to $${-5\!\left(t-4\right)}$$ that has no grouping symbols.

in-context


text:
Accelerating the pace of engineering and science

Overmodulation

Add third harmonic or triplen harmonic zero-sequence signal to three-phase signal

Library

Control and Measurements/Pulse & Signal Generators

Description

The Overmodulation block increases the linear region of a three-phase PWM generator by adding a third harmonic or triplen harmonic zero-sequence signal V0 to the three-phase original reference signal Uref. This zero-sequence signal does not appear in the line-to-line voltages.

A modulation index of up to 1.1547 (exact value = 2/sqrt(3)) can be used without pulse dropping.

The Overmodulation block implements three overmodulation techniques:

• The Third Harmonic overmodulation technique. In this technique the third-harmonic signal V0 subtracted from the original signal is calculated as

${V}_{0}=\frac{|U|}{6}×\mathrm{sin}\left[3\cdot \left(wt+\angle U\right)\right]$

• The Flat Top overmodulation technique. In this technique the portion of the three-phase input signal exceeding values +/−1 is computed. The three resulting signals are then summed and removed from the original signal Uref. The resulting modified signal Uref* is therefore a flat-top three-phase signal that contains zero-sequence triplen-harmonics. The block outputs a value between −1 and 1.

• The Min-Max overmodulation technique. In this technique the minimum and maximum values of the three components of input signal Uref are summed and divided by two, and then subtracted from the input signal. The resulting modified signal Uref* also contains zero-sequence triplen-harmonics. The block outputs a value between −1 and 1.

Dialog Box and Parameters

Overmodulation type

Select the overmodulation technique you want to apply to the Uref signal: Third harmonic, Flat top, or Min-Max.

Inputs and Outputs

Uref

The three-phase reference signal of three-phase PWM generator.

Uref*

The overmodulated three-phase signal of three-phase PWM generator.

Characteristics

 Sample Time Inherited Scalar Expansion No Dimensionalized Yes

Example

The power_OverModulationpower_OverModulation example compares the three overmodulation techniques implemented in the Overmodulation block. Choose the overmodulation technique (type 1, 2, or 3 on the first input of the Multiport Switch) and run the simulation. Observe the resulting waveforms in Scope 1.

The model sample time is parameterized with variable Ts (default value of 5e-6). To run a continuous simulation, specify Ts = 0 in the MATLAB Command Window and change the Simulation type parameter of the Powergui block to Continuous before starting the simulation.


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# Creating a cloud of blocks that form I ♥ U in different viewpoints

This picture shown bellow has three shapes with different view angle (I ♥ U). How to do this type of model? A few tries:

With[{p=Position[ImageData[Binarize[Rasterize[#, ImageSize -> 10]]], 0]},
Graphics3D[Cuboid /@ PadRight[p, {Length@p, 3}, 0]]
] & /@ {"I", "♥", "U"}

tl;dr

Playing with your code I found that using Rasterize is a mayor pain in the behind because of various reasons (RasterSize is, at least for me, not working as advertised in the documentation; I found no easy way to remove padding around letters and get a nice $n \times n$ representation of each letter, get each letter the appropriate size, issues with different fonts etc.)

So instead I googled for a 8-bit letter generator and created letters of my own. A bit of copy-paste and string-manipulation in Mathematica and I got the following letter representation:

i = {{1,1,1,1,1,1,1,1},
{0,0,0,1,1,0,0,0},
{0,0,0,1,1,0,0,0},
{0,0,0,1,1,0,0,0},
{0,0,0,1,1,0,0,0},
{0,0,0,1,1,0,0,0},
{0,0,0,1,1,0,0,0},
{1,1,1,1,1,1,1,1}}

u = {{1,1,1,0,0,1,1,1},
{0,1,0,0,0,1,1,0},
{0,1,0,0,0,1,1,0},
{0,1,0,0,0,1,1,0},
{0,1,0,0,0,1,1,0},
{0,1,0,0,0,1,1,0},
{0,1,1,0,0,1,1,0},
{0,0,1,1,1,1,0,0}}

and (with some artistic freedom):

h = {{0,0,1,0,0,0,1,0},
{0,1,1,1,0,1,1,0},
{1,1,1,1,1,1,1,1},
{0,1,1,1,1,1,1,1},
{0,0,1,1,1,1,1,1},
{0,0,0,1,1,1,1,0},
{0,0,0,0,1,1,1,0},
{0,0,0,0,0,1,0,0}}

Helper functions:

1. The following function transforms a matrix of zeros and ones into positions where the matrix contains a one

toIndex[matrix_]:= matrix
//MapIndexed[If[#1==1,#2, Nothing[] ]&, #, {2}]&
//Flatten[#,1]&

2. This function transforms a list of form {{x1, y1},...{x8, y8}} into {{x1, y1, z1}, {x1, y1, z2}} ... {x8, y8, z8}

to3D[pos_]:= Table[pos /. {x_,y_}-> {x,y,z}, {z, 1, 8}] //Flatten[#,1]&

3. A wrapper for Cuboid to accept the center of the unit cube instead of the lower left corner as parameter

toCube[center_]:= Cuboid[center-{0.5, 0.5, 0.5}]

Now lets create some graphics and save the position definitions in some symbols box1, box2 and box3 :

(box1=(i//toIndex//to3D) /.{x_,y_,z_}-> {y, z, x} ) //Map[toCube] //Graphics3D

(box2=(u//toIndex //to3D) /.{x_,y_,z_}-> {z, y, -x+9} ) //Map[toCube] //Graphics3D

(box3=(heart //toIndex//to3D) /.{x_,y_,z_}-> {x, y, z} ) //Map[toCube] //Graphics3D

Note the coordinate transformation via replacement rules to get the right orientation

Finally we can assemble the whole thing via intersection:

Intersection[box1, box2, box3] //Map[toCube] //Graphics3D

Edit:

As one can see my solution does not result in the kind of "random box-cloud" that the OP showed in his question. This is due to the intersection method that was used to "sculpt" the geometry. Here is an improved version that amends this:

The following function neighboursQ tests wether a block has neighbours. A neighbour is a block that shares one axis with the block in question. Only when it has neighbouring blocks to all sides can it be deleted without compromising the projection.

neighboursQ[list_][{x_,y_,z_}]:= {
Cases[list, {_,y,z}],
Cases[list, {x,_,z}],
Cases[list, {x,y,_}]
} //Map[Length]
//Subtract[#, 1]&
//MatchQ[{___,0,___}]
//Not

Let's test on two simple examples:

testbox = Table[{i, j, k}, {i, 1, 3}, {j, 1, 3}, {k, 1, 3}]  //Flatten[#,2]&
testbox2= DeleteCases[testcube, {2,1,1}|{3,1,1}|{1,1,2}]

testbox is a full $9 \times 9$ box so that we expect each block to have neighbours. We can confirm that via testbox //Map[neighboursQ[testbox]] which yields a whole bunch of True. Now on to testbox2:

 neighboursQ[testcube2][{1,1,1}] (*False*)
neighboursQ[testcube2][{1,1,3}] (*True*)

neighboursQ works as we saw in the examples so we just have to repeatedly (and randomly) apply it our blocks. I wrote the following function to do just that. (It gets applied 500 times via Nest so that it has a high chance to have found any remaining block with neighbours, your milage may vary depending on number of blocks to start with)

deleteDuplicates[list_]:= Nest[With[{this=RandomChoice@#},
If[neighboursQ[#][this], DeleteCases[#,this], #]]&, list, 500]

Some other random examples:

Grid[{{x,x},{x,x}}]
/. x:>(Intersection[box1, box2, box3]  //deleteDuplicates //Map[toCube] //Graphics3D)


text:
# 1.1.21 simplify rational expression

• MHB
Gold Member
MHB
simplify
$\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2=$
OK this could get confusing quickly
but I don't think we want to square it first

Last edited:

## Answers and Replies

HOI
Yes, it can be simplified by cancelling "a" before squaring.
(The "2" in the denominator is already simple.)
$\frac{\left(ab^3-2a^{-4}b^3\right)^2}{4}$

Of course we can also factor out that $b^3$ that is in each term. Squaring it gives $b^6$.

$\frac{b^6(a- 2a^{-4})^2}{4}$

Now it is relatively easy to square $a- 2a^{-4}$.
$(a- 2a^{-4})^2= a^2- 2(a)(2a^{-4})+ (2a^{-4})^2$
$= a^2- 4a^{-3}+ 4a^{-8}$

$\frac{b^6(a^2-4a^{-3}+4a^{-8}}{4}$.

Gold Member
MHB
ok we still have negative exponents in the numerator??

HOI
Is that a problem? Different people have different ideas of what "simple" is. If you don't want negative exponents, remember what a negative exponent means:
$\frac{b^2- 4a^{-3}+ 4a^{-8}}{4}= \frac{b^2- 4\frac{1}{a^{3}}+ 4\frac{1}{a^{8}}}{4}$

Now, to eliminate the "fractions in fractions", multiply both numerator and denominator by $a^8$
$\frac{a^8b^3- 4a^5+ 4}{4a^8}$

Gold Member
MHB
well the few examples I saw they brought the terms with negative exponents in the numerator to the denumerator

HOI
That is exactly what I did!

Gold Member
MHB
That is exactly what I did!

yes but you created fractions over over a fraction $\dfrac{a^{-3}}{4}=\dfrac{1}{4a^3}$

anyway,, just being cranky

HOI
No, $$a^{-3}$$ is not a fraction!

Gold Member
MHB
why not the exponent is positive

Gold Member
MHB
No, $$a^{-3}$$ is not a fraction!

It is when it's written as \displaystyle \begin{align*} \frac{1}{a^3} \end{align*}...

HOI
Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.

Gold Member
MHB
numerals?

Last edited:
Gold Member
MHB
Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.

Do you realize that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?

Gold Member
MHB
they just said move the terms with negative exponents in the numerator to the denominator
the bell rang and I left
no fancy stuff

HOI
Do you realize that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?
Yes, I am. They are two different numerals that repreent the same number. The term "fraction" has to do with the way a number is represented, it is not a property of the number itself.


text:
# How to move around a player in a curved tunnel?

In a scenario where the player moves freely along the walls of a tunnel (like this), what is the best way to move the player along the walls of the tunnel?

I'm using Unity3d game engine. I've tried to remove all gravity and apply relative downward force on the object on every tick. Pressing right or left key applies relative force on the x-axis. But the player object will not rotate along with the walls of the tunnel and hence wont climb. I can calculate tangent to the center of the circle inside the tunnel and rotate the player myself but that wont work with curved tunnels.

You could do a Physics.Raycastin the down direction of the player. The normal of the RaycastHit this gives you, can then be used to determine the new orientation of the player. Then to "stick" the player to the surface, the new position can be calculated from the hit point, the surface normal and the desired distance from the surface.

For smoother results, do this twice, once a bit left of the player and once a bit to the right, then average.

This should work even when the tunnel cross section is not circular, should you choose so at some point.

The cross section of the part of the tunnel the player is in is always a circle. Simply use the center point of the tunnel as a position and the direction the tunnel is going as a rotation axis for rotating your player. Essentially, if the camera is exactly following the centerline of the tunnel, you'll rotate the player around the Z axis of the camera.

You may be able to achieve this by setting the origin of the player above the player by the radius of the tunnel. Then simply move the player along the center of the tunnel, and rotate about the player Z axis. Or make a parent object called "tunnelfollower" and put the player as a child object, offset with the tunnel radius.

That'll take care of curves in the tunnel and you won't be dealing with gravity. This rotation doesn't change the rotation of the player relative to the center of the tunnel. This means relative to the tunnel walls, the player up is aligned to the normal.

• I don't understand how I can calculate the center point of the cross-sectional circle the player is in at any moment, and the z-rotation of that circle. The tunnels would be curved. What I have at any moment is the player position (origin) and the origin of the tunnel. I'm new to game dev so I might be missing something obvious. – Arsalan Ahmad Aug 13 '13 at 19:21

I think the best solution is to turn off physics. Make it work in not curved tunnels, then use a vertex shader to create the curves. If you use the same transformation both in tunnels and the player, you achieve the effect you desire.


text:
# Clifford Algebra is driving me crazy

How do we determine the Clifford Algebra of a quadratic form $q$? Thanks!

-
The downvotes are probably because it looks like you are pretty lost, and haven't added much of your own thoughts. I think I know where you are coming from, though. Without altering the original question, it would be good if you typed in specific things that you are unsure about, or would like to know more about. –  rschwieb Jan 15 '13 at 18:30
It is also probably too broad. Specific requests about the algebra might help to turn that around. People considering voting to close might also want to take into account the author has not had much time to enact changes. (Right now, at least.) –  rschwieb Jan 15 '13 at 18:39

I don't want to reprove the existence of Clifford algebras in general here. You will have to go look that up, for a rigorous explanation.

I can give a heuristic explanation, though, on how to think of it.

The quadratic form $Q:V\rightarrow F$ gives rise to a bilinear form $B:V\times V\rightarrow F$ that works like an inner product on $V$. From $Q$ and $V$, one can prove the existence of the algebra $C\ell(V,Q)$.

To get a grip on what it looks like, begin with any orthonormal basis $b_1\dots b_n$ for $V$. (It doesn't have to be finite dimensional, and it doesn't have to be orthonormal, but I'll assume so for convenience.) By the way the algebra was built, you are guaranteed that the multiplication works this way on the basis: $b_i\otimes b_i=Q(b_i)$ and $b_i\otimes b_j=-b_j\otimes b_i$ for $i\neq j$.

As you probably know, $C\ell(V,Q)$ has dimension $2^n$, but we only have $n$ things in the set of $b_i$, so it is clearly not a basis. To get a complete basis you form every product of the $b_i$ possible with ascending indices. By doing this, we ensure that the set of $2^n$ elements formed is a linearly independent set. We don't want to include $b_2\otimes b_1\otimes b_3$ for example, because that is the same as $-b_1\otimes b_2\otimes b_3$, and that is linearly dependent with $b_1\otimes b_2\otimes b_3$. Similarly you don't want to include $b_2\otimes b_3\otimes b_3$, because that is the same as $b_2\cdot Q(b_3)$ which is linearly dependent with $b_2$.

Using the anticommutativity and squaring rules, along with the associative and distributive property of multiplication, you can multiply any two elements of the Clifford algebra.

-
There could be something special said about fields of characteristic two, but I had really written thinking of $char(F)\neq 2$. –  rschwieb Jan 15 '13 at 18:38


text:
Coulomb's law vs Electric Field

Homework Statement

I just wanted to make sure I understood the difference between the two.
Coulombs law is the force between two charges. Two charges that are physical(maybe not the right word) located on some axes. We can use the equation
F=k(q1 * q2 * r(hat))/r^2

r^2=|r| = |x-x'|
where r hat equal= r/|r|= (x-x')/|x-x')|

if we only have one charge and we want to know what the force would be at some random point q. We know that F=qE(x). So, F/q=E(x). which would give us
E(x)=k(q1* r(hat))/r^2

They both calculate the force, coulombs is the force between two charges and the electric field is the force a some point q. (q would be a field point.)
As the distance grows on both the force gets weaker and weaker.

The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
I know that you can sum the forces with Coulombs Law but can you sum up the electric fields?

[C]oulombs is the force between two charges and the electric field is the force at some point q
A coulomb is a unit of measure (quantity of electricity).
Force is a physical quantity.
An electric field models the network of electrostatic forces charged bodies would exert on each other.

whats the difference between an electric field an electric force? They seem synonymous

gneill
Mentor
whats the difference between an electric field an electric force? They seem synonymous
The electric field is what is in the surrounding environment that produces a force on a charge. One way to express the units associated with an electric field is N/C (Newtons per Coulomb). The electric field and force have different units.

Compare with the gravitational field: The acceleration due to gravity (meters per second2) can also be expressed in units of N/kg (Newtons per kilogram).

Chandra Prayaga
if we only have one charge and we want to know what the force would be at some random point q. We know that F=qE(x). So, F/q=E(x). which would give us
E(x)=k(q1* r(hat))/r^2
If there is only one charge, call it q1, there is no force at any random point q. A force acts on a material object. There is no force "at" an empty point in space. It is only when you place another charge q at some point in space, that the charge q has a force acting on it, because of your original charge q1.
If there is only one charge, it generates an electric field, not force, at every point in space, given by your formula:
E(x)=k(q1* r(hat))/r^2
This is different from a force. If now you place a charge q2 at the point x, then there is a force on q2 given by
F = q2 E(x)

Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

I just wanted to make sure I understood the difference between the two.
Coulombs law is the force between two charges. Two charges that are physical(maybe not the right word) located on some axes. We can use the equation
F=k(q1 * q2 * r(hat))/r^2

r^2=|r| = |x-x'|
where r hat equal= r/|r|= (x-x')/|x-x')|

if we only have one charge and we want to know what the force would be at some random point q. We know that F=qE(x). So, F/q=E(x). which would give us
E(x)=k(q1* r(hat))/r^2

They both calculate the force, coulombs is the force between two charges and the electric field is the force a some point q. (q would be a field point.)
As the distance grows on both the force gets weaker and weaker.

The Attempt at a Solution

The electric (and magnetic) fields have existence that is (essentially) separated from the charges that created them. For example, take
the force expression
$$\mathbf{F} = \frac{k q_1 q_2 \: \hat{r}}{r^2}.$$
If the particles are quite far apart, then when you suddenly wiggle $q_2$, the expression above for $\mathbf{F}$ changes (because $\hat{r}$ and $r$ change), but for a while the actual force on particle #1 does not change. Rather, the influence of particle #2 is propagated at the speed of light, so there would be a bit of a delay before the wiggling effects are felt by particle #1. The effects arise through the influence of the electromagnetic field. The field is very real, the electrostatic force is very real, and certainly the two are closely linked; but they are not really the same.

Got it. Thanks. Another question super basic, but entertain me.

If E=M. Which is agreed upon. How is it possible that E=Mc^2?

Ray Vickson
Homework Helper
Dearly Missed
Got it. Thanks. Another question super basic, but entertain me.

If E=M. Which is agreed upon. How is it possible that E=Mc^2?
Nobody agrees that E=M. Why do you think such a thing could be true?

Chandra Prayaga
Got it. Thanks. Another question super basic, but entertain me.

If E=M. Which is agreed upon. How is it possible that E=Mc^2?
What is E and what is M?

Nobody agrees that E=M. Why do you think such a thing could be true?
Thats the basis of quantum mechanics.

Chandra Prayaga


text:
5 Dec 2013 16:29

### FYI Emacs-24.4 survey

http://www.emacswiki.org/emacs/FrequentlyEnabledPackages_Emacs244_Survey

5 Dec 2013 14:31

### Export all radio tables off a document

Hi all,

this is my first post to this list, so please be patient with me

I have many radio tables in my document with this structure:

\begin{comment}
#+TBLNAME: sec-10
#+ORGTBL: SEND sec-10 orgtbl-to-latex :skip 3 :splice t
|--------------------+---------------+---------------|
| Anschaffungskosten | Nutzungsdauer | AfA in €/Std. |
|--------------------+---------------+---------------|
|             720000 |         36000 |            20 |
|--------------------+---------------+---------------|
\end{comment}

A C-c C-c exports the table to Latex like this:

\begin{tabular}{lll}

% BEGIN RECEIVE ORGTBL sec-10
720000 & 36000 & 20\\
% END RECEIVE ORGTBL sec-10

\end{tabular}

It would be very nice to automatically export *all* tables of the
document to their corresponding latex tables in the same document. I
found a function that does this for calculations in org-tables:

5 Dec 2013 14:17

### org-babel-next-src-block respecting #+call lines

Hi all,

how could I make org-babel-next-src-block (bound to 'C-c C-v n' by
default) jump to the next src block or #+call line?

In its current form it skips over call lines.

Many thanks in advance,
Andreas

5 Dec 2013 11:33

### Can't write accents in R graphics

#+TITLE:     ECM Process Coding System
#+PROPERTY:  eval yes
#+PROPERTY:  exports both
#+PROPERTY:  cache no

* Problem

I can't write accents in graphic titles (nor legends).

* Settings

After Googling (a lot), I always come back to the same set of solutions:

- Adding the following in my ~.emacs~ file:

#+begin_src emacs-lisp
(setq process-coding-system-alist '(("R.*" . utf-8-unix)))
#+end_src

#+begin_src emacs-lisp
(lambda () (set-buffer-process-coding-system
'utf-8-unix 'utf-8-unix)))
#+end_src

- Setting the process coding system in the =*R*= buffer:

#+begin_src emacs-lisp
M-x set-buffer-process-coding-system RET utf-8-unix RET utf-8-unix RET
#+end_src

5 Dec 2013 10:58

### #+PROPERTY: for babel header args worked... then it didn't

Wee bit of frustration with Babel just now.

Some weeks ago, I wrote a Beamer presentation with LilyPond source code
blocks. I needed to add a little code to each example to suppress the
"tagline" from each results file. In that presentation, setting the
:prologue header argument in a #+PROPERTY line worked perfectly:

Now I'm working on a short article-class document, and I included the same
#+PROPERTY at the top, and... no effect. But... *the lines are identical*.
??

":prologue \header{tagline=##f}" does take effect if I include it in every
#+begin_src line. That's an acceptable workaround for this file, since
there won't be many examples.

But it bothers me when I figure something out, and then I try to do it
again, and it doesn't work. It's like 1 + 1 = 2 in one place, and 1 + 1 = 0
in another.

Why would "#+PROPERTY" work in one file and be ignored in the other?

hjh

Header of the working file:
~~
#+BIND: org-latex-listings-options (("basicstyle" "\\ttfamily\\scriptsize")
("captionpos" "b") ("tabsize" "3"))

5 Dec 2013 09:56

### how to prevent html export from including the default javascript?

Hello

I have trouble finding the correct explanation in ox-html.el.gz and
searching on the web. I'm trying to tell the html exporter *not* to
include the "<![CDATA[/*>