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## densely defined operators of Fock space

Could someone prove the following (if it is precisely correct):

Since Fock space is the closure of the finite linear span of finite excitations of the vacuum state $\Omega$, then the operator $\hat{O}$ is densely defined if and only if $\hat{O} \Omega$ has finite norm.

Or more generally, an operator is densely defined on a separable space if it is defined on a given single vector in that space.

thanks
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Recognitions:
 Quote by julian Could someone prove the following (if it is precisely correct):
I'm not an expert on Functional Analysis, but since no one else has
answered yet I'll have a go...

(BTW, $....$ doesn't work here for latex strings. I've fixed them
below. Click on any one of them to see the latex source, and a link
to some latex info relevant to PhysicsForums.)

 Since Fock space is the closure of the finite linear span of finite excitations of the vacuum state $\Omega$, then the operator $\hat{O}$ is densely defined if and only if $\hat{O} \Omega$ has finite norm.
I think you can use the axiom that Fock space is generated by
cyclic operations of the annihilation and creation operators
upon a distinguished vacuum vector (and a subsequent
closure). Under this construction the a/c operators form an
irreducible set, meaning that *any* operator on the Fock
space can be expressed as a sum of polynomials of these
operators. (Since you only want "densely defined",
polynomials are enough -- you don't need to worry about
infinite series and associated existence issues.)

Then you take an arbitrary product of a/c operators.
Eg, for a quadratic in the creation operators it might
be something like:
$$O_2 ~:=~ \int dp dq ~ f(p,q) a^*(p) a^*(q)$$
More generally, there might also be annihilation operators
but you can use the CCRs to normal-order them, and if
the result is any naked delta fns (which cannot be
integrated away) then such an operator doesn't give a
finite-norm vector when acting on the vacuum.

I think that's enough to do the proof in one direction
(though you'll have to clean it up and fill in more details).
Once you've seen it in one direction you can probably figure
out the reverse direction.

And if I've got anything wrong in the above, maybe someone
else will enjoy correcting it. :-)

 Or more generally, an operator is densely defined on a separable space if it is defined on a given single vector in that space.
I don't think that's true in general without further
assumptions (such as the cyclicity of the vacuum and
maybe some other stuff). "Separability" in this context
only means the space has a countable orthonormal basis,
so you've got to show that the operator acting on any
basis vector results in a finite-norm vector. (Actually,
that's kinda what I did above in a clumsy way for
the Fock, since one gets a countable basis there by
acting on the vacuum with products of creation
operators.)

HTH.
 Thanks for the response sorry i've been a bit busy to reply. Yes, a lot of operators have the same action on a given vector (in a separable space) but some will give non-normalizable vectors when acting on other vectors - unless addition assumptions are made. I didn't read the paper carefully anough, te separable space in question is isomorphic to Fock space Thanks julian

 Similar discussions for: densely defined operators of Fock space Thread Forum Replies Atomic, Solid State, Comp. Physics 20 Atomic, Solid State, Comp. Physics 6 Quantum Physics 9 Quantum Physics 13 Atomic, Solid State, Comp. Physics 3


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# symmetric systems of polynomial equations

Suppose I have a polynomial $p(x_1,...,x_N)$ in $N$ complex variables, and I wish to solve $p(x_{\pi(1)},...,x_{\pi(N)})=0$ for all permutations $\pi \in S_N$. Clearly this is overdetermined for generic $p$, but suppose $p$ is symmetric under exchange of all but one variable. Then this gives $N$ distinct equations, and so generically one expects a discrete set of solutions. Are there any general techniques for solving such a system of equations? I’m most interested in simply counting the number of solutions (up to permutations).

format: plain


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Cell Representation of a galvanic cell:

I know basic cell representations of galvanic cells and can identify the anode/cathode. But here's a weird representation that I encountered and couldn't understand it:

$\ce{Pt|H{2}(g)}(1.0 atm)\ce{|HX||KCl(1M)|Hg_{2}Cl_{2}(s)|Hg|Pt}$

What does this actually translate into?

• Could you elaborate which type you could understand – DSinghvi Sep 1 '14 at 16:57


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123 views

$P, Q$, and $R$ are the decimal integers corresponding to the $4$-bit binary number $1100$ considered in signed magnitude, $1$’s complement, and $2$’s complement representations, respectively. The $6$-bit $2$’s complement representation of $(P+Q+R)$ is

1. $110101$
2. $110010$
3. $111101$
4. $111001$


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# DIODE LASER VELOCITY MODULATION SPECTROSCOPY OF CARBANIONS: THE C=C STRETCHING: MODE OF ACETYLIDE ($C_{2}H^{-}$)

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/17291

Files Size Format View
1987-RE-08.jpg 74.87Kb JPEG image

 Title: DIODE LASER VELOCITY MODULATION SPECTROSCOPY OF CARBANIONS: THE C=C STRETCHING: MODE OF ACETYLIDE ($C_{2}H^{-}$) Creators: Gruebele, M.; Polak, M.; Saykally, R. J. Issue Date: 1987 Publisher: Ohio State University Abstract: Ten lines in the $C=C$ stretching vibration of $C_{2}H^{-}$ have been assigned and fit with a linear molecule Hamiltonian. The molecular constants are remarkably close to those of neutral $C_{2}H$, except for the vibrational band origin, which lies $87 cm^{-1}$ below the neutral radical band origin. This is in keeping with the expected lower electron density between the carbons in $C_{2}H^{-}$. Description: Author Institution: Department of Chemistry, University of California URI: http://hdl.handle.net/1811/17291 Other Identifiers: 1987-RE-8


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# How to check dependency on nominal variables?

I have data that map one real (float) and several nominal arguments to one real value:

y = f(r, n1, n2, n3, n4)

If I check dependency of y on r (r is my real-valued variable), I see that y depends linearly on r with a lot of noise on top of this dependency.

Now I want to check if there is any dependency of y on nominal variables: n1, n2, n3, n4. I also would like to know what nominal variable has the largest influence on y. So, my question is: What methods can I used to do that?

-
Can you show us the data (using graphs)? – Roman Luštrik Feb 27 '13 at 8:52
@Roman Luštrik, I am not sure what graph you need. How should I build it? The only graph that I have is y as function of x (it ignores values of other variables). It looks like y linearly depends on x with a lot of noise. The second problem, I think I am not allowed to post the data on line, since it is property of the company that I am working on. – Roman Feb 27 '13 at 8:59

## 1 Answer

Regression with dummy variables for the levels of the nominal variables (factors) would probably be an obvious first thought, but as @RomanLuštrik points out the first thing to do is plot the data.

An example plot with one linear variable and one factor:

(That data is from this page)

There are various other plots that can be done.

Here's another example:

(Same data, somewhat different kind of plot)

Here is a document that discusses some other plots for this kind of data (like boxplots).

-
thank you for the answer. Unfortunately, I do not understand the data. There are four columns which of them is function and which are arguments. If I assume that the last (or fist) columns is the function than I have one real-valued argument and two (not one as you say) nominal-valued variables (or factors). I also do not understand the first plot. It looks like there are four plots. Why four? And what is x and what is y. What is "priorlmp" and what is "Improv"? – Roman Feb 27 '13 at 10:23
Improv is the $y$, PriorImp is the $x$. RDExp is a factor (nominal). The other one is actually a replicate number but I have treated it kind of like it was a factor in the second plot. As for what the data measure, I have no clue. The fourth plot (lower right) combines the two plots on the left. Don't worry about it - just try to comprehend the general idea of doing a separate y vs x for each level of a factor (that's called a coplot). The second plot is called an interaction plot. If you gave some sample data, someone might be able to generate one for you. – Glen_b Feb 27 '13 at 10:36


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# Changing Percents to Decimals

Need some basic practice with decimals and percents first? Changing Decimals to Fractions and Multiplying and Dividing Decimals by Powers of Ten and Changing Decimals to Percents

Whenever computations need to be done with percents, the percents are first renamed as decimals.

Here are examples of changing percents to decimals. Notice that the $\,\%\,$ symbol is replaced with a factor of $\,\frac{1}{100}\,$:

## Examples

$3.5\% = 3.5\cdot\frac{1}{100} = \frac{3.5}{100} = 0.035$

$25\% = 25\cdot \frac{1}{100} = \frac{25}{100} = 0.25$

$50\% = 50\cdot \frac{1}{100} = \frac{50}{100} = 0.5$

$100\% = 100\cdot \frac{1}{100} = \frac{100}{100} = 1$

$250\% = 250\cdot \frac{1}{100} = \frac{250}{100} = 2.5$

As these examples illustrate, the percent symbol instructs multiplication by $\,\frac{1}{100}\,$ (or, equivalently, division by $\,100\,$), which is accomplished by moving the decimal point two places to the left. Remember that if you don't see a decimal point, then it gets inserted just to the right of the ones place.

Now, you can go from percents to decimals in one easy step,
by moving the decimal point two places to the left:

## Examples

$3\% = 0.03$

$2.37\% = 0.0237$

$0.01\% = 0.0001$

$5032\% = 50.32$

Some of my students find this memory device helpful: PuDdLe   DiPpeR   (Imagine a little kid, barefoot, dipping piggy-toes in puddles!)

Percent to Decimal, two places to the Left

Decimal to Percent, two places to the Right

## Practice

Here, you will practice renaming percents as decimals. For example, $\,54\%\,$ gets renamed as $\,0.54\,$.


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home

# Logarithmic functions

Logarithmic functions have the form $$f(x)=\log_b x$$, where $$b$$ is the base.

They are the inverse of Expontential functions such that $$\log_b x = y \iff x = b^y$$.

## Identities

$\log_b m \times n = \log_b m + \log_b n$ $\log_b \frac m n = \log_b m - \log_b n$ $\log_b m^n = n \log_b m$ $\log_b 1 = 0$ $\log_b b = 1$


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## Calculus 8th Edition

$r= \frac{6}{1 + sin (θ)}$
$r= \frac{ed}{1 + e sin (θ)}$ (sine since directrix is vertical line) e = 1 since it is a parabola, vertedx is at (3, $\pi /2$) $r= \frac{d}{1 + sin (θ)}$ Substitute (3, $\pi /2$) into the equation $3= \frac{d}{1 + sin (\pi /2) }$ $3= \frac{d}{2}$ Thus d=6 $r= \frac{6}{1 + sin (θ)}$


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# Compress a sparse matrix

Compress a sparse matrix using Compressed sparse row (CSR, CRS or Yale format).

These are all the same form of compression (ignore new Yale).

Input may be any 2d data structure (list of lists, etc): e.g

[[0 0 0 0],
[5 8 0 0],
[0 0 3 0],
[0 6 0 0]]

And the output should be three 1d data structures (list etc), that denote the outputs A, IA and JA, for example

[5, 8, 3, 6]
[0, 0, 2, 3, 4]
[0, 1, 2, 1,]

The process is described by wikipedia:

• The array A is of length NNZ and holds all the nonzero entries of M in left-to-right top-to-bottom ("row-major") order.

• The array IA is of length m + 1. It is defined by this recursive definition:

• IA[0] = 0

IA[i] = IA[i − 1] + (number of nonzero elements on the (i − 1)-th row in the original matrix)

• Thus, the first m elements of IA store the index into A of the first nonzero element in each row of M, and the last element IA[m] stores NNZ, the number of elements in A, which can be also thought of as the index in A of first element of a phantom row just beyond the end of the matrix M. The values of the i-th row of the original matrix is read from the elements A[IA[i]] to A[IA[i + 1] − 1] (inclusive on both ends), i.e. from the start of one row to the last index just before the start of the next.[5]

• The third array, JA, contains the column index in M of each element of A and hence is of length NNZ as well.

If your language doesn't support actual data structures, input and output may be text.

# Test cases

Input 1:

[[0 0 0 0],
[5 8 0 0],
[0 0 3 0],
[0 6 0 0]]

Output 1:

[ 5, 8, 3, 6 ]
[ 0, 0, 2, 3, 4 ]
[ 0, 1, 2, 1, ]

Input 2

[[10 20 0 0 0 0],
[0 30 0 40 0 0],
[0 0 50 60 70 0],
[0 0 0 0 0 80]]

Output 2:

[ 10 20 30 40 50 60 70 80 ]
[  0  2  4  7  8 ]
[  0  1  1  3  2  3  4  5 ]

Input 3:

[[0 0 0],
[0 0 0],
[0 0 0]]

Output 3:

[ ]
[ 0 0 0 0 ]
[ ]

Input 4:

[[1 1 1],
[1 1 1],
[1 1 1]]

Output 4:

[ 1 1 1 1 1 1 1 1 1 ]
[ 0 3 6 9 ]
[ 0 1 2 0 1 2 0 1 2 ]

Input 5:

[[0 0 0 0],
[5 -9 0 0],
[0 0 0.3 0],
[0 -400 0 0]]

Output 5:

[ 5, -9, 0.3, -400 ]
[ 0, 0, 2, 3, 4 ]
[ 0, 1, 2, 1, ]

Assume inputs may contain any real number, you need not consider mathematical symbols or exponential representation (e.g. 5,000 will never be entered as 5e3). You will not need to handle inf, -inf, NaN or any other 'pseudo-numbers'. You may output a different representation of the number (5,000 may be output as 5e3 if you so choose).

# Scoring

This is a , fewest bytes wins.

• Could 1-based indices be used for the last row?
– Leo
Jul 5 '17 at 16:30
• @Leo for JA? No. Jul 5 '17 at 16:50
• Isn't IA[0] = 0 completely unnecessary? It's only needed to define IA[i] = IA[i − 1]..., yet we could simply state that if i-1 < 0 to use 0. That is, IA[0] is always equal to 0, therefor it can be compressed out (yes, I realize that this is a critique of the algorithm, not this challenge). Jul 5 '17 at 21:01
• Will we have the inverse challenge too?
Jul 5 '17 at 21:42
• Neat! Hadn't run into either format before, but I'm glad to see someone else did see that before (I shouldn't be the kind of person who spots trivial optimizations in algorithms this old). Jul 6 '17 at 13:03

# MATL, 19 bytes

!3#f!Dx0Gg!XsYshDq!

Input uses ; as row separator.

Try it online! Or verify all test cases: 1, 2, 3, 4, 5.

### Explanation

!     % Implicit input. Transpose
3#f   % 3-output version of find: it takes all nonzero values and pushes
% their column indices, row indices, and values, as column vectors
!     % Transpose into a row vector
D     % Display (and pop) vector of values
x     % Delete vector of row values
0     % Push 0
G     % Push input
g     % Convert to logical: nonzeros become 1
!     % Transpose
Xs    % Sum of columns. Gives a row vector
Ys    % Cumulative sum
h     % Prepend the 0 that's below on the stack
D     % Display (and pop) that vector
q     % Subtract 1 from the vector of row indices
!     % Transpose into a row vector. Implicitly display

# Mathematica, 78 bytes

{a=SparseArray@#;a@"NonzeroValues",a@"RowPointers",Join@@a@"ColumnIndices"-1}&

# Jelly, 24 bytes

n0S€0;;\S€,T€Ẏ$’$
Ẏḟ0W;Ç

Try it online!

# APL (Dyalog), 31 28 chars or 36 33 bytes*

Requires ⎕IO←0 for zero based indexing. I/O is list of lists.

{(∊d)(0,+\≢¨d←⍵~¨0)(∊⍸¨⍵≠0)}

Try it online!

{} anonymous function where the argument is represented by

()()() return a list of three things:

⍵≠0 Boolean where the argument differs from 0
⍸¨ɩndices of those for each sub-list
∊ϵnlist (flatten) to combine into single list

⍵~¨0 remove zeros from each sub-list of the argument
d← store that as d
≢¨ tally each
+\ cumulative sum
0, prepend a zero

∊dϵnlist (flatten) d to combine into single list

* To run in Dyalog Classic, simply replace ⍸ with ⎕U2378.

• Nice, I don't understand the input format though? f 4 4⍴ and then the values? Jul 5 '17 at 12:08
• @Pureferret the Code defines the function f. The Input is really a REPL, which calls f on the result of 4 4⍴… which reshapes the data into a 4×4 matrix.
Jul 5 '17 at 12:09
• Rho for reshapes. I get it! Jul 5 '17 at 12:12
• @Pureferret I've updated the Try it online! link to better show test cases.
Jul 5 '17 at 12:21

f s|a<-filter(/=0)<$>s=(id=<<a,scanl(+)0$length<$>a,s>>= \t->[i|(i,e)<-zip[0..]t,e/=0])  Try it online! How it works: a<-filter(/=0)<$>s           -- let a be the list of lists with all 0 removed]
-- e.g. [[1,0,0],[0,3,4]] -> [[1],[3,4]]

-- return a triple of

id=< A

scanl(+)0$length<$>a         -- partial sums of the length of the sublists of a
-- strating with an additional 0 -> IA

s>>=                         -- map the lambda over the sublists of s and concatenate
-- into a single list
\t->[i|(i,e)<-zip[0..]t,e/=0]  -- the indices of the non-zero elements -> JA

# Python+SciPy, 79 bytes

i guess built-ins were not forbidden

from scipy.sparse import*
A=csr_matrix(input())
print A.data,A.indptr,A.indices

Accepts input in the format [[0, 0, 0, 0],[5, 8, 0, 0],[0, 0, 3, 0],[0, 6, 0, 0]]

# R, 70 bytes

function(m,M=t(m))list(M[x<-!!M],diffinv(colSums(x)),which(x,T)[,1]-1)

Try it online!

This is an old, challenge that didn't get an R answer for 3 years! Having to do this in row-major rather than column-major order costs 8 bytes (,M=t(m))).

# Python 3, 96 bytes

lambda m:list(map(list,zip(*[[e,i,j] for i,r in enumerate(m) for j,e in enumerate(r) if e!=0])))

Try it online!

# PHP, 107 bytes

<?for($y=[$c=0];$r=$_GET[+$l++];)foreach($r as$k=>$v)!$v?:[$x[]=$v,$z[]=$k,$y[$l]=++$c];var_dump($x,$y,$z);  Try it online! # PHP, 109 bytes <?$y=[$c=0];foreach($_GET as$r){foreach($r as$k=>$v)if($v){$x[]=$v;$z[]=$k;$c++;}$y[]=$c;}var_dump($x,$y,$z);  Try it online! • Does this need the numbers to be strings? Jul 5 '17 at 14:06 • @Pureferret Any Input in PHP is a string or an array of strings. I have not casted the input so if you wish that the output is purely int replace $x[]=$v with $x[]=+$v Jul 5 '17 at 14:21 # JavaScript (ES6), 117 bytes a=>[a.map((b,i)=>(b=b.filter((x,c)=>x&&o.push(c)),m[i+1]=m[i]+b.length,b),m=[0],o=[]).reduce((x,y)=>x.concat(y)),m,o]  Input is a 2D array of numbers and output is an array of [A, IA, JA]. ## Explained a=>[ a.map((b,i) => ( // map each matrix row b = b.filter((x,c) => x // filter to only non-zero elements && o.push(c) // and add this index to JA ) m[i+1] = m[i] + b.length, // set next value of IA b // and return filtered row ), m=[0],o=[] // initialize IA (m) and JA (o) ).reduce((x,y) => x.concat(y)), // flatten the non-zero matrix m,o] // append IA and JA  ## Tests let f= a=>[a.map((b,i)=>(b=b.filter((x,c)=>x&&o.push(c)),m[i+1]=m[i]+b.length,b),m=[0],o=[]).reduce((x,y)=>x.concat(y)),m,o] let run=x=>O.innerHTML+=f(x).map(s=>[${s.join, }]).join\n+"\n\n"
run([[0,0,0,0],[5,8,0,0],[0,0,3,0],[0,6,0,0]])
run([[10,20,0,0,0,0],[0,30,0,40,0,0],[0,0,50,60,70,0],[0,0,0,0,0,80]])
run([[0,0,0],[0,0,0],[0,0,0]])
run([[1,1,1],[1,1,1],[1,1,1]])
run([[0,0,0,0],[5,-9,0,0],[0,0,0.3,0],[0,-400,0,0]])

# Japt, 3127 17 bytes

Hopefully this output format is OK.

cf pUmè iT å+ Ucð

Try it

cf pUmè iT å+ Ucð     :Implicit input of 2D array U
c                     :Flat map
f                    :  Filter
p                  :Push the following two elements
Um                :  First element: Map U
è               :    Count the truthy elements in each
i             :  Prepend
T            :    Zero
å+         :  Cumulatively reduce by addition
Uc      :  Second element: Flat map U
ð     :    0-based indices of truthy elements

• I just ran the other examples though and it works Jul 5 '17 at 12:08

# Python 2, 115 bytes

lambda m:zip(*[[v,i]for k in m for i,v in enumerate(k)if v])+[reduce(lambda a,b:a+[len(b)-b.count(0)+a[-1]],m,[0])]

Try it online!

Output is [A, JA, IA]

# Perl 6, 84 bytes

{.flatmap(*.grep(+*)),(0,|[\+] .map(+*.grep(+*))),.flat.kv.flatmap:{$^a%.[0]xx?$^b}}

Try it online!

The single matrix argument is in $_. • .flatmap(*.grep(+*)) selects the nonzero elements of the entire matrix. • [\+] .map(+*.grep(+*)) is the triangular reduction of the number of elements in each row (which some languages call scan). (0,|...) prepends a zero to that list. • .flat.kv produces an indexed list of all elements of the matrix. .flatmap: {$^a % .[0] xx ?\$^b } flat-maps over the modulus of each index by the number of columns in the array (.[0], the number of elements in the first row), replicated by the element itself, interpreted as a boolean. That is, nonzero elements are replicated once, and zero elements are replicated zero times (ie, removed).

# Julia, 66 63 bytes

using SparseArrays
n*a=getfield(sparse(a'),n)
!a=5a,3a.-1,4a.-1

SparseArrays is a standard library that stores with CSC instead of CSR (column instead of row or something), that's why we need to transpose (a').

-6 bytes if 1-indexing was allowed

overloads * with getfield to save a few bytes

Try it online!

# Scala, 82 bytes

x=>x.flatMap(_.zipWithIndex).filter(_._1!=0).unzip->x.scanLeft(0)(_+_.count(_!=0))

Try it in Scastie!

Returns ((A, JA), IA)

Explanation:

x =>     //The sparse matrix to be smooshed
x.flatMap(      //Do the following operation to each row, then flatten the result
_.zipWithIndex //Zip each element with its column (makes tuples)
).filter(      //Keep the tuples where
_._1!=0      //the first element isn't 0
).unzip        //Unzip into a tuple of two lists, A and JA
->             //Make that the first element of another 2-tuple with IA:
x.scanLeft(0)(  //Starting with 0
_+_.count(_!=0)  //Add the number of nonzero elements in each row
)               //Keep the intermediate results, giving us IA


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Appropriate Decimal Units for Metric Measurement

## Choose among millimeters, centimeters, meters, and kilometers for given situations.

Estimated2 minsto complete
%
Progress
Practice Appropriate Decimal Units for Metric Measurement
Progress
Estimated2 minsto complete
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Appropriate Decimal Units for Metric Measurement

### Vocabulary Language: English

Centimeter

Centimeter

A centimeter is a very small unit of metric measure. A centimeter is one hundredth of one meter.
Kilometer

Kilometer

A kilometer is a common metric unit of length that is equal to 1,000 meters.
Length

Length

Length is a measurement of how long something is. Examples of customary units of length are inches, feet, yards and miles.
Meter

Meter

A meter is a common metric unit of length that is slightly larger than a yard.
Metric System

Metric System

The metric system is a system of measurement commonly used outside of the United States. It contains units such as meters, liters, and grams, all in multiples of ten.
Millimeter

Millimeter

A millimeter is the smallest common metric unit of length.


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# When do infinite descending chains of prime ideals exist in commutative rings?

I was wondering for what kind of commutative rings we can always construct an infinite descending chain of distinct prime ideals?

-
One example is $R[X_1,X_2,\ldots]$ for a domain $R$. There we have the infinite descending chain of prime ideals $I_1 \supsetneq I_2 \supsetneq \ldots$ with $I_k = (X_k, X_{k+1},\ldots)$. –  marlu Mar 14 '12 at 16:07

## 2 Answers

I doubt that there is a crisp definitive answer to your question, but here are a few implications and, maybe more importantly, non-implications.
For the sake of concision, let us call strongly infinite dimensional a ring in which there exists an infinite strictly decreasing sequence of prime ideals $\;\mathfrak p_0 \supsetneq \mathfrak p_1 \supsetneq \mathfrak p_2\supsetneq...$

a) If a ring is strongly infinite dimensional, it is infinite dimensional.
This is obvious.

b) If a ring is noetherian it cannot be strongly infinite dimensional.
Indeed, by definition every ideal (prime or not) in a noetherian ring is generated by a finite number $r$ of elements.
And by a generalization of Krull's Hauptidealsatz, if a prime ideal $\mathfrak p$ is generated by $r$ elements, then $height(\mathfrak p)\leq r$.
Hence no prime ideal $\mathfrak p$ can be the beginning of an infinite strictly descending chain of prime ideals.

c) If a ring is infinite dimensional, it needn't be strongly infinite dimensional.
Indeed there exist examples (due to Nagata) of infinite dimensional noetherian rings, and we have just seen in b) that a noetherian ring cannot be strongly infinite dimensional.

d) If a ring is non-noetherian it needn't be strongly infinite dimensional.
For example consider a field $k$ and the non-noetherian ring $k[X_i\mid i\in \mathbb N]/\langle X_i\cdot X_j\mid i,j\in \mathbb N\rangle$.
This ring is non-noetherian but since its only prime ideal is $\mathfrak m=\langle \bar X_i\mid i\in \mathbb N\rangle$, there is no risk that it could be strongly infinite dimensional!

-

This is a correction of my previous post.

The statement "For a local ring $O$ to be infinite dimensional and to be strongly infinite dimensional are equivalent." seems to be wrong.

What remains of my previous post seems to be this:

1. Let $R$ be a strictly infinite dimensional ring, and let $S$ be an extension ring of $R$ such that for $S/R$ the Lying-Over-Theorem and the Going-Down-Theorem hold. Then $S$ is strictly infinite dimensional. In particular this holds if $R$ is an integrally closed domain, $S$ is a domain, and $S/R$ is an integral extension. For $R$ we can for example take a polynomial ring in infinitely many variables over a field.

2. Kang (Proc. AMS 126 (3), 1998) has shown that for any infinite chain of primes in a domain $R$, there exists a valuation ring $O\supseteq R$ of the field of fractions $K$ of $R$ which contains a chain of primes lying over the given one. We can conclude that if $R$ is a strictly infinite dimensional domain, then there exists a strictly infinite dimensional valuation domain $O$ of the field of fractions $K$ of $R$. General valuation theory then yields that the transcendence degree of the extension $K/k$, where $k$ is the prime field of $K$, is infinite. Consequently $R$ contains a polynomial ring in infinitely many variables over $k$ or over $\mathbb{Z}$. This last conclusion is rather weak, because we only use the fact that $O$ has infinite dimension, not the particular structure of the spectrum. There seems to be room for improvement here ...

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Dear @Hagen, what you write is very interesting, but I don't quite understand the relationship between your assertions (I must admit that I only just had a quick look at them as of now). In particular : is the result in the first line proved in the next ones? Even if this sounds childish to you, I would be very grateful if you could take a few minutes to write your answer in a somewhat more structured format: Statement1: "blah" Reason : " blah" (or: "see following literature"), Statement 2: "blah" or "follows from statement 1", etc. But, let me say it again: your answer is very informative:+1 –  Georges Elencwajg Mar 14 '12 at 23:47
Dear Georges. Nothing childish about your critics (in fact your feedback in the past has always been valuable) - even more because I think I was too quick and things are not as simple as I thought they are. So I will probably correct or at least improve my post in the next few hours. Right now I am busy with something else. –  Hagen Mar 15 '12 at 8:39


text:
# Giants Were Sleeping in This Place For 12,000 Years- Iran

Glass capsules were used to keep such giants in a “deathlike” condition which is similar to death, although they were not dead. At the time of its disappearance, this civilization had glass technology which is known to have altered the flow of time. The ancient technology had far more capacities than we believe. 30,000 years seemed like 30 minutes for those who were placed in the glass capsules. Not only humans but also extraterrestrials coveted this multidimensional technology.

The place when this glass capsules were located was covered with dark material up to 90 feet and was discovered during 19th century.


text:
# Thread: Find H on this triangle

1. ## Find H on this triangle

here's the pic I made... (find H)

now, I'm confused on how to set it up, should it be

$tan(18) = h / 5+x$
$tan(32) = h / x$

or

$tan(18) = h / b$
$tan(32) = h / b-5$

(b = base)

or neither?

please tell me which set-up it is and why, i can't get it

2. Originally Posted by Arez
here's the pic I made... (find H)

now, I'm confused on how to set it up, should it be

$tan(18) = h / 5+x$
$tan(32) = h / x$

or

$tan(18) = h / b$
$tan(32) = h / b-5$

(b = base)

or neither?

please tell me which set-up it is and why, i can't get it
Both are correct, except that you should use parentheses for the denominators so that
h/ 5+x would appear h / (5+x)
h / b -5 would appear h /(b-5)

tan is opposite side / adjacent side .......that's it.
No matter how you come up with the measurements of the sides, as long as you carefully follow tan = opp/adj, then you're okay.

---
EDIT:

Opsss, I'm wrong!
After reading masters' reply (which is correct), I am wrong with mine.
tan(32 deg) cannot be h /(5 +x) nor h /(b -5)....because h is not the opposite side of 32 degrees.

Must be that I just woke up.
Still, I shouldn't have missed that tan(32 deg)....

Sometimes, though, I like being wrong. :-)

3. Originally Posted by Arez
here's the pic I made... (find H)

now, I'm confused on how to set it up, should it be

$tan(18) = h / 5+x$
$tan(32) = h / x$

or

$tan(18) = h / b$
$tan(32) = h / b-5$

(b = base)

or neither?

please tell me which set-up it is and why, i can't get it
Not enough info, I fear. According to your diagram:

$\tan 18=\frac{H}{5+X}$ is correct!

The 32 degree angle has no opposite side defined. H is the opposite side of the 18 degree angle. So,

$\tan 32 \neq \frac{H}{X}$

4. Originally Posted by ticbol
Opsss, I'm wrong!
After reading masters' reply (which is correct), I am wrong with mine.

Sometimes, though, I like being wrong. :-)
It keeps us humble, you know.


text:
# ABC conjecture and Fermat's last theorem

I have frequently read and heard that given the ABC-conjecture a number of important unsolved problems of number theory can be solved (with relatively simple proofs). Among them, the celebrated Fermat's Last theorem is frequently mentioned.

So, my question is: Given that the $ABC$ conjecture is valid, can we prove that it implies Fermat's Last theorem ?

P.S.: I can understand that ABC conjecture "easily" implies the asymptotic FLT (stating that: "the equ-ation $x^n+y^n=z^n$ can have solutions in positive integers only for $n< n_0$, where $n_0$ is some finite number"). This is outlined in Lang's Algebra (p.196, 1994 edition), see also here and here.

• No. The largeness of exponents in the asymptotic version depends crucially on what constant emerges (for a given specific epsilon) in an ABC proof. In particular, if ABC were to be proved with ineffective constants then the "largeness" would be totally mysterious, and Vesselin Dimitrov's "effective" version of Mochizuki's result (assuming the latter as a black box!) gives horrifically gigantic constants (massive exponentials, etc.). For this and other reasons, the proof of FLT via modularity methods seems likely to remain the only way to prove FLT and to understand why it is true. Sep 18, 2016 at 5:08
• No, abc implies at most finitely many counterexamples to FLT, but it allows counterexamples.
– joro
Sep 18, 2016 at 12:02
• Actually, it isn't that inconceivable that ABC implies FLT, even though not in the obvious way - we might e.g. be able to prove that if there is a solution with exponent $n$, then there is a solution with an exponent $m>n$, in which case we would get a contradiction with ABC. I am not claiming I know how to prove FLT from ABC this way, just sharing a thought. Sep 18, 2016 at 12:46
• Here's one way to think about it. $ABC$ easily implies that if $n$ is sufficiently large, then the equation $X^n+Y^n=94151567435Z^n$ has no solutions in non-zero integers. But it could not imply that there are no solutions for all $n$, since in fact $(2,3,1)$ is a solution for $n=23$. So $ABC$ will, as others have said, rule out all sufficiently large $n$ (even for more general equations such as $aX^n+bY^n=cZ^n$), but then one will need some method of dealing with the remaining values of $n$. Sep 22, 2016 at 0:50
• Subsumed by mathoverflow.net/q/130980/41291 (imo) Sep 22, 2016 at 6:56

For all exponents $n > 3$, abc implies at most finitely many counterexamples to FLT, but it allows counterexamples to FLT.
For exponent $n=3$ it allows infinitely many counterexamples and the Fermat-like equation $x^3+y^3=a z^3$ has infinitely many coprime solutions for some $a$ via the group law on the elliptic curve.
Over number fields FLT fails, e.g. $1^n+1^n=(\sqrt[n]{2})^n$.


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# Big O notation of max?

I'm coding a few set comparisons and noting their big O's using different algorithms and set implementations. I got to one particular function and I decided that it is $O(max(n,m))$ runtime. Is that the proper way to express this?

The expression $O(\max(n,m))$ is meaningful, and we even have $O(n) + O(m) = O(n+m) = O(\max(n,m))$.

• Just to clarify: $O(n) + O(m)$ is the maximum of n and m? I've only really used expressions like $O(1)$, $O(n log n)$. I haven't seen any 'operators' or anything before. – Levi Morrison May 1 '13 at 19:05
• @LeviMorrison What do you mean by the maximum of n and m? If you're wondering about the $=$ there, yes, these do hold. – phant0m May 1 '13 at 19:08
• Meaning, the runtime is equal to $O(n)$ if n > m, and $O(m)$ otherwise. – Levi Morrison May 1 '13 at 19:12
• @LeviMorrison Yes, the argument is quite simple: Suppose $n > m$ and your function is in $\mathcal{O}(n + m)$: Then your function is also in $\mathcal{O}(n + n)$, which is the same as $\mathcal{O}(n)$ Combine that with the argument for $m > n$ and you can see that it also being in $\mathcal{O}(\text{max}(n, m))$ makes sense. Note that adding $\mathcal{O}$s together is a slight abuse of notation. – phant0m May 1 '13 at 19:15
• @Yuval Filmus. What "expression is meaningful" means? And BTW, I have my own, open similar question in StackOverflow. You are welcome to claim 200 bounty points there. – Leonid Volnitsky Nov 4 '16 at 9:06


text:
Find coefficient of $x^ny^m$

how can I find the coefficient of $x^ny^m$ of following series $$\frac{\log\big(\frac{1}{1-xy}\big)}{(1-x)(1-y)(1-xy)}$$ and $$\frac{\log(\frac{1}{1-x})}{(1-x)(1-y)(1-xy)}$$ where $1 \leq m \leq n$

-
You can write this functions as a Cauchy product of four series. Why do you want $m\leq n$? – Davide Giraudo Jul 22 '12 at 16:37
I am trying to understand this paper on partial quick sort. lsi.upc.edu/~conrado/research/reports/ALCOMFT-TR-03-50.pdf . Page no 3 (F(z,u)*=) – ami Jul 22 '12 at 16:41

Use $\log\left(\frac{1}{1-x y}\right) = -\log(1- xy) = - \sum_{n=1}^\infty \frac{1}{n} x^n y^n$, and $$-\frac{\log(1-x y)}{1-x y} = \sum_{n=1}^\infty x^n y^n \sum_{m=1}^n \frac{1}{n} = \sum_{n=1}^\infty x^n y^n H_n$$ Then use $$\frac{ \sum_{n \geqslant 1, m\geqslant 1} c_{n,m} x^n y^m}{(1-x)(1-y)} = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \sum_{k=1}^n \sum_{\ell=1}^m c_{k,\ell}$$ In our case $c_{k,\ell} = H_k \delta_{k,\ell}$, and $\sum_{k=1}^n \sum_{\ell=1}^m H_k \delta_{k,\ell} = \sum_{k=1}^{\min(n,m)} H_k$ $$-\frac{\log(1-x y)}{1-x y} \cdot \frac{1}{(1-x)(1-y)} = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \sum_{k=1}^{\min(n,m)} H_k = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \left( \left(\min(n,m) +1\right)\left( H_{\min(n,m)+1}-1\right) \right)$$

For the second series, we start with $$[x]^n \frac{-\log(1-x)}{1-x} = H_n \quad [x]^m \frac{1}{1-x y} = y^m$$ Therefore $$[x]^n \frac{-\log(1-x)}{1-x} \cdot \frac{1}{1-x y} = \sum_{k=0}^{n-1} H_{n-k} y^k$$ Then $$[y]^m [x]^n \frac{-\log(1-x)}{1-x} \cdot \frac{1}{1-x y} \cdot \frac{1}{1-y} = \sum_{k=0}^{n-1}H_{n-k} \mathbf{1}\left( k \leqslant m \right) = \sum_{k=0}^{\min(m, n-1)}H_{n-k}$$

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Thanks. please solve series no. 2. – ami Jul 22 '12 at 16:54
Thanks @Sasha. your solution is more generic. – ami Jul 22 '12 at 17:37
I think you're off by $1$ in the first series. You seem to be saying (in the last line) $\sum_{k=1}^K H_k = (K+1) (H_K-1)$ but it should be $\sum_{k=1}^K H_k = (K+1) (H_K-1) + 1$. So the answer should be $$\sum_{n=1}^\infty\sum_{m=1}^\infty x^n y^m \left((\min(n,m)+1)(H_{\min(n,m)}-1)+1\right)$$ – Robert Israel Jul 22 '12 at 18:25
@RobertIsrael Thanks, I mean to say $\sum_{k=1}^n H_k = (n+1)(H_{n+1}-1)$. – Sasha Jul 22 '12 at 19:29

Using geometric series and $\displaystyle\frac{1}{1-r}\log\frac{1}{1-r}=\sum_{k\ge1}H_kr^k$, with $H_k$ the $k$th harmonic number,

$$\sum_{a=0}^\infty x^a~\sum_{b=0}^\infty y^b~\sum_{k=1}^\infty H_k(xy)^k=\sum_{a,b\ge0}^{k\ge1}H_kx^{a+k}y^{b+k}.$$

The coefficient of $x^ny^m$ will be the sum of all $H_k$ with $k\le n,m$. With $m\le n$, this is

$$\sum_{k=1}^mH_k=(m+1)(H_{m+1}-1)$$

by induction. Similarly for the second,

$$\sum_{a=0}^\infty y^a\sum_{b=0}^\infty (xy)^b \sum_{k=1}^\infty H_kx^k=\sum_{a,b\ge0}^{k\ge1} H_k x^{b+k}y^{a+b}.$$

The coefficient of $x^ny^m$ will be the sum of all $k\le n$ such that $0\le n-k\le m$, i.e.

$$\sum_{k=n-m}^{n}H_k=\left(\sum_{k=1}^n H_k\right)-\left(\sum_{k=1}^{n-m-1}H_k\right)$$

$$=(n+1)\big(H_{n+1}-1\big)-(n-m)\big(H_{n-m}-1\big).$$

(Or $0$ if $m>n$.)

-
Thanks @anon. I learned something today. :) – ami Jul 22 '12 at 17:35


text:
Double Integral Using Polar Coordinates

Heavydoody

Joined Jul 31, 2009
140
"R is the region in the first quadrant enclosed by x^2+y^2=25. Use polar coordinates to evaluate the double integral of ye^x dA."

So, I see that the base area of integration is a quarter circle from 0 to ∏/2 and radius 0 to 5 (the limits of integration). I cannot see how to convert the integrand to polar coordinates. Everything else we have done has had X^2+y^2 in some form in the integrand. We did get a hint that we should integrate w.r.t. theta first. Can someone get me going on this?

Heavydoody

Joined Jul 31, 2009
140
Took some huntin' on the www, but x = r cos θ, y = r sin θ. Piece of cake from there.


text:
Browse Questions

# If a luminous point is moving at a speed $V_0$ towards spherical mirror of focal length f along its axis, Find the speed of image.

$(a)\;\frac{-f V_0}{u- f} \\ (b)\;\bigg[\frac{f}{u+ f}\bigg]^2 V_0 \\ (c)\;\bigg[\frac{f}{u- f}\bigg]^2. v_0 \\ (d)\;- \bigg[\frac{f}{u- f}\bigg]^2 V_0$

As we know,
$\large\frac{1}{v}+\frac{1}{u} =\frac{1}{f}$
differentiating with respect to time
we obtain , $-\large\frac{-1}{V^2} \frac{dv}{dt} -\frac{1}{u^2} \frac{du}{dt}$$=0$
or $\large\frac{-1}{V^2} V_{image}- \large\frac{1}{u^2} V_{object}=0$
But from mirror equation $v= \large\frac{uf}{u-f}$
So $V_{image}= - \bigg(\large\frac{V}{u} \bigg)^2 V_{object}$
$\qquad= - \bigg[ \large\frac{f}{u-f}\bigg]^2 V_0$
Hence d is the correct answer.


text:
Recommended Posts

I've noticed a lot of questions regarding xfiles and how to manually load them in. So I took the liberty of writing 2 tutorials about it. One loads xfiles and displays it, the other manually parses the xfile and loads the mesh information in. Tutorials can be found in my sig, and look forward to more mesh related topics...

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This is a very good idea. Back when i first started out, I looked all around for a tut like this. Btw, love that scene graph tutorial.

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thanks. Couldnt have done it though without the community here and the various books people have written :)

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Excellent, excellent, excellent! Finally someone has done something about it.
Temporarily sticked until I put this stuff in the FAQ. (I know I've not updated it in a long time - I apologize)

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Great stuff! Just what iv'e been looking for with DirectX9.
Thanks a lot-;)

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Alf, I haven't seen you in ages on MSN, but you should submit these tutorials to GDNet aswell, and have them end up on the frontpage. People will see them quicker than.

Toolmaker

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Quote:
 Original post by ToolmakerAlf, I haven't seen you in ages on MSN, but you should submit these tutorials to GDNet aswell, and have them end up on the frontpage. People will see them quicker than.

That's becuase I don't sign in too much. But anyway, it's easier and quicker to just stick the tutorials up on my site. And besides I have a whole collection now, so one tutorial outside wouldn't really make sense.

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Cool stuff, love it ..

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great tutorial!
the document in the sdk help is just not enough to explain how this works!
are you going to update the tutorial with the new ID3DXFile interface in dx9?

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Quote:
 Original post by clipare you going to update the tutorial with the new ID3DXFile interface in dx9?

That's what they used from the start!

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Great tutorial -:)
Just a small request,would you be able to update this tut to also load the textures and to demo the use of the anim controller please?

keep up the good work.
One more thing(before i forget)...the bumpmapping tut on your website is missing some files,or it needs or uses your GameSDK?

would it be possible to seperate this tutorial from your SDK if that is the case.

Thanks -:)

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Quote:
 Great tutorial -:)Just a small request,would you be able to update this tut to also load the textures and to demo the use of the anim controller please?

Thanks,

Animation is on the todo list. It'll get there eventually.

Quote:
 One more thing(before i forget)...the bumpmapping tut on your website is missing some files,or it needs or uses your GameSDK?would it be possible to seperate this tutorial from your SDK if that is the case.

It needs the mini sdk, yes. Which can be downloaded from the main tutorials page (before you enter the graphics section). It would be possible to seperate the tutorial from the SDK, but that SDK is just a bunch of really stripped down initialization routines (and input routines IIRC). So that would really be unnecessary.

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Thanks -:)
The reason i asked was becuase i'm still using VC++ 6,not .NET,
and i remember a while ago i tried to compile it under 6 and couldn't.I remember seeing on your forums someone posted a VS6 workspace,but the link is broken now.

I don't suppose you have a backup copy of that download i could get from you?

Thanks.

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To compile dx9 with vc6 you're going to have to download the vc6 libs. it's in the sdk extras download. If you've already done that, then you can post specific errors on the sites forums and we'll see what we can do.

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can you tell me how to load animated X file.
I am using Panda Exporter to load my X file and there's an option for the animation too but i dont know how can i pick up the frames of animation in DirectX. pretty hard though, I know but if anyone can help me it'll be very nice.

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You can either parse the xfile your self and build the animation data by getting the TID_D3DRMAnimationSet, TID_D3DRMAnimation and TID_D3DRMAnimationKey. But you'll need to build a hierarchy by parsing TID_D3DRMFrame objects. OR you can can set up a few classes and use another way to do all this which is described here in a few docs and samples. Specifically the mesh hierarchy and skinned mesh docs is what you're looking for. If you're having specific problems however then you should probably create a topic for that so everyone can see and help you. Everyone is not going to be checking this thread since it's just an announcement.

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Is there any tutorial about how to load LightWave 3D objects into D3D mesh format?

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not that i know of.

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Check out this SDK sample. It loads a Lightwave OBJ file into an ID3DXMesh.

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If we don't want to use the ID3DXMesh, how to interpret the X file data manually?

Specifically regarding the Mesh template, it contains:

NumVertices
Array of vertices
NumFaces
Array of MeshFace

Optionals are MeshFaceWraps, MeshTextureCoords, MeshNormals, MeshVertexColors, MeshMaterialList.

For simple meshes I could get by with loading the vertices in the array to a vertex buffer. As for the MeshFace array, I ignore the first item, which is the Number of FaceVertexIndices (seems to be always 3, I could be wrong since I don't really understand); then I just load the rest into an index buffer.

I could anyone provide a better explaination to the inter-working of this? What about the MeshNormals, seems to have the same dilema. Thanks

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Hey IFooBar,thanks for your tutorials...very good :)
I have a couple of questions about the .x file format in general for you,as you seem to know your stuff about the format.

Someone recently told me that i would not be able to use the .x file format for characters or maps,if i wanted to use bump-mapping and light-mapping in my project.

Is this true? if so why is this,and what does he mean about the format being crap?

Can you(or anyone one) shed some light on this for me please?
I'm really starting to get into shaders now,and want to use this format for both my characters and map data.

So far,most of the books on directX iv'e read have said how flexible .xfiles are,so what this person said to me doesn't really make sense as he is a pro coder.

thanks.

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Quote:
 If we don't want to use the ID3DXMesh, how to interpret the X file data manually?Specifically regarding the Mesh template, it contains:

The Mesh templates contains the following data:

- number of faces
- face vertices
- number of indexed faces
- index data as arrays

That's the basics. You multiply the number of faces by three and you'll get how many vertices you have. You then loop through for that many vertices and get the vertex info. Then you get the indices by looping thorugh and reading arrays of 3. so, you'd go about it something like this:

SIZE_T size;DWORD* pDword;// Get num facespData->Lock( &size, &pDword );DWORD numFaces = *pDword;pData->Unlock();// Create vertex buffer bug enough for numFaces * 3 vertices.// get vertex datafor( int i = 0; i < numFaces; ++i ){  float* c1, *c2, *c3;  pData->Lock( &size, &c1 );  pData->Lock( &size, &c2 );  pData->Lock( &size, &c3 );  D3DXVECTOR pos( *c1, *c2, *c3 );  // store pos directly in vertex buffer or put away in a std::vector  // or something to store later.  pData->Unlock();  pData->Unlock();  pData->Unlock();}  // Get num indicespData->Lock( &size, &pDword );DWORD numIdxFaces = *pDword;pData->Unlock();// Create index buffer big enough for 3 * numIdxFaces// get index datafor( int i = 0; i < numIdxFaces; ++i ){  float* i1, *i2, *i3;  pData->Lock( &size, &pDword ); // array size comes first.  pData->Lock( &size, &i1 );  pData->Lock( &size, &i2 );  pData->Lock( &size, &i3 );  D3DXVECTOR idx( *i1, *i2, *i3 );  pData->Unlock();  pData->Unlock();  pData->Unlock();  pData->Unlock();}

The MeshFace array (the last loaded thing in the for loop in the above code) is just an array of indices. Now I've never manually laoded in data using the new SDK yet, so Im not sure if the data retrieval is done correctly. Going by the documentation, teh above should work. The problem you seem to be having is with MeshFace. Its not just a group of 3 values. It's an array. It can have an arbitrary number of values in it. Almost all the time, it's 3 though. For every MeshFace, you fist have to read in the number of value sit contains, then you can read in the actual values.

Quote:
 Someone recently told me that i would not be able to use the .x file format for characters or maps,if i wanted to use bump-mapping and light-mapping in my project.

you definetly can. lightmapping just requires texture coordinates which is available with standard xfiles. Although you cant store tangent and binormal data for bumpmapping in xfiles, you can compute it at run time which is what is usually done as far as I know. Cloning comes in handy over here. So it is very possible to use.

Quote:
 if so why is this,and what does he mean about the format being crap?

I dunno, ask him. Maybe because it's very annoying to work with - which it is, but its like anything else once you get the hang of it. Best thing for you to do is experiment and decide for yourself whether you want to use it or not.

Quote:
 So far,most of the books on directX iv'e read have said how flexible .xfiles are,so what this person said to me doesn't really make sense as he is a pro coder.

The person never implied it wasnt flexible by saying the format was crap. It is very flexible as Im sure this person will agree with you on that. But with flexibility comes complexity - unfortuanetly, and the complexity part is probably what he dosen't like.

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Is the number of indices for the face there for flexibility's sake, in which it's is almost always 3? For the MeshNormals, for the time being is it pretty much safe to ignore the index part? For example a simple geometry with 2 faces, 3 indices each. There may be syntax errors, but hopefully you'll get the idea:

Mesh Rect{
4;
0.0, 0.0, 0.0,
0.0, 1.0, 0.0,
1.0, 0.0, 0.0,
1.0, 1.0, 0.0;

2;
3;0, 1, 2;
3;2, 1, 3;

MeshNormals
{
4;
0.0, 0.0, 0.0,
0.0, 1.0, 0.0,
1.0, 0.0, 0.0,
1.0, 1.0, 0.0;

2;
3;0, 1, 2;
3;2, 1, 3;
}
}

The only case I sometimes see is that one vertex may have many normals. Is there any examples out there where it loads meshes correctly and still providing the flexibility?

What about the idea of a "subset". I have a feeling that a subset is something( a group of faces) that uses the same material. How to correctly divide subsets while loading meshes?

Thanks.

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We are writing a Java loader for the DirectX "X" format. I am trying to find info on exactly what information is output from the animatian arrays in file specificly in the rotation end.
Here's my example
AnimationKey {
0; //rotation
41; //41 frames
0;4;-0.696364,-0.696364,-0.122788,-0.122788;;,
1;4;-0.696768,-0.696768,-0.120478,-0.120478;;,
2;4;-0.697852,-0.697852,-0.114028,-0.114028;;,
3;4; ? , ? , ? , ? ;;,
ect....
The question marks inicate the section we are having trouble finding info on.
The output type is keyframes. We have managed to get just about everything else loaded but this area eludes us. The X file template in the msdn is pretty vague.

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# Inverse Trig Functions and Reciprocals

Gold Member

## Homework Statement

Evaluate and express your answer in radians:

$$cot^{-1}\left(1\right)$$

## The Attempt at a Solution

I start by identifying that the domain of Arccotangent is all real numbers. So 1 is in the domain.

From here, I looked at the unit circle and saw that ##\frac{π}{4}## has a cotangent of 1, but this was incorrect.

So I pulled out my calculator and evaluated ##\frac{1}{arctan\left(1\right)}##

However, if you'll look at my attached image, the step-by-step instructions say that ##cot\left(x\right)=\frac{1}{tan\left(x\right)}## BUT ##arccot\left(x\right)≠\frac{1}{arctan\left(x\right)}##
Why is this true?

I'm not sure what this page is telling me.

#### Attachments

• Screen Shot 2018-06-01 at 4.21.14 PM.png
19.4 KB · Views: 351

## Answers and Replies

jedishrfu
Mentor
Think about it for a moment using sine or cosine.

The sin(30) = 0.5 but so does the sin(390) = sin(30 + 360) = sin(30 + n*360) for n=0,1,2,3...

(EDIT: fixed sin(210) typo error)

so now given the sin of an angle you will get a whole bunch of angles with the same sine value.

and so the ##cot^{-1} (x) = tan^{-1}(1/x)##

(EDIT:## arccot## is often written as ##cot^{-1}## in math books implying an inverse but not a multiplicative inverse.)

In your case, they're warning you not to make that mistake in general considering the x=1 and 1/1 is 1.

Last edited:
opus
Gold Member
Ok so in terms of all the values that a function can take on, I understand that there is more than one angle that can produce a specific value. This is why we restrict the domain of the inverse trig functions- to make them invertible.

So in terms of reciprocals, the cotangent function is equal to the reciprocal of the tangent function. But the page says that the inverse cotangent function is NOT equal to the reciprocal of the inverse tangent function. I guess I still don't see why the inverses cannot be reciprocals but the basic functions can. The only difference is that for the inverses, they have a restricted domain.

Mark44
Mentor
This is a terrrible problem, at least from what I can see in the screen shot you posted. At the beginning, it asks you to evaluate ##\cot^{-1}(1)## rounding the answer to four decimal places. The exact answer is ##\frac \pi 4##, or rounded, it is approx. .7854.

The answer they show at the bottom of the screen shot is not the answer to the question above, which is at least misleading.

If ##x = \cot^{-1}(1)##
then ##\cot(x) = 1##
so ##\tan(x) = 1##, and it is this 1 that they expect you to put in the box, at least based on what they show is the correct answer.

However that is not the value of ##\cot^{-1}(1)## -- the equation ##\tan(x) = 1## is just an intermediate step.

Last edited:
opus
Gold Member
Getting tan by itself as not a reciprocal is giving me trouble as well. Say it wasn't a 1, but rather a -3. Then ##\frac{1}{tan\left(x\right)}=-3##
How do I go about getting tan(x) rather than just the reciprocal? I would say multiply both sides by tan(x), but that would leave me with ##1=-3tan\left(x\right)##

Mark44
Mentor
Getting tan by itself as not a reciprocal is giving me trouble as well. Say it wasn't a 1, but rather a -3. Then ##\frac{1}{tan\left(x\right)}=-3##
How do I go about getting tan(x) rather than just the reciprocal? I would say multiply both sides by tan(x), but that would leave me with ##1=-3tan\left(x\right)##
Sure that would work, but then divide both sides by -3, so tan(x) = -1/3.

Something simpler:
1/tan(x) = -3
=> tan(x) = -1/3
Just take the reciprocal of each side.

If a = b, then 1/a = 1/b, provided that neither a nor b is zero.

opus
Gold Member
As an example to my question, I wrote out my train of thought here.

#### Attachments

• 37CA9811-5F26-4C5B-9659-AB8D487E41E3.jpeg
33.5 KB · Views: 318
Gold Member
Ahh ok. I see. The 1 in the tangent problem threw me off because 1tan(x) is just tan(x) and I was assuming the coefficient in front of the function should be 1.

Gold Member
So here is what is still confusing me, and I'll throw in a new but similar problem because I've been trying different ones trying to get it to click:

##sec^{-1}\left(3\right)##

Right off the bat, I see if it's in the restricted domain, and it is.

Next, I think if there are any common angles that I know that would give a secant value of 3. There are none.

Now I know that I have to pop out the calculator. What I want to do (and what I apparently am not allowed to do) is plug ##\frac{1}{arccos\left(3\right)}## into the calculator. This is where I am confused, as I don't know why this is not allowed.

If ##sec\left(x\right)=\frac{1}{cos\left(x\right)}## why is it also NOT true that ##arcsec\left(x\right)=\frac{1}{arccos\left(x\right)}## ?

Why would I have to even put this in terms of cosine?

Gold Member
Thank you jedishrfu. I'll have a look!

Gold Member
Just found on that website a problem almost identical to the last one I posted. Please see attached image. This seems like a contradiction to the screenshot of my first post.

So let me ask this question- Is the following a true statement?
##arcsec\left(3\right)=arccos\left(\frac{1}{3}\right)##

#### Attachments

• Screen Shot 2018-06-01 at 6.20.41 PM.png
104.9 KB · Views: 318
jedishrfu
Mentor
##sec(x)=2## then ##1/cos(x)=2## then ##cos(x)=1/2##

##x = sec^{-1}(2)## and from the second one ##x = cos^{-1}(1/2)##

And you can see how it works.

opus
Mark44
Mentor
Just found on that website a problem almost identical to the last one I posted. Please see attached image. This seems like a contradiction to the screenshot of my first post.

So let me ask this question- Is the following a true statement?
##arcsec\left(3\right)=arccos\left(\frac{1}{3}\right)##
Here's a drawing to help you out, using a right triangle.

cos(a) = 1/3, so sec(a) = 3
From this, ##a = \sec^{-1}(3) = \text{arcsec}(3)##
From cos(a) = 1/3, we also have ##a = \cos^{-1}(\frac 1 3) = \text{arccos}(\frac 1 3)##
Since a is equal to itself, we have ##\text{arcse}c(3) = \text{arccos}(\frac 1 3)##
So yeah, yours is a true statement.

BTW, the third side's length is ##\sqrt 8 = 2\sqrt{2}##, but it doesn't enter the picture with these functions.

#### Attachments

• secant.png
1,023 bytes · Views: 234
opus
Gold Member
That makes great sense thinking about it like that! Thanks Mark!

Gold Member
Thanks guys for the help. I think I got a grip on it now.


text:
# How do I find the count of a particular column, based on another column(date) using pandas?

I have a dataframe with 3 columns, such as SoldDate,Model and TotalSoldCount. How do I create a new column, 'CountSoldbyMonth' which will give the count of each of the many models sold monthly?

Date        Model  TotalSoldCount
Jan 19        A          4
Jan 19        A          4
Jan 19        A          4
Jan 19        B          6
Jan 19        C          2
Jan 19        C          2
Feb 19        A          4
Feb 19        B          6
Feb 19        B          6
Feb 19        B          6
Mar 19        B          6
Mar 19        B          6

The new df should look like this.

Date      Model     TotalSoldCount     CountSoldbyMonth
Jan 19     A               4                    3
Jan 19     A               4                    3
Jan 19     A               4                    3
Jan 19     B               6                    1
Jan 19     C               2                    2
Jan 19     C               2                    2
Feb 19     A               4                    1
Feb 19     B               6                    3
Feb 19     B               6                    3
Feb 19     B               6                    3
Mar 19     B               6                    2
Mar 19     B               6                    2

I tried doing df['CountSoldbyMonth'] = df.groupby(['date','model']).totalsoldcount.transform('sum') but it is generating a different value.

df['CountSoldbyMonth'] = df.groupby(['Date','Model']).transform('count')

data['CountSoldbyMonth']= data.groupby(['Date','Model']).TotalSoldCount.transform('count') is working perfectly.


text:
# [OS X TeX] LaTeXiT 1.2 beta

Pierre Chatelier pierre.chatelier at club-internet.fr
Sun Jul 3 12:08:54 CEST 2005

> LaTeXiT is very cool and it is invaluable with Keynote.
Thanks,

> One thing that has plagued me about it and it still seems to be a
> problem with the most recent beta is that changes to the Default
> Preamble don't take immediate effect. It seems that I have to quit
> and restart LaTeXiT for them to take effect. Can this be corrected
> for the next release?
I am surprised, since it should work. When you type in a new default
preamble, the preamble of opened documents is not changed, since it
could have been tuned by the user. But if you open a new document
(Command-N), the new default preamble will be used.

Here is a crosspost with toolbox-talk, because there are some
questions for which I would like to get the opinion of as many meople
as possible. But if you think that such cross-post is useless, just
tell me, I'll understand.

> [in the error manager] so if it could be reliably determine what
> the offending token is it would be nice to see that as well.
For now, I would prefer not to. I prefer to be less precise, but
ensure that the error-line is ok, rather than trying to deduce the
faulty token, with the risk of getting wrong and give a bad
indication. However, since LaTeXiT is open source and will be a
toolbox project, if anybody has the knowledge to improve that, this
will be no problem I think. Do you agree with this opinion ?

>  - I think I like the use of those things that look like tabs but
> are actually buttons. The old school way would be to use radio
> buttons, but I'm undecided if these things are actually better than
> radio buttons.
This thing is called a "segmented control"; it was already available
with Panther, but with Tiger, it is directly usable from
InterfaceBuilder. At the very beginning of LaTeXiT, I was using radio
buttons, because I did not know the existence of segmented controls.
A friend of mine told me about that, I and find that in this context,
it is much more beautiful.

> - Since we're talking LaTeX, it would be better to write
>    Display ($...$)
>   because $$...$$ is, to be strict, incorrect.
> I guess this is a debatable problem. I did never use $...$
> myself, since I find  more handy. But I am certainly not a LaTeX
> expert. What does the majority of people expect here (this is a
> question for the mailing list) ?
>
>   But I think users tend to know enough of what's going on to be
> able to write   | Display | Inline | Text |
>   in those buttons and put $...$ , etc., in a tooltip.
Here I disagree a little. For people not familiar with LaTeX's
vocabulary, the use of the button becomes obvious. Here again, what
do people think about that ?

> - For a future version, it'd be good to add a toolbar for the LaTexise
A toolbar ? For what purpose ?

> (don't suppose we've got enough of an anti-American majority to get
> rid of the "z"? Can this be localised?)
Oh, do you mean that the "z" is a mistake (regardless of the fact
that the word does no exist) ? English is not my mother language, so
I did not pay great attention to this. It really does not distrub me
to put an "s" instead.

> [...for the LaTexise] button as well as a button for the drawer, etc.
Do you mean, to open/close the drawers ? Oh, I understand, now :
putting this buttons in the toolbar. Mmmh, it seems to me that my
toolbar would be very "poor" (only two button), so it would not be
very nice.

>  - I'm not sure if it's necessary to use \usepackage[applemac]
> {inputenc}. For a start, I think [utf8] would be better for
> forwards compatibility, and second -- since this thing will
> *mostly* be used for maths, I think if people *do* want accented
> characters then they can add this line to the preamble themselves.
Haha, you are forgetting french people like me; I do use accentuated
characters, even in an equation label ! And [utf8] does not work,
since the encoding of the input in the textfield is MacOS Roman.

>  - European decimal mark :) Can this be localised?
This should be... I will have a look.

>  - The cropping seems slightly over-enthusiastic. But not as bad as
> TeX itself! For example:
> \documentclass{article}
> \begin{document}
> \parindent=0pt
> \voffset=-1in
> \hoffset=-1in
> \setbox0=\hbox{$1=3$}
> \dimen0=\ht0
> \advance\dimen0\dp0
> \pdfpageheight=\dimen0
> \pdfpagewidth=\wd0
> \shipout\box0
> \end{document}
> % (Stolen ages ago from Maarten, perhaps)
You are cheating, using Maarten's pathological latex source ( ;-) no
offense!)
I have an immediate solution, that concerns Herbert's remark too : I
could add a control to add some margin to the default bounding box.
If you need great precision, it would be a textfield. On the
contrary, it would be a slider. But it would need a re-latexisation
(nice word). Any thoughts ?

The preview package seems to give the same results as LaTeXiT:

> \documentclass{article}
> \usepackage[active,pdftex,textmath,tightpage]{preview}
> \begin{document}
> $1=3$
> \end{document}
> Is this actually what is used to generate the output? I couldn't
> work it out from looking at the log file.
I do not use the preview package. If you want to get the generated
files, you can look into /private/tmp/folders.[your uid]/Temporary Items
(this is under Tiger the result of NSTemporaryDirectory(). The files
are put there.

Regards

Pierre
--------------------- Info ---------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Post: <mailto:MacOSX-TeX at email.esm.psu.edu>

More information about the macostex-archives mailing list


text:
시간 제한메모리 제한제출정답맞힌 사람정답 비율
10 초 (추가 시간 없음) 1024 MB (추가 메모리 없음)54333380.488%

## 문제

We have a list of integers X1, X2, …, XN. We would like them to be in strictly increasing order, but unfortunately, we cannot reorder them. This means that usual sorting algorithms will not work.

Our only option is to change them by appending digits 0 through 9 to their right (in base 10). For example, if one of the integers is 10, you can turn it into 100 or 109 with a single append operation, or into 1034 with two operations (as seen in the image below).

Given the current list, what is the minimum number of single digit append operations that are necessary for the list to be in strictly increasing order?

For example, if the list is 100,7,10, we can use 4 total operations to make it into a sorted list, as the following image shows.

## 입력

The first line of the input gives the number of test cases, T. T test cases follow. Each test case is described in two lines. The first line of a test case contains a single integer N, the number of integers in the list. The second line contains N integers X1, X2, …, XN, the members of the list.

## 출력

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimum number of single digit append operations needed for the list to be in strictly increasing order.

• 1 ≤ T ≤ 100.

## Test Set 1 (12점)

• 2 ≤ N ≤ 3.
• 1 ≤ Xi ≤ 100, for all i.

## Test Set 2 (14점)

• 2 ≤ N ≤ 100.
• 1 ≤ Xi ≤ 109, for all i.

## 예제 입력 1

4
3
100 7 10
2
10 10
3
4 19 1
3
1 2 3

## 예제 출력 1

Case #1: 4
Case #2: 1
Case #3: 2
Case #4: 0

## 힌트

In Sample Case #1, the input is the same as in the example given in the problem statement. As the image shows, the list can be turned into a sorted list with 4 operations. Notice that the last two integers need to end up with at least 3 digits (requiring at least 3 append operations in total). If all of the final numbers had exactly three digits, the second would be larger than the third because it starts with a 7 instead of a 1. This means we cannot do it with fewer than 4 operations.

In Sample Case #2, notice that the list needs to be in strictly increasing order, so we have to do at least one operation. In this case, any valid append operation to the second integer works.

In Sample Case #3, we can use two append operations to get the list to 4,19,193.

In Sample Case #4, the given list is already in strictly increasing order, so no operations are necessary.

## 채점 및 기타 정보

• 예제는 채점하지 않는다.


text:
## University Calculus: Early Transcendentals (3rd Edition)

$(\dfrac{3}{2},\dfrac{-3}{2},\dfrac{1}{2})$
Parametric equations are: $x=1-t; y=3t; z=1+t$ We have the equation of the plane: $2x-y+3z=6$ Now, we plug in all of the parametric equations into the above equation of a plane: $t=\dfrac{-1}{2}$ Thus, our parametric equations become: $x=1-(\dfrac{-1}{2})=\dfrac{3}{2}; y=3(\dfrac{-1}{2})=\dfrac{-3}{2}; z=1+(\dfrac{-1}{2})=\dfrac{1}{2}$ Hence, the line will meet at the point: $(\dfrac{3}{2},\dfrac{-3}{2},\dfrac{1}{2})$


text:
• ### Measurements of $\pi^\pm$, K$^\pm$, p and $\bar{\textrm{p}}$ spectra in proton-proton interactions at 20, 31, 40, 80 and 158 GeV/c with the NA61/SHINE spectrometer at the CERN SPS(1705.02467)

Sept. 27, 2017 nucl-ex
Measurements of inclusive spectra and mean multiplicities of $\pi^\pm$, K$^\pm$, p and $\bar{\textrm{p}}$ produced in inelastic p+p interactions at incident projectile momenta of 20, 31, 40, 80 and 158 GeV/c ($\sqrt{s} =$ 6.3, 7.7, 8.8, 12.3 and 17.3 GeV, respectively) were performed at the CERN Super Proton Synchrotron using the large acceptance NA61/SHINE hadron spectrometer. Spectra are presented as function of rapidity and transverse momentum and are compared to predictions of current models. The measurements serve as the baseline in the NA61/SHINE study of the properties of the onset of deconfinement and search for the critical point of strongly interacting matter.
• ### Measurement of Meson Resonance Production in $\pi^{-} +$C Interactions at SPS energies(1705.08206)

May 23, 2017 nucl-ex, astro-ph.HE
We present measurements of $\rho^0$, $\omega$ and K$^{*0}$ spectra in $\pi^{-} +$C production interactions at 158 GeV/c and $\rho^0$ spectra at 350 GeV/c using the NA61/SHINE spectrometer at the CERN SPS. Spectra are presented as a function of the Feynman's variable $x_\text{F}$ in the range $0 < x_\text{F} < 1$ and $0 < x_\text{F} < 0.5$ for 158 GeV/c and 350 GeV/c respectively. Furthermore, we show comparisons with previous measurements and predictions of several hadronic interaction models. These measurements are essential for a better understanding of hadronic shower development and for improving the modeling of cosmic ray air showers.
• ### Two-particle correlations in azimuthal angle and pseudorapidity in inelastic p+p interactions at the CERN Super Proton Synchrotron(1610.00482)

Feb. 7, 2017 nucl-ex
Results on two-particle $\Delta\eta\Delta\phi$ correlations in inelastic p+p interactions at 20, 31, 40, 80, and 158~GeV/c are presented. The measurements were performed using the large acceptance NA61/SHINE hadron spectrometer at the CERN Super Proton Synchrotron. The data show structures which can be attributed mainly to effects of resonance decays, momentum conservation, and quantum statistics. The results are compared with the EPOS and UrQMD models.
• ### Measurements of $\pi^{\pm}$ differential yields from the surface of the T2K replica target for incoming 31 GeV/c protons with the NA61/SHINE spectrometer at the CERN SPS(1603.06774)

Nov. 29, 2016 hep-ex, physics.ins-det
Measurements of particle emission from a replica of the T2K 90 cm-long carbon target were performed in the NA61/SHINE experiment at CERN SPS, using data collected during a high-statistics run in 2009. An efficient use of the long-target measurements for neutrino flux predictions in T2K requires dedicated reconstruction and analysis techniques. Fully-corrected differential yields of $\pi^\pm$-mesons from the surface of the T2K replica target for incoming 31 GeV/c protons are presented. A possible strategy to implement these results into the T2K neutrino beam predictions is discussed and the propagation of the uncertainties of these results to the final neutrino flux is performed.
• ### Multiplicity and transverse momentum fluctuations in inelastic proton-proton interactions at the CERN Super Proton Synchrotron(1510.00163)

Aug. 30, 2016 hep-ex, nucl-ex
Measurements of multiplicity and transverse momentum fluctuations of charged particles were performed in inelastic p+p interactions at 20, 31, 40, 80 and 158 GeV/c beam momentum. Results for the scaled variance of the multiplicity distribution and for three strongly intensive measures of multiplicity and transverse momentum fluctuations \$\Delta[P_{T},N]\$, \$\Sigma[P_{T},N]\$ and \$\Phi_{p_T}\$ are presented. For the first time the results on fluctuations are fully corrected for experimental biases. The results on multiplicity and transverse momentum fluctuations significantly deviate from expectations for the independent particle production. They also depend on charges of selected hadrons. The string-resonance Monte Carlo models EPOS and UrQMD do not describe the data. The scaled variance of multiplicity fluctuations is significantly higher in inelastic p+p interactions than in central Pb+Pb collisions measured by NA49 at the same energy per nucleon. This is in qualitative disagreement with the predictions of the Wounded Nucleon Model. Within the statistical framework the enhanced multiplicity fluctuations in inelastic p+p interactions can be interpreted as due to event-by-event fluctuations of the fireball energy and/or volume.
• ### Production of $\Lambda$ hyperons in inelastic p+p interactions at 158 GeV/$c$(1510.03720)

April 25, 2016 hep-ex, nucl-ex
Inclusive production of $\Lambda$-hyperons was measured with the large acceptance NA61/SHINE spectrometer at the CERN SPS in inelastic p+p interactions at beam momentum of 158~\GeVc. Spectra of transverse momentum and transverse mass as well as distributions of rapidity and x$_{_F}$ are presented. The mean multiplicity was estimated to be $0.120\,\pm0.006\;(stat.)\,\pm 0.010\;(sys.)$. The results are compared with previous measurements and predictions of the EPOS, UrQMD and FRITIOF models.
• ### Measurements of $\pi^\pm$, $K^\pm$, $K^0_S$, $\Lambda$ and proton production in proton-carbon interactions at 31 GeV/$c$ with the NA61/SHINE spectrometer at the CERN SPS(1510.02703)

Feb. 24, 2016 hep-ex, nucl-ex
Measurements of hadron production in p+C interactions at 31 GeV/c are performed using the NA61/ SHINE spectrometer at the CERN SPS. The analysis is based on the full set of data collected in 2009 using a graphite target with a thickness of 4% of a nuclear interaction length. Inelastic and production cross sections as well as spectra of $\pi^\pm$, $K^\pm$, p, $K^0_S$ and $\Lambda$ are measured with high precision. These measurements are essential for improved calculations of the initial neutrino fluxes in the T2K long-baseline neutrino oscillation experiment in Japan. A comparison of the NA61/SHINE measurements with predictions of several hadroproduction models is presented.
• ### Measurement of event-by-event transverse momentum and multiplicity fluctuations using strongly intensive measures $\Delta[P_T, N]$ and $\Sigma[P_T, N]$ in nucleus-nucleus collisions at the CERN Super Proton Synchrotron(1509.04633)

Sept. 15, 2015 nucl-ex
Results from the NA49 experiment at the CERN SPS are presented on event-by-event transverse momentum and multiplicity fluctuations of charged particles, produced at forward rapidities in central Pb+Pb interactions at beam momenta 20$A$, 30$A$, 40$A$, 80$A$, and 158$A$ GeV/c, as well as in systems of different size ($p+p$, C+C, Si+Si, and Pb+Pb) at 158$A$ GeV/c. This publication extends the previous NA49 measurements of the strongly intensive measure $\Phi_{p_T}$ by a study of the recently proposed strongly intensive measures of fluctuations $\Delta[P_T, N]$ and $\Sigma[P_T, N]$. In the explored kinematic region transverse momentum and multiplicity fluctuations show no significant energy dependence in the SPS energy range. However, a remarkable system size dependence is observed for both $\Delta[P_T, N]$ and $\Sigma[P_T, N]$, with the largest values measured in peripheral Pb+Pb interactions. The results are compared with NA61/SHINE measurements in $p+p$ collisions, as well as with predictions of the UrQMD and EPOS models.
• ### Phase-space dependence of particle-ratio fluctuations in Pb+Pb collisions from 20A to 158A GeV beam energy(1310.3428)

Oct. 12, 2013 hep-ex, nucl-ex
A novel approach, the identity method, was used for particle identification and the study of fluctuations of particle yield ratios in Pb+Pb collisions at the CERN Super Proton Synchrotron (SPS). This procedure allows to unfold the moments of the unknown multiplicity distributions of protons (p), kaons (K), pions ($\pi$) and electrons (e). Using these moments the excitation function of the fluctuation measure $\nu_{\text{\text{dyn}}}$[A,B] was measured, with A and B denoting different particle types. The obtained energy dependence of $\nu_{\text{dyn}}$ agrees with previously published NA49 results on the related measure $\sigma_{\text{dyn}}$. Moreover, $\nu_{\text{dyn}}$ was found to depend on the phase space coverage for [K,p] and [K,$\pi$] pairs. This feature most likely explains the reported differences between measurements of NA49 and those of STAR in central Au+Au collisions.
• ### Centrality dependence of proton and antiproton spectra in Pb+Pb collisions at 40A GeV and 158A GeV measured at the CERN SPS(1009.1747)

Sept. 10, 2010 nucl-ex
The yields of (anti-)protons were measured by the NA49 Collaboration in centrality selected Pb+Pb collisions at 40A GeV and 158A GeV. Particle identification was obtained in the laboratory momentum range from 5 to 63 GeV/c by the measurement of the energy loss dE/dx in the TPC detector gas. The corresponding rapidity coverage extends 1.6 units from mid-rapidity into the forward hemisphere. Transverse mass spectra, the rapidity dependences of the average transverse mass, and rapidity density distributions were studied as a function of collision centrality. The values of the average transverse mass as well as the midrapidity yields of protons when normalized to the number of wounded nucleons show only modest centrality dependences. In contrast, the shape of the rapidity distribution changes significantly with collision centrality, especially at 40A GeV. The experimental results are compared to calculations of the HSD and UrQMD transport models.
• ### Inclusive production of protons, anti-protons and neutrons in p+p collisions at 158 GeV/c beam momentum(0904.2708)

April 17, 2009 hep-ex
New data on the production of protons, anti-protons and neutrons in p+p interactions are presented. The data come from a sample of 4.8 million inelastic events obtained with the NA49 detector at the CERN SPS at 158 GeV/c beam momentum. The charged baryons are identified by energy loss measurement in a large TPC tracking system. Neutrons are detected in a forward hadronic calorimeter. Inclusive invariant cross sections are obtained in intervals from 0 to 1.9 GeV/c (0 to 1.5 GeV/c) in transverse momentum and from -0.05 to 0.95 (-0.05 to 0.4) in Feynman x for protons (anti-protons), respectively. pT integrated neutron cross sections are given in the interval from 0.1 to 0.9 in Feynman x. The data are compared to a wide sample of existing results in the SPS and ISR energy ranges as well as to proton and neutron measurements from HERA and RHIC.
• ### Reheating after Supercooling in the Chiral Phase Transition(nucl-th/9507033)

Sept. 12, 1995 hep-ph, nucl-th
The chirally symmetric quark-gluon plasma produced in energetic heavy-ion collisions is predicted to supercool at the late stages of its evolution. The thermal energy is then transformed into the potential energy associated with an energetically unfavorable field configuration. Since the system is in an unstable state it eventually rolls down to the true minimum of the effective chiral potential. When this motion is described in terms of the sigma-model, we find that the energy of the coherent $\sigma-$field is very efficiently converted into pionic excitations due to anharmonic oscillations around this minimum. The system is expected to partially thermalize before its disintegration.
• ### Wigner Functional Approach to Quantum Field Dynamics(hep-th/9405036)

Sept. 2, 1994 hep-th, hep-ph
We introduce the Wigner functional representing a quantum field in terms of the field amplitudes and their conjugate momenta. The equation of motion for the functional of a scalar field point out the relevance of solutions of the classical field equations to the time evolution of the quantum field. We discuss the field in thermodynamical equilibrium and find the explicit solution of the equations of motion for the so-called roll-over'' phase transition. Finally, we briefly discuss the approximate methods for the evaluation of the Wigner functional that may be used to numerically simulate the initial value problem..


text:
d364. 10637 - Coprimes
Tags :
Accepted rate : 136人/156人 ( 87% ) [非即時]

Content

How many times must a man look up
Before he can see the sky?
Yes, 'n' how many ears must one man have
Before he can hear people cry?
Yes, 'n' how many deaths will it take till he knows
That too many people have died?
The answer, my friend, is blowin' in the wind,
The answer is blowin' in the wind.

Input

Output

Sample Input #1
4
40 6
8 3
5 2
3 1

Sample Output #1
Case 1:
1 1 1 1 1 35
1 1 1 1 5 31
1 1 1 1 7 29
1 1 1 1 11 25
1 1 1 1 13 23
1 1 1 1 17 19
1 1 1 3 5 29
1 1 1 3 11 23
1 1 1 5 9 23
1 1 1 5 11 21
1 1 1 5 13 19
1 1 1 7 11 19
1 1 1 7 13 17
1 1 1 9 11 17
1 1 3 5 7 23
1 1 3 5 11 19
1 1 3 5 13 17
1 1 3 7 11 17
1 1 5 7 9 17
1 1 5 9 11 13
1 3 5 7 11 13
Case 2:
1 1 6
1 2 5
1 3 4
Case 3:
1 4
2 3
Case 4:
3

Hint ：

DFS

Lucky 貓 翻譯

Tags:

[管理者： morris1028(碼畜) ]

 ID User Problem Subject Hit Post Date 25547 allllllan123...(God of Computer...) d364 484 2021-05-31 17:54 24726 hwjnctucsie9...(Barney Huang) d364 517 2021-03-18 16:48


text:
Publications

### Server    domino.mpi-inf.mpg.de

 Proceedings Article, Paper @InProceedings Beitrag in Tagungsband, Workshop
 Author, Editor
 Author(s): Brodal, Gerth Stølting Kaligosi, Kanela Katriel, Irit Kutz, Martin dblp dblp dblp dblp Not MPG Author(s): Brodal, Gerth Stølting Katriel, Irit
 Editor(s): Moshe, Lewenstein Gabriel, Valiente dblp dblp Not MPII Editor(s): Moshe, Lewenstein Gabriel, Valiente
 BibTeX cite key*: Kaligosi2006b
 Title, Booktitle
 Title*: Faster Algorithms for Computing Longest Common Increasing Subsequences Booktitle*: Combinatorial Pattern Matching, 17th Annual Symposium, CPM 2006
 Event, URLs
 Conference URL:: http://www.lsi.upc.edu/~cpm2006/ Downloading URL: http://www.springerlink.com/content/m24861t2425k0251/fulltext.pdf Event Address*: Barcelona, Spain Language: English Event Date* (no longer used): Organization: Event Start Date: 5 July 2006 Event End Date: 7 July 2006
 Publisher
 Name*: Springer URL: http://www.springer.com/west/home?SGWID=4-102-0-0-0 Address*: Berlin, Germany Type:
 Vol, No, Year, pp.
 Series: Lecture Notes in Computer Science
 Volume: 4009 Number: Month: Pages: 330-341 Year*: 2006 VG Wort Pages: ISBN/ISSN: 3-540-35455-7 Sequence Number: DOI:
 (LaTeX) Abstract: We present algorithms for finding a longest common increasing subsequence of two or more input sequences. For two sequences of lengths $m$ and $n$, where $m\ge n$, we present an algorithm with an output-dependent expected running time of $O((m+n\ell) \log\log \sigma + \cleanSort)$ and $O(m)$ space, where $\ell$ is the length of an LCIS, $\sigma$ is the size of the alphabet, and $\cleanSort$ is the time to sort each input sequence. For $k\ge 3$ length-$n$ sequences we present an algorithm which improves the previous best bound by more than a factor $k$ for many inputs. In both cases, our algorithms are conceptually quite simple but rely on existing sophisticated data structures. Finally, we introduce the problem of longest common weakly-increasing (or non-decreasing) subsequences (LCWIS), for which we present an $O(m+n\log n)$-time algorithm for the 3-letter alphabet case. For the extensively studied longest common subsequence problem, comparable speedups have not been achieved for small alphabets. URL for the Abstract: http://dx.doi.org/10.1007/11780441_30 Personal Comments: We present algorithms for finding a longest common increasing subsequence of two or more input sequences. For two sequences of lengths m and n, where m≥n, we present an algorithm with an output-dependent expected running time of and O(m) space, where ℓ is the length of an LCIS, σ is the size of the alphabet, and is the time to sort each input sequence. For k≥3 length-n sequences we present an algorithm which improves the previous best bound by more than a factor k for many inputs. In both cases, our algorithms are conceptually quite simple but rely on existing sophisticated data structures. Finally, we introduce the problem of longest common weakly-increasing (or non-decreasing) subsequences (LCWIS), for which we present an O(m+nlogn)-time algorithm for the 3-letter alphabet case. For the extensively studied longest common subsequence problem, comparable speedups have not been achieved for small alphabets. Download Access Level: Public

 Correlation
 MPG Unit: Max-Planck-Institut für Informatik MPG Subunit: Algorithms and Complexity Group Appearance: MPII WWW Server, MPII FTP Server, MPG publications list, university publications list, working group publication list, Fachbeirat, VG Wort

BibTeX Entry:

@INPROCEEDINGS{Kaligosi2006b,
AUTHOR = {Brodal, Gerth Stølting and Kaligosi, Kanela and Katriel, Irit and Kutz, Martin},
EDITOR = {Moshe, Lewenstein and Gabriel, Valiente},
TITLE = {Faster Algorithms for Computing Longest Common Increasing Subsequences},
BOOKTITLE = {Combinatorial Pattern Matching, 17th Annual Symposium, CPM 2006},
PUBLISHER = {Springer},
YEAR = {2006},
VOLUME = {4009},
PAGES = {330--341},
SERIES = {Lecture Notes in Computer Science},


text:
# How do you find the value of tan(csc^-1(2))?

Apr 24, 2018

$\sqrt{3}$

#### Explanation:

First you need to find the angle that corresponds to a csc value of 2, which is what inverse csc is doing.
${\csc}^{-} 1 = \frac{1}{{\cos}^{-} 1} = 2$
${\cos}^{-} 1 = \frac{1}{2}$
So find when cos is 1/2, which is at $\frac{\pi}{3}$
Then plug in $\frac{\pi}{3}$ for ${\csc}^{-} 1 \left(2\right)$
Evaluate $\tan \left(\frac{\pi}{3}\right)$
$\tan \left(\frac{\pi}{3}\right) = \sqrt{3}$

Apr 24, 2018

Use the identity:

$\tan \left({\csc}^{-} 1 \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} - 1}}$

#### Explanation:

Please see this reference section Relationships between trigonometric functions and inverse trigonometric functions. I am referring you to this section because it contains a table that will help you if your current studies require you to do many problems of this type.

The table gives the following identity:

$\tan \left({\csc}^{-} 1 \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} - 1}}$

Please notice that there is a nice triangle drawing to the right within the table:

Substitute $x = 2$ into the identity:

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{1}{\sqrt{{2}^{2} - 1}}$

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{1}{\sqrt{4 - 1}}$

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{1}{\sqrt{3}}$

Rationalize the denominator:

$\tan \left({\csc}^{-} 1 \left(2\right)\right) = \frac{\sqrt{3}}{3}$


text:
J. Korean Ceram. Soc. > Volume 48(5); 2011 > Article
 Journal of the Korean Ceramic Society 2011;48(5): 390. doi: https://doi.org/10.4191/kcers.2011.48.5.390
 Fabrication of Hydroxyapatite Ceramics to Mimic the Natural Bone Structure Dae-Hee Moon, Su-Chak Ryu Department of Nanomedical Engineering, Pusan National University ABSTRACT The objective of our study was to produce an imitation bone material consisting of hydroxyapatite with a compact and spongy structure. This study shows the ideal content of $SiO_2$ and the sintering temperature to produce imitation bone that has the mechanical properties of natural bone. On the basis of our determination of the ideal conditions, a compact part was produced and its mechanical properties were tested. A compact part made of 0.5 wt% $SiO_2$ and sintered at $1350^{circ}C$ showed excellent mechanical properties. The bioactivity of the compact part under this condition was tested, and it was found to be bioactive. The porous part was produced by controlling the powder size, and the dual structure was manufactured by combining the compact and porous parts. A water permeability test confirmed that the dual structure had an interconnected pore structure. Therefore, this dual-body structure is feasible for use in the creation of implants. Key words: Hydroxyapatite, Dual, Porous, Compact, Implant
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text:
# Splitting a caption and cross-referencing

I'm currently writing my thesis and I'm struggling with my redefinition of the caption command.

I've got large captions which I don't want to show up in my list of figures. Of each caption I want only to use the first sentence. In some captions I would like to refer to other floats using the cleveref package. Below a minimal example is given.

\documentclass{article}
\usepackage{cleveref}
\def\mcaption#1{%
\begingroup
\edef\@tempa{#1}%
\expandafter\endgroup
\expandafter\split\@tempa\relax
}
\def\split#1.#2\relax{\caption[#1]{#1.#2}}
\begin{document}
\listoffigures
\begin{figure}
\centering
Insert figure here
\mcaption{This is a working caption. One can see that ......}
\label{fig:abc}
\end{figure}
\begin{figure}
\centering
Insert figure here
\end{figure}
\end{document}

This fails to produce a PDF with the following error message:

./temp.tex:27: Undefined control sequence. \@cref ...oup =1\count@group =1\def \cref@variant {#1}\newif \if@secondref ... l.27 ... not work. See also Figure \cref{fig:abc}}

./temp.tex:27: Undefined control sequence. \if@secondref

Is it possible to do what I want or should I forgo the idea of referencing other floats in a caption?

• Welcome to TeX.SX! I'd prefer \mcaption{First sentence}{Rest of the caption}: just as easy to type and much easier to parse for TeX. So \newcommand{\mcaption}[2]{\caption[#1]{#1. #2}} Apr 10 '14 at 7:54

You're doing a useless \edef, which is also dangerous if commands such as \textbf appear in the caption text.

Just say

\documentclass{article}
\usepackage{cleveref}

\makeatletter
\def\mcaption#1{\split@caption#1\@nil}
\def\split@caption#1.#2\@nil{\caption[#1]{#1.#2}}
\makeatother

\begin{document}
\listoffigures
\begin{figure}
\centering
Insert figure here
\mcaption{This is a working caption. One can see that ......}
\label{fig:abc}
\end{figure}
\begin{figure}
\centering
Insert figure here
\end{figure}
\end{document}

Better not using \split as a name, because this is a command used by amsmath. However, I'd prefer a simpler approach:

\documentclass{article}
\usepackage{cleveref}

\newcommand\mcaption[2]{\caption[#1]{#1. #2}}

\begin{document}
\listoffigures
\begin{figure}
\centering
Insert figure here
\mcaption{This is a working caption}{One can see that ......}
\label{fig:abc}
\end{figure}
\begin{figure}
\centering
Insert figure here


text:
# Use global/local coordinates in supports settings

Dit artikel is ook beschikbaar in
While defining supports of the model, you will now have an option to choose between the local system of coordinates (LCS) and the global system of coordinates (GCS). The local system is set as default.
$$This aims to remedy some difficulties with supporting the model. The feature is available for steel and concrete members.IDEA StatiCa Member for steel members is available in Enhanced edition of IDEA StatiCa Steel.$$