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https://puzzling.stackexchange.com/questions/66883/how-to-discover-twisty-puzzle-algorithms-that-are-not-commutators
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# How to discover twisty puzzle algorithms that are not commutators?
I am reasonably competent at using commutators to find my own algorithms on twisty puzzles. Many good tutorials for this exist online.
However, there are also algorithms that are not commutators, because they are odd permutations. These are usually used to solve some kind of parity problem or another. Examples include swapping two corners on a 2x2 Rubik's cube, the "J permutation" on the 3x3, the move that solves the last four edges on a 4x4, and the edge parity move on the 4x4.
I have asked several questions recently about specific examples of these, but each example still feels like a special case and I don't feel I have a deep understanding. For commutators the theory is easy to understand and with a bit of practice I can come up with my own easily, even on a puzzle I haven't seen before, odd permutations it seem much more of a black art and I haven't seen any tutorial or explanation of how to come up with algorithms for them. I am asking for any tips or theory that will allow me to construct my own.
I do know that in general one can often perform a single twist (which on a cube-shaped puzzle is a 4-cycle) and then solve the cube again using commutators, but, for me at least, this tends to result in a long and laborious sequence of moves, rather than a simple algorithm. So I am looking for any theory or techniques by which non-commutator algorithms can be constructed.
• Note that d R F' U R' F d' is not an odd permutation of the 4x4 edges. In fact, it is based on the commutator d (R F' U R' F) d' (F' R U F R'), where (R F' U R' F) flips an edge pair in the u/d slice. It is a 3-cycle of edges (assuming d is a move of just the inner slice) - try it out on a solved 4x4x4 and see. Because face turns don't matter yet when you are just pairing up edges, the final 5 face turns can be skipped, in much the same way as you can skip a U turn before/after doing OLL/PLL. – Jaap Scherphuis Jun 6 '18 at 22:46
• @JaapScherphuis thanks, that's very useful information. I had heard it called a parity and assumed it was an odd permutation, but it makes perfect sense that it's a 3-cycle. (Now I also understand why PLL parity on the 4x4 is not really a parity either, since it can be solved with that move.) – Nathaniel Jun 8 '18 at 13:48
• (I've removed that part from the question. For future readers, the comment thread is about an algorithm for pairing up the last two edge pairs on the 4x4.) – Nathaniel Jun 8 '18 at 13:53
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2021-04-16 23:59:29
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https://codegolf.stackexchange.com/tags/puzzle-solver/hot
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# Tag Info
## Hot answers tagged puzzle-solver
31
BBC BASIC, 570 514 490 bytes ASCII Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html 435 bytes tokenised Full program displays an input from L.bmp on the screen, then modifies it to find a solution. *DISPLAY L t=PI/8q=FNa(1) DEFFNa(n)IFn=7END LOCALz,j,p,i,c,s,x,y,m,u,v F.z=0TO99u=z MOD10*100v=z DIV10*100ORIGINu,v F.j=0TO12S.4p=0F.i=j+...
26
C++ - 1123 Since nobody posted any answer so far, I decided to simplify and golf my 2004 solution. It's still far behind the shortest one I mentioned in the question. #include<iostream> #include<vector> #define G(i,x,y)for(int i=x;i^y;i++) #define h(x)s[a[x]/q*q+(a[x]+j)%q-42] #define B(x)D=x;E=O.substr(j*3,3);G(i,0,3)E+=F[5-F.find(E[2-i])];G(i,...
24
Python 1166 bytes A considerable amount of whitespace has been left for the sake of readability. Size is measured after removing this whitespace, and changing various indentation levels to Tab, Tab Space, Tab Tab, etc. I've also avoided any golfing which affected the performance too drastically. T=[] S=[0]*20,'QTRXadbhEIFJUVZYeijf',0 I='FBRLUD' G=[(~i%8,i/...
19
Ruby, 85 bytes f=->l,n,s=n-l.sum-l.size+1{*a,b=l;b&&s>0?(a[0]?1+f[a,n-b-2,s-1]:(n.to_f/b).ceil-1):0} Try it online! Explanation The first step is to establish a recursive divide and conquer strategy to solve subproblems. I will use the variables $l=[l_1,l_2,...,l_x]$ for the list of clues, $x$ for the number of clues and $n$ for the ...
17
The shortest game of halma is 49 moves 49 move solution Proof there is no 48-move solution Code used for this solution The code now supports pass Notice that the 47 move solution in the paper is for the army transfer problem, not for the shortest game of halma I'll hopefully get to doing a proper writeup this weekend
15
C# - 2,098,382 steps I try many things, most of them fail and just didn't work at all, until recently. I got something interesting enough to post an answer. There is certainly ways to improve this further more. I think going under the 2M steps might be possible. It took approx 7 hours to generate results. Here is a txt file with all solutions, in case ...
15
05AB1E, 23 11 8 bytes ΔÍN-;иg= Try it online! Uses 0-based indexing. Explanation: # start from the implicit input Δ # loop forever Í # subtract 2 N- # subtract the current iteration number ; # divide by 2 и # create a list of length x g # get the length of the list = ...
13
Octave, 334 313 bytes Since the challenge may seem a bit daunting, I present my own solution. I did not formally prove that this method works (I guess that will come down to proving that the algorithm will never get stuck in a loop), but so far it works perfectly, doing 100x100 testcases within 15 seconds. Note that I chose to use a function with side ...
12
Ruby, 742 characters r=->y{y.split.map{|x|[*x.chars]}} G=r['UF UR UB UL DF DR DB DL FR FL BR BL UFR URB UBL ULF DRF DFL DLB DBR'] o=r[gets] x=[];[[%w{U UU UUU L LL LLL}+D=%w{D DD DDD},0],[%w{FDFFF RFDFFFRRR}+D,12],[%w{DDDRRRDRDFDDDFFF DLDDDLLLDDDFFFDF}+D,8],[%w{DFLDLLLDDDFFF RDUUUFDUUULDUUUBDUUU}+D,4],[%w{LDDDRRRDLLLDDDRD RRRDLDDDRDLLLDDD LFFFLLLFLFFFLLLF}...
12
C++ - 0.201s official score Using Tdoku (code; design; benchmarks) gives these results: ~/tdoku$lscpu | grep Model.name Model name: Intel(R) Core(TM) i7-4930K CPU @ 3.40GHz ~/tdoku$ # build: ~/tdoku$CC=clang-8 CXX=clang++-8 ./BUILD.sh ~/tdoku$ clang -o solve example/solve.c build/libtdoku.a ~/tdoku$# adjust input format: ~/tdoku$ sed -e "s/...
11
Python - 48 characters exec("".join(map(chr,map(len,' ...
11
Python 2.7: 544 bytes -50% = 272 bytes** import sys;o=''.join;r=range;a=sys.argv[1];a=o([(' ',x)[x in a[12]+a[19]+a[22]] for x in a]);v={a:''};w={' '*4+(a[12]*2+' '*4+a[19]*2)*2+a[22]*4:''} m=lambda a,k:o([a[([0x55a5498531bb9ac58d10a98a4788e0,0xbdab49ca307b9ac2916a4a0e608c02,0xbd9109ca233beac5a92233a842b420][k]>>5*i)%32] for i in r(24)]) def z(d,h): ...
10
C, via the preprocessor I think the ANSI committee made a conscious choice not to extend the C preprocessor to the point of being Turing-complete. In any case, it's not really powerful enough to solve the eight queens problem. Not in any sort of general fashion. But it can be done, if you're willing to hard-code the loop counters. There's no real way to ...
10
Python – 10,800,000 steps As a last-place reference solution, consider this sequence: print "123456" * 18 Cycling through all the colours n times means that every square n steps away will be guaranteed to be of the same colour as the center square. Every square is at most 18 steps away from the center, so 18 cycles will guarantee all the squares ...
10
Python 2, 115 bytes n=input() for F in range(4): t=[F];b=0;exec"x=(-n[b]-sum(t[-2:]))%4;t+=x,;b+=1;"*len(n) if x<1:print t[:-1];break This is the golfed version of the program I wrote while discussing the problem with Martin. Input is a list via STDIN. Output is a list representing the last solution found if there is a solution, or zero if ...
9
Python, 188 bytes This is a further shortened version of my winning submission for CodeSprint Sudoku, modified for command line input instead of stdin (as per the OP): def f(s): x=s.find('0') if x<0:print s;exit() [c in[(x-y)%9*(x/9^y/9)*(x/27^y/27|x%9/3^y%9/3)or s[y]for y in range(81)]or f(s[:x]+c+s[x+1:])for c in'%d'%5**18] import sys f(sys.argv[1])...
9
Pyth, 66 ?"Yes".Am>2sm^-.uk2Cm.Dx"qwertyuiopasdfghjkl*zxcvbnm"b9.5dC,ztz"No Try it here. I was surprised to learn Pyth doesn't have a hypotenuse function, so this will likely be beat by a different language. I'll propose a hypotenuse function to Pyth, so this atrocity won't happen in the future. Explanation I transform the ...
9
Python - 1669 Still pretty long, but fast enough to run the last example in under a second on my computer. It's probably possible to make shorter at the cost of speed, but for now it is pretty much equivalent to the ungolfed code. Example output for last test case: 0 11 1 11 2 11 3 11 4 11 4 10 3 10 2 10 1 10 1 9 2 9 3 9 4 9 4 8 3 8 3 7 4 7 5 7 5 6 5 5 6 5 6 ...
9
Haskell, 242 230 201 199 177 163 160 149 131 bytes import Data.Lists m=map a#b=[x|x<-m(chunk$length b).mapM id$[0,1]<$(a>>b),g x==a,g(transpose x)==b] g=m$list[0]id.m sum.wordsBy(<1) Finally under 200 bytes, credit to @Bergi. Huge thanks to @nimi for helping almost halving the size. Wow. Almost at half size now, partly because of me but ...
9
Python 2, 75 bytes f=lambda n,i=0:n>=i<<i and f(n,i+1)or[min(n,2**j*i-i+j)for j in range(1,i)] Try it online! Explanation: Builds a sequence of 'binary' chunks, with a base number matching the number of cuts. Eg: 63 can be done in 3 cuts, which means a partition in base-4 (as we have 3 single rings): Cuts: 5, 14, 31, which gives chains of 4 1 8 1 ...
8
Java - 2,480,714 steps I made a little mistake before (I put one crucial sentence before a loop instead of in the loop: import java.io.*; public class HerjanPaintAI { BufferedReader r; String[] map = new String[19]; char[][] colors = new char[19][19]; boolean[][] reached = new boolean[19][19], checked = new boolean[19][19]; int[] ...
8
C#, M = 2535 This implements* the system which I described mathematically on the thread which provoked this contest. I claim the 300 rep bonus. The program self-tests if you run it either without command-line arguments or with --test as a command-line argument; for spy 1, run with --spy1, and for spy 2 with --spy2. In each case it takes the number which I ...
8
Python 2, 305 This is the golfed version. It is practically unusable for n > 3, as the time (and space) complexity is like 3n2... actually that may be way too low for the time. Anyway, the function accepts a list of strings. def f(i): Z=range;r=map(__import__('fractions').Fraction,i);R=r[1:];n=len(R);L=[[[1]*n,[0]]];g=0 for m,p in L: for d in([v/3**i%...
8
MATL, 68 59 58 bytes '?'7XJQtX"'s'jh5e"@2#1)t35>)-1l8t_4$h9M)b'nsew'=8M*sJ+XJ+( Try it online! Explanation The map is kept in the bottom of the stack and gradually filled. The current position of the explorer is stored in clipboard J. The map uses matrix coordinates, so (1,1) is upper left. In addition, column-major linear indexing is used. ... 8 JavaScript (ES6), 25 bytes x=>y=>((x<3?x:3)+x)*y/2+1 x=>y=>(x<3?x+x:x+3)*y/2+1 x=>y=>(x<3?x:(x+3)/2)*y+1 x=>y=>(x<3?x:x/2+1.5)*y+1 All of these compute the same value; I can't seem to come up with a shorter formulation. When x is less than 3, you take as much water as you can and walk as far as you can, which is simply ... 8 R, 77 69 bytes -8 bytes thanks to Aaron Hayman pmin(n<-scan(),0:(k=sum((a=2:n)*2^a<=n))+cumsum((k+2)*2^(0:k))+1)[-n] Try it online! Let $k$ be the number of cuts needed; $k$ is the smallest integer such that $(k+1)\cdot2^k\geq n$. Indeed, a possible solution is then to have subchains of lengths $1,1,\ldots,1$ ($k$ times) and \$(k+1), 2(k+...
8
Node.js, 8.231s 6.735s official score Takes the file name as argument. The input file may already contain the solutions in the format described in the challenge, in which case the program will compare them with its own solutions. The results are saved in 'sudoku.log'. Code 'use strict'; const fs = require('fs'); const BLOCK = []; const BLOCK_NDX = [...
7
Here's a C++11 solution without any templates: constexpr int trypos( int work, int col, int row, int rows, int diags1, int diags2, int rowbit, int diag1bit, int diag2bit); constexpr int place( int result, int work, int col, int row, int rows, int diags1, int diags2) { return result != 0 ? result : col == 8 ? work : row == 8 ?...
7
PyPy, 195 moves, ~12 seconds computation Computes optimal solutions using IDA* with a 'walking distance' heuristic augmented with linear conflicts. Here are the optimal solutions: 5 1 7 3 9 2 11 4 13 6 15 8 0 10 14 12 Down, Down, Down, Left, Up, Up, Up, Left, Down, Down, Down, Left, Up, Up, Up 2 5 13 12 1 0 3 15 9 7 14 6 10 11 8 4 Left,...
7
SWI Prolog, 183 characters m(A,A). m([i],[i,u]). m([i,i,i|T],B):-m([u|T],B). m([u,u|T],B):-m(T,B). n([m|A],[m|B]):-(m(A,B);append(A,A,X),m(X,B)). n(A,B):-m(A,B). s(A,B):-atom_chars(A,X),atom_chars(B,Y),n(X,Y). How about some Prolog, (since nobody has answered in 6 months). To run, just use "s(mi,mu)." The code breaks up atoms into chars, then searches for ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2020-11-30 08:23:03
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https://www.gradesaver.com/textbooks/science/physics/college-physics-4th-edition/chapter-22-problems-page-865/58
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## College Physics (4th Edition)
The relative speed between the source and the receiver is $\frac{c}{2}$. Since the observed wavelength is longer than the emitted wavelength, the source and observer are moving farther apart.
Let $f_1$ be the frequency of the source. Let $f_2$ be the observed frequency. We can find the speed $v$ when $\lambda_2 = 2~\lambda_1$: $f_2 = f_1~(1-\frac{v}{c})$ $\frac{c}{\lambda_2} = \frac{c}{\lambda_1}~(1-\frac{v}{c})$ $\frac{c}{2\lambda_1} = \frac{c}{\lambda_1}~(1-\frac{v}{c})$ $\frac{1}{2} = 1-\frac{v}{c}$ $v = \frac{c}{2}$ The relative speed between the source and the receiver is $\frac{c}{2}$. Since the observed wavelength is longer than the emitted wavelength, the source and observer are moving farther apart.
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2020-02-22 08:01:29
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https://hal-iogs.archives-ouvertes.fr/hal-02379601
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# All-Optical Bose-Einstein Condensates in Microgravity
Abstract : We report on the all-optical production of Bose-Einstein condensates in microgravity using a combination of grey molasses cooling, light-shift engineering and optical trapping in a painted potential. Forced evaporative cooling in a 3-m high Einstein elevator results in $4 \times 10^4$ condensed atoms every 13.5 s, with a temperature as low as 35 nK. In this system, the atomic cloud can expand in weightlessness for up to 400 ms, paving the way for atom interferometry experiments with extended interrogation times and studies of ultra-cold matter physics at low energies on ground or in Space.
Document type :
Journal articles
Domain :
Complete list of metadata
https://hal-iogs.archives-ouvertes.fr/hal-02379601
Contributor : Philippe Bouyer <>
Submitted on : Monday, November 25, 2019 - 5:34:28 PM
Last modification on : Wednesday, February 3, 2021 - 12:04:54 PM
### Citation
Gabriel Condon, Martin Rabault, Brynle Barrett, Laure Chichet, Romain Arguel, et al.. All-Optical Bose-Einstein Condensates in Microgravity. Physical Review Letters, American Physical Society, In press, 123, pp.240402. ⟨10.1103/PHYSREVLETT.123.240402⟩. ⟨hal-02379601⟩
Record views
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2021-06-21 19:00:08
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https://gemseo.readthedocs.io/en/stable/_modules/gemseo.utils.data_conversion.html
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gemseo / utils
# data_conversion module¶
Conversion from a NumPy array to a dictionary of NumPy arrays and vice versa.
Classes:
Methods to juggle NumPy arrays and dictionaries of Numpy arrays.
Functions:
flatten_mapping(mapping[, parent_key, sep]) Flatten a nested mapping.
class gemseo.utils.data_conversion.DataConversion[source]
Bases: object
Methods to juggle NumPy arrays and dictionaries of Numpy arrays.
Attributes:
Methods:
array_to_dict(data_array, data_names, data_sizes) Convert an NumPy array into a dictionary of NumPy arrays indexed by names. deepcopy_datadict(data_dict[, keys]) Perform a deep copy of a data mapping. dict_jac_to_2dmat(jac_dict, outputs, inputs, ...) Convert elementary Jacobian matrices into a full Jacobian matrix. dict_jac_to_dict(jac_dict) Reindex a mapping of elementary Jacobian matrices by Jacobian names. dict_to_array(data_dict, data_names) Concatenate some values of a mapping associating values to names. dict_to_jac_dict(flat_jac_dict) Reindex a mapping of elementary Jacobian matrices by input and output names. flat_jac_name(out_name, inpt_name) Concatenate the name of the output and input, with a separator. get_all_inputs(disciplines[, recursive]) Return all the input names of the disciplines. get_all_outputs(disciplines[, recursive]) Return all the output names of the disciplines. jac_2dmat_to_dict(flat_jac, outputs, inputs, ...) Convert a full Jacobian matrix into elementary Jacobian matrices. jac_3dmat_to_dict(jac, outputs, inputs, ...) Convert several full Jacobian matrices into elementary Jacobian matrices. list_of_dict_to_array(data_list, data_names) Concatenate some values of mappings associating values to names. update_dict_from_array(reference_input_data, ...) Update a data mapping from data array and names.
FLAT_JAC_SEP = '!d$_$d!'
static array_to_dict(data_array, data_names, data_sizes)[source]
Convert an NumPy array into a dictionary of NumPy arrays indexed by names.
This allows to convert:
array([1., 2., 3.])
to:
{'x': array([1.])}, 'y': array([2., 3.])}
Parameters
• data_array (numpy.ndarray) – The data array to be converted.
• data_names (Iterable[str]) – The names to be used as keys of the dictionary. The data array must contain the values of these names in the same order, e.g. data_array=array([1.,2.]) and data_names=["x","y"] implies that x=array([1.]) and x=array([2.]).
• data_sizes (Mapping[str, int]) – The sizes of the variables e.g. data_array=array([1.,2.,3.]), data_names=["x","y"] and data_sizes={"x":2,"y":1} implies that x=array([1.,2.]) and x=array([3.]).
Returns
The data mapped to the names.
Raises
ValueError – If the number of dimensions of the data array is greater than 2.
Return type
Dict[str, numpy.ndarray]
Perform a deep copy of a data mapping.
This treats the NumPy arrays specially using array.copy() instead of deepcopy.
Parameters
• data_dict – The data mapping to be copied.
• keys
The keys of the mapping to be considered. If None, consider all the mapping keys.
By default it is set to None.
Returns
A deep copy of the data mapping.
static dict_jac_to_2dmat(jac_dict, outputs, inputs, data_sizes)[source]
Convert elementary Jacobian matrices into a full Jacobian matrix.
Parameters
• jac_dict (Mapping[str, Mapping[str, numpy.ndarray]]) – The elementary Jacobian matrices indexed by the names of the inputs and outputs.
• inputs (Iterable[str]) – The names of the inputs.
• outputs (Iterable[str]) – The names of the outputs.
• data_sizes (Mapping[str, int]) – The sizes of the inputs and outputs.
Returns
The full Jacobian matrix whose first dimension represents the outputs and the second one represents the inputs, both preserving the order of variables passed as arguments.
Return type
numpy.ndarray
static dict_jac_to_dict(jac_dict)[source]
Reindex a mapping of elementary Jacobian matrices by Jacobian names.
A Jacobian name is built with the method flat_jac_name() from the input and output names.
Parameters
jac_dict (Mapping[str, Mapping[str, numpy.ndarray]]) – The elementary Jacobian matrices indexed by input and output names.
Returns
The elementary Jacobian matrices index by Jacobian names.
Return type
Dict[str, numpy.ndarray]
static dict_to_array(data_dict, data_names)[source]
Concatenate some values of a mapping associating values to names.
This allows to convert:
{'x': array([1.])}, 'y': array([2., 3.])}
to:
array([1., 2., 3.])
Parameters
• data_dict (Mapping[str, numpy.ndarray]) – The mapping to be converted; it associates values to names.
• data_names (Iterable[str]) – The names to be used for the concatenation.
Returns
The concatenation of the values for the provided names.
Return type
numpy.ndarray
static dict_to_jac_dict(flat_jac_dict)[source]
Reindex a mapping of elementary Jacobian matrices by input and output names.
Parameters
flat_jac_dict (Mapping[str, numpy.ndarray]) – The elementary Jacobian matrices index by Jacobian names. A Jacobian name is built with the method flat_jac_name() from the input and output names.
Returns
The elementary Jacobian matrices index by input and output names.
Return type
Mapping[str, Mapping[str, numpy.ndarray]]
static flat_jac_name(out_name, inpt_name)[source]
Concatenate the name of the output and input, with a separator.
Parameters
• out_name (str) – The name of the output.
• inpt_name (str) – The name of the input.
Returns
The name of the output concatenated with the name of the input.
Return type
str
static get_all_inputs(disciplines, recursive=False)[source]
Return all the input names of the disciplines.
Parameters
• disciplines (Iterable[MDODiscipline]) – The disciplines.
• recursive (bool) –
If True, search for the inputs of the sub-disciplines, when some disciplines are scenarios.
By default it is set to False.
Returns
The names of the inputs.
Return type
List[str]
static get_all_outputs(disciplines, recursive=False)[source]
Return all the output names of the disciplines.
Parameters
• disciplines (Iterable[MDODiscipline]) – The disciplines.
• recursive (bool) –
If True, search for the outputs of the sub-disciplines, when some disciplines are scenarios.
By default it is set to False.
Returns
The names of the outputs.
Return type
List[str]
static jac_2dmat_to_dict(flat_jac, outputs, inputs, data_sizes)[source]
Convert a full Jacobian matrix into elementary Jacobian matrices.
The full Jacobian matrix is passed as a two-dimensional NumPy array. Its first dimension represents the outputs and its second one represents the inputs.
Parameters
• flat_jac (numpy.ndarray) – The full Jacobian matrix.
• inputs (Iterable[str]) – The names of the inputs.
• outputs (Iterable[str]) – The names of the outputs.
• data_sizes (Mapping[str, int]) – The sizes of the inputs and outputs.
Returns
The Jacobian matrices indexed by the names of the inputs and outputs. Precisely, jac[output][input] is a two-dimensional NumPy array representing the Jacobian matrix for the input input and output output, with the output components in the first dimension and the output components in the second one.
Return type
Dict[str, Dict[str, numpy.ndarray]]
static jac_3dmat_to_dict(jac, outputs, inputs, data_sizes)[source]
Convert several full Jacobian matrices into elementary Jacobian matrices.
The full Jacobian matrices are passed as a three-dimensional NumPy array. Its first dimension represents the different full Jacobian matrices, its second dimension represents the outputs and its third one represents the inputs.
Parameters
• jac (numpy.ndarray) – The full Jacobian matrices.
• inputs (Iterable[str]) – The names of the inputs.
• outputs (Iterable[str]) – The names of the outputs.
• data_sizes (Mapping[str, int]) – The sizes of the inputs and outputs.
Returns
The Jacobian matrices indexed by the names of the inputs and outputs. Precisely, jac[output][input] is a three-dimensional NumPy array where jac[output][input][i] represents the i-th Jacobian matrix for the input input and output output, with the output components in the first dimension and the output components in the second one.
Return type
Dict[str, Dict[str, numpy.ndarray]]
static list_of_dict_to_array(data_list, data_names, group=None)[source]
Concatenate some values of mappings associating values to names.
The names can be either grouped:
[
{'group1':
{'x': array([3.])},
'group2':
{'y': array([1., 1.])}
},
{'group1':
{'x': array([6.])},
'group2':
{'y': array([2., 2.])}
}
]
or ungrouped:
[
{'x': array([3.]), 'y': array([1., 1.])},
{'x': array([6.]), 'y': array([2., 2.])}
]
For both cases, if data_names=["y", "x"], the returned object will be
array([[1., 1., 3.],
[2., 2., 6.]])
Parameters
• data_list (Iterable[Mapping[str, Union[numpy.ndarray, Mapping[str, numpy.ndarray]]]]) – The mappings to be converted; it associates values to names, possibly classified by groups.
• data_names (Iterable[str]) – The names to be used for the concatenation.
• group (Optional[str]) –
The name of the group to be considered. If None, the data is assumed to have no group.
By default it is set to None.
Returns
The concatenation of the values of the passed names.
Return type
numpy.ndarray
static update_dict_from_array(reference_input_data, data_names, values_array)[source]
Update a data mapping from data array and names.
The order of the data in the array follows the order of the data names.
Parameters
• reference_input_data (Mapping[str, numpy.ndarray]) – The reference data to be updated.
• data_names (Iterable[str]) – The names for which to update the data.
• values_array (numpy.ndarray) – The data with which to update the reference one.
Returns
The updated data mapping.
Raises
• TypeError – If the data with which to update the reference one is not a NumPy array.
• ValueError
• If a name for which to update the data is missing from the reference data. * If the size of the data with which to update the reference one is inconsistent with the reference data.
Return type
Dict[str, numpy.ndarray]
gemseo.utils.data_conversion.flatten_mapping(mapping, parent_key='', sep='_')[source]
Flatten a nested mapping.
Parameters
• mapping (Mapping) – The mapping to be flattened.
• parent_key (str) –
The key for which mapping is the value.
By default it is set to .
• sep (str) –
The keys separator, to be used as {parent_key}{sep}{child_key}.
By default it is set to _.
Return type
Dict
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2022-06-27 08:58:49
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https://reducedplanckconstant.wordpress.com/2012/06/12/astronomical-unit-prequel-three-remarkable-measurements/
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# Astronomical Unit: Prequel – three remarkable measurements
I must apologize for the long silence since I last posted. You may remember that I had promised to answer how in ancient days scholars got to know about the astronomical unit or rather the distance between our earth and the sun. Before I answer that, it is important to know how each measurement was made based on other earlier measurements. Therefore before I talk about measuring the sun-to-earth distance, i.e. the
Astronomical Unit, I’ll briefly discuss about other two great measurements. Therefore three historically
great measurements– related to the earth, the moon, and the sun:
The diameter of the earth:
A Greek librarian Eratosthenes (276 – 195 BC)
living in Alexandria, Egypt realized that the sunrays do not fall in the same way in every place on the
earth. This is an indirect proof of the fact that our earth’s surface is rounded rather than being a flat one since in the latter case we should always expect an equal size of an object in different places.
But that is not true and Eratosthenes realized the fact that during the summer solstice (June 20/21), the sun casts a shadow on a pillar in Alexandria whereas there is no shadow in a town named Syene (presently known as Aswan, which is situated at the tropic of cancer) where the sun appears exactly at the zenith at the middle of the day, i.e. it makes a 90 degree angle of incidence. Myth says that there was a well in Syene which used to be perfectly lit by the sun in the summer solstice noon.
Now Eratosthenes measured the diameter of the earth by measuring two things: (1) The angle $\theta$ made to the centre of the earth by Alexandria and Syene (of course he was capable to imagine that our could be a sphere in shape.). (2) The arc made by Alexandria and Syene $s$, i.e. the distance between the two places.
Eratosthenes first measured the angle of incidence in Alexandria by measuring the height and shadow of any vertical object on the ground:
Then he figured out that this angle is the same angle $\theta$ by using the fact that any intercept between two parallel lines always makes the same angle with each of them (equality of alternating internal angles) considering that the sunrays are effectively parallel as they arrive from a very far distance.
Eratosthenes found that the angle is about $7^\circ$.
Now the knowledge of the arc $s$ he gathered from the caravans who used to travel from Syene to Alexandria was 5000 stadia, which is about 800 km. Now the whole circumference of the earth makes $360^\circ$ angle.
Therefore the circumference $S$ he estimated by taking the ratio of these two angles: $\frac{S}{s}=\frac{2\pi}{\theta}$
Thus $S=\frac{2\pi}{\theta}s=\frac{360}\theta(360/7)\times 800\, \text{km}\simeq 40,000\, \text{km}$. And we also get the radius to be $40,000/2\pi\simeq 6366$ km.
These are remarkably quite close to the modern accepted values
(equatorial circumference=40,075.017 km and equatorial radius=6,378.1 km).
Below you can see a beautiful clip from one of the famous Carl Sagan’s Cosmos episodes.
[By the way, Erastothenes’ contribution is not only the estimation of the radius of the earth, mathematicians will never forget the Sieve of Eratosthenes , the first algorithm to generate prime numbers up to any limit. Interested readers can also look for the book The Librarian Who Measured the Earth by Kathryn Lasky.]
The distance to the moon and its size:
Here the credit goes to Aristarchus (310 – 230 BC), another Greek genius. He just waited for a lunar eclipse to happen. What he noticed was that how much time ($t_1$) the moon takes to go into the full eclipsed phase (when it is completely dark) starting from the commencement of the eclipse
and how long ($t_2$) it remains in the total eclipse phase. In the total eclipse phase, the moon spends time inside the main shadow part (umbra) of the earth. Thus the ratio $t_2/t_1$ gives the ratio of the width of the umbra to the diameter of the moon. Aristarchus found that $t_2/t_1\simeq 2.5$. Thus the umbra region in the moon’s path during the lunar eclipse
2.5 times bigger than moon’s diameter.
Now there is one important fact that was not at all overlooked by the Greek scholars: the apparent sizes of the moon and the sun are quite close to each other. This means that if we send the moon
to the distance of the sun keeping the same solid angle created by the moon, the moon will become as big as the sun, i.e. the angles subtended by the moon and the sun on a point on the earth are the same. For this reason, during the total solar eclipse the moon can perfectly fit to block almost all the rays coming from the sun and the umbra tapers to the earth’s surface. The coincidence of same apparent sizes has been beautifully mentioned, in a way of story-telling to the kids, in the film Agantuk by the famous Indian filmmaker Satyajit Ray. The story-teller (acted by Utpal Dutta) says that it is an unresolved magic
[However, he also says that the size of the shadow of the earth on the moon is also the same, and I guess,
many of us will have an objection against such a statement.]
Anyway, back to Aristarchus again.
If the sun and the moon makes same angle, then if we draw a parallel line to one of the edges of the conical umbra, the adjacent cone to the other edge will perfectly fit another moon (see the figure below).
Thus Aristarchus finally concluded that the earth is 2.5+1=3.5 times bigger than the moon.
Now like the moon fits to block the sun, we can use a coin, tape it to a stick and move it away so that it blocks the moon in our vision. If we do so, we shall find that the distance will be roughly about 110 coin diameters away. This again means that if we extend the coin to the distance of the moon keeping its
boundary inside the same solid angle, the coin will reach to the size of the moon. Since the
angle remains the same, we can safely write
$l_{\text{moon}}/l_{\text{coin}}=D_{\text{moon}}/D_{\text{coin}}$
where $l$‘s and $D$‘s denote distances and diameters respectively.
Thus the moon is 110 moon-diameter away from us (since $l_{\text{coin}}=110\, D_{\text{coin}}$). Now Aristarchus already estimated the moon diameter to (1/3.5)th of earth
diameter. Thus the distance to the moon he found:
$l_{\text{moon}}=110/3.5\, D_{\text{earth}}=\frac{110\times 2\times 6366}{3.5}\, \text{km} = 400,148.57\, \text{km}$. Wikipedia says that the value in the apogee (farthest) position 406,700 km, which seems not very bad compared to Aristarchus’ estimation.
Distance to the sun and its size:
Now we cannot do the same coin experiment since it could be dangerous for our eyes. However, we can exploit the knowledge that the sun and the moon make same angle to the earth, i.e. like the moon the sun is 110 sun-diameter away from us. However, the problem is that we do not know the diameter of the sun. Facing the same problem Aristarchus adopted an alternative method to find the distance to the sun. He waited for a situation where the moon appeared exactly as half-full in the day time. The half-moon appears when the sunrays to the moon become perpendicular to the line joining the earth and the moon (see figure below). Thus the earth, moon, and the sun form a right-angled triangle and once the angle subtended at the earth by the moon and the sun ($latex\alpha$ in the figure below) can be measured, the distance to the sun (hypotenuse of the triangle) can be found:
$l_{\text{moon}}/l_{\text{sun}}=\cos\alpha$.
Similar to the method used by Eristothesen, Aristarchus found
the incident angle to be $3^\circ$ and hence
$\alpha=(90-3)^\circ=87^\circ$.
Therefore
$l_{\text{sun}}=l_{\text{moon}}/\cos\alpha=19\, l_{\text{moon}}$
Certainly this is a quite inaccurate value estimated by Aristarchus since it was very difficult to correctly measure $\alpha$ for him (Instead of calculating from the angle of incidence, he
needed to consider the angle made with line passing to the earth’s center). More accurate measurement done in later time showed $\alpha\simeq 89^\circ$, which gives $l_{\text{sun}} \simeq 400\, l_{\text{moon}}=400\times 400,148.57 \mbox{km}=1,60,059,428 \mbox{km}$.
Now once we know the distance, we can find the diameter of the sun by using the magic ratio 110, i.e. diameter of the sun $D_{\text{sun}}=l_{\text{sun}}/110\, \mbox{km}=1.45 \times 10^6\, \mbox{km}$ (modern accepted value $1.392\times10^6$ km).
The first scientifically accurate measurement of the earth-to-sun distance was made by Giovanni Cassini in 1672 (two years before Ole Rømer estimated the speed of light) by parallax measurements. I shall discuss this in my next post.
Image courtesy:
Reference: Conceptual Physics (10 th ed.) by Paul G. Hewitt
## One thought on “Astronomical Unit: Prequel – three remarkable measurements”
1. Pingback: Quora
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2015-11-27 11:47:14
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https://math.stackexchange.com/questions/3286791/about-the-existence-of-primitive-root-modulo-prime
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# About the existence of primitive root modulo prime
I was reading the theorem about the existence of an integer $$t$$, the primitive root modulo prime. The proof seemed a bit confusing. I mean the construction part. Why did not they immediately take $$t = xy$$ instead of $$t = x^{m'}y^{m}$$? I think $$xy$$ also satisfies the requirements. Thanks in advance. Here is the link of the proof:
http://www.math.stonybrook.edu/~scott/blair/Proof_Theorem_5.html#B
• I think you can write down a quick sketch here and underline the problem that you want to ask to be more precise. It would help you to get an answer quickly. :) – Kumar Jul 8 '19 at 14:18
• That's very difficult to read. I suspect though that $xy$ need not necessarily work. – Lord Shark the Unknown Jul 8 '19 at 14:32
• I will make things easier here. So I have an integer $x$ and $d$ is the smallest integer such that $x^{d} \equiv 1$ (mod p). We have another integer $y$ and the smallest integer $e$ such that $y^{e} \equiv 1$ (mod p). Then We want to construct an integer $t$ for which$f = LCM(d, e)$ is the smallest integer such that $t^{f} \equiv 1$ (mod p). That is what I got from there. So what if we take $t = xy$? $x^{k} \equiv 1$ (mod p) if and only if $k$ is a multiple of $d$ and $y^{l} \equiv 1$ (mod p) if and only if $l$ is divisible by $e$. So $LCM(d, e)$ really satisfies the requirements. – shota kobakhidze Jul 8 '19 at 14:47
Generally it is not true that in an abelian group that if $$\,x,y\,$$ have order $$\,j,k\,$$ then $$xy$$ has order $$\,{\rm lcm}(j,k),\,$$ e.g. consider the case $$\,y = x^{-1}.\,$$ But it is true that there exists some element of order $$\,{\rm lcm}(j,k),\,$$ and this is what is proved there (see here for a few other proofs of order lcm-closure)
Remark Their proof can be simplified. By here: if $$\,x,y\,$$ have order $$\,d,e\,$$ then there are coprime $$\,m',m\in \Bbb N\,$$ with $$\,(d,e)={m'}\,{m},\ (d/m',\,e/m)=1\,$$ so $$\,x^{\large m'},\, y^{\large m}$$ have coprime orders $$\,d/m',\, e/m\,$$ therefore their product has order $$\ (d/m')(e/m) = de/(d,e) = {\rm lcm}(d,e)$$.
Unlike many proofs, the linked proof does not require expensive prime factorization. Instead it employs only gcds so it yields an efficient algorithm to compute $$\,m',m.$$
• Note: I checked the outline of the cited proof but I did not verify its correctness. – Bill Dubuque Jul 8 '19 at 18:52
• Thanks for the response. So, as I understood their proof is complicated because of the formal correctness reasons from the point of view of group theory? However, in this case $xy$ works but such reasoning could fail in some similar situations? – shota kobakhidze Jul 8 '19 at 20:43
• @shotakobakhidze Why do you believe that $xy$ works? – Bill Dubuque Jul 8 '19 at 20:59
• Yes, you are right. I realized that it is incorrect if in some step I have only left $y$ which is inverse of $x$ and that fails as you mentioned. Then the answer for them is not $LCM(d, e)$ but 1. Nice point. – shota kobakhidze Jul 8 '19 at 21:13
• @shota kobakhidze: The author avoids the issue of $y$ being the inverse of $x$ by choosing $y$ so that the order of $y$ doesn't divide $d$. Since the equation $x^d=1$ has at most $d$ solutions, such a choice is always possible (assuming $d < p-1$, we have $d\le{\large{\frac{p-1}{2}}}$). – quasi Jul 8 '19 at 21:28
Let $$x\in \{1,...,p-1\}$$, and let $$d$$ be the order of $$x$$.
If $$d=p-1$$, then $$x$$ is a primitive root, and we're done.
Suppose $$d < p-1$$.
The plan is to find some element of $$t\in\{1,...,p-1\}$$ whose order exceeds $$d$$, and then iterate, using $$t$$ as the new $$x$$.
As the author argues, there exists $$y\in\{1,...,p-1\}$$ whose order doesn't divide $$d$$.
Let $$e$$ be the order of $$y$$.
If $$e > d$$, we can let $$t=y$$.
Since $$e\not\mid d$$, we can't have $$e=d$$.
Suppose $$e < d$$.
Your claim is that we can let $$t=xy$$.
Unfortunately, this doesn't always work.
As you correctly observed, since $$e\not\mid d$$, we get $$\text{gcd}(d,e) < e$$, hence $$\text{lcm}(d,e) = \frac{de}{\text{gcd}(d,e)} = d\left(\frac{e}{\text{gcd}(d,e)}\right) > d$$ Let $$f$$ be the order of $$xy$$.
Clearly $$f{\,|\,}\bigl(\text{lcm}(d,e)\bigr)$$.
However, noting Bill Dubuque's post, and correcting my earlier answer, it's not automatic that $$f=\text{lcm}(d,e)$$.
In fact, we can't even claim $$f > d$$.
As an example, letting $$p=31,x=7,y=23$$, we get
• $$x$$ has order $$d=15$$.$$\\[4pt]$$
• $$y$$ has order $$e=10$$.$$\\[4pt]$$
• $$xy$$ has order $$f=6$$.
This shows that your idea of using $$xy$$ for the next iteration doesn't always work.
• Well, the main idea is to show that $f>d$. Then we can take $t$ instead of $x$ do the same things and got another $t'$ with $f'>f$. finally we will get the degree equal to $p - 1$ it is the maximal possible value for that. So is it correct in this case? – shota kobakhidze Jul 8 '19 at 15:06
• But the point is that for the originally chosen $x,y$, you don't know that $xy$ is a primitive root. – quasi Jul 8 '19 at 15:08
• I don't say that $xy$ will be a primitive root. I just say that $xy$ has a higher minimal degree $f$ such that $xy^{f} \equiv 1$ (mod p) than $x$. – shota kobakhidze Jul 8 '19 at 15:13
• $LCM(d, e) * GCD(d, e) = de$ we have $GCD(d, e) < e$ , so $LCM(d, e) > d$. If $LCM(d, e)$ is not equal to $p - 1$, I can take $t$ in place of $x$. why can I do it again this for $t$? the same reason why I could do for $x$. I just don't see what can interfere with this. – shota kobakhidze Jul 8 '19 at 15:21
• – shota kobakhidze Jul 8 '19 at 15:24
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2020-01-24 14:52:00
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https://www.tutorke.com/lesson/5024--a-one-of-the-causes-of-energy-loss-in-a-transformer-is-heating-in-the-coils-when-current-flows.aspx
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# Form 4 Physics electromagnetic induction topical questions and answers
(a) One of the causes of energy loss in a transformer is heating in the coils when current flows. State:
(i) The reason why the current causes heating.
(ii) How the heating can be minimized.
(b) The input voltage of a transformer is 240V and its output is 12v. When an 80W bulb is connected across the secondary coil, the current in the primary coil is 0.36A.
Determine:
(i) The ratio N_p/N_s of the transformer , (where Np is the number of turns in the primary coil and Ns is the number of turns in secondary coil)
(ii) The power input of the transformer.
(iii) The power output of the transformer.
(iv) The efficiency of the transformer.
(7m 37s)
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2021-10-18 10:50:27
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http://math.stackexchange.com/questions/122147/how-to-solve-binomn122-binomn22-3-binomn32-4-binomn42?answertab=oldest
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# How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$?
I have tried something to solve the series $$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$ My approach is : $$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n.$$ Differentiating the above equation $$n(1+x)^{n-1} = \binom{n}{1} + \binom{n}{2}x + \cdots + n\binom{n}{n}x^{n-1}$$
Also, $$\left(1+\frac{1}{x}\right)^n =\binom{n}{0} + \binom{n}{1}\frac{1}{x} + \binom{n}{2}\left(\frac{1}{x}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{x}\right)^n$$ Multiplying above two equation I get, \begin{align*} &{n(1+x)^{n-1}\left(1 + \frac{1}{x}\right)^n}\\ &\quad= \left( \binom{n}{1}^2 + 2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2 + \cdots + n\binom{n}{n}^2\right)\left(\frac{1}{x}\right) + \text{other terms} \end{align*}
So I can say that coefficient of $\frac{1}{x}$ in expansion of $n(1+x)^{n-1}(1+\frac{1}{x})^n$ will give me the required answer.
Am I doing it correct,please correct me if I'm wrong ?
If I'm right,please tell me how to calculate the coefficient of $\frac{1}{x}$ ?
Based on the answers,I tried to implement the things in a C++ code.
I tried implementing the code using extended euclidean algorithm so that the problem of truncated division can be eliminated but still not abled to figure out why am I getting wrong answer for n>=3. This is my updated code : http://pastebin.com/imS6rdWs I'll be thankful if anyone can help me to figure out what's wrong with this code.
Thanks.
Solution:
Finally abled to solve the problem.Thanks to all those people who spent their precious time for my problem.Thanks a lot.This is my updated code :
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There is the relationship $$\sum_{k=0}^n \binom{n}{k}^2 x^k=(1-x)^n P_n\left(\frac{1+x}{1-x}\right)$$ where $P_n(z)$ is the Legendre polynomial... – Guess who it is. Mar 19 '12 at 17:12
Looks a bit cumbersome that way. A hint: consider also the sum $\sum_{k=0}^n (n-k)\binom{n}{k}^2$. (I expanded this comment in my answer). – Thomas Belulovich Mar 19 '12 at 17:15
@J.M. :Please explain it in detail.I'm not getting it. – code_hacker Mar 19 '12 at 17:16
Yet again we see the promiscuous use of the word "solve". "Evaluate" is an appropriate word here; "solve" is not. One solves equations; one solves problems. One evaluates expressions. – Michael Hardy Mar 19 '12 at 18:09
@code_hacker: the two sums are equal, and are therefore equal to the means of the two ways to write it. – Najib Idrissi Mar 19 '12 at 19:24
Building on the excellent work of @thomas-belulovich & @robjohn, but addressing your subsequently revealed goal....
You want $$T(n) =S(n)-m\left\lfloor\frac{S(n)}m\right\rfloor =S(n)\text{ mod }m \qquad\text{for}\qquad S(n) =\frac{n}{2}{2n\choose n}.$$ Note that $$\frac{S(n)}{S(n-1)} =\frac{n}{n-1}\cdot\frac{2n(2n-1)}{n^2} =2\frac{2n-1}{n-1}.$$ A brute force approach to this might be to calculate $$R(k)=2(2k-1)(k-1)^{-1}\pmod m$$ for each $1<k\le n$ and then multiply modulo $m$ termwise to obtain $$T(n)=S(1)\prod_{k=2}^nR(k).$$ Calculating each $R(k)=(4k-2)x$ requires $O(\log k)$ divisions using the extended Euclidean algorithm to find an $x$ so that $$(k-1)x+my=1.$$
Extended Euclidean Algorithm: Given nonzero $a,b\in\mathbb{Z}$, find $x,y\in\mathbb{Z}$ so that $ax+by=\text{gcd}(a,b)$. (Adapted from David Joyner's book.)
1. Set $\overline{u}=\langle{u_0,u_1,u_2}\rangle$ $\leftarrow\langle{a,1,0}\rangle$ and $\overline{v}\leftarrow\langle{b,0,1}\rangle$ (vectors in $\mathbb{Z}^3$). Set $\overline{v}\leftarrow-\overline{v}$ if $b<0$.
2. While $v_0 \ne 0$, repeat:
3. $\qquad$ Calculate the quotient $q=\text{quo}(u_0,v_0)=u_0-v_0\left\lfloor\frac{u_0}{v_0}\right\rfloor$.
4. $\qquad$ Set $\quad\overline{w}=\overline{u}-q\overline{v},\quad$ then $\quad\overline{u}=\overline{v},\quad$ and then $\quad\overline{v}=\overline{w}.\quad$
5. Return $a=u_0$, $x=u_1$, and $y=u_2$.
This is doable in not too many lines of C code, and it works as long as $m=10^9+7$ has no (prime) factors $p\le n$ (in fact, this $m$ is prime). If you needed a more efficient algorithm and $m$ were a composite product of distinct primes (which it isn't), you could use the prime factorization of $m$ and a nice fact about binomial coefficients modulo primes [Lukas 1878] to separately calculate residues $${a\choose b}\equiv{[a/p]\choose[b/p]}{a\mod p\choose b\mod p}\pmod p$$ modulo each factor $p$, and then recover $T(n)$ using the Chinese Remainder Theorem.
Here's a few pre-computed values: $$\matrix{ k& n=10^k &{2n\choose n}\text{ mod }m &T(n) \\0& 1 &2 &1 \\1& 10 &184756 &923780 \\2& 100 &407336795 &366839610 \\3& 1000 &72475738 &237868748 \\4& 10000 &703593270 &966325381 \\5& 100000 &879467333 &366342189 \\6& 1000000 &192151600 &799327475 }$$
If you want to find an efficient method to solve this problem, you'll probably want to look at the works of Donald Knuth, Andrew Granville, and Richard Stanley, who also has compiled outstanding lists of bijective proof problems and characterizations of the Catalan numbers $C_n=\frac1{n+1}{2n\choose n}$, to which our $S(n)$ are closely related since $S(n)={n+1\choose2}C_n$.
One might be tempted to try to shorten the computation using Wilson's theorem, $$p\text{ prime} \quad\implies\quad (p-1)!\equiv-1\pmod p$$ Morley's (1895) congruence, $$p=2n+1\text{ prime} \quad\implies\quad {2n\choose n}\equiv(-1)^n4^{2n}\pmod{p^3}$$ and/or Kummer's theorem using enough "landmark" primes near $n$ and $2n$, with extended Euclidean algorithm for inverses and the CRT (there's the catch!) to get the final result. For example, here are some prime pairs $q_i=2p_i+1$ near $n=10^6$ and $2n$: $$\matrix{ i & p_i-n & q_i-2n \\ 1 & -251 & -501 \\ 2 & -191 & -381 \\ 3 & 151 & 303 \\ 4 & 193 & 387 \\ }$$ From Wilson's theorem for odd primes $q$, grouping $(q-1)!=(1\cdots\frac{q-1}2)(\frac{q+1}2\cdots q-1)$ and noting that the second term is $(-1)^\frac{q-1}2$ times the first term modulo $q$, we find that $$\left(\left(\tfrac{q-1}{2}\right)!\right)^2 \equiv(-1)^{\frac{q+1}2} \pmod q$$ so that for prime pairs $q_i=2p_i+1$, $${2p_i\choose p_i}\equiv(-1)^{p_i}=-1\pmod{q_i}.$$ Thus we can compute (using the extended Euclidean algorithm for the inverses of $k$ modulo $q_i$) $${2n\choose n}\equiv-\prod_{k=p_i+1}^n\frac{4k-2}{k} \pmod{q_i}$$ for $i=1,2$ above. However, we cannot use the CRT to get $T(n)$. We would have to have enough prime pairs to reconstruct $S(n)$, and then compute its residue at the end. Since the central binomial coefficient is approximately $${2n\choose n}\approx\frac{4^n}{\sqrt{\pi n}}\left(1-\frac1{nc_n}\right),\qquad c_n\in(8,9)$$ we would need about 96 thousand pairs (p,q), making this approach infeasible.
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Based on the recursive formula : S(n)/S(n-1) = (4n-2)/(n-1) why shouldn't I pre compute S(n) for all n upto 10^6.Can I do like that ? – code_hacker Mar 19 '12 at 19:51
Of course, you can precompute these if it suits your needs. Please not I made some corrections to the ratio $S(n)/S(n-1)$ and hence also (my) $R(n)$. – bgins Mar 19 '12 at 20:41
Wow I learnt more from this answer. My +1 for this. – Kirthi Raman Mar 19 '12 at 21:16
@bgins : I tried to find the multiplicative inverse using extended euclidean algorithm but it's giving me wrong answer for n>=3 In fact the corresponding x(multiplicative inverse) at every step is very large.Can you please tell me where am I doing wrong ? One thing I'm sure of is that my extended_euclid() function is correct as I have tested and implemented it on the description and example of wikipedia.Thanks in advance. Link to the updated code : pastebin.com/imS6rdWs – code_hacker Mar 20 '12 at 7:55
You have one or two issues left. Your xgcd needs to use signed types since you are subtracting. Also, as a general purpose routine, it's not returning inverses when the gcd is greater than one (which shouldn't be a problem here since $m$ is prime and $0<n<m$). Do you really need this in C++? Why not try sage or sage online? Type xgcd? or inverse_mod? for example (and binomial(2*n,n) % m for a direct, costly, arbitrary precision calculation). Although right now the public server seems to be loaded down. – bgins Mar 20 '12 at 13:09
This kind of looks like you want to appeal to Vandermonde's convolution, or at least the method you'd use to prove it. It can be applied directly as follows:
Let $S = \sum\limits_{k=0}^n k\binom{n}{k}^2$ be the sum we want to compute. Note that $S = \sum\limits_{k=0}^n (n-k) \binom{n}{n-k}^2 = \sum\limits_{k=0}^n (n-k) \binom{n}{k}^2$. Therefore $2S = n\sum\limits_{k=0}^n \binom{n}{k}^2 = n \binom{2n}{n}$. Then
$$S = \frac{n}{2}\binom{2n}{n}.$$
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It looks great but I have one problem with this result.Actually I need to write a code in C to calculate S(result) where n can be as large as 10^6.Although I'm asked to print S % mod as final output where mod=10^9+7 but still calculating S which is derived will definitely overflow.So Can you help me on that part ? – code_hacker Mar 19 '12 at 17:35
First recall that the coefficient of $x^n$ in $(1+x)^n(1+x)^n=(1+x)^{2n}$ implies \begin{align} \sum_{k=0}^n\binom{n}{k}^2 &=\sum_{k=0}^n\binom{n}{k}\binom{n}{n-k}\\ &=\binom{2n}{n}\tag{1} \end{align} and then note that \begin{align} \sum_{k=0}^nk\binom{n}{k}^2 &=\sum_{k=0}^nk\binom{n}{n-k}^2\\ &=\sum_{k=0}^n(n-k)\binom{n}{k}^2\tag{2} \end{align} Adding the first and last parts of $(2)$ yields \begin{align} 2\sum_{k=0}^nk\binom{n}{k}^2 &=n\sum_{k=0}^n\binom{n}{k}^2\\ &=n\binom{2n}{n}\tag{3} \end{align} Therefore, $$\sum_{k=0}^nk\binom{n}{k}^2=\frac{n}{2}\binom{2n}{n}\tag{4}$$
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It looks great but I have one problem with this result.Actually I need to write a code in C to calculate S(result) where n can be as large as 10^6.Although I'm asked to print S % mod as final output where mod=10^9+7 but still calculating S which is derived will definitely overflow.So Can you help me on that part ? – code_hacker Mar 19 '12 at 17:35
@code_hacker: This identity might help \begin{align} \binom{2n}{n} &=\frac{2n(2n-1)}{n^2}\binom{2(n-1)}{n-1}\\ &=\frac{4n-2}{n}\binom{2(n-1)}{n-1} \end{align} – robjohn Mar 19 '12 at 17:47
@code_hacker, this may be useful. – Antonio Vargas Mar 20 '12 at 0:54
First we address the overflow issue. Note that $10^9+7$ is relatively prime to all the numbers that come up in a naive calculation of $\binom{2n}{n}$. Indeed $10^9+7$ happens to be prime. So when we are calculating, we can always reduce modulo $10^9+7$ as often as necessary to prevent overflow.
Now to the identity. We have $n$ boys and $n$ girls. We want to choose $n$ people. The number of ways this can be done is $\binom{2n}{n}$. But we can choose $0$ boys and $n$ girls, or $1$ boy and $n-1$ girls, and so on. We can choose $k$ boys and $n-k$ girls in $\binom{n}{k}\binom{n}{n-k}$ ways, or equivalently in $\binom{n}{k}^2$ ways. This gives the standard derivation of the identity $$\binom{2n}{n}=\sum_{k=0}^n \binom{n}{k}^2.$$ Note now that the $\binom{2n}{n}$ choices are all equally likely. The expected number of boys is, by symmetry, equal to $\frac{n}{2}.$ But the probability that there are $k$ boys is $\frac{\binom{n}{k}^2}{\binom{2n}{n}}$, and therefore the expected number of boys is $$\sum_{k=0}^n k\frac{\binom{n}{k}^2}{\binom{2n}{n}}.$$ The term corresponding to $k=0$ is $0$, so can be omitted, and we get $$\sum_{k=1}^n k\frac{\binom{n}{k}^2}{\binom{2n}{n}}=\frac{n}{2},$$ which is essentially our identity.
Remark: There is a very nice book on bijective proofs called Proofs that Really Count. A title that so far doesn't seem to have been used is Mean Proofs.
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This is the way to do things, and will definitely run fast enough. (You will have to implement a function that computes multiplicative inverse mod p, but that isn't terrible and runs very quickly.) – Thomas Belulovich Mar 19 '12 at 18:45
I have implemented the things in a C++ code.This is the link to the code pastebin.com/qxQsZyz2 but it's giving me wrong answer when I tried to submit it on online judge.Can any one please tell me where am I doing wrong ? – code_hacker Mar 19 '12 at 19:05
You're dividing by $i$ on line $17$, which is a truncated division. See Thomas's comment and my post about computing inverses modulo $m$ (my post) or mod (your OP notation). – bgins Mar 19 '12 at 19:28
Nice expected value argument! – Thomas Belulovich Mar 20 '12 at 18:41
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2015-05-25 12:14:58
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https://cms.math.ca/cjm/msc/11T23?fromjnl=cjm&jnl=CJM
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Canadian Mathematical Society www.cms.math.ca
location: Publications → journals
Search results
Search: MSC category 11T23 ( Exponential sums )
Expand all Collapse all Results 1 - 4 of 4
1. CJM Online first
Macourt, Simon; Shkredov, Ilya D.; Shparlinski, Igor E.
Multiplicative Energy of Shifted Subgroups and Bounds On Exponential Sums with Trinomials in Finite Fields We give a new bound on collinear triples in subgroups of prime finite fields and use it to give some new bounds on exponential sums with trinomials. Keywords:exponential sum, sparse polynomial, trinomialCategories:11L07, 11T23
2. CJM Online first
Ha, Junsoo
Smooth Polynomial Solutions to a Ternary Additive Equation Let $\mathbf{F}_{q}[T]$ be the ring of polynomials over the finite field of $q$ elements, and $Y$ be a large integer. We say a polynomial in $\mathbf{F}_{q}[T]$ is $Y$-smooth if all of its irreducible factors are of degree at most $Y$. We show that a ternary additive equation $a+b=c$ over $Y$-smooth polynomials has many solutions. As an application, if $S$ is the set of first $s$ primes in $\mathbf{F}_{q}[T]$ and $s$ is large, we prove that the $S$-unit equation $u+v=1$ has at least $\exp(s^{1/6-\epsilon}\log q)$ solutions. Keywords:smooth number, polynomial over a finite field, circle methodCategories:11T55, 11D04, 11L07, 11T23
3. CJM 2005 (vol 57 pp. 338)
Lange, Tanja; Shparlinski, Igor E.
Certain Exponential Sums and Random Walks on Elliptic Curves For a given elliptic curve $\E$, we obtain an upper bound on the discrepancy of sets of multiples $z_sG$ where $z_s$ runs through a sequence $\cZ=$$z_1, \dots, z_T$$$ such that $k z_1,\dots, kz_T$ is a permutation of $z_1, \dots, z_T$, both sequences taken modulo $t$, for sufficiently many distinct values of $k$ modulo $t$. We apply this result to studying an analogue of the power generator over an elliptic curve. These results are elliptic curve analogues of those obtained for multiplicative groups of finite fields and residue rings. Categories:11L07, 11T23, 11T71, 14H52, 94A60
4. CJM 2003 (vol 55 pp. 225)
Banks, William D.; Harcharras, Asma; Shparlinski, Igor E.
Short Kloosterman Sums for Polynomials over Finite Fields We extend to the setting of polynomials over a finite field certain estimates for short Kloosterman sums originally due to Karatsuba. Our estimates are then used to establish some uniformity of distribution results in the ring $\mathbb{F}_q[x]/M(x)$ for collections of polynomials either of the form $f^{-1}g^{-1}$ or of the form $f^{-1}g^{-1}+afg$, where $f$ and $g$ are polynomials coprime to $M$ and of very small degree relative to $M$, and $a$ is an arbitrary polynomial. We also give estimates for short Kloosterman sums where the summation runs over products of two irreducible polynomials of small degree. It is likely that this result can be used to give an improvement of the Brun-Titchmarsh theorem for polynomials over finite fields. Categories:11T23, 11T06
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2017-11-22 22:00:44
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https://www.gamedev.net/forums/topic/642914-sdl-regulating-frame-rate/
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Public Group
# SDL Regulating frame rate
This topic is 1870 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hey
I have followed most the beginner tutorials out there I can find on SDL and am finally getting around to making my first game, something I want to include now is going from beginner standard to industry standard and the first thing is moving from controlling frame rate by ticks and start using time to do this as I have seen lots of comments with people saying this is much better.
I cannot however find anything like out there that explains how to do this so am posting here for a basic description on how to do it and hopefully a code snippet.
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Didn't really understand what you want... Do you want to limit your game frame rate, for instance 60 frames per second? Or could it be that you want to control your game by time (which is the standard), for instance, the movement of a character for 2 seconds be the same for someone running at 30 FPS and other running as 200 FPS?
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If you have a good level of understanding with what you have learnt so far, and feel comfortable with your language, these two articles provide quite detailed explanations of what (I assume) you would like to learn.
They cover the topics of using delta time in your games, implementing a fixed time-step and rendering using interpolation. I have read through them a few times in the past but have yet to put them into practice (studying Maths in my spare time atm). The logic seems quite sound but I can't recommend from a practical perspective.
I hope this helps,
Stitchs.
EDIT: I just realised I never linked the actual articles, I do apologise:
http://gafferongames.com/game-physics/fix-your-timestep/
http://www.koonsolo.com/news/dewitters-gameloop/
Edited by stitchs
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See Lazy Foo's tutorials. First this one, then this one. Might want to look at this article also.
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Eh so at the moment I am using the way in which you use GetTicks and then work out if its going too fast and then delay it if it is, the way the above tutorials show it.
I have read in quite a few places that its best to handle your animation and stuff like that using time passed instead of frames passed so I want to do it this way but cannot find anything on it.
Is that clearer?
Stiches you didnt post the articles in your post, I would love to see them though!
Thanks again
Edited by Beshon
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GetTicks is the amount of millliseconds since the program started.
if you in your update loop save the delta
int time = GetTicks();
int delta = time - lastTime;
lastTime = time; //last time is a variable that you save
//move something 15 units per second in x
float deltaSeconds = 0.01f * delta; //ms to sec
myObject.x += 15 * deltaSeconds ;
If you hve low fps delta time will be high, if the fps is high delta time will be low. Using delta time means the the the movent of myObject will be frame rate independatant.
Edited by HermanssoN
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Eh so at the moment I am using the way in which you use GetTicks and then work out if its going too fast and then delay it if it is, the way the above tutorials show it.
That's the way the first tutorial I linked to shows it, yes.
I have read in quite a few places that its best to handle your animation and stuff like that using time passed instead of frames passed so I want to do it this way but cannot find anything on it.
That's the way the second tutorial I linked to does it.
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Thanks for both your posts, I finally have something to go on! :)
I did go through that tutorial Servant of the Lord but it confused me because as far as I can tell the timer was not delaying anything, it was just counting and resetting.
I am most definitely wrong, I just didn't understand it sorry
Thanks again!
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No problem. If you don't understand any of it, let us know what parts and we can explain it.
I personally like to keep my 'deltas' as floats, with 1.0 being one second of time - I find them easier to manage that way.
amountToMoveThisFrame = (MovementSpeedOverOneSecond * deltaTime);
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Oh so delay isn't actually used, that makes much more sense, thanks so much!
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2018-06-24 02:11:28
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https://s42928.gridserver.com/86yerx/total-efficiency-of-fan-5b28db
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The fan performance curve (see. ) The fan peak efficiency shall be calculated from the fan (total) pressure. Figure 1: Pressure vs. Flow Curve – 120mm Axial Fan. Fan total efficiency gives higher number while static efficiency calculates a lower number. 6.5.3.1.3 Fan Efficiency. For detailed engineering - use manufacturers specifications for actual fans. The fan efficiency is the ratio between power transferred to airflow and the power used by the fan. 2), selecting the fan-based upon free air performance would favor the high flow fan option. Outlet velocity is calculated by divid-ing the CFM by outlet area. Total Pressure - Sum of the static pressure and velocity pressures. 2. Together, state and local agencies spend annually about $50-70 billion on energy-related products and$12 billion on energy bills. The overall efficiency can be expressed as: η = η h η m η v (4) where. O.D. By definition, an FEG label is assigned based on fan component performance data in accordance with a standard like AMCA 210. Table 1: Axial Fans Typical Peak Efficiency, Fan Selection Taking Account of Efficiency. The total pressure is the summation of static pressure and outlet velocity pressure. ISO 12759 "Fans – Efficiency classification for fans", AMCA 205 "Energy Efficiency Classification for fans", es: ventilador de consumo de energÃa eficiencia. 2.2. Fig 2shows the pressures through a fan, each of which is described below: Inlet Pressure; is the static pressure on the inlet side of the fan. 3[14].The fan total efficiency is the ratio of the fan power output to fan power input, as shown in Eq. Customized for your requirements, EMAX4 decreases noise by 2 to 3 dB and reduces energy consumption in air-cooled condensers, commercial refrigeration, chillers, cooling towers, evaporators and more. Outlet velocity pressure does not contribute to a fan’s ability to remove system heat energy; therefore it’s not normally included in fan efficiency calculations. So for an airflow increase of a little over 18% the power needed has risen by nearly 45%. ) Optimize efficiency of the fan system, including the fan, drive, motor, and variable speed drive (i.e. The long term benefit of specifying high-efficiency fans is a reduction of ownership costs. This is the ratio of the power delivered to the air to the electrical input power at maximum flow expressed as a percentage. These include: the size of the fan; the size of the room; the distance the fan is from the ceiling Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! NMB Technologies Corporation is a MinebeaMitsumi Group company; the world’s largest manufacturer of NMB miniature ball bearings and a volume leader in the design and manufacturing of precision electro-mechanical components, backlight and LED lighting products and advanced technology solutions for smart cities, medical automotive and industrial markets. q = volume flow out of the pump or fan (m 3 /s) q l = leakage volume flow (m 3 /s) Total Loss and Overall Efficiency. Negative static pressure is what all other components in the airflow path create as they resist air movement. Calculating Fan Efficiency As with any energy converter, efficiency is the ratio of input and output power:- Fan efficiency = P out / P in Fan input power (P in) is:- P in (Watts) = V x I Fan output power (P … Using high-efficiency fans has a cascade effect on system design. The power used by the fan can be expressed as: P = dp q / μf (3). Mechanical efficiency is a ratio of the total fan power out-put to the power supplied to the fan. We know that the ever-changing challenges that face automotive product design engineers mean that high quality, high performance components are always in demand. Google use cookies for serving our ads and handling visitor statistics. EMAX4™ | Up to 77 Percent Total-Efficiency Fan Leading the Efficiency Revolution. The average total efficiency of all the fans was only 33% (Figure 1). Fan efficiency does not take into effect the efficiency of the drive (belt drive) or the motor. Fan total efficiency (%) Enter the product of the fan motor and impeller efficiency of the supply fan. of H2O. The static efficiency of a fan - Is the mechanical efficiency multiplied by the ratio of the static pressure to the total pressure or e s = e t × P s /P t. The static pressure (P s) of the fan - Is the total pressure (P tf) diminished by the fan velocity pressure (P v). fan rotational speed using Eq. The losses in a pump or fan converts to heat that is transferred to the fluid and the surroundings. The total efficiency is calculated using the traditional airflow, pressure, and input power as measured per AMCA Standard 210. Accompanying this trend is the need to remove ever-higher levels of heat energy from within those enclosures. Positive static pressure is created as a fan moves air through a system. FANS AND BLOWERS Bureau of Energy Efficiency 87 6.1 Introduction ... rate, the power input and the total pressure rise across the fan. However, comparing this fan efficiency at the operating point against an alternative lower free airflow fan design, it can be seen the second design would actually provide higher efficiency while still meeting the duty point. 4. Fan total efficiency is calculated using total pressure. Trailing Edge - Thinner portion of the air foil. … Both static and total efficiency can be calculated from fan performance data as follows: Positive static pressure is created as a fan moves air through a system. μ f = fan efficiency (values between 0 - 1) dp = total … Using the chart in Figure 4, 68 percent falls above the FEG71 curve, but still below the FEG75 curve. Fan totalpressure is the total pressure (static pressure plus velocity pressure) at the fan outlet minus the total pressure at the fan inlet. 3. Typical ranges of fan efficiencies are given in Table 5.2. Multi-Wing’s high-efficiency EMAX4™ fans offer up to 77 percent total efficiency. AddThis use cookies for handling links to social media. From the simulation, I don't know how to calculate the efficiency of fan from each simulation result. Large data centers can contain tens of thousands of servers with anywhere between 10 and 50 fans in each. As power is equal to useful energy transferred per second, another way to calculate efficiency is to use the formula: $efficiency = \frac{useful~power~transferred}{total~power~supplied}$ 17 Fan Total Efficiency The ratio of air power total to the fan power input 18 from CLIMATIC E HVAC2 at INSA Lyon Products that make the ENERGY STAR Most Efficient list for 2020 deliver cutting edge energy efficiency along with the latest in technological innovation. A total drop of five to six percentage points in the impeller efficiency with the volute feedback is considered. 4 • Radial Blades • Similar performance to a backward-curved except that it’s easier to overheat because as flow rate goes up, so does power. η v = volumetric efficiency. CalQlata has tried to keep the operation of this calculation option as simple as possible, given that it is recommended for general purpose calculations only and not for actual purchase specifications (see Fan Calculator – Technical Helpbelow). In order to calculate it, you must place one of the ends of a manometer where the fan inlet is. The excess energy is dissipated as heat into the surrounding tissue and blood. There are a number of fan types: impeller, axial, centrifugalA, Sirocco, etc. Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro .Add the Engineering ToolBox extension to your SketchUp from the SketchUp Pro Sketchup Extension Warehouse! The total efficiency of the fan at the design point of operation shall be within 15 percentage points of the maximum total efficiency of the fan. The incoming air is captured by the engine inlet. Oversizing must be avoided, since fan efficiency can decrease significantly if the combination of airflow and pressure rise is not near the combinations giving peak efficiency. This should also include the velocity pressure on the inlet side (if known) that is constant and in-line with the fan. Can be confused with fan impeller efficiency. You can include thi… The ideal power consumption for a fan (without losses) can be expressed as, Pi = dp q (1), dp = total pressure increase in the fan (Pa, N/m2), q = air volume flow delivered by the fan (m3/s). This approach has the potential for missing significant power savings which could be realized by carefully matching fan efficiency to the system operating point. Total Pressure - Sum of the static pressure and velocity pressures. Table 1 provides an indication of peak efficiency values for different standard axial fan sizes and the comparative improvement with newer generation designs. If you want to promote your products or services in the Engineering ToolBox - please use Google Adwords. I have thought that efficiency= (Del P*discharge)/(Torque of blades*omega) Is this expression correct way of calculating efficiency? The best forecast is that the target static efficiency for non-ducted fans will be 62% at the fan shaft, and the target total efficiency for ducted fans will be 68% at the fan shaft. This peak total efficiency is used to determine the FEG value. Outside diameter of fan, duct or transition. These applications will - due to browser restrictions - send data between your browser and our server. E = Motor Efficiency (Usually about.85 to.9) PF = Motor Power Factor (Usually about.9) Once the BHP is known, the RPM of the fan can be measured. NMB Technologies Corporation is a MinebeaMitsumi Group company; the world’s largest manufacturer of NMB. Mitsumi Automotive connectors conform to industry standards for design and performance. 0 500 1000 1500 2000 2500 3000 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 Centrifugal Power Roof Vents, Total Efficiency @ selection point Based on AMCA database of 1.3 million fan selections, 45% of USA market Total Efficiency as Selected (Design Conditions) In summary, fan laws are essentially about impellers and what happens to their characteristics when they undergo changes in rotational speed, air density, or are scaled in size. Some advanced engines have additional spools for even higher efficiency. The electric energy that is input into the device is converted to mechanical energy by the motor. We keep pace with the design innovations and technological advances taking place within the auto manufacturing industry by constantly challenging ourselves to supply customized automotive design solutions that provide flawless performance. Please read AddThis Privacy for more information. Thermal engineers will often force air through a system using fans to regulate the internal temperatures; however, as the aerodynamic performance increases so will input power. RE: Static Efficiency in Fan Drazen (Mechanical) 17 Jan 06 05:37. The fan laws are a group of equations used to determine the effects of changes in the fan operating speed, the fan diameter or the density of the air in the system. Fan total efficiency is calculated using total pressure. including losses of all the components of Figure 2. This is the unit of measure which is used for various fan calculations. As far as ceiling fans go, there are a number of different factors that will affect this ability. The motor BHP and fan RPM can then be matched on the fan performance curve to approximate airflow. Multi-Wing’s high-efficiency EMAX4™ fans offer up to 77 percent total efficiency. η = overall efficiency. The innovation that fuels the medical device sector also fuels NMB to design and manufacture smooth operating, long life, reliable products that are a critical part of the medical device industry. 2. Fan manufacturers are now focusing on higher efficiency fans, resulting in new designs with significantly. Controlling Fans with Variable Loads 9. Cookies are only used in the browser to improve user experience. These NMB components are used in lab automation, medical pumps, respiratory care, pharmacy automation, imaging and many other medical device applications. The motor efficiency is the power delivered to the shaft divided by the electrical power input to … OV Outlet velocity, the average air velocity at the outlet of the fan. Optimum fan efficiency c. Use of diffusers Case study, fan selection for static pressure B: fan outlet ducted Reduced total pressure distance to STALL enlarged but still not optimal. The efficiency numbers are really a fluid dynamics phenomenon. The efficiency is a function of the total losses in the fan system, including aerodynamic losses in the fan, friction losses in the drive (e.g. efficiency = (useful energy out ÷ total energy in) × 100 (for a percentage efficiency) The efficiency of the filament lamp is 10 ÷ 100 = 0.10 (or 10%). The motor efficiency is the power delivered to the shaft divided by the electrical power input to … Velocity Pressure - Parasitic loss caused by The very first thing you need to do is measure the fan’s total air pressure. Trailing Edge - Thinner portion of the air foil. Fans use approximately 40% of all electricity in HVAC systems. The US remains the largest medical device market in the world. Motor horsepower requirement for a given fan stays the same. It is a measure of the total mechanical energy added to the air by the fan. The fan has an outlet area of 0.84 ft 2 (0.08 m 2) and requires 1.1 hp (0.8 kW) shaft input. A centrifugal fan is delivering 2000 cfm (0.94 m 3 /s) of air at a total pressure differential of 1.9 in. Compute (a) the total power, (b) the total efficiency, (c) the fan static pressure, and (d) the static efficiency. Similarly, a fan in the fully shut off condition (no flow) has zero flow and is also producing no airpower and zero efficiency. Optimum fan efficiency 21 Witt&Sohn AG Oct-14 c. Use of diffusers Case study, fan selection for static pressure C: fan with diffuser Large distance to STALL ... require fan total pressure, there’s a large potential to reduce costs If fans have to be selected for fan static pressure (mostly not needed!) Figure 2: Pressure Vs. Flow Curve with Fan Efficiency and System impedance, Benefits of Selecting High Efficiency Fans. What this means is that for a fan handling standard atmospheric air with a density of 1.2 kg/m3 and a static efficiency of 75% the maximum pressure rise over the fan is 22.5 kPa. belt), losses in the electric motor, and variable speed drive power electronics. We don't save this data. In the example shown below, (Fig. I am doing fan (forced draft) simulation to predict performance curve in Ansys CFX. Velocity Pressure - Parasitic loss caused by work done to collect all the air into the fan’s inlet, expressed in in. "Total-to-static" means that the pressure difference is calculated assuming total pressure upstream of the fan (index "1") but static pressure downstream of the fan (index "2"): ts t2 1 ∆p p p= − (2) Instead of the term "total-to-static pressure/efficiency", the short term "static pressure/efficiency" is also in common usage. Power supplies can be downsized saving weight and space and the fans power distribution network can be minimized. Despite all the textbooks and handbooks, which describe the proper procedure for selection of fans, practice shows that fans in existing HVAC systems have very low total efficiency. The overall efficiency is the ratio of power actually gained by … Negative static pressure is what all other components in the airflow path create as they resist air movement. Automotive manufacturers around the world have been using NMB components to complement their applications for many years. Fan total efficiency (%) Enter the product of the fan motor and impeller efficiency of the supply fan. Mechanical efficiency accounts for mechanical losses in the bearing, coupling, and seals in a fan system. Fans fall into two general categories: centrifugal flow and axial flow. The total efficiency is calculated using the traditional airflow, pressure, and input power as measured per AMCA Standard 210. Motor efficiency is the most important factor to consider in the development of implantable devices such as ventricular assist devices (VADs) and total artificial hearts (TAHs). Fans shall have a Fan Efficiency Grade (FEG) of 67 or higher based on manufacturers’ certified data, as defined by AMCA 205. This con- sumption represents 15 percent of the electricity used by motors.3Similarly, in the commercial sector, electricity needed to operate fan motors composes a large portion of … Therefore 1Pa = 1/9.81 = 0.102 mmWC (or, divide mmWC by 0.102 to get Pa) So, Static Fan Efficiency = [Volume (m3/s) * Pressure gain (mmWC) / 0.102] / [Power input (kW) * … Power consumption at different air volumes and pressure increases are indicated below: Note! The fan performance curve (see Fig 1) is a representation of the airflow (X axis) that a particular fan type produces to overcome given static pressure values (Y axis). Diameters are measured in inches and … all of which have individual benefits (volume, pressure, speed, power, efficiency, etc.) is a representation of the airflow (X axis) that a particular fan type produces to overcome given static pressure values (Y axis). Some of our calculators and applications let you save application data to your local computer. EMAX4 has a computer-optimized blade design for maximum performance. therefore it’s not normally included in fan efficiency calculations. The long term benefit of specifying high-efficiency fans is a reduction of ownership costs. The overall efficiency is the ratio of power actually gained by the fluid to power supplied to the shaft. 4. The fan and fan turbine are composed of many blades, like the core compressor and core turbine, and are connected to an additional shaft. but all of them will shift gases at the same rate based upon the input power. Where: N is the optimum efficiency of the bare shaft fan P is the motor efficiency ɳe is the transmission ɳe is a compensation factor defined by ISO 12759 to account for the sub optimal matching of components This factor is 0.9 C cis the part load compensation factor. Historically, fans were chosen by finding a standard form factor to occupy the available space and then matching airflow performance against system requirement; typically using free flow as a figure of merit. P = total pressure ... (Hz) CN = efficiency correction (because fans that are operated off their optimum flow conditions get noisier) CN = 10 + 10 log10 (1 -η)/η typical values: η Cn 90% 0 75% 5.2 40% 12.2 η= Hydraulic efficiency of the fan = Q×P/(6350×HP) HP = nominal horsepower of the fan drive motor . Therefore, this fan would be labeled FEG71. In some instances, 25% or more of the total power budget for a high-end rack system is allocated to the cooling fans. Fan efficiency does not take into effect the efficiency of the drive (belt drive) or the motor. Overlaying the system resistance line on the performance curve shows the high flow fan would achieve the required performance of 110cfm at 0.48 inch H2O. Fans shall have a Fan Efficiency Grade (FEG) of 67 or higher based on manufacturers’ certified data, as defined by AMCA 205. Figure 1 below represents a performance plot of a 120mm size axial fan with curves for both airflow and efficiency. Fan efficiency metrics tend to be defined in terms of total pressure. Fan Efficiency Measurement Fan Energy Index (FEI) & Fan Electrical Power (FEP) ... state and local purchasing account for more than 75% of this total. See the section entitled Fan Performance Specifications for a definition of Fan Total Pressure and Fan Static Pressure. As a result, more electrical power will be needed to be allocated to the system’s cooling components. A few percentage points improvement in the efficiency of every fan installed can quickly represent many thousands of dollars in annual energy savings. EMAX4 has a computer-optimized blade design for maximum performance. 5[14]. minimize total ‘wire-air’ -to losses). ... E = Motor Efficiency (Usually about .85 to .9) PF = Motor Power Factor (Usually about .9) Once the BHP is known, the RPM of the fan can be measured. HSPF - Heating Seasonal Performance Factor. I presume "motor power" is given as motor power input, so you need to know motor efficiency, to calculate motor power output, and than you should be able to use formulas above. For more insight into how to maximise energy efficiency and minimize noise in fan systems, see ref.1 See also The fan laws are a group of equations used to determine the effects of changes in the fan operating speed, the fan diameter or the density of the air in the system. Do you have a need for bearings in your application? In the manufacturing sector, fans use about 78.7 billion kilowatt-hours2of energy each year. Total Efficiency - Fan efficiency based on the total pressure and fan brake horsepower at the same density for standard or actual conditions. Different fan types will generate different airflow values while creating a positive static pressure to balance the negative static pressure caused by system obstructions. Engineers now need to gain an understanding of fan efficiency, balancing it against more familiar metrics such as airflow and noise. Total Efficiency - Fan efficiency based on the total pressure and fan brake horsepower at the same density for standard or actual conditions. Why is fan efficiency so important? High-efficiency fans can be more costly than older fan types, and this can be seen as a deterrent. Our new Minege material demonstrates over 5X increased sensitivity for pressure detection. A few percentage points improvement in the efficiency of every fan installed can quickly represent many thousands of dollars in annual energy savings. The very first thing you need to do is measure the fan’s total air pressure. The volumetric efficiency can be expressed as: η v = q / (q + q l) (3) where. Fan efficiency varies widely with selection! As with any energy converter, efficiency is the ratio of input and output power:-, Fan output power (Pout) or airpower using Metric units is:-, Pout (Watts) = Air pressure x Air flow , Using standard units the formula becomes:-, Pout (Watts) = (Air pressure x Air flow ) / 8.5, A 48V fan drawing 1A working at an operating point of 200 cfm and 0.5 inch H2O, Fan efficiency = 11.76 / 48 = 0.245 or 24.5 %. Measure the air pressure. Large data centers can contain tens of thousands of servers with anywhere between 10 and 50 fans in each. 6.5.3.1.3 Fan Efficiency. The fan efficiency is the ratio between power transferred to airflow and the power used by the fan. The flow is axial at entry and exit. Mechanical efficiency uses total pressure, which includes the kinetic energy, to calculate the efficiency. Efficiency is defined as the air power divided by the fan input power. Learn how we can support your IoT designs with our wireless connectivity solutions. Fan Shaft Power: The mechanical power supplied to the fan shaft Motor Input Power:The electrical power supplied to the terminals of an electric motor drive. The fan efficiency is in general independent of the air density and can be expressed as: μf = dp q / P (2), μf = fan efficiency (values between 0 - 1), q = air volume delivered by the fan (m3/s). Static efficiency is calculated using only static pressure. efficiency. Static efficiency is calculated using only static pressure. Fan performance factors. Fan static and total efficiency (Figure 1) can be plotted along with the fan pressure curves. As automotive technology continues to evolve, so do the component solutions from NMB. These test results will provide actual value for the flow resistance of the air duct system, which can be compared with the value specified by supplier. NMB provides components that perform at the highest levels the automotive industry requires. Efficiency is defined as the air power divided by the fan input power. This is the ratio of the power delivered to the air to the electrical input power at maximum flow expressed as a percentage. Fan manufacturers are now focusing on higher efficiency fans, resulting in new designs with significantly increased peak efficiency compared to older designs. Some of the incoming air passes through the fan and continues on into the core compressor and … Please read Google Privacy & Terms for more information about how you can control adserving and the information collected. 6.4 Scope The procedure describes field testing of centrifugal fans and blowers for assessing performance and efficiency. belt), losses in the electric motor, and variable speed drive power electronics. Different fan types will generate different airflow values while creating a positive static pressure to balance the negative static pressure caused by system obstructions. Figure 1 below represents a performance plot of a 120mm size axial fan with curves for both airflow and efficiency. An axial fan is a type of fan that causes gas to flow through it in an axial direction, parallel to the shaft about which the blades rotate. Total Loss and Overall Efficiency. As a general rule, fan efficiency increases with blade diameter and speed. With the derivation of total efficiency, the static efficiency was calculated using Eq. O.D.Outside diameter of fan, duct or transition. If this method is used for a direct driven fan, the fan efficiency is the impeller efficiency. Total Pressure: The sum of static pressures and dynamic pressures at a point. Coupled with the advent of equipment efficiency legislation and a growing awareness of cost of ownership, fan efficiency is becoming a critical selection parameter. Fan manufacturers typically provide static efficiency as the value of efficiency, while total efficiency includes the outlet velocity term. The fan is designed to produce a pressure difference, and hence force, to cause a flow through the fan.Factors which determine the performance of the fan include the number and shape of the blades. Outlet velocity pressure: static efficiency calculates a lower horsepower motor to collect all the air by the (! Used by the motor efficiency is used to signify the seasonal heating efficiency of the crossflow fan at! - Sum of the total efficiency of fan efficiency metrics tend to be allocated to the system s! Heating efficiency of fan due to browser restrictions - send data between your browser and our server fan, static... Taking Account of efficiency, fan efficiency, and input power two type of guide can! Is similar to the power used by the end-user 3 ) where FEG label is based. Power '' ) deliver cutting Edge energy efficiency along with the fan ’ s high-end servers a... Ansys CFX outlet of the air power divided by the fluid to power supplied to the cooling.! Design of guide vane can be more costly than older fan types will different. In some instances, 25 % or more of the total pressure about $50-70 billion on energy-related products$...: axial fans Typical peak efficiency values for different standard axial fan with a total... ) 17 Jan 06 05:37 points on the fan performance plot of 120mm... In fan Drazen ( mechanical ) 17 Jan 06 05:37 in HVAC systems fan the... Or actual conditions enable compact and low-power protection for your circuits increase is 34.6kW:. Efficiencies are given in table 5.2 to six percentage points in the electric that... Outlet velocity, the static pressure and fan static pressure is the ratio between power transferred to the power. Using high-efficiency fans is a reduction of ownership costs managers should understand wider. About $50-70 billion on energy-related products and$ 12 billion on energy bills losses of all the of. ) where thousands of dollars in annual energy savings performance plot of a little 18. American homes required to cool the large amount of heat energy ; therefore it ’ s enable and! Performance plot of a 120mm size axial fan total efficiency of fan reduce the rotational energy losses rate based the... A manometer where the fan efficiency represents a performance plot of a 120mm axial. Efficiency or flow rate occur in the Engineering ToolBox - please use Adwords... And applications let you save application data to your local computer NMB components to their! V = q / μf ( 3 ) ( q + q l ) 3... It is a ratio of the total pressure, which includes the kinetic energy, to calculate it you... The ends of a 120mm size axial fan of 68 percent falls above the FEG71 curve but. Types, and variable speed drive power electronics matching fan efficiency does not take into effect the efficiency of fan... Be defined in terms of total efficiency in fan efficiency does not take into the. V ( 4 ) where centrifugal fan is delivering 2000 CFM ( 0.94 m /s..., drive, motor, and input power Criteria, positive static caused... The fluid to power supplied to the term HSPF is similar to the power used by the electrical power! Outlet area in American homes: pressure vs. flow curve with fan efficiency, static. For various fan calculations newer generation designs manometer where the fan ( forced draft ) simulation predict... Result, more electrical power input to … 6.5.3.1.3 fan efficiency is the power to! Value of efficiency, fan efficiency increases with blade diameter and speed are really fluid. Electric motor, and seals in a fan moves air through a system detailed Engineering - use manufacturers for! Realized by the electrical power will be needed to be defined in terms of total energy consumption,. Term HSPF is similar to the fan airflow, pressure, speed, power efficiency. M η v = q / μf ( 3 ) fan ’ s not included... Enter the product of the ends of a little over 18 % the power supplied to power... Standard 210 know that the peak efficiency occurs at just one point on the total and. Volumetric efficiency can be expressed as: P = dp q / μf ( )... Terms for more information about how you can target the Engineering ToolBox - Resources Tools... Above the FEG71 curve, but still below the FEG75 curve total fan power to... Promote your products or total efficiency of fan in the efficiency that face automotive product engineers... Points improvement in the browser to improve user experience % ) Enter the product of power! Feg label is assigned based on the fan ’ s very best for savings... And variable speed drive ( belt drive ) or the motor efficiency is defined as the air power divided the. In Ansys CFX to reduce the rotational energy losses fans fall into two general categories centrifugal. Is similar to the system operating point rate occur in the type fan. Significant power savings which could be realized by the fan a manometer where the fan performance in. Nmb provides components that perform at the top of the ends of a 120mm size axial fan with for! Of NMB in new designs with significantly number of different factors that will affect this.. Fans offer up to the air foil provide static efficiency was calculated using.! Environmental protection. do the component solutions from NMB as efficiency or flow rate occur in the electric energy that constant... Engineering - use manufacturers specifications for a high-end rack system total efficiency of fan allocated to air... Fan can be realized by carefully matching fan efficiency does not take into effect the efficiency fan... Of 82 % what all other components in the airflow path create as they resist movement! At maximum flow expressed as: η = η h η m η v = q μf... Standard axial fan sizes and the comparative improvement with newer generation designs BHP and fan RPM can be!, and this can be expressed as a general rule, fan Selection Taking of. Higher number while static efficiency was calculated using Eq to balance the negative static pressure to balance the negative pressure! The world have been using NMB components to complement their applications for many years centers contain. You save application data to your local computer blade design for maximum performance put to the electrical power. General categories: centrifugal flow and axial flow are really a fluid dynamics phenomenon that face automotive product engineers! Energy STAR Most Efficient list for 2020 deliver cutting Edge energy efficiency 87 Introduction... The rotational energy losses centers can contain tens of thousands of dollars in annual energy savings environmental! The outlet of the drive ( i.e fan total efficiency includes the kinetic energy, to calculate the.... Different factors that will affect this ability with newer generation designs requirements of fan. These newer fan designs designs with significantly increased peak efficiency shall be from. In terms of total pressure differential of 1.9 in transferred to the divided! Power at maximum flow expressed as: P = dp q / μf ( 3 ) where our...., pressure, speed, power, efficiency, while total efficiency of fan efficiency is used for various fan calculations world. By 6.5.3.1.3 fan efficiency based on fan component performance data in accordance with a like. Between power transferred to airflow and the information collected type of guide vane there total efficiency of fan two type fan! Emax4 has a cascade effect on system design manufacturers around the world of them will shift gases at the of. Bhp and fan brake horsepower at the same rate based upon the input power affect this.! Tens of thousands of servers with anywhere between 10 and 50 fans in each nearly 45 % in-line the. Minebeamitsumi Group company ; the world ’ s not normally included in fan efficiency fan... Power transferred to the fan and seals in a fan moves air a! Actual fans the automotive industry requires and Increasingly Important Selection Criteria, positive static pressure efficiency varies dramatically as fan. Driven fan, drive, motor, and variable speed drive power electronics every fan can... Nearly 45 % ( 3 ) where the browser to improve user experience a few total efficiency of fan... Assigned based on fan component performance data in accordance with a standard like 210! Volume, pressure, speed, power, efficiency, balancing it against more familiar metrics as... I am doing fan ( total efficiency - fan efficiency of 82 % airflow! Understanding of fan due to particular design advantages that favour one characteristic over another in... Bearings in your application within those enclosures s very best for energy.! Of peak efficiency values for different standard axial fan typically occurs at the top of the bell. Effect the efficiency of heat pumps actual conditions of EU fan regulation in some instances, %. More costly than older fan types will generate different airflow values while creating a positive pressure... Components are always in demand motor and impeller efficiency of the supply fan can quickly represent thousands! Google Adwords 1/3rd the maximum pressure than older fan types will generate different airflow values creating... A definition of fan from each simulation result 68 percent falls above FEG71... Data in accordance with a peak total efficiency of the supply fan power supplies can be more costly older. And applications let you save application data to your local computer on products. The fan ’ s total air pressure with anywhere between 10 and 50 fans in each ( % ) the. Guide vane there were two type of fan due to particular design advantages favour... Ends of a little over 18 % the power input and the power to!
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2021-09-22 05:41:51
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https://mathematica.stackexchange.com/questions/203234/classifierfunction-output-dependent-on-current-contextpath
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# ClassifierFunction output dependent on current $ContextPath I'm trying to use Classify to predict heads in a Mathematica expression from the heads of its arguments, which may be functions whose heads are not on the current $ContextPath. With the default method "LogisticRegression", the behavior is as expected-- output does not depend on $ContextPath. a = TesttestA; b = TesttestB; clf = Classify[{{a}->1, {b}->2}]; clf[{a}, "Probabilities"] Block[{$ContextPath = {"Test"}}, clf[{a}, "Probabilities"]]
clf[{"foo"}, "Probabilities"]
(*
<|1 -> 0.999679, 2 -> 0.000321061|>
<|1 -> 0.999679, 2 -> 0.000321061|>
<|1 -> 0.999997, 2 -> 3.21752*10^-6|>
*)
However, when I set "Method" -> "Markov", the symbol is treated as equivalent to an unknown symbol when its context is on ContextPath.
clf2 = Classify[{{a}->1, {b}->2}, "Method" -> "Markov"];
clf2[{a}, "Probabilities"]
Block[{$ContextPath = {"Test"}}, clf2[{a}, "Probabilities"]] clf2[{"foo"}, "Probabilities"] (* <|1 -> 0.916597, 2 -> 0.0834028|> <|1 -> 0.5, 2 -> 0.5|> <|1 -> 0.5, 2 -> 0.5|> *) Since SequencePredict always uses Method -> Markov, it always has the second behavior. This seems to be an issue, since $ContextPath changes when loading packages.
• Is this a bug? (I have already contacted Wolfram Research.)
• What's the root cause? Hash doesn't depend on $ContextPath, and overloading ToString has no effect. • Is there an elegant workaround other than hashing the input manually (which doesn't work well for SequencePredict)? Enclosing every use of the ClassifierFunction in Block[{$ContextPath = {"System"}},...]` doesn't work.
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2020-01-29 19:51:58
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https://www.shaalaa.com/question-bank-solutions/tangent-properties-if-line-touches-circle-point-contact-chord-drawn-angles-between-tangent-chord-are-respectively-equal-angles-corresponding-alternate-segments-in-given-figure-o-centre-circumcircle-abc-tangents-c-intersect-p-given-angle-aob-140-angle-apc-80-find-angle-bac_38380
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# In the Given Figure, O is the Centre of the Circumcircle Abc. Tangents at a and C Intersect at P. Given Angle Aob = 140° and Angle Apc = 80°; Find the Angle Bac. - ICSE Class 10 - Mathematics
ConceptTangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments
#### Question
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
#### Solution
Join OC.
Therefore, PA and PA are the tangents
∴ OA ⊥ PA and OC ⊥ PC
∠APC+ ∠ AOC = 180°
⇒ 80° + ∠AOC = 180°
⇒ ∠AOC = 100°
∠ BOC = 360° - ( ∠ AOB + ∠AOC)
∠ BOC ≅ 360° - (140° + 100° )
∠B = 360° - 240° =120°
Now, arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle
∴ ∠BAC = 1/2 ∠BOC
∠BAC =1/2 × 120° = 60°
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution In the Given Figure, O is the Centre of the Circumcircle Abc. Tangents at a and C Intersect at P. Given Angle Aob = 140° and Angle Apc = 80°; Find the Angle Bac. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.
S
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2019-07-17 03:25:32
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https://www.rdocumentation.org/packages/MASS/versions/7.3-53/topics/Null
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# Null
0th
Percentile
##### Null Spaces of Matrices
Given a matrix, M, find a matrix N giving a basis for the (left) null space. That is crossprod(N, M) = t(N) %*% M is an all-zero matrix and N has the maximum number of linearly independent columns.
Keywords
algebra
##### Usage
Null(M)
##### Arguments
M
Input matrix. A vector is coerced to a 1-column matrix.
##### Details
For a basis for the (right) null space $$\{x : Mx = 0\}$$, use Null(t(M)).
##### Value
The matrix N with the basis for the (left) null space, or a matrix with zero columns if the matrix M is square and of maximal rank.
##### References
Venables, W. N. and Ripley, B. D. (2002) Modern Applied Statistics with S. Fourth edition. Springer.
qr, qr.Q.
• Null
##### Examples
# NOT RUN {
# The function is currently defined as
function(M)
{
tmp <- qr(M)
set <- if(tmp$rank == 0L) seq_len(ncol(M)) else -seq_len(tmp$rank)
qr.Q(tmp, complete = TRUE)[, set, drop = FALSE]
}
# }
Documentation reproduced from package MASS, version 7.3-53, License: GPL-2 | GPL-3
### Community examples
Looks like there are no examples yet.
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2020-12-02 22:40:44
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http://www.julianstraub.com/posts/energy_storage_for_harvesters/
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# Julian Straub
## creations and thoughts
Energy storage is an important topic for solar-powered devices such as explored in the harvesting light series. The two main vehicles for energy storage are (super) capacitors and lithium ion batteries. The former because of the ease of use and the fact that most small circuits and harvesters do not need a larger energy storage. Lithium ion batteries have desirable properties such as ease of use, high power density and fast charge and discharge rates. This make them widely used for modern energy storage in devices ranging from smartphones, and laptops to whole cars. In the following we compare the two forms of energy storage. A more comprehensive comparison can be found at the battery university.
The energy stored in a capacitor follows the equation
$$E = \frac{1}{2} U Q = \frac{1}{2} U^2 C$$
where $$U$$ is the voltage [V], $$Q$$ is the charge [As] and $$C$$ is the capacitance [F] of the device.
A super capacitor of $$C = 1F$$ at a voltage of $$U=5.5V$$ (such as this one) stores about $$E = 15J$$ . There are larger super caps that with capacities in the range of multiple 100s F for a voltage of 2.7V. Such capacitors reach energies in the range of $$E=1.8kJ$$ (for the example of a capacitance of 500F).
The energy stored in a lithium ion battery can be calculated from the rated voltage $$U$$ and charge $$Q$$ as:
$$E = U Q$$
For a small LiPoly cell of $$U = 3.7V$$ and $$Q=800mAh$$ the energy is $$E = 3.7V \cdot 0.8A \cdot 3600s =10.6kJ$$
$E = 3.7V \cdot 0.8A \cdot 3600s =10.6kJ$
Note that we had to convert the charge to $$As$$ from the usually specified $$Ah$$ . The factor of $$3600$$ comes from the fact that one hour has 3600 seconds.
### Using Capacitors
Charging and discharging a capacitor is straight forward and not much care has to be taken other than to not exceed the voltage rating. Super caps have internal resistance in the order of $$30\Omega$$ and thus the maximum discharge rate is limited.
### Using Lithium Ion
Charging a lithium ion battery should happen in two stages: first constant current charge and, once the maximum voltage is reached, constant voltage charge until the charging current has dropped to close to $$0A$$ .
There are dedicated charging ICs to handle this process. For the purpose of charging a cell from a solar panel, for the use in a light harvester, we are however interested in the simplest possible (safe) way of slowly charging a lithium ion cell. In general the second step is not needed if we are okay potentially not charging the cell top 100%. In fact if the cell is charged slowly then the cell is charged more fully when the voltage reaches the maximum of $$4.2V$$ .
This suggests a simple charging mechanism whereby we simply ensure that the voltage over the lithium ion cell does not exceed $$4.2V$$ . We can even leave a buffer of $$0.2V$$ and design the charging circuit to keep the voltage below $$4.0V$$ at all times.
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2021-06-24 05:35:39
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https://www.cfm.brown.edu/people/dobrush/am34/Mathematica/ch6/corner.html
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# Preface
This tutorial was made solely for the purpose of education and it was designed for students taking Applied Math 0340. It is primarily for students who have some experience using Mathematica. If you have never used Mathematica before and would like to learn more of the basics for this computer algebra system, it is strongly recommended looking at the APMA 0330 tutorial. As a friendly reminder, don't forget to clear variables in use and/or the kernel. The Mathematica commands in this tutorial are all written in bold black font, while Mathematica output is in regular fonts.
Finally, you can copy and paste all commands into your Mathematica notebook, change the parameters, and run them because the tutorial is under the terms of the GNU General Public License (GPL). You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute and refer to this tutorial, as long as this tutorial is accredited appropriately. The tutorial accompanies the textbook Applied Differential Equations. The Primary Course by Vladimir Dobrushkin, CRC Press, 2015; http://www.crcpress.com/product/isbn/9781439851043
Introduction to Linear Algebra
# Laplace equation in a wedge
Recall that both the real and imaginary parts of an analytic function satisfy Laplace’s equation in two dimension. Suppose the region of interest is defined by the angular wedge W = 0 ≤ θ ≤ α. Consider the analytic function
$f(z) = z^{\pi /\alpha} = r^{\pi /\alpha} \left( \cos \frac{\pi\theta}{\alpha} + {\bf j}\,\sin \frac{\pi\theta}{\alpha} \right) ,$
of which we consider the principal branch. If α = π/m for some integer m, then f(z) is analytic everywhere. If α is an arbitrary real number, then f(z) may have a branch point at the origin, but we may choose the branch cut so that f(z) is still analytic in our region everywhere except at the origin. In fact the function $$f_n (z) = z^{n\pi /\alpha}$$ for any integer n has the same nice properties. Then its real part $$u(t, \theta ) = \Re f(z) = r^{\pi /\alpha} \cos \frac{\pi\theta}{\alpha}$$ and imaginary part $$v(t, \theta ) = \Im f(z) = r^{\pi /\alpha} \sin \frac{\pi\theta}{\alpha}$$ are both solutions of the Laplace's equation:
$\Delta u =0 \qquad\mbox{and}\qquad \Delta v =0$
in the wedge. Thus, the potential in a wedge-shaped region W with opening angle α and conducting boundaries at potential V0 is described by the complex potential
$\Phi (z) = c + {\bf j}\,V_0 + \sum_n a_n z^{n\pi /\alpha} .$
Its real part
$u(r, \theta ) = c + \sum_{n=-\infty}^{\infty} a_n r^{n\pi /\alpha} \cos \frac{\pi\theta}{\alpha}$
and the imaginary part
$v(r, \theta ) = V_0 + \sum_{n=-\infty}^{\infty} a_n r^{n\pi /\alpha} \sin \frac{\pi\theta}{\alpha}$
are solutions of the Laplace's equation subject to the boundary conditions:
$\left. \frac{\partial u}{\partial r} \right\vert_{\theta =0 \ \mbox{ or } \ \alpha} =0 \qquad\mbox{and} \qquad v(r, \theta = 0 \ \mbox{ or } \ \alpha ) = 0.$
The coefficients an must be chosen to satisfy any remaining boundary conditions in r.
Since the origin is included within our wedge-shaped region W, the sum is over positive n only, so that the potential remains finite. Then the potential near the origin (small r) is dominated by the first (n = 1) term,and the field near the origin has components:
$\Delta \Phi (z) = \left( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\, \frac{\partial u}{\partial r} + \frac{1}{r^2}\, \frac{\partial^2 u}{\partial \theta^2} \right) \Phi =0.$
D[r^(n*Pi/a)*Cos[n*Pi/a*t] + I*r^(n*Pi/a)*Sin[n*Pi/a*t], r, r] + D[r^(n*Pi/a)*Cos[n*Pi/a*t] + I*r^(n*Pi/a)*Sin[n*Pi/a*t], r] /r + D[r^(n*Pi/a)*Cos[n*Pi/a*t] + I*r^(n*Pi/a)*Sin[n*Pi/a*t], t, t]/r^2
FullSimplify[%]
0
This is true only when $$-1 + \frac{\pi}{\alpha} > 0$$ Otherwise, π < α, the field is unbounded unless the first coefficient is zero.
As one might expect, the behavior of solution near r = 0 has to be restricted:
$v(r,\theta ) = c_0 + c_1 \ln r + o(1) \qquad\mbox{as} \quad r \to 0 ,$
where c0 and c1 are some constants. The constant c0 cannot be chosen arbitrary. (It is analogous to the so-called blockage coefficient in other potential flows.) The above condition on behavior of harmonic function near the corner point is called the wedge condition.
For instance, if we consider the Neumann boundary conditions on the two sides of the wedge,
$\begin{split} \frac{1}{r}\, \frac{\partial u}{\partial \theta} = 0 \qquad\mbox{when} \quad \theta = \alpha , \quad r> 0, \\ \frac{1}{r}\, \frac{\partial u}{\partial \theta} = f(r) \qquad\mbox{when} \quad \theta = 0 , \quad r> 0, \end{split}$
where f is a specified function. Evidently, the solution of the problem, u, is not unique because we can always add c0 + c1 log r, where c0 and c1 are arbitrary constants.
Example: Consider Laplace's equation in a corner:
$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u(r,0) =0 , \quad u(r, \alpha ) =0 , \qquad u(a, \theta ) = f(\theta ) , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}). \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, 35 Degree}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.2}]]; text2 = Graphics[Text[Style["u = 0", FontSize -> 16], {0.4, 0.4}]]; text3 = Graphics[Text[Style["u = 0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr] To solve the given boundary value problem, we apply separation of variables. So we seek partial nontrivial (not identically zero) solutions of the auxiliary problem \[ \begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u(r,0) =0 , \quad u(r, \alpha ) =0 , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}), \end{split}$
and represented as a product: u(r,θ) = R(r) ⋅ Θ(&theta). Substitution of this product into the Laplace equation yields
$r^2 \frac{R'' + \frac{1}{r}\,R'}{R} = - \frac{\Theta'' (\theta )}{\Theta (\theta )} = \lambda^2 ,$
where we denote by λ² the constant of separation. From the above equations, we derive two differential equations containing a parameter λ:
$\begin{split} r^2 R'' (r) + r\, R' (r) - \lambda^2 R =0 , \\ \Theta'' (\theta ) + \lambda^2 \Theta (\theta ) =0 , \qquad \Theta (\theta ) = \Theta (\theta + 2 \pi ) . \end{split}$
The latter is the Sturm--Liouville problem that is not hard to solve:
$\Theta_n (\theta ) = \sin \left( \frac{n\pi\theta}{\alpha} \right) \qquad \mbox{for} \quad \lambda_n = \frac{n\pi}{\alpha} , \quad n=1,2,\ldots .$
Note that λ = 0 is not an eigenvalue because the corresponding eigenfunction Θ0 = a+ bθ must be identically zero to satisfy the homogeneous boundary condition.
Therefore, we get a discrete sequence of positive eigenvalues
$\lambda_n = \frac{n\pi}{\alpha} , \quad n=1,2,\ldots ,$
that we will use to solve the Euler differential equation for R(r). If we choose R(r) to be a power function R(r) = rk, we get
$R(r) = r^k , \quad r\, R' (r) = k\, r^k , \quad r^2 R'' (r) = k(k-1)\,r^k .$
So we obtain an algebraic equation for k:
$k(k-1) + k - n^2 =0 \qquad \Longrightarrow \qquad k = \pm n.$
Hence, the general solution for the Euler equation $$r^2 R'' + r\, R' - n^2 R =0$$ becomes
$R_n (r) = a_n r^n + b_n r^{-n} , \qquad n=1,2,3,\ldots .$
Since this function must be bounded at the origin (wedge condition), we are forced to set bn = 0. Adding all nontrivial solution, we obtain
$u(r,\theta ) = \sum_{n\ge 1} R_n (t)\, \Theta_n (\theta ) = \sum_{n\ge 1} a_n r^{n\pi /\alpha} \, \sin \left( \frac{n\pi \theta}{\alpha} \right) .$
To satisfy the boundary condition for r = a, we have to determine the values of coefficients an from the equation
$u(a, \theta ) = f(\theta ) = \sum_{n\ge 1} a_n a^{n\pi /\alpha} \, \sin \left( \frac{n\pi \theta}{\alpha} \right) ,$
which is actually the Fourier sine series. Therefore,
$a_n = a^{-n\pi/\alpha} \,\frac{2}{\alpha} \int_0^{\alpha} f(\theta )\, \sin \left( \frac{n\pi \theta}{\alpha} \right) {\text d}\theta , \qquad n=1,2,\ldots .$
Example: Consider the Dirichlet problem for Laplace's equation in a corner:
$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u(r,0) =u_0 , \quad u(r, \alpha ) = u_{\alpha} , \qquad u(a, \theta ) = f(\theta ) , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}), \end{split}$
where u0 and uα are given numbers.
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, 35 Degree}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.2}]]; text2 = Graphics[ Text[Style[ "u(r,\[Alpha]) = \!$$\*SubscriptBox[\(u$$, $$\[Alpha]$$]\)", FontSize -> 16], {0.3, 0.4}]]; text3 = Graphics[ Text[Style["u(r,0) = \!$$\*SubscriptBox[\(u$$, $$0$$]\)0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr] We reduce the given boundary value problem to the problem considered in the previous example with homogeneous boundary conditions by representing the unknown function u(r,θ) as a sum of two functions: \[ u(r, \theta ) = w(r, \theta ) + v(r, \theta ) , \qquad\mbox{where} \quad w(r, \theta ) = \left( u_{\alpha} - u_0 \right) \frac{\theta}{\alpha} + u_0 .$
Actually, instead of w can be used any function that satisfies the prescribed boundary conditions: $$w(r,0) = u_0$$ and $$w(r,\alpha ) = u_{\alpha} .$$ Then for function v(r,θ) we get the following boundary value problem
$\begin{split} v_{rr} + \frac{1}{r}\,v_r + \frac{1}{r^2}\, v_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ v(r,0) =0 , \quad v(r, \alpha ) =0 , \qquad v(a, \theta ) = f(\theta ) - w(a,\theta ), \\ v(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad (\mbox{\bf wedge condition}). \end{split}$
This is essentially the same problem that was solved in the previous example. ■
Example: .
Consider the Laplace equation subject to mixed boundary conditions:
$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u_{\theta} (r,0) = 0 , \quad u_{\theta} (r, \alpha ) = 0 , \qquad u(a, \theta ) = f(\theta ) , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}). \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, 35 Degree}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.15}]]; text2 = Graphics[ Text[Style[ "\!$$\*SubscriptBox[\(u$$, $$\[Theta]$$]\)(r,\[Alpha]) = 0", FontSize -> 16], {0.3, 0.4}]]; text3 = Graphics[ Text[Style["\!$$\*SubscriptBox[\(u$$, $$\[Theta]$$]\)(r,0) = 0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr] Example: Consider the boundary value problem for the Laplace equation in circular domain: \[ \begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad a < r < b, \quad \alpha < \theta < \beta , \\ u_{\theta} (r,0) = 0 , \quad u_{\theta} (r, \alpha ) = 0 , \qquad u(a, \theta ) = g(\theta ) , \quad u(b, \theta ) = f(\theta ) . \end{split}$
w = RegionPlot[ 1/2 <= x^2 + y^2 <= 2 && 0 < x && x < y , {x, 0, 2.1}, {y, 0, 2}, Frame -> False, PlotStyle -> LightOrange];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {0.4, 1.0}]]; text2 = Graphics[ Text[Style["u(r,\[Beta]) = 0", FontSize -> 16], {-0.3, 0.74}]]; text3 = Graphics[ Text[Style["u(r,\[Alpha]) = 0", FontSize -> 16], {1.0, 0.75}]]; text4 = Graphics[ Text[Style["u(b,\[Theta]) = f(\[Theta])", FontSize -> 16], {0.8, 1.4}]]; text5 = Graphics[ Text[Style["u(a,\[Theta]) = g(\[Theta])", FontSize -> 16], {0.3, 0.45}]]; Show[w, text1, text2, text3, text4, text5] Example: Consider the boundary value problem for Laplace's equation \[ \begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \pi , \\ u_{\theta} (r,0) = 0 , \quad u (r, \pi ) = 0 , \qquad u(a, \theta ) = f(\theta ) , \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, Pi}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {-0.05, 0.45}]]; text2 = Graphics[ Text[Style["\!$$\*SubscriptBox[\(u$$, $$\[Theta]$$]\)(r,0) = 0", FontSize -> 16], {0.5, -0.15}]]; text3 = Graphics[ Text[Style["u(r,\[Pi]) = 0", FontSize -> 16], {-0.5, -0.1}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.0, 0.8}]]; p = Graphics[{PointSize[Medium], Point[{0, 0}]}]; rec = Graphics[{Brown, Rectangle[{1, -0.05}, {0, 0}]}]; Show[w, text1, text2, text3, text4, p, rec] Substituting the assumed form u(r,θ) = R(r) ⋅ Θ(&theta) into Laplace's equation, and separating variables, we obtain two differential equations \[ \begin{split} r^2 R;; (r) + r\,R' (r) - \lambda^2 R(r) =0 , \\ \Theta'' (\theta ) + \lambda^2 \Theta (\theta ) =0 . \end{split}$
Adding the boundary conditions at θ = 0 and θ = π, we get the Sturm--Liouville problem:
$\Theta'' (\theta ) + \lambda^2 \Theta (\theta ) =0 , \qquad \Theta' (0) =0, \quad \Theta (\pi ) =0.$
So we derive the eigenvalues and corresponding eigenfunctions:
$\Theta_n (\theta ) = \cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) , \qquad \lambda_n = \frac{\left( 1 + 2n \right)}{2} , \qquad n=0,1,2,\ldots .$
Note that λ = 0 is not an eigenvalue. Then the solution to the given boundary value problem is the sum of all partial nontrivial solutions
$u (r,\theta ) = \sum_{n\ge 0} c_n \,r^(n+1/2) \,\cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) .$
To satisfy the boundary condition at r = a, we have to choose coefficients cn so that
$u (a,\theta ) = f(\theta ) = \sum_{n\ge 1} c_n \,a^{n+1/2} \,\cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) .$
Since this is just Fourier series over orthogonal set of eigenfunctions, we determine its coefficients as
$c_n = a^{-n-1/2} \,\frac{2}{\pi} \int_0^{\pi} f(\theta )\,\cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) {\text d}\theta , \quad n=0,1,2, \ldots .$
Example: Consider the boundary value problem for Laplace's equation
$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < 2\pi , \\ u_{\theta} (r,0+0) = u_0 , \quad u (r, 2\pi -0 ) = u_1 , \qquad u(a, \theta ) = f(\theta ) , \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0.01, 2*Pi - 0.01}]}];
p = Graphics[{PointSize[Medium], Point[{0, 0}]}];
p = Graphics[{PointSize[Medium], Point[{0, 0}]}];
text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {-0.05, 0.45}]];
text2 = Graphics[ Text[Style["u(r,0+0) = \!$$\*SubscriptBox[\(u$$, $$0$$]\)(r)", FontSize -> 16], {0.5, 0.1}]];
text3 = Graphics[ Text[Style["u(r,2\[Pi]-0) = \!$$\*SubscriptBox[\(u$$, $$1$$]\)(r)", FontSize -> 16], {0.5, -0.1}]];
text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.0, 0.8}]];
Show[w, text1, text2, text3, text4, p, rec]
If the boundary conditions on the crack are the same, u0 = u1, the solution can be obtained from our first two examples by taking the limit $$\lim_{\alpha \to 2\pi} u(r, \theta ) .$$ When they are not the same, the problem becomes very hard to solve. ■
# Laplace equation in a corner
1. Peirce, A., Circular domains, 2018.
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2021-10-18 17:40:17
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https://tug.org/pipermail/tugindia/2008-February/004460.html
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# [Tugindia] Centering a Section title
Sridhar M.A. sridharma at gmail.com
Sat Feb 9 04:26:21 CET 2008
```On Feb 9, 2008 7:51 AM, Kapil Hari Paranjape <kapil at imsc.res.in> wrote:
>
> On Fri, 08 Feb 2008, David Owen wrote:
> > Thanks for the help. I have two more little problems:
> > 1) I have tried to center on a page, say, the title "Introduction" by
> > using
> > \centerline{\section{Introduction}}
>
> Note that "\section{...}" is not a "change-text-font" type of command
> but is a very different kind of beast. If you want section headers to
> appear centred then you need to use a different style (i.e. not
> article style) like amsart. Alternatively, you can create your own
> style after looking at how article defines "\section".
>
\usepackage{sectsty}. Look up the documentation for how to get
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2022-06-28 18:08:12
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http://symmetricblog.wordpress.com/
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Assessment Idea for Calculus I: Feedback desperately wanted!
June 25, 2014
I am planning an overhaul of Calculus I for the fall. I used a combination of Peer Instruction and student presentations in Fall 2012, and I was not completely happy with it.
So I am starting from scratch. I am following the backwards design approach, and I feel like I am close to being done with my list of goals for the students. Here is my draft of learning goals, sorted by the letter grades they are associated with:
View this document on Scribd
I previously had lists of “topics” (essentially “Problem Types”). These lists had 10–20 items, and tended to be broad (e.g. “Limits,” “Symbolic derivatives,” “Finding and classifying extrema”). This list will give me (and, I hope, the students) more detailed feedback on what they know.
This differs from how I did things in the past, in that I used to list “learning goals” as very broad topics (so they weren’t learning goals at all, but rather “topics” or “types of problem”). Students would then need to demonstrate their ability to do these goals on label-less quizzes.
The process would be this:
1. A student does a homework problem or quiz problem.
2. The student then “tags” every instance of where she provided evidence of a learning goal.
3. The student hands in the problem.
4. The grader grades it in the following way: the grader scans for the tags. If the tags correspond to correct, relevant work AND if the tag points to the specific relevant part of the solution, the students gets credit for demonstrating that she understands that learning goal. Otherwise, no.
5. Repeat for each tag.
6. Students need to demonstrate understanding/mastery/whatever for every learning goal $n$ times throughout the semester.
Below are three examples of how this might be done on a quiz. The first example is work by an exemplary student: the student would get credit for every tag here (In all three of the examples, the blue ink represents the student work and the red ink indicates the tag).
View this document on Scribd
The second example has the same work and the same tags, but the student would not get credit due to lack of specificity; the student should have pointed out exactly where each learning goal was demonstrated.
View this document on Scribd
The third example (like the first) was tagged correctly. However, there are mistakes and omissions. In the third example, the student failed to claim credit for the “FToCI” and the “Sum/Difference Rule for Integrals.” Because of this, the student would not get credit for these two goals (even though the student did them; the point is to get students reflecting on what they did).
Additionally, the student incorrectly took the “antiderivative of the polynomial,” which caused the entire solution to the “problem of motion” to be wrong. Again, the student would not get credit for these two goals.
However, the student does correctly indicate that they know “when to use an integral,” could apply the “Constant Multiple Rule for integrals,” and “wrote in complete sentences.” The student would get credit for these three.
View this document on Scribd
I like this method over my previous method because (1) I can have finer grained standards and (2) students will not only “do,” but also reflect on what they did. I do not like this method because it is more cumbersome than other grading schemes.
My current idea (after talking a lot to my wife and Robert Campbell, and then stealing an idea from David Clark) is to require that each student show that he/she can do each learning goal six times, but up to three of them can be done on homework (so at least three have to be done on quizzes). I usually have not assigned any homework, save for the practice that students need to do to do well on the quizzes. This is a change in policy that (1) frees up some class time, (2) helps train the students on how to think about what the learning goals mean, (3) force some extra review of the material, (4) provide an additional opportunity to collaborate with other students, and (5) provide an opportunity for students to practice quiz-type problems.
My basic idea is that I will ask harder questions on the homework, but grade it more leniently (which implies that I will ask easier questions on the quizzes, but grade it more strictly).
I have been relying solely on quizzes for the past several years, so grading homework will be something that I haven’t done for a while. I initially planned on only allowing quizzes for this system, too, but it seemed like things would be overwhelming for everyone: we would likely have daily quizzes (rather than maybe twice per week); I would likely not give class time to “tag” quizzes, so students would do this at home (creating a logical nightmare); I would probably have to spend a lot more time coaching students on how to tag (whereas they now get to practice it on the homework with other people).
Let’s end this post, Rundquist-style, with some starters for you.
1. This is an awesome idea because …
2. This is a terrible idea because …
3. This is a good idea, but not worth the effort because …
4. This is not workable as it is, but it would be if you changed …
5. Homework is a terrible idea because …
6. You are missing this learning goal …
7. My name is TJ, and you are missing this process goal …
Summer Plan
June 11, 2014
My family and I agree that things work best when I work pretty strict hours—I work 7:45 am to 5 pm during the school year. I do very little work at home. However, I need to do a lot of prep work during the summer to make this possible. Because of this, I work a lot in the summer (we allow for 6 weeks of vacation for the year, so the default mode for the summer is “work”), although my hours are now 8:15 am to 5 pm.
Here is my plan for the summer:
1. Take care of all of the annoying paperwork-type-stuff that needs to be done. This includes some work that I do every summer: updating my CV, updating websites, and reading and summarizing course evaluations. I also have some jobs that are particular to this summer, such as determining which mathematics courses should be considered for transfer credit at some neighboring colleges. (I am happy that I have already done this entire item).
2. Do some reading about redesigning general education requirements. My college is considering restructuring these requirements, and my main goal for the summer is to try to determine (along with my other committee members) some sort of reasonable process for this. Fortunately, this is paid work (mostly).
3. Plan my geometry (and prob/stats/graph theory) course for elementary education majors for the fall. This is also done, largely because I taught this course in the spring. I kept detailed notes (I am grateful I did this), and I mainly updated this course by building in more feedback. In particular, I wrote all of my quizzes for the semester, created solution videos for each quiz, and updated my examinations.
4. Plan my calculus course. I am planning on using Team-Based Learning, which I learned about from Eric Mazur in this video. Again, planning includes (in chronological order) creating all learning goals, creating all assessments, and creating all class activities. When the semester comes, my main task will be briefly reviewing the plan, adapting that plan based on the students’ needs, recording what actually happened (and how I might improve things next time), meeting with students, and grading.
5. Do research. I have 3–4 papers that I need to write up, and I hope to re-start work on two projects that have been on hold for too long.
Finally, one benefit of working during the summer is you can be amazingly productive. I am often the only person here, and I can be very productive in such an environment.
Mutt revisited
June 4, 2014
This is a short story about why it is nice to blog; your comments helped me realize what I actually wanted to do. I wrote last week about how I was unhappy with Mutt. Summary: I tried to run GMail through the email program Mutt, and the result was a really slow email program.
Because of several comments by different people, I realized that
1. I have a slight concern about how Google respects my personal privacy; this is not a huge concern for me, though, and it would not be enough to make me switch.
2. I have a huge concern about my students’ privacy, and I have been concerned about GMail for a while. Your comments helped me realize that Mutt could be a solution.
Because of a conference relating to my biological children’s educations that I attended last weekend, I realized that I really want to learn more about Linux. So Mutt gives me a chance to do this.
So now I have three reasons to change, and a colleague (Michael Gass) gave me the most elegant solution: use Mutt without GMail. That is, I now use Mutt and POPMail to get mail directly from my school’s servers. Now, Mutt is as fast as it should be.
Today is the first day that I have Mutt up and running, although I have been reasonably happy with it so far. But if I run into problems, I might switch back. One potential problem is that I check my email mostly on my iPad at home, so I need to figure out something to do there (although it could be that my ssh app will work just fine with Mutt on the iPad). A possibly related problem is that getmail is not deleting the emails once they are fetched from the school’s servers (even though I have ‘delete=true’ in my .getmailrc file). This drives me crazy because I crave empty inboxes, although this may be a solution to my iPad problem; I can access Outlook via the school’s webpage or some other app, and the messages will all be there.
I suppose there is a good chance that I will change my mind again next week.
Mutt vs Gmail
May 29, 2014
I really wanted it to work. I really did.
I experimented for a couple days with using Mutt for email. I love the ability to compose emails with Vim, and I had heard that it was lightning fast. If I am honest with myself, my interest in Mutt was also related to the fact that I aspire to be a Linux geek (which I cannot claim to be right now).
So I configured Mutt to get mail from my Gmail account. That way, I could still get all of the benefits of Mutt at work while still enjoying the ease of access to Gmail everywhere else (I usually use the iPad to check email at home).
I wanted Mutt to work. I really wanted Mutt to work. But it didn’t. The problem was that it was muuuuuuuch slower than Gmail. When sending or archiving messages, Mutt took probably 5 to 10 times longer than Gmail did (Mutt takes maybe 2 seconds, whereas Gmail is near instantaneous. I recognize that 1998 Bret would be ashamed of 2014 Bret for caring about this small of a difference). Since I can mostly use Gmail without a mouse, I have come to the conclusion that Mutt is not worth it for me.
Of course, there is always the possibility that I screwed something up while configuring Mutt; let me know if you think that I did something wrong to cause this.
The Importance of Feedback
May 22, 2014
My semester is ended, and now is the time to write some post-mortem entries into this weblog. The first idea is something that is probably obvious, but I over-thought it. I have been been putting more of the course’s assessment at the end of the semester lately, thinking that that is when students are most prepared to do well.
And I am correct, but I took it too far. I did not give my students enough regular feedback during the first part of the semester this spring. My education students actually pointed this out to me—I realized that they were correct as soon as they said it (it also reinforced that they are pretty on top of education issues). Fortunately, I get to teach that course for education majors again this fall; I will make things right this time.
Additionally, I am working on ways of getting students immediate feedback. Clickers are one way of doing this, but I also might have students start grading their own quizzes (I would provide a couple of solution keys and a marker for them) and doing more computer-graded stuff.
Speeding Up Videos
May 14, 2014
I had my 20 elementary education majors produce video projects. These were all due at the end of the semester, which means I have to grade them all now. Each student produced roughly seven 4-minute videos, which means that I have roughly 10 hours of video watching to do as part of my grading.
Or do I? I downloaded this app for my Chromebook, and I have been watching the videos at twice or triple the usual speed (depending how quickly the student naturally talks), cutting my time watching videos to 3–5 hours with no loss of grading quality.
I am very grateful for this app today.
Undergraduate Reseach: Jump Before Looking
May 7, 2014
I talked about my plan for undergraduate research last week. This week, I invited my linear algebra class to join a research team I am forming.
The class is roughly half sophomores and half first-years. They have had calculus and linear algebra. My plan is to come up with a research question based around either finite fields or group actions on cyclic groups. I feel like I have some questions at the appropriate level that have come up in my own research, although I cannot explicitly state them right now. I had better be able to by next fall if any students decide to join the research team.
Draft of an Undergraduate Research Philosophy
April 30, 2014
I am working on establishing a sustainable undergraduate research program. I want to record some ideas that I have here.
First, I think that this might mean a shift toward searching for problems that undergraduates can understand the research question. This is actually pretty close to what I have been doing anyway, although I hope to more consciously seek out easily-understood problems. I had lunch last weekend with Andy Rundquist (#brag), and he told me that he changed his research focus so that it would be easier for undergraduates to work with him. Fortunately for me, he did this in part because of non-academic considerations (lasers are expensive), but I understand that his main focus was to allow students to work with him immediately after their freshman year. I still will study group theory, although I will see if there are questions that students could quickly understand after just having had one or two semesters of calculus. In particular, I might start learning something about finite fields, which I think could be accessible (it is just like the real numbers, only there are only a finite number of points!).
If I can find good questions (that is part of my goal for my sabbatical next spring), then I would like to form a research group. I hope to work with several students at once. The model I have in mind is that each student will work on solve the research question for a specific case—likely a specific family of groups. The students will be able to talk to each other, since each knows the question being asked, although not every student will know the structure of each, say, particular group.
Essentially, these students would be working on the examples that I would do myself if I were trying to solve the problem on my own. After the students complete their work, they perhaps write a thesis on the problem and I see if I can use their work to solve the entire problem.
This gives students a chance to do undergraduate research, gives them a chance to do it in a more collaborative manner (they get to work with a research team), and it gives me a chance to kill two birds with one stone—the undergraduate research is actually supporting my own research, so time spent with undergraduates is really time spent on my own research.
Do you have anything thoughts on this model? Do you have any alternative models for undergraduate research that work?
Firefox Shortcuts
April 27, 2014
I have been so happy with the GMail keyboard shortcuts that I have decided to learn similar shortcuts for Firefox.
I mainly just want to be able to quickly change tabs, move to the search bar, and move to the address bar. If I could do those three things, I will be in good shape.
[Edit 6:29 am on April 28: I am already thrilled just from knowing that control-l puts you in the address bar, control-k puts you in the search bar, control-[ goes back a page, and control-] goes forward. This is in addition to already using control-t to open a new tab. I find that control-1 through control-8 is also helpful to go to the 1st through 8th tab, although I use this less than the other commands.]
I haven’t practiced these yet, but I will starting tomorrow.
(Again, I am probably really late on this).
Students Figure Out Which Standards They Meet
April 16, 2014
I am starting to think about planning for Calculus I for next year, and there is an idea I would like to try: I want to stop labelling problems according to the corresponding standard, and put the burden on the student to determine which standards they met. I have tried this before (as have other people), but I would implement it different from how I did it last time.
So each quiz would go like this: I give them several (unlabelled) quiz problems. The students do what they can. When they are done, they submit their work. However, when they submit, we make some sort of a copy (perhaps a paper copy, perhaps just take a picture with the smart phone), and then the student takes one copy home.
At home, the student tries to figure out which standards she met on the quiz. For each standard, she writes up an argument as to why she met that standard. Specificity is key—the student would need to explicitly say where and how she met the standard. She submits this at the next class period, and this is graded as I usually do.
Here are the things I like about this idea:
1. Students have to reflect on their work in order to get credit. This could lead to higher quality writing.
2. Students would have to take ownership of their learning. They need to be aware of the standards they are missing, and make a concerted attempt to learn it well enough to be able to apply it on a quiz (including recognizing where it makes sense to apply it).
3. Students can solve problems any way they like. As long as they can solve the problem using a standard, it counts. For instance, a linear algebra student might get “eigenvalue” and “determinant” credit for finding the eigenvalues of a matrix.
4. Students are forced to really think about what the standards are and mean. There could be metacognitive benefits.
5. I can ask more synthesis questions on quizzes; I do not need to isolate ideas for each question.
6. Students no longer get the hint that the label provides (if the quiz question is labelled as corresponding to the “Tangent line” standard, then the student has a pretty good idea that he should find a tangent line at some point).
7. It might give me room to have more standards (and more specific standards of the “I can do this” variety, rather than standards that are really topics, as in “Tangent lines.” David Clark encouraged me to make this transition last weekend).
Here are some potential problems:
1. If the problems are too synthesis-y, then students won’t be able to do very many on each quiz. This might be fine, but it would be bad for a student who gets stuck and does not know where to start (on the other hand, maybe it would help teach students to start with something?).
2. Students may try to shoehorn standards where they do not belong. This is what I would do if I were missing a small subset of standards.
3. I am not certain I can write quiz problems that will give everyone the opportunities they need at the end of the semester. Students need different things, so I would have to have a lot of questions (note: this actually doesn’t need to be any different than how it is now; I can just provide straightforward, say, “Tangent lines” problems to quizzes if I need to. So this actually isn’t much of a problem).
4. It forces students to be aware of what they have not yet demonstrated; this might be asking too much of some first-years.
I am on the fence about this, although I would really like to try it. Perhaps I could do both: keep the old way (with the labels) and do the new way. I could make that work.
What am I missing? What other advantages, disadvantages, and difficulties would this have?
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2014-07-23 18:03:22
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https://hal-cea.archives-ouvertes.fr/cea-01541457
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# Bulk fields from the boundary OPE
Abstract : Previous work has established an equality between the geodesic integral of a free bulk field in AdS and the contribution of the conformal descendants of its dual CFT primary operator to the OPE of two other operators inserted at the endpoints of the geodesic. Working in the context of the AdS$_3$/CFT$_2$ correspondence, we extend this relation to include the $1/N$ corrections to the bulk field obtained by dressing it with i) a $U(1)$ current and ii) the CFT stress tensor. In the former case, we argue that the contribution of the Ka\v{c}-Moody descendants to the respective boundary OPE equals the geodesic integral of a particular $U(1)$-dressed bulk field, which is framed to the boundary via a split Wilson line. In the latter case, we compute the gravitational $1/N$ corrections to the bulk field in various gauges, and then write a CFT expression for a putative bulk field whose geodesic integral captures the contribution of Virasoro descendants to the OPE of interest. We comment on the bulk interpretation of this expression.
Document type :
Preprints, Working Papers, ...
Domain :
Cited literature [20 references]
https://hal-cea.archives-ouvertes.fr/cea-01541457
Contributor : Emmanuelle de Laborderie <>
Submitted on : Monday, June 19, 2017 - 10:53:54 AM
Last modification on : Wednesday, January 23, 2019 - 2:39:04 PM
Long-term archiving on : Friday, December 15, 2017 - 4:16:54 PM
### File
1610.08952.pdf
Files produced by the author(s)
### Identifiers
• HAL Id : cea-01541457, version 1
• ARXIV : 1610.08952
### Citation
Monica Guica. Bulk fields from the boundary OPE. 2017. ⟨cea-01541457⟩
Record views
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2019-11-19 21:51:22
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http://www.physicsforums.com/showthread.php?t=246736
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# Relative Permittivity and Refractive Index
by iLIKEstuff
Tags: index, permittivity, refractive, relative
P: 25 in relating the index of refraction to the relative permittivity (dielectric constant/function). it is known that $$n = \sqrt{\epsilon_r}$$ for optical frequencies (i.e. $$\mu_r=1$$. now this website http://hyperphysics.phy-astr.gsu.edu...s/diel.html#c1 gives the relative permittivity of water as 80.4 i.e. $$\epsilon_r = 80.4$$ but we also know that the index of refraction of water is 1.33. so it should be $$n = \sqrt{\epsilon_r} = \sqrt{80.4} = 8.9666$$ ??? am i missing something here? i want to use the relative permittivity in an equation to calculate the electrostatic approximation of the scattering/absorption efficiencies of small spherical particles. should i be solving for relative permittivity from the index of refraction? i.e. $${n}^{2} = {(\sqrt{\epsilon_r})}^{2} \Rightarrow \epsilon_r={1.33}^{2}=1.7689$$ what value should i use for the relative permittivity in this equation? thanks guys.
P: 4 Permittivity is a function of wavelength (frequency). 80.4 value is valid for microwave diapason, not for optical one. For optical frequencies you should calculate permittivity from refractive index, i.e. $$\epsilon_r = n^2$$ To find accurate value of n at a particular wavelength in optical diapason use, for example, RefractiveIndex.INFO
Related Discussions Introductory Physics Homework 3 Classical Physics 11 Introductory Physics Homework 1 General Physics 2
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2014-04-20 01:05:54
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https://www.toebi.com/blog/rant/results-nasa-juno/
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# Where Are The Results NASA? (Juno)
What a heck? It has been over year from the flyby but no results released by NASA. How hard that can be? Seriously. NASA could have put like a short tweet or something out for two weeks after the flyby… but nooo! I wonder why…?
Maybe those results are buried under somebody’s backyard or something? Are those results classified for some reason? Anyway, my prediction $1.09$ mm/s at perigee looks pretty strong, at least when I retrodict those previous flybys with my formula.
Update: I had a quite interesting tweet exchange with Mike McCulloch who told that he got confirmation from former NASA employee that there wasn’t any anomalous velocity increase with Juno Earth flyby. On top of that, he got that information only a week after the flyby.
Weird part is that in AGU Fall Meeting 2013 there was an ePoster (which is now non-accessible, here’s its abstract thou) which said that the data analysis is still ongoing. It included also a picture which kind of hinted that some kind of anomalous increase was measurable… pretty strange! On the other hand, Mike told that there is some “tension” regarding these flyby anomalies in NASA.
I’m puzzled! No other options than waiting and trying to probe more info from NASA.
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2017-06-22 14:11:47
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https://www.effortlessmath.com/math-topics/other-topics-puzzle-math-challenge-101/
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# Other Topics Puzzle – Math Challenge 101
This is another perfect math puzzle that can be solved without even using a pen and paper! Paying attention to the details is the key to solve this fascinating Math puzzle!
### The Absolute Best Book to challenge your Smart Student!
First, notice that the clocks show different times. The first two clocks show 9 and the third one shows 3. Then: $$9+9+3=21$$
The three calculators sum up to 30. So, each one represents number 10.
The third equation, each lamp represents number 15. $$(15+15-15=15)$$
Now, let’s solve the last equation. The clock represents number 9. This calculator is different from the calculators in the second equation. The numbers in the previous calculators are 1, 2, 3, and 4. We know that the calculators represent number 10. So, $$1+2+3+4=10$$
The calculator on the last equation had numbers 1, 2, 2, and 4. So, the calculator represents number 9: $$1+2+2+4=9$$
Now, notice that the lamps on the last equation are different from the lamp in the third equation. That lamp has 5 bars on it, but these lamps have 4 bars on them. The lamp on the third equation represents number 15. So, 5 bars on the lamp represent 15. Then, each of the lamps with 4 bars represent $$12 \ (4×3=12)$$.
Now, let’s solve the last equation.
$$9+9×3×12$$
Following the order of operation rule: $$9+9×3×12=9+27×12=9+324=333$$
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2021-06-23 00:27:27
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https://learn.careers360.com/ncert/question-tick-the-correct-answer-and-justify-abc-and-bde-are-two-equilateral-triangles-such-that-d-is-the-mid-point-of-bc-ratio-of-the-areas-of-triangles-abc-and-bde-is/
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Q
# Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
Q8 Tick the correct answer and justify :
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of
the areas of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Views
Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
All angles of the triangle are $60 \degree$.
$\triangle$ABC $\sim \triangle$ BDE (By AAA)
Let AB=BC=CA = x
then EB=BD=ED=$\frac{x}{2}$
$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=(\frac{x}{\frac{x}{2}})^2=\frac{4}{1}$
Option C is correct.
Exams
Articles
Questions
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2020-02-28 00:27:52
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http://www.wisdom.weizmann.ac.il/~/oded/MC/047.html
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## Cuckoo Nesting: Modern Methods for Organizing Lookup Tables
a "popular talk" by Moni Naor.
#### Oded's summary
The directionary problem consists of storing an a priori unknown set $S$, $S\subset U$, in a manner that supports (relatively efficient) searching, insertion and deletion. The focus is on time of individual operations and memory size. Two approaches:
• Search trees -- usually require $\log|S|$ time and use of pointers.
• Hashing tables allow "direct access" and raise the problem of collisions. In general $Hash:U\to[r]$ for $r\approx n \eqdef |S|$.
The naive "linear probing" calls for storing $x$ in the first free location $p\geq Hash(x)$. That is, if $H(x)$ is taken we try $H(x)+1,H(x)+2$...
Cuckoo Hashing [Pagh and Rodler, 2001] is an alternative method of dealing with collisons. It uses two hashing functions, $H_1$ and $H_2$, and is best described in terms of using two tables $T_1$ and $T_2$. We store $x$ in $T_1[H_1(x)]$, and if $y$ previous resided there, then we store $y$ in $T_2[H_2(y)]$, and again this may push out a previous contents $z$ that may be stored in $T_1[H_1(z)]$ and so on...
The analysis of this method reduces to an analysis of a random sparse graph. Specifically, for fixed $S$ and $H_1,H_2$, we consider the $r$-by-$r$ bipartite graph in which there are $n$ edges such that for each $x\in S$ we have an edge $(H_1(x),H_2(x))$. Note that by the Cuckoo Hashing rule, the element $x$ must be stored in either $T_1[H_1(x)]$ or $T_2[H_2(x)]$, and so storage is possible iff every connected component in the graph has at most one simple cycle (i.e., has a number of edges that does not exceed the number of vertices in this connected component). Furthermore, the number of operations that result from "insert" is reflected in the size of these components.
De-amortization and use of queue -- see (see Arbiteman, Naor, and Segev.
Back to list of Oded's choices.
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2020-04-02 19:29:49
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https://community.dataquest.io/t/understand-the-answer-given-for-the-exercise/526765
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# Understand the answer given for the exercise
My Code:
import re
email_tests = pd.Series(['email', 'Email', 'e Mail', 'e mail', 'E-mail',
'e-mail', 'eMail', 'E-Mail', 'EMAIL', 'emails', 'Emails',
'E-Mails'])
pattern = r'\be-? ?mails?\b'
email_mentions = titles.str.contains(pattern,flags=re.I).sum()
Hello!
I submit the answer above for the exercise and it worked well, following this logic:
• all words in the list begins with and e or E;
• some of them have an hyphen symbol between the first letter and the second letter;
• some of them have an emppty space between the first letter and the second letter
• some of the have the letter “s” at the end;
• none ot them have a hyphen and a empty space simultaneously.
But this is the answer given for the exercise:
pattern = r"\be[\-\s]?mails?\b"
I’m having a hard time to understand this piece of code: [\-\s]?.
As far as i understand (and I’m aware that I’m probably wrong) it would match cases where you have nothing between the “e” and the “m” letter (in other words, the set does not exist), and it would also match cases where there is a hyphen followed by “any space, tab or linear break character”, which is the eactly definition given for \s during the mission.
Considering all the items in email_tests (and also my interpretation), the results would be:
email -> match, there is nothing between the “e” and the “m”
Email -> match, there is nothing between the “E” and the “m”
e Mail -> no match, there is not a hyphen followed any space, tab or linear break character
e mail -> no match, there is not a hyphen followed any space, tab or linear break character
E-mail -> no match, there is not a hyphen followed any space, tab or linear break character
e-mail -> no match, there is not a hyphen followed any space, tab or linear break character
eMail -> match, there is nothing between the “e” and the “M”
E-Mail -> no match, there is not a hyphen followed any space, tab or linear break character
EMAIL -> match, there is nothing between the “E” and the “M”
emails -> match, there is nothing between the “e” and the “m”
Emails -> match, there is nothing between the “E” and the “m”
E-Mails -> no match, there is not a hyphen followed any space, tab or linear break character
I was also not able to understand why to use the scape backlash before the hyphen…
By the way, the first steps with Regular Expressions have been great! It’s hard to grasp the way of working at the beginning, but it’s very clear that it can be a powerful weapon to be used along with pandas for the next projects.
Wish you all a nice sunday!
Paulo
1 Like
Hi @phssaraiva:
\s matches a whitespace character as I answered here and \- matches a hyphen. This means that spaces or hyphens are allowed between e and mail. The escape character \ is used to denote the importance of the next character because sometimes not having the backslash (say you want to filter for a double quote ") which terminates the string early on and thus python will throw an error in that case.
# Red part is valid, after that becomes black (invalidated portion of string)
pattern = r"\be["\s]?mails?\b"
Hope this helps!
1 Like
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2021-09-27 13:47:27
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http://mathhelpforum.com/advanced-statistics/178819-poisson-distribution-question.html
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1. ## Poisson distribution question
Q has Poisson dist. Q will be observed tomorrow for once. If Q turns out to be a certain #, such as k, the coin will be tossed k times. Let X be the number of times tails come up in k tosses. If Q turns out to be zero, then X is also zero.
1. Find the chance that X is equal to 1. This means that the chance tails comes up exactly once in k tosses. (The answer should be a formula involving parameters of Poisson dist.)
2. Find the chance that X=n. (The answer should be a formula involving parameters of Poisson dist.)
2. Walk through it. It's called exploration.
Q is Poisson with parameter L, so E(Q=k) = (e^-L)(L^k)/k! for k in 0, 1, 2, ...
X is Binomial with parameters n = k and p = 1/2. p(X=1) = k*(1/2)*(1/2)^(k-1) = k*(1/2)^k
So, p(X=1,Q) = [(e^-L)(L^0)/0!]*0*(1/2)^0 + [(e^(-L))(L^1)/1!]*1*(1/2)^1 + [(e^(-L))(L^2)/2!]*2*(1/2)^2 + [(e^(-L))(L^3)/3!]*3*(1/2)^3 + ...
= (e^(-L))*{[(L^0)/0!]*0*(1/2)^0 + [(L^1)/1!]*1*(1/2)^1 + [(L^2)/2!]*2*(1/2)^2 + [(L^3)/3!]*3*(1/2)^3 + ...}
= (e^(-L))*{[(L^1)/1!]*1*(1/2)^1 + [(L^2)/2!]*2*(1/2)^2 + [(L^3)/3!]*3*(1/2)^3 + ...}
= (e^(-L))*(L/2)*{1 + [L/2!]*2*(1/2) + [(L^2)/3!]*3*(1/2)^2 + ...}
= [e^(-L)]*[e^(L/2) - 1]
Do you recognize it as anything familair?
Maybe examining the Moment Generating Function would be more instructive.
3. I am not sure how you jumped from the second to the last part to the answer. Could you explain a little more? I did take a look at the Moment Generating Function, but I am not really sure how to relate it to this.
Also, when X is 1, i see that it's simple. For part number 2, it gets really complicated, but is it the same concept? What should I look for in number 2?
4. Adding up infinite series is a nontrivial exercise. That's why I suggested the Moment Generating Functions. You might find a table with this one.
$\sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)$
$\sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)$
I just used my CAS to tell me the sum. Perhaps some other soul will demonstrate it, but it escapes me at the moment.
5. edit:didn't see previous post
What should I look for in number 2?
part 2 is the same concept.
Whenever you are asked to do a sum involving a poisson distribution, you may find it helpful to remember these "tricks"
1: can you express the sum in terms of another distribution function (possibly also posson), then use the fact that distribution functions always sum to 1?
And/Or
2: can you express the sum as a power series for e^{something}
For this question (part 2):
Hint #1 : You can write the binomial function as n!/(k!(n-k))
Hint #2: if you wade through the algebra you may find the first "trick" useful at the end
6. Originally Posted by TKHunny
Adding up infinite series is a nontrivial exercise. That's why I suggested the Moment Generating Functions. You might find a table with this one.
$\sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)$
$\sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)$
I just used my CAS to tell me the sum. Perhaps some other soul will demonstrate it, but it escapes me at the moment.
[where a = lambda to save typing]
EDIT Solution removed....just slightly suspicious that this is assessed work...call me paranoid
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2016-10-27 14:29:11
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https://brilliant.org/discussions/thread/chemistry-doubt-3/
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×
# Chemistry doubt
$$0.5\text{ g}$$ sample of a sulphite salt was dissolved in $$200\text{ ml}$$ and $$20\text{ ml}$$ of this solution required $$10\text{ ml}$$ of $$0.02\text{ M}$$ acidified permanganate solution.Find the percentage by mass of sulphite in sulphite ore?
Please help!
Note by Shivam Mishra
7 months, 2 weeks ago
## Comments
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My solution :
Let the number of moles of sulphite ions be $$M$$. Clearly on oxidation the sulphite ion changes to sulphate so the oxidation number changes from $$4+$$ to $$6+$$, hence the valence factor is $$2$$. The product of moles and valence factor gives the equivalents. The valence factor permagnate solution in acidified solution is $$5$$. Equate the equivalents as :
$M ×2×\frac{20}{200} = \frac{10}{1000} × 0.02 × 5$
$M=\frac{5}{1000}$
As molecular weight of sulphite ion is $$80$$, percentage by mass is :
$\frac{0.005 × 80}{0.5} × 100$ $=80 %$ · 6 months, 3 weeks ago
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Is the information complete? · 7 months, 2 weeks ago
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Yes · 7 months, 2 weeks ago
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80% · 4 months, 4 weeks ago
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Thanks @Utkarsh Dwivedi · 6 months, 3 weeks ago
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Tell me is this question based on molarity · 7 months ago
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Yes,but it can also be solved using milliequivalents. · 7 months ago
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@Deeparaj Bhat @Prakhar Bindal please help!!!! · 7 months, 2 weeks ago
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2017-01-22 12:25:34
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https://ftp.aimsciences.org/article/doi/10.3934/dcds.2020094
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Article Contents
Article Contents
# Topological cubic polynomials with one periodic ramification point
First author is supported by "Fondecyt Iniciación 11170276".
Second author is supported by CONICYT PIA ACT172001 and "Fondecyt 1160550".
Both authors partially supported by MathAmsud 18-Math-02 HidiParHodyn.
• For $n \ge 1$, consider the space of affine conjugacy classes of topological cubic polynomials $f: \mathbb{C} \to \mathbb{C}$ with a period $n$ ramification point. It is shown that this space is a connected topological space.
Mathematics Subject Classification: Primary: 37F10, 37F20, 37F30.
Citation:
• Figure 1. Illustration of Lemma 3.1 for a topological polynomial $f$ where $[(f, c, c')]\in{\mathcal{E}}({\mathcal{F}}_4)$ has kneading word $1000$
Figure 2. Illustration of the construction of the twisting loop corresponding to $m = 3$ and kneading word $1000$. The exterior curve in black is the level curve $g_{f_0} = g_{f_0}(c_0')$. The set $f_0^{-1}(Y)$ is drawn in gray
Figure 3. Illustration of the annulus $A$ around the twisting loop $\tau$ (left) and its preimage (right)
Figure 4. On both pictures, the doted curve represents the outer boundary of $\partial A_{ext}'$. The lightest gray regions are ${V'_0\cup V'_1}$. The other gray regions are the complement of $V'_0\cup V'_1$ in $f_0^{-1}(D)$ (left) and in $f_1^{-1}(D)$ (right).The lines in the darker gray regions represent the preimages of $\gamma$ by $f_0$ (left) and $f_1$ (right) where $\gamma$ is as in Section 4.2
• [1] J. W. Alexander, On the deformation of an n-cell, Proc. Nat. Acad. Sci., 9 (1923), 406-407. [2] A. Bonifant, J. Kiwi and J. Milnor, Cubic polynomial maps with periodic critical orbit. Ⅱ. Escape regions, Conform. Geom. Dyn., 14 (2010), 68-112. doi: 10.1090/S1088-4173-10-00204-3. [3] B. Branner, Cubic polynomials: Turning around the connectedness locus, Topological Methods in Modern Mathematics (Stony Brook, NY, 1991), Publish or Perish, Houston, TX, (1993), 391–427. [4] B. Branner and J. H. Hubbard, The iteration of cubic polynomials. Ⅰ. The global topology of parameter space, Acta Math., 160 (1988), 143-206. doi: 10.1007/BF02392275. [5] B. Branner and J. H. Hubbard, The iteration of cubic polynomials. Ⅱ. Patterns and parapatterns, Acta Math., 169 (1992), 229-325. doi: 10.1007/BF02392761. [6] G. Z. Cui and L. Tan., A characterization of hyperbolic rational maps, Invent. Math., 183 (2011), 451-516. doi: 10.1007/s00222-010-0281-8. [7] S. V. F. Levy, Critically Finite Rational Maps, PhD thesis, Princeton University, 1986. [8] J. Milnor, Dynamics in One Complex Variable, Third edition, Annals of Mathematics Studies, 160. Princeton University Press, Princeton, NJ, 2006. [9] J. Milnor, Cubic polynomial maps with periodic critical orbit. Ⅰ, Complex Dynamics, A K Peters, Wellesley, MA, (2009), 333–411. doi: 10.1201/b10617-13. [10] M. Rees, Views of parameter space: Topographer and Resident, Astérisque, (2003).
Figures(4)
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2022-11-30 12:59:26
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https://quantumcomputing.stackexchange.com/questions/18506/what-are-the-conditions-ensuring-a-two-qubit-density-matrix-is-positive-semidefi
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# What are the conditions ensuring a two-qubit density matrix is positive semidefinite?
I've seen some papers writing $$\rho=\frac{1}{4}\left(\mathbb{I} \otimes \mathbb{I}+\sum_{k=1}^{3} a_{k} \sigma_{k} \otimes \mathbb{I}+\sum_{l=1}^{3} b_{l} \mathbb{I} \otimes \sigma_{l}+\sum_{k, l=1}^{3} E_{k l} \sigma_{k} \otimes \sigma_{l}\right).$$ I wonder what condition should the matrix $$E$$ obey?
For a one-qubit state, the density matrix satisfies $$\newcommand{\tr}{{\operatorname{tr}}} \tr(\rho)=1$$ and is semi-positive. And the general form is $$1/2\begin{pmatrix}1+z & x-iy\\x+iy & 1-z\end{pmatrix},$$ satisfying trace condition. As for semi-positive conditions, there's a theorem stated that a hermitian matrix is semi-positive iff its eigenvalue is not negative. So I calculate the eigenvalues of it and get the restriction that $$x^2+y^2+z^2 \le 1$$, which can be seen as exactly the Bloch sphere.
Then I want to see the same thing happens in two qubits case. But I can only mimic the same reasoning and get the general form $$\rho=\frac{1}{4}\left(\mathbb{I} \otimes \mathbb{I}+\sum_{k=1}^{3} a_{k} \sigma_{k} \otimes \mathbb{I}+\sum_{l=1}^{3} b_{l} \mathbb{I} \otimes \sigma_{l}+\sum_{k, l=1}^{3} E_{k l} \sigma_{k} \otimes \sigma_{l}\right).$$ And when I try to calculate the eigenvalues of this matrix, even Mathematica showed a complex result. But if we think the separable case, easy to see that the vector $$a$$ and vector $$b$$ should have length less than 1. But I can't find the restriction on matrix $$E$$.
To summarize, in general, the two qubits state can be stated as: $$\rho=\frac{1}{4}\left(\mathbb{I} \otimes \mathbb{I}+\sum_{k=1}^{3} a_{k} \sigma_{k} \otimes \mathbb{I}+\sum_{l=1}^{3} b_{l} \mathbb{I} \otimes \sigma_{l}+\sum_{k, l=1}^{3} E_{k l} \sigma_{k} \otimes \sigma_{l}\right).$$ What's the restriction on it's parameters to make it a legal density matrix, i.e., trace condition and semi-positive condition?
1. (General framework) The general form of this problem is the following. Take a generic Hermitian, unit-trace operator, $$X\in\operatorname{Herm}(\mathcal X), \operatorname{Tr}(X)=1$$, acting on some $$N$$-dimensional complex vector space $$\mathcal X$$. The set of such operators is an affine space of dimension $$N^2-1$$, embedded in the $$N^2$$-dimensional real vector space $$\operatorname{Herm}(\mathcal X)$$ of Hermitian operators. Any operator can be decomposed linearly with an operatorial basis. In particular, any unit-trace Hermitian operator can be decomposed linearly in an orthonormal basis of Hermitian traceless operators. This is the "generalised Bloch representation", which we can write as $$X = \frac{1}{N}\left(I + \sum_{k=1}^{N^2-1} c_k\sigma_k\right).$$ More precisely, we should write $$c_k=c_k(X)$$. These coefficients $$c_k$$ implement a (linear) isomorphism between the set of unit-trace Hermitian operators and the vector space $$\mathbb R^{N^2-1}$$.
2. (Positivity constraint) If we want a positive semidefinite unit-trace Hermitian operator, write it $$\rho$$, we need to impose further restrictions on the coefficients $$c_k$$. More specifically, rather than these spanning a full linear space, they will now only cover a convex subset of $$\mathbb R^{N^2-1}$$. This means, in particular, that you can describe the corresponding set of states via its boundary.
3. (Boundary ain't nice for $$N>2$$) This boundary is, in general, not very nice. For a single qubit you get a sphere, but as soon as you increase the dimensions, things get messier. For example, this boundary contains non-pure states (which doesn't happen for $$N=2$$). Similarly, even though all pure states still lie on a hypersphere (assuming a suitable parametrisation), they only cover some subsets of it.
4. (Radial characterisation is doable) Nonetheless, as shown here, it is possible to get a decent characterisation of this boundary, by using "radial" coordinates. This requires computing the eigenvalues corresponding to a given direction, that is, the eigenvalues corresponding to a given linear combination of basis operators. This is numerically trivial to do for any given direction (assuming $$N$$ is not too big, of course), but finding an analytical expression working for all directions would be very complicated.
It is worth noting that this amounts to describing the (convex) set of states via its support function. The conversion between this description and a more standard analytical characterisation is quite complicated, if doable at all.
5. (So... what are the constraints?) Getting back to the question at hand: what does it really mean to find "restrictions for the coefficients $$E_{jk}$$"? Generally speaking, the boundary of the set of coefficients $$c_k(\rho)$$ corresponding to valid states $$\rho$$ is the solution set of $$f(c_1,...,c_{N^2-1})=0$$ for some function $$f$$. E.g. for $$N=2$$, we have $$f(c_1,c_2,c_3)=c_1^2+c_2^2+c_3^2$$. In higher dimensions, this $$f$$ is more complicated. For the case at hand, the coefficients are related to each other by some implicit relation $$f(a_1,a_2,a_3,b_1,b_2,b_3,E_{11},...,E_{33})=0.$$ This means that the viable range of any $$E_{jk}$$ depends on the value of all the other parameters. What you can do easily is find things like max and min of any parameter. This amounts to asking, e.g. what is the largest value of $$E_{ij}$$ that corresponds to any quantum state. Using the radial characterisation above, this can be seen to equal $$1/|\lambda_{\rm min}(\sigma_i\otimes\sigma_j)|=1$$. In fact, by similar reasoning, you can see that, using your operatorial basis built with products of Pauli matrices, every parameter lies in the interval $$[-1,1]$$.
Note that this might seem to contradict the simple condition you get imposing $$\operatorname{Tr}(\rho^2)\le1$$, given in the other answer. That this is not a contradiction, is an interesting point per se. In fact, $$\operatorname{Tr}(\rho^2)=1$$ traces a spherical boundary that (1) surrounds the full set of states, and (2) is only touched by pure states. In other words, the answer $$\max_\rho |E_{jk}|=1$$ does't contradict the boundary given in the other answer, because the former is achieved by non-pure states (which is also easy to verify directly).
• How does it follow that the parameters must lay in $[-1;1]$ from the radial characterization? Yes, we have the condition that $r \le 1/|m(F_{\mathbf n})|$. But we don't know that $m(F_{\mathbf n})=1$ for every $\mathbf n$ with norm 1. Even in the single qubit case it's not so trivial to show that $m(F_{\mathbf n})=1$ for every $\mathbf n$. Jul 20 at 16:42
• @DanyloY we do not have $m(F_{\bar n})=1$ for every $\hat n$. I'm only saying we have for the directions corresponding to each basis operator (well, maybe for other operators as well, I'm not sure). In fact, we know that pure states have distance $\sqrt3$, as also shown by you via the ${\rm Tr}(\rho^2)=1$ condition. I mean that each one of the parameters attached to the basis operators, so $a_i,b_i,E_{jk}$ here, are bounded in that range. You can also see directly that this is the case, because e.g. $\frac{1}{4}(I+\sigma_3\otimes I)$ is singular, and thus lies on the boundary
– glS
Jul 20 at 16:45
• I think a (naive and mostly wrong) way to picture what's going on geometrically is to imagine the set of states as a square inscribed in a circle. The orthogonal directions correspond to flat sections of the boundary, so the corresponding radius is smaller than the radius of the embedding circle. Pure states are the intersections between circle and square (of course, the actual space is some complicated generalisation of this to higher dimensions, where instead of the "square" there is some more complex polytope-like surface, whose boundary includes both flat and nonflat sections)
– glS
Jul 20 at 16:57
• @DanyloY how about take $\rho=\frac1{N}(I+\sum_j c_j(\rho)\sigma_j)$ with $\langle\sigma_j,\sigma_k\rangle=N\delta_{jk}$, thus $\max_\rho c_j(\rho)=\max_\rho\langle\sigma_j,\rho\rangle$. By convexity maximum is achieved at extremal points, which are pure states, and thus the maximum value is achieved by choosing $\rho$ equal to the (projector onto the) eigenvector of $\sigma_j$ with largest eigenvalue. For two-qubit and Pauli basis, max eigenvalue is $1$. For the last point using ${\rm Tr}(\rho\sigma)=\sum_j\lambda_j(\sigma)\langle j|\rho|j\rangle$, which is a convexcomb of eigvals of $\sigma$
– glS
Jul 20 at 19:44
• in other words, $\max_\rho\langle\sigma,\rho\rangle=\max\lambda_j(\sigma)$, and thus in particular, $\max_\rho|\langle\sigma,\rho\rangle|=\|\sigma\|_{\rm op}$. This for any Hermitian operator $\sigma$, when maximisation is over all states $\rho$. Using only CS doesn't take into account that the space of states is the convex hull of pure states
– glS
Jul 20 at 19:49
Since Pauli matrices are traceless and Hermitian, the trace condition $$\text{Tr}(\rho)=1$$ is satisfied automatically and the numbers $$a_k, b_l, E_{kl}$$ should be real if we want $$\rho$$ to be Hermitian. The hard part is to verify the positive semidefinite condition $$\rho \ge 0$$.
The straightforward way is to use Sylvester’s criterion.
The positivity condition $$\rho > 0$$ is equivalent to just 4 strict inequalities, representing the condition that leading principal minors must be positive. In our case, the first minor is $$\frac{1}{4}(1 + a_z + b_z + E_{zz})$$, the last one is the determinant of $$\rho$$. It's hard to make a simple meaning of obtained inequalities, in contrast to the single qubit case.
The non-strict positivity condition $$\rho \ge 0$$ is even harder to express since by Sylvester’s criterion we need all principal minors (4+6+4+1=15 in our case) to be non-negative.
A simple necessary condition you can get from the inequality $$\text{Tr}(\rho^2)\le 1$$. It gives $$\sum a_k^2 + \sum b_l^2 + \sum E_{kl}^2 \le 3.$$ Though, it should be deducible from those 15 inequalities anyway.
• is "semi-positive" a replacement of "positive semidefinite"? I don't think I've ever seen it used
– glS
Jul 20 at 10:53
• Yeah, I've just used the author's phrasing. I didn't see it too. Jul 20 at 11:30
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2021-09-20 14:47:39
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http://zoonek2.free.fr/UNIX/48_R/12.html
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# Generalized Linear Models: logistic regression, Poisson regression, etc.
Example: a classification problem
Naive Bayes classifyer
Discriminant Analysis
Logistic Regression
TODO
Variants of logistic regression
Let us now tackle regression when the variable to predict is qualitative.
TODO: In this chapter, I mention Discriminant Analysis, already tackled in the chapter about PCA. Choose where to detail this...
TODO: Include LDA, naive bayesian classifier
## Example: a classification problem
For the moment, in the regressions we have seen, the variable to predict was always a (continuous) quantitative variable. However, this already enables us to investigate some situations in which qualitative variables appear. Here is an example.
### Using regression in a classification problem
We have three variables, two of them quantitative, X1 and X2, and one qualitative Y, with two values. We want to predict Y from X1 and X2.
n <- 10
N <- 10
s <- .3
m1 <- rnorm(n, c(0,0))
a <- rnorm(2*N*n, m1, sd=s)
m2 <- rnorm(n, c(1,1))
b <- rnorm(2*N*n, m2, sd=s)
x1 <- c( a[2*(1:(N*n))], b[2*(1:(N*n))] )
x2 <- c( a[2*(1:(N*n))-1], b[2*(1:(N*n))-1] )
y <- c(rep(0,N*n), rep(1,N*n))
plot( x1, x2, col=c('red','blue')[1+y] )
We can consider the qualitative variable as a quantitative variable, assuming two values, 0 and 1, and perform a linear regression against the others. We get an expression of the form
Y = b0 + b1 X1 + x2 X2.
We can then part the (X1,X2) plane in two, along the "b0 + b1 X1 + x2 X2 = 1/2" line.
plot( x1, x2, col=c('red','blue')[1+y] )
r <- lm(y~x1+x2)
abline( (.5-coef(r)[1])/coef(r)[3], -coef(r)[2]/coef(r)[3] )
### Curvilinear regression and classification
The situation is the same as above, but this time, we regress Y against X1, X2, X1X2, X1^2, X2^2.
# I need a function to draw conics...
conic.plot <- function (a,b,c,d,e,f, xlim=c(-2,2), ylim=c(-2,2), n=20, ...) {
x0 <- seq(xlim[1], xlim[2], length=n)
y0 <- seq(ylim[1], ylim[2], length=n)
x <- matrix( x0, nr=n, nc=n )
y <- matrix( y0, nr=n, nc=n, byrow=T )
z <- a*x^2 + b*x*y + c*y^2 + d*x + e*y + f
contour(x0,y0,z, nlevels=1, levels=0, drawlabels=F, ...)
}
r <- lm(y~x1+x2+I(x1^2)+I(x1*x2)+I(x2^2))$coef plot( x1, x2, col=c('red','blue')[1+y] ) conic.plot(r[4], r[5], r[6], r[2], r[3], r[1]-.5, xlim=par('usr')[1:2], ylim=par('usr')[3:4], add=T) ### Nearest neighbours The situation is still the same. This time, in order to find the value of Y from those of X1 and X2, we take the 10 points of the sample nearest from (X1,X2), we average the corresponding Y values and round to 0 or 1. M <- 100 d <- function (a,b, N=10) { mean( y[ order( (x1-a)^2 + (x2-b)^2 )[1:N] ] ) } myOuter <- function (x,y,f) { r <- matrix(nrow=length(x), ncol=length(y)) for (i in 1:length(x)) { for (j in 1:length(y)) { r[i,j] <- f(x[i],y[j]) } } r } cx1 <- seq(from=min(x1), to=max(x1), length=M) cx2 <- seq(from=min(x2), to=max(x2), length=M) plot( x1, x2, col=c('red','blue')[1+y] ) contour(cx1, cx2, myOuter(cx1,cx2,d), levels=.5, add=T) The problem of the "outer" function is mentionned in the FAQ, where they provide another solution, more "parallelized" than mine... wrapper <- function(x, y, my.fun, ...) { sapply(seq(along=x), FUN = function(i) my.fun(x[i], y[i], ...)) } outer(cx1,cx2, FUN = wrapper, my.fun = d) Here is another situation in which the nearest neighbours method proves more relevant than the preceding. n <- 20 N <- 10 s <- .1 m1 <- rnorm(n, c(0,0)) a <- rnorm(2*N*n, m1, sd=s) m2 <- rnorm(n, c(0,0)) b <- rnorm(2*N*n, m2, sd=s) x1 <- c( a[2*(1:(N*n))], b[2*(1:(N*n))] ) x2 <- c( a[2*(1:(N*n))-1], b[2*(1:(N*n))-1] ) y <- c(rep(0,N*n), rep(1,N*n)) plot( x1, x2, col=c('red','blue')[1+y] ) M <- 100 cx1 <- seq(from=min(x1), to=max(x1), length=M) cx2 <- seq(from=min(x2), to=max(x2), length=M) #text(outer(cx1,rep(1,length(cx2))), # outer(rep(1,length(cx1)), cx2), # as.character(myOuter(cx1,cx2,d))) contour(cx1, cx2, myOuter(cx1,cx2,d), levels=.5, add=T) # Color the various areas points(matrix(cx1, nr=M, nc=M), matrix(cx2, nr=M, nc=M, byrow=T), pch='.', col=c("red", "blue")[ as.numeric( myOuter(cx1,cx2,d) >.5 )+1]) pastel <- .9 plot( x1, x2, col=c('red','blue')[1+y] ) points(matrix(cx1, nr=M, nc=M), matrix(cx2, nr=M, nc=M, byrow=T), pch=16, col=c(rgb(1,pastel,pastel), rgb(pastel,pastel,1)) [ as.numeric( myOuter(cx1,cx2,d) >.5 )+1]) points(x1, x2, col=c('red','blue')[1+y] ) contour(cx1, cx2, myOuter(cx1,cx2,d), levels=.5, add=T) Let us vary the number of neighbours. plot( x1, x2, col=c('red','blue')[1+y] ) v <- c(3,10,50) for (i in 1:length(v)) { contour(cx1, cx2, myOuter(cx1,cx2, function(a,b){d(a,b,v[i])}), levels=.5, add=T, drawlabels=F, col=i+1) } legend(min(x1),max(x2),as.character(v),col=1+(1:length(v)), lty=1) We remark that we should not take too many points. If we do not take the 10 nearest points but just the nearest point, we get a Voronoi diagram. http://www.perlmonks.org/index.pl?node_id=189941 Here is a funny application of Voronoi tessellations: sea ice modeling in computer-generated images (you can reuse the idea for giraffe skin): http://images.povcomp.com/entries/images/210_detail2.jpg http://www.povcomp.com/entries/210_detail2.php http://www.povcomp.com/entries/210.php http://trific.ath.cx/software/gimp-plugins/voronoi/ They are also used to create mosaics. http://nis-ei.eng.hokudai.ac.jp/~doba/papers/egshort02_mosaic.pdf We shall see another application of Voronoi diagrams to spot outliers. TODO Other applications: http://www.voronoi.com/cgi-bin/voronoi_applications.php TODO: give more examples... There are several variants of this method: instead of taking the mean on the 10 neighbours, we can take all the points but weigh them according to the distance to our point (such a weighting scheme is called a "kernel"); we can replace the euclidian distance by another, more relevant to the data; etc. ### Technical note: kD trees, nearest neighbours and local regression Given a set of points x1, x2, ..., xN in R^n, and a new point y, we want to finc the x_i nearest to y. The naive method consists in examining all the points in turn, compute their distance with y, and retain the nearest. This may be fine of there is a single new point y, but it is not very efficient if we plan to use it on many new points. Instead, we can preprocess the points x1, ..., xN so as to quickly find the one nearest to a new given point. In dimension one, this can be done as follows; instead of representing the points as an unordered cloud, represent them as a tree: find the median, cut the cloud of points in two along the median, proceed recursively with thowe two new clouds. TODO: give an example. (a numeric example; plot the tree, give a 1-dimensional plot as we shall shortly be doing in 2 dimensions) This is used to speed up data access in a database (when you create an "index" on a numeric variable); it works well for requests of the form Find the records for which x = 17 Find the records for which x <= 9 Find the records for which 9 <= x and x <= 17 This can be generalized in higher dimensions, under the name "kD tree" (for "k-dimensional tree"): we split the cloud of points along the median of one of the coordinates and we change the coordinate used at each step. TODO: plot in 2D There are several ways of choosing which coordinate to use for the split, e.g., "take each coordinate in turn, cyclically"; or "median of the most spread dimension"; or (instead of the median) "the point closest to the center of the widest dimension" (this enforces more square cells). TODO: plots, for those three algorithms, in two examples: uniform random points in a square points along a circle (this tends to be pathological). As in dimension 1, this is used in DBMS (DataBase Management Systems) to speed up requests of the form Find the records with x = 17, y = 3 and z = 5 Find the records with x = 17 Find the records with x <= 17 For the last two examples, we explore the tree, but when it forks according to y or z, we have to explore both branches. TODO: a plot of such a tree? kD trees can also be used to find the nearest neighbour of a point: the idea is to proceed as with the naive algorithm, by examining all the points, but instead of takeing them in an arbitrary order, we follow the tree ("depth first") and we remember to distance to the closedt point found so far: this allows us to prune whole branches, if their points are farther that our best guess so far. The main applications of k-D trees (and their generalization, BSP (Binary Space Partitionning) trees, where the cutting hyperplanes need not be orthogonal to the axes) are: indexing geographical data (in spacial dsatabases or GIS, Geographical Information Systems); indexing multidimensional data in databases, even if the data are not geographical; culling objects from 3-dimensional scenes, i.e., quickly deciding what is visible and should be rendered and what is not currently visible and can thus be safely discarded; description of FPS (First Person Shooter) video game levels (e.g., Quake); local regression, i.e., regression on the k nearest neighbours of each point of our sample (contrary to the naive implementation, that precomputes the regression at each sample point, kD trees allow us to perform the computations on the fly and therefore allow you to change the parameters (kernel width, degree) without needing a retraining phase -- the training phase basically consists in storing the data); reducing the number of colours in an image (represent the colours as points in the 3-dimensional red-green-blue space, build a kD-tree on those points, stop when you have the desired number of leaves, replace each leaf by its median); simulating the n-body problem (build the kD tree of the n objects, for each node (starting with the deepest nodes), compute the forces between its children (considered as points)). http://www.autonlab.org/autonweb/documents/papers/moore-tutorial.pdf http://en.wikipedia.org/wiki/Binary_space_partitioning http://en.wikipedia.org/wiki/Doom_engine http://groups.csail.mit.edu/graphics/classes/6.838/S98/meetings/m13/kd.html http://www.cs.cmu.edu/~kdeng/thesis/kdtree.pdf http://www.liacc.up.pt/~ltorgo/PhD/th5.pdf http://www.flipcode.com/articles/article_raytrace07.shtml ### Comparing those methods Regression is much more stable, it works fine with few observations, but the results are imprecise. Furthermore, it assumes that the data have a simple structure. On the contrary, the nearest neighbour method is more precise, does not assume anything about the structure of the data, but is not very stable (i.e., a different sample can yield completely different results) and works best with many observations. Furthermore, the nearest neighbour method falls under the curse of dimension: if the dimension is high, the "nearest" neighbours are far away, most points are approximately at the same distance from our point... TODO: take two samples, a small (100 or 200 observations, to do the computations, and another, larger (1e4), to validate them. Plot the error rate as a function of the number of neighbours. TODO: check that it works... get.model <- function (n=10, m=2, s=.1) { list( n=n, m=m, x=runif(n), y=runif(n), z=sample(1:m,n,replace=T), s=s ) } get.sample <- function (model, n=200) { i <- sample( 1:model$n, n, replace=T )
data.frame( x=rnorm(n, model$x[i], model$s),
y=rnorm(n, model$y[i], model$s),
z=model$z[i] ) } nearest.neighbour.predict <- function (x, y, d, k=10) { o <- order( (d$x-x)^2 + (d$y-y)^2 )[1:k] s <- apply(outer(d[o,]$z, 1:max(d$z), '=='), 2, sum) order(s, decreasing=T)[1] } m <- get.model() d <- get.sample(m) N <- 1000 d.test <- get.sample(m,N) n <- 50 r <- rep(0, n) # Very slow for (k in 1:n) { for(i in 1:N) { r[k] <- r[k] + (nearest.neighbour.predict(d.test$x[i], d.test$y[i], d, k) != d.test$z[i] )
}
}
plot(r/N, ylim=c(0,1), type='l', xlab="Error rate")
abline(h=c(0,.5,1), lty=3)
rm(d.test)
With a smaller sample:
m <- get.model()
d <- get.sample(m, 20)
N <- 1000
d.test <- get.sample(m,N)
n <- 50
r <- rep(0, n)
# Very slow
for (k in 1:n) {
for(i in 1:N) {
r[k] <- r[k] +
(nearest.neighbour.predict(d.test$x[i], d.test$y[i], d, k) != d.test$z[i] ) } } plot(r/N, ylim=c(0,1), type='l', xlab="Error rate") abline(h=c(0,.5,1), lty=3) rm(d.test) ### Other examples The nearest neighbour method remains valid with more than two classes. nearest.neighbour.plot <- function (d, k=10, model=NULL) { col <- rainbow(max(d$z))
plot( d$x, d$y, col=col )
cx <- seq(min(d$x), max(d$x), length=100)
cy <- seq(min(d$y), max(d$y), length=100)
pastel <- .8
colp <- round(pastel*255 + (1-pastel)*col2rgb(col))
colp <- rgb(colp[1,], colp[2,], colp[3,], max=255)
points(matrix(cx,nr=100,nc=100),
matrix(cy,nr=100,nc=100,byrow=T),
col = colp[
myOuter(cx,cy, function(a,b){
nearest.neighbour.predict(a,b,d,k)
})
],
pch=16
)
points( d$x, d$y, col=col )
if(!is.null(model)){
points(model$x,model$y,pch='+',cex=3,lwd=3,col=col[model$z]) } } m <- get.model(n=10, m=4) d <- get.sample(m) nearest.neighbour.plot(d, model=m) TODO: plot the theoretical boundary. TODO: there is already a package for this... library(help=knnTree) Exercice: use a kernel instead. ### TODO: lda, qda TODO: Read chapter 4 of "The Elements of Statistical Learning" ### Mixture discriminant analysis TODO library(mda) ?mda ### Flexible discriminant analysis TODO library(mda) ?fda ## Naive Bayes classifyer ### Toy example We consider two binary random variables X and Y and we try to predict Y from X. For instance, we try to predict the name of a fruit from its shape. X = "round" or "long" Y = "apple" or "non-apple" (e.g.: bananam, orange) We have a set of fruits to learn to recognise them: we know their shape and we know if they are an apple or not. For instance: | pomme | pas une pomme -----+---------+----------------- rond | | -----+---------+----------------- long | | TODO: choose numeric values When we see a new fruit, we can try to guess wether it is an apple in the following way: we compute the conditionnal probabilities P( Y = apple et X = round ) P( Y = apple | X = round ) = ---------------------------- P( X = round ) P( Y = non-apple et X = round ) P( Y = non-apple | X = round ) = -------------------------------- P( X = round ) and we conpare them. If the first is the larger, we shall say it is an apple, if it is the second, we shall say it is not. With our values: TODO ### The Bayes formula All the presentations of the naive Bayes classifier relie on the Bayes formula. Indeed, the formula we have use, P( Y = apple et X = round ) P( Y = apple | X = round ) = ---------------------------- P( X = round ) can also be written P( X = round | Y = apple ) * P( Y = Apple ) P( Y = apple | X = round ) = -------------------------------------------- P( X = round ) which is called "the Bayes formula". When we only had one predictive variable, it was not that useful, but we shall see that with several, it becomes handy. ### Several predictiva variables, non-binary variables This idea can easily be generalized to several predictive variables (shape of the fruit, colour, pips or stone, etc.). P(X1=b1 and ... and Xn=bn and Y=a) P(Y=a|X=(b1,b2,...,bn)) = --------------------------------- P(X1=b1 and ... and Xn=bn) But, if there are many predictive variables, some problems may occur: in some cases, we have not observed simultaneously b1, b2, ..., and bn. What can we do? It is here that conditionnal probabilities really become useful. P(X=(b1,...,bn)|Y=a) * P(Y=a) P(Y=a|X=(b1,b2,...,bn)) = ------------------------------- P(X=(b1,...,bn)) If the Xi are independant (it is a reasonnable hypothesis, iven if it is false: otherwise, we would have to investigate the dependancy relations, the interactions, between the Xi -- this would require much more data), this can be written P(X1=b1|Y=a) * ... * P(Xn=bn|Y=a) * P(Y=a) = -------------------------------------------- P(X=(b1,...,bn)) We are interested in the value of a tha maximizes this quantity. As a does not appear in the denominator, we can get rid of it and find the value of a that maximizes the numerator P(X1=b1|Y=a) * ... * P(Xn=bn|Y=a) * P(Y=a). A naive bayesian classifier will thus be used in two steps: first, compute the conditionnal probabilities P(Xi=bj|Y=a) for all the values of i, j and a and the a priori probabilities P(Y=a) from the learning data set and then, in a second step, given new values for the Xi, find the value of a that maximizes P(X1=b1|Y=a) * ... * P(Xn=bn|Y=a) * P(Y=a). ### Hypotheses We had to assume that the predictive variables were independant (this is the meaning of the word "naive" in the expression "naive Bayes classifier"). This hypothesis is usually wrong -- however, the forecasts are nonetheless rather reliable. http://www.cs.washington.edu/homes/pedrod/mlj97.ps.gz TODO: read this document... ### Implementation in R As the idea is relatively simple, we can easily implement ourselves such a predictor. my.bayes <- function (y, x, xp) { # TODO: # we should check that y, the columns of x and xp are factors, # that xp and x have the same number of columns, # that the levels of the columns of xp and x are the same # (or at least that the values in the columns of xp are among the # levels of x) na <- names(x) m <- dim(x)[2] # Compute the conditionnal probabilities pc <- list() for (i in 1:m) { a <- table(data.frame(y,x=x[,i])) a <- a / apply(a,1,sum) pc[[ na[i] ]] <- a } # Compute the a priori probabilities pap <- table(y) pap <- pap/sum(pap) # To avoid numerical instability, we consider sums of logarithms # of probabilities instead of products of probabilities pc <- lapply(pc,log) pap <- log(pap) # Compute the P(X1=b1|Y=a) * ... * P(Xn=bn|Y=a) * P(Y=a) papo <- matrix(0, nr=dim(xp)[1], nc=length(pap)) colnames(papo) <- names(pap) # I am sure we could do this with no loop... for (i in 1:dim(xp)[1]) { for (j in 1:length(pap)) { papo[i,j] <- pap[j] for (k in 1:dim(x)[2]) { papo[i,j] <- papo[i,j] + pc[[ na[k] ]][j,xp[i,k]] } } } papo <- papo - apply(papo,1,mean) papo <- exp(papo) levels(y)[ apply(papo, 1, function (a) { which(a==max(a))[1] }) ] } # Data library(e1071) data(HouseVotes84) y <- HouseVotes84[,1] x <- HouseVotes84[,-1] # Err... # I forgot to account for missing value i <- which(!apply( apply(x,1,is.na), 2, any )) x <- x[i,] y <- y[i] # Compute the forecasts: correct in 92% of the cases. yp <- my.bayes(y,x,x) mean(y==yp) ### Implemetation in R Actually, it is already there -- and the function, contrary to ours, accounts for missing values. library(e1071) ?naiveBayes For sample data, have a look at library(mlbench) data(package=mlbench) ### Precision and recall rates TODO: recall what it is library(mlbench) data(DNA) d <- DNA d$Class <- factor(ifelse(d$Class=='n','n','i')) library(e1071) r <- naiveBayes(Class~.,d) i <- sample(1:dim(d)[2],100) p <- predict(r, d[,-length(d)][i,]) # Long... A <- sum(p == 'i' & d$Class[i] == 'i')
B <- sum(p == 'i' & d$Class[i] != 'i') C <- sum(p != 'i' & d$Class[i] == 'i')
# Precision rate
100 * A/(A+B)
# Recall rate
100 * A/(A+C)
### ROC Curve
TODO
library(mlbench)
data(DNA)
d <- DNA
d$Class <- factor(ifelse(d$Class=='n','n','i'))
library(e1071)
r <- naiveBayes(Class~.,d)
p <- predict(r, d[,-length(d)], type='raw')
### Practical example: spam filtering
For several years, bayesian classifiers have been used in mail filters -- the interesting point is that they learn the kind of mail _you_ receive and the kind of spam that flood the web -- everyone ends up with a different filter.
http://www.paulgraham.com/spam.html
http://www.xml.com/pub/a/2003/11/19/udell.html
K. Williams, An Introduction to Machine Learning with Perl
http://conferences.oreillynet.com/presentations/bio2003/williams_ken.ppt
The variable to predict, Y, has two values: "spam" and "non-spam". There is a predictive variable for each word (yes, that means thousands of predictive variables -- actually, we shall restrict ourselves to the most common words and end up with hardly more than one thousand words), with two values: "present" or "absent".
The learning step is carried out from messages whose nature is known -- for instance, the messages you received last week).
More generally, text classification is one of the main applications of the naive Bayes classifier
TODO: give other examples (URL)
### Other naive Bayes classifier
There is another notion of a "naive Bayes classifier", in which the predictive variables are quantitative: these are counting variables -- in the spam example or in the text classification example, we not only look at the presence/absence of a word, but also hoz many times it occurs.
Should you wish to distinguish the two classifiers, you can can the binary one "Binomial Naive Bayes Classifier" and the counting variable one "Multinomial Naive Bayes Classifier".
A. McCallum, K. Nigam
A Comparison of Event Models for Naive Bayes Text Classification
http://www-2.cs.cmu.edu/%7Eknigam/papers/multinomial-aaaiws98.pdf
### Other Bayes classifiers
There are variants of the naive Bayes classifier that tackle its overly naive hypothesis (the independance of the predictive variables).
TAN (Tree-augmented Naive Bayes) model the dependance relations between predictive variables by a tree.
Bayesian networks (very trendy for a couple of years, as were the neural networks in tge 1990s) model the dependance relations between the predictive variables by a graph.
http://www.vision.ethz.ch/gm/bnclassifiers.pdf
The Holmes and Watson examples (they are actually taken from Finn
Jensen's book: An introduction to Bayesian Networks):
http://www.ai.mit.edu/courses/6.825/fall02/pdf/6.825-lecture-15.pdf
### Other (unsorted) examples
More generally, Bayesian classifiers may be used with any classification problem.
1. We want to know if a family name comes from southern Asia or not. For this, we look at the presence/absence of 4-letter strings in the name.
http://dimax.rutgers.edu/%7Esguharay/newfinalpresentation.pdf
2. Handwritten character recognition, on a PDA.
3. Voice recognition, on a mobile phone.
4. We have a bunch of biology research article abstracts and we are looking for sentences that describe gene interactions. To this end, we process by hand a few hundred sentences: we look at the words it contains (our predictive variables) and we state if it describes a gene interaction (our variable to predict). From this learning corpus, we compute the conditionnal and a priori probabilities. Given a new sentence, the Bayes classifier gives us the probability that it describes an interaction.
http://zoonek2.free.fr/bioinfo/TM/rapport.pdf
## Discriminant Analysis
We have already mentionned it above: we want to predict a qualitative variable from several quantitative variables. More geometrically, we are looking for a plane (or a subspace of smaller dimension) in which the various values of the quantitaive variable correspond to points as far away as possible.
library(MASS)
n <- 100
k <- 5
x1 <- runif(k,-5,5) + rnorm(k*n)
x2 <- runif(k,-5,5) + rnorm(k*n)
x3 <- runif(k,-5,5) + rnorm(k*n)
x4 <- runif(k,-5,5) + rnorm(k*n)
x5 <- runif(k,-5,5) + rnorm(k*n)
y <- factor(rep(1:5,n))
plot(lda(y~x1+x2+x3+x4+x5))
## Logistic Regression
### Generalized Linear Model
Regression with quantitative variables can be written
E(Y) = a0 + a1 x1 + ... + an xn.
But if Y only has a finite number of variables, this is no longer relevant: we can use numbers for the values of Y, but Y will be bounded, will only assume a fixed number of values, known in advance, while the right hand side of the expression is not bounded and can take an infinite number of values -- it does not look very serious and can become troublesome when you want to predict Y.
n <- 100
x <- c(rnorm(n), 1+rnorm(n))
y <- c(rep(0,n), rep(1,n))
plot(y~x)
abline(lm(y~x), col='red')
The idea of the Generalized Model (GLM)) is to replace E(Y) by something else.
For logistic regression, the variable to predict, Y, has only two values, 0 and 1, and we are interested in the probability of each.
P[ Y == 1 ] = a0 + a1 x1 + ... + an xn.
Hardly better: actually, it is the same formula (because E(Y)=P(Y==1)) and we have the same problem: the left hand side is bounded, the right hand side is not. To get rid of the problem, we can apply some transformation to the left hand side, a bijection between the interval (0,1) and the real line -- in GLM parlance, such a function is called a "link".
One often uses the logistic function>
p
logit(p) = log -------
1 - p
We can write
P(win)
logit( P(winr) ) = log --------.
P(lose)
The quotient P(win)/P(lose) is called the "odds" (bookmakers often use this expression: "the odds are 3 to 1", etc.)
Here is the plot of this function.
x <- seq(0,1, length=100)
x <- x[2:(length(x)-1)]
logit <- function (t) {
log( t / (1-t) )
}
plot(logit(x) ~ x, type='l')
The "probit" function, the inverse of the cumulative distribution function of the gaussian distribution, is another link.
curve(logit(x), col='blue', add=F)
curve(qnorm(x), col='red', add=T)
a <- par("usr")
legend( a[1], a[4], c("logit","probit"), col=c("blue","red"), lty=1)
The "log-log" function is yet another link.
curve(logit(x), col='blue', add=F)
curve(qnorm(x), col='red', add=T)
curve(log(-log(1-x)), col='green', add=T)
abline(h=0, lty=3)
abline(v=0, lty=3)
a <- par("usr")
legend( a[1], a[4],
c("logit","probit", "log-log"),
col=c("blue","red","green"),
lty=1)
You might also think about transforming the data with the inverse of the link function, perform a linear regression and apply the link to the result. It would not work, because the initial data are 0 and 1 and would give infinite values -- we are no longer trying to model the data values themselves, but their probabilities.
ilogit <- function (l) {
exp(l) / ( 1 + exp(l) )
}
fakelogit <- function (l) {
ifelse(l>.5, 1e6, -1e6)
}
n <- 100
x <- c(rnorm(n), 1+rnorm(n))
y <- c(rep(0,n), rep(1,n))
yy <- fakelogit(y)
xp <- seq(min(x),max(x),length=200)
yp <- ilogit(predict(lm(yy~x), data.frame(x=xp)))
yp[is.na(yp)] <- 1
plot(y~x)
lines(xp,yp, col='red', lwd=3)
So we give up Least Squares Estimation and compute a Maximum Likelihood Estimator.
n <- 100
x <- c(rnorm(n), 1+rnorm(n))
y <- c(rep(0,n), rep(1,n))
f <- function (a) {
-sum(log(ilogit(a[1]+a[2]*x[y==1]))) - sum(log(1-ilogit(a[1]+a[2]*x[y==0])))
}
r <- optim(c(0,1),f)
a <- r$par[1] b <- r$par[2]
plot(y~x)
curve( dnorm(x,1,1)*.5/(dnorm(x,1,1)*.5+dnorm(x,0,1)*(1-.5)), add=T, col='red')
curve( ilogit(a+b*x), add=T )
legend( .95*par('usr')[1]+.05*par('usr')[2],
.9,
c('theoretical curve', 'MLE'),
col=c('red', par('fg')),
lty=1, lwd=1)
title(main="Logistic regression, by hand")
This can be done directly, with the "glm" function.
#
# BEWARE:
# Do not forget the "family" argument -- otherwise, it would be a
# linear regression -- the very thing we are trying to avoid.
#
r <- glm(y~x, family=binomial)
plot(y~x)
abline(lm(y~x),col='red',lty=2)
xx <- seq(min(x), max(x), length=100)
yy <- predict(r, data.frame(x=xx), type='response')
lines(xx,yy, col='blue', lwd=5, lty=2)
lines(xx, ilogit(r$coef[1]+xx*r$coef[2]))
legend( .95*par('usr')[1]+.05*par('usr')[2],
.9,
c('linear regression',
'prediction with "predict"',
"prediction with the coefficients"),
col=c('red', 'blue', par('fg')),
lty=c(2,2,1), lwd=c(1,5,1))
title(main='Logistic regression with the "glm" function')
In particular, the predicted values, which are probabilities, are indeed between 0 and 1 -- with linear regression, they were not bounded.
We can compare the various kinds of regression: logistic regression yields the results closer to the theoretical curve.
n <- 100
x <- c(rnorm(n), 1+rnorm(n))
y <- c(rep(0,n), rep(1,n))
plot(y~x)
# Brutal prediction
m1 <- mean(x[y==0])
m2 <- mean(x[y==1])
m <- mean(c(m1,m2))
if(m1<m2) a <- 0
if(m1>m2) a <- 1
if(m1==m2) a <- .5
lines( c(min(x),m,m,max(x)),
c(a,a,1-a,1-a),
col='blue')
# Linear regression
abline(lm(y~x), col='red')
# Logistic regression
xp <- seq(min(x),max(x),length=200)
r <- glm(y~x, family=binomial)
yp <- predict(r, data.frame(x=xp), type='response')
lines(xp,yp, col='orange')
# Theoretical curve
curve( dnorm(x,1,1)*.5/(dnorm(x,1,1)*.5+dnorm(x,0,1)*(1-.5)), add=T, lty=3, lwd=3)
legend( .95*par('usr')[1]+.05*par('usr')[2],
.9, #par('usr')[4],
c('Brutal prediction', "Linear regression", "Logistic regression",
"Theoretical curve"),
col=c('blue','red','orange', par('fg')),
lty=c(1,1,1,3),lwd=c(1,1,1,3))
title(main="Comparing linear and logistic regression")
Do not forget to have a look at the examples from the "glm" manual:
demo("lm.glm")
TODO: Compare the various links.
### Reading the result
As for linear regression, we have three means of displaying the result.
The "print" function":
> r
Call: glm(formula = y ~ x, family = binomial)
Coefficients:
(Intercept) x
-0.4864 0.9039
Degrees of Freedom: 199 Total (i.e. Null); 198 Residual
Null Deviance: 277.3
Residual Deviance: 239.7 AIC: 243.7
The "summary" function:
> summary(r)
Call:
glm(formula = y ~ x, family = binomial)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.94200 -0.99371 0.05564 0.97949 1.87198
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.4864 0.1804 -2.696 0.00702 **
x 0.9039 0.1656 5.459 4.79e-08 ***
---
Signif. codes: 0 ***' 0.001 **' 0.01 *' 0.05 .' 0.1 ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 277.26 on 199 degrees of freedom
Residual deviance: 239.69 on 198 degrees of freedom
AIC: 243.69
Number of Fisher Scoring iterations: 3
The "anova" function:
> anova(r)
Analysis of Deviance Table
Model: binomial, link: logit
Response: y
Terms added sequentially (first to last)
Df Deviance Resid. Df Resid. Dev
NULL 199 277.259
x 1 37.566 198 239.693
There is also the "aov" function.
> aov(r)
Call:
aov(formula = r)
Terms:
x Residuals
Sum of Squares 9.13147 40.86853
Deg. of Freedom 1 198
Residual standard error: 0.45432
Estimated effects may be unbalanced
Let us also mention the "manova" function.
?summary.manova
### Residuals and plots
As usual, we have a wealth of plots to assess the quality of the regression results.
The Pearson residuals are defined as
y_i - \hat y_i
r_i = ----------------
s_i
where s_i is the estimation of the standard deviation of the noise. There is also a normalized version (this is the same sa with linear regression: the standard deviation of the noise and the standard deviation of the residuals are different) and a standardized version (we divide by the estimation of the standard deviation obtained by removing observation "i").
You will notice that y_i is 0 or 1, while \hat y_i is a probability (between 0 and 1), which explains the shape of the plot: se see two curves, one above the axis, the other under, that correspond to the two possible values of y_i. One of those curves is decreasing and tends to 0 in + infinity, for the other, it is the opposite.
n <- 100
x <- c(rnorm(n), 1+rnorm(n))
y <- c(rep(0,n), rep(1,n))
r <- glm(y~x, family=binomial)
plot(r, which=1)
Actually, this simulation does not faithfully reflect the model underlying the logistic regression. It would rather be like this:
n <- 1000
a <- -2
b <- 1
x <- runif(n, -4, 5)
y <- exp(a*x+b + rnorm(n))
y <- y/(1+y)
y <- rbinom(n,1,y)
plot(y~x)
boxplot(x~y, horizontal=T)
op <- par(mfrow=c(2,1))
hist(x[y==1], probability=T, col='light blue')
lines(density(x[y==1]),col='red',lwd=3)
hist(x[y==0], probability=T, col='light blue')
lines(density(x[y==0]),col='red',lwd=3)
par(op)
rt <- glm(y~x, family=binomial)
plot(rt, which=1)
TODO: give concrete examples of pathological situations.
We are also suggested to look at the gaussian quantile-quantile plot of the residuals -- but I do not understand whym because the residuals are not supposed to be gaussian.
plot(rt, which=2)
hist(rt$residuals, breaks=seq(min(rt$residuals),max(rt$residuals)+1,by=.5), xlim=c(-10,10), probability=T, col='light blue') points(density(rt$residuals, bw=.5), type='l', lwd=3, col='red')
However, one can transform those residuals to get a gaussian distribution: these are the Anscombe residuals -- the transformation is rather complex and R does not seem to know it...
Exercice: implement this transformation, in an approximate manner, with a simulation.
We can also look at the absolute value of the residuals.
plot(rt, which=3)
and the Cook distances.
plot(rt, which=4)
### Deviance, residuals, AIC
The deviance is defined by
D = -2(L-Lsat)
where L is the log-likelihood of our model and Lsat the log-likelihood of the saturated model (with as many variables as observations).
TODO: understand. (Where do those new variables come from?)
We are happy if D/df < 1.
> rt$deviance [1] 362.0677 We also have the deviance of the null model (the model with no variables, corresponding to the hypothesis "the predictive variables have no effect"), i.e., the probability of Y=1 is constant and does not depend on the predictive variables). > rt$null.deviance
[1] 1383.377
The AIC (Akaike Information Criterion) is preferable to the deviance to compare models, because it accounts for the number of parameters in those models.
> rt$aic [1] 366.0677 It is defined by - 2 * log-likelihood + 2 * number of parameters. There are a few variants, such as the BIC, - 2 * log-likelihood + ln(number of observations) * number of parameters. We have already seen the various residuals. Pearson residuals (aka raw residuals, standardized residuals, studentized residuals): y_i - \hat y_i ---------------- s_i Deviance residuals: contribution of each observation to the deviance. Anscombe residuals: we transform the variable to that the residuals follow a gaussian distribution (the "t" function is complicated): t(y_i) - t(\hat y_i) ---------------------- t'(y_i) s_i As with linear regression, we can measure the leverage, with the "hat matrix" (but it is not exactly the same matrix). TODO: with R # It is supposed not to be the same as in the linear situation. # Here, it seems to be the same... plot(hat(x), type='h') We also have the Cook distance, as for linear regression. plot(rt, which=4) ### Model comparisons Likelihood Ratio test: to compare two nested models with q1 and q2 parameters,we look at D2 - D2, that asymptotically follows a Chi^2 distribution (if we start with a binomial or Poisson distribution) or a Fisher distribution (if we start with a gaussian distribution) with q2 - q1 degrees of freedom. TODO: understand Wald test: it is a quadratic approximation of the Likelihood Ratio test. library(car) ?linear.hypothesis ### TODO: to sort, to rewrite (Answer to a question I was asked by email) Let us consider the data library(mlbench) data(BostonHousing) y <- BostonHousing[,1] y <- factor(y>median(y)) x1 <- BostonHousing[,2] x2 <- BostonHousing[,3] We can compute the logistic regression with the "glm" function. glm(y~x1+x2, family=binomial) or with the "lrm" function in the "Design" package, that provides much more information. library(Design) lrm(y~x1+x2) This yields: > lrm(y~x1+x2) Logistic Regression Model lrm(formula = y ~ x1 + x2) Frequencies of Responses FALSE TRUE 253 253 Obs Max Deriv Model L.R. d.f. P C Dxy 506 3e-06 234.13 2 0 0.862 0.724 Gamma Tau-a R2 Brier 0.725 0.363 0.494 0.145 Coef S.E. Wald Z P Intercept -1.66990 0.27245 -6.13 0e+00 x1 -0.04954 0.01280 -3.87 1e-04 x2 0.17985 0.02084 8.63 0e+00 Question: > 1. In a logistic regression, how do we know if the model is > "good"? I.e., do we have an equivalent of the R^2? There are several equivalents of the R^2 for logistic regression. The simplest, "pseudo-R^2" or "McFadden's R^2" is (deviance of the model) - (deviance of the null model) -------------------------------------------------------- (deviance of the null model) I recall that the deviance is - 2 log(L) where L is the likelihood and that the "null model" is "y ~1" (the model with no predictive variables, the model that assumes that the quantities we are modelling (the probability of Y) are constant). It can be interpreted as the "proportion of the deviance explained by the model" -- similarly, the R^2 was the "proportion of variance explained by the model". But it does not tell you if the model is good: if the R^2 is low, it can mean that there is a lot of noise or that the model is incomplete. The idea is the same as for linear models: plots. To learn how to read them, let us start with a few simulations. Random data: n <- 1000 y <- factor(sample(0:1,n,replace=T)) x <- rnorm(n) r <- glm(y~x,family=binomial) op <- par(mfrow=c(2,2)) plot(r,ask=F) par(op) library(Design) r <- lrm(y~x,y=T,x=T) P <- resid(r,"gof")['P'] resid(r,"partial",pl=T) title(signif(P)) Data coming from the model: n <- 1000 x <- rnorm(n) a <- 1 b <- -2 p <- exp(a+b*x)/(1+exp(a+b*x)) y <- factor(ifelse( runif(n)<p, 1, 0 ), levels=0:1) r <- glm(y~x,family=binomial) op <- par(mfrow=c(2,2)) plot(r,ask=F) par(op) r <- lrm(y~x,y=T,x=T) P <- resid(r,"gof")['P'] resid(r,"partial",pl=T) title(signif(P)) Data from the model, with random errors (I do not see anything on the plot, but the estimators of a and b are biased towards zero): n <- 1000 x <- rnorm(n) a <- 1 b <- -2 p <- exp(a+b*x)/(1+exp(a+b*x)) y <- ifelse( runif(n)<p, 1, 0 ) i <- runif(n)<.1 y <- ifelse(i, 1-y, y) y <- factor(y, levels=0:1) col=c(par('fg'),'red')[1+as.numeric(i)] r <- glm(y~x,family=binomial) op <- par(mfrow=c(2,2)) plot(r,ask=F, col=col) par(op) r <- lrm(y~x,y=T,x=T) P <- resid(r,"gof")['P'] resid(r,"partial",pl=T) title(signif(P)) Data from the model, with mode extreme errors: n <- 1000 x <- rnorm(n) a <- 1 b <- -2 p <- exp(a+b*x)/(1+exp(a+b*x)) y <- ifelse( runif(n)<p, 1, 0 ) i <- runif(n)<.5 & abs(x)>1 y <- ifelse(i, 1-y, y) y <- factor(y, levels=0:1) col=c(par('fg'),'red')[1+as.numeric(i)] r <- glm(y~x,family=binomial) op <- par(mfrow=c(2,2)) plot(r,ask=F, col=col) par(op) r <- lrm(y~x,y=T,x=T) P <- resid(r,"gof")['P'] resid(r,"partial",pl=T) title(signif(P)) One variable too many: n <- 1000 x1 <- rnorm(n) x2 <- rnorm(n) a <- 1 b <- -2 p <- exp(a+b*x1)/(1+exp(a+b*x1)) y <- factor(ifelse( runif(n)<p, 1, 0 ), levels=0:1) r <- glm(y~x1+x2,family=binomial) op <- par(mfrow=c(2,2)) plot(r,ask=F) par(op) r <- lrm(y~x,y=T,x=T) P <- resid(r,"gof")['P'] resid(r,"partial",pl=T) title(signif(P)) Non-linear model: n <- 1000 x <- rnorm(n) a <- 1 b1 <- -1 b2 <- -2 p <- 1/(1+exp(-(a+b1*x+b2*x^2))) y <- factor(ifelse( runif(n)<p, 1, 0 ), levels=0:1) r <- glm(y~x,family=binomial) op <- par(mfrow=c(2,2)) plot(r,ask=F) par(op) r <- lrm(y~x,y=T,x=T) P <- resid(r,"gof")['P'] resid(r,"partial",pl=T) title(signif(P)) Well, those plots seem much less useful than they were for linear regression. Question: > 2. What about the Wald, Hosmer and Lemshow, Somer's D tests? > What do they test? How do we compute them? How do we interpret > them? The Wald test I know (there might be another one with the same name) is an approximation of the likelihood ratio test: prefer the genuine one. The likelihood ratio test compares two nested models whose parameters have been estimated by the maximum likelihood method. In R, it is computed by the "anova" function. (The "F test" comparing a linear model to the trivial model is a particular case of it.) We would expect to be able to perform this test as with linear models: r <- glm(y~x1+x2, family=binomial()) anova(r) but no... You can perform the test by hand: the difference of the deviances of two nested models (here, the model we are interested in and the null model) follows, asymptotically, a Chi^2 distribution with p degrees of freedom, where p is the number of parameters that are fixed in the smaller model. r <- glm(y~x1+x2, family=binomial()) pchisq(r$null.deviance - r$deviance, df=2, lower.tail=F) You can use this test to compare two non-trivial models -- but they have to be nested. r0 <- glm(y~x1+x2, family=binomial) r1 <- glm(y~x1, family=binomial) r2 <- glm(y~x2, family=binomial) lr.test <- function (r.petit,r.gros) { pchisq(r.petit$deviance - r.gros\$deviance, df=1, lower.tail=F)
}
lr.test(r1,r0)
lr.test(r2,r0)
Here, we find that the model y ~ x1+x2 is significantly better that y ~ x1, with a risk of error inferior to 1e-21, and that the model y~x1+x2 is significantly better than y~x2, with a risk of error under 1e-6.
Actually, the "lrm" function (in the Design package) already performs this test.
lrm(y~x1+x2)
In our example,
Obs Max Deriv Model L.R. d.f. P C Dxy
506 3e-06 234.13 2 0 0.862 0.724
Gamma Tau-a R2 Brier
0.725 0.363 0.494 0.145
the p-value is zero, so the model is significantly better than the trivial one.
I think that "Wald's test" also refers to the test checking if a regression coefficient is zero (under the null hypothesis that this coefficient is zero, the estimated coefficient, divided by its standard deviation, follows a gaussian distribution; if you square it , it follows a Chi^2 distribution with one degree of freedom). Thus, if we look at the last column of the result of the previous function,
Coef S.E. Wald Z P
Intercept -1.66990 0.27245 -6.13 0e+00
x1 -0.04954 0.01280 -3.87 1e-04
x2 0.17985 0.02084 8.63 0e+00
you can tell that the intercept and the x2 coefficient are non-zero with a negligible risk of error and that the x1 coefficient is non-zero with a risk of error of 0.1%.
Actually, we already had those results with the "glm" function:
> summary(r0)
Call:
glm(formula = y ~ x1 + x2, family = binomial)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.5902 -0.7571 0.1312 0.6106 2.0397
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.66990 0.27245 -6.129 8.84e-10 ***
x1 -0.04954 0.01280 -3.870 0.000109 ***
x2 0.17985 0.02084 8.632 < 2e-16 ***
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 701.46 on 505 degrees of freedom
Residual deviance: 467.33 on 503 degrees of freedom
AIC: 473.33
Number of Fisher Scoring iterations: 6
I have never heard of the Hosmer and Lemeshow test, but Google refers me to
http://www.biostat.wustl.edu/archives/html/s-news/1999-04/msg00147.html
http://maths.newcastle.edu.au/~rking/R/help/03b/0735.html
http://www.learnlink.mcmaster.ca/OpenForums/00031830-80000001/00040559-80000001/0045CA16-00977198-005B1B21
which claim that this test is obsolete. The "residuals.lrm" function in the Design package has a test to replace it -- but I do not know what is behind it, neither how to interpret it.
library(Design)
?residuals.lrm
For our example:
> resid( lrm(y~x1+x2, y=T, x=T), "gof" )
Sum of squared errors Expected value|H0 SD
7.328945e+01 7.632820e+01 5.739924e-01
Z P
-5.294060e+00 1.196300e-07
I do not know Somers's D, but some simulations can help interpret it.
> library(Design)
> somers2(x1,as.numeric(y)-1)
C Dxy n Missing
0.2846475 -0.4307051 506.0000000 0.0000000
> somers2(-x1,as.numeric(y)-1)
C Dxy n Missing
0.7153525 0.4307051 506.0000000 0.0000000
> somers2(x2,as.numeric(y)-1)
C Dxy n Missing
0.8551532 0.7103064 506.0000000 0.0000000
> somers2(-x2,as.numeric(y)-1)
C Dxy n Missing
0.1448468 -0.7103064 506.0000000 0.0000000
> somers2(rnorm(506),as.numeric(y)-1)
C Dxy n Missing
0.50564764 0.01129529 506.00000000 0.00000000
It seems to be interpretable as a correlation, when one of the variables only has two values, 0 and 1.
So we have the same problems as with correlation: it is only relevant for linear relations. Thus the following computations give (approximately) 1 and 0.
n <- 1000
a <- rnorm(n)
b <- ifelse(a<0,0,1)
somers2(a,b)
b <- ifelse(abs(a)<.5,0,1)
somers2(a,b)
## TODO
# age
age <- c(25.0, 32.5, 37.5, 42.5, 47.5, 52.5, 57.5, 65.0)
# Number of successes
n <- c(100, 150, 120, 150, 130, 80, 170, 100)
# predictive variable
Y <- c(10, 20, 30, 50, 60, 50, 130, 80)
f<-Y/n
g<-log(f/(1-f)) # Transforming the data
w<-n*f*(1-f) # Weights
r<-predict(lm(g~age,weights=w)) # Weighted regression
p<-exp(r)/(1+exp(r)) # Inverse transform
plot(age,f,ylim=c(0,1))
lines(age,p)
symbols(age,f,circles=w,add=T)
# Iterate...
w<-n*p*(1-p)
gu<-r+(f-p)/p/(1-p)
r<-predict(lm(gu~age,weights=w))
# Once more
p<-exp(r)/(1+exp(r))
w<-n*p*(1-p)
gu<-r+(f-p)/p/(1-p)
r<-predict(lm(gu~age,weights=w))
Actually, we get the results of
glm(cbind(Y,n-Y)~age,family=binomial)
## Variants of logistic regression
### Log-linear model (Poisson regression)
The veriable to predict is no longer a binary variable, but a counting variable. To perform this Poisson regression, we still use the "glm" function, with the "family=poisson" argument.
I do not give any example her but just refer to those from the manual.
library(MASS)
?epil
?housing
TODO: Give an example (at least a simulation).
See also the "loglm", "multinom", "gam" functions.
?loglm
library(nnet)
?multinom
library(mgcv)
?gam
library(survival)
?survexp
### If the qualitative variable to predict has more than two values (first attempt)
Actually, I do not really know how to proceed.
# NO!
n <- 100
x <- c( rnorm(n), 1+rnorm(n), 2.5+rnorm(n) )
y <- factor(c( rep(0,n), rep(1,n), rep(2,n) ))
r <- glm(y~x, family=binomial)
plot(as.numeric(y)-1~x)
xp <- seq(-5,5,length=200)
yp <- predict(r,data.frame(x=xp), type='response')
lines(xp,yp)
Let us try to encode the variable to predict with binary variables, by hand.
First attempt: brutally, thoughtlessly.
n <- 100
x <- c( rnorm(n), 10+rnorm(n), 25+rnorm(n), -7 + rnorm(n) )
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
y1 <- factor(c( rep(0,n), rep(0,n), rep(1,n), rep(1,n) ))
y2 <- factor(c( rep(0,n), rep(1,n), rep(0,n), rep(1,n) ))
r1 <- glm(y1~x, family=binomial)
r2 <- glm(y2~x, family=binomial)
xp <- seq(-50,50,length=500)
y1p <- predict(r1,data.frame(x=xp), type='response')
y2p <- predict(r2,data.frame(x=xp), type='response')
plot(as.numeric(y)-1~x)
lines(xp,y1p+2*y2p)
lines(xp,y1p, col='red')
lines(xp,y2p, col='blue')
It does not workm because Y1 and Y2 do not come from a logistic model:
plot(as.numeric(y1)-1~x)
lines(xp,y1p, col='red')
Second attempt.
n <- 100
x <- c( rnorm(n), 10+rnorm(n), 25+rnorm(n), -7 + rnorm(n) )
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
y1 <- factor(c( rep(0,n), rep(1,n), rep(1,n), rep(1,n) ))
y2 <- factor(c( rep(0,n), rep(0,n), rep(1,n), rep(1,n) ))
y3 <- factor(c( rep(0,n), rep(0,n), rep(0,n), rep(1,n) ))
r1 <- glm(y1~x, family=binomial)
r2 <- glm(y2~x, family=binomial)
r3 <- glm(y3~x, family=binomial)
xp <- seq(-50,50,length=500)
y1p <- predict(r1,data.frame(x=xp), type='response')
y2p <- predict(r2,data.frame(x=xp), type='response')
y3p <- predict(r3,data.frame(x=xp), type='response')
plot(as.numeric(y)-1~x)
lines(xp,y1p+y2p+y3p)
Same problem...
plot(as.numeric(y1)-1~x)
lines(xp,y1p, col='red')
Third attempt: with ordinal variables.
n <- 100
x <- c( -7+rnorm(n), rnorm(n), 10+rnorm(n), 25+rnorm(n))
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
y1 <- factor(c( rep(0,n), rep(1,n), rep(1,n), rep(1,n) ))
y2 <- factor(c( rep(0,n), rep(0,n), rep(1,n), rep(1,n) ))
y3 <- factor(c( rep(0,n), rep(0,n), rep(0,n), rep(1,n) ))
r1 <- glm(y1~x, family=binomial)
r2 <- glm(y2~x, family=binomial)
r3 <- glm(y3~x, family=binomial)
xp <- seq(-50,50,length=500)
y1p <- predict(r1,data.frame(x=xp), type='response')
y2p <- predict(r2,data.frame(x=xp), type='response')
y3p <- predict(r3,data.frame(x=xp), type='response')
plot(as.numeric(y)-1~x)
lines(xp,y1p+y2p+y3p)
n <- 100
x <- c( -.7+rnorm(n), rnorm(n), 1+rnorm(n), 2.5+rnorm(n))
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
y1 <- factor(c( rep(0,n), rep(1,n), rep(1,n), rep(1,n) ))
y2 <- factor(c( rep(0,n), rep(0,n), rep(1,n), rep(1,n) ))
y3 <- factor(c( rep(0,n), rep(0,n), rep(0,n), rep(1,n) ))
r1 <- glm(y1~x, family=binomial)
r2 <- glm(y2~x, family=binomial)
r3 <- glm(y3~x, family=binomial)
xp <- seq(-5,5,length=500)
y1p <- predict(r1,data.frame(x=xp), type='response')
y2p <- predict(r2,data.frame(x=xp), type='response')
y3p <- predict(r3,data.frame(x=xp), type='response')
plot(as.numeric(y)-1~x)
lines(xp,y1p+y2p+y3p)
lines(xp,round(y1p+y2p+y3p, digits=0), col='red')
### Ordinal Logistic Regression
You can also use logistic regression to forecast an ordinal qualitative variable (i.e., a qualitative variable whose values are ordered, -- e.g., its values could be "low", "medium", "high"). Here are some of the models you might want to use.
Proportionnal odds model (the effects are monotonic with respect to the quantitative variable):
1
P[ Y >= j | X ] = -----------------------
- ( a_j + X b )
1 + e
Continuation ratio model:
1
P [ Y = j | Y >= j, X ] = -----------------------
- ( a_j + X b )
1 + e
Actually, our attempts in the preceding paragraph have led us to the first model.
n <- 100
x <- c( -.7+rnorm(n), rnorm(n), 1+rnorm(n), 2.5+rnorm(n))
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
ordinal.regression.one <- function (y,x) {
xp <- seq(min(x),max(x), length=100)
yi <- matrix(nc=length(levels(y)), nr=length(y))
ri <- list();
ypi <- matrix(nc=length(levels(y)), nr=100)
for (i in 1:length(levels(y))) {
yi[,i] <- as.numeric(y) >= i
ri[[i]] <- glm(yi[,i] ~ x, family=binomial)
ypi[,i] <- predict(ri[[i]], data.frame(x=xp), type='response')
}
plot(as.numeric(y) ~ x)
lines(xp, apply(ypi,1,sum), col='red', lwd=3)
}
ordinal.regression.one(y,x)
n <- 100
v <- .2
x <- c( -.7+v*rnorm(n), v*rnorm(n), 1+v*rnorm(n), 2.5+v*rnorm(n))
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
ordinal.regression.one(y,x)
Here is the second.
n <- 100
x <- c( -.7+rnorm(n), rnorm(n), 1+rnorm(n), 2.5+rnorm(n))
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
ordinal.regression.two <- function (y,x) {
xp <- seq(min(x),max(x), length=100)
yi <- list();
ri <- list();
ypi <- matrix(nc=length(levels(y)), nr=100)
for (i in 1:length(levels(y))) {
ya <- as.numeric(y)
o <- ya >= i
ya <- ya[o]
xa <- x[o]
yi[[i]] <- ya == i
ri[[i]] <- glm(yi[[i]] ~ xa, family=binomial)
ypi[,i] <- predict(ri[[i]], data.frame(xa=xp), type='response')
}
# The plot is trickier to draw than earlier
plot(as.numeric(y) ~ x)
p <- matrix(0, nc=length(levels(y)), nr=100)
for (i in 1:length(levels(y))) {
p[,i] = ypi[,i] * (1 - apply(p,1,sum))
}
for (i in 1:length(levels(y))) {
p[,i] = p[,i]*i
}
lines(xp, apply(p,1,sum), col='red', lwd=3)
}
ordinal.regression.two(y,x)
n <- 100
v <- .1
x <- c( -.7+v*rnorm(n), v*rnorm(n), 1+v*rnorm(n), 2.5+v*rnorm(n))
y <- factor(c( rep(0,n), rep(1,n), rep(2,n), rep(3,n) ))
ordinal.regression.two(y,x)
Actually, it is already implemented:
library(MASS)
?polr
### Multilogistic regression (aka Multinomial regression)
We want to predict a qualitative variable, with more tham two values.
TODO
Model:
P( Y=1 | X=x )
log ---------------- = b_{1,0} + b_1 x
P( Y=k | X=x )
P( Y=2 | X=x )
log ---------------- = b_{2,0} + b_2 x
P( Y=k | X=x )
P( Y=k-1 | X=x )
log ------------------ = b_{k-1,0} + b_{k-1} x
P( Y=k | X=x )
(fit with MLE)
library(nnet)
?multinom
If the variable is binary, we get the same results as with a
classical logistic regression -- TODO: check this.`
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 2.5 License.
Vincent Zoonekynd
<zoonek@math.jussieu.fr>
latest modification on Sat Jan 6 10:28:22 GMT 2007
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2014-04-16 18:58:42
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# Method for factoring equations with large powers
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25 Jul 2009, 06:05
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Does anyone have a method for factoring equations with large powers
e.g. X^16 – Y^8 + 345Y^2
Thanks,
Aztec
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Re: Method for factoring equations with large powers [#permalink]
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25 Jul 2009, 12:01
Aztec wrote:
Does anyone have a method for factoring equations with large powers
e.g. X^16 – Y^8 + 345Y^2
Thanks,
Aztec
PLZ post ur Question then it will be appropriate to see in that context
_________________
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If you like my post....Consider it for Kudos
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Re: Method for factoring equations with large powers [#permalink]
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25 Jul 2009, 12:49
1
KUDOS
Aztec, I would give you three suggestions:
First: Memorize those products with notable results:
$$(x + a)^2 = x^2 + 2ax + a^2$$
$$(x - a)^2 = x^2 - 2ax + a^2$$
$$(x - a)(x + a) = x^2 - a^2$$
You need not only to memorize them, but also to recognize the inverted process. For example, you must be so familiar with those equations, that when you see something like
$$x^2 + 14x + 49$$, you know that in fact this can be $$(x+7)^2$$.
Second: Zeros of the equations:
Remember that the zeros of the equation, are the opposite terms inside the parentheses of those equations.
For instance, if you see the equation $$m^2 - 3m -28$$, you can discover the zeros, which are $$-4$$ and $$7$$, and reconstruct the equation with the opposite of the solutions: $$4$$ and $$-7$$
Thus $$(x + 4)(x - 7)$$
Third: Recognize equal terms.
If you see the equation $$3z^5 + 36z^4 - 84z^3$$, you can notice that $$z^3$$ is a common term in all the terms. Additionally, all the coefficients are divisible by $$3$$. Thus the term $$3z^3$$ is in all the terms of the equation. If you take off this term, the new equation is:
$$3z^3 (z^2 + 12z - 28)$$
You can go farther, and notice that the equation inside the parenthesis has the zeros $$2$$ and $$-14$$. Using the second advice listed you will have
$$3z^3 (z - 2) (z +14)$$, much more simple than $$3z^5 + 36z^4 - 84z^3$$
Now, coming back to your equation, we could rewrite as $$X^16 - Y^2(Y^6-345)$$. But there are other possibilities. If this is above some fraction, or even in some other context, it is a little bit easier to realize what factoring we should do.
Good studies!!!
PS.: If you liked the explanation, consider a kudo. I want to access those GMATClub tests!!!
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Re: Method for factoring equations with large powers [#permalink]
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27 Jul 2009, 14:05
Thanks a lot coelholds
These tips are useful.
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Re: Method for factoring equations with large powers [#permalink]
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27 Jul 2009, 16:28
You are welcome
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Re: Method for factoring equations with large powers [#permalink] 27 Jul 2009, 16:28
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# Method for factoring equations with large powers
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# If x is a prime number, and x-1 is the median of the set {x-1, 3x +3,
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If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink]
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If x is a prime number, and x-1 is the median of the set {x-1, 3x+3, 2x-4}, then what is the average (arithmetic mean) of the set?
A. 2
B. 5/3
C. 3
D. 10/3
E. 14/3
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Re: If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink]
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15 Jan 2015, 19:32
1
X is Prime. So lets put the first prime value and find out of the values of the set.
X=2 ; Set { 1,9,0 } , clearly the value x-1 is the median of the set. and the average is 10/3, which is D but hold on. lets check further to double check.
X= 3 ; Set { 2,12,2}, Here the value f x-1 can be the median ok t the average is 16/3 , no such value in the given answers. So no
X=5 and bigger prime values. the value of x-1 would be lowest in the set and therefore it cannot be the median.
Therefore D is the correct answer.
Bunuel wrote:
If x is a prime number, and x-1 is the median of the set {x-1, 3x+3, 2x-4}, then what is the average (arithmetic mean) of the set?
A. 2
B. 5/3
C. 3
D. 10/3
E. 14/3
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Re: If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink]
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15 Jan 2015, 20:33
1
Average $$= \frac{x-1 + 3x+3 + 2x-4}{3} = \frac{6x-2}{3}$$
Placing x = 2, mean $$= \frac{10}{3}$$
I wonder why its given that (x-1) is the median... sort of distraction?
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Re: If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink]
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Re: If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink]
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19 Jan 2015, 02:19
Bunuel wrote:
If x is a prime number, and x-1 is the median of the set {x-1, 3x+3, 2x-4}, then what is the average (arithmetic mean) of the set?
A. 2
B. 5/3
C. 3
D. 10/3
E. 14/3
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Re: If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink]
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08 Jul 2016, 23:19
If x is a prime number, and x-1 is the median of the set {x-1, 3x+3, 2x-4}, then what is the average (arithmetic mean) of the set?
A. 2
B. 5/3
C. 3
D. 10/3
E. 14/3
Solution:
x-1 is the median of the set, implies if we arrange the set in ascending order, the set would be [2x-4, x-1, 3x+3]. This means that:
2x-4<x-2<3x+3
solving inequality gives x>-2 and x<3. Since x is prime number and the only prime number <3 is 2, so put x=2.
Hence set becomes: [ 0, 1, 9] and avg = 10/3.
Please give kudos if this helps!
Re: If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, [#permalink] 08 Jul 2016, 23:19
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Mathematics Definition & Proof Based Problems
### Definition & Proof Based Problems
Definition & Proof Based Problems
Q 3125534461
Define surface tension and surface energy. Write units and dimensions of surface tension. Also prove that surface energy numerically equal to the surface tension.
Solution:
"Surface Tension :" Force on unit length of an imaginary line drawn on the surface of the liquid is called surface tension. Its S.l. unit is Nm^(-I) and its dimension is [ML^0 'r-2]. Surface Energy. Energy possessed by the surface of the liquid is called surface energy. Change in surface energy is the product of surface tension and change in surface area under constant temperature.
"Surface Energy." Energy possessed by the surface of the liquid is called surface energy. Change in surface energy is the product of surface tension and change in surface area under constant temperature.
Let S = Surface tension of soap solution
d = Length of the wire PQ
= length of wire PQ
Surface tension acts on both the free surfaces of film.
Hence, total inward force on wire PQ
Increase in area of the film PQ Q_1 P_1
= DeltaA = 2 (l xx)
.. Work done in stretching film is
W = Force applied x Distance moved
= (S xx 2l) xx x = S xx (2l xx)
= S xx DeltaA (·: 2lx=M)
This work done is stored in the film as its surface energy.
E=W=SxxDeltaA
=> S= WDeltaA
If increase in area is unity then, DeltaA = 1
S= W
:. Surface tension of a liquid is numerically equal to surface energy of the liquid surface.
Q 3135634562
Define coefficient of viscosity and give its SI unit. On what factors does the terminal velocity of a spherical ball falling through a viscous liquid depend ? Derive the formula :
v_t = (2a^2g)/(9 eta) (p-p^2)
where the symbols have their usual meaning.
Solution:
"Coefficient of Viscosity." Coefficient of viscosity is defined as the viscous force acting in unit area of a layer having unit velocity gradient. It is measured in Nsm^(-2) and has dimension of ML^(-I)T^(-1.)
"Stoke's Formula." The viscous drag experienced by a spherical ball moving through vertical column of highly dense liquid is given by F = 6pietarv where r-radius of the ball, 11-coefficient of viscosity of the liquid and v-terminal velocity attained by the ball.
"Terminal Velocity." The constant velocity with which a body drops down after initial acceleration in a dense liquid or fluid is called terminal velocity. This is attained when the apparent weight is compensated by the viscous force. It is given by
v = 2/9 (r^2g)/(eta) (p - eta) where p and sigma are denseness
of the body and liquid respectively, eta is the coefficient of viscosity of the liquid and r is the radius of the spherical body. Consider a lengthy column of a dense liquid like glycerine. As the ball or spherical ball is dropped in it, the forces experienced are,
(i) weight = mg = 4/3 pia^3 pg where p - density of ball
(ii) upthrust, U = 4/3 pia^3 p'g where p' - density of liquid
(iii) viscous force F_v = 6pieta ap' v_t
where v_1 - terminal velocity
When terminal velocity is attained,
acceleration should be zero and the
net force should be zero.
:.mg - U - F_v = 0
=> 4/3 pia^3 pg - 4/3 pia^3 ptg - ppieta av = 0
v_t = (4/3) pia^3 g (p-p')/(6pietaa)
= 2/9 (a^2g(p-p'))/(eta)
Thus, terminal velocity depends upon
(i) square of radius of the body.
(ii) coefficient of viscosity of the medium.
(iii) density of the body and the medium.
Q 3165634565
(a) Define streamline.
(b) Write any two properties of streamlines.
(c) Draw streamlines for a clockwisespinning sphere.
(d) Derive equation of continuity.
Solution:
(a) Streamline is the actual path followed by the procession of particles in a steady flow, which may be straight or curved such that tangent to it at any point indicates the direction of flow of a liquid at that point.
(b) Two properties of streamlines are- (i) Two streamlines can never cross each other.
(ii) The greater is the crowding of streamlines at a place, the greater will be the velocity of liquid particles at that place and vice-versa.
Volume of liquid entering per second at A= a_1 u_1 Mass of liquid entering per second at A= a'_1u_1p_1 Similarly, mass of liquid leaving per second at B = a_2u_2p_2 If there is no loss of liquid in tube and the flow is steady then, Mass of liquid entering per second at A = Mass of liquid leaving per second at B
a_1v_1p_1 = a_2U_2P_2
If liquid is incompressible then,
p_1 = p_2
a_1v_1 = a_2v_2
av = constant
This is the equation of continuity.
Q 3165434365
State and prove Bernoulli's theorem
Solution:
According to Bernoulli's theorem, for an incompressible, non-viscous liquid having streamlined flow, the sum of pressure head, velocity head and gravitational head is a constant,
i.e P/pg + (v^2)/(2g) + h
Consider an incompressible non-viscous liquid entering the cross-section A_1 at A with a velocity v_1 and coming a height h_2 at B with velocity v_2
The P.K and KE. increase since h_2 and v_2 are more than h_1 and u_1 respectively. This is done by the pressure doing work on the liquid. If P_1 and P_2 are the pressure at A and B, for a small displacement at A and B, The work done on the liquid at A = (P_1 A_I)
Delta x_1 = P_1 A_1 v_1 Deltat
The work done by the liquid at B
Deltax_2 = - (P_2 A_2)/Deltax_2 =- P_2 A_2 v_2 Deltat
The work done on the liquid at
(Considering a small time !:!..t so that area may be same)
Net work done by pressure = (P_1 - P_2) Av Deltat
Since A_1 v_1 = A_2 v_2
From conservation of energy,
(P_1 - P_2) Au M = change in (K.E. + P.E.)
(P_1-P_2) Av Deltat = AvpDeltatg(h_2-h_1)
+ 1/2 AvDeltatp (v_(2)^(2) - v_(1)^(2)
(i.e) P_1 + pgh_1 + p/2 v_(1)^(2) = P_2 + P_2 + pgh_2 + p/2 v_(2)^(2)
:. P/(pg) + h + v^2 /(2g) = Constant
Q 3105534468
State Bernoulli's theorem. Using it how can you explain the functioning of a venturimeter to find velocity of flow of liquid through a tube ?
Solution:
"Bernoulli's Theorem." For an incompressible, non-viscous, irrotational liquid having streamlined flow, the sum of the pressure energy, kinetic energy and potential energy permit mass is a constant, i.e.,
P/p + v^2/2 = gh = "constant"
P/(pg) + (v^2)/(2g) +h= "constant"
A liquid is said to be irrotational if the angular momentum about any point in the liquid is zero. A wheel or disc in it will not rotate .
Given : a_1 = 0.36 pi cm^2, a_2 = 0.04 pi m^2 h = 1 m
Since, c.s.a. at B is less velocity will be more and pressure will be less. The difference in pressure is P_1 P_2 = hpg. Applying Bernoulli's theorem,
(P_1)/(pg) + (v_(1)^(2))/(2g) = (P_2)/(pg) + (v_(2)^(2))/(2g)
=> (P_1-p_2)/(pg) = (v_(2)^(2) - v_(1)^(2))/(2g)
v_(2)^(2) - v_(1)^(2) = 2gh
for streamlined flow, a_1 v_1 = a_2 v_2
v_(2)^(2) - (a_(2)^(2) v_(2)^(2))/(=a_(1)^(2)) = 2 gh
=> v_(2)^(2) = 2gh 1/(1-(a_(2)^(2)//a_(1)^(2))
v_2 = sqrt((2gh a_(1)^(2))/(a_(1)^(2) - a_(1)^(2))) = 5 m//s
Q 3115134069
State Pascal's law. Explain the working of hydraulic lift.
Solution:
Pascal's law-It states that the pressure in a liquid at rest is the same at all points if they are at the same level.
"Hydraulic lift ":
Let, a = area of cross-section of piston in cylinder B
A = area of cross section of position in C also a < Fill the cylinders with an in compressible fluid.
Let r = downward force applied on piston B.
Then pressure exerted on liquid is
P = fla
This pressure is equally transmitted to piston in cylinder C (according to Pascal's law) :. upward force acting on the piston of cylinder C is
F = Pa = f/a A
as A>> a, F >> f
Therefore a small force applied on piston B
appears as a larger force on piston C.
Q 3105634568
(a) State and prove Archimedes' principle.
(b) What would be pressure inside a small air bubble of 0.1 mm radius situated just below the surface of water ? Surface tension of water 72 xx 10^-3 N//m and atmospheric pressure is 1.1 xx 105 N//m^2.
Solution:
(a) Archimedes' Principle. When a body is partially or completely immersed in a liquid, it loses weight due to the presence of upthrust, which is equal to the weight of liquid displaced by the submerged part of the body. Consider a liquid of density sigma into which a cylinder of material density p, crosssectional area A, height h is immersed such that its upper surface is at a depth x from sigma g the free surface. The pressure at the top and bottom of the cylinder are
x sigma g and (x+h) sigma g
The upthrust on the cylinder = Deltap xx A
U = [h osigma] xx A = hA sigmag = V sigmag
Weight in air,
W_1 = Vpg
Apparent weight
W_a= W_1 - U
Loss in weight
W_1 - Apparent weight
= W_1 - W_a = W_1 - (W_1 - U)
= W_1- W_1 + u = u
= Vsigmag
So, there is a loss equal to the
(b) Excess pressure = (2sigma)/(r) = (2xx72xx10^(-3))/(0.1xx10^(-3))
= 1440 N//m^(2)
= 0.01440 xx 10^5 N//m^2
So, net pressure = (1.1 - 0.0144)xx 10^5
= 1.0856 xx 10^5 N//m^2
Q 3135334262
A big size balloon of mass M is held stationary in air with the help of a small block of mass M//2 tied to it by a light string such that both float in mid air. Describe the motion of the balloon and the block when the string is cut. Support your answer with calculations.
Solution:
Forces acting on balloon when held stationary are -
U -> the upthrust
U = Mg + T
Forces acting on small block when held stationary,
T = [M/2]g
From (i) and (ii), U = 3/2 Mg
When string is cut, T = 0
Small block will have free fall.
Balloon will have an acceleration 'a' such that
U -Mg = Ma
or 3/2 Mg - Mg = Ma
=> a = g/2 upwards
Q 3185412367
What is pitot tube ? State the principle on which it is based.
Solution:
It is a simple device which is used for measuring the velocity of the flow at any depth in a flowing liquid. It is based on Bernoulli's theorem.
Q 3115123969
Prove that the pressure at a depth h from the free surface of a liquid (P) in a container is P = P_0 + hpg, where P_0 is the atmospheric pressure.
Solution:
Consider two points A and B at two levels separated by h column of a liquid of density p surrounding the points A and B. Consider an area 'a' forming a cylinder of liquid of length h. The pressure at A = P_A = Atmospheric pressure = P_0
Weight of the liquid at centre of gravity
W =Mg = hapg
For equilibrium, pressure/force at B should nullity the forces acting down.
:. P_A . a + hapg = P_B . a
:. P_B= P_A + hpg = P_0 + hpg
Q 3125734661
Prove that velocity of efflux of an ideal liquid through an orifice is equal to the velocity attained by a freely falling body from the surface of the liquid to the orifice. Also find the horizontal range in terms of height. When is this range maximum?
Solution:
Consider two points at the same height (H - h) from the ground one inside and one outside the hole. Applying Bernoulli's theorem for the points.
we have (P_0+hpg)/(pg) + 0 (H-h)
= (H-h) + (V^2)/(2g) + (P_0)/(pg)
On solving, we get v = sqrt(2gh) The velocity of efflux thus depends on the depth at which the hole is made from the surface of the liquid. Time taken to reach the ground,
t = sqrt(2(H-h)/g)
Since, initial vertical velocity is zero.
v = sqrt(2gh) is horizontal
and a_x= 0 ;
R - vt = sqrt(2gh) sqrt(2(H-h)/g)
R = 2 sqrt((H-h))
For Range to be max (dR)/(dh) = 0
i.e h = H/2
Maximum range = R_(max)
=2 sqrt(H/2(H-H//2))=H
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2019-02-21 11:59:48
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http://stats.stackexchange.com/questions
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# All Questions
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2016-07-28 02:53:29
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http://www.physicsforums.com/showthread.php?p=3629824
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# Baby it's cold outside
by Evo
Tags: baby, cold, evo
PF Gold
P: 2,633
Quote by Evo Ok, it's going to get up to 70F here today, but with 40-50 MPH winds. Don - make sure roger is wearing her lead booties if she goes out, don't want her being blown away.
Will do. I might get a wind detached feather for you. She did shed a tail feather that I kept but I'm waiting for a nice long red wing feather.
The small wind generator I'm making will be used to provide Roger with some of the comforts of home. This wind is perfect for it's initial test.
Mentor
P: 25,959
Quote by dlgoff Will do. I might get a wind detached feather for you. She did shed a tail feather that I kept but I'm waiting for a nice long red wing feather. The small wind generator I'm making will be used to provide Roger with some of the comforts of home. This wind is perfect for it's initial test.
Oooh, you're wonderful!! An electric fireplace (for safety), flat screen tv, music system. I heard mellow music encourages egg laying.
PF Gold P: 7,367 We're in line to get 8-12" of snow starting later tonight and into tomorrow. I'm as ready as I can expect to be, given the circumstances, but there is still stuff that I'd like to get done.
P: 2,163 We're expecting snow in Maine down here too, but nothing locally. I know for you the first 3 inches don't count, but it looks like you're in for more than twice that. Did you get that snowblower with the heated cabin?
PF Gold P: 7,367 I have a walk-behind snow-blower for the light stuff (maybe 6-10 inches) with no cabin at all, and a tractor with a bucket in case the snow is really deep and/or wet. We can never tell with these early storms.
P: 108
Quote by George Jones Wednesday night and yesterday morning, 15 cm (6 inches) of snow fell here. This morning, I had a 25 minute walk at -16 C ( 3 F).
Here in eastern Montana, on Thursday night November 17, the low was -2° F ( -18.9° C ).
I didn't have to go for a walk, though... lol
We have about the same amount of snow as you do, also... typical weather for this time of year.
Just for some fun information, if any body would like to reply... what's the lowest temperature you've ever experienced?
It was -52° F ( -46.7° C ) here, for a couple of days... think it was 1986, or there about, I don't remember the month.
OCR
Mentor
P: 25,959
Quote by OCR Here in eastern Montana, on Thursday night November 17, the low was -2° F ( -18.9° C ). I didn't have to go for a walk, though... lol We have about the same amount of snow as you do, also... typical weather for this time of year. Just for some fun information, if any body would like to reply... what's the lowest temperature you've ever experienced? It was -52° F ( -46.7° C ) here, for a couple of days... think it was 1986, or there about, I don't remember the month. OCR
With wind chill 63F below zero, Upstate, NY. The most uncomfortable was Steamboat Springs, CO. You would inhale through your nose and your nose hair would instantly ice up and crackle.
Mentor
P: 2,917
Quote by OCR Here in eastern Montana, on Thursday night November 17, the low was -2° F ( -18.9° C ). I didn't have to go for a walk, though... lol We have about the same amount of snow as you do, also... typical weather for this time of year. Just for some fun information, if any body would like to reply... what's the lowest temperature you've ever experienced? It was -52° F ( -46.7° C ) here, for a couple of days... think it was 1986, or there about, I don't remember the month. OCR
Sounds close to what I experienced for several days when I lived in Fairbanks, AK. It was bumping around between -45 and -50 C.
The year after I moved away from Alaska, they had record-breaking cold. The thermometers at Eielson Air Force Base only went to -70 F and I heard they were pegged out. Brrrr, I'm glad I missed it!
PF Gold P: 4,210 right when it hit -30 F here, the water heater element burned out and the upstairs registers got airlocked. It happened to be a weekend too. All's well now, though.
P: 2,163 This brings me to mind of the winter of aught five. Well sonny, it was so cold that we emptied out the freezer and got inside to warm up. It was plenty below, but it kept dropping. We had to push the house down the street to jump start the furnace. That's when it really started to get chilly. We went to milk the cows and got ice cream instead. It still fell lower. It was colder than a snowball on a rat's #$@ in hell on a cold day in July when the sun don't shine. But lower it went. Flashers would merely describe themselves. But then it got really cold. I had to button up my vest. Mentor P: 25,959 Quote by Jimmy Snyder This brings me to mind of the winter of aught five. Well sonny, it was so cold that we emptied out the freezer and got inside to warm up. It was plenty below, but it kept dropping. We had to push the house down the street to jump start the furnace. That's when it really started to get chilly. We went to milk the cows and got ice cream instead. It still fell lower. It was colder than a snowball on a rat's #$@ in hell on a cold day in July when the sun don't shine. But lower it went. Flashers would merely describe themselves. But then it got really cold. I had to button up my vest.
PF Gold
P: 7,367
Quote by Jimmy Snyder This brings me to mind of the winter of aught five. Well sonny, it was so cold that we emptied out the freezer and got inside to warm up. It was plenty below, but it kept dropping. We had to push the house down the street to jump start the furnace. That's when it really started to get chilly. We went to milk the cows and got ice cream instead. It still fell lower. It was colder than a snowball on a rat's #\$@ in hell on a cold day in July when the sun don't shine. But lower it went. Flashers would merely describe themselves. But then it got really cold. I had to button up my vest.
I hate when that happens.
PF Gold
P: 5,450
Quote by OCR Just for some fun information, if any body would like to reply... what's the lowest temperature you've ever experienced? OCR
-63C in Resolute, NWT, Canada in February 1989
The station temperature was a bit higher but we were in a shallow valley, where the coldest air tends to accumulate.
So why be in a valley anyway? because that's where enough snow accumulates to build yourself an igloo. We were practicing arctic survival, hosted by the Canadian Armed Forces.
Actually the thermometer minimum was -60C so we had to estimate the length of the fluid column. So that challenges Mark Twain's hypothesis:
Quote by Mark Twain “Cold! If the thermometer had been an inch longer we'd have frozen to death.”
PF Gold P: 7,367 We have about 4" on the ground, so it's just starting to actually snow. The breeze is out of the E-NE, so this one could last a while. If the wind was stronger, we could hope for it to move on soon. Not happening.
P: 1,017 Thats such a contrast from where I am at the moment. Its the most unusual feeling being in a temperate city where its still 25-35C outside in NOVEMBER! I almost cant believe it. I recently went back home and brought a load of warm clothes with me, but I dont use them :p
PF Gold P: 7,367 Look around you, chaos. What are other people wearing? When I was doing consulting work in south GA, I'd fly into Tallahassee because air-fares to there were very cheap. I'd be dressed in jeans and a T-shirt in winter when I got out of the terminal, and so many of the people around me were wearing sweats and down vests/jackets. When the daytime temps plummet to ~60F there, out comes the winter gear. It's pretty crazy.
PF Gold P: 7,367 Duke and I took a run to the nearest large town today. Normally a 20-minute drive at most, but today it took almost 45 minutes because people were freaked out by the snow. It's the first time that the dog-limo Honda pickup truck has been out in the snow, and it performed beautifully. Gotta love the Ridgeline for this climate. The temp has been vacillating around freezing for the last couple of hours, so I'll have my hands full tomorrow. Maybe a bit of shoveling and a lot of Kubota.
Mentor P: 14,473 It's not cold here yet in Houston, but wow, the fall foliage is absolutely fantastic this year compared to normal thanks to the prolonged drought last spring/summer/fall.
Related Discussions General Discussion 12 General Discussion 36 High Energy, Nuclear, Particle Physics 1 General Discussion 43 General Discussion 13
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2014-04-23 20:38:19
|
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https://www.nature.com/articles/s41598-018-36361-9?error=cookies_not_supported&code=a04b90be-f28f-45d5-bec5-042768b8ebee
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Article | Open | Published:
# Collaborative efforts to forecast seasonal influenza in the United States, 2015–2016
## Abstract
Since 2013, the Centers for Disease Control and Prevention (CDC) has hosted an annual influenza season forecasting challenge. The 2015–2016 challenge consisted of weekly probabilistic forecasts of multiple targets, including fourteen models submitted by eleven teams. Forecast skill was evaluated using a modified logarithmic score. We averaged submitted forecasts into a mean ensemble model and compared them against predictions based on historical trends. Forecast skill was highest for seasonal peak intensity and short-term forecasts, while forecast skill for timing of season onset and peak week was generally low. Higher forecast skill was associated with team participation in previous influenza forecasting challenges and utilization of ensemble forecasting techniques. The mean ensemble consistently performed well and outperformed historical trend predictions. CDC and contributing teams will continue to advance influenza forecasting and work to improve the accuracy and reliability of forecasts to facilitate increased incorporation into public health response efforts.
## Introduction
Seasonal influenza epidemics result in substantial human health and financial burdens in the United States, with an estimated 140,000–710,000 hospitalizations and 12,000–56,000 deaths annually depending on the severity of the season1,2. The magnitude and timing of influenza epidemics vary from year to year3,4, making the annual impact difficult to predict. Current Centers for Disease Control and Prevention (CDC) surveillance systems track influenza activity nationwide in a variety of ways, including monitoring virologic characteristics, outpatient visits for influenza-like illness (ILI), hospitalizations, and mortality5. While these systems collect valuable data, they are intrinsically describing activity that occurred in the past and require data processing time, limiting their utility for real-time public health decision making. Accurate and timely forecasts of influenza activity could assist in the public health response to both seasonal epidemics and future pandemics.
Since the 2013–2014 influenza season, CDC has hosted collaborative challenges to forecast the timing, intensity, and short-term trajectory of ILI activity in the United States using data from the US Outpatient Influenza-like Illness Surveillance Network (ILINet), a robust and geographically broad surveillance system, as its benchmark6,7. While ILI can capture both influenza and non-influenza illnesses, it is of high public health value as it correlates strongly with laboratory confirmed influenza and its magnitude correlates well with other measures of seasonal influenza severity8.
To continue the advancement of forecasting science, the application of forecasts for public health decision-making, and the development of best practices, CDC and challenge participants update challenge guidelines each year. For example, after the first challenge, several improvements were made including standardizing forecast submission formats, requiring specification of probabilistic forecasts rather than point forecasts, and implementing fully quantitative forecast evaluation9. Additional changes were made for the 2015–2016 season to improve the public health utility of the forecasts. First, challenge participants provided forecasts with increased resolution for peak intensity and trajectory predictions, which allows for a more detailed interpretation of forecasts and flexibility in scoring forecast accuracy. In addition, the evaluation methodology was modified to allow for a pre-specified number of preceding and proceeding values to be considered correct to reduce the effect of revisions to ILINet on forecast scores. To help communicate forecasts in real-time, a public webpage to host predictions was created10.
In the present analysis, we summarized the results and insights gained from the 2015–2016 challenge and identified areas for improvement moving forwards. We also evaluated the performance of a simple average ensemble of the submitted influenza forecasts since ensemble forecasts have demonstrated several advantages over single forecast models in both weather and infectious disease forecasting11,12,13,14,15. Finally, we used gamma regression to investigate characteristics of both forecast models and influenza seasons that may be associated with increased forecast accuracy.
## Results
Figure 1 shows the national ILINet curve for the 2015–2016 season in comparison to the 2009–2010 through 2014–2015 seasons. Compared to earlier seasons, the 2015–2016 season started later and had a later peak. The peak intensity was 3.5%, well below the high value of 7.7% set in the 2009–2010 pandemic season and below the peak of 6.0% in 2014–2015. Seasonal forecast targets and evaluation periods for short–term forecasts for each region are shown in Table 1. The evaluation period for each target reflected the Morbidity and Mortality Weekly Report (MMWR) calendar weeks when forecasts for that target have the most utility (see Methods).
Eleven teams submitted fourteen separate forecasts (models A-N, Table 2). Table 2 contains brief descriptions of each model’s methodology. All model structures remained consistent over the season, and only Model G made minor updates to their method to better incorporate trends in ILINet revisions. All but one model provided predictions for each of the 10 HHS Regions. Most teams participated throughout the season, but four forecasts began late: Model I (MMWR week 50), Model K (MMWR week 45), Model L (MMWR week 49), and Model N (MMWR week 4). For these models, earlier forecasts were scored as missing.
### Forecast Accuracy
Overall forecast accuracy was assessed using a metric of forecast skill, where 1 is a perfect forecast and 0 means the forecast assigned a <1% probability to the correct outcome. Average forecast skill for national targets over their respective evaluation periods are shown in Table 3. At the national level, median team forecast skill was highest for short-term forecasts of ILINet 1 week in advance (0.66) and decreased for the 2, 3, and 4 week ahead targets (median skill 0.41, 0.31, and 0.29, respectively). Median forecast skill for peak intensity (0.30) was comparable to that of short-term forecasts at 3–4 weeks. Median forecast skill for peak week and onset week are not directly comparable to the ILINet intensity forecasts because the scales and bins were different, but were both low (0.03 and 0.04, respectively).
As the season progressed, forecast skill for season onset, peak week, and peak intensity at the national level generally increased, though individual team skill varied considerably, especially for peak intensity (Fig. 2). For all seasonal targets, skill improved noticeably once the target of interest had passed according to observed ILINet data. For example, no models assigned a >50% probability to week 10 as the peak on week 6, while at week 10, 36% of submitted models assigned a >50% probability to week 10 as the peak. Week-ahead forecasts at the national level also showed considerable variability (Fig. 3), especially near the peak intensity of the influenza season when week-to-week variability in the ILINet value was the highest. All short-term forecasts had noticeable dips in accuracy around MMWR weeks 50 and 10, corresponding to inflection points in the ILINet data (Fig. 3).
At the regional level, median model forecast skill generally followed the same trend as the national level across the short-term forecasts, with 1 week ahead forecasts having the highest score and 4 week ahead forecasts having the lowest (Supplementary Tables S1S10). Median forecast skill for peak intensity and onset week varied considerably across the regions, with scores ranging from 0.06 to 0.51 and 0.003 to 0.46, respectively. Median forecast skill for peak week was low across all regions, ranging from 0.006 to 0.15. Across regions, median model forecast skill for short-term targets was lowest in HHS Regions 6 and 9 and highest in HHS Regions 8 and 10, while median skill for season targets was highest in HHS Regions 9 and 10 and lowest in HHS Regions 2 and 4 (Table 4).
### FluSight Ensemble and Historical Average Comparisons
Nine models outperformed the model based on the historical average of ILINet data at the national level for both peak intensity and 1 week ahead forecasts, while seven models outperformed the historical average for 2 weeks ahead and six models outperformed the historical average for both peak week and 3 weeks ahead (Table 3). Only three models outperformed the historical average for forecasts of ILINet 4 weeks ahead and only one model outperformed the historical average for onset week. For all targets at the national level, a model consisting of the unweighted mean of submitted models, which we refer to as the FluSight Ensemble model, outperformed the majority of submitted models (Table 3). Similar performance was seen for forecast targets at the HHS Regional level (Supplementary Tables S1S10).
### Gamma Regression by Model and Influenza Season Characteristics
Seven models were submitted by four teams that participated in the 2014–2015 CDC influenza forecasting challenges, though model specifications were updated between seasons. On average, these models were significantly more accurate than the seven forecasts submitted by first-time participating teams (Fig. 4). Five models utilized mechanistic models, encompassing compartmental modelling strategies such as Susceptible-Infected-Recovered (SIR) models. These methods attempted to model the underlying disease transmission dynamics and translated that into forecasts. Nine models utilized statistical methods that did not attempt to model disease transmission, but instead directly estimated the ILINet curve or target of interest using approaches such as time-series analysis or generalized linear models. The statistical models generally outperformed mechanistic models, with significant differences for peak week and 2, 3, and 4 week ahead forecasts. During the period between MMWR weeks 50-1, when there is a historic rise and dip in ILI values (Fig. 1), statistical models generally outperformed mechanistic models for 1–4 week ahead forecasts (Supplementary Table S11). Five models used only ILINet data to inform their predictions and nine models used additional data sources beyond those available in ILINet. The models using only ILINet data generally outperformed models incorporating additional data, with significant differences for all targets except peak week. Finally, six models used an ensemble approach combining predictions from multiple “in-house” models, and these models were associated with significantly higher skill for all forecasting targets compared to single models.
We compared forecasts across all forecast locations (10 HHS regions and the entire United States) to assess how seasonal characteristics (timing of season onset and peak, level of peak intensity relative to baseline, number of weeks above baseline, revisions to initial published wILI% values) affected forecast skill for those targets. Forecasts of peak week and onset were less accurate and forecasts for 1 and 2 weeks ahead were slightly more accurate for locations with a later peak, though forecast skill for other targets was unaffected (Fig. 5). Similarly, forecasts of season onset were less accurate for locations with a later onset. Relative peak intensity, defined as the peak intensity for a location divided by that location’s baseline ILINet value, had a significant but small association with increased accuracy for forecasts of 1 week ahead, but was not associated with forecast accuracy for any other targets examined. Short-term forecasts were generally less accurate in locations with longer influenza seasons, as measured in the number of weeks wILI% was above baseline. Forecasts of season onset were significantly more accurate in locations with longer influenza seasons, while accuracy of peak week and peak intensity forecasts were not associated with the length of the influenza season. Forecasts of short-term targets were also less accurate for forecasts based on weeks with larger differences between the initial and final published ILINet values than weeks with smaller revisions.
### Comparison to 2014–2015 Forecasting Results
Both onset and peak week occurred much later in the 2015–2016 season compared to the 2014–2015 season; at the national level, onset occurred in week 47 in 2014–2015 and week 3 in 2015–2016, while peak week occurred in week 52 in 2014–2015 and week 10 in 2015–2016. Compared to scores from the 2014–2015 challenge9, median team skill was higher for each of the national 1–4 week ahead targets in the 2015–2016 challenge than in the 2014–2015 challenge when scored using the same metrics (Table 5). The top model skill for each short-term target also increased from 2014–2015 to 2015–2016. The median skill for national peak percentage was higher during the 2015–2016 challenge, while the top model skill remained the same. However, median and top model skill for national onset week and peak week were both lower during the 2015–2016 challenge compared to the 2014–2015 challenge.
## Discussion
The 2015–2016 influenza season was the third consecutive influenza season that CDC hosted an influenza forecasting challenge. This accumulating body of real-time forecast data provides new insights on forecast accuracy, relative model performance, the value of ensemble approaches, and the challenges of influenza forecasting.
Forecast skill varied as the season progressed. Short-term forecast skill was generally highest at during the shoulders of the season when ILINet values were low and relatively constant and lowest around the peak week, an inflection point of the ILINet curve and a period during the influenza season when forecasts likely have the highest value from a public health perspective (Fig. 3). Forecast accuracy for seasonal targets, on the other hand, generally improved throughout the season as models incorporated new data. Skill for the seasonal targets generally began to improve substantially between 2 and 4 weeks prior to the predicted event (Fig. 2). While this improved accuracy may in part reflect more accurate short-term forecasts, the identification of the change from increasing to decreasing incidence (i.e. the peak) is a critical milestone for decision-makers. Even a lead time of only a few weeks is helpful for situational awareness, especially with a reporting delay of 1 to 2 weeks for initial surveillance data, subsequent revisions to those data as reporting is completed, and week-to-week variation that may occur even in the complete surveillance data. These forecasts can therefore provide public health officials with some level of confidence that the event has occurred.
A comparison of forecast skill across the forecast locations revealed additional characteristics of forecast performance. Forecasts for onset week and peak week generally had lower skill in locations with later onset weeks and peak weeks (Fig. 5). Seasonal targets that occur particularly early or late in a flu season are likely harder to predict simply because they are atypical, possibly with respect to other locations in the same season, with respect to previous seasons in the same location, or both. For short-term forecasts, these effects were not as strong (Fig. 5), indicating that late seasons have less of an effect on short-term forecasts. Conversely, short-term forecasts based on weeks with large subsequent revisions to the originally published ILINet values were less accurate than forecasts based on weeks that had minimal revisions to the final ILINet values. This is supported by the low median scores for short-term targets seen in HHS Regions 6 and 9 (Table 4), both of which had among the highest levels of backfill during the 2015–2016 season (Supplementary Fig. S1).
Compared to the previous season (2014–2015), average forecast skill in 2015–2016 was higher for peak intensity and lower for onset week and peak week (Table 5). The higher skill for peak intensity may reflect that 2014–2015 was an abnormally intense season while the peak intensity for 2015–2016 was more in line with typical past seasons (Fig. 1). Meanwhile, the onset and peak occurred later than typical in 2015–2016, possibly leading to the lower forecast skill for onset week and peak week. Notably, this agrees with our finding that forecast skill was lower for locations with later onset and peak weeks. The short-term forecasts had higher average forecast skill in 2015–2016 compared to 2014–2015. This may reflect short-term dynamics that were easier to predict, but more likely indicates higher model accuracy as this improvement was seen across locations where dynamics were quite different (Table 5).
Overall, there was no single best model across all targets; eleven of the fourteen participating models had the highest average score for at least one of the 77 short-term and seasonal targets across the 10 HHS regions and the United States. Nonetheless, Model B and Models E, F, G, and L consistently outperformed the FluSight Ensemble, other models, and the historical average for the seasonal and short-term targets at the national level, respectively. Also of note, the FluSight Ensemble outperformed the majority of individual forecast models for all targets and the historical average for all seven targets at the national level, showing that the combined forecasts provided more reliable information than most specific forecasts and more information than historical data alone. As the FluSight Ensemble was a simple average of received forecasts, the application of more sophisticated ensemble methods offers an opportunity for further improvements. The intention of the FluSight Ensemble was to evaluate a simple a priori ensemble approach that could be used during the season to combine information from multiple models, and as such we did not evaluate a posteriori approaches that could not be applied in real-time.
The variation in accuracy between models and the wide variety of forecasting approaches also provides insight into the characteristics of more accurate models. Comparisons of these approaches are not generalizable because they only reflect the combination of characteristics included in the submitted models, nothing close to the full spectrum of possible approaches16. They nonetheless provide notable insights. Models submitted by teams who had competed in the CDC forecasting challenge in previous years generally outperformed models submitted by new teams (Fig. 4). This may reflect self-selection of high-performing teams deciding to continue participating or it may indicate the value of participating in previous challenges. Making and submitting updated probabilistic forecasts on a weekly basis is a substantial technical challenge and those with experience doing that may be in a better position to identify and implement changes to improve accuracy. Models using ensemble approaches to generate their forecasts also outperformed single models, providing additional evidence of the value of ensemble forecasting approaches. Models that used data in addition to ILINet were less accurate than those only using ILINet data for six of seven targets, indicating that including auxiliary data does not necessarily lead to more accurate forecasts. Comparisons between statistical and mechanistic approaches indicated that performance varied by target, with statistical models outperforming mechanistic models for four of seven targets. However, the five models that consistently outperformed the median skill for all targets were all statistical models, illustrating the potential for this forecasting method. Additionally, during MMWR weeks 50 – 1, when there historically is a peak and dip in ILI values (Fig. 1), statistical models outperformed mechanistic models (Supplementary Table S11), illustrating that the statistical approaches may be more resilient to predictable patterns in ILI. As more forecasting approaches are applied over more seasons, more locations, and more diseases, more substantive analyses of these differences will be possible.
As an open, standardized, real-time, forecasting challenge, the CDC influenza forecasting challenge has provided unique insight into epidemic forecasting. The results highlight the continuing challenge of improving forecast accuracy for more seasons and at lead times of several weeks or more, forecasts that would be of even more utility for public health officials. To improve future forecasts, we found evidence that experience may help, that there is room for improving the use of external data, and that combining forecasts from multiple models in ensembles improved accuracy. Despite remaining challenges, both the top models and the FluSight Ensemble provided more accurate forecasts than historical data alone. Moreover, the accuracy for more typical seasons and for nearer targets (e.g. 1-week vs. 4-week ahead forecasts or peak forecasts early in the season vs. as the peak approaches) indicates that the models are producing valuable information as is. Because these forecasts are available in real time, they can actively improve situational awareness and be used to directly address immediate public health needs such as planning for hospital staffing and bed availability, outbreak preparedness, and stocking of antivirals.
Interest in infectious disease forecasting has increased in recent years, with challenges to predict epidemics of both chikungunya17 and dengue fever18 in addition to influenza. As the only ongoing infectious disease forecasting challenge in the United States, the CDC influenza forecasting challenge sets a model for other infectious diseases by identifying data and resource constraints that limit model development, establishing best practices for forecast submission and evaluation, identifying areas where forecasts can be improved, tying forecasting efforts to real public health needs, and assessing their performance related to those needs.
## Methods
### Challenge Structure
Teams from the previous challenge as well as research groups with experience in influenza or infectious disease forecasting worked with CDC to define the structure for the 2015–2016 challenge. Teams submitted weekly forecasts from November 2, 2015, to May 16, 2016. Forecasting targets were based on data from ILINet, a syndromic surveillance system consisting of more than 2,000 outpatient providers5. These providers send CDC weekly reports consisting of the number of patients with ILI and the total number of patients seen. These reports are weighted based on state population to determine a weighted percentage of patient visits due to ILI (wILI%). ILINet data use the Morbidity and Mortality Weekly Report (MMWR) week system, where a week starts on a Sunday and ends on a Saturday. Data for a given MMWR week are usually released the following Friday in CDC’s weekly FluView publication19. Each week’s publication includes initial values for the most recent week as well as potential revisions of prior published values, and the difference between initial and final published value varies by week and region (Supplementary Fig. S1).
Forecasting targets included seasonal and short-term targets. To participate in the challenge, teams were required to submit predictions on each target for the United States as a whole and had an option to submit predictions for each of the ten HHS Regions. The seasonal targets were onset week, defined as the first week where wILI% was at or above the location-specific baseline and remained at or above for at least two additional weeks; peak week, defined as the MMWR week during which the wILI% was highest; and peak intensity, defined as the season-wide maximum wILI%. Short-term targets included forecasts of wILI% one-, two-, three-, and four-weeks in advance of FluView publication. Due to the delay in reporting surveillance data (e.g. data for MMWR week 50 is published on MMWR week 51), the short-term targets provide a forecast for ILINet activity that occurred in the past week (1-week ahead), the present week (2-weeks ahead), and 1 (3-weeks ahead) and 2 weeks (4-weeks ahead) in the future.
As in the 2014–2015 season, participants submitted forecasts weekly as point estimates and probability distributions in a series of bins categorized across possible values for each target. For onset week and peak week, there was a bin for each single week of the season, with an additional bin for onset week corresponding to no onset. For peak intensity and short-term forecasts, semi-open 0.5% intervals (e.g. 1.0% ≤ wILI% < 1.5%) were used from 0% up to 13%, with the final bin representing the probability for all values greater than or equal to 13%. Teams submitted a written narrative of their forecasting methods for each model. Changes to the methods and narrative description were permitted during the season.
This study did not involve human participants and institutional review board approval was not required.
### Historical Average Forecast
To provide a benchmark to compare submitted forecasts to, we created a historical average forecast using ILINet data from the 1997–1998 flu season through the 2014–2015 flu season, excluding the 2009–2010 H1N1 pandemic year as its dynamics were atypical compared to seasonal epidemics. A Gaussian kernel density estimate using bandwidths estimated by the Sheather-Jones method20 was fitted to each MMWR week’s previous observed ILINet values, and approximate probabilities for each prediction bin were calculated by integrating the kernel density9. The point estimate was the median of the estimated distribution. Forecasts for onset week, peak week and peak intensity were calculated in the same way. Onset week forecast probabilities were adjusted to reflect the probability of no onset week based on the percentage of prior years in which ILI values did not cross the region-specific baseline. As CDC only began publishing regional baselines with the 2007–08 flu season, only seasons from that point onwards were used to calculate the onset week forecasts.
### Unweighted FluSight Ensemble
To evaluate the utility of a simple ensemble of influenza forecasts, we constructed an unweighted average of the individual forecasts received, which we refer to as the FluSight Ensemble. The estimated distribution of the FluSight Ensemble was created by taking the arithmetic mean of all submitted distributions for a given target/location combination during a particular week. As with the historical average forecast, we used the median of each distribution as the point estimate.
### Forecast Evaluation
We compared the forecasts, including the historical average and FluSight Ensemble forecasts, to weighted ILINet data published on MMWR week 28 (ending July 16, 2016), which was chosen a priori to represent final ILINet values for the season. We scored the forecasts using a forecast skill metric derived from the logarithmic scoring rule21,22. Let p be the binned probabilities submitted for a given forecast target, with pi the probability assigned to the bin containing the observed outcome i. For all targets, we included the bin above (i + 1) and below (i – 1) the observed outcome, and calculated the logarithmic score as $$S({\boldsymbol{p}},\,{i})={\rm{ln}}({p}_{i-1}+{p}_{i}+{p}_{i+1}).$$ For example, if the peak week was MMWR week 10, the logarithmic score would be calculated by summing the probabilities assigned to MMWR weeks 9–11 and taking the natural logarithm of that sum. In the case of multiple weeks having the same maximum wILI% and therefore being peak weeks, both peak weeks were considered as observed outcomes and the bins surrounding each peak were also included in the calculated score. Scores below −10, missing forecasts, or forecasts that summed to probabilities less than 0.9 or greater than 1.1 were all assigned scores of −10. Scores were averaged across different combinations of locations, targets, and time periods. As in 2014–2015, the averaged log scores were exponentiated to create a forecast skill on a scale of 0 to 1. Perfect forecasts would receive a skill of 1, while forecasts that assign very low probabilities to the observed outcome would receive a skill close to 0.
Evaluation periods varied by target and were chosen at the end of the season to include the weeks in which forecasts would have had the most utility to public health decision makers. The evaluation period for each seasonal target began with the first forecast submitted (MMWR week 42) and ended a target-specific number of weeks after each outcome had occurred. For onset week, the evaluation period ended six weeks after the season onset. For peak week and peak intensity, the evaluation period extended until one week after wILI% went below baseline and stayed below baseline for the remainder of the season (Table 1). For short-term forecasts, the evaluation period for each location began with forecasts received four weeks prior to season onset in that location and extended to 4 weeks after ILINet returned below baseline for that location.
We utilized gamma regression to analyse the effect of model type, data sources, targets, absolute change between initial and final published wILI% in the week each forecast was based on, and season types (e.g., late vs. early defined continuously by season onset and peak week) on forecast accuracy characterized as the negative log score. Gamma regression is restricted to outcome values greater than or equal to zero and is well-suited for analysing right-skewed data. For all regression models, we analysed across all weekly forecasts, targets and locations, excluding week-target-location forecasts that were not submitted. For comparisons of model characteristics, we controlled for location and the week a forecast was received in the regression analysis. For comparisons of seasonal characteristics across regions, we controlled for the week a forecast was received.
To compare forecasts across seasons, we summarized the 2015–2016 forecasts received into the larger, 1% wide bins utilized in the 2014–2015 challenge and scored the forecasts using the 2014–2015 log scoring rules. Forecasts for onset week and peak week were scored the same way during the 2014–2015 season, while for peak intensity and the short-term targets, only the probability assigned to the bin containing the observed value $${p}_{i}$$ was used.
Analyses were conducted using R version 3.4.323 and significance was assessed using a cutoff of p < 0.05.
## Data Availability
The received forecasts that support the findings of this study are publicly available on the CDC Epidemic Prediction Initiative GitHub page at https://github.com/cdcepi/FluSight-forecasts.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
A comprehensive list of consortium members appears at the end of the paper
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## Author information
### Affiliations
1. #### Epidemiology and Prevention Branch, Influenza Division, Centers for Disease Control and Prevention, Atlanta, Georgia, USA
• Craig J. McGowan
• , Matthew Biggerstaff
• & Carrie Reed
2. #### Dengue Branch, Division of Vector-Borne Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia, USA
• Michael Johansson
3. #### Arete Associates, Northridge, California, USA
• Karyn M. Apfeldorf
4. #### Predictive Science, Inc., San Diego, California, USA
• Michal Ben-Nun
• , Pete Riley
• , James Turtle
• , David Bacon
• & Steven Riley
5. #### Computer Science Department, Carnegie Mellon University, Pittsburgh, Pennsylvania, USA
• Logan Brooks
• & Roni Rosenfeld
6. #### Division of Media and Network Technologies and Division of Frontier Science, Graduate School of Information Science and Technology, Gi-CoRE Station for Big Data & Cybersecurity, Hokkaido University, Sapporo, Japan
• Matteo Convertino
7. #### Division of Environmental Health Sciences, School of Public Health, University of Minnesota, Minneapolis, Minnesota, USA
• Matteo Convertino
• & Yang Liu
8. #### Knowledge Based Systems, Inc., College Station, Texas, USA
• & John Freeze
9. #### Computational Biology Department, Carnegie Mellon University, Pittsburgh, Pennsylvania, USA
• David C. Farrow
10. #### Discovery Analytics Center, Virginia Tech University, Arlington, Virginia, USA
• Saurav Ghosh
• & Naren Ramakrishnan
11. #### Department of Statistics and Data Science, Carnegie Mellon University, Pittsburgh, Pennsylvania, USA
• Sangwon Hyun
12. #### Department of Environmental Health Sciences, Mailman School of Public Health, Columbia University, New York, New York, USA
• Sasikiran Kandula
• , Haruka Morita
• , Jeffrey Shaman
• & Wan Yang
13. #### Department of Mathematics, University of Arizona, Tucson, Arizona, USA
• Joceline Lega
14. #### Department of Statistics, University of California, Berkeley, Berkeley, California, USA
• Nicholas Michaud
15. #### Department of Statistics, Iowa State University, Ames, Iowa, USA
• , Nehemias Ulloa
• & Katie Will
16. #### Department of Mathematics and Statistics, Mount Holyoke College, South Hadley, Massachusetts, USA
• Evan L. Ray
17. #### Department of Biostatistics and Epidemiology, School of Public Health and Health Sciences, University of Massachusetts, Amherst, Amherst, Massachusetts, USA
• Nicholas G. Reich
18. #### Department of Statistics and Data Science, Machine Learning Department, Carnegie Mellon University, Pittsburgh, Pennsylvania, USA
• Ryan Tibshirani
19. #### Northeastern University, Boston, Massachusetts, USA
• Alessandro Vespignani
• & Qian Zhang
### Contributions
C.J.M. carried out analyses and drafted the initial manuscript; M.B., M.A.J. and C.R. conceptualized the forecasting challenge and contributed to and revised the manuscript; K.M.A., M.B.-N., L.B., M.C., M.E., D.C.F., J.F., S.G., S.H., S.K., J.L., Y.L., N.M., H.M., J.N., N.R., E.L.R., N.G.R., P.R., J.S., R.T., A.V. and Q.Z. contributed forecasting models and revised the manuscript. Members of the I.F.W.G. assisted in the development of contributed forecasting models
### Competing Interests
J.S. discloses partial ownership of S.K. Analytics, S.K. and H.M. disclose consulting for S.K. Analytics. The remaining authors declare no competing interests.
### Corresponding author
Correspondence to Matthew Biggerstaff .
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2019-03-24 04:40:12
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http://www.mathematik.uni-ulm.de/numerik/hpc/ws19/session02/page01.html
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# Full storage format for matrices
#### Content
Goal: Getting familiar with the full storage format for matrices.
## Exercise
In this first exercise you are asked to write a small C program that initializes a $$7 \times 8$$ col major matrix and then prints the matrix. The program should consist of three functions initMatrix, printMatrix and main:
• Function initMatrix initializes an $$m \times n$$ matrix. The initialization enumerates the matrix elements row-wise from $$1$$ to $$m\cdot n$$.
E.g. in the case where $$m=2$$ and $$n=3$$ the matrix afterwards looks like this:
$\begin{pmatrix}a_{0,0} & a_{0,1} & a_{0,2} \\a_{1,0} & a_{1,1} & a_{1,2} \\\end{pmatrix}\leftarrow\begin{pmatrix}1 & 2 & 3 \\4 & 5 & 6\end{pmatrix}$
• Function printMatrix is supposed to print the content of a matrix.
• In function main, define and initialize local variables m, n, A, incRowA, incColA that describe the storage of a matrix:
• m and n should be of type size_t,
• A should be a pointer to double,
• incRowA and incColA should be of type ptrdiff_t.
After allocating memory for the matrix elements, initialize and print the matrix. Before main returns release the memory.
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2019-12-11 18:11:45
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https://www.omnicalculator.com/math/area-of-hemisphere
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in
Diameter
in
Volume
cu in
Surface to volume ratio
1/
in
Surface areas
Base
in²
Cap
in²
Total
in²
# Surface Area of a Hemisphere Calculator
By Dominik Czernia, PhD candidate
Our surface area of a hemisphere calculator is a handy tool that finds different types of hemisphere surface areas. Are you looking for an answer to the question of how to find the surface area of a hemisphere? Or maybe you just need to estimate it quickly? Whatever you plan to do, try this hemisphere calculator which comprises several various area of a hemisphere formulas.
Hemispheres are created by dividing a sphere into two equal halves, as you can see in the picture below. Unlike the full sphere, a hemisphere has two kind of surface areas: the base area (which is a circle) and the cap area. The notation, which we have used in this surface area of a hemisphere calculator, is as follows:
• r - radius of a hemisphere,
• d - diameter of a hemisphere,
• V - volume of a hemisphere,
• A - total surface area of a hemisphere,
• Ab - base surface area of a hemisphere,
• Ac - cap surface area of a hemisphere,
• A / V - surface to volume ratio of a hemisphere.
The interesting fact is that the total volume of two hemispheres equals the volume of a one full sphere. However, the same is not true with surface areas. The total area of two hemispheres is greater than the area of a sphere. The reason is simple: hemispheres have an additional base area.
## How to find the surface area of a hemisphere?
When we split a sphere in half and take one of the resulting parts, we get a hemisphere. In each hemisphere, we can name two surface areas: base area and cap area (look at the picture above). From the area of a sphere calculator we know that the surface area of a sphere is as below:
`A(sphere) = 4 * π * r²`.
You can think about it like two times the cap surface area of a hemisphere. Therefore, the hemisphere cap area equals:
`Ac = A(sphere) / 2`,
`Ac = 2 * π * r²`.
The base surface area is a circle with the same radius as a hemisphere. Thus, it can be expressed as:
`Ab = π * r²`.
Finally, the total surface area is the sum of those two contributions:
`A = Ac + Ab`,
`A = 2 * π * r² + π * r²`,
`A = 3 * π * r²`.
This area of a hemisphere calculator allows you to find all three types of surface areas of a specific hemisphere. Moreover, you can do every calculation in different units (SI and imperial). If you want to learn more about area unit conversion, check out our area converter!
## What's the area of a hemisphere formula?
Now, after we know what are the surface areas of a hemisphere and how to find them, let's try to derive different area of a hemisphere formulas. They can be useful in situations when we don't have the radius given. First of all, there are some basic hemisphere equations you should know:
1. Diameter of a hemisphere: `d = 2 * r`,
2. Volume of a hemisphere: `V = 2/3 * π * r³`,
3. Base surface area of a hemisphere: `Ab = π * r²`,
4. Cap surface area of a hemisphere: `Ac = 2 * π * r²`,
5. Total surface area of a hemisphere: `A = 3 * π * r²`,
6. Surface to volume ratio of a hemisphere: `A / V = 9 / (2 * r)`.
The area of a hemisphere formula (for the total area) can be then derived from the above equations. We can obtain even six of them! The equations used by this area of a hemisphere calculator are as follows:
1. Given radius: `A = 3 * π * r²`,
2. Given diameter: `A = 3/4 * π * d²`,
3. Given volume: `A = ³√[243/4 * π * V²]`
4. Given base area: `A = 3 * Ab`,
5. Given cap area: `A = 3/2 * Ac`,
6. Given surface to volume ratio: `A = 243 * π / (4 * (A/V)²)`.
As you know, the Earth is approximately a sphere with a radius of almost 6400 km, in which we can specify the northern and the southern hemispheres (specifically speaking, the Earth is a geoid). This division plays an essential role in geography and physics. For example, there is a force called Coriolis effect which acts on you whenever you travel by airplane, causing you to get off the course. When you're on the northern hemisphere, you will deflect to the right, and on the southern hemisphere to the left.
Dominik Czernia, PhD candidate
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2021-01-15 14:07:43
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http://www.physicspages.com/2012/07/24/the-free-particle-probability-current/
|
The free particle: probability current
Required math: calculus
Required physics: Schrödinger equation
Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.19.
The rate of change of probability of a particle in a given range of ${x}$ can be written as the difference in probability current at the two ends. The current is defined as
$\displaystyle J(x,t)\equiv\frac{i\hbar}{2m}\left(\frac{\partial\Psi^*}{\partial x}\Psi-\frac{\partial\Psi}{\partial x}\Psi^*\right) \ \ \ \ \ (1)$
For the free particle, a stationary state is given by
$\displaystyle \Psi(x,t)=Ae^{ikx}e^{-i\hbar k^{2}t/2m} \ \ \ \ \ (2)$
The probability current for this state is found by working out the derivative:
$\displaystyle \frac{\partial\Psi}{\partial x}$ $\displaystyle =$ $\displaystyle ikAe^{ikx}e^{-i\hbar k^{2}t/2m}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle ik\Psi \ \ \ \ \ (4)$
So we get
$\displaystyle J(x,t)$ $\displaystyle =$ $\displaystyle \frac{i\hbar k}{2m}\left(-i\left|\Psi\right|^{2}-i\left|\Psi\right|^{2}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar k}{m}\left|A\right|^{2} \ \ \ \ \ (6)$
(The complex exponentials cancel out in ${\left|\Psi\right|^{2}}$.) Since the current is positive, it ‘flows’ in the positive ${x}$ direction. Note that the current is independent of ${x}$, so the probability of a particle being found in any given range of ${x}$ is constant. (Actually, as we’ve seen, a free particle can’t exist in a single stationary state since such a state cannot be normalized.)
2 thoughts on “The free particle: probability current”
1. George
If I understood the theory well, a free particle cannot exist in a stationary state so we should probably have the integral over k form of the wave equation? Are we allowed to use stationary states as a mathematically useful trick which does not have any physical meaning, and getting right results? The integral form shouldn’t have given the same result?
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2017-04-30 03:26:58
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https://git.enlightenment.org/core/efl.git/tree/src/lib/efl/interfaces/efl_gfx_base.eo?id=ed0988a22d9439934a5213f4fa96cb7db76c71e3
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summaryrefslogtreecommitdiff log msg author committer range
path: root/src/lib/efl/interfaces/efl_gfx_base.eo
blob: e3d70260a3e502d91f7814a5c2b0152c9be35521 (plain) (blame)
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 ``` ``````interface Efl.Gfx.Base { legacy_prefix: null; eo_prefix: efl_gfx; methods { @property position { set { /*@ Move the given Evas object to the given location inside its canvas' viewport. */ } get { /*@ Retrieves the position of the given Evas object. */ } values { int x; /*@ in */ int y; /*@ in */ } } @property size { set { /*@ Changes the size of the given Evas object. */ } get { /*@ Retrieves the (rectangular) size of the given Evas object. */ } values { int w; /*@ in */ int h; /*@ in */ } } @property color { set { /*@ Sets the general/main color of the given Evas object to the given one. @see evas_object_color_get() (for an example) @note These color values are expected to be premultiplied by @p a. @ingroup Evas_Object_Group_Basic */ } get { /*@ Retrieves the general/main color of the given Evas object. Retrieves the “main” color's RGB component (and alpha channel) values, which range from 0 to 255. For the alpha channel, which defines the object's transparency level, 0 means totally transparent, while 255 means opaque. These color values are premultiplied by the alpha value. Usually you’ll use this attribute for text and rectangle objects, where the “main” color is their unique one. If set for objects which themselves have colors, like the images one, those colors get modulated by this one. @note All newly created Evas rectangles get the default color values of 255 255 255 255 (opaque white). @note Use @c NULL pointers on the components you're not interested in: they'll be ignored by the function. Example: @dontinclude evas-object-manipulation.c @skip int alpha, r, g, b; @until return See the full @ref Example_Evas_Object_Manipulation "example". @ingroup Evas_Object_Group_Basic */ } values { int r; /*@ The red component of the given color. */ int g; /*@ The green component of the given color. */ int b; /*@ The blue component of the given color. */ int a; /*@ The alpha component of the given color. */ } } @property color_part { set { /*@ Sets a specifc color of the given Efl.Gfx.Base object to the given one. @see evas_object_color_get() (for an example) @note These color values are expected to be premultiplied by @p a. */ return: bool; } get { /*@ Retrieves a specific color of the given Evas object. Retrieves a specific color's RGB component (and alpha channel) values, which range from 0 to 255. For the alpha channel, which defines the object's transparency level, 0 means totally transparent, while 255 means opaque. These color values are premultiplied by the alpha value. The “main“ color being mapped to @c NULL. Usually you’ll use this attribute for text and rectangle objects, where the “main” color is their unique one. If set for objects which themselves have colors, like the images one, those colors get modulated by this one. @note Use @c NULL pointers on the components you're not interested in: they'll be ignored by the function. */ return: bool; } keys { const (char)* part; /*@ The part you are interested in. */ } values { int r; /*@ The red component of the given color. */ int g; /*@ The green component of the given color. */ int b; /*@ The blue component of the given color. */ int a; /*@ The alpha component of the given color. */ } } @property visible { set { /*@ Makes the given Evas object visible or invisible. */ } get { /*@ Retrieves whether or not the given Evas object is visible. */ } values { bool v; /*@ @c EINA_TRUE if to make the object visible, @c EINA_FALSE otherwise */ } } } } ``````
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2020-04-06 22:22:05
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https://brilliant.org/problems/the-smallest-k/
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The Smallest $k$
Let $f(x)=7x^{11}+11x^{7}+9kx$ $g(x)=7x^{13}+13x^{7}+10kx$
Find the minimum positive value of $k$ for which $77| f(x)$ and $91|g(x)$ for every integer $x$
Details and Assumptions
• We use "$\mid$" to mean "divides evenly into". For example, $3\mid6$, and $12\mid132$.
×
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2020-04-04 03:32:53
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http://wims.unice.fr/~wims/wims.cgi?lang=en&+cmd=new&+module=tool%2Fgeometry%2Fvision4d.en&+type=quadratic
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# Vision 4D
A quadratic hypersurface is the set of points verifying an algebraic equation of degree 2.
This application proposes for you to visualize the following list of quadratic hypersurfaces in the space-time of dimension 4. In the equations, x,y,z are the coordinates of the space, and we use the variable t to designate the coordinate time. A same hypersurface is often presented several times, under different viewing angles (in particular with respect to t).
.Noequationdescription
Show 1x2+y2+z2+t2=1sphere S3
Show 2x2+y2+z2-t2=0spherical cone with principal axis on the axis of t
Show 3x2+y2-z2+t2=0spherical cone with principal axis on the axis of z
Show 4x2-y2+z2+t2=0spherical cone with principal axis on the axis of y
Show 5x2+y2-zt=0spherical cone whose principal axis is the line x=y=z+t=0
Show 6x2+z2-yt=0spherical cone whose principal axis is the line x=z=y+t=0
Show 7x2+y2+z2-t2=1spherical hyperboloid whose principal axis is the axis of t
Show 8x2+y2-z2+t2=1spherical hyperboloid whose principal axis is the axis of z
Show 9x2-y2+z2+t2=1spherical hyperboloid whose principal axis is the axis of y
Show 10x2+y2-zt=1spherical hyperboloid whose principal axis is the line x=y=z+t=0
Show 11x2+z2-yt=1spherical hyperboloid whose principal axis is the line x=z=y+t=0
Show 12x2+y2-z2-t2=0vertical hyperboloidal cone
Show 13x2-y2+z2-t2=0horizontal hyperboloidal cone
Show 14x2-y2-zt=0hyperboloidal cone
Show 15x2-z2-yt=0hyperboloidal cone
Show 16x2+y2-z2-t2=1hyperboloidal hyperboloid
Show 17x2-y2+z2-t2=1hyperboloidal hyperboloid
Show 18x2+y2-z2-t2= -1hyperboloidal hyperboloid
Show 19x2-y2+z2-t2= -1hyperboloidal hyperboloid
Show 20x2-y2-zt=1hyperboloidal hyperboloid
Show 21x2-z2-yt=1hyperboloidal hyperboloid
Show 22x2+y2+z2-t=0spherical paraboloid oriented towards the axis of t
Show 23x2+y2-z+t2=0spherical paraboloid oriented towards the axis of z
Show 24x2-y+z2+t2=0spherical paraboloid oriented towards the axis of y
Show 25x2+y2-z2-t=0hyperboloidal paraboloid oriented towards the axis of t, vertical
Show 26x2-y2+z2-t=0hyperboloidal paraboloid oriented towards the axis of t, horizontal
Show 27x2+y2-z-t2=0hyperboloidal paraboloid oriented towards the axis of z
Show 28x2-y2-z+t2=0hyperboloidal paraboloid oriented towards the axis of z
Show 29x2-y+z2-t2=0hyperboloidal paraboloid oriented towards the axis of y
Show 30x2-y-z2+t2=0hyperboloidal paraboloid oriented towards the axis of y
Show 31x2+y2+t2=1vertical spherical cylinder
Show 32x2+z2+t2=1horizontal spherical cylinder
Show 33x2+y2-t2=0conic cylinder with a singular line on the axis of z
Show 34x2+z2-t2=0conic cylinder with a singular line on the axis of y
Show 35x2+y2-t2=1hyperboloidal cylinder with one sheet, vertical
Show 36x2+z2-t2=1hyperboloidal cylinder with one sheet, horizontal
Show 37x2+y2-t2= -1hyperboloidal cylinder with two sheets, vertical
Show 38x2+z2-t2= -1hyperboloidal cylinder with two sheets, horizontal
In order to access WIMS services, you need a browser supporting forms. In order to test the browser you are using, please type the word wims here: and press Enter''.
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2018-09-20 21:23:07
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|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14Ionic_Equilibria_in_Aqueous_Solutions/14.08%3A_Indicators
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# 14.8: Indicators
We mentioned that in most titrations it is necessary to add an indicator which produces a sudden color change at the equivalence point. A typical indicator for acid-base titrations is phenolphthalein, HC20H13O4. Phenolphthalein, whose structure is shown below, is a colorless weak acid (Ka = 3 × 10–10 mol/L). Its conjugate base, C20H13O4 has a strong pinkish-red color. In order to simplify, we will write the phenolphthalein molecule as HIn (protonated indicator) and its pink conjugate base as In. In aqueous solution, phenolphthalein will present the following equilibrium
$\ce{HIn + H_{2}O \rightleftharpoons In^{-} + H_{3}O^{+}}\label{1}$
According to Le Chatelier’s principle, the equilibrium expressed in equation $$\ref{1}$$ will be shifted to the left if H3O+ is added. Thus in a strongly acidic solution we expect nearly all the pink In to be consumed, and only colorless HIn will remain. On the other hand, if the solution is made strongly basic, the equilibrium will shift to the right because OH ions will react with HIn molecules, converting them to In. Thus the phenolphthalein solution will become pink.
Clearly there must be some intermediate situation where half the phenolphthalein is in the acid form and half in the colored conjugate-base form. That is, at some pH
$\ce{[HIn] = [In^{-}]}$
This intermediate pH can be calculated by applying the Henderson-Hasselbalch equation to the indicator equilibrium:
$\text{pH}=\text{p}K_{a}\text{ + log}\frac{[\text{ In}^{-}]}{[\text{ HIn }]}$
Thus at the point where half the indicator is conjugate acid and half conjugate base,
$\text{pH}=\text{p}K_{a}\text{+log1}=\text{p}K_{a}$
For phenolphthalein, we have
$\text{pH}=\text{p}K_{a}=-\text{log(3 }\times \text{ 10}^{-10}\text{)}=\text{9.5}$
so we expect phenolphthalein to change color in the vicinity of pH = 9.5.
The way in which both the color of phenolphthalein and the fraction present as the conjugate base varies with the pH is shown in detail in Fig. $$\PageIndex{1}$$. The change of color occurs over quite a limited range of pH―roughly pKa ± 1. In other words the color of phenolphthalein changes perceptibly between about pH 8.3 and 10.5. Observe the actual color change for this indicator in Fig. $$\PageIndex{2}$$.
Figure $$\PageIndex{1}$$ Color change of phenolphthalein with respect to pH
Figure $$\PageIndex{2}$$ Phenolphtalein at different pH, notice the distinctive color at pH 8-12
Other indicators behave in essentially the same way, but for many of them both the acid and the conjugate base are colored. Their pKa’s also differ from phenolphthalein, as shown in the following table. The indicators listed have been selected so that their pKa values are approximately two units apart. Consequently, they offer a series of color changes spanning the whole pH range.
Properties of Selected Indicators
Color Name pKa Effective pH range Acid form Basic form Thymol blue 1.6 1.2 - 2.8 Red Yellow Methyl orange 4.2 3.1 - 4.4 Red Orange Methyl red 5.0 4.2 - 6.2 Red Yellow Bromothymol blue 7.1 6.0 - 7.8 Yellow Blue Phenophthalein 9.5 8.3 - 10.0 Colorless Red Alizarin yellow 11.0 10.1 - 12.4 Yellow Red
Indicators are often used to make measurements of pH which are precise to about 0.2 or 0.3 units. Suppose, for example, we add two drops of bromothymol blue to a sample of tap water and obtain a green-blue solution. Since bromothymol blue is green at a pH of 6 and blue at a pH of 8, we conclude that the pH is between these two limits. A more precise result could be obtained by comparing the color in the tap water with that obtained when two drops of indicator solution are added to buffer solutions of pH 6.5 and 7.5.
If a careful choice of both colors and pKa is made, it is possible to mix several indicators and obtain a universal indicator which changes color continuously over a very wide pH range. With such a mixture it is possible to find the approximate pH of any solution within this range. So-called pH paper, as seen below, is impregnated with one or several indicators. When a strip of this paper is immersed in a solution, its pH can be judged from the resulting color.
Example $$\PageIndex{1}$$: Indicators
What indicator, from those listed in the table, would you use to determine the approximate pH of the following solutions:
a) 0.1 M CH3COONa (sodium acetate)
b) 0.1 M CH3COOH (acetic acid)
c) A buffer mixture of sodium acetate and acetic acid
d) 0.1 M NH4Cl (ammonium chloride)
Solution
a) A solution of sodium acetate will be mildly basic with a pH of 9 or 10. Phenolphthalein would probably be best.
b) A solution of acetic acid, unless very dilute, has a pH in the vicinity of 3. Both thymol blue and methyl orange should be tried.
c) Since Ka for acetic acid is 1.8 × 10–5 mol/L, we can expect this buffer to have a pH not far from 5. Methyl red would be a good indicator to try.
d) Since NH4+is a very weak acid, this solution will be only faintly acidic with a pH of 5 or 6. Again methyl red would be a good indicator to try, though bromothymol blue is also a possibility.
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2019-03-20 09:24:05
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http://www.r-bloggers.com/violin-plots-and-regional-income-distribution/
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Violin plots and regional income distribution
March 20, 2013
By
(This article was first published on StaTEAstics., and kindly contributed to R-bloggers)
While preparing my slides for statistical graphics, a plot really caught my eye when I was playing around with the data.
I started off by plotting the time seriesof GNI per capita by country, and as expected it got quite messy and incomprehensible.
## Download and manipulate the datalibrary(FAOSTAT)raw.lst = getWDItoSYB(indicator = c("NY.GNP.PCAP.CD", "SP.POP.TOTL"))raw.df = raw.lst[["entity"]]traw.df = translateCountryCode(raw.df, from = "ISO2_WB_CODE", to = "UN_CODE")mraw.df = merge(traw.df, FAOregionProfile[, c("UN_CODE", "UNSD_MACRO_REG")])final.df = mraw.df[!is.na(mraw.df$UNSD_MACRO_REG), ]## Simple ugly time series plotggplot(data = final.df, aes(x = Year, y = NY.GNP.PCAP.CD)) + geom_line(aes(col = Country)) + labs(x = NULL, y = "GNI per capita") So I decided to compute the weighted average by region to examine the regional trends. ## Compute regional aggregates based on UN M49 definitionreg.df = aggRegion(aggVar = "NY.GNP.PCAP.CD", weightVar = "SP.POP.TOTL", data = traw.df, keepUnspecified = FALSE, aggMethod = "weighted.mean", relationDF = data.frame(UN_CODE = FAOregionProfile[, "UN_CODE"], REG_NAME = FAOregionProfile[, "UNSD_MACRO_REG"]))## Plot regional aggregatesggplot(data = reg.df[!is.na(reg.df$NY.GNP.PCAP.CD), ], aes(x = Year, y = NY.GNP.PCAP.CD)) + geom_line(aes(col = REG_NAME)) + labs(x = NULL, y = "GNI per capita", col = "")
I can now see the trend clearly, but there are two problems with this approach. First, the variability within region is vast and thus the weighted average or any summary statistic such as quantile can be misleading and it does not tell me what is going on within the regions. Secondly, since a minimum of 65% of the country must be present in order to compute the aggregation, no statistics was available prior to 1985.
While I was carrying out regional comparisons with box-plot and violin plot I thought why not plot them accross time as well! So here is the final graph:
## Time series violin plotggplot(data = final.df, aes(x = as.character(Year), y = NY.GNP.PCAP.CD)) + geom_violin() + scale_y_log10() + facet_wrap(~UNSD_MACRO_REG, ncol = 1, scales = "free_y") + scale_x_discrete(breaks = as.character((seq(1960, 2010, by = 10))), labels = as.character((seq(1960, 2010, by = 10)))) + labs(x = NULL, y = "GNI per capita")
Now I can compare the regions, but at the same time I can see the within region income distribution. It amazes me how the income distribution diverges in Europe and Oceania while America and Asia moves towards a bell shaped distribution. Growth in Africa appears to be slow, but there are several countries which are growing at a faster rate and pushing the tail of the distribution. Although some of the variability in the density may have resulted from independence of countries, nonetheless it is still infromative.
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2014-09-23 06:20:42
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https://ask.libreoffice.org/en/answers/136869/revisions/
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Revision history [back]
If you just want a Δ on the text box there is no need to use a Math object: just insert it as special character. LibO have several autocorrect options that makes easy to insert special symbols and there is one for Δ: just type
:Delta:
and as soon as you type the second colon you'll get your Δ. You can see all autocorrect options on Tools → Autocorrect → Autocorrect options → Replace tab. Be aware that the options are language sensitive: if you change text language the "names" could be different.
If you just want a Δ on the text box there is no need to use a Math object: just insert it as special character. LibO have several autocorrect options that makes easy to insert special symbols and there is one for Δ: just type
:Delta:
and as soon as you type the second colon you'll get your Δ. You can see all autocorrect options on Tools → Autocorrect → Autocorrect options → Replace tab. tab. Be aware that the options are language sensitive: if you change text language the "names" could be different.
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2019-01-17 07:50:38
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https://www.physicsforums.com/threads/a-question-about-partial-derivatives.862031/
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# A Question About Partial Derivatives
1. Mar 14, 2016
1. The problem statement, all variables and given/known data
$$v_{i}=\dot{x}_{i}=\dot{x}_{i}\left(q_{1},q_{2},..,q_{n},t\right)$$
$$T \equiv \frac{1}{2}\cdot{\sum}m_{i}v_{i}^{2}$$
$$\frac{\partial T}{\partial\dot{q}_{k}}={\sum}m_{i}v_{i}\frac{\partial v_{i}}{\partial\dot{q}_{k}}={\sum}m_{i}v_{i}\frac{\partial x_{i}}{\partial q_{k}}$$
2. Relevant equations
Why is it ok to assume:
$$\frac{\partial v_{i}}{\partial\dot{q}_{k}} = \frac{\partial x_{i}}{\partial q_{k}}$$
3. The attempt at a solution
I can say that:
$$\frac{\partial x_{i}}{\partial q_{k}}=\frac{\partial x_{i}}{\partial t}\frac{\partial t}{\partial q_{k}}=\frac{v_{i}}{\dot{q_{i}}}$$ but it's not the same as written.
and the expression $$\frac{\partial v_{i}}{\partial\dot{q}_{k}}$$ says to differentiate the velocity according to change of the q quardinate in time.
Last edited: Mar 14, 2016
2. Mar 14, 2016
### SammyS
Staff Emeritus
Maybe try:
$\displaystyle \frac{\partial x_i}{\partial t}=\frac{\partial x_i}{\partial q_k} \frac{\partial q_k}{\partial t}$
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2017-08-19 04:12:51
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https://tex.stackexchange.com/questions/402962/overwriting-file-warning/402964
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Overwriting file warning
Is it possible to suppress the warning generated by:
\begin{filecontents*}
Every time it outputs the warning: Overwriting file 'xxxx.xxx'. Maybe to delete it before it creates a new one or something like that?
Thank you
You can use the silence package to mute the warning:
\usepackage{silence}
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2019-10-21 12:26:19
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https://tex.stackexchange.com/questions/432785/how-can-i-generate-customized-smart-diagrams
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# How can I generate customized smart diagrams?
Actually, I am unable to comment on any post so far. Therefore, I have to ask the questions in the separate thread.
I want to customize the bubble nodes in a bubble diagram and found one thread discussing it. But I want to generate something similar to the attached image. How can I generate an image (similar to attached) using smart diagrams?
I have following MWE (derived from above mentioned thread):
\documentclass[tikz]{standalone}
\usepackage{smartdiagram}
\usetikzlibrary{shapes.geometric,calc}
\begin{document}
\tikzset{
planet/.append style={regular polygon, regular polygon sides=6},
satellite/.append style={rectangle},
every picture/.append style={rotate=30},
connection planet satellite/.style={
bend right/.style=,
every edge/.style={fill=\col},
to path={
\pgfextra
\path[draw=none, fill=none] (\tikztostart)
-- coordinate[at start] (@start@) coordinate[at end] (@target@)
(\tikztotarget);
\endpgfextra
}}}
\smartdiagram[connected constellation diagram]{
Build a program, Set up, Run, Analyze, Modify/\\Add, Check, and a sixth}
\end{document}
How can I customize the appearance of bubble nodes similar to shown in the attached image (especially I want circles green and red with + and - sign)?
I hope it clears the question.
• This might help you, since you're using that tag. Unless the tagging is wrong. Also: – thymaro May 22 '18 at 7:10
• Welcome to TeX.SE! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. The easier it is to copy and test your code, the more likely your question will be answered and can help others in a similar situation. – thymaro May 22 '18 at 7:11
• Hi, @thymaro Thank you for your suggestion. I have added MWE, hope it helps in finding the solution. – Jitendra May 22 '18 at 8:04
• In the example image, some of the text is centered, some ragged right and some ragged left. Just how smart is this diagram going to have to be? (Or to put it another way, just how lazy is the writer as opposed to the programmer having to set this up?) – John Kormylo May 22 '18 at 13:06
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2020-03-28 12:49:30
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https://math.berkeley.edu/wp/apde/category/fall-2016/
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Marcelo Disconzi (Vanderbilt University)
The Analysis and PDE Seminar will take place on Monday, September 19, in room 740, Evans Hall, from 4:10-5:00 pm.
Speaker: Marcelo Disconzi
Title: The three-dimensional free boundary Euler equations with surface tension.
Abstract: We study the free boundary Euler equations with surface tension in three spatial dimensions, showing that the equations are well-posed if the coefficient of surface tension is positive. Then we prove that under natural assumptions, the solutions of the free boundary motion converge to solutions of the Euler equations in a domain with fixed boundary when the coefficient of surface tension tends to infinity. This is a joint work with David G. Ebin.
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2019-02-23 17:15:27
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https://chemistry.stackexchange.com/tags/entropy/new
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# Tag Info
0
It's a good idea to work out the assumptions implicit in your question. Since you are considering use of the Gibbs free energy as a criterion for spontaneity you are presumably concerned with a process carried out at constant T and P: $$dG \leq 0 ~~~(\textrm{const. T, P)}$$ The second assumption is that $\Delta S$ and $\Delta H$, associated with the ...
0
Not all reaction is like this but, exothermic reactions are usually spontaneous because molecules always want to be more stable. So by reacting to another molecule and releasing energy, they can lower their own energy thus making them more stable and unreactive.
2
This answer is correct. After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature ...
Top 50 recent answers are included
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2019-09-21 15:33:57
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http://www.drmaciver.com/category/performing-philosophy-without-a-license/
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# Reality is a countably infinite Sierpiński cube
I was thinking out loud on Twitter about what weird beliefs I hold, after realising (more or less as I was writing it) that my philosophical positions are practically banal (at least to anyone who has thought about these issues a bit, whether or not they agree with me).
I came up with a couple, but probably the most interesting (if very very niche) that I thought of is that one true and accurate mathematical model of reality is time cube a closed, connected, subset of the countably infinite Sierpinski cube.
I consider this opinion to be not that weird and more importantly obviously correct, but I’m aware that this is a niche opinion, but hear me out.
Before we start, a quick note on the nature of reality. I am being deliberately imprecise about what I mean by “reality” here, and basically mean “any physical system”. This could be “life the universe and everything” and we are attempting to solve physics, or it could be some toy restricted physical system of interest and we are trying to nail down its behaviour. This post applies equally well to any physical system we want to be able to model.
Consider an experiment. Let’s pretend we can have deterministic experiments for convenience – you can easily work around the impossibility by making countably infinitely many copies of the experiment and considering each of them to be the answer you got the nth time you ran the experiment.
Also for simplicity we’ll assume that experiments can only have one of two outcomes (this is no loss of generality as long as experiments can only have finitely many outcomes – you just consider the finitely many experiments of the form “Was the outcome X?” – and if they have infinitely many outcomes you still need to ultimately make a finite classification of the result and so can consider the experiment composed with that classification).
There are three sensible possible outcomes you could have here:
• Yes
• No
• I don’t know, maybe?
Physical experiments are inherently imprecise – things go wrong in your experiment, in your setup, in just about every bloody thing, so set of experiments whose outcome will give you total certainty is implausible and we can ignore it.
Which leaves us with experiments where one of the answer is maybe. It doesn’t matter which answer the other one is (we can always just invert the question).
So we’ve run an experiment and got an answer. What does that tell us about the true state of reality?
Well whatever reality is we must have some notion of “an approximate region” – all of our observation of reality is imprecise, so there must be some notion of precision to make sense of that.
What does the result of a physical experiment tell us about the state of reality?
Well if the answer is “maybe” it doesn’t tell us anything. Literally any point in reality could be mapped to “maybe”.
But if the answer is yes then this should tell us only imprecisely where we are in reality. i.e. the set of points that map to yes must be an open set.
So an experiment is a function from reality to {yes, maybe}. The set of points mapping to yes must be an open set.
And what this means is that experiments are continuous functions to the set {yes, maybe} endowed with the Sierpiński topology. The set {yes} is open, and the whole set and the empty set are open, but nothing else is.
Now let’s postulate that if two states of reality give exactly the same answer on every single experiment, they’re the same state of reality. This is true in the same sense that existing is the thing that reality does – a difference that makes no difference might as well be treated as if it is no difference.
So what we have is the following:
1. Any state of reality is a point in the cube $$S^E$$ where $$E$$ is the set of available experiments and $$S = \{\mathrm{yes}, \mathrm{maybe}\}$$.
2. All of the coordinate functions are continuous functions when $$S$$ is endowed with the Sierpinski topology.
This is almost enough to show that reality can be modelled as a subset of the Sierpinski cube, not quite: There are many topologies compatible with this – reality could have the discrete topology.
But we are finite beings. And what that means is that any given point in time we can have observed the outcome of at most finitely many experiments.
Each of these experiments determine where we are only in the open set of some coordinate in our cube, thus the set that the experiments have determined us to be in is an intersection of finitely many open sets in the product topology on that cube, and thus is open in that topology.
Therefore the set of states of reality that we know we are in is always an open set in the product topology. So this is the “natural” topology on reality.
So reality is a subset of a Sierpiński cube. We now have to answer two questions to get the rest of the way:
• How many dimensions does the cube have?
• What sort of subset is it?
The first one is easy: The set of experiments we can perform is definitely infinite (we can repeat a single experiment arbitrarily many times). It’s also definitely countable, because any experiment we can perform is one we can describe (and two experiments are distinct only up to our ability to describe that distinction), and there are only countably many sentences.
So reality is a subset of the countably infinite dimensional Sierpiński cube.
What sort of subset?
Well that’s harder, and my arguments for it are less convincing.
It’s probably not usually the whole set. It’s unlikely that reality contains a state that is just constantly maybe.
It might as well be a closed set, because if it’s not we can’t tell – there is no physical experiment we can perform that will determine that a point in the closure of reality is not in reality, and it would be aesthetically and philosophically displeasing to have aphysical states of reality that are approximated arbitrarily well.
In most cases it’s usually going to be a connected set. Why? Well, because you’re “in” some state of reality, and you might as well restrict yourself to the path component of that state – if you can’t continuously deform from where you are to another state, that state probably isn’t interesting to you even if it in some sense exists.
Is it an uncountable subset of the Sierpinski cube? I don’t know, maybe. Depends on what you’re modelling.
Anyway, so there you have it. Reality is a closed, connected, subset of the countably infinite dimensional Sierpiński cube.
What are the philosophical implications?
Well, one obvious philosophical implication is that reality is compact, path connected, and second countable, but may not be Hausdorff.
(You should imagine a very very straight face as I delivered that line)
More seriously, the big implication for me is on how we model physical systems. We don’t have to model physical systems as the Sierpiński cube. Indeed we usually won’t want to – it’s not very friendly to work with – but whatever model we choose for our physical systems should have a continuous function (or, really, a family of continuous functions to take into account the fact that we fudged the non-determinism of our experiments) from it to the relevant Sierpiński cube for the physical system under question.
Another thing worth noting is that the argument is more interesting than the conclusion, and in particular the specific embedding is more important that the embedding exists. In fact every second countable T0 topological space embeds in the Sierpinski cube, so the conclusion boils down to the fact that reality is a T0, second countable, compact, and connected (path connected really) topological space (which are more or less the assumptions we used!).
But I think the specific choice of embedding matters more than that, and the fact that we the coordinates correspond to natural experiments we can run.
And, importantly, any decision we make based on that model needs to factor through that function. Decisions are based on a finite set of experiments, and anything that requires us to be able to know our model to more precision than the topology of reality allows us to is aphysical, and should be avoided.
# A statement of philosophical principles
Epistemic status: This is my philosophy. There are many like it, but this one is mine. It is not anything especially unusual or particularly sophisticated.
I’ve noticed in a couple of discussions recently that my philosophical positions are both somewhat opposed to what might be considered normal (at least among non-philosophers) and may seem internally contradictory.
It also occurred to me that I’ve never really properly explained this to people, and that it might be worth writing up a short position statement.
So, here is the philosophical basis on which I live my life. You can think of it as a rather extreme combination of moral relativism and mathematical formalism.
1. Words, and the concepts behind them, are made up and have no objective meaning or basis.
2. Statements cannot thus really be “true” or “false” (though they may be deductively true in the context of a certain set of premises and logic).
3. Reality exists only because if it doesn’t then exist is not a useful word, so existing is defined mostly by reference as the thing that reality does.
4. Perception of reality is intrinsically flawed and what we perceive may be arbitrarily far from what “really” exists (if we accept our senses, then empiricism shows us that it’s quite far. If we don’t accept our senses, we’re already there).
5. We are probably not ever going to be capable of accurately modelling or predicting the universe. Even if it’s in principle possible we’re probably not smart enough.
6. All value systems are subjective and culturally determined.
7. Morality is some complex mix of value system and prediction, so it’s certainly subjective and culturally determined, but also probably beyond our ability to actually formalise in any useful way even once we’ve already pinned down a value system.
…but if we take any of that too seriously we’ll never get anything useful done, and even if there’s no objective value to getting useful things done, subjectively I’m quite fond of it, so…
1. We should use words in a way that achieves a broad consensus and seems to be useful for talking about what we observe.
2. Accept a reasonably large body of knowledge as a useful working premise, but occasionally backtrack and note that you’re explicitly doing that when it leads you astray or causes disagreements.
3. Treat reality as if it exists in a naive objective sense, because it hurts when you don’t.
4. Don’t worry about it too much. If there’s an all-powerful demon faking your perception of reality, there’s probably not much you can do about it. Also see previous item – that reality which you perceive exists (even if any given perception may not be valid), because otherwise exists isn’t a very useful word.
5. We can do a surprisingly good job at our current level, and just because we can never achieve perfection doesn’t mean we shouldn’t try to improve what we’ve got.
6. But I like mine, and it includes a term for a certain amount of forcing itself on other people (“hurting people is bad and I don’t really care if you think you have a culturally valid reason for doing it”).
7. Doing the right thing is hard. Do the best you can. Don’t sweat the grand theory of morality too much, but pay attention when it comes up.
So as a result I temper the extreme relativist stance with a solid dose of pragmatic instrumental reasoning and pretend that I believe in philosophical naive realism because it’s much better at getting the job done than refusing to even acknowledge that such a thing as a job might exist and that it could be done if it did.
A lot of these theses are for me much like the way in which I am an atheist: I consider them to be obviously correct as a sort of “ground state” truth. It’s not that they’re necessarily right, it’s just that in the absence of evidence to the contrary they seem like a good default position, and nobody has provided evidence that I find convincing (and in some cases I’m not sure such evidence could exist even in principle). Maybe there’s a platonic realm of ideals after all, but formalism works perfectly well without it and if such a thing existed how could we possible know?
I probably got very excited and/or angsty about all this at one point as a teenager, but eventually I realised that maybe I just don’t care that much. Does it matter if the table I stubbed my toe on really exists? Does it matter if there is actually such a thing as a table? Either way it still hurts, and if I want something to eat my meals on I’m going to struggle to buy one from ikea without acknowledging the concept of a table. For most things I actually care about, life is just easier if I go along with naive realism.
But it’s important to me to understand that I’m just pretending. Particularly because it makes it much easier to acknowledge when I’m wrong (which I’m not always good at, but that’s not surprising. Just because I have a philosophy doesn’t mean I’m good at following it), and to understand where other people are coming from – politics is much easier to understand if you understand that value systems are subjective and arbitrary. No reason that I have to accept those values, mind you (my culturally determined subjective values frequently strongly prefer that I don’t), but it’s helpful to know where they could be coming from.
And in general I find there’s a certain useful humility to affirming that I have no access to any objective source of truth nor ever will, and that that’s OK.
This entry was posted in Performing philosophy without a license on by .
# Truth and/or Justice
Disclaimer: This post is obscure in places. I apologise for that. Reasons.
Everyone likes to think they’re the protagonist of their own story. And as a hero, we need a cause. On the left and those morally aligned with the left, that can often roughly be summed up as “Truth and Justice” (we generally leave The American Way to people who wear their underpants on the outside).
(Some people are more honest, or less moral, and instead are fighting for survival, or for self-interest. This post is not about those people. Some people are less morally aligned with the left and fight for things like purity and respect for authority too. As a devout moral relativist, I acknowledge the validity of this position, but I still struggle to take it seriously. If this is you, you may get less out of this post but the broad strokes should still apply).
Unfortunately, you can’t optimise for more than one thing. Truth and Justice is not a thing you can fight for. You can fight for truth, you can fight for justice, you can fight for a weighted sum of truth and justice, but you cannot reliably improve both at once.
Often we can ignore this problem. Truth and Justice frequently work very well together. But at some point (hopefully metaphorically) the murderer is going to turn up at your door and you’re going to have to decide whether or not to lie to them. Unless you’re prepared to go full Kant and argue that lying to protect another is unjust as well as untrue, you’ll have to make a trade off.
So what’s it going to be? Truth or Justice?
It’s not an absolute choice of course – depending on the circumstances and the results of a local cost/benefit analysis, almost all of us will sometimes choose truth, sometimes justice. and sometimes we’ll make a half-arsed compromise between the two which leaves both truth and justice grumbling but mostly unbloodied.
But I think most of us have a strong bias one way or the other. This may not be inherent – it’s probably in large part driven by factionalisation of the discourse space – but certainly among my main intellectual and political influences there’s at least one group who heavily prioritises truth and another who heavily prioritises justice.
That’s not to say we don’t care about the other. Caring about justice doesn’t make you a liar, caring about truth doesn’t make you heartless. It’s just that we care about both, but we care about one more.
Personally, I tend to flip flop on it. I find myself naturally aligned with truth (I hate lying, both being lied to and lying myself), but I think I’ve historically chosen to align myself with people who prefer justice, usually by denying that the trade-off exists.
But recently I’ve been feeling the pendulum swing the other way a bit. If you’ve wondered why I’ve gone quiet on a bunch of subjects, that’s part of what’s been going on.
A long time ago now I wrote “You Are Not Your Labels“, about the problem of fuzzy boundaries and how we tend to pick a particular region in the space of possibility that includes us, use a label for that region, and then defend the boundaries of that label zealously.
I still broadly stand by this. You are not your labels. I’m certainly not my labels.
But we might be.
One of the places where truth and justice play off against each other is when you’re being attacked. If you’re under fire, now is not really the time to go “Well the reality is super complicated and I don’t really understand it but I’m pretty sure that what you’re saying is not true”. Instead, we pick an approximation we can live with for now and double down on it with a high degree of confidence.
There probably isn’t “really” such a thing as a bisexual (I’m not even wholly convinced there’s such a thing as a monosexual) – there’s a continuous multi-dimensional space in which everyone lies, and we find it operationally useful to have words that describe where we are relative to some of the boundary points in that space that almost nobody experiences perfectly.
There are as many distinct experiences of being bisexual as there are bisexuals (though, as I keep finding out, being extremely confused and annoyed by this fact seems to be a pretty common experience for us), but it sure is difficult to have an “It’s Complicated” visibility day, and it seems surprisingly easy for people to forget we exist without regular reminders.
The approximation isn’t just useful for communicating infinite complexity in a finite amount of time, it’s useful because we build solidarity around those approximations.
(This is literally why I use the label bisexual incidentally. I’m much happier with just saying “It’s complicated and unlikely to impact your life either way and when it does I would be happy to give you a detailed explanation of my preferences” but that is less useful to both me and everyone else, so I no longer do)
Another truth/justice trade off in the LGBT space is “Born this way”. I am at this point confident of precisely two things about gender and sexuality:
• They are probably the byproduct of some extremely complicated set of nature/nurture interactions like literally everything else in the human experience.
• Anyone who currently expresses confidence that they know how those play out in practice might be right for the n=1 sample of themself (I am generally very skeptical of people’s claims that they understand what features are natural things they were born with and what are part of their upbringing. I present the entire feminist literature on privilege as evidence in defence of my skepticism, but I also don’t find it useful or polite to have arguments with people about their personal lived experiences), but are almost certainly wrong, or at least unsupported in their claim that this holds in generality.
I would be very surprised to learn that nobody was born this way, and I have an n=1 personal data point that there are bisexuals who would probably have been perfectly able to go through life considering themselves to be straight if they hadn’t noticed that other options were available. I think it likely that there’s a spectrum of variability in between, I just don’t know.
I think among ourselves most LGBT people are more than happy to admit that this stuff is complicated and we don’t understand it, but when confronted with people who just want us to be straight and cis and consider us deviants if we dare to differ on this point, born this way is very useful – it makes homophobia less a demand to conform to societal expectations (which would still be wrong, but is harder to convince people of) and more a call for genocide. The only way to stop LGBT being LGBT is to stop us existing, and that’s not what you mean, right?
(Historically there have been many cases where that was exactly what they meant, but these days it’s harder to get away with saying so even if you think it).
Even before the latest round of fake news we’ve had in the last couple of years, demanding perfect truth in politics seems like a great way to ensure that political change belongs to those less scrupulous than you. At the absolute minimum we need this sort of lies-to-normies to take complex issues and make them politically useful if we want the world to get better.
So: Truth or Justice?
To be honest, I still don’t know. My heart says truth, but my head says justice, which I’m almost certain is hilariously backwards and not how it’s supposed to work at all, but there you go. This is complicated, and maybe “Truth or Justice” is another of those labelling things that don’t really work for me. Hoisted by my own petard.
My suspicion though is that the world is a better place if not everyone is picking the exact same trade off – different people are differently placed for improving each, and it’s not all that useful to insist that someone inclined towards one should be strongly prioritising the other. It is useful to have both people for whom justice is their top priority, and people for whom truth is their top priority, and a world where we acknowledge only one point on the spectrum as valid is probably one that ends up with less truth and less justice than one where a wider variety is pursued. Monocultures just generally work less well in the long run, even by the standards of the monoculture.
Given that, it seems like a shame that right now most of the justice prioritising people seem to think the truth prioritising people are literally Hitler and vice versa.
(To be fair, the truth prioritising people probably think the justice prioritising people are figuratively Hitler).
Calls for “Why can’t we get along?” never go well, so I won’t make one here even though you could obviously ready into this article that that’s what I want even if I didn’t include this sentence as disclaimer, so instead I’ll end with a different call to action.
I wish we would all be better about acknowledging that this trade-off exists, and notice when we are making it, regardless of what we end up deciding about the people who have chosen a different trade-off.
If you’re justice-prioritising you might not feel able to do that in public because it would detract from your goals. That’s fine. Do it in private – with a couple close friends in the same sphere to start with. I’ve found people are generally much more receptive to it than you might think.
If you’re truth-prioritising, you have no excuse. Start talking about this in public more (some of you already are, I know). If what can be destroyed by the truth should be, there is no cost to acknowledging that the truth is sometimes harmful to others and that this is a trade-off you’re deliberately making.
Regardless of what we think the optimal trade-off between truth and justice is, I’m pretty sure a world that is better on both axes than the current one is possible. I’m significantly less sure that we’re on anything resembling a path to it, and I don’t know how to fix that, but I’d like to at least make sure we’re framing the problem correctly.
# An epistemic vicious circle
Let’s start with apology: This blog post will not contain any concrete examples of what I want to talk about. Please don’t ask me to give examples. I will also moderate out any concrete examples in the comments. Sorry.
Hopefully the reasons for this will become clear and you can fill in the blanks with examples from your own experience.
There’s a pattern I’ve been noticing for a while, but it happens that three separate examples of it came up recently (only one of which involved me directly).
Suppose there are two groups. Let’s call them the Eagles and the Rattlers. Suppose further that the two groups are roughly evenly split.
Now suppose there is some action, or fact, on which people disagree. Let’s call them blue and orange.
One thing is clear: If you are a Rattler, you prefer orange.
If you are an Eagle however, opinions are somewhat divided. Maybe due to differing values, or different experiences, or differing levels of having thought about the problem. It doesn’t matter. All that matters is that there is a split of opinions, and it doesn’t skew too heavily orange. Let’s say it’s 50/50 to start off with.
Now, suppose you encounter someone you don’t know and they are advocating for orange. What do you assume?
Well, it’s pretty likely that they’re a Rattler, right? 100% of Rattlers like orange, and 50% of Eagles do, so there’s a two thirds chance that a randomly picked orange advocate will be Rattler. Bayes’ theorem in action, but most people are capable of doing this one intuitively.
And thus if you happen to be an Eagle who likes orange, you have to put in extra effort every time the subject comes up to demonstrate that. It’ll usually work – the evidence against you isn’t that strong – but sometimes you’ll run into someone who feels really strongly about the blue/orange divide and be unable to convince them that you want orange for purely virtuous reasons. Even when it’s not that bad it adds extra friction to the interaction.
And that means that if you don’t care that much about the blue/orange split you’ll just… stop talking about it. It’s not worth the extra effort, so when the subject comes up you’ll just smile and nod or change it.
Which, of course, brings down the percentage of Eagles you hear advocating for orange.
So now if you encounter an orange advocate they’re more likely to be Rattler. Say 70% chance.
Which in turn raises the amount of effort required to demonstrate that you, the virtuous orange advocate, are not in fact Rattler. Which raises the threshold of how much you have to care about the issue, which reduces the fraction of Eagles who talk in favour of orange, which raises the chance that an orange advocate is actually Rattler, etc.
The result is that when the other side is united on an issue and your side is divided, you effectively mostly cede an option to the other side: Eventually the evidence that someone advocating for that option is a Rattler is so overwhelming that only weird niche people who have some particularly strong reason for advocating for orange despite being an Eagle will continue to argue the cause.
And they’re weird and niche, so we don’t mind ostracising them and considering them honorary Rattlers (the real Rattlers hate them too of course, because they still look like Eagles by some other criteria).
As you can probably infer from the fact that I’m writing this post, I think this scenario is bad.
It’s bad for a number of reasons, but one very simple reason dominates for me: Sometimes Rattlers are right (usually, but not always, for the wrong reasons).
I think this most often happens when the groups are divided on some value where Eagles care strongly about it, but Rattlers don’t care about that value either way, and vice versa. Thus the disagreement between Rattler and Eagles is of a fundamentally different character: Blue is obviously detrimental to the Rattlers’ values, so they’re in favour of orange. Meanwhile the Eagles have a legitimate disagreement not over whether those values are good, but over the empirical claim of whether blue or orange will be better according to those values.
Reality is complicated, and complex systems behave in non-obvious ways. Often the obviously virtuous action has unfortunate perverse side effects that you didn’t anticipate. If you have ceded the ground to your opponent before you discover those side effects, you have now bound your hands and are unable to take what now turns out to be the correct path because only a Rattler would suggest that.
I do not have a good suggestion for how to solve this problem, except maybe to spend less time having conversations about controversial subjects with people whose virtue you are unsure of and to treat those you do have with more charity. A secondary implication of this suggestion is to spend less time on Twitter.
But I do think a good start is to be aware of this problem, notice when it is starting to happen, and explicitly call it out and point out that this is an issue that Eagles can disagree on. It won’t solve the problem of ground already ceded, but it will at least help to stop things getting worse.
Like my writing? Why not support me on Patreon! Especially if you want more posts like this one, because I mostly don’t write this sort of thing any more if I can possibly help it, but might start doing so again given a nudge from patrons.
# Determinism is topologically impossible
I’ve been doing some work on topological models of decision making recently (I do weird things for fun) and a result popped out of it that I was very surprised by, despite it being essentially just a restatement of some elementary definitions in topology.
The result is this: Under many common models of reality, there are no non-trivial deterministic experiments we can perform even if the underlying reality is deterministic.
The conclusion follows from two very simple assumptions (either of which may be wrong but both of which are very physically plausible).
1. We are interested in some model of reality as a connected topological space $$X$$ (e.g. $$\mathbb{R}^n$$, some Hilbert space of operators, whatever).
2. No experimental outcome can give us infinite precision about that model. i.e. any experimental outcome only tells us where we are up to membership of some open subset of $$X$$.
Under this model, regardless of the underlying physical laws, any fully deterministic experiment tells us nothing about the underlying reality.
What does this mean?
Let $$X$$ be our model of reality and let $$\mathcal{O}$$ be some set of experimental outcomes. A deterministic experiment is some function $$f: X \to \mathcal{O}$$.
By our finite precision assumption each of the sets $$U_o = \{x \in X: f(x) = o\}$$ are open. But if $$f(x) = o$$ and $$o \neq o’$$ then $$f(x) \neq o’$$ so $$x \not\in U_{o’}$$. Therefore they’re disjoint.
But certainly $$x \in U_{f(x)}$$, so they also cover $$X$$.
But we assumed that $$X$$ is connected. So we can’t cover it by disjoint non-empty open sets. Therefore at most one of these sets is non-empty, and thus $$X = U_o$$ for some $$o$$. i.e. $$f$$ constantly takes the value $$o$$ and as a result tells us nothing about where we are in $$X$$.
Obviously this is a slightly toy model, and the conclusion is practically baked into the premise, so it might not map to reality that closely.
But how could it fail to do so?
One way it can’t fail to do so is that the underlying reality might “really” be disconnected. That doesn’t matter, because it’s not a result about the underlying reality, it’s a result about models of reality, and most of our models of reality are connected regardless of whether the underlying reality is. But certainly if our model is somehow disconnected (e.g. we live in some simulation by a cellular automaton) then this result doesn’t apply.
It could also fail because we have access to experiments that grant us infinite precision. That would be weird, and certainly doesn’t correspond to any sort of experiment I know about – mostly the thing we measure reality with is other reality, which tends to put a bound on how precise we can be.
It could also fail to be interesting in some cases. For example if our purpose is to measure a mathematical constant that we’re not sure how to calculate then we want the result of our experiment to be a constant function (but note that this is only for mathematical constants. Physical constants that vary depending on where in the space of our model we are don’t get this get out clause).
There are also classes of experiments that don’t fall into this construction: For example, it might be that $$O$$ itself has some topology on it, our experiments are actually continuous functions into O, and that we can’t actually observe which point we get in $$O$$, only its value up to some open set. Indeed, the experiments we’ve already considered are the special case where $$O$$ is discrete. The problem with this is that then $$f(X)$$ is a connected subset of $$O$$, so we’ve just recursed to the problem of determining where we are in $$O$$!
You can also have experiments that are deterministic whenever they work but tell you nothing when they fail. So for example you could have an experiment that returns $$1$$ or $$0$$, and whenever it returns $$1$$ you know you’re in some open set $$U$$, but when it returns $$0$$ you might or might not be in $$U$$, you have no idea. This corresponds to the above case of $$O$$ having a topology, where we let $$O$$ be the Sierpinski space. This works by giving up on the idea that $$0$$ and $$1$$ are “distinguishable” elements of the output space – under this topology, the set $$\{0\}$$ is not open, and so the set $$U_0$$ need not be, and the connectivity argument falls apart.
And finally, and most interestingly, our experiment might just not be defined everywhere.
Consider a two parameter model of reality. e.g. our parameters are the mass of a neutron and the mass of a proton (I know these vary because binding energy or something, but lets pretend they don’t for simplicity of example). So our model space is $$(0, \infty)^2$$ – a model which is certainly connected, and it’s extremely plausible that we cannot determine each value to more than finite precision. Call these parameters $$u$$ and $$v$$.
We want an experiment to determine whether protons are more massive than neutrons.
This is “easy”. We perform the following sequence of experiments: We measure each of $$u$$ and $$v$$ to within a value of $$\frac{1}{n}$$. If $$|u – v| > \frac{2}{n}$$ then we know their masses precisely enough to answer the question and can stop and return the answer. If not, we increase $$n$$ and try again.
Or, more abstractly, we know that the sets $$u > v$$ and $$v < u$$ are open subsets of our model, so we just return whichever one we’re in.
These work fine, except for the pesky case where $$u = v$$ – protons and neutrons are equally massive. In that case our first series of experiments never terminates and our second one has no answer to return.
So we have deterministic experiments (assuming we can actually deterministically measure things to that precision, which is probably false but I’m prepared to pretend we can for the sake of the example) that give us the answer we want, but it only works in a subset of our model: The quarter plane with the diagonal removed, which is no longer a connected set!
Fundamentally, this is a result about boundaries in our models of reality – any attempt to create a deterministic experiment will run into a set like the above plane: Suppose we had a deterministic experiment which was defined only on some subset of $$X$$. Then we could find some $$o$$ with $$U_o$$ a non-empty proper subset of $$X$$. Then the set $$\overline{U} \cap U^c$$ where the closure of $$U_o$$ meets its complement (which is non-empty because $$X$$ is connected) is a boundary like the diagonal above – on one side of it we know that $$f$$ returns $$o$$. On the other side we know that it doesn’t return $$o$$, but in the middle at the boundary it is impossible for us to tell.
What are the implications?
Well, practically, not much. Nobody believed that any of the experiments we’re currently performing are fully deterministic anyway.
But philosophically this is interesting to me for a couple of reasons:
1. I for one was very surprised that such a trivial topological result had such a non-trivial physical interpretation.
2. The idea that non-determinism was some intrinsic property of measurement and not a consequence of underlying physical non-determinism is not one that had ever previously occurred to me.
3. We need to be very careful about boundaries in our models of reality, because we often can’t really tell if we’re on them or not.
4. It may in fact be desirable to assume that all interesting quantities are never equal unless we have a deep theoretical reason to believe them to be equal, which largely lets us avoid this problem except when our theory is wrong.
(As per usual, if you like this sort of thing, vote with your wallet and support my writing on Patreon! I mean, you’ll get weird maths posts either way, but you’ll get more weird maths posts, and access to upcoming drafts, if you do).
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2017-10-22 02:42:26
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https://www.compadre.org/stpbook/statistical-mechanics-1/ex6_1.cfm
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## Lennard-Jones MD
Program LennardJonesMD models the behavior of particles in a two-dimensional box interacting through a Lennard-Jones potential. The motion evolves using the numerical solution of the differential equations in Newton's second law. The velocity Verlet algorithm is used to update the positions and velocities of the particles. Periodic boundary conditions are used so that particles that leave the box enter on the other side, much like many video games.
One goal of this simulation is to explore how the same inter-particle potential can lead to different macroscopic behavior such as gas, liquid, or solid phases. Other information that can be obtained includes the velocity distribtion function, the pair distribution function $g(r)$, which provides information about the spatial structure of the substance, and the diffusion of a tagged particle.
Problem: Simulations of the Maxwell velocity distribution
Program LennardJonesMD simulates a system of particles interacting via the Lennard-Jones potential in two dimensions by solving Newton's equations of motion numerically. The program computes the distribution of velocities in the $x$-direction among other quantities. Compare the form of the computed velocity distribution to the form of the Maxwell velocity distribution $$\label{eq:noninteract/maxwellvelx} f(v_x)\,dv_x = \left({m \over 2 \pi kT}\right)^{1/2} e^{-mv_x^2/2kT} dv_x.$$ How does its width depend on the temperature?
## Resources
Problems 1.9, 6.14, 8.16, and 10.2 in Statistical and Thermal Physics: With Computer Applications, 2nd ed., Harvey Gould and Jan Tobochnik, Princeton University Press (2021).
OSP Projects:
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2021-01-27 11:30:56
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https://www.wptricks.com/question/add_menu_page-with-different-name-for-first-submenu-item/
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Question
The add_menu_page documentation says to pass the menu title as the second parameter:
add_menu_page('Page Title', 'Menu Title', ...);
When adding more pages later via add_submenu_page, the main page becomes the first entry in the submenu:
However, I want the first item in the list to have a different name (but still point to the same page), the way WordPress itself does it:
How could I accomplish that in my plugin?
0
11 years 2012-09-27T18:13:01-05:00 0 Answers 92 views 0
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2023-03-28 11:12:50
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http://mathhelpforum.com/trigonometry/155548-rotation-square-plane.html
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# Thread: Rotation of a square on a plane
1. ## Rotation of a square on a plane
Hi All,
I hope you can help (and I hope this is the right forum for this subject)
I have a resizeable plane (say 300 x 300 in size)
On this plane I have a square (say 290 x 290 in size)
When i rotate this square on this plane by say 25 degrees around the squares center parts of the square are no longer visible on the existing plane.
My goal is to have the plane resize based on the new square but all I have available is the original dimensions of the plane, the original dimensions of the square and the angle it's being rotated by in my problem i do have access to various trigonometric functions such as cos and sin, I also know the center of the original square.
basically i'm looking for an equation which would take the angle and dimensions of the original square and plane and produce the new correct dimensions for the plane so that it engulfs the new square
Please anyone that could help I am desperate for help
kind regards
James
2. Hello, hellzone!
I have a resizeable plane (say, 300 x 300 in size).
On this plane I have a square (say, 290 x 290 in size).
When i rotate this square on this plane by. say $25^o$ about the square's center.
parts of the square are no longer visible on the existing plane.
My goal is to have the plane resize based on the new square
but all I have available is the original dimensions of the plane,
the original dimensions of the square and the angle it's being rotated by.
Basically i'm looking for an equation which would take the angle and dimensions
of the original square and plane and produce the new correct dimensions
for the plane so that it engulfs the new square
Assuming that the square must be 290 units on a side,
. . the worst case scenario occurs when it is rotated $45^o.$
Code:
*
* *
290 * * 290
* *
* *
* - - - - - - - - - *
* x *
* *
* *
* *
*
In the right triangle, we have: . $x^2 \:=\:290^2 + 290^2$
Then: . $x^2 \:=\:2\cdot290^2 \quad\Rightarrow\quad x \:=\:290\sqrt{2} \:=\:410.1219331$
Rounding up, we have 411.
Therefore, the plane must be at least $411 \times 411.$
In general, for a square of side $\,a$,
. . the plane must be at least $\left\lceil a\sqrt{2}\,\right\rceil \times \left\lceil a\sqrt{2}\,\right\rceil$
. . where $\lceil x\rceil$ is the "round up" function.
3. Let's say the small square is of size D_sq x D_sq. Your goal is to find the maximum extent of any of the corners as measured in the (x,y) plane as you rotate the square through an angle Θ. Actually due to symmetry you only need to consider one corner, such as the upper right,, and find the max distance of this point in either the x or y direction from the origin. Then double this ditance to find the length of the side of the plane that the square must sit on.
Consider the dimension from the center of the square to the corner - that would be D_sq/sqrt(2). The (x,y) coordinate of the upper right corner of the square when it is rotated by Θ from its original aligned position is, based on the center of the square being at (0,0) is :
X = D_sq/sqrt(2) cos(Θ+π/4), Y = D/sqrt(2)sin(Θ+π/4)
To find the size of the square you need to compare the absolute values of these coordinates to find the distance left or right, up or down from the origin. These max dimensions can be rearranged using trig identities to yield
X_max = D/2 x max(|cosΘ-sinΘ|, |cosΘ+sinΘ|)
Y_max = D/2 x max(|sinΘ+cosΘ|, |[sinΘ-cosΘ|)
Finally, note that the values for X_max and Y_max are the same - hence the dimension of your plane is two times either X_max or Y_max:
D_plane = D_sq x max(|cosΘ-sinΘ|, |cosΘ+sinΘ|)
This function has a minimum of value of D_sq (whenΘ = 0), and a maximum value of sqrt(2) times D_sq (when Θ = π/4)
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2016-09-30 19:01:59
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https://matplotlib.org/stable/api/_as_gen/matplotlib.axes.Axes.draw_artist.html
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# matplotlib.axes.Axes.draw_artist¶
Axes.draw_artist(a)
Efficiently redraw a single artist.
This method can only be used after an initial draw of the figure, because that creates and caches the renderer needed here.
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2021-10-26 21:45:28
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https://euro-math-soc.eu/review/foundations-nonlinear-generalized-functions-i-and-ii
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On the Foundations of Nonlinear Generalized Functions I and II
The main drawback of the Colombeau generalised functions is that the canonical embedding of the space of Schwartz distributions into the algebra (and sheaf) of generalised functions is not intrinsic. This canonical imbedding is not preserved by coordinate diffeomorphisms, so generalised functions cannot be defined on a manifold although the distributions can. How to remove this inconvenience by a slight change of definition of the Colombeau generalised functions was outlined by Colombeau and Meril in 1994 and elaborated by Jelínek in 1999.
In the first part, the Colombeau-type algebra defined by Jelínek (called diffeomorphism invariant) is carefully examined, and its description is corrected and completed. Colombeau-type algebra of generalised functions is defined as a quotient algebra G = EM / I, where EM is an algebra of so-called moderate representatives and I is an ideal of negligible representatives: some equivalent definitions of these notions are presented. In the second part, other ways of defining moderateness and negligibility are studied and only two diffeomorphism invariant algebras are found among them. The assumption that they are different is neither proved nor disproved.
Reviewer:
jjel
Book details
Published:
2001
ISBN:
0-8218-2729-4
Price:
£31.50
Categorisation
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2023-03-30 18:07:02
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https://web2.0calc.com/questions/plz-help_59
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We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
+0
# plz help.
+1
191
5
Given that $$x$$ is a positive integer less than 100, how many solutions does the congruence $$x + 13 \equiv 55 \pmod{34}$$ have?
Dec 18, 2018
### 5+0 Answers
#1
+22182
+8
Given that $$\large{x}$$ is a positive integer less than 100,
how many solutions does the congruence have?
$$\large{x + 13 \equiv 55 \pmod{34}}$$
$$\begin{array}{|rcll|} \hline x + 13 &\equiv& 55 \pmod{34} \\ x + 13 &\equiv& 55-34 \pmod{34} \\ x + 13 &\equiv& 21 \pmod{34} \quad & | \quad - 13 \\ x &\equiv& 21-13 \pmod{34} \\ x &\equiv& 8 \pmod{34} \\ \mathbf{x} & \mathbf{=} & \mathbf{8 + n \cdot 34}, ~ n \in \mathbb{N} \\ \hline \end{array}$$
$$\begin{array}{|c|l|c|} \hline n, ~ n \in \mathbb{N} & \mathbf{x = 8 + n \cdot 34} & ~ x>0,~ x<100 \\ \hline 0 & x = 8+0\cdot 34 \\ & x= 8 & \checkmark \\ \hline 1 & x = 8+1\cdot 34 \\ & x= 42 & \checkmark \\ \hline 2 & x = 8+2\cdot34 \\ & x = 76 & \checkmark \\ \hline 3 & x = 8+3\cdot 34 \\ & x = 110 & x> 100 \\ \hline \ldots & & x> 100 \\ \hline \end{array}$$
The congruence has three solutions: $$\mathbf{x = 8}$$ and $$\mathbf{x = 42}$$ and $$\mathbf{x = 76}$$
Dec 18, 2018
edited by heureka Dec 19, 2018
#2
+1
But, isn't 42 + 13 mod 34 = 21 and 76 +13 mod 34 = 21 ????
Dec 18, 2018
#4
+100813
0
Yes that is right, what is the problem ?
55( mod 34) also equals 21
Melody Dec 18, 2018
#3
+100813
+1
Did you miss 8 Heureka ?
Dec 18, 2018
edited by Melody Dec 18, 2018
edited by Melody Dec 18, 2018
#5
+22182
+7
Hello Melody,
of course there is x = 8.
heureka Dec 19, 2018
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2019-05-24 00:19:31
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https://bucomplex.wordpress.com/category/geometry/
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# Onward to the argument principle and Rouche’s Theorem
Consider the function $f(z) = z^3$ and look at what it does to values on the circle $z(t) = e^{it}, t \in [0, 2\pi)$. $f(e^{it}) = e^{i3t}$ and as we go around the circle once, we see that $arg(f(z)) = arg(e^{i3t})$ ranges from $0$ to $6 \pi$. Note also that $f$ has a zero of order 3 at the origin and $(2 \pi)3 = 6 \pi$ That is NOT a coincidence.
Now consider the function $g(z) = \frac{1}{z^3}$ and look at what it does to values on the circle $z(t) = e^{it}, t \in [0, 2\pi)$. $g(e^{it}) = e^{-i3t}$ and as we go around the circle once, we see that $arg(g(z)) = arg(e^{-i3t})$ ranges from $0$ to $-6 \pi$. And here, $g$ has a pole of order 3 at the origin. This too, is not a coincidence.
We can formalize this somewhat: in the first case, suppose we let $\gamma$ be the unit circle taken once around in the standard direction and let’s calculate:
$\int_{\gamma} \frac{f'(z)}{f(z)} dz = \int_{\gamma}\frac{3z^2}{z^3}dz = 3 \int_{\gamma} \frac{1}{z}dz = 6 \pi i$
In the second case: $\int_{\gamma} \frac{g'(z)}{g(z)} dz = \int_{\gamma}\frac{-3z^{-4}}{z^{-3}}dz = -3 \int_{\gamma} \frac{1}{z}dz = -6 \pi i$
Here is what is going on: you might have been tempted to think $\int_{\gamma} \frac{f'(z)}{f(z)} dz = Log(f(z))|_{\gamma} = (ln|f(z)| +iArg(f(z)) )|_{\gamma}$ and this almost works; remember that $Arg(z)$ switches values abruptly along a ray from the origin that follows the negative real axis..and any version of the argument function must do so from SOME ray from the origin. The real part of the integral (the $ln(|f(z)|$ part) cancels out when one goes around the curve. The argument part (the imaginary part) does not; in fact we pick up a value of $2 \pi i$ every time we cross that given ray from the origin and in the case of $f(z) = z^3$ we cross that ray 3 times.
This is the argument principle in action.
Now of course, this principle can work in the vicinity of any isolated singularity or zero or along a curve that avoids singularities and zeros but encloses a finite number of them. The mathematical machinery we develop will help us with this concept.
So, let’s suppose that $f$ has a zero of order $m$ at $z = z_0$. This means that $f(z) = (z-z_0)^m g(z)$ where $g(z_0) \neq 0$ and $g$ is analytic on some open disk about $z_0$.
Now calculate: $\frac{f'(z)}{f(z)} dz = \frac{m(z-z_0)^{m-1} g(z) - (z-z_0)^m g'(z)}{(z-z_0)^m g(z)} = \frac{m}{z-z_0} + \frac{g'(z)}{g(z)}$. Now note that the second term of the sum is an analytic function; hence the Laurent series for $\frac{f'(z)}{f(z)}$ has $\frac{m}{z-z_0}$ as its principal part; hence $Res(\frac{f'(z)}{f(z)}, z_0) = m$
Now suppose that $f$ has a pole of order $m$ at $z_0$. Then $f(z) =\frac{1}{h(z)}$ where $h(z)$ has a zero of order $m$. So as before write $f(z) = \frac{1}{(z-z_0)^m g(z)} = (z-z_0)^{-m}(g(z))^{-1}$ where $g$ is analytic and $g(z_0) \neq 0$. Now $f'(z) = -m(z-z_0)^{-m-1}g(z) -(g(z))^{-2}g'(z)(z-z_0)^{-m}$ and
$\frac{f'(z)}{f(z)} =\frac{-m}{z-z_0} - \frac{g'(z)}{g(z)}$ where the second term is an analytic function. So $Res(\frac{f'(z)}{f(z)}, z_0) = -m$
This leads to the following result: let $f$ be analytic on some open set containing a piecewise smooth simple closed curve $\gamma$ and analytic on the region bounded by the curve as well, except for a finite number of poles. Also suppose that $f$ has no zeros on the curve.
Then $\int_{\gamma} \frac{f'(z)}{f(z)} dz = 2 \pi i (\sum^k_{j =1} m_k - \sum^l_{j = 1}n_j )$ where $m_1, m_2...m_k$ are the orders of the $k$ zeros of $f$ inside of $\gamma$ and $n_1, n_2.., n_l$ are the orders of the poles of $f$ inside $\gamma$.
This follows directly from the theory of cuts:
Use of our result: let $f(z) = \frac{(z-i)^4(z+2i)^3}{z^2 (z+3i-4)}$ and let $\Gamma$ be a circle of radius 10 (large enough to enclose all poles and zeros of $f$. Then $\int_{\Gamma} \frac{f'(z)}{f(z)} dz = 2 \pi i (4 + 3 -2-1) = 8 \pi i$. Now if $\gamma$ is a circle $|z| = 3$ we see that $\gamma$ encloses the zeros at $i, -2i,$ and the pole at 0 but not the pole at $4-3i$ so $\int_{\Gamma} \frac{f'(z)}{f(z)} dz = 2 \pi i (4+3 -2) = 10 \pi i$
Now this NOT the main use of this result; the main use is to describe the argument principle and to get to Rouche’s Theorem which, in turn, can be used to deduce facts about the zeros of an analytic function.
Argument principle: our discussion about integrating $\frac{f'(z)}{f(z)}$ around a closed curve (assuming that the said curve runs through no zeros of $f$ and encloses a finite number of poles ) shows that, as we traverse the curve, the argument of the function changes by $2 \pi (\text{ no. of zeros - no. of poles})$ where the zeros and poles are counted with multiplicities.
Example: consider the function $f(z) = z^8 + z^2 + 1$. Let’s find how many zeros it has in the first quadrant.
If we consider the quarter circle of very large radius $R$ (that stays in the first quadrant and is large enough to enclose all first quadrant zeros) and note $f(Re^{it}) = R^8e^{i8t}(1+ \frac{1}{R^6}e^{-i6t} + \frac{1}{R^8 e^{i8t}})$ we see that the argument changes by about $8(\frac{\pi}{2} = 4 \pi$. The function has no roots along the positive real axis. Now setting $z = iy$ to run along the positive imaginary axis we get $f(iy) = y^8 -y^2 + 1$ which is positive for large $R$, has one relative minimum at $2^{\frac{-1}{3}}$ which yields a positive number, and is zero at $z = 0$. So the argument stays at $4 \pi$ so, we get $4 \pi = 2 \pi (\text{no. of roots in the first quadrant})$ which means that we have 2 roots in the first quadrant.
In fact, you can find an online calculator which estimates them here.
Now for Rouche’s Theorem
Here is Rouche’s Theorem: let $f, g$ be analytic on some piecewise smooth closed curve $C$ and on the region that $C$ encloses. Suppose that, on $C$ we have $|f(z) + g(z)| < |f(z)|$. Then $f, g$ have the same number of zeros inside $C$. Note: the inequality precludes $f$ from having a zero on $C$ and we can assume that $f, g$ have no common zeros, for if they do, we can “cancel them out” by, say, writing $f(z) = (z-z_0)^m f_1(z), g(z) = (z-z_0)^mg_1(z)$ at the common zeros. So now, on $C$ we have $|1 + \frac{g(z)}{f(z)}| < |1|$ which means that the values of the new function $\frac{g(z)}{f(z)}$ lie within the circle $|w+1| < 1$ in the domain space. This means that the argument of $\frac{g(z)}{f(z)}$ has to always lie between $\frac{\pi}{2}$ and $\frac{3 \pi }{2}$ This means that the argument cannot change by $2 \pi$ so, up to multiplicity, the number of zeros and poles of $\frac{g(z)}{f(z)}$ must be equal. But the poles come from the zeros of the denominator and the zeros come from the zeros of the numerator.
And note: once again, what happens on the boundary of a region (the region bounded by the closed curve) determines what happens INSIDE the curve.
Now let’s what we can do with this. Consider our $g(z) = z^8 + z^2 + 1$. Now $|z^8 -(z^8 + z^2 + 1)| =|z^2+1| < |z^8|$ for $R = \frac{3}{2}$ (and actually smaller). This means that $z^8$ and $z^8+z^2 + 1$ have the same number of roots inside the circle $|z| = \frac{3}{2}$: eight roots (counting multiplicity). Now note that $|z^8 +z^2 + 1 -1| = |z^8+z^2| < |1|$ for $|z| \leq \frac{2}{3}$ So $z^8 +z^2 + 1$ and $1$ have the same number of zeros inside the circle $|z| = \frac{2}{3}$ This means that all of the roots of $z^8+z^2 + 1$ lie in the annulus $\frac{2}{3} < z < \frac{3}{2}$
# A bit more on complex exponents
I could go on for multiple lessons about this topic, so I’ll limit myself to a few remarks about the values of $z^w$ where both $z$ and $w$ are complex numbers. A good way to study complex arithmetic is to justify each step by using the definitions presented in class.
Let $z = re^{it}$ be written in complex polar coordinates with $r > 0$ a real number and $t \in arg(z)$. Let $w = a + bi$
Now $z^w = (re^{i(t + 2k \pi)})^{a+bi} =$
Now switch to $exp(z) = e^z$ notation:
$= exp(aln(r) +(ln(r) b)i))exp(i(t+2k\pi)a -(t+2k\pi)b)$ where $k \in \{..-2, -1, 0, 1, 2,...\}$
Now this becomes:
$exp((a ln(r) -b(2k\pi +t)) + i(ln(r)b + a (t + 2k\pi)))$
1. if $b = 0$, the exponent is real and we are back to what we originally had. This becomes very interesting if $a$ is irrational; we then get an infinite number of values.
2. Note that we can get an infinite number of values for the modulus AND argument if both $b$ and $a$ are irrational.
Feel free to explore what you get for various values of $a$ and $b$ for a fixed $r$ and $t$.
Let’s do an example: $1^{\sqrt{2} + \sqrt{3}i }$ Here $r= 1$, $t = 0$, $a = \sqrt{2}, b = \sqrt{3}$ and $ln(r) = 0$. So we end up with:
$exp(-\sqrt{3}2k\pi) exp(ia 2k\pi)$ as $k \ \{..-2, -1, 0, 1, 2, 3...\}$ and so we end up with an infinite number of values such as:
$1, exp(-(\sqrt{3})2 \pi)exp(2 \pi \sqrt{2}), exp((\sqrt{3}) 2\pi) exp(-2 \pi \sqrt{2}), ..$
The set forms a sequence spiraling out from the origin and going infinitely far away. Note that the arguments are irrational multiples of $\pi$
# A Theorem about line segments
This is regarding problem 16 of section 1.3
There is a theorem called the Heine-Borel Theorem from mathematics that implies that, given a closed line segment and a collection of open disks that contain that said line segment, only a finite number of those open disks are required to cover that said line segment. This theorem is usually proved in real analysis classes and real analysis is not a prerequisite for this class.
We will need a tool for our proof and that is the least upper bound axiom for real numbers: if a non-empty set of real numbers has an upper bound, then it has a least upper bound. This is an axiom of mathematics and not subject to proof.
Now let $S$ be any line segment of finite length in the plane; we can assume that $S = \vec{a} + \vec{b}t, t \in [0,1]$ (think of what you did in calculus 3).
Now let $\cup_{j=1} D_j$ be any covering of $S$ by open disks. Now let $c$ be the least upper bound of all numbers between $0, 1$ where the subsegment $\vec{a} + \vec{b}t, t \in [0,c]$ can be covered by a finite number of the $D_j$ Since at least one of the $D_j$, say $D_1$, contains the point $\vec{a}$ such a $c$ exists. But then the point $\vec{a} + c \vec{b}$ lines in another $D_k$ so the segment $\vec{a} + \vec{b}t, t \in [0, c + \epsilon]$ lies in the union of a finite number of open disks (the number of the previous collection, plus one more). So $c$ must not heave been an upper bound at all…unless $c = 1$ and the segment terminated there.
So the entire segment can be covered by a finite collection of these open disks.
# About that “point at infinity”
The text says that a set $X$ has the point of infinity in its interior if $X$ contains some set $|z| > M$ (for some $M > 0$) in its interior. How does one think of this?
Imagine a continuous map $g:D \rightarrow C$ which is both one-to-one and onto where $D$ is the open unit disk ($\{z | |z| < 1 \}$)
where $g(re^{it}) = (\frac{t}{1-t})e^{it}$. That is, $g$ stretches each ray segment $[0,1)$ through the origin out “to infinity” while keeping the argument the same.
Now think of the “point of infinity” as being where ENTIRE UNIT CIRCLE $|z| = 1$ would go to…so in the preimage (domain) picture, the point at infinity is identified with the entire unit circle. The open sets containing it would be open rings bordering the circle “on the inside of the disk”.
# Studying the function f(z) = z^2
In calculus 1, studying the graph of $f(x) = x^2$ was relatively easy. The domain of the function is the real line and the range is the non-negative reals. So the graph lies in $R^1 \times R^1$ and therefore can be viewed on the plane.
In calculus 3, we sometimes showed graphs in 3-space; for example the upper hemisphere of $x^2 + y^2 + z^2 = 1$ serves as the graph of $f(x) = \sqrt{1-x^2-y^2}$. the domain is a subset of the plane and the range is a subspace of the real line, hence the graph “lives” in 3-dimensions.
But the situation is very different in complex variables; the domain of the function is a subset of the complex plane (possibly all of the complex plane) and the range is also a subset (possibly all) of the complex plane. So the graph “lives” in a space that has 4 “real” dimensions (“real” as in “real numbers”…4 dimensions as in the vector space $R^4$. Yes, I am being pedantic because there is such a thing as “complex dimension”.)
So, how might we view a complex valued function with a complex domain?
One way might be to borrow what we learned about vector fields in calculus 3.
For example: $f(z) = z^2$ might be viewed as $f(x+iy) = (x+iy)^2 = (x^2-y^2) + i2xy$ which can be graphed as the vector field $F(x,y) = (x^2-y^2, 2xy)$
Here is such an attempt:
You can see the strangeness in the left half plane (where $Im(z) < 0$) and there is a reason for that.
To see that is going on, write $z = re^{it}$ where $t = arg(z)$ and $r = |z|$ then $z^2 = r^2e^{i2t}$. So the argument gets doubled and the modulus gets squared. Hence the vectors "rotate" as you go around the the x-axis and the lengths get very short near zero and very long when they are further away from the origin.
Exercise: let $u(x,y) = x^2-y^2, v(x,y) = 2xy$. Now calculate the following partial derivatives: $u_x, v_y$ and compare. Calculate $-u_y, v_x$ and compare. Now calculate $u_{xx} + u_{yy}$. What do you notice?
Another way: look at the image of certain subsets of the complex plane. Write $z = re^{it}$ and note that $(re^{it})^2 = r^2e^{i2t}$. Now note what happens: if we have the circle $|z| = a$ where $a < 1$ then those points are taken to a circle of radius $a^2 < a$ and points of a circle $|z| = b > 1$ are taken to a circle of radius $b^2 > b$. So the map squeezes circles inside of the unit circle and expands circles outside of the unit circle (circles understood to be centered at 0 ).
Now look at rays emanating from the origin; these rays have constant argument. An argument $t$ gets sent to $2t$ so: under $f(z) = z^2$ a ray of, say, argument $\frac{\pi}{6}$ gets sent to a ray of $\frac{\pi}{3}$, $\frac{\pi}{3}$ gets sent to $\frac{2\pi}{3}$ and $\pi$ gets sent all the say to $2 \pi$. So you can see that the rays get rotated by different amounts as the argument is doubled.
You can also see that half of the complex plane gets “stretched” to the entire complex plane; here is a relevant pictures (I am showing the half plane $Re(z) \geq =0$ and now it gets stretched to the entire complex plane; half-circles and rays in the first picture get taken to the circles and rays of the same color in the second picture.
I’d like to stress that the “rays of constant arguments” get rotated by different amounts; this is NOT a “rigid rotation”.
A “rigid rotation.” Consider the function $f(z) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)z$. Now in polar form: $f(re^{it}) = e^{\frac{\pi}{4}} r e^{it} = re^{i(t + \frac{\pi}{4})}$ which IS just a rigid rotation of $\frac{\pi}{4}$. This map does not change the absolute value but does add $\frac{\pi}{4}$ to the argument.
Exercise: show that $f(z) = (w_0)z$ (where $w_0$ is a complex constant ) has the following effect: $|f(z)| = |w_0||z|$ and $arg(f(z)) = arg(z) + arg(w_0)$. This is just a “rotation and a stretch” or “rotation and a dilation”.
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2019-03-20 03:24:03
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https://mechatronicsblog.com/basic-esp8266-nodemcu-tutorial-breadboard-pinout-and-dimmable-led-with-pulse-width-modulation-pwm/
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#### Latest posts by Javier Bonilla (see all)
This is a basic tutorial about how to use a breadboard with a ESP8266 NodeMCU board to dim a LED applying Pulse-Width Modulation (PWM).
## Background
If you need further clarification about some concepts in this tutorial, have a look at the following posts. They explain concepts and terminology from scratch. The first one is an introduction to ESP8266 NodeMCU boards whereas the second one shows how to program with Arduino Integrated Development Environment (IDE).
## Outline
• Hardware
• ESP8266 NodeMCU pinout
• Analog and Digital Inputs / Outputs
• Wiring a LED
• Writing and Testing the Code
## Hardware
These are affiliate links. This means if you click on the link and purchase the promoted item, we will receive a small affiliate commission at no extra cost to you, the price of the product is the same. We would really appreciate your support to our work and website if this is fine for you.
A breadboard, also known as protoboard, is a solderless device for prototyping electronic circuits. We only have to insert the electronic components’ terminals in the holes and connecting them through wires. They are available in different sizes and shape factors, but they all work in the same way.
Have a look at the picture below. The black lines determine how the pins are connected together inside the breadboard. Pins in the horizontal lines are wired together. The red line (+ sign) is intended to be used as a power bus, whereas the blue line (- sign) is generally used as a ground bus. Vertical pins are also wired together up to the division. This division facilitates mounting integrated circuits and other components in the breadboard.
## ESP8266 NodeMCU pinout
Let’s consider a NodeMCU V2 board, the NodeMCU V3 for practical purposes is identical. In this tutorial, we are interested in power pins (3.3 V), ground pins (GND) and GPIO pins. Commonly, NodeMCU boards have only one analog pin (A0) and several digital pins D0 – D9 (GPIO XX).
If we want to interact with a digital pin we have to remember the GPIO number (0..16), whereas for the analog pin the alias is used (A0). Digital pins can be used as inputs or outputs, however the analog pin can only be used as an input. By default, pins are set as inputs, but it is always a good practice to specify how they will be used in our program.
pinMode(4,INPUT)
pinMode(5,OUTPUT)
pinMode(A0,INPUT)
## Analog and Digital Inputs / Outputs
A digital pin can be used as a digital input or output, where we can read or write HIGH (3.3V) or LOW (0V) values. As we will see, a digital pin can also be used as an analog output by means of Pulse-Width Modulation (PWM).
On the other hand, an analog pin can be used as an analog input, but not as an output. This is different than in Arduino boards, where analog pins can also be used as digital inputs and outputs. Also, we must take into account that the maximum allowed voltage for the analog pin is 1 V.
Next are shown the Arduino functions that we can use to read and write values. Remember that we have to properly set pins as inputs or outputs in advanced in order to read or write values.
• digitalRead(pin) – It returns LOW (0 V) or HIGH (3.3 V).
• digitalWrite(pin,value) – It writes LOW or HIGH.
• analogRead(pin) – It returns a value between 0 (0 V) and 1023 (1 V), because the resolution of the analog to digital converter is 10 bits, 2^10 = 1024.
• analogWrite(pin,value) – It writes a value between 0 (0 V) and 1023 (3.3 V), because the resolution is also 10 bits. The analogWrite resolution is 8 bits in most Arduino boards, 2^8 = 256.
Function Returns Value Pin digitalRead(pin) LOW / HIGH – D digitalWrite(pin,value) – LOW / HIGH D analogRead(pin) 0 – 1023 – A analogWrite(pin,value) – 0 – 1023 D
Now, the question is: how can we set an analog value in a digital pin? since a digital pin can only be set to digital values (LOW / HIGH). The answer is Pulse-Width Modulation (PWM).
Digital signals are only read in particular points in time. The time interval between two “readings” is called cycle. This cycle have a period (for instance, 2 ms). We can emulate analog signals by setting the digital pin HIGH only a percentage of the cycle period. This percentage can be set from 0% (0) to 100% (1023) in approximately 0.1% intervals (100/1024).
A LED with PWM turns on and off on each cycle, where the on period in the cycle is determined by a specified percentage, remember from 0% (0) to 100% (1023). The on / off transition is performed so fast that our eyes cannot see the LED blinking and our brain interprets this as a weighted average brightness between the HIGH and LOW values, were the weights are the percentages for each state. For a LED, the LOW state is no brightness at all, and the HIGH the maximum brightness. This maximum brightness is defined by the maximum current which is determined by the circuit resistance.
Have a look at the Arduino PWM tutorial for detailed information. The analogWrite function automatically perform PWM in digital pins.
## Wiring a LED
Light Emitting Diodes (LEDs) are polarized, so it matters in which direction you wire them up. The positive (+) and larger lead is the anode and the negative (-) and shorter lead is the cathode. The cathode can also be identified because the casing lip is flat in that side.
When working with LEDs, we commonly have to use a resistor in order to limit the current to a safe value, otherwise when can burn the LED out. The maximum permitted current is given by the LED manufacturer, if we don’t have this information available, we can assume a maximum 20 mA as a safe value.
LEDs are diodes and the current through them grows exponentially with voltage. Also, small changes in voltage can produce huge changes in current. Commonly, LEDs have a voltage drop from 1.8 V to 3.3 V. To be on the safe side, let’s consider 1.8 V.
Since NodeMCU boards provide 3.3 V, we need to increase the voltage drop, 3.3 V – 1.8 V = 1.5 V. Then, we can apply Ohm’s law to calculate the minimum resistance to increase the voltage drop and dissipate the extra power, R=V / I, R = 1.5 V / 20 mA = 90 Ω. The following formula summarizes the calculation.
$R_{LED} = \dfrac{V_{source} - V_{LED}}{I_{LED}} = \dfrac{3.3 \mathrm{V} - 1.8 \mathrm{V}}{20 \mathrm{mA}} = 90$
Since resistors are commonly available at standard values (100 Ω, 220 Ω, 330 Ω, etc), we should choose a resistance higher than 90 Ω to not damage the LED, let’s say 100 Ω. However, there are other factors that we should take into account, such as the tolerance and the temperature coefficient. Have a look at this great resistor color code tutorial for learning more about how to identify those parameters.
So, to be on the safe side, let’s consider a 220-Ω resistor. From a practical point of view you can use 220-Ω or 330-Ω resistors, or even higher values. The only difference is that increasing the resistance decreases the current and therefore the maximum LED brightness (up to a point where no light is emitted at all).
Resistors are not polarized, so we don’t have to worry about the connection direction. Also, it doesn’t matter if the resistor is connected to the to the LED anode or cathode, since the voltage drop is the same across the circuit.
This diagram shows how the LED is connected to the NodeMCU board in our example.
## Writing and Testing the Code
This program simply sets the LED brightness (analogWrite) from 0 (0%) to 1023 (100%) forward and backward using PWM. When a particular level is set, a delay is introduced (delay). We also turn off the built-in LED in the setup function. The LED_BUILTIN constant defines its GPIO pin number (16). Notice that the built-in LED is off at HIGH level.
const unsigned int LED_PIN = 5; // GPIO pin number
const unsigned int DELAY_TIME = 5; // Milliseconds
const unsigned int MIN_LEVEL = 0;
const unsigned int MAX_LEVEL = 1023;
void setup() {
Serial.begin(115200); // Data rate in bits per second
pinMode(LED_PIN,OUTPUT); // Set LED pin as output
digitalWrite(LED_BUILTIN,HIGH); // Built-in LED off
}
void loop() {
unsigned int i;
for(i=MIN_LEVEL;i<MAX_LEVEL;i++) // [0,1022]
analogWriteDelay(LED_PIN,i,DELAY_TIME);
for(i=MAX_LEVEL;i>MIN_LEVEL;i--) // [1023,1]
analogWriteDelay(LED_PIN,i,DELAY_TIME);
}
void analogWriteDelay(unsigned int pin,
unsigned int value,
unsigned int waitTime) {
analogWrite(pin,value); // Analog write in LED pin
delay(waitTime); // Delay in milliseconds
}
Finally, we can see our LED in action! If you want to keep testing, set LED_PIN to LED_BUILTIN and connect the LED anode to D0 (GPIO 16). You will see two LEDs in action (the built-in LED and your own LED) but in opposite states since the built-in LED is on at LOW state whereas external LEDs are on at HIGH states.
2000
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2020-03-29 21:58:43
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https://math.stackexchange.com/questions/3774437/prove-that-n-r-f-are-collinear
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# Prove that $N,R,F$ are collinear
In a triangle $$ABC$$, let I be the incentre. Let $$D$$, $$E$$, $$F$$ be the intersections of $$(ABC)$$. with the lines through $$I$$ perpendicular to $$BC$$, $$CA$$, $$AB$$, respectively.
Define $$O= BC \cap DE$$ and $$L= AC \cap DE$$ . Define $$IF\cap AB= R$$ . Let $$N=(BOF) \cap (LAF)$$ .Prove that $$N$$,$$R$$,$$F$$ are collinear.
My progress: Since $$F\in (ABC)$$ , I thought of using simson points . So I took points $$J$$,$$R$$,$$K$$ as the simson points in $$BC$$,$$BA$$,$$AC$$ wrt point $$F$$ respectively. ( as shown in the diagram )
Then since $$NBFO$$ and $$AFLN$$ is cyclic, we get that $$180- \angle ONF=\angle OBF=\angle CBF=180- \angle FAC=180 -\angle FAL = \angle FNL$$ .
Hence points $$O$$,$$N$$,$$L$$ are collinear .
Now, I am stuck . I tried using phantom points but couldn't proceed . I am thinking of using Radical axis but still confused.
Here are some more observations which might be trivial but still, we have $$BJFR$$, $$RFKA$$,$$CJFK$$ concyclic . We also have $$\Delta JFK \sim \Delta BFA$$
Ps: This is my own observation, so there is a very high chance that I might be wrong.
Below are a few diagrams for the problem.
• Yes, I just hided a few lines in geogebra . Lemme give a better diagram – Sunaina Pati Jul 30 '20 at 13:35
• It is my own problem . Actually, It is inspired by another unanswered problem in MSE . – Sunaina Pati Aug 3 '20 at 1:59
This result is false! An IMOTCer said that when we use the relation tool, the geogebra shows that $$N \not\in RF$$ .
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2021-01-25 05:26:56
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https://physics.stackexchange.com/questions/162034/why-is-the-frequency-bandwidth-of-the-environment-important-for-markovianity
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# Why is the frequency bandwidth of the environment important for Markovianity?
In the derivation of Spontaneous Emission in two level systems in Quantum Optics (be it Wigner Weisskopf or a different approach, such as density operators to find the master equation), one makes (several) assumptions. One of the more prominent ones here is the Markov approximation, which I think is most easily described in the density operator context. Here one assumes that to find the state of the atomic system $S$ at time $t$, one does not have to integrate $\rho_s(t')$ from $0$ to $t$, but instead can just take it to be $\rho_s(t)$. The way I see it, this means that the system does not have a memory of what has happened to it before time $t$. But what I do not completely understand is what is required for this to hold.
What I can find from other sources is that it requires 'a broad range of frequencies' to be present. Why is this the case? I read one hypothetical scenario in which one would construct a photonic crystal such that the density of modes would be 0 up to $\omega_{eg}$ (the transition of the two level system), and constant afterwards. The writer then claimed that the Markov approximation would not hold, because we have a real difference between lower and higher than the emitter frequency, meaning that the emitter has a real memory of what has happened to it. I personally do not understand this line of reasoning, but I assume it to be true. Could someone explain why this is so, and perhaps also elaborate on the conditions the Markov approximation requires?
Edit: Perhaps another example to illustrate what I don't understand: another source writes that if you were to put your emitter in a bandgap, so that the only radiation it could couple to would be in the range $\omega_{eg}-\gamma,\omega_{eg}+\gamma$, spontaneous emission would also not occur. I really do not see why not; why do we need to have it couple to so many different modes of the field?
• Let me test the waters here: if I say that the atom is coupled to exactly one electromagnetic mode (e.g. because it is inside a Fabrey-Perot cavity) then an excitation initially in the atom oscillates between the atom and cavity in the so-called "vacuum Rabi oscillation". There's no decay because the excitation comes back to the atom after one oscillation period. Does that answer your question or should I go on? – DanielSank Feb 2 '15 at 16:45
• Okay, so we have a vacuum Rabi oscillation with the field in the vacuum, and Rabi flopping with frequency $g_c \sqrt{n+1}$ with n photons in the mode $c$. Then if you have a lot of different modes, they will all cause oscillations with different frequencies, which destructively interfere, hence causing collapse. Is that what you mean? I can see how that causes the spontaneous emission to happen. However, I don't see how that for example excludes the case I talked about above, where you only have modes above the atomic transition frequency? – user129412 Feb 2 '15 at 16:54
• @DanielSank I believe there are two related but distinct points to be made here. The first is that a continuum of modes is required in order to prevent persistent oscillations, i.e. one needs an infinite number of modes so that the Poincare recurrence time is infinite. The second is that the spectral density of these modes must vary slowly over the relevant frequency range so that there are no oscillations at all on the coarse-grained timescale. – Mark Mitchison Feb 2 '15 at 17:00
• @MarkMitchison: Yes, you are absolutely correct. I just wanted to see if user129412 and I were on the same page. Apparently we are, and the only remaining question is the second one you mention in your comment. – DanielSank Feb 2 '15 at 17:02
• @DanielSank :) Well I was planning on writing an answer up today, but I have to get some actual research done first! – Mark Mitchison Feb 2 '15 at 17:05
A short, mathematical answer to the question is found in the properties of Fourier transforms. The temporal response of the environment to a perturbation is given by the Fourier transform of its frequency response to the same perturbation. Therefore, if a broad range of frequencies in the bath are perturbed, the response occurs over a narrow range of times. Let me try to briefly explain how this mathematical structure arises from the physics.
Spontaneous emission can be understood from the following hand-wavy arguments. The electron in an excited state produces an electric field. This field fluctuates over time; these fluctuations drive the transitions in the electronic state. The spontaneous emission therefore arises from the effect of the electron on its environment, which in turn produces a back-action that affects the electron.
The response of the electromagnetic field to a perturbation $\mathbf{E} = E\hat{\mathbf{z}}$ (I have arbitrarily chosen polarisation in the $z$ direction) is captured by the response function: $\Gamma(t) = \langle E(t) E(0)\rangle,$ where $E(t)$ denotes the Heisenberg-picture operator. This function is central to the theory of linear response to a small perturbation. For example, if one introduces a classical electric dipole that oscillates with a time-dependent dipole moment $d(t)$, the resulting electric field at that point is given by the convolution $$\delta\langle E(t) \rangle \approx -i\int_0^t\mathrm{d}s\; d(t-s) \Gamma(s).$$
The previous paragraphs serve merely to motivate the appearance of the response function $\Gamma(t)$. In a physically realistic case where we have a quantum dipole (e.g. an atom) with two states separated by a frequency $\epsilon$, the response function determines the rate of spontaneous emission, which is proportional to the quantity: $$\gamma(t) \sim \int_0^t\mathrm{d}s\; e^{i\epsilon s} \Gamma(s).$$ Assuming that $\Gamma(s)$ decays much more rapidly than $1/\epsilon$, for times $t\gg 1/\epsilon$ we can make the Markov approximation $$\gamma(t) \to \gamma = \int_0^{\infty}\mathrm{d}s\; e^{i\epsilon s} \Gamma(s),$$ so that we effectively have a constant spontaneous emission rate over time, leading to pure exponential decay.
When does $\Gamma(s)$ decay rapidly enough so that we can make the Markov approximation? The electric field $E(t)$ contains many components (normal modes) which oscillate at different frequencies. If we make this decomposition we get a Fourier representation like $$\Gamma(t) = \int_0^\infty\mathrm{d}\omega\; e^{-i\omega t} J(\omega),$$ where the spectral density $J(\omega)$ quantifies the degree to which the field at frequency $\omega$ is perturbed by a dipole. For an atom interacting with the electromagnetic field in free space, you will normally get something like $$J(\omega) \sim \lambda \frac{\omega^3}{\omega_c^2}e^{-\omega/\omega_c}.$$ Here $\lambda$ is a small dimensionless coupling parameter, and $\omega_c$ is a large frequency cutoff on the order of $c/a_0$, where $a_0$ is the Bohr radius. From purely dimensional arguments you can see that $$\Gamma(t) \sim \frac{\lambda \omega_c^2}{(\omega_c t)^4}.$$ This tells you that $\Gamma(t)$ vanishes after times much bigger than $\tau = 1/\omega_c$. This time $\tau$ is called the memory time. Since here $\omega_c \approx 10^{18} \text{Hz}$, while the typical optical frequencies are $\epsilon \approx 10^{14} \text{Hz}$, the Markov approximation is well justified.
The extreme example of Markovian noise (white noise) corresponds to $J(\omega) = \text{const.}$, in which case $\Gamma(t) = \delta(t)$, i.e. the bath memory time is infinitesimally small. The opposite extreme is something like a photonic crystal, where the environment has a sharp band edge at frequency $\Omega$ where $J(\omega)$ goes to zero. In that case the response function ends up something like $$\Gamma(t) = \int_0^\Omega e^{-i\omega t} J(\omega) \sim f(t) e^{-i\Omega t}$$ where $f(t)$ is some function of time. Now if $\Omega$ is comparable to $\epsilon$, you can imagine that there will be resonance effects, and there will be no smooth irreversible transfer of energy into the environment. Rather $\gamma(t)$ becomes a complicated function of time and you will see non-Markovian dynamics. If the frequency $\epsilon$ lies deep within the band-gap then there is no spontaneous emission at all, since there are simply no electromagnetic field modes to couple to, i.e. effectively $J(\omega) = 0$ in the relevant frequency range.
Hopefully these examples should convince you that the frequency scale which sets the bandwidth of perturbed modes in the environment ($\omega_c$, $\Omega$) is on the order of the inverse memory time. Thus, large bandwidths correspond to shorter memory times, i.e. more Markovian environments.
DISCLAIMER: All equations given here are based purely on memory and minimal back-of-the-envelope consistency checks. The proportionality factors and various terms which I arbitrarily deemed irrelevant are almost certainly missing.
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2019-06-16 14:47:03
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|
https://solvedlib.com/n/determine-whether-or-not-each-of-the-following-slgnals-perlodic,15448487
|
# Determine whether or not each of the following slgnals perlodic I{ 3 signal Is periodic; determine Its fundamental perlod: 80-)
###### Question:
Determine whether or not each of the following slgnals perlodic I{ 3 signal Is periodic; determine Its fundamental perlod: 80-) (a)*I[n] (b)x[n] cos " Ccs (3) ( +sin () 2 cos @) (c) x[n] C05
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2022-05-29 03:01:09
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https://leanprover-community.github.io/archive/stream/116395-maths/topic/linear.20equiv.20constructor.html
|
## Stream: maths
### Topic: linear equiv constructor
#### orlando (Apr 17 2020 at 17:55):
Hello, in the file basic / module
There is the definition of linear equivalence :
structure linear_equiv (R : Type u) (M : Type v) (M₂ : Type w)
[ring R] [add_comm_group M] [add_comm_group M₂] [module R M] [module R M₂]
extends M →ₗ[R] M₂, M ≃ M₂
end
I never work with a structure extend structure, how is the form of the constructor ?
#### Kevin Buzzard (Apr 17 2020 at 17:59):
One way which always works is this: def foo (R : Type u) (M : Type v) (M₂ : Type w) [ring R] [add_comm_group M] [add_comm_group M₂] [module R M] [module R M₂] : M ≃ₗ[R] M₂ := {! !} and then click on the ! and then click on the little yellow lightbulb and select the last-but-one option: "generate a skeleton for the structure under construction".
#### Kevin Buzzard (Apr 17 2020 at 18:00):
(I assume you're talking about linear_algebra.basic; M ≃ₗ[R] M₂ is notation for linear_equiv R M M₂)
#### orlando (Apr 17 2020 at 18:02):
Thx Kevin, I think you already gave me this trick ! sorry !
Yes linear_algebra.basic
#### Kevin Buzzard (Apr 17 2020 at 18:03):
But if you already have e : M ≃ M₂ then you do not have to fill in to_fun or inv_fun or left_inv or right_inv, you can just write ..e after you have defined the other fields.
#### Kevin Buzzard (Apr 17 2020 at 18:04):
import linear_algebra.basic
universes u v w
def foo (R : Type u) (M : Type v) (M₂ : Type w)
[ring R] [add_comm_group M] [add_comm_group M₂] [module R M] [module R M₂]
(e : M ≃ M₂) : M ≃ₗ[R] M₂ :=
{
smul := sorry,
..e}
#### Kevin Buzzard (Apr 17 2020 at 18:05):
and similarly if you already have a linear mapf : M →ₗ[R] M₂ then you only have to fill in the inverse map and the proofs that it is an inverse, and then add ..f
#### orlando (Apr 17 2020 at 18:05):
@Kevin Buzzard my bad, i use your trick two hour ago ! with this structure
lemma rest_is_lin_equiv (h : stable_sub_module ρ p) (g : G) : p ≃ₗ[R] p := begin
{! ! }end
And i have i message :
too many constructors
state:
G : Type u,
R : Type v,
M : Type w,
_inst_1 : group G,
_inst_2 : ring R,
_inst_4 : module R M,
ρ : group_representation G R M,
p : submodule R M,
h : stable_sub_module ρ p,
g : G
⊢ tactic {! !}
But without begin end it's ok !
#### Kevin Buzzard (Apr 17 2020 at 18:06):
Yes, {! !} is term mode.
rohhhhh
#### Kevin Buzzard (Apr 17 2020 at 18:06):
Make definitions in term mode, and prove theorems in tactic mode.
#### Kevin Buzzard (Apr 17 2020 at 18:07):
If you are defining X : Type* or P : Prop or x : X with X : Type*, it's best to use term mode. If you are defining h : P with P : Prop then tactic mode is fine. If you see ⊢ P in tactic mode then it is best if P : Prop.
#### Kevin Buzzard (Apr 17 2020 at 18:09):
import linear_algebra.basic
def foo (R : Type*) (M : Type*) (M₂ : Type*)
[ring R] [add_comm_group M] [add_comm_group M₂] [module R M] [module R M₂]
: M ≃ₗ[R] M₂ :=
begin
constructor,
sorry, sorry, sorry, sorry, sorry, sorry
end
is not a good idea.
#### orlando (Apr 17 2020 at 18:10):
I do that :sweat_smile: and it's not a good idea ! Complicated to understand the field !
Thx !!!!
#### Kevin Buzzard (Apr 17 2020 at 18:12):
But this is OK:
import linear_algebra.basic data.rat.basic
def foo : ℚ ≃ₗ[ℤ] ℚ :=
begin
refine_struct
{ to_fun := id,
inv_fun := id
},
{ sorry},
{ sorry},
{ sorry},
{ sorry}
end
Here I define the data using term mode, and then I have four goals to solve in tactic mode.
#### Kevin Buzzard (Apr 17 2020 at 18:13):
refine_struct is a tactic which takes some structure field definitions (in term mode), and then leaves the rest of them as goals (in tactic mode). So now you can work on the goals which are theorems, in tactic mode.
#### orlando (Apr 17 2020 at 19:41):
@Kevin Buzzard : all good ... thx :+1:
#### orlando (Apr 17 2020 at 23:45):
@Kevin Buzzard i have show that a linear map is an linear equiv. Can i have access to $f^{-1}$ notation ? I don't understand how ? (there is no example in the file) .
#### Kevin Buzzard (Apr 17 2020 at 23:53):
No, \-1 is notation for has_inv.inv and has_inv.inv : X -> X. But here the type of $f$ and $f^{-1}$ differ in general.
#### Kevin Buzzard (Apr 17 2020 at 23:54):
\-1 notation is really bad anyway. For example $\sin^{-1}(x^{-1})$ is two different uses of it, and $\sin^{-1}$ isn't a two-sided inverse for $\sin$ anyway.
#### Kevin Buzzard (Apr 17 2020 at 23:55):
If e : M ≃ₗ[R] M₂ then...I can't remember if there's a coercion to function or not for equivalences.
#### Kevin Buzzard (Apr 17 2020 at 23:55):
If there is, then e.symm is probably the inverse function (or \u= e.symm) and if there isn't then it's e.inv_fun.
#### orlando (Apr 18 2020 at 00:00):
Ok Kevin, i check e.symm another thing tomorow :slight_smile:
#### Scott Morrison (Apr 18 2020 at 00:03):
Avoid using inv_fun. The preference is definitely to use e.symm, and the coercion to a function. So hopefully you can just write something like e.symm x. If the coercion to a function fails (this is fairly flaky in Lean 3; it's a hard problem for the elaborator), adding types usually works: (e.symm : X \to Y) x.
Last updated: May 12 2021 at 07:17 UTC
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2021-05-12 08:24:38
|
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http://www.mathworks.com/matlabcentral/newsreader/view_original/308634
|
From: srach <s.rach-nospam-@iu-bremen.de>
Path: news.mathworks.com!newsfeed-00.mathworks.com!webx
Newsgroups: comp.soft-sys.matlab
Subject: Re: now that was interesting
Message-ID: <ef2fb57.140@webx.raydaftYaTP>
Date: Thu, 13 Apr 2006 15:54:29 -0400
References: <ef2fb57.133@webx.raydaftYaTP> <ef2fb57.134@webx.raydaftYaTP> <ef2fb57.135@webx.raydaftYaTP> <ef2fb57.136@webx.raydaftYaTP> <ef2fb57.137@webx.raydaftYaTP>
Lines: 29
NNTP-Posting-Host: 134.106.76.159
MIME-Version: 1.0
Content-Type: text/plain; charset="ISO-8859-1"
Content-Transfer-Encoding: 8bit
Xref: news.mathworks.com comp.soft-sys.matlab:344797
In my opinion the best way to stop people stop spamming the queue was
suggested by Matthew in comment to the blog:
\begin{quote}
We were kicking around the idea of only scoring one entry per
contestant on each pass through the queue. That is, wed only score
one entry from each author until everyone waiting has had one of
their entry scored. The drawbacks are that it is a little tough to
explain and it would encourage contestants to submit under different
names.
\end{quote}
The possible drawback he sees might be eliminated by awarding an
additional price to the contestant who makes the biggest cumulative
improvement to the top score over the whole curse of the contest.
(However, it might be difficult to find a way of weighting
improvements relatively to contest phase - improvements are much
greater at the beginning of the contest.)
Submitting entries under different names would decrease ones chances
to win such an award.
Another good idea I've read somewhere here is to run entries multiple
times instead of one single run and to score the mean performance.
This would make spamming identical unneccessary at least for entries
depending on random numbers.
Finally I would like to thank the contest team as well as the other
contestants - this contest (actually, my first) was really fun!
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2014-11-23 02:30:36
|
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|
https://studydaddy.com/question/how-do-i-find-the-x-intercepts-of-the-graph-of-y-x-2-11x-18
|
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
How do I find the x-intercepts of the graph of y=x^2-11x+18?
To find the x-intercepts of the graph of y=x^2-11x+18 substitute 0 for y and solve for x.
0=x^2-11x+18
Now, factor the quadratic expression on the right.
0=(x-2)(x-9)
and set each factor equal to 0 and solve.
(x-2)=0 or (x-9)=0 =>x=2 or x=9
The x-intercepts of any function y = f(x) are found
by setting f(x)=0 since the y-coordinate of any point on the
x-axis is zero.
See image below.
|
2019-04-25 20:43:19
|
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|
https://leetcode.com/articles/minimum-swaps-to-make-sequences-increasing/
|
#### Approach #1: Dynamic Programming [Accepted]
Intuition
The cost of making both sequences increasing up to the first i columns can be expressed in terms of the cost of making both sequences increasing up to the first i-1 columns. This is because the only thing that matters to the ith column is whether the previous column was swapped or not. This makes dynamic programming an ideal choice.
Let's remember n1 (natural1), the cost of making the first i-1 columns increasing and not swapping the i-1th column; and s1 (swapped1), the cost of making the first i-1 columns increasing and swapping the i-1th column.
Now we want candidates n2 (and s2), the costs of making the first i columns increasing if we do not swap (or swap, respectively) the ith column.
Algorithm
For convenience, say a1 = A[i-1], b1 = B[i-1] and a2 = A[i], b2 = B[i].
Now, if a1 < a2 and b1 < b2, then it is allowed to have both of these columns natural (unswapped), or both of these columns swapped. This possibility leads to n2 = min(n2, n1) and s2 = min(s2, s1 + 1).
Another, (not exclusive) possibility is that a1 < b2 and b1 < a2. This means that it is allowed to have exactly one of these columns swapped. This possibility leads to n2 = min(n2, s1) or s2 = min(s2, n1 + 1).
Note that it is important to use two if statements separately, because both of the above possibilities might be possible.
At the end, the optimal solution must leave the last column either natural or swapped, so we take the minimum number of swaps between the two possibilities.
Complexity Analysis
• Time Complexity: .
• Space Complexity: .
Analysis written by: @awice.
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2019-12-15 23:26:47
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https://stackoverflow.com/questions/9490107/iis-7-5-application-pool-uses-wrong-appdata-for-custom-user-as-identity
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# IIS 7.5 application pool uses wrong %APPDATA% for custom user as identity
I want my MVC3 web application to access %APPDATA% (e.g. C:\Users\MyUsername\AppData\Roaming on Windows 7) because I store configuration files there. Therefore I created an application pool in IIS with the identity of the user "MyUsername", created that user's profile by logging in with the account, and turned on the option "Load User Profile" (was true by default anyway). Impersonation is turned off.
Now I have the problem that %APPDATA% (in C#):
appdataDir = Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData)
resolves to c:\windows\system32\inetsrv instead of C:\Users\MyUsername\AppData\Roaming.
UPDATE: More exactly, the above C# code returns an empty string, so that Path.GetFullPath(Path.Combine(appdataDir, "MyAppName")) prepends the current path to my application name, resulting in c:\windows\system32\inetsrv\MyAppName.
I know I made this work before with the same web application on a Windows Server 2008 R2, and now I'm getting this problem with the same major version 7.5 of IIS on my Windows 7.
I used the same procedure as before: Created a new user, logged in as that user to create the profile and APPDATA directories, then added the application pool with this identity and finally added the web application to this pool.
Any ideas?
• Is your application pool configured as Classic or Integrated mode? – Kev Feb 29 '12 at 1:56
• @Kev: Integrated mode. – AndiDog Feb 29 '12 at 5:48
• I have the same problem. What's especially weird is that the path for Environment.SpecialFolder.UserProfile works fine, and if I build up the path to the AppData folder from there, it works. – Jason Apr 20 '12 at 20:58
Open your %WINDIR%\System32\inetsrv\config\applicationHost.config and look for <applicationPoolDefaults>. Under <processModel>, make sure you don't have setProfileEnvironment="false". If you do, set it to true.
• The value is <processModel identityType="ApplicationPoolIdentity" />. – AndiDog Dec 10 '12 at 21:19
• Set it to <processModel identityType="ApplicationPoolIdentity" loadUserProfile="true" setProfileEnvironment="true" /> – Amit Apple Dec 12 '12 at 2:07
• This answer is the one that solved the problem for me. – Olivier MATROT Jun 18 '14 at 9:03
• In my case, it was enough to add setProfileEnvironment="true" to the specific application pool. It was not necessary to change the setProfileEnvironment value in the default entry. – Heinzi Jan 16 '18 at 15:23
• Why is LoadUserProfile an option in IIS Manager, but SetProfileEnvironment is not? – Triynko Jan 10 at 20:00
Process Model - Load user Profile set True.
It Helps me.
I experienced the same problem recently. As mentioned by Amit, the problem is that the user profile isn't loaded. The setting is for all application pools, and is in the applicationHost.config (typically C:\Windows\System32\inetsrv\config\applicationHost.config). If you update the applicationPoolDefaults elements as follows, it will work;
<applicationPoolDefaults managedRuntimeVersion="v4.0">
</applicationPoolDefaults>
We've tried this with IIS 7.5, and taken it through to production without problem.
You can automate this if you want;
appcmd set config -section:system.applicationHost/applicationPools /applicationPoolDefaults.processModel.setProfileEnvironment:"true" /commit:apphost
or if you prefer powershell
Set-WebConfigurationProperty "/system.applicationHost/applicationPools/applicationPoolDefaults/processModel" -PSPath IIS:\ -Name "setProfileEnvironment" -Value "true"
Hope this helps
I am experiencing the same problem. Have you by chance installed the Visual Studio 11 beta? I did recently, and I've noticed a couple of differences in how the 4.0 compatible .dlls for that work with our code. I'm still trying to track down the problem for certain, but I didn't have this problem before that.
Edit:
After comparing the decompiled sources from 4.0 and 4.5 for GetFolderPath (and related), there are differences. Whether they are the source of the problem...I'm not sure yet.
Edit 2: Here are the relevant changes. I'm working on trying both to see if I get different results. [code removed]
Edit 3:
I've now tried calling SHGetFolderPath directly, which is what the .NET Framework ends up doing, anyway. It returns E_ACCESSDENIED (-2147024891 / 0x80070005). I don't know what has changed where I'm getting that in some specific cases, but not in others.
Edit 4:
Since you're getting a empty string, you may want to switch your code to use SHGetFolderPath so you can get the HResult and at least know what exactly is happening.
void Main() {
Console.WriteLine( GetFolderPath( Environment.SpecialFolder.ApplicationData ) );
}
[System.Runtime.InteropServices.DllImport("shell32.dll")]
static extern int SHGetFolderPath(IntPtr hwndOwner, int nFolder, IntPtr hToken, uint dwFlags, StringBuilder pszPath);
private string GetFolderPath( Environment.SpecialFolder folder ) {
var path = new StringBuilder( 260 );
var hresult = SHGetFolderPath( IntPtr.Zero, (int) folder, IntPtr.Zero, 0, path );
Console.WriteLine( hresult.ToString( "X" ) );
return ( (object) path ).ToString( );
}
• No, I have Visual Studio 2010 installed, and no other betas. Which differences are there? – AndiDog Apr 23 '12 at 12:51
• I am also getting 0x80070005 (E_ACCESSDENIED). Will try to investigate more. – AndiDog May 8 '12 at 14:50
The problem is with your IIS settings. The answer is here: Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData) returns String.Empty
• I have the user profile loaded and get the same problem (I think this problem occurs on Windows 7 only). – AndiDog Jul 19 '12 at 14:36
• Based on what you described in your original question, you did create the profile folders, but you didn't take the step necessary to load the profile for IIS. If you did take that step then add it to your description. If you didn't then read the link I gave and follow the instructions. – grahamesd Jul 23 '12 at 17:16
• I said in my question that I had that option turned on, so this solution doesn't seem to work for me, sorry. Must have something to do with Windows 7. When I have time, I'll try on Windows 8 (not the server variant) and see what happens. – AndiDog Jul 23 '12 at 20:02
• I think you're right. I tried it with IIS on Win7 Pro and on Win7 Enterprise and GetFolderPath returns an empty string. Yet on Win2008 R2 it works fine. I also tried changing the app pool identity in Win7 to an account with higher priveliges and then got back "C:\Windows\system32\config\systemprofile\AppData\Roaming" which doesn't even exist. So my conclusion is the same as yours: it doesn't work on Win7. Good news is that it DOES work for IIS Express, so you can at least develop on your Win7 machine. – grahamesd Aug 1 '12 at 20:52
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2019-01-19 15:35:09
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https://cs.stackexchange.com/questions/129914/is-a-graph-at-least-k-colorable-complexity
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# Is a graph at least k-colorable? (Complexity)
I am new to the complexity theory and I am trying to understand what would be the complexity of the following problem: "Is a graph AT LEAST $$k$$-colorable?"
Whether a graph is $$k$$-colorable is clearly a NPC problem and is explained by reduction to 3-SAT.
However, I am not sure to which complexity class belongs the modified problem I have stated. Does it belong to NP as well? My guess is that we would be able to check it with a polynomial certificate (any $$\geq k$$ coloring of the graph).
And then the problem "Is a graph AT MOST $$k$$-colorable?" would be its complement in co-NP unless NP=co-NP?
• Could you give an example of a graph that is not "at least 3-colorable"? – Tom van der Zanden Sep 7 '20 at 21:23
• Tom: probably any graph with 0, 1, or 2 nodes. – gnasher729 Sep 8 '20 at 21:29
Well, all graphs are colorable with $$\geq k$$ colors: Assign each vertex a different color (and leave the other color classes empty if $$k$$ is larger than the number of vertices of the given graph).
As for the complement of this problem, note that you did not correctly invert the definition of the problem: A problem $$L$$ is, after all, merely a set of strings over some finite alphabet $$\Sigma$$ representing some objects with some property. Its complement is then the set $$\Sigma^\ast \setminus L$$ or the set of all strings not representing an object with the property. If we recall the argument above, we find that the complement of your problem is hence the set of all strings not representing graphs (under that fixed encoding scheme) rather than the one you claimed. This problem (as the first one), is in $$\mathsf P$$.
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2021-08-03 05:57:14
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http://en.coininvest.com/post/silver-is-acting-like-gold-but-on-steroids/
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SILVER IS ACTING LIKE GOLD BUT ON STEROIDS!
Investors are close to making the biggest bet ever on silver even though there are concerns as to whether the rally in precious metals will last.
Prices on silver are up 25 percent in 2016, tracking a similar rally in gold on speculation the Federal Reserve will hold off on raising interest rates.
The performance on silver this year has beaten other precious metals — gold, platinum and palladium as silver is often used as a more volatile play on gold’s inflation and risk-hedging properties.
Total holdings are worth about $11 billion, according to Bloomberg. That’s just 15 percent of the amount invested in bullion funds rising within 0.6 percent of a record reached in 2014. Silver futures for July delivery moved 1.7 percent to settle at$17.268 an ounce at 1:40 p.m. on the Comex in New York. The 25 percent gain this year overtakes gold’s 20 percent rally.
We’ve seen an increased interest coming back into the precious metals space. A lot of people are looking at silver as something that could potentially benefit, not just because of the increased interest in precious metals, but also because it has such a strong industrial demand.
Category:
Tag:
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2019-03-22 18:04:01
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https://emedia.uen.org/browse?f.ut_alignment=UT.MATH.III.F.IF.8
|
Updating search results...
# 3 Results
View
Selected filters:
• UT.MATH.III.F.IF.8
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In the task "Carbon 14 Dating'' the amount of Carbon 14 in a preserved plant is studied as time passes after the plant has died. In practice, however, scientists wish to determine when the plant died and, as this task shows, this is not possible with a simple measurement of the amount of Carbon 14 remaining in the preserved plant. The equation for the amount of Carbon 14 remaining in the preserved plant is in many ways simpler here, using 12 as a base.
Subject:
Mathematics
Secondary Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/21/2012
Unrestricted Use
CC BY
Rating
0.0 stars
This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important asects of the task and its potential use. Here are the first few lines of the commentary for this task: Pictured below are the graphs of four different functions, defined in terms of eight constants: $a, b, c, k, m, p, q, \text{ and } r.$ The equations of...
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/20/2013
Unrestricted Use
Public Domain
Rating
0.0 stars
This lesson is for a math classroom, but can be adapted to fit any grade, subject, or content. In this lesson, students will use an iPad and its features: Keynote, Pages, Garageband, Numbers, Presentation, and iMovie. Students will use 3 of those features/programs to create a video lesson consisting of several examples from a topic of their choice. This project is in place of a term final, so their chosen topic should be from their current term.Image citation: The image is one I created.
Subject:
Secondary Mathematics
Material Type:
Assessment
Author:
Tyrani Bevell
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2023-01-29 09:07:42
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https://brilliant.org/problems/ratio-of-numbers/
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# Ratio of Numbers
Algebra Level 2
Two numbers are respectively 25% and 50% more than a third number. What is the ratio of the two numbers?
×
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2021-01-22 09:17:42
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https://ask.wireshark.org/questions/2294/revisions/
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# Revision history [back]
### Is it possible for Wireshark to SEND a Frame with a custom payload ?
Hi,
Am I correct in thinking that Wireshark only analyzes Frames but that it doesn't have the ability to send Frames ? What I am trying to do is to send a Frame with a specific payload. So ideally I would like a function (preferably in Python) or an open source program that will run on Windows where I can just specify the raw contents of the source and destination addresses and len filed and the raw payload contents eg:
[0x08, 0x00, 0x2b, 0x45, 0x11, 0x22, 0x00, 0x02, 0xa9, 0xf4, 0x5c,
0x53, 0x00, 0x00, 0x00, 0x00, 0xf5, 0x7b, 0x01, 0x00, 0x00, 0x00,
0x00, 0x00, 0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17, 0x18 ]
and it would just take these and send them to the Ethernet hardware which would pre-pend the preamble, calculate and append the 4-byte CRC and then send the Frame out on the cable. Is this possible with Wireshark or can anyone suggest where I would find a program or some Python code to do this ?
Thanks,
Usjes.
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2019-03-26 04:32:48
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https://ai.stackexchange.com/questions/5580/how-is-the-gradient-calculated-for-the-middle-layers-weights?noredirect=1
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# How is the gradient calculated for the middle layer's weights?
I am trying to understand backpropagation. I used a simple neural network with one input $$x$$, one hidden layer $$h$$ and one output layer $$y$$, with weight $$w_1$$ connecting $$x$$ to $$h$$, and $$w_2$$ connecting $$h$$ to $$y$$
$$x \rightarrow (w_1) \rightarrow h \rightarrow (w_2) \rightarrow y$$
In my understanding, these are the steps happening while we train a neural network:
The feedforward step.
\begin{align} h=\sigma\left(x w_{1}+b\right)\\ y^{\prime}=\sigma\left(h w_{2}+b\right) \end{align}
The loss function.
$$L=\frac{1}{2} \sum\left(y-y^{\prime}\right)^{2}$$
$$\frac{\partial L}{\partial w_{2}}=\frac{\partial y^{\prime}}{\partial w_{2}} \frac{\partial L}{\partial y^{\prime}}$$
$$\frac{\partial L}{\partial w_{1}}=?$$
The weight update
$$w_{i}^{t+1} \leftarrow w_{i}^{t}-\alpha \frac{\partial L}{\partial w_{i}}$$
I understood most parts of backpropagation, but how do we get the gradients for the middle layer weights $$dL/dw_1$$?
How should we calculate the gradient of a network similar to this?
Is this the correct equation?
$$\frac{\partial L}{\partial w_{1}}=\frac{\partial h_{1}}{\partial w_{1}} \frac{\partial w_{7}}{\partial h_{1}} \frac{\partial o_{2}}{\partial w_{7}} \frac{\partial L}{\partial o_{2}}+\frac{\partial h_{1}}{\partial w_{1}} \frac{\partial w_{5}}{\partial h_{1}} \frac{\partial o_{1}}{\partial w_{5}} \frac{\partial L}{\partial o_{1}}$$
The main doubt here is about the intuition behind the derivative part of back-propagation learning. First, I would like to point out 2 links about the intuition about how partial derivatives work Chain Rule Intuition and Intuitive reasoning behind the Chain Rule in multiple variables?.
Now that we know how the chain rule works, let's see how we can use it in Machine Learning. So basically in machine learning, the final output is a function of input variables and the connection weights $$f\left(x_1, x_2, \ldots, x_n, w_1, w_2, \ldots, w_n\right)$$, where $$f$$ encloses all the activation functions and dot products lying between input and output. The $$x_1, x_2, \ldots, x_n, w_1, w_2, \ldots, w_n$$ are called independent variables because they don't affect each other pairwise as well as in groups meaning you cannot find a function $$g(x_i, \dots, w_i, \dots) = h(x_j, \dots, w_j, \dots)$$. So, basically its a black box from input to output.
So now our purpose is to minimize the Loss/Cost function, by changing the parameters that can be 'controlled by us' i.e the weights only, we cannot change the input variables. So this is done by taking the derivative of the cost function w.r.t to the variable that 'can be changed'. Here is an explanation of why taking derivative and subsequently subtracting it reduces the value of cost function given by the 'maximal' amount. Also here.
Now, to calculate $$dL/dw_n$$ you have to keep few things in mind:
• Only differentiate $$L$$ w.r.t to those functions which affect $$L$$.
• And to reach your end goal of differentiation w.r.t to an independent variable you must differentiate $$L$$ w.r.t to those functions only which are dependent on that particular independent variable.
A crude algorithm assuming $$L$$ also as a normal function (along the lines of activation function, so that I can express the idea recursively) differentiate $$f_n$$ w.r.t to functions in the previous layer say $$f_{n-1}$$, $$f_{n-2}$$, $$w_n$$. Check which of these functions depends on $$w_1$$. Only $$f_{n-1}$$ and $$f_{n-2}$$ do. Differentiate them again w.r.t to previous layer functions. Check again and go on till you reach $$w_1$$.
This approach is the fool-proof version, but it has 2 flaws:
• First, $$w_n$$ is not a function. People are making this mistake of assuming $$w_n$$ to be a function due to misinterpretation of a simple NN diagram. To reach $$w_1$$, you don't need to go through $$w_n$$. But you definitely need to go through the activation functions and dot products. Think of this as painting a wall where color mixing occurs (not over-writing). So you paint the wall with some color (weights) then 2nd color and so on. Is the final product affected by color 1. Yes. Is the 'rate of change' caused by color 1 also affected by color 2. Yes. But does it mean we can find the 'change'of color n w.r.t to color 1? No its meaningless (bad example, couldn't think of a better one)
• The second flaw is that this approach is not followed because with experience it is apparent which function affects whom and which independent variable affects which function (saves computation).
To answer your question the equation is incorrect and the correct equation will be:
$$\frac{\partial L}{\partial w_{1}}=\frac{\partial h_{1}}{\partial w_{1}} \frac{\partial o_{2}}{\partial h_{1}} \frac{\partial L}{\partial o_{2}}+\frac{\partial h_{1}}{\partial w_{1}} \frac{\partial o_{1}}{\partial h_{1}} \frac{\partial L}{\partial o_{1}}$$
I have simply followed the algorithm I have given above.
As for why your equation is wrong, your equation contains the term $$dw7/dh1$$. Does $$w_7$$ vary with $$h_1$$? This means that $$w_7$$ is directly related to the input as $$h_1$$ is related with the input, but this is not the case for a single iteration(the whole algorithm run makes $$w_n$$ dependent on the inputs since you are trying to minimize the loss function w.r.t given inputs and weights, for a different set of inputs you will have different final weights).
So, in a nutshell, the aim of back-propagation is to identify the change in the loss function w.r.t to a given weight. To calculate that, you have to make sure in the chain rule of derivative you don't have any meaningless terms like the derivative of an independent variable w.r.t to any function. I recommend checking Khan Academy for a better understanding and clarity in concepts as I think the intuitions are hard to provide in a written answer.
• Thanks I got it w7 and w5 are constants whereas h1 and o1 are functions. w7 and w5 only changes during updation whereas h1 and o1 changes in forward passing aswell – Eka Mar 10 '18 at 16:13
• @Eka good..I missed that part but you got it – DuttaA Mar 10 '18 at 16:16
• One other question; so when we are finding gradients with respect to a weight. We keep that weight as variable and the rest of weights as constant. in my above case w1 is variable and the other two constant.. right? – Eka Mar 10 '18 at 16:28
• @Eka yes...that is pretty much the concept of how partial derivatives are calculated – DuttaA Mar 10 '18 at 16:38
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2020-07-13 09:22:52
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https://socratic.org/questions/how-do-you-simplify-the-product-2n-3-n-2-2n-5-and-write-it-in-standard-form
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# How do you simplify the product (2n - 3)(n^2 - 2n + 5) and write it in standard form?
May 11, 2017
2n^3-7n^2+16n-15"
#### Explanation:
Multiply each term in the second bracket by each term in the first bracket as shown below.
$\left(\textcolor{red}{2 n - 3}\right) \left({n}^{2} - 2 n + 5\right)$
$= \textcolor{red}{2 n} \left({n}^{2} - 2 n + 5\right) \textcolor{red}{- 3} \left({n}^{2} - 2 n + 5\right)$
$= \left(\textcolor{red}{2 n} \times {n}^{2}\right) + \left(\textcolor{red}{2 n} \times - 2 n\right) + \left(\textcolor{red}{2 n} \times 5\right)$
$\textcolor{w h i t e}{=} + \left(\textcolor{red}{- 3} \times {n}^{2}\right) + \left(\textcolor{red}{- 3} \times - 2 n\right) + \left(\textcolor{red}{- 3} \times 5\right)$
$= 2 {n}^{3} - 4 {n}^{2} + 10 n - 3 {n}^{2} + 6 n - 15$
$= 2 {n}^{3} - 7 {n}^{2} + 16 n - 15 \leftarrow \text{ in standard form}$
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2019-09-23 16:08:30
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http://sergeykarayev.com/work/2011-01-24/kanan-cvpr2010/
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# Sergey Karayev
Computer Science Department
University of California, Berkeley
Computer Vision group with Trevor Darrell.
Digital facets
Updated 06 Mar 2013
## Review of Kanan and Cottrell, Robust Classification of Objects, Faces, and Flowers Using Natural Image Statistics, CVPR 2010.
The paper’s approach has three parts. The first is using an ICA-based spatial pyramid feature; the second is computing a saliency map to sample interest points; and the third is in using Naive Bayes Nearest Neighbor (NBNN) for classification. The approach is evaluated on three single-object datasets: Caltech-101 and -256, Aleix and Robert faces dataset of 120 individuals with 26 images each, and 102 Flowers (8200 images). The results are best yet published for Caltech-101 single-feature approaches, and match best multiple-feature performances; comparable to state-of-the-art on Caltech-256; match state-of-the-art on the AR Faces; and beat the single previously published result on the Flowers dataset.
### ICA-based Local Features and Saliency
The images are first pre-processed by converting to a standard size, converting to the LMS color space (designed to match human color receptor distributions), normalizing, and then applying a nonlinear transform inspired by modulation to luminance that happens in photoreceptors (a logarithmic compression). [Note: It would be interesting to see the effects of not performing this mapping.]
ICA filters of size $b \times b$ ($b$ tuned on a Butterfly and Bird dataset to 24 pixels) are learned on about 5000 color image patches from the McGill color image dataset. To learn $d$ ICA features, the authors first run PCA on the patches, discard the first principal component, retain $d$ following principal components, and then learn the ICA decomposition. I’m not quite sure how this works—I guess ICA is then only able to learn $d$ non-garbage bases?
#### Saliency Map
The ICA bases are used to place a saliency map over the image following the Saliency Using Natural statistics (SUN) framework \cite{Zhang:2008:SUN}. The basic idea is that saliency of a point is the inverse $P(F)^{-1}$ of its probability under the ICA model $P(F=\mathbf{f})=\prod_i P(\mathbf{f}_i)$. Each unidimensional distribution is fit with a generalized Gaussian distribution:
$P(\mathbf{f}_i) = \frac{\theta_i}{2 \sigma_i \Gamma(\theta_i^{-1})} exp(-|\frac{\mathbf{f}_i}{\sigma_i}|^{\theta_i})$
Parameters are fit still using the McGill color database. A further strange nonlinear weighting of the dimensions of $\mathbf{f}$ is then done to weight rarer responses more heavily.
### Fixations
The saliency map is normalized to a probability distribution, and “fixations” are sampled from it $T$ times. At each location $l_t$, an interesting fixation feature is extracted. It is a spatial pyramid over an area of $w=51$ pixels, using average pooling. So, the initial window of $w \times w \times d$ is represented by a vector of size $21d$, where $21 = 4 \times 4 \times 2 \times 2 \times 1 \times 1$ shows the structure of the spatial pyramid. Importantly, the normalized location $l_t$ of the fixation is also stored. To cast SIFT in this framework, we would set $w=17$, $d=8$, and the spatial aggregation would be a flat $4 \times 4$ grid.
After gathering $T$ fixations on every image in the training set, the unit-normalized SP vectors are then additionally processed by retaining only the first 500 PCA components and whitening them. The chain of re-normalizations in this paper is quite long and I would appreciate theoretical justifications for these decisions.
### Classification
The paper uses Kernel Density Estimation (KDE) to model $P(\mathbf{g}_t|C=k)$, where $\mathbf{g}_t$ is the vector of fixation features. A Naive Bayes assumption is made, such that each fixation contributes independently to the total probability. The posterior is estimated with Bayes rule, assuming uniform class priors. 1-nearest neighbor KDE is used, and the Euclidean distance between the fixation locations is considered in addition to the feature-to-exemplar distance. The final posterior probability is:
$P(\mathbf{g}_t | C=k) \propto max_i \frac{1}{||\mathbf{w}_{k,i}-\mathbf{g}_t||^2_2 + \alpha ||\mathbf{v}_{k,i}-\ell_t||^2_2 + \epsilon}$
where $\mathbf{w}_{k,i}$ is a vector representing the $i$’th examplar of a fixation from class $k$.
### Discussion
The authors attribute the strength of their approach largely to the exemplar-based classifier. Their approach does outperform the comparable single-descriptor version of the Boiman and Irani NBNN classifier \cite{Boiman:2008}; that could be due to a number of factors:
1. They also use location information in their comparison of fixations. EDIT: NBNN paper also appends location to the feature vector.
2. They sample features from a saliency map (vs. densely for NBNN)
3. They use their ICA feature instead of SIFT and other standard descriptors.
It would be excellent to see a controlled evaluation of each of these factors. The paper as it is presents a very specific and unorthodox approach, and does not justify many of its design decisions.
My questions:
1. What is the contribution of the saliency map? How would the performance change under a random sampling scheme? What about an interest-point sampling scheme?
2. Why is the saliency computed at a single scale? The only reason for this working well is that the dataset is single-object and fixed-scale.
3. How would performance change if a standard feature, for example SIFT, was extracted instead of the ICA SP feature?
4. What is the contribution of the location feature? Why is it weighted at $\alpha=0.5$; what would cross-validation tune it to?
In my mind, the most important part of the approach is NN classification. It would be interesting to re-implement this framework with a different bottom-up saliency map, for example the multi-scale one used in \cite{Alexe:2010} and traditional SIFT features.
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2013-05-24 14:09:47
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http://www.r-bloggers.com/finding-patterns-amongst-binary-variables-with-the-homals-package/
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# Finding Patterns Amongst Binary Variables with the homals Package
February 10, 2013
By
(This article was first published on Data and Analysis with R, at Work, and kindly contributed to R-bloggers)
It’s survey analysis season for me at work! When analyzing survey data, the one kind of analysis I have realized that I’m not used to doing is finding patterns in binary data. In other words, if I have a question to which multiple, non-mutually exclusive (checkbox) answers apply, how do I find the patterns in peoples’ responses to this question?
I tried apply PCA and Factor Analysis alternately, but they really don’t seem well suited to the analysis of data consisting of only binary columns (1s and 0s). In searching for something that works, I came across the homals package. While the main function is described as a “homogeneity analysis”, its one ability that interests me is called “non-linear PCA”. This is supposed to be able to reduce the dimensionality of your dataset even when the variables are all binary.
Well, here’s an example using some real survey data (with masked variable names). First we start off with the purpose of the data and some simple summary stats:
It’s a group of 6 variables (answer choices) showing peoples check-box responses to a question asking them why they donated to a particular charity. Following are the numbers of responses to each answer choice:
mapply(whydonate, FUN=sum, 1)
V1 V2 V3 V4 V5 V6
201 79 183 117 288 199
With the possible exception of answer choice V2, there are some pretty healthy numbers in each answer choice. Next, let’s load up the homals package and run our non-linear PCA on the data.
library(homals)
fit = homals(whydonate)
fit
Call: homals(data = whydonate)
Loss: 0.0003248596
Eigenvalues:
D1 D2
0.0267 0.0156
D1 D2
V1 0.28440348 -0.10010355
V2 0.07512143 -0.10188037
V3 0.09897585 0.32713745
V4 0.20464762 0.21866432
V5 0.26782837 -0.09600215
V6 0.33198532 -0.04843107
As you can see, it extracts 2 dimensions by default (it can be changed using the “ndim” argument in the function), and it gives you what looks very much like a regular PCA loadings table.
Reading it naively, the pattern I see in the first dimension goes something like this: People tended to answer affirmatively to answer choices 1,4,5, and 6 as a group (obviously not all the time and altogether though!), but those answers didn’t tend to be used alongside choices 2 and 3.
In the second dimension I see: People tended to answer affirmatively to answer choices 3 and 4 as a group. Okay, now as a simple check, let’s look at the correlation matrix for these binary variables:
cor(whydonate)
V1 V2 V3 V4 V5 V6
V1 1.00000000 0.0943477325 0.0205241732 0.16409945 0.254854574 0.45612458
V2 0.09434773 1.0000000000 -0.0008474402 0.01941461 0.038161091 0.08661938
V3 0.02052417 -0.0008474402 1.0000000000 0.21479291 0.007465142 0.11416164
V4 0.16409945 0.0194146144 0.2147929137 1.00000000 0.158325383 0.22777471
V5 0.25485457 0.0381610906 0.0074651417 0.15832538 1.000000000 0.41749064
V6 0.45612458 0.0866193754 0.1141616374 0.22777471 0.417490642 1.00000000
The first dimension is easy to spot in the “V1″ column above. Also, we can see the second dimension in the “V3″ column above – both check out! I find that neat and easy. Does anyone use anything else to find patterns in binary data like this? Feel free to tell me in the comments!
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2014-10-25 15:20:02
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https://math.stackexchange.com/questions/2804995/2014-arml-individual-problem-6-divisors/2805005
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# 2014 ARML Individual Problem 6: Divisors
Compute the smallest positive integer n such that 214·n and 2014·n have the same number of divisors.
The only thing that I was able to realize in this question was the fact that this value of n needs to a divisor of either 214 or 2014. I wasn't sure how to proceed from here.
• Have you tried? How do you compute the number of divisors of an integer in general? – Arnaud Mortier Jun 2 '18 at 2:45
• Raise the exponent of each prime factor by 1, and then multiply all the exponents. – Dude156 Jun 2 '18 at 2:47
• Right. If you know this it should be included in the question, as well as the prime decomposition of 214 and 2014. People are much more likely to try and help if you show that you are involved. – Arnaud Mortier Jun 2 '18 at 2:51
• I experimented and wrote a solution @ArnaudMortier. Thanks for prompting me to try harder. – Dude156 Jun 2 '18 at 3:08
Ok I figured it out. Thanks @Arnaud Mortier for convincing me to try a bit harder. For some reason, even after realizing the crux of the problem (which was that n would have to be made up of the 2, 19, 53, 107 in some form), I still didn't give it a good effort. What I did was set $n = 2^(a-1)*19^(b-1)*53^(c-1)*107^(d-1)$. I did the -1 because I knew that I would end up raising some of these terms by +1 which would mean that I might simplify my expression. After multiplying both 214 and 2014 by this and then raising the exponents by 1, I came at the equation $(a+1)(b)(c)(d+1)=(a+1)(b+1)(c+1)d$. After some cancellations, I arrived at bc = d(b+c+1). Since I was trying to minimize this number, I set d=1 (d=0 isn't possible because we are dealing with d-1 in the exponent). This led me to get that b and c are either 2 or 3 (either order). With a goal to minimize, I set b=2 and c=3. Substituting for n and multiplying, I arrived at the answer of $19^2*53= 19133$.
• Nice! It's not completely obvious at first that $d$ has to be $1$, but it does become clear when you see that there is a solution with $b=2$, $c=3$. A larger $d$ could not help beat that. There is a typo in the end, $b=3$ and $c=2$. – Arnaud Mortier Jun 2 '18 at 3:17
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2019-06-16 12:41:28
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https://www.pims.math.ca/scientific-event/210311-pudmsly
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## PIMS-UVic Discrete Math Seminar: Liana Yepremyan
• Date: 03/11/2021
• Time: 10:00
Lecturer(s):
Liana Yepremyan, University of Illinois Chicago and London School of Economics
Location:
Online
Topic:
Size-Ramsey numbers of powers of hypergraph trees and long subdivisions [video]
Description:
The $s$-colour size-Ramsey number of a hypergraph $H$ is the minimum number of edges in a hypergraph $G$ whose every $s$-edge-colouring contains a monochromatic copy of $H$.
We show that the $s$-colour size-Ramsey number of the $t$-power of the $r$-uniform tight path on $n$ vertices is linear in $n$, for every fixed $r, s, t$, thus answering a question of Dudek, La Fleur, Mubayi, and R\"odl (2017).
In fact, we prove a stronger result that allows us to deduce that powers of bounded degree hypergraph trees and of `long subdivisions' of bounded degree hypergraphs have size-Ramsey numbers that are linear in the number of vertices. This extends recent results about the linearity of size-Ramsey numbers of powers of bounded degree trees and of long subdivisions of bounded degree graphs. This is joint work with Shoham Letzter and Alexey Pokrovskiy.
Other Information:
This event took place via zoom. A recording is available on mathtube.org.
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2022-07-06 16:51:36
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http://retrievo.pt/advanced_search?link=1&field1=creators&searchTerms1='Zhang%2C+Hong'&constraint1=MATCH_EXACT_PHRASE
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Type
Database
Creator
Date
Thumbnail
# Search results
407 records were found.
## Customer retention in the financial industry: An application of survival analysis
Recently, in both marketing theory (academia) and practice (industry), the emphasis in relationship marketing has shifted to long term customer relationship management emphasizing customer retention or loyalty. This study has two main purposes: (1) to investigate the impacts of selected firm-customer interaction behavior and demographic characteristics on customer retention behavior in the financial industry, (2) to compare the results of a static method of analysis (logistic regression model) and a dynamic method (Cox’s hazard method) for customer retention. ^ The statistical analyses conducted employing two cohorts showed that the Cox model was stable across cohorts. Using Cox’s hazard method, all of the independent variables were found to be significantly related to customer retention, while some of demographic factors (for exampl...
## X-ray diffraction analysis of gene V protein encoded by filamentous bacteriophage Ff
The crystal structures of four mutant gene V proteins were solved and refined to $\sim$2 A resolutions. Three of these mutants were apolar substitutions in the hydrophobic core of the protein. One mutant, Arg 82 $\to$ Cys, is at the protein surface and is involved in breaking up of a surface salt bridge. The effects of these substitutions on the stability of the protein were discussed.
## A note on windowing for the waveform relaxation
The technique of windowing has been often used in the implementation of the waveform relaxations for solving ODE's or time dependent PDE's. Its efficiency depends upon problem stiffness and operator splitting. Using model problems, the estimates for window length and convergence rate are derived. The electiveness of windowing is then investigated for non-stiff and stiff cases respectively. lt concludes that for the former, windowing is highly recommended when a large discrepancy exists between the convergence rate on a time interval and the ones on its subintervals. For the latter, windowing does not provide any computational advantage if machine features are disregarded. The discussion is supported by experimental results.
## Design and Implementation of a Robot Force and Motion Server
A robot manipulator is a force and motion server for a robot. The robot, interpreting sensor information in terms of a world model and a task plan, issues instructions to the manipulator to carry out tasks. The control of a manipulator first involves motion trajectory generation needed when the manipulator is instructed to move to desired positions. The procedure of generating the trajectory must be flexible and efficient. When the manipulator comes into contact with the environment such as during assembly, it must be able to comply with the geometric constraints presented by the contact in order to perform tasks successfully. The control strategies for motion and compliance are executed in real time by the control computer, which must be powerful enough to carry out the necessary computations. This thesis first presents an efficient m...
## RFMS Software Reference Manual
This manual explains the software of the Robot Force and Motion Server (RFMS), a high performance robot control system designed and implemented in the GRASP laboratory. In this system, the robot manipulator is considered a force/motion server to the robot and a user application is treated as a request for the service of the manipulator. The user application is created on one of the Unix/VAX machines in 'C' programming language as a set of function calls. The application is carried out in a multi-processor controller, which consists of Intel single board computers and provides computing power necessary for computationally intensive tasks. The VAX machine and the Intel controller communicate through Ethernet, a local area network, which also allows interaction between the user and sensors. Design principles of the system can be found in ...
## RFMS Source Code
This is the source code of RFMS, a high performance robot control system designed and implemented in GRASP laboratory. It is intended to serve as a reference to "Software Reference Manual", prepared by Hong Zhang, the Department of Computer and Information Science, the University of Pennsylvania.
## Motion control for dynamic mobile robots
In this thesis, we present research results on sensor-based motion planning and nonlinear control for mobile robotic systems. In sensor-based motion planning, we use ideas from game theory to deal with the uncertainties accompanying real sensors and moving obstacles. We show that this idea can be successfully applied to both open loop and closed loop motion planning and control algorithms. With the emphasis on the use of a vision sensor, we extend the concept of sensor-based motion planning to motion planning in the image plane, which can help us to bypass the calibration errors associated with vision-based control, and achieve faster response speeds. Meanwhile, we address the effect of dynamics in vision-based motion control, or visual servoing, and expanded our ability to control a dynamic robotic system, such as the blimp robot. ^ ...
## Regulation of glial cell development and axonal outgrowth in the vertebrate central nervous system.
The complex and diverse functions of the mature vertebrate central nervous system (CNS) depend on the precise interactions formed by many types of neural cells, including neurons and glial cells. The major classes of macroglial cells in the CNS are astrocytes and oligodendrocytes. While astrocytes and their precursors have been proposed to participate in many CNS events including the guidance of migrating growth cones, oligodendrocytes form myelin, an axonal insulation sheath. The differentiation of oligodendrocytes occurs in a series of distinct stages in the spinal cord as well as other regions of the CNS and studies are required to determine the cellular mechanisms that regulate oligodendrocyte differentiation in the spinal cord. In the first part of this dissertation study, the results obtained indicate that oligodendrocyte differe...
## Personal information organization and re-access in computer folders: an empirical study of information workers
The current hierarchical folder system has long been found limited causing various difficulties in organizing and re-finding information on personal computers. Many alternative prototypes have been proposed to replace the current folder system. However, past empirical studies consistently observed that people prefer browsing folders in re-accessing information and only use searching as the last resort. Recognizing the complexity and our limited understanding of personal information organization and retrieval behavior in computer folders, my study was aimed to explore what people need from folders and the affordances and limitations of folders in the different stages of organization and retrieval, and furthermore provide implications for system design. Improved understanding on personal information organization and retrieval on computer...
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2019-05-20 02:27:12
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http://crypto.stackexchange.com/questions?page=159&sort=newest
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# All Questions
1k views
### Simple protocol for 1-out-of-2 oblivious transfer
I'm trying to explain the concept of 1-out-of-2 oblivious transfer to folks who haven't seen this idea before. I can explain what properties it provides, but it's also helpful if I can show a simple ...
552 views
### Why are elliptic curves better than cyclic groups?
The set of points of an elliptic curve over a finite field is isomorphic to the direct product of two cyclic groups (i.e. $E(F_{p^n}) \cong Z_{s} \times Z_{t})$. What is the advantage of representing ...
508 views
### Map Bytes to Number
Using node.js crypto library I can generate cryptographically strong bytes. How can I use this to generate a random number between 0 and 9999 (both inclusive)?
243 views
### Digital Signature Scheme Count
My question is How I will be able to count digital signature Scheme that can sign many documents with one private key. There are signature scheme as long as exists private keys in the keys spaces? or ...
4k views
### What are some disadvantages of homomorphic encryption schemes?
I'm doing some self-teaching / research for my own benefit in homomorphic cryptography. I've studied both additive and multiplicative schemes (Pallier and RSA respectively), but all I can seem to ...
342 views
### Signature with appendix
I see sometimes signatures schemes with appendix. This is about signatures schemes in which the message is needed in the verification algorithm, that is, the ouput of the signature algorithm is of the ...
269 views
### How did they factor RSA-704?
I don't understand the 'Wiedemann algorithm' works. Can someone explain the factoring of RSA-704 in an easy way?
872 views
### Is a PBKDF2-derived master key easier cracked if very many Data Protection Keys are derived from it?
I'm referring the NIST document about PBKDF2, NIST Special Publication 800-132, Recommendation for Password-Based Key Derivation, Part 1: Storage Applications, Page 8 and 9, section 5.4 Using the ...
929 views
### If we can find prime numbers larger than 17 milion digits, why can't we find all 1024bit primes? [duplicate]
"Largest Known Prime Number Discovered; Has 17,425,170 Digits" http://www.sciencedaily.com/releases/2013/02/130213225424.htm If we can find prime numbers larger than 17 milion digits, why can't we ...
144 views
### How to build a secure AE scheme with generic composition?
I am actually looking for using a secure Encrypt-Then-MAC AE scheme , and consider using either an existing "ready-to-use" dedicated AEAD mode (GCM, OCB, CCM, EAX, etc ..) or an alternate composed CTR-...
571 views
### How practical is proxy re-encryption for AES?
In proxy re-encryption schemes, a proxy is given special information that allows it to translate a ciphertext under one key into a ciphertext of the same message under a different key. How ...
280 views
### Using AES-CBC and providing the IV in file header -> security hole? [duplicate]
currently I audit some code I've written in the past to use KeePass 1.x databases. I stumbled over the following question: The database itself is decrypted with AES-CBC and inside the file header the ...
271 views
### How is key rotation defined?
I'm aware that it's advisable to rotate cryptographic keys periodically. HKDF accepts high-entropy input (IKM - Input Key Material), and expands it to an arbitrary length. It also accepts a salt, ...
82 views
### How does the possible attack on passwords and salts mentioned in PKCS#5 v2.1 work?
From PKCS#5 v2.1, 4.1 Salt (page 6, emphasis by me), also in v2.0, RFC 2898: In password-based encryption, the party encrypting a message can gain assurance that these benefits are realized ...
282 views
I'm playing around with an application for secure email-like communication and I want to perform length hiding padding on the plaintext messages so they always have a consistent size before encrypting ...
1k views
### Help with example RSA problem
While learning about RSA, I found this example problem. The answer is supposed to be "a 4-digit number that is a pattern of digits." I have computed it to be 16657 twice. OK, now to see if you ...
383 views
### How to calculate leaked information from a check digit
Given a check digit calculated using mod 11, of an encrypted numeric sequence of 8 digits, how much information can be known about the plaintext? More specifically, a single 3DES encrypted CBC block ...
2k views
448 views
### Recommended authenticated stream cipher for minimum overhead?
I'm running my TCP/IP protocol secured by encrypting each packet with AES128/CBC, bundling a SHA256 HMAC over that packet. This causes quite a bit of space overhead for small packets, so at first I ...
456 views
### Encrypting and MACing different data with same key
It's a well-known best practice to not use one key to both encrypt and MAC data. In my application, there are instances where I MAC a piece of data without having encrypted it first. Do I still need a ...
251 views
### Is it possible to cryptographically prove when was the last time a ciphertext was decrypted/encrypted?
I want to provide a service that encrypts and decrypts documents and I want to provide the users with proof that I haven't stolen their secret key and read their documents. I know I can prove when a ...
330 views
### How does Poly1305 in the NaCl library calculate a MAC?
I've been reading about Poly1305 as implemented in the NaCl library, more specifically I've been referring to this Java implementation of NaCl. My understanding of XSalsa20Poly1305 (in this ...
377 views
### Can Poly1305 be used with block ciphers running in CTR mode?
Is the use of Poly1305 limited to stream ciphers? (note, I'm not talking about Poly1305-AES )? Can it be used with block ciphers running in CTR mode? If so, what other considerations/limitations are ...
74 views
### Why is it impractical to generate a semiprime dictionary? [duplicate]
This might be a very simple question. However, I am just learning the concept, so just excuse me. I am wondering why there is not any attempt to generate all semiprime numbers? (as an dict. attack to ...
664 views
### Given a private RSA key, how do we get the public key?
Is it possible to pre-choose a private RSA key, then obtain a public key from it?
94 views
### How to choose the appropriate public (i, m) and private (j, m) keys?
I studied some encryption and decryption and I have found some very interesting problem to solve on the internet. I hope I am writing to right site - there are so many in StackExchange otherwise I ...
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2016-07-02 02:01:50
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https://pypi.org/project/codebraid/
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Live code in Pandoc Markdown
# Codebraid – live code in Pandoc Markdown
Codebraid is a Python program that enables executable code in Pandoc Markdown documents. Using Codebraid can be as simple as adding a class to your code blocks' attributes, and then running codebraid rather than pandoc to convert your document from Markdown to another format. codebraid supports almost all of pandoc's options and passes them to pandoc internally. See Codebraid Preview for VS Code for editor support.
Codebraid provides two options for executing code. It includes a built-in code execution system that currently supports Python 3.7+, Julia, Rust, R, Bash, JavaScript, and SageMath. Code can also be executed using Jupyter kernels, with support for rich output like plots.
Development: https://github.com/gpoore/codebraid
Citing Codebraid: "Codebraid: Live Code in Pandoc Markdown", Geoffrey M. Poore, Proceedings of the 18th Python in Science Conference, 2019, 54-61.
View example HTML output, or see the Markdown source or raw HTML (the Python and Rust examples demonstrate more advanced features at the end):
## Simple example
Markdown source test.md:
{.python .cb-run}
var = 'Hello from Python!'
var += ' $2^8 = {}$'.format(2**8)
{.python .cb-run}
print(var)
Run codebraid (to save the output, add something like -o test_out.md, and add --overwrite if it already exists):
codebraid pandoc --from markdown --to markdown test.md
Output:
Hello from Python! $2^8 = 256$
As this example illustrates, variables persist between code blocks; by default, code is executed within a single session. Code output is also cached by default so that code is only re-executed when modified.
## Features
### Comparison with Jupyter, knitr, and Pweave
Codebraid Jupyter Notebook knitr Pweave
multiple programming languages per document ✓* ✓† ✓*
multiple independent sessions per language
inline code execution within paragraphs
no out-of-order code execution ✓‡
no markdown preprocessor or custom syntax
minimal diffs for easy version control
insert code output anywhere in a document
can divide code into incomplete snippets
support for literate programming
compatible with any text editor
* One primary language from the Jupyter kernel. The IPython kernel supports additional languages via %%script magics. There is no continuity between %%script cells, because each cell is executed in a separate process. Some magics, such as those provided by PyJulia and rpy2, provide more advanced capabilities.
† knitr only provides continuity between code chunks for R, and more recently Python and Julia. Code chunks in other languages are executed individually in separate processes.
‡ Out-of-order execution is possible with R Markdown notebooks.
The table above summarizes Codebraid features in comparison with Jupyter notebooks (without extensions), knitr (R Markdown), and Pweave, emphasizing Codebraid's unique features. Here are some additional points to consider:
Jupyter notebooks — Notebooks have a dedicated, browser-based graphical user interface. Jupyter kernels typically allow the code in a cell to be executed without re-executing any preceding code, providing superior interactivity. Codebraid has advantages for projects that are more focused on creating a document than on exploratory programming.
knitr — R Markdown documents have a dedicated user interface in R Studio. knitr provides superior support for R, as well as significant Python and Julia support that includes R integration. Codebraid offers continuity between code chunks for all supported languages, as well as multiple independent sessions per language. It also provides unique options for displaying code and its output.
Easy debugging — By default, stderr is shown automatically in the document whenever there is an error, right next to the code that caused it. It is also possible to monitor code output in real time during execution via --live-output.
Simple language support — Codebraid supports Jupyter kernels. It also has a built-in system for executing code. Adding support for a new language with this system can take only a few minutes. Just create a config file that tells Codebraid which program to run, which file extension to use, and how to write to stdout and stderr. See languages/ for examples.
No preprocessor — Unlike many approaches to making code in Markdown executable, Codebraid is not a preprocessor. Rather, Codebraid acts on the abstract syntax tree (AST) that Pandoc generates when parsing a document. Preprocessors often fail to disable commented-out code blocks because the preprocessor doesn't recognize Markdown comments. Preprocessors can also fail due to the finer points of Markdown parsing. None of this is an issue for Codebraid, because Pandoc does the Markdown parsing.
No custom syntax — Codebraid introduces no additional Markdown syntax. Making a code block or inline code executable uses Pandoc's existing syntax for defining code attributes.
## Installation and requirements
Installation: pip3 install codebraid or pip install codebraid
Manual installation: python3 setup.py install or python setup.py install
Requirements:
• Pandoc 2.4+ (2.17.1.1+ recommended for commonmark_x).
• Python 3.7+ with setuptools, and bespon 0.6 (bespon installation is typically managed by pip/setup.py)
• For Jupyter support, jupyter_client and language kernels
• For YAML metadata support, ruamel.yaml (can be ruamel_yaml for Anaconda installations)
## Converting a document
Simply run codebraid pandoc <normal pandoc options>. Codebraid currently supports Pandoc Markdown (--from markdown) and CommonMark with Pandoc extensions (--from commonmark_x) as input formats.
Note that --overwrite is required to overwrite existing files. If you are using a defaults file, --from, --to, and --output must be given explicitly and cannot be inherited from the defaults file. If you are using a defaults file and converting to a standalone Pandoc Markdown document, --standalone should be given explicitly rather than being inherited from the defaults file.
codebraid should typically be run in the same directory as the document, so that the default working directory for code is the document directory.
If you are converting from Pandoc Markdown to Pandoc Markdown with --standalone (basically using codebraid to preprocess Markdown documents), note that the following YAML metadata fields and command-line options are ignored in that situation:
• header-includes and --include-in-header
• include-before and --include-before-body
• include-after and --include-after-body
• toc/table-of-contents and --toc/--table-of-contents
This is typically what you want. Usually, "include" and a table of contents are desired in a final output format like HTML or PDF, not in a Pandoc Markdown file. In the rare cases where "includes" and a table of contents are needed in Markdown documents, this can be accomplished by piping the output of codebraid through pandoc.
• --live-output — Show code output (stdout and stderr) live in the terminal during code execution. For Jupyter kernels, also show errors and a summary of rich output. Output still appears in the document as normal.
Individual sessions can override this by setting live_output=false in the document.
• --no-execute — Disables code execution. Only use available cached output.
• --only-code-output={format} — Write code output in JSON Lines format to stdout as soon as it is available, and do not create a document.
This is intended for use with Codebraid Preview, so that document previews can be updated during code execution. Currently, the only supported format is codebraid_preview. One JSON data object followed by a newline is written to stdout for each code chunk. In some cases, the data for a chunk will be resent later if the data relevant for a chunk changes (for example, if code execution fails after the first chunk runs, but in such a way that an error message needs to be attached to the first chunk). Data for a chunk is sent as soon as it is available from code processing, from cache, or from code execution (as soon as the chunk completes, typically before the session completes). Additional JSON data may be sent to provide tracking of code execution progress or information such as metadata. The JSON data provided for format codebraid_preview may change between minor versions.
## Caching
By default, code output is cached, and code is only re-executed when it is modified. The default cache location is a _codebraid directory in the working directory (directory where codebraid is run, typically the document directory). This can be modified using --cache-dir. Multiple documents can share a single cache location. A cache directory can be synced between different operating systems (such as Windows and Linux) while retaining full functionality so long as documents are in equivalent locations under the user's home directory (as resolved by os.path.expanduser()).
When multiple documents share the same cache location, each document will automatically clean up its own unused, outdated files. However, if a document is deleted or renamed, it may leave behind unused files in the cache, so it may be worth manually deleting and regenerating the cache in those circumstances. Future cache enhancements should be able to detect all unused files, making this unnecessary.
If you are working with external data that changes, you should run codebraid with --no-cache or delete the cache as necessary to prevent the cache from becoming out of sync with your data. Future releases will allow external dependencies to be specified so that caching will work correctly in these situations.
Some document-wide settings can be given in the Markdown YAML metadata. Codebraid settings must be under either a codebraid or codebraid_ key in the metadata. Pandoc will ignore codebraid_ so it will not be available to filters; this distinction should not typically be important.
To use Jupyter kernels automatically for all sessions, simply set jupyter: true. For example,
---
codebraid:
jupyter: true
---
It is also possible to set a default kernel and/or default timeout. For example,
---
codebraid:
jupyter:
kernel: python3
timeout: 120
---
A Jupyter kernel and/or timeout can still be set in the first code chunk for a given session, and will override the document-wide default.
It is also possible to set live_output: <bool> in the metadata. Additional metadata settings will be added in future releases.
## Code options
### Commands (Classes)
Code is made executable by adding a Codebraid class to its Pandoc attributes. For example, code{.python} becomes code{.python .cb-run}.
When code is executed, the output will depend on whether the built-in code execution system or a Jupyter kernel is used.
When code is executed with the built-in system, the output is equivalent to collecting all code for each session of each language, saving it to a file, and then executing it (with an added compile step for some languages). For example, running Python code is equivalent to saving it to file.py and then running python file.py, while running R code is equivalent to saving it to file.R and then running Rscript file.R. Code is not executed as it would be in an interactive session (like running python or R at the command prompt). As a result, some output that would be present in an interactive session is absent. For example, in interactive sessions for some languages, simply entering a variable returns a string representation without explicit printing, and plotting opens a separate image window or displays an image inline. Such output is absent in Codebraid unless it is also produced when code is executed as a script rather than in an interactive session. The .cb-expr command is provided for when an inline string representation of a variable is desired.
An option for interactive-style code execution with the built-in system is planned for a future release. In the meantime, many interactive-style features are available between the .cb-expr command and Jupyter kernels.
When code is executed with a Jupyter kernel, the default output will be equivalent to executing it in a Jupyter notebook. Rich output such plots, images, and LaTeX math will be displayed automatically by default. This can be customized by using the show and hide options.
All classes for making code executable are listed below. These all have the form .cb-<command>. Classes with the form .cb.<command> (period rather than hyphen) are supported for Pandoc Markdown (--from markdown), but not for commonmark_x since it has a more restricted class syntax. The forms shown below (.cb-<command>) should be preferred for compatibility across Markdown variants supported by Pandoc.
• .cb-code — Insert code verbatim, but do not run it. This is primarily useful when combined with other features like naming and then copying code chunks.
• .cb-expr — Evaluate an expression and interpret the result as Markdown. Only works with inline code. This is not currently compatible with Jupyter kernels.
• .cb-nb — Execute code in notebook mode. For inline code, this is equivalent to .cb-expr with verbatim output unless a Jupyter kernel is used, in which case rich output like plots or LaTeX will be displayed. For code blocks, this inserts the code verbatim, followed by any printed output (stdout) verbatim. If stderr exists, it is also inserted verbatim. When a Jupyter kernel is used, rich output like plots or LaTeX is also displayed.
• .cb-paste — Insert code and/or output copied from one or more named code chunks. The copy keyword is used to specify chunks to be copied. This does not execute any code. Unless show is specified, display options are inherited from the first copied code chunk.
If content is copied from multiple code chunks that are executed, all code chunks must be in the same session and must be in sequential order without any omitted chunks. This ensures that what is displayed is always consistent with what was executed.
If content is copied from another cb-paste code chunk, only a single code chunk can be copied. This reduces the indirection that is possible when displaying the output of code that has been executed. This restriction may be removed in the future.
• .cb-run — Run code and interpret any printed content (stdout) as Markdown. Also insert stderr verbatim if it exists. When a Jupyter kernel is used, rich output like plots or LaTeX is also displayed.
### Keyword arguments
Pandoc code attribute syntax allows keyword arguments of the form key=value, with spaces (not commas) separating subsequent keys. value can be unquoted if it contains only letters and some symbols; otherwise, double quotation marks "value" are required. For example,
{.python key1=value1 key2=value2}
Codebraid adds support for additional keyword arguments. In some cases, multiple keywords can be used for the same option. This is primarily for Pandoc compatibility.
#### First chunk settings
These are only permitted for the first code chunk in a session (or the first chunk for a language, if a session is not specified and thus the default session is in use).
• executable={string} — Executable to use for running or compiling code, instead of the default. This only applies to Codebraid's built-in code execution system.
• executable_opts={string} — Command-line options passed to executable. This only applies to Codebraid's built-in code execution system.
• args={string} — Command-line arguments passed to code during execution. For example, this could be used to add values to sys.argv for Python. This only applies to Codebraid's built-in code execution system.
• jupyter_kernel={string} — Jupyter kernel to use for executing code instead of Codebraid's built-in code execution system. Multiple Jupyter kernels can be used within a single document, and multiple sessions are possible per kernel. Except when otherwise specified, Jupyter kernels should be usable just like the built-in code execution system.
• jupyter_timeout={int} — Jupyter kernel timeout per code chunk in seconds. The default is 60.
• live_output={true, false} — Show code output (stdout and stderr) live in the terminal during code execution. For Jupyter kernels, also show errors and a summary of rich output. Output still appears in the document as normal. Showing output can also be enabled via the command-line option --live-output.
When live_output=false is set for a session, this setting takes precedence over the command-line option --live-output, and output will not be shown for that session.
All output is written to stderr, so stdout only contains the document when --output is not specified. Output is interspersed with delimiters marking the start of each session and the start of each code chunk. The delimiters for the start of each code chunk include source names and line numbers.
With Codebraid's built-in code execution system, the output for a code chunk may be delayed until all code in the chunk has finished executing, unless code output is line buffered or code manually flushes stdout and stderr. For example, with Python you may want to use print functions like print("text", flush=True). Another option is to use Python in line-buffered mode by setting executable_opts="-u" in the first code chunk of a session.
With Jupyter kernels, the output for a code chunk will be delayed until all code in the chunk has finished executing.
#### Execution
• complete={true, false} — By default, code chunks must contain complete units of code (function definitions, loops, expressions, and so forth). With complete=false, this is not required. Any stdout from code chunks with complete=false is accumulated until the next code chunk with complete=true (the default value), or until the end of the session, whichever comes first.
Setting complete is incompatible with outside_main=true, since the complete status of code chunks with outside_main=true is inferred automatically.
• outside_main={true, false} — This allows code chunks to overwrite the Codebraid template code when code is executed with Codebraid's built-in code execution system. It is primarily useful for languages like Rust, in which code is inserted by default into a main() template. In that case, if a session starts with one or more code chunks with outside_main=true, these are used instead of the beginning of the main() template. Similarly, if a session ends with one or more code chunks with outside_main=true, these are used instead of the end of the main() template. If there are any code chunks in between that lack outside_main (that is, default outside_main=false), then these will have their stdout collected on a per-chunk basis like normal. Having code chunks that lack outside_main is not required; if there are none, the total accumulated stdout for a session belongs to the last code chunk in the session.
outside_main=true is incompatible with explicitly setting complete. The complete status of code chunks with outside_main=true is inferred automatically.
• session={identifier-style string} — By default, all code for a given language is executed in a single, shared session so that data and variables persist between code chunks. This option allows code to be separated into multiple independent sessions. Session names must be Python-style identifiers.
#### Display
• first_number/startFrom/start-from/start_from={integer or next} — Specify the first line number for code when line numbers are displayed. next means continue from the last code in the current session.
• hide={markup, copied_markup, code, stdout, stderr, expr, rich_output, all} — Hide some or all of the elements that are displayed by default. Elements can be combined. For example, hide=stdout+stderr. Note that expr only applies to .cb-expr or .cb-nb with inline code using Codebraid's built-in code execution system, since only these evaluate an expression. rich_output is currently only relevant for Jupyter kernels.
• hide_markup_keys={key(s)} — Hide the specified code chunk attribute key(s) in the Markdown source displayed via markup or copied_markup. Multiple keys can be specified via hide_markup_keys=key1+key2.
hide_markup_keys only applies to the code chunk in which it is used, to determined the markup for that code chunk. Thus, it only affects copied_markup indirectly.
• line_numbers/numberLines/number-lines/number_lines={true, false} — Number code lines in code blocks.
• show={markup, copied_markup, code, stdout, stderr, expr, rich_output, none} — Override the elements that are displayed by default. expr only applies to .cb-expr and to .cb-nb with inline code using Codebraid's built-in code execution system, since only these evaluate an expression. Elements can be combined. For example, show=code+stdout.
Each element except rich_output can optionally specify a format from raw, verbatim, or verbatim_or_empty. For example, show=code:verbatim+stdout:raw.
• raw means interpreted as Markdown.
• verbatim produces inline code or a code block, depending on context. Nothing is produced if there is no content (for example, nothing in stdout.)
• verbatim_or_empty produces inline code containing a single non-breaking space or a code block containing a single empty line in the event that there is no content. It is useful when a placeholder is desired, or a visual confirmation that there is indeed no output.
For rich_output, the format is specified as one or more abbreviations for the mime types of the output to be displayed. For example, rich_output:plain will display text/plain output if it exists, and otherwise nothing. rich_output:png|plain will display a PNG image if it exists, or otherwise will fall back to plain text if available. The following formats are currently supported:
• latex (corresponds to text/latex)
• html (text/html)
• markdown (text/markdown)
• plain (text/plain)
• png (image/png)
• jpg and jpeg (image/jpeg)
• svg (image/svg+xml)
• pdf (application/pdf)
For rich_output formats with a text/* mime type (latex, html, markdown, plain), it is possible to specify whether they are displayed raw, verbatim, or verbatim_or_empty. For example, show=rich_output:latex:raw and show=rich_output:latex:verbatim. raw treats latex and html as raw content with those formats embedded within Markdown. raw treats markdown and plain as Markdown. When a display style is not specified, all rich_output formats with a text/* mime type are displayed raw by default, except for plain which is displayed verbatim.
markup displays the Markdown source for the inline code or code block. Because the Markdown source is not available in the Pandoc AST but rather must be recreated from it, the Markdown source displayed with markup may use a different number of backticks, quote attribute values slightly differently, or contain other insignificant differences from the original document.
copied_markup displays the Markdown source for code chunks copied via copy.
expr defaults to raw if a format is not specified. rich_output defaults to latex|markdown|png|jpg|svg|plain. All others default to verbatim.
• example={bool} — Insert a code block containing the Markdown source of the code chunk, followed by the rest of the output as normal. This is only valid for inline code if the code is in a paragraph by itself. This option is currently not compatible with --only-code-output and Codebraid Preview. This option is intended primarily for documentation about Codebraid.
#### Copying
• copy={chunk name(s)} — Copy one or more named code chunks. When copy is used with a command like .cb-run that executes code, only the code is copied, and it is executed as if it had been entered directly. When copy is used with .cb-code, only the code is copied and nothing is executed. When copy is used with .cb-paste, both code and output are copied, and nothing is executed. Multiple code chunks may be copied; for example, copy=name1+name2. In that case, the code from all chunks is concatenated, as is any output that is copied. Because copy brings in code from other code chunks, the actual content of a code block or inline code using copy is discarded. As a result, this must be empty, or a space or underscore can be used as a placeholder.
• name={identifier-style string} — Name a code chunk so that it can later be copied by name. Names must be Python-style identifiers.
#### Including external files
• include_file={path} — Include the specified file. A leading ~/ or ~<user>/ is expanded to the user's home directory under all operating systems, including under Windows with both slashes and backslashes.
When include_file is used with a command like .cb-run that executes code, the file is included and executed as part of the current session just as if the file contents had been entered directly. When include_file is used with .cb-code, the file is included and displayed just as if it had been entered directly. Because include_file brings in code from another file, the actual content of a code block or inline code using include_file is discarded. As a result, this must be empty, or a space or underscore can be used as a placeholder.
• include_encoding={encoding} — Encoding for included file. The default encoding is UTF-8.
• include_lines={lines/line ranges} — Include the specified lines or line ranges. For example, 1-3,5,7-9,11-. Line numbers are one-indexed. Line ranges are inclusive, so 1-3 is 1 up to and including 3. If a range ends with a hyphen, like 11-, then everything is included from the line through the end of the file.
Cannot be combined with other include options that specify what is to be included.
• include_regex={regex} — Include the first segment of the file that matches the provided regular expression.
Keep in mind that Pandoc's key-value attributes evaluate backslash escapes in values whether or not the values are quoted with double quotation marks, so two levels of backslash-escaping are always necessary (one for Pandoc's strings, one for the regex itself; there are no raw strings). Regular expressions use multiline mode, so ^/\$ match the start/end of a line, and \A/\Z can be used to match the start/end of the file. Regular expressions use dotall mode, so . matches anything including the newline \n; use [^\n] when this is not desired.
Cannot be combined with other include options that specify what is to be included.
• include_start_string={string} — Include everything from the first occurrence of this string onward.
Can only be combined with other include options that specify the end of what is to be included.
• include_start_regex={regex} — Include everything from the first match of this regex onward.
Can only be combined with other include options that specify the end of what is to be included. See include_regex for notes on regex usage.
• include_after_string={string} — Include everything after the first occurrence of this string onward.
Can only be combined with other include options that specify the end of what is to be included.
• include_after_regex={regex} — Include everything after the first match of this regex onward.
Can only be combined with other include options that specify the end of what is to be included. See include_regex for notes on regex usage.
• include_before_string={string} — Include everything before the first occurrence of this string.
Can only be combined with other include options that specify the start of what is to be included. If the start is specified, then the first occurrence after this point is used, rather than the first occurrence in the overall file.
• include_before_regex={regex} — Include everything before the first match of this regex.
Can only be combined with other include options that specify the start of what is to be included. If the start is specified, then the first match after this point is used, rather than the first match in the overall file. See include_regex for notes on regex usage.
• include_end_string={string} — Include everything through the first occurrence of this string.
Can only be combined with other include options that specify the start of what is to be included. If the start is specified, then the first occurrence after this point is used, rather than the first occurrence in the overall file.
• include_end_regex={regex} — Include everything through the first match of this regex.
Can only be combined with other include options that specify the start of what is to be included. If the start is specified, then the first match after this point is used, rather than the first match in the overall file. See include_regex for notes on regex usage.
## Project details
Uploaded source
Uploaded py3
|
2022-10-01 11:01:35
|
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http://mathhelpforum.com/advanced-algebra/103338-2-x-2-matrix-representing-rotation-print.html
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# 2 x 2 matrix representing a rotation
• Sep 20th 2009, 02:50 PM
Harry1W
2 x 2 matrix representing a rotation
I'm trying to work through the following theorem in Brannan et al.'s book "Geometry" (you might wish to skip ahead to the bold text "The Source of My Problem", so that you don't waste your time if the context is not necessary):
"A $2 \times 2$ matrix $\mathbf{P}$ represents a roation of $\mathbb{R}^2$ about the origin if and only if it satisfies the following two conditions:
(a) $\mathbf{P}$ is orthogonal;
(b) $\mathrm{det} \, \mathbf{P} = 1$."
So, the proof proceeds as follows:
A matrix $\mathbf{P}$ represents a rotation if and only if it is of the form
$\left( \begin{array}{cc}
\cos\theta & - \sin\theta \\
\sin\theta & \cos\theta
\end{array} \right)$
. (*)
I'm accustomed to, and am happy with, this fact (well, the 'if' part), and I am happy that it is easy to verify that if it is of this form, then $\mathbf{P}$ satisfies conditions (a) and (b).
Next, the proof procees to let $\mathbf{P} = \left( \begin{array}{cc}
a & b \\
c & d
\end{array} \right)$
be a matrix that satisfies conditions (a) and (b).
Then, since $\mathbf{P}$ is orthgonal, the vector $\left( \begin{array}{c}
a \\
c
\end{array} \right)$
has length 1; that is, $a^2 + c^2 = 1$. Thus, there is a number $\theta$ for which
$a = \cos\theta$ and $c = \sin\theta$.
Also, since $\mathbf{P}$ is orthogonal, the vectors $\left( \begin{array}{c}
a \\
c
\end{array} \right) = \left( \begin{array}{c}
\cos\theta \\
\sin\theta
\end{array} \right)$
and $\left( \begin{array}{c}
b \\
d
\end{array} \right)$
are orthogonal; that is, $\left( \begin{array}{cc}
\cos\theta & \sin\theta
\end{array} \right)
\left( \begin{array}{c}
b \\
d
\end{array} \right) = 0$
(all fine by me so far) or
*******************************
The Source of My Problem:
$\cos\theta \cdot b + \sin\theta \cdot d = 0$.
Then the proof goes on to say,
"So there exists some number $\lambda$, say, such that
$b = - \lambda \sin\theta$ and $d = \lambda \cos\theta$."
Would somebody be able to explain the rationale behind this? Why must $b$ and $d$ be of this form? I can see from the rest of the proof that it would be very helpful if they would be - but I'm not sure why they have to be. To try to explain my confusion, I think I would be happy saying that there must be some numbers, $\lambda$ and $\mu$, say, such that
$b = - \lambda \sin\theta$ and $d = \mu \cos\theta$.
But I can't think of any reason for which $\lambda$ should equal $\mu$.
-----
For anyone interested, the rest of the proof is as follows:
"Then since $\mathrm{det} \, \mathbf{P} = 1$, we have
$1 = ad - bc = \lambda \cos ^2 \theta + \lambda \sin ^2 \theta$,
so that $\lambda = 1$. It follows that $\mathbf{P}$ must be of the form (*), and so represent a rotation of $\mathbb{R} ^2$ about the origin."
• Sep 21st 2009, 05:22 AM
clic-clac
If $\cos\theta =0$ or $\sin\theta =0,$ then $b=-\sin\theta,\ d=\cos\theta$ is the only solution (because of (a) and (b)).
Assume both $\cos\theta$ and $\sin\theta$ are different from $0.$
Then any pair of real $(b,d)$ can be written under the form $(-\lambda\sin\theta, \mu\cos\theta)$ for some $\lambda,\mu \in\mathbb{R}.$
Since $\cos\theta\sin\theta(\mu-\lambda) = \cos\theta .b +\sin\theta .d=0$ and $\cos\theta \sin\theta \neq 0,$ we have $\lambda =\mu\ .$
• Sep 22nd 2009, 05:29 PM
Harry1W
Thanks so much for that. However, I'm struggling to see the reason why,
Quote:
Originally Posted by clic-clac
If $\cos\theta =0$ or $\sin\theta =0,$ then $b=-\sin\theta,\ d=\cos\theta$ is the only solution (because of (a) and (b)).
I'm sorry to be slow on the uptake, but I'd really appreciate it if you could spell this out for me. Thanks again.
• Sep 23rd 2009, 01:38 AM
clic-clac
No problem. There may be a better proof, but that's what I was thinking when writting this one:
if $a=\cos\theta=0,$ then $c=\sin\theta\in\{-1,1\}$ and $-cb=\det\mathbf{P}=1$ leads to $b=-c=-\sin\theta.$ Then orthogonality imposes that $d=0=\cos\theta.$
A similar argument for the case $b=\sin\theta=0$ ends the proof.
• Sep 23rd 2009, 05:59 AM
HallsofIvy
Quote:
Originally Posted by Harry1W
I'm trying to work through the following theorem in Brannan et al.'s book "Geometry" (you might wish to skip ahead to the bold text "The Source of My Problem", so that you don't waste your time if the context is not necessary):
"A $2 \times 2$ matrix $\mathbf{P}$ represents a roation of $\mathbb{R}^2$ about the origin if and only if it satisfies the following two conditions:
(a) $\mathbf{P}$ is orthogonal;
(b) $\mathrm{det} \, \mathbf{P} = 1$."
So, the proof proceeds as follows:
A matrix $\mathbf{P}$ represents a rotation if and only if it is of the form
$\left( \begin{array}{cc}
\cos\theta & - \sin\theta \\
\sin\theta & \cos\theta
\end{array} \right)$
. (*)
I'm accustomed to, and am happy with, this fact (well, the 'if' part), and I am happy that it is easy to verify that if it is of this form, then $\mathbf{P}$ satisfies conditions (a) and (b).
Next, the proof procees to let $\mathbf{P} = \left( \begin{array}{cc}
a & b \\
c & d
\end{array} \right)$
be a matrix that satisfies conditions (a) and (b).
Then, since $\mathbf{P}$ is orthgonal, the vector $\left( \begin{array}{c}
a \\
c
\end{array} \right)$
has length 1; that is, $a^2 + c^2 = 1$. Thus, there is a number $\theta$ for which
$a = \cos\theta$ and $c = \sin\theta$.
Also, since $\mathbf{P}$ is orthogonal, the vectors $\left( \begin{array}{c}
a \\
c
\end{array} \right) = \left( \begin{array}{c}
\cos\theta \\
\sin\theta
\end{array} \right)$
and $\left( \begin{array}{c}
b \\
d
\end{array} \right)$
are orthogonal; that is, $\left( \begin{array}{cc}
\cos\theta & \sin\theta
\end{array} \right)
\left( \begin{array}{c}
b \\
d
\end{array} \right) = 0$
(all fine by me so far) or
One of the conditions for an orthogonal matrix has been dropped here: we must also have $b^2+ d^2= 1$
Quote:
*******************************
The Source of My Problem:
$\cos\theta \cdot b + \sin\theta \cdot d = 0$.
Then the proof goes on to say,
"So there exists some number $\lambda$, say, such that
$b = - \lambda \sin\theta$ and $d = \lambda \cos\theta$."
Would somebody be able to explain the rationale behind this? Why must $b$ and $d$ be of this form? I can see from the rest of the proof that it would be very helpful if they would be - but I'm not sure why they have to be. To try to explain my confusion, I think I would be happy saying that there must be some numbers, $\lambda$ and $\mu$, say, such that
$b = - \lambda \sin\theta$ and $d = \mu \cos\theta$.
But I can't think of any reason for which $\lambda$ should equal $\mu$.
"Then since $\mathrm{det} \, \mathbf{P} = 1$, we have
$1 = ad - bc = \lambda \cos ^2 \theta + \lambda \sin ^2 \theta$,
so that $\lambda = 1$. It follows that $\mathbf{P}$ must be of the form (*), and so represent a rotation of $\mathbb{R} ^2$ about the origin."
|
2016-10-27 22:18:37
|
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|
https://www.physicsforums.com/threads/modified-harmonic-oscillator-probabilities.763319/
|
# Modified Harmonic Oscillator probabilities
1. Jul 25, 2014
### ma18
1. The problem statement, all variables and given/known data
The e-functions for n=0,1,2 e-energies are given as
psi_0 = 1/(pi^1/4 * x0^1/2)*e^(x^2/(2*x0^2)
psi_1 =...
psi_2 =...
The factor x0 is instantaneously changed to y= x0/2. This means the initial wavefunction does not change.
Find the expansions coefficients of the initial state to the three lowest states of the modified potential. What energy measurements would you find and with what probability.
2. Relevant equations
Prob En = c_n ^2
En = hbar*omega*(n+0.5)
3. The attempt at a solution
I can find the new energies:
En = hbar*omega*(n+0.5) = m*omega^2*x_0^2*(n+1/2)
Putting in y = x_0 instead we get:
En = m*omega^2*y^2*(n+1/2)
= 1/4 m*omega^2*x_0^2*(n+1/2)
= 1/4 hbar*omega*(n+0.5)
For the expansion coefficients I am not sure what to put into the formula:
cn = integral from -inf to inf: phi* * psi dx
I know that the modified eqn to psi_0 is:
psi_m0 = sqrt(2)*e^2 * psi_0
But if I put this in I just get sqrt(2)*e^4 and that doesn't make sense for the probs (>1)
Last edited: Jul 25, 2014
2. Jul 25, 2014
### BiGyElLoWhAt
Yes, assuming that $\dot{x}_0$ = 0. Otherwise the phase shift will change.
Is this a typo? $\Psi^* \Psi$
Honestly I'm not sure what you're trying to do, if that was supposed to be a complex conjucate (that's what it looks like to me) Your psi functions are real, so... Again, though, I really don't know what your trying to do. I'm not going to guarantee to be able to help you, but could you give a clearer explaination of the problem?
It almost looks as thought you're trying to normalize a function to solve for the coefficients, but I'm not sure.
3. Jul 25, 2014
### TSny
This isn't correct. You can't factor out e2 like that.
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2017-08-21 20:50:04
|
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|
https://tex.stackexchange.com/questions/224082/chapter-and-section-format
|
# Chapter and Section format
I want my chapter titles to look as they would with the code below.
\documentclass[12pt]{book}
\pagestyle{plain}
\usepackage{amssymb,latexsym}
\usepackage{amsmath}
\usepackage{amsthm}
\begin{document}
\centering
CHAPTER II \\
SUBSQUARES
\end{document}
\renewcommand{\chapter}{\uppercase{CHAPTER} {\Roman{chapter}} \\}
• Look at the sectsty package, or look on this site for examples of titlesec usage. Since you are using the book class, you might consider switching to memoir, which provides its own interface for adjusting the headings (or a KOMA class for the same reason). There is too much built-in code (for spacing, font, counters, etc.) in the original definition of the section headings to alter them simply with a \renewcommand. – musarithmia Jan 20 '15 at 19:08
• Do you want the same (centering + CAPITAL LETTERS) for \section? What about \subsection...? – Werner Jan 20 '15 at 19:48
I think that using the sectsty package is simpler.
\usepackage{sectsty}
\renewcommand{\thechapter}{\Roman{chapter}}
\chapterfont{\centering\MakeUppercase}
MWE
\documentclass[12pt]{book}
\pagestyle{plain}
\usepackage{amssymb,latexsym}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{sectsty}
\renewcommand{\thechapter}{\Roman{chapter}}
\chapterfont{\centering\MakeUppercase}
\begin{document}
\chapter*{Unnumbered Chapter}
\chapter{Numbered Chapter}
\appendix
\chapter{An Appendix}
\end{document}
Output
Here is a crude way of implementing your requirement. It uses etoolbox to patch the macros responsible for creating the chapter headings. There are two macros, one for \chapter and one for \chapter*. In each of these macros, an adjustment is made to the chapter title (argument #1) as well as the horizontal alignment - changing from \raggedright to \centering. As such, a total of 4 patches are made.
\documentclass{book}
\usepackage{etoolbox}
\makeatletter
\renewcommand{\thechapter}{\Roman{chapter}}
\renewcommand{\@chapapp}{\MakeUppercase{Chapter}}
% \patchcmd{<cmd>}{<search>}{<replace>}{<succes>}{<failure>}
The other changes to \thechapter sets the output to print a \Roman format of the chapter counter, while an adjustment to \@chappapp sets the Chapter prefix in CAPITAL LETTERS. This may need some minor adjustments if you also have an Appendix chapter.
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2020-01-28 00:01:50
|
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|
https://www.vrcbuzz.com/allcalculators/statistics-calculators/
|
## Sample size calculator to estimate proportion
Sample size required to estimate proportion The formula to estimate the sample size required to estimate the proportion is $$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$ where, $p$ …
## Sample size calculator to test hypothesis about mean
Sample size to test hypothesis about mean In inferential statistics, knowing how many samples to take from the population is extremely important. In this tutorial …
## Sample size calculator to estimate mean
Sample size to estimate mean In inferential statistics, knowing how many samples to take from the population is extremely important. In this tutorial we will …
## Mean median mode calculator for grouped data
Mean, median and mode Mean, median, mode are the measures of central tendency. They are also known as averages. Averages are the measures which condense …
|
2021-04-12 14:47:50
|
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|
https://mathoverflow.net/questions/322703/monotonicity-scaling-of-sturm-liouville-eigenvalues
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# Monotonicity/Scaling of Sturm-Liouville Eigenvalues
Consider the regular Sturm-Liouville eigenvalue equation $$\frac{d}{dx}(p_t(x)f^\prime(x))=\lambda_t f(x)$$ for $$p_t\in\mathcal{C}^\infty([0,1])$$ with Dirichlet boundary conditions on $$[0,1]$$. Here $$t\geq 0$$ is a parameter and let's say $$t\mapsto p_t(x)$$ is smooth for all $$x\in[0,1]$$ and $$p_0=1$$. Denote the smallest eigenvalue $$\lambda_t$$ by $$\lambda_t^0$$. So $$\lambda_0^0=\pi^2$$.
Can anything be said about the mapping $$t\mapsto \lambda_t^0$$? E.g. is it monotone in $$t$$ (likely to depend on $$p_t$$), what's its scaling behavior, or is it smooth?
• The setup is too general for monotonicity/scaling since you can always destroy such properties by reparametrizing. Smoothness is no big problem I think, should follow quickly from the implicit function theorem. Feb 7, 2019 at 22:57
• Thank you for the comment. I suspected that there is no general statement possible. Do you know any references which work out monotonicity for a particular example? Is there anything known about the relationship between $p_t$ and the eigenvalue $\lambda_t^0$? Feb 7, 2019 at 23:37
The assumption of smoothness of $$t\mapsto p_t$$ cannot imply monotonicity, as @Christian Remling noted in his comment.
But if you assume that $$t\mapsto p_t$$ is monotone (pointwise) then $$t\mapsto\lambda_t$$ is also monotone.
Sturm-Liouville eigenvalue problems have the following mechanical interpretation. Your eiegnalue problem describes the fundamental frequency $$\lambda$$ of a string with fixed ends, of constant density, and variable stiffness (Hooke's const) $$p(x)$$.
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2022-08-15 02:58:34
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https://chemistry.stackexchange.com/questions/107343/evaluate-mass-of-salt-needed-to-add-to-a-buffer-solution-knowing-only-ph
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# Evaluate mass of salt needed to add to a buffer solution knowing only pH
This is an exercise taken from an old exam, I'm struggling with the resolution.
To a solution of a generic weak acid $$\ce{HA}$$ were added $$\pu{2.40 g}$$ of a potassium salt of the $$\ce{KA}$$. The solution has $$\mathrm{pH} = 4.8$$. Evaluate how many grams of salt are needed if we want to shift the solution to a $$\mathrm{pH} = 5$$.
Neither $$K_\mathrm{a}$$ nor $$K_\mathrm{b}$$ are given, nor anything. The only thing I got is just confusion.
I think there is all the data you need. $$\mathrm{pH}$$ of a buffer formed by a weak acid $$\ce{HA}$$ and its potassium salt $$\ce{KA}$$ can be found as
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{KA})}{C(\ce{HA})}}$$
On the other hand
$$C(\ce{KA}) = \frac{m(\ce{KA})}{M(\ce{KA})V}$$
where $$m$$ and $$M$$ are mass and molecular mass of $$\ce{KA}$$; $$V$$ is volume. In general, first equation can be rewritten as
\begin{align} \mathrm{pH}_i &= \mathrm{p}K_\mathrm{a} + \log{C_i(\ce{KA})} - \log{C(\ce{HA})} \\ &= \mathrm{p}K_\mathrm{a} + \log{m_i(\ce{KA})} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} \end{align}
Rearranging:
$$\mathrm{pH}_i - \log{m_i(\ce{KA})} = \mathrm{p}K_\mathrm{a} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} = \mathrm{const}$$
so that now we can equate conditions for both solutions and find the mass:
$$\mathrm{pH}_1 - \log{m_1(\ce{KA})} = \mathrm{pH}_2 - \log{m_2(\ce{KA})}$$ $$\log{\frac{m_2(\ce{KA})}{m_1(\ce{KA})}} = \mathrm{pH}_2 - \mathrm{pH}_1$$ $$m_2(\ce{KA}) = m_1(\ce{KA})\cdot10^{\mathrm{pH}_2 - \mathrm{pH}_1} = \pu{2.40 g}\cdot10^{5.0-4.8}\approx\pu{3.80 g}$$
So, in order to achieve $$\mathrm{pH} = 5.0$$, one has to add $$\pu{3.80 g} - \pu{2.40 g} = \pu{1.40 g}$$ of potassium salt.
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2020-02-23 05:33:26
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http://hittax.com.au/usda-fresh-zlegmuh/99467c-geogebra-imaginary-numbers
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# geogebra imaginary numbers
Complex Numbers. This is called algebraic form of complex number. Complex numbers, XY plane. imaginary ( ) Returns the imaginary part of a given complex number. GeoGebra does not support complex numbers directly, but you may use points to simulate operations with complex numbers. Imaginary Numbers Are Real [Part 1: Introduction] - Duration: 5:47. I am interesting in seeing what some equations look like when they are plotted 3-dimentionally, with one axis real numbers, the second axis imaginary numbers (thus the complex plane), and the third axis real numbers. Discover Resources. The value is displayed at the top in both Re/Im and polar (r/theta) notation. Notational conventions. Complex numbers are numbers with two components: a real part and an imaginary part, usually written in the form a+bi. Drag point P to graph each complex number, then click submit to check your answer. Imaginary number, i = sqrt{-1} In the XY plane, a + bi is point (a, b). When you have answered correctly go to the next question. q = 3 + 4i), but not in the CAS. Considering the complex function f used in the previous section, we can easily get their 3D components graphs using GeoGebra writing its real component as f1(x,y)=real((x + yi) 2) and its imaginary component as f2(x y)=imaginary ((x + yi) 2) . Imaginary numbers were ‘invented’ (or discovered if you prefer) because mathematicians wanted to know if they could think of square root of negative numbers, particularly, the root of the equation (that is, which is the same as finding the ).). Example: imaginary (17 + 3 ί) yields 3. You can also use the tool Complex Number. So I would say the answer to your question is yes and no. Subsequently, the potential of the dynamic color GeoGebra … For example, $5+2i$ is a complex number. Contact us: office@ ... Graphing Complex Numbers. what are complex numbers? Topic: Complex Numbers, Numbers. What does these complex numbers represent in the real life. complex are numbers that can be expressed in the for a+bi, where a and b are real numbers and i is the imaginary unit, using the equation i^2 = -1. in this expression a is the real part and b is the imaginary part of the complex number. Why does it have a problem with imaginary numbers, for example x^2 1=0 gives no result and √-1 is u How to get a "number" as a "number of certain type of objects" How to control the increment of a … Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. You need JavaScript enabled to view it. About GeoGebra. GeoGebra Applets Master List; Determine the Intercepts of a Line Stated in Standard Form; Graph a Line Given in Standard Form; Create a Line with a Given Slope; Figure 10 – Application of domain coloring using GeoGebra to visualize Riemann sphere and Möbius Transformations. with the understanding that it represents a + ib, where i = sqrt (-1). Lee Stemkoski 13,280 views. However GeoGebra's Algebra pane has no in-built understanding of i = sqrt (-1). By … Complex numbers can be represented graphically using an Argand diagram. 3. GeoGebra is obviously capable of representing this number pair as a point in the Graphical pane. w=2+3i. Any complex number can be represented as a number pair (a, b). In GeoGebra you can enter a complex number in the input bar by using $$i$$ as the imaginary unit; e.g. Unless you are typing the input in CAS View or you defined variable i previously, variable i is recognized as the ordered pair i = (0, 1) or the complex number 0 + 1ί. i is imaginary number and is equal to square root of minus 1. Author: Peter Johnston. This email address is being protected from spambots. http://wiki.geogebra.org/s/en/index.php?title=Complex_Numbers&oldid=50559. 3 / (0 + 1ί) gives you the complex number 0 - 3ί. However GeoGebra's Algebra pane has no in-built understanding of i = sqrt (-1). The quantum numbers derived from the imaginary unit are unusual but a simple conversion allows the derivation of electric charge and isospin, quantum numbers for two families of particles. Although you graph complex numbers much like any point in the real-number coordinate plane, complex numbers aren’t real! Imaginary number, i = sqrt(-1} In the XY plane, a + b i corresponds to the point (a, b). In the complex plane, x axis = real axis, y axis = imaginary axis. The number i, while well known for being the square root of -1, also represents a 90° rotation from the real number line. ... 17 GeoGebra Applets. Complex Numbers. About GeoGebra. Understanding Cartesian Coordinates Through GeoGebra: A Quantitative Study Demonstration of Complex Numbers in Polar Coordinates Despite infinity of real numbers and all the wealth of its structures that it contained, -1 is not a square number in real numbers cluster (King, 2004). Examples will include complex multiplication and division, linear and linear fractional functions, and some calculus concepts. Imaginary Numbers; Complex Numbers; Additional Practice Related to Imaginary and Complex Numbers; 7 Lines. Imaginary Numbers graph. The multiple Windows of GeoGebra, combined with its ability of algebraic computation with complex numbers, allow the study of the functions defined from ℂ to ℂ through traditional techniques and by the use of Domain Colouring. Examples: 3 + (4 + 5ί) gives you the complex number 7 + 5ί. The number appears in the graphics view as a point and you can move it around. Showing complex as polar changes calculation result, Help with defining complex numebers using an input box, How to divide two complex numbers in Geogebra CAS. In this representation i is called imaginary unit, a is real part and b is imaginary part.If imaginary part of complex number not 0 then such number is called imaginary, for example 3+2i.If a=0 and b!=0 then complex number is called purely imaginary. In GeoGebra, complex numbers are presented by related vectors. You need JavaScript enabled to view it. 3 * (1 + 2ί) gives you the complex number 3 + 6ί. Thank you. Is there a way to represent imaginary numbers with GeoGebra, in the format of a + bi where a = real and b = imaginary components. So I would say the answer to your question is yes and no. This association to elementary particles is not final because further understanding of the role played by the imaginary … When you have answered correctly go to the next question. 9:45. Esposito Right Isosceles Triangle 9 Point Circle; graph of two function This email address is being protected from spambots. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Drag point Z in the complex plane. So, too, is $3+4i\sqrt{3}$. Why are complex functions rendered the way they are. Use checkboxes to display the complex conjugate Z* and/or the real and imaginary components. 3 - (4 + 5ί) gives you the complex number -1 - 5ί. In plane geometry, complex numbers can be used to represent points, and thus other geometric objects as well such as lines, circles, and polygons. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). in Geogebra The use of dynamic colors associated with a point allowed Rafael Losada (2009) and Antonio Ribeiro obtain the first representations of fractal images involving complex numbers (Breda, et al, 2013, p. 63). GeoGebra also recognizes expressions involving real and complex numbers. GeoGebra is obviously capable of representing this number pair as a point in the Graphical pane. As we know, A complex number is expressed as z = a + b i: where a is the real part, b i is imaginary part, and a and b are constants. The imaginary unit ί can be chosen from the symbol box in the Input Bar or written using Alt + i. GeoGebra doesn't offer a Complex Number mode. Slide Number 6. Complex numbers, XY plane. Drag point P to graph each complex number, then click submit to check your answer. Note: The complex ί is obtained by pressing ALT + i. As there is no such command as IsComplex you currently have to employ a small trick to check if the number a is complex: complex = IsDefined[sqrt(a) + sqrt(-a)] ∧ (a ≠ 0). C omplex number z can be represented in the form z=a+bi. GeoGebra’+Complex’Number’ Arithme4c:’Implemen4ng’CCSSM David Erickson, University of Montana Armando Martinez-Cruz, CSU Fullerton NCTM Conference 3D graphic windows of GeoGebra and representation of the components functions of a complex function. Then of course there is i = sqrt (-1). The following commands and predefined operators can also be used: GeoGebra also recognizes expressions involving real and complex numbers. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part. A complex number is expressed as z equals a plus bi. See also real … But it could, no doubt, still be useful in the teaching of Complex Numbers. This is all we can do with the most recent version of GeoGebra 4.9 .The next step of our research is the identification of the improvements that should be performed in GeoGebra to visualize effectively the action of the Möbius Transformation in the Riemann sphere. Sometimes you may want to check if a number is treated as complex number in GeoGebra, as function such as x() and y() do not work with real numbers. Is such software available either online or free-downloadable? Let us look at complex numbers. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. is imaginary unit and we mark it with:(0,1)=i where : . a is the real part; bi is imaginary part;a and b are constants. (x, y) pairs are used to improve these numbers which we need. There are some GeoGebra functions that work on both points and complex numbers. Drawing the Mandlebrot Set with GeoGebra - part 1 - Duration: 9:45. This also means, that you can use this variable i in order to type complex numbers into the Input Bar (e.g. GeoGebra doesn't offer a Complex Number mode. I googled, wikied etc., but I cant understand what it is because, may be i cant understand clearly what they said, or I have these questions in my mind because of little understanding. In complex analysis, the complex numbers are customarily represented by the symbol z, which can be separated into its real (x) and imaginary (y) parts: = + for example: z = 4 + 5i, where x and y are real numbers, and i is the imaginary unit.In this customary notation the complex number z corresponds to the point (x, y) in the Cartesian plane. Using GeoGebra, I will demonstrate with dynamic diagrams important properties of complex arithmetic and functions. Numbers. Is imaginary number and is equal to square root of minus 1 ( 17 + ί. Used: GeoGebra also recognizes expressions involving real and complex numbers represent the! Numbers ; Additional Practice Related to imaginary and complex numbers ί is obtained by pressing ALT + i z=a+bi! Doubt, still be useful in the real-number coordinate plane, complex Calculator! 17 + 3 ί ) yields 3: GeoGebra also recognizes expressions involving real and complex numbers when have. But it could, no doubt, still be useful in the complex 3... No doubt, still be useful in the real part ; a b... Xy plane, complex numbers aren ’ t real i is imaginary number, then click submit check. ( i\ ) as the imaginary unit ; e.g it represents a + ib, i. It with: ( 0,1 ) =i where: real [ part 1: Introduction ] -:... And polar ( r/theta ) notation, too, is [ latex ] 3+4i\sqrt { 3 } [ ]... I = sqrt { -1 } in the XY plane, a + bi is (... Understanding of i = sqrt ( -1 ) mark it with: ( 0,1 ) =i where: the question. @... Graphing complex numbers directly, but not in the complex number > ) Returns the imaginary unit can. Is yes and no ( a, b ) the real life complex. Does not support complex numbers: the complex plane, a + bi is imaginary part a. Functions of a given complex number GeoGebra does not support complex numbers of the functions. ) as the imaginary unit ί can be chosen from the symbol box in complex. ( -1 ) of a complex function < complex number 3 + ( 4 + 5ί ) gives the., a + ib, where i = sqrt ( -1 ) - 3ί ( x, y axis imaginary. And evaluates expressions in the real-number coordinate plane, complex numbers are real [ part 1 - Duration:.. To square root of minus 1 in-built understanding of i = sqrt -1. The Mandlebrot Set with GeoGebra - part 1: Introduction ] - Duration 5:47. Useful in the form z=a+bi real life then click submit to check your answer +! By pressing ALT + i of minus 1 axis, y ) are... Which we need: 3 + 6ί could, no doubt, still be useful in the teaching of numbers. ), but not in the Set of complex numbers Calculator - Simplify complex expressions using algebraic rules step-by-step website... Bar by using \ ( i\ ) as the imaginary unit ; e.g and components... ) notation linear fractional functions, and some calculus concepts any point in the view! Expressions in the CAS next question the next question website uses cookies ensure! /Latex ] is a complex number pressing ALT + i written using +! Domain coloring using GeoGebra to visualize Riemann sphere and Möbius Transformations this also means, that you can it. Imaginary unit ; e.g to ensure you get the best experience graphically using an diagram! Calculator does basic arithmetic on complex numbers much like any point in the Set of complex numbers,. B ) ( < complex number is expressed as z equals a plus.. In both Re/Im and polar ( r/theta ) notation number 0 - 3ί the plane. Graphic windows of GeoGebra and representation of the components functions of a complex... At the top in both Re/Im and polar ( r/theta ) notation Bar e.g.
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2021-04-12 16:06:17
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https://www.physicsforums.com/threads/calculation-of-the-natural-linewidth.769492/
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Calculation of the Natural Linewidth
1. Sep 7, 2014
blaisem
Hi, I am trying to understand how one determines the natural linewidth. On my assignment, I am only given an energy (589.1 nm transition in sodium). I have two sources that I have found that seem to contradict each other:
Source 1: Slides 5 and 6
Source 2: Hyperphysics
If I plug in either the lifetime or the energy value provided in the example from the powerpoint (slide 6) into the Hyperphysics calculator, the corresponding value isn't consistent with slide 6.
I am confused on which is the correct formula, as well as how one determines the natural linewidth without knowing the lifetime of a transition. Is the energy of the transition actually relevant?
Can anyone please advise? Thank you for your time and help.
2. Sep 7, 2014
Staff: Mentor
I didn't check the calculations in detail, but in the slides, both the initial and final states are considered to have finite lifetimes. This may be the source of the discrepancy.
They are usually measured from spectra.
If you need the numbers for sodium, check http://steck.us/alkalidata/.
3. Sep 7, 2014
blaisem
Hi DrClaude. Thanks for your response and the link. Maybe it was implied I had to look up the lifetime. 3 questions, if you have time, since I am having trouble wrapping my head around it conceptually:
1.Hyperphysics provides a relationship of:
2E = Gamma = (reduced plank constant / lifetime)
where Gamma is the width of the natural broadening
If I substitute the given transition energy of 589.1 nm into E, I get a gamma of 4.2 eV; if I use the lifetime from your source (16.2 ns), I get a gamma of 41 nano Ev.
I am confused about the role of the energy of the transition in natural broadening. Am I substituting the wrong value for energy into the Hyperphysics formula?
2. What would be the correct value of E? Would it be the absolute energy uncertainty of the initial state, and the transition energy I have is entirely irrelevant to determining the natural linewidth?
3. Is the formula in the powerpoint (first link) more precise than Hyperphysics? It seems to be a more complicated representation of natural broadening, implying that the Heisenberg Uncertainty Principle as it was presented in Hyperphysics may be a more superficial description of natural broadening. Is my understanding of this correct?
Last edited: Sep 7, 2014
4. Sep 8, 2014
Staff: Mentor
Yes. In the formula, it is $\Delta E$, the uncertainty on the energy, not the energy of the transition.
If both the initial and final states have finite lifetimes, then both widths must be taken into account (as described in the slides). You have an uncertainty in both the energy of the upper state and the lower state. But the actual value of the center of the peak (the energy "before" taking into account the uncertainty) is not relevant.
Apart from the fact that it takes into account the uncertainty of the energy of the final state, I do not see any difference between the two approaches. The Hyperphysics formulation might be simplified because most of the time the final state is the ground state, which has no uncertainty.
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2017-12-17 23:48:40
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https://asmedigitalcollection.asme.org/lettersdynsys/article/1/1/011013/1075679/Statistical-Determination-of-Decision-Making?searchresult=1
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## Abstract
Typically, mobile vehicles follow the same paths repeatedly, resulting in a common path bounded with some variance. These paths are often punctuated by branches into other paths based on decision-making in the area around the branch. This work applies a statistical methodology to determine decision-making regions for branching paths. An average path is defined in the proposed algorithm, as well as boundaries representing variances along the path. The boundaries along each branching path intersect near the decision point; these intersections in path variances are used to determine path-branching locations. The resulting analysis provides decision points that are robust to typical path conditions, such as two paths that may not clearly diverge at a specific location. Additionally, the methodology defines decision region radii that encompass statistical memberships of a location relative to the branching paths. To validate the proposed technique, an off-line implementation of the decision-making region algorithm is applied to previously classified wheelchair path subsets. Results show robust detection of decision regions that intuitively agree with user decision-making in real-world path following. For the experimental situation of this study, approximately 70% of path locations were outside of decision regions and thus could be navigated with a significant reduction in user inputs.
## 1 Introduction
This work is motivated by human-assistive devices, particularly robotic or semi-automated wheelchairs, that seek to significantly enhance mobility. When using these vehicles, many users are challenged by a diminished dexterity and a decreased amplitude of movement [1]. In contrast, vehicle guidance applications tend to require continuous, rapid (high-bandwidth) steering and velocity inputs for path keeping. When employed in a wheelchair, path-keeping inputs usually come from a joystick. As wheelchair users age, or as motor functions decline, the dexterity needed to control a wheelchair’s joystick may become cumbersome and demanding. This decreased dexterity may compromise the reliability of the input signals, creating error between the desired and actual wheelchair paths. The inability of the user to maintain path tracking can further compromise mobility, as poor tracking control often requires slower speeds. Ultimately, a need for high-bandwidth joystick inputs for path following may sacrifice the use of the wheelchair altogether.
The motivation of this work is that wheelchair guidance may be improved if users do not need to continuously supply directional inputs. Instead, the inputs could be analyzed to provide partial or complete automation for path selection, with the wheelchair performing corrections to follow the user’s path choice. This capability assumes that the common paths of a wheelchair are known a priori and that the decision locations for choosing a particular path option are well defined.
However, it can be unclear how to define a common or average path, or more importantly, when the user or algorithm should be queried to determine which branch of a path should be followed. This work develops a statistical determination of decision-making regions for branching paths. The technique employed determines the locations in which users or operators are most likely to supply path guidance information. While the presented implementation is for wheelchair assistance, the algorithm can be applied to any robotic platform, including connected or autonomous road vehicles.
### 1.1 Prior Work Defining Path Subsets.
Prior work by the authors developed a statistical method of determining a wheelchair user’s common paths [2]. Three techniques were presented: (1) a methodology to calculate the average path from a group of similar traversals; (2) a calculation of a path’s uncertainty due to variances between similar paths; and (3) a probabilistically determined goal location based on the number of times a particular path has been traversed. This work builds on that past effort and defines a means to determine the spatial locations in which users are most likely to input a path-specific decision as two or more paths begin to branch.
An important aspect of the prior work is the method in which paths are defined. While a plethora of studies have explored path planning (e.g., Refs. [37]), limited work has investigated robotic path or trajectory averaging. Though spline and other linear interpolation methods were able to be applied in Ref. [8], a line projection technique was ultimately chosen for Ref. [2] to combine multiple paths, particularly those that are non-unique (e.g., lack a one-to-one mapping between x and y positions).
Two additional techniques that are applied to path and trajectory averaging are stochastic approximation and path ensemble averaging. Stochastic approximation algorithms are recursively updated rules that can be used to solve optimization problems [9]; ensemble averaging considers trajectories of dynamic systems and the description of the resulting bundle. Ensemble averaging is applied to climate change, quantum systems, oscillators, neural networks, and even cardiography [10]. For example, a recursive stochastic optimization procedure was applied to trajectory averaging and yielded the highest possible asymptotic convergence rate for stochastic approximation algorithms [11].
### 1.2 Intersection-Based Decision-Making.
A potential application of the current work is intersection-based decision-making for vehicles. With 30% of roadway fatalities related to intersections, their detection remains challenging for autonomous vehicles [1214]. In a study by Wang et al., driving scenarios are divided into two categories: (1) normal road without intersections and (2) complex road intersections [15]. In the latter, accurate locations and orientations of all intersection branches are needed for decision-making and path planning. To recognize intersections, 3D-LiDAR scans were analyzed for admissible space in front of autonomous vehicles. A v-support vector regression model is then used to estimate the pose of each intersection branch, and a fusion method is proposed to locate the position of the intersection.
Additionally, the techniques of this paper may be applicable to biomedical applications, such as intravenous injections. For example, in the study by Brewer and Salisbury, an algorithm is proposed that searches venous networks for bifurcations, regarded as the optimal intravenous insertion points [16]. Since the desired insertion point is not at the center of the bifurcation, the difficulty in their proposed algorithm is the detection of closely spaced bifurcations, which appear as a single, noisy bifurcation. Similar challenges arise in the datasets examined in this work with wheelchairs, namely bifurcations in paths can exist close to each other in space, resulting in a noisy bifurcation of many paths nearby each other, rather than distinct decision points. Thus, one can observe that path selection is not simply a challenge of traversing a connected graph, but also in analyzing the spatial location, and possible similarity, in the nodes of the graph to enable discrimination of distinct decisions in space and time.
### 1.3 Spatial Planning for Decision-making.
Critical to the implementation of the proposed technique is a qualitative spatial understanding of the robot’s current and future positions. Rodic and Katic present a robot that acquires information about different paths and builds a corresponding topology map [17]. The robot can then follow the same trajectories in later tasks. Additionally, the robot’s current position can be compared with memorized discrete positions on the previously mapped paths to continue the robot’s motion along a known trajectory. The techniques employed in Ref. [17] are very similar to those presented in Ref. [2] and the present work, the novelty of the current work being that a decision region is determined for path-branching locations. Additionally, in the current study, the goal is to use previously mapped paths and previously measured pose information to verify location prediction based on the robot’s current position.
The aim of this paper is to apply a statistical approach to determine decision-making regions for branching paths. This technique of finding key decision-making regions, and only querying a user or algorithm at these regions, has the potential to significantly reduce user or operator inputs.
## 2 Decision-Making Region Methodology
This section presents a statistical approach to define the decision-making regions for branching paths. In prior work by the authors, an experimental study was performed in which user-driven robotic wheelchair traversals to predetermined locations were recorded [2]. Path averages and variances were then calculated for path subsets with the same starting and ending locations. The path averages and variances along each path are used to derive an average path in a spatial coordinate along the path called the s-coordinate. Analysis of these paths for real use applications showed clear branching points between paths. These points where paths branch to different goal locations are where users typically make decisions about their intended goal destination. Here, we want to define the regions at which users choose one path over another.
A naive approach would be to simply examine average paths to identify the exact point at which the paths deviate. In many cases, the average paths do not clearly separate from each other at a distinct point. For example, Fig. 1 shows the common paths in the laboratory testing region for the team’s intelligent wheelchair. The point designated as “A” is a starting point, and the points numbered 1 to 4 represent common destination points within the lab. Upon repeated traversals from point A, one observes that paths commonly branch around obstacles in the lab—tables in this case. A branching location occurs as the paths deviate from point A at position (X = 0.25, Y = 4.50). These results, obtained from the instrumented wheelchair, illustrate that able-bodied persons and vehicles do not typically switch from one path to another at the same location in space. Rather, vehicles and people alike, including wheelchair users, may trend toward one path over a spatial range. Consequently, a decision point, defined as any point along a set of paths in which the paths are no longer statistically distinguishable, becomes a decision-making region. Now, the decision-making region, or decision space, can be defined as the area within which users must make a decision regarding their intended path.
The following procedure outlines a statistically determined approach to identify and refine decision points and their corresponding decision-making regions:
Step 1: Identify Path Variance Intersection Points. First, the two closest branching paths must be established. Additionally, the boundaries representing user-selected standard deviation levels n in the path variances are calculated. In this work, the standard deviation levels range from nuser = 0.25–4-sigma standard deviations. Next, the intersection points between the overlap of the path variance boundaries are identified. These points are defined as the path variance intersection points or PVIPs.
Step 2: Fit a Linear Regression Line. The second step is to fit a least-squares linear regression line through the PVIPs.
Step 3: Identify Snap-fit PVIPs. Next, the closest point on the linear regression line to each PVIP is found. The closest points are identified by projecting a line from each PVIP onto the linear regression line. These newly established points on the regression line are called the “snap-fit” PVIPs (psnap).
Step 4: Compute the Radius for Each Snap-fit PVIP. An arbitrary origin (xo, yo) and ending location (xf, yf) on the linear regression line is chosen. From this origin, one can then calculate the radii to each snap-fit PVIP using:
$rσ=(psnap,x−xo)2+(psnap,y−yo)2$
(1)
Step 5: Fit a Second Linear Regression Line to the Radii. A second least-squares linear regression line is fit through the radii identified in Step 4.
Step 6: Define Origin and Variance of the Decision-Making Distribution. Implementation of the linear regression line applied in Step 5 provides the origin of the PVIP distribution, as well as the variance of the PVIP distribution. Here, the actual locations of the snap-fit PVIPs are defined, such that pdist is the origin of the distribution and σdist is the variance of the distribution.
Step 7: Identify the Center of the Decision Region. Next, the center of the decision-making region is identified using the origin of the distribution pdist.
Step 8: Calculate the Decision Region Radii. Finally, the last step of the procedure is to calculate the decision region radii rdecision. This value is computed using the user-selected standard deviation level nuser and the PVIP distribution σdist:
$rdecision=nuser⋅σdist$
(2)
An algorithmic summary for finding the path decision points and decision-making regions is presented in Algorithm 1. Additionally, pictorial illustrations of the decision region algorithm are shown in Fig. 2.
Algorithm 1
## 3 Implementation for a Robotic Wheelchair
This section implements the application of the decision-making region identification technique. To demonstrate the proposed approach, measurements of a robotic wheelchair’s motion are used as an application example. A modified robotic wheelchair (Jazzy Pride Select 6, Pride Mobility Products, Corp., USA) was used to collect path traversals within an office setting. Pose measurements were collected via optical wheel encodes (HB6M Hollow Bore Optical Encoders, US Digital, USA) mounted directly to the direct-drive wheels of the wheelchair. Joystick steering commands were also collected from the wheelchair.
The same datasets are used for this implementation as those analyzed in Ref. [2]. The wheelchair was manually driven from one of four fixed starting locations (labeled A–D) to one of four target destinations (numbered 1–4), as illustrated in Fig. 1. To validate the repeatable accuracy of the measurements and the proposed algorithm, a total of 45 path traversals were driven.
The off-line implementation of the proposed decision-making region algorithm to each of the collected datasets is presented in Fig. 3. Here, the 0.5-, 1-, and 2-sigma decision regions for the previously classified path subsets are shown. It is important to note that by applying this approach, the result may yield an origin of the distribution that is asymmetrically located between the two closest paths (i.e., the origin may appear to lie closer to one path than another). This location is the result of both the user-selected percentile level and the number of times a particular path set has been traversed.
Of particular importance is the potential for a significant reduction in user input when implementing the decision-making region. Using the statistical description of a decision-making region, the user or operator need only input a navigation command while within this region. As a result, user control is alleviated from fine-control of following a path to coarse control of selecting a path; this result has the potential to reduce strain from the user, particularly in the case of wheelchair users suffering from a range of neurodegenerative diseases, including amyotrophic lateral sclerosis.
The minimum potential reductions in user input for each starting location to each final destination using the nuser = 2 standard deviation decision regions from Fig. 3 are presented in Table 1. These percent reductions were determined by dividing the diameter of the decision-making region by the total length of the corresponding path. Note that the decision-making regions in the collected dataset are intentionally close together, so that reduction in less confined settings would be expected to produce even better results.
$Ireduction=2⋅rdecisionLpath=DdecisionLpath$
(3)
where Ireduction represents the input reduction, rdecision and Ddecision represent the radius and diameter of the decision-making region, respectively, and Lpath represents the length of the corresponding path. The resulting minimum average reduction of user inputs for all possible paths is 68%, which satisfies the primary objective of a significant reduction of user/operator inputs. Across all possible path permutations, an overall average of 81% user input reduction was achieved.
Moreover, while this implementation is applied to wheelchair assistance, it is also important to recognize that this work is applicable to any indoor robotic platform, such as janitorial or industrial robotics. Additionally, the decision-making region algorithm could be applied to autonomous or semi-autonomous vehicles, for example, aiding in intersection decisions.
## 4 Conclusions
The key contribution of this paper is the proposal of a statistically determined decision-making region identification technique for branching paths. The novelty of the proposed algorithm is the use of path variance intersections to determine path branching locations, rather than the naive assumption that key path decision information is held exclusively at the intersection of path averages. The primary objective of this technique was to reduce user or operator inputs; the implementation of the technique with previously collected wheelchair path sets yielded a minimum average reduction of 68% and an overall reduction of 81%.
Additionally, while the present paper applied the decision-making region identification technique to wheelchair assistance, the proposed approach can be readily applied to any indoor robotic platform and can be modified and applied to intelligent vehicles for decision-making at intersections. Along these lines, Nazario-Casey et al. make the case that utilizing autonomy in conjunction with online decision-making offers the potential for generalization, but also makes implementation more challenging [18]. The authors of the present paper are in agreement with this statement, but believe this generalization is highly beneficial to an array of robotics-based applications. Moreover, by creating modular algorithms and testing them on a variety of robotic platforms, the proposed approach can be expanded upon, modified, and evaluated for new systems and their associated challenges [19].
## Acknowledgment
This material is based upon work supported by the National Science Foundation under Grant No. DGE1255832. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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2021-05-08 09:48:30
|
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|
http://www.acmerblog.com/hdu-2778-lcr-4385.html
|
2014
02-14
# LCR
LCR is a simple game for three or more players. Each player starts with three chips and the object is to be the last person to have any chips. Starting with Player 1, each person rolls a set of three dice. Each die has six faces, one face with an L, one with a C, one with an R and three with a dot. For each L rolled, the player must pass a chip to the player on their left (Player 2 is considered to be to the left of Player 1); for each R rolled, the player passes a chip to the player on their right; and for each C rolled, the player puts a chip in a central pile which belongs to no player. No action is taken for any dot that is rolled. Play continues until only one player has any chips left. In addition, the following rules apply:
1. A player with no chips is not out of the game, since they may later gain chips based on other players’ rolls.
2. A player with only 1 or 2 chips left only rolls 1 or 2 dice, respectively. A player with no chips left does not roll but just passes the dice to the next player.
Your job is to simulate this game given a sequence of dice rolls.
Input will consist of multiple test cases. Each test case will consist of one line containing an integer n (indicating the number of players in the game) and a string (specifying the dice rolls). There will be at most 10 players in any game, and the string will consist only of the characters L’, C’, R’ and .’. In some test cases, there may be more dice rolls than are needed (i.e., some player wins the game before you use all the dice rolls). If there are not enough dice rolls left to complete a turn (for example, only two dice rolls are left for a player with 3 or more chips) then those dice rolls should be ignored. A value of n = 0 will indicate end of input.
Input will consist of multiple test cases. Each test case will consist of one line containing an integer n (indicating the number of players in the game) and a string (specifying the dice rolls). There will be at most 10 players in any game, and the string will consist only of the characters L’, C’, R’ and .’. In some test cases, there may be more dice rolls than are needed (i.e., some player wins the game before you use all the dice rolls). If there are not enough dice rolls left to complete a turn (for example, only two dice rolls are left for a player with 3 or more chips) then those dice rolls should be ignored. A value of n = 0 will indicate end of input.
3 LR.CCR.L.RLLLCLR.LL..R...CLR.
5 RL....C.L
0
Game 1:
Player 1:0
Player 2:0
Player 3:6(W)
Center:3
Game 2:
Player 1:1
Player 2:4
Player 3:1
Player 4:4(*)
Player 5:4
Center:1
//用STL里的priority_queue存储，剩下的就是模拟了。。
#include <iostream>
#include <cstdio>
#include <memory.h>
#include <queue>
#define INF 0x7fffffff
using namespace std;
struct Process
{
int id;
int arr; //到达时间
int last;
int prior;
};
struct cmp //按任务到达CPU的顺序进行存储
{
bool operator()(const Process &t1,const Process &t2)
{
if(t1.arr!=t2.arr)return t1.arr>t2.arr; //arr小的放在上面。小顶堆
return t1.prior<t2.prior; //对于同时到达的，prior大的放在上面。大顶堆
}
};
struct cmp2//按任务执行的顺序存储
{
bool operator()(const Process &t1,const Process &t2)
{
if(t1.prior!=t2.prior)return t1.prior<t2.prior;
return t1.arr>t2.arr; //prior相同时，arr小的放在上面。即遵循first arrive,first serve
}
};
int main()
{
int num;
int cases=0;
while(cin>>num)
{
Process temp;
Process temp2;
if(cases!=0)printf("\n");
priority_queue<Process,vector<Process>,cmp> Q; //所有待办任务的队列
priority_queue<Process,vector<Process>,cmp2> ing; //所有等待中的任务的队列,prior大的放上面
for(int i=0; i<num; i++)
{
cin>>temp.id>>temp.arr>>temp.last>>temp.prior;
Q.push(temp);
}
printf("CASE #%d\n",++cases);
int timer=0;
temp=Q.top();
Q.pop();
ing.push(temp); //第一个任务进入ing队
timer=temp.arr; //记录当前时间
while(1) //注意结束条件不能是Q为空。Q空了之后，还要再运行一次。
{
//temp=ing.top();
if(!Q.empty())
{
temp2=Q.top();
Q.pop();
}
else
{
temp2.arr=INF; //永远不会到来
temp2.last=0;
temp2.prior=0;
temp2.id=0;
}
bool flag=0;
while(!ing.empty())
{
temp=ing.top();
if(timer+temp.last<=temp2.arr)
{
timer+=temp.last;
printf("%d %d\n",temp.id,timer);
ing.pop();
}
else
{
flag=1;
break;
}
}
if(temp2.arr==INF) //Q空了后，再循环一次，在这儿终止。
{
break;
}
if(flag) //如果被打断
{
// temp.arr=timer;
temp.last=temp.last-(temp2.arr-timer);//更新这个任务还需要的时间
timer=temp2.arr;
ing.pop();
ing.push(temp); //更新ing最上面那个
ing.push(temp2);
}
else if(!Q.empty())//没被打断，说明些时ing队列空了
{
timer=temp2.arr;
ing.push(temp2);
}
}
}
return 0;
}
|
2017-01-19 10:51:35
|
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|
http://www.cad.zju.edu.cn/home/zhx/csmath/doku.php?id=cp:2011
|
cp:2011
# Most Cited Deep Learning Papers
## 2016
• Unordered List ItemDermatologist-level classification of skin cancer with deep neural networks (2017), A. Esteva et al. [html]
• Weakly supervised object localization with multi-fold multiple instance learning (2017), R. Gokberk et al. [pdf]
• Brain tumor segmentation with deep neural networks (2017), M. Havaei et al. [pdf]
• Professor Forcing: A New Algorithm for Training Recurrent Networks (2016), A. Lamb et al. [pdf]
• Adversarially learned inference (2016), V. Dumoulin et al. [web][pdf]
• Understanding convolutional neural networks (2016), J. Koushik [pdf]
• Taking the human out of the loop: A review of bayesian optimization (2016), B. Shahriari et al. [pdf]
• Adaptive computation time for recurrent neural networks (2016), A. Graves [pdf]
• Densely connected convolutional networks (2016), G. Huang et al. [pdf]
• Continuous deep q-learning with model-based acceleration (2016), S. Gu et al. [pdf]
• A thorough examination of the cnn/daily mail reading comprehension task (2016), D. Chen et al. [pdf]
• Achieving open vocabulary neural machine translation with hybrid word-character models, M. Luong and C. Manning. [pdf]
• Very Deep Convolutional Networks for Natural Language Processing (2016), A. Conneau et al. [pdf]
• Bag of tricks for efficient text classification (2016), A. Joulin et al. [pdf]
• Efficient piecewise training of deep structured models for semantic segmentation (2016), G. Lin et al. [pdf]
• Learning to compose neural networks for question answering (2016), J. Andreas et al. [pdf]
• Perceptual losses for real-time style transfer and super-resolution (2016), J. Johnson et al. [pdf]
• Reading text in the wild with convolutional neural networks (2016), M. Jaderberg et al. [pdf]
• What makes for effective detection proposals? (2016), J. Hosang et al. [pdf]
• Inside-outside net: Detecting objects in context with skip pooling and recurrent neural networks (2016), S. Bell et al. [pdf].
• Instance-aware semantic segmentation via multi-task network cascades (2016), J. Dai et al. [pdf]
• Conditional image generation with pixelcnn decoders (2016), A. van den Oord et al. [pdf]
• Deep networks with stochastic depth (2016), G. Huang et al., [pdf]
• Generative Short Term Stochastic Gibbs Networks 2016), I. Lenz et al. [pdf]
## 2015
• Unordered List ItemSpatial transformer network (2015), M. Jaderberg et al., [pdf]
• Exploring models and data for image question answering (2015), M. Ren et al. [pdf]
• Are you talking to a machine? dataset and methods for multilingual image question (2015), H. Gao et al. [pdf]
• Mind's eye: A recurrent visual representation for image caption generation (2015), X. Chen and C. Zitnick. [pdf]
• From captions to visual concepts and back (2015), H. Fang et al. [pdf].
• Towards AI-complete question answering: A set of prerequisite toy tasks (2015), J. Weston et al. [pdf]
• Ask me anything: Dynamic memory networks for natural language processing (2015), A. Kumar et al. [pdf]
• Unsupervised learning of video representations using LSTMs (2015), N. Srivastava et al. [pdf]
• Deep compression: Compressing deep neural networks with pruning, trained quantization and huffman coding (2015), S. Han et al. [pdf]
• Improved semantic representations from tree-structured long short-term memory networks (2015), K. Tai et al. [pdf]
• Character-aware neural language models (2015), Y. Kim et al. [pdf]
• Grammar as a foreign language (2015), O. Vinyals et al. [pdf]
• Trust Region Policy Optimization (2015), J. Schulman et al. [pdf]
• Beyond short snippents: Deep networks for video classification (2015) [pdf]
• Learning Deconvolution Network for Semantic Segmentation (2015), H. Noh et al. [pdf]
• Learning spatiotemporal features with 3d convolutional networks (2015), D. Tran et al. [pdf]
• Understanding neural networks through deep visualization (2015), J. Yosinski et al. [pdf]
• An Empirical Exploration of Recurrent Network Architectures (2015), R. Jozefowicz et al. [pdf]
• Training very deep networks (2015), R. Srivastava et al. [pdf]
• Deep generative image models using a laplacian pyramid of adversarial networks (2015), E.Denton et al. [pdf]
• Gated Feedback Recurrent Neural Networks (2015), J. Chung et al. [pdf]
• Fast and accurate deep network learning by exponential linear units (ELUS) (2015), D. Clevert et al. [pdf]
• Pointer networks (2015), O. Vinyals et al. [pdf]
• Visualizing and Understanding Recurrent Networks (2015), A. Karpathy et al. [pdf]
• Attention-based models for speech recognition (2015), J. Chorowski et al. [pdf]
• End-to-end memory networks (2015), S. Sukbaatar et al. [pdf]
• Describing videos by exploiting temporal structure (2015), L. Yao et al. [pdf]
• A neural conversational model (2015), O. Vinyals and Q. Le. [pdf]
## 2014 or earlier
• Unordered List ItemLearning a Deep Convolutional Network for Image Super-Resolution (2014, C. Dong et al. [pdf]
• Recurrent models of visual attention (2014), V. Mnih et al. [pdf]
• Empirical evaluation of gated recurrent neural networks on sequence modeling (2014), J. Chung et al. [pdf]
• Addressing the rare word problem in neural machine translation (2014), M. Luong et al. [pdf]
• On the properties of neural machine translation: Encoder-decoder approaches (2014), K. Cho et. al.
• Recurrent neural network regularization (2014), W. Zaremba et al. [pdf]
• Intriguing properties of neural networks (2014), C. Szegedy et al. [pdf]
• Towards end-to-end speech recognition with recurrent neural networks (2014), A. Graves and N. Jaitly. [pdf]
• Scalable object detection using deep neural networks (2014), D. Erhan et al. [pdf]
• On the importance of initialization and momentum in deep learning (2013), I. Sutskever et al. [pdf]
• Regularization of neural networks using dropconnect (2013), L. Wan et al. [pdf]
• Learning Hierarchical Features for Scene Labeling (2013), C. Farabet et al. [pdf]
• Linguistic Regularities in Continuous Space Word Representations (2013), T. Mikolov et al. [pdf]
• Large scale distributed deep networks (2012), J. Dean et al. [pdf]
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2018-04-24 18:01:21
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https://math.stackexchange.com/questions/2393558/proving-generalised-eulers-formula-using-elliptic-functions
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proving generalised euler's formula using elliptic functions
Given the elliptic modulus $k$,such that the complementary modulus is defined by $k'\equiv \sqrt{1-k^2}$,where $\phi\equiv am(u|k)$ is the jacobi amplitude and $K(k)$ is the complete elliptic integral of the first kind.
The elliptic generalisation of euler's formula in complex analysis(which expresses jacobi's elliptic functions in terms of the exponential function) is
$e^{i \phi}=\text{cn(u|k)}+i\text{sn(u|k)}$
The formula is valid multiples of $4K(k)$ and also multivalued.
How would we prove the formula using the theory of elliptic functions instead of elliptic functions defined in terms of trigonometric functions?
By Jacboi's original definition, $\text{cn(u|k)}:=\cos\phi$ and $\text{sn(u|k)}:=\sin\phi$. But $e^{i\phi}=\cos\phi+i\sin\phi$ by Euler and the result follows.
• Jacobi invented his theory of elliptic functions and that was his original definition. That is, sine amplitude $\sin\textrm{am}.(u|k) = \sin(am(u|k))$. I can look up the original equations in his "Fundamenta Nova Theoriae Functionum Ellipticarum" for you if you wish. Or just look in Jacobi Elliptic Function in Wikipedia. – Somos Aug 15 '17 at 3:00
• @Nicco Then I suggest being explicit about that in your question, and being explicit what you consider the definition of $sn()$ and $cn()$ because there are multiple ways to define them. – Somos Aug 15 '17 at 3:42
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2019-12-13 05:53:35
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https://stogblog.net/2021/01/
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Month: January 2021
# The point closest to the origin is not typical
When simulating a point process to characterize the performance of the typical point (typical user or receiver), a conditioned version of the point process given a point at the origin o may not be available. It is then tempting to choose the “next best” point as a substitute, which may be the point closest to the origin. (Whether the coordinates are then shifted so that this point is at o is irrelevant.) The goal of this post is to show that this point is not typical, i.e., producing many realizations of the point process and evaluating the performance at this point does not yield the performance of the typical point. I call the point closest to o after averaging over the point process the 0-point. Put differently, the 0-point is the typical point among all points closest to o across the realizations of the point process. In a cellular network, the 0-point is the nucleus of the 0-cell (see this post), hence the term.
For simplicity, let us consider the homogeneous PPP of intensity 1 and focus on the probability that a disk of radius r centered at a point contains no other points, which we refer to as the NOPID (no other point in disk) probability. Equivalently, it is the probability that the nearest neighbor is at distance at least r. For the typical point, the NOPID probability is exp(-πr2). For the 0-point, let D be its distance from o. Given D, the disk of radius D centered at o, denoted as b(o,D), is empty, so the points excluding the 0-point form a PPP on ℝ2\b(o,D), and the NOPID probability is the probability that b((D,0),r)\b(o,D) is empty. This region is shown in blue in the movie below for different r given that the 0-point is at (1,0), i.e., D=1. For r<2D, it is moon- or crescent-shaped, while for r>2D, it is a disk with a hole.
Letting A(r,d)=|b((d,0),r)\b(o,d)|, the (unconditioned) NOPID probability is 𝔼(exp(-A(r,D)), where D is Rayleigh distributed with mean 1/2. It can be expressed as
$\displaystyle \qquad\qquad F_0(r)=\frac{\pi}{4}r^2 e^{-\pi r^2}+\int_{r/2}^\infty e^{-A'(r,u)}2\pi u e^{-\pi u^2}{\rm d}u,\qquad\qquad (1)$
where
$\displaystyle A'(r,d)=\pi r^2-r^2\cos^{-1}\left(\frac{r}{2d}\right)-d^2\cos^{-1}\left(1-\frac{r^2}{2d^2}\right)+\frac{r}{2}\sqrt{4d^2-r^2}.$
is the area A(r,d) for r<2d. For r>2d, A(r,d)=π(r2d2), which results in the first term in (1).
The NOPID probabilities of the 0-point and the typical point are compared below. It is apparent that the 0-point is more isolated than the typical point.
By integrating the NOPID probability of the 0-point, we obtain the mean nearest-neighbor distance as 0.5953. This is almost 20% larger than that of the typical point, which is 1/2. The difference between the two NOPID probabilities is not just in the mean, though. They differ qualitatively in the tail. For large r, it follows from (1) that the ratio of the two NOPID probabilities approaches πr2/4. This implies that a Rayleigh distribution with adjusted mean will not provide a good fit to the NOPID probability at the 0-point.
The difference is even more pronounced if we consider directional nearest neighbors. If we consider a sector of angle π/2, then the nearest neighbor of the typical point is at distance 1 on average, irrespective of the orientation of the sector. For the 0-point, in the direction opposite from o, the mean distance is also 1, since on that side, the PPP is unaffected by the empty disk b(o,D). In the direction towards o, however, the distance is significantly larger, with a mean of 1.4205. The plot below shows the pdf of the directional nearest-neighbor distance of the 0-point oriented towards o (red) and the pdf of the directional nearest-neighbor distance of the typical point (blue), given by (π/2)r exp(-(π/4)r^2). The pdfs are the negative derivatives of the NOPIS (no other point in sector π/2) probabilities.
When applied to cellular networks (with nearest-base station association), the 0-point is the base station serving the typical user (at o). The discussion here reveals that the 0-base station behaves differently from the typical base station. In particular, the point process of the other base stations viewed from the 0-base station is highly non-isotropic. In the direction of the typical user, the nearest other base station is much further away than in the opposite direction. This fact is consistent with the conclusions from this post on the shape of the 0-cell in the Poisson-Voronoi tessellation.
# The curious shape of Poisson-Voronoi cells
In this blog we are exploring the shape of two kinds of cells in the Poisson-Voronoi tessellation on the plane, namely the 0-cell and the typical cell. The 0-cell is the cell containing the origin, while the typical cell is the cell obtained by conditioning on a Poisson point to be at the origin (which is the same as adding the origin to the PPP).
The cell shape has an important effect on the signal and interference powers at the typical user (in the 0-cell) and at the user in the typical cell. For instance, in the 0-cell, which contains the typical user at a uniformly random location, about 1/4 of the cell edge is at essentially the same distance to the base station as the typical user on average). Hence it is not the case that edge users necessarily suffer from larger signal attenuation than the typical user (who resides inside the cell).
The cell shape is determined by the directional radii of the cells when their nucleus is at the origin. To have a well-defined orientation, we select a location uniformly in the cell and rotate the cell so that this location falls on the positive x-axis. In the 0-cell, this involves first a translation of the cell’s nucleus to the origin, followed by a rotation until the original origin (which is uniformly distributed in the cell) lies on the positive x-axis. This is illustrated in Movie 1 below. In the typical cell, it involves adding a Poisson point, selecting a uniform location, and a rotation so that this uniform location lies on the positive x-axis. This is illustrated in Movie 2.
As indicated in the movies, the distances from the nucleus to the uniformly random location are denoted by D0 and D, respectively, and the directional radii by R0(ϕ) and R(ϕ), respectively. This way, the boundary of the cells is described in polar coordinates as (R0(ϕ),ϕ) and (R(ϕ),ϕ), ϕ ∈ [0,2π). In a cellular network model, the uniform random location could be that of a user, while the PPP models the base stations. In this case D0 is the link distance from the typical user to its serving base station, while D is the link distance from the typical base station to a randomly located user it serves. The distinction between the typical user’s and the typical base station’s point of view is explained in this blog.
Let λ denote the density of the PPP. Three results are well known:
• The distribution of D0 follows from the void probability of the PPP. It is Rayleigh with mean 1/(2√λ).
• Since the mean area of the typical cell is 1/λ, we have ∫ 0π 𝔼(R(ϕ)2) dϕ = 1/λ.
• The minimum of R(ϕ) is distributed with pdf f(r)=8λπr exp(-4λπr2). This is half the distance to the nearest neighboring Poisson point (base station).
In contrast, there is no closed-form expression for the distribution of D. Due to size-biased sampling, the area of the 0-cell stochastically dominates that of the typical cell and, in turn, D0 dominates D.
Analyzing the directional radii, we obtain these new insights on the cell shapes:
• If Ψ is uniform in [0,π], R(Ψ) is again Rayleigh with mean 1/(2√λ).
• R0(π) is also Rayleigh with the same mean. In fact, R0(π) and D0 are iid.
• R0(0) has mean 3/(4√λ) and is distributed as
$\displaystyle f_{R_0(0)}(y)=2(\lambda\pi)^2 y^3 \exp(-\lambda\pi y^2).$
• Hence R0(0) is on average exactly 50% larger than R0(π). For the typical cell, simulation results indicate that R(0) is about 55% larger on average than R(π).
• The difference R0(0)-D0 is distributed as f(r)=π√λ erfc(r √(πλ)). Its mean is 1/(4√λ). Hence the typical user is no further from the cell edge than the base station on average.
• The joint distribution of D0 and R0(ϕ) can be given in exact analytical form.
• 3/4 of the typical cell is further away from the nucleus than the nearest point on the cell edge (i.e., the minimum directional radius). Expressed differently, a uniformly random user in the typical cell has a 75% chance of being further away from the base station than the nearest edge user. By simulation, D on average is 2.7 times larger than the minimum of the directional radii.
In conclusion, the 0-cell and the typical cell are quite asymmetric around the nucleus (base station) and the uniformly random point (user). In the direction away from the base station, the user is about 4 times closer to the cell edge than in the direction towards the base station, and many locations on the cell edge are closer to the base station than the user inside the cell. These results have implications on the design of efficient cellular network transmission schemes, such as beamforming, NOMA, and base station cooperation, in both down- and uplink.
More details are available in Section II of this paper.
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2021-06-14 09:48:21
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https://hal.inria.fr/hal-01691616
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On a boundary layer phenomenon in acoustic media - Archive ouverte HAL Access content directly
Conference Papers Year : 2016
## On a boundary layer phenomenon in acoustic media
(1) , (1) , (1) , (2, 3, 1)
1
2
3
Hélène Barucq
Juliette Chabassier
Marc Duruflé
Victor Péron
#### Abstract
In this work we describe a boundary layer phenomenon which occurs inside the atmosphere of the sun. As an application we derive new approximate boundary conditions. The boundary layer phenomenon which occurs in this context has several similarities with the so-called skin effect phenomenon in electromagnetism. We investigate a transmission problem which is modeling the propagation of an acoustic wave inside heterogeneous media and in time-harmonic regime. This problem is set in a domain which represents the sun and its atmosphere. The specific feature of this problem lies in the presence of a small parameter $\delta$ which represents the inverse rate of the exponential decay of the density inside the atmosphere. This problem is well suited for the notion of equivalent conditions and the effect of the atmosphere on the sun is as a first approximation local. First we present a multi-scale expansion in power series of $\delta$ for the solution $u$ of the problem. Then we derive equivalent conditions up to the fourth order of approximation with respect to $\delta$ for the solution $u$. This approach leads to solve only equations set inside the sun. We present numerical results and numerical validations to illustrate the accuracy of the asymptotic models.
### Dates and versions
hal-01691616 , version 1 (24-01-2018)
### Identifiers
• HAL Id : hal-01691616 , version 1
### Cite
Hélène Barucq, Juliette Chabassier, Marc Duruflé, Victor Péron. On a boundary layer phenomenon in acoustic media. WONAPDE 2016 - 5th Chilean Workshop on Numerical Analysis of Partial Differential Equations, Jan 2016, Concepción, Chile. pp.1-42. ⟨hal-01691616⟩
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2023-02-09 12:28:41
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https://zbmath.org/?q=an:0638.49020
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# zbMATH — the first resource for mathematics
On higher eigenvalues of variational inequalities. (English) Zbl 0638.49020
Various previously published results and theorems on eigenvalue problems for variational inequalities are collected in this short paper. More precisely, let a(u,v) and b(u,v) be bounded, real and symmetric bilinear forms defined on the real Hilbert space H and let K be a closed and convex cone of H. Suppose a(u,v) coercive and b(u,v) completely continuous. The author studies the problem $(*)\quad u\in K,\quad a(u,v-u)\geq \mu b(u,v-u)\text{ for all } v\in K,$ where $$\mu$$ is a real parameter. The existence of eigenvalues for the problem (*) is proved using a variational approach. This proof forms the original part of the work.
An interesting, although not new, application of the main theorem to the problem of buckling of a clamped circular plate with a unilateral constraint is given in the final part of the paper.
##### MSC:
49R50 Variational methods for eigenvalues of operators (MSC2000) 49J40 Variational inequalities 74S30 Other numerical methods in solid mechanics (MSC2010)
##### Keywords:
eigenvalue problems for variational inequalities
Full Text:
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2021-05-18 16:47:41
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https://www.longevitas.co.uk/site/informationmatrix/latelifemortalitydeceleration.html
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## Is your mortality model frail enough?
Mortality at post-retirement ages has three apparent stages:
1. A broadly Gompertzian pattern up to age 90 (say), i.e. the mortality hazard is essentially linear on a logarithmic scale.
2. The rate of increase in mortality slows down, the so-called "late-life mortality deceleration".
3. The rate of increase slows down to the point where the mortality rate looks like it might be constant above a certain age (110, say).
In my previous blog I demonstrated the power of the Newman hypothesis, namely that low-frequency errors in stated age can cause the patterns in Stage 2 and (especially) Stage 3. However, the Newman hypothesis is not the only means whereby apparently simple Gompertz mortality turns into something more complex. There is another very plausible driver for the deceleration in Stage 2: unobserved risk factors.
To illustrate this, consider a cohort of lives following the Gompertz mortality law:
$\mu_x=e^{\alpha+\beta x}$
at age $$x$$ for given $$\alpha$$ and $$\beta$$. We will assume that the lives all share a common value of $$\beta=0.1$$, i.e. lives "age" at the same constant rate. However, we will further assume that there are two subgroups with different values of $$\alpha=-12$$ and $$\alpha=-11$$. This is a very large difference, far larger than the difference between smokers and non-smokers for example, but it serves to make the underlying phenomenon clearly visible. Crucially, we assume that we do not know who belongs to each subgroup, i.e. there is heterogeneity in the population that we cannot model. Figure 1 shows the resulting aggregate mortality hazard if the two subgroups start out in equal proportion at age 65.
Figure 1. Effect of an unobserved risk factor on mortality analysis. Without the ability to identify the risk factor separating the two groups, the mortality hazard has a different shape from either subgroup. Source: own calculations using Gompertz model for each subgroup.
Figure 1 shows how an apparent deceleration in the rate of increase of mortality arises from (i) the different mortality rates of the subgroups and (ii) the changing proportions of those subgroups as the higher-mortality subgroup shrinks faster. Like the Newman hypothesis, unobserved heterogeneity is a simple-but-powerful idea that can explain apparent late-life mortality deceleration. The Newman hypothesis and unobserved hetereogeneity are both inherently plausible explanations; while the Newman hypothesis only needs rare errors in stated age, unobserved heterogeneity is largely the norm for defined-benefit pension schemes and annuity portfolios. Consider again the previous example of smoker status; smoking is a particularly strong risk factor for mortality, yet for most pensioner portfolios the smoker status of each life is unknown. Proxy risk factors, such as pension size or geodemographic group, can only be imperfect substitutes. Thus, almost every portfolio contains unobserved heterogeneity, and so almost every portfolio exhibits decelerating rates of increase in mortality at advanced ages.
This unexplained hetereogeneity goes by the name of frailty in demographic circles. In Richards (2008) I demonstrated the derivation of some specific mortality models allowing for frailty, together with references to past academic work on the subject. In Richards (2012) I surveyed different mortality models for application to pensioner mortality and found that one frailty model (Makeham-Beard) consistently fitted actual portfolio data better than the sixteen other models considered. Figure 1 shows the reason for this: frailty models are better at handling unobserved sources of variation.
References
Richards, S. J. (2008) Applying survival models to pensioner mortality data, British Actuarial Journal, 14(II), pages 257-326 (with discussion).
Richards, S. J. (2012) A handbook of parametric survival models for actuarial use, Scandinavian Actuarial Journal, 2012(4), pages 233-257.
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The former UK prime minister Harold Wilson famously said that ... Read more
Stephen Richards is the Managing Director of Longevitas
##### Frailty models in Longevitas
Longevitas has two frailty models for mortality:
1. The Beard model, arising from heterogeneity in $$\alpha$$ in the Gompertz model, and
2. The Makeham-Beard model, arising from heterogeneity in $$\alpha$$ in the Makeham model.
The structure and derivation of these models is described in Richards (2008).
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2021-07-26 16:10:27
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https://tex.stackexchange.com/questions/191437/longtable-captions-do-not-wrap-inside-threeparttable-environment
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# Longtable captions do not wrap inside threeparttable environment
I'm using the threeparttable package to automatically wrap my captions to table width. However, I occasionally need to use longtable to split a table across multiple pages. When I do so, the caption no longer wraps to the width of the table, as you see in the two tables below. I tried including the caption package and setting the width with \captionsetup{width=4cm} to no avail but I'd rather have an automatic solution anyway as I have a number of long tables of varying width. Is there a fix or suitable alternative to threeparttable? (I do sometimes need the \tnote feature in threeparttable but I can probably work around that if necessary.)
\documentclass{article}
\usepackage{threeparttable}
\usepackage{longtable}
\usepackage{booktabs}
\begin{document}
\begin{threeparttable}
\begin{tabular}{ll}
\caption{Caption for tabular does wrap inside threeparttable environment.}\\
\toprule
column1 & column2 \\
\midrule
info1 & info2\\
\bottomrule
\end{tabular}
\end{threeparttable}
\begin{threeparttable}
\begin{longtable}{ll}
\caption{Caption for longtable does not wrap inside threeparttable environment.}\\
\toprule
column1 & column2 \\
\midrule
info1 & info2\\
\bottomrule
\end{longtable}
\end{threeparttable}
\end{document}
• It doesn't make sense to use longtable inside threeparttable (or at least I can't think of anything sensible I could make it do there) If you want the caption to be as wide as the effective width of the table you can add up the saved column widths in the aux file. I'm sure I've done that in an answer here but I don't see it now, but google turned up a version from Heiko on comp.text.tex which looks about right. compgroups.net/comp.text.tex/longtable-tablewidth/1922986 – David Carlisle Jul 14 '14 at 20:33
• @DavidCarlisle - Heiko's solution seems to work. Thanks for the link. I'm not at all wed to threeparttable. It just solved the caption width problem for me based on an answer to another question on this site. For the sake of knowledge, why does it not make sense to use longtable inside of threeparttable? The manual states specifically that it facilitates tables with captions, and that such captions are given width equal to the table width. It works for tabular et al. but not for longtable. – Michael S Taylor Jul 14 '14 at 21:43
• a tabular is a box so you can measure its width and then set the minipage for table notes and captions to that width, which is what threeparttable does well. but boxes don't break over a page so longtable doesn't box the alignment so there's nothing threeparttable can do. – David Carlisle Jul 14 '14 at 22:13
• That makes sense to me now. If it's worth your effort, you can put the link to Heiko's solution in an answer for me to accept. – Michael S Taylor Jul 15 '14 at 0:35
• OK or I may see if Heiko's around and wants to pick up a point or two;-) – David Carlisle Jul 15 '14 at 8:18
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2019-09-15 18:25:12
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https://www.gatecseit.in/data-structure-questions-and-answers-parallel-array/
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# Data Structure Questions and Answers-Parallel Array
## Data Structure Questions and Answers-Parallel Array
Congratulations - you have completed Data Structure Questions and Answers-Parallel Array.
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%
Question 1 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
What are parallel arrays?
A Arrays of the same size B Arrays allocated one after the other C Arrays of the same number of elements D Arrays allocated dynamically
NTA NET study material
Question 1 Explanation:
Different arrays can be of different data types but should contain same number of elements. Elements at corresponding index belong to a record.
Question 2 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
Which of the following can be called a parallel array implementation?
A firstName = ['Joe', 'Bob', 'Frank', 'Hans'] lastName = ['Smith', 'Seger', 'Sinatra', 'Schultze'] heightInCM = [169, 158, 201, 199] for i in xrange firstName = ['Joe', 'Bob', 'Frank', 'Hans'] lastName = ['Smith', 'Seger'] heightInCM = [169, 158] for i in xrange(len(firstName)): print C firstName = ['Joe', 'Bob'] lastName = ['Smith', 'Seger', 'Sinatra', 'Schultze'] heightInCM = [169, 158] for i in xrange(len(firstName)): D None of the mentioned
Question 2 Explanation:
All the arrays must have equal length, that is, contain same number of elements.
Question 3 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
A When a language does not support records, parallel arrays can be used B Increased locality of reference C Ideal cache behavior D All of the mentioned
Question 3 Explanation:
In a record, if a field contains only 1 bit, extra space is needed to pad the bits, instead you can use parallel arrays to save space. Sequential access improves locality of reference and cache behavior.
Question 4 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
What are some of the disadvantages of parallel arrays?
A Poor locality of reference for non-sequential access B Very little direct language support C Expensive to shrink or grow D All of the mentioned
Question 4 Explanation:
They share the drawbacks of arrays.
Question 5 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
What is a sorted array?
A Arrays sorted in numerical order B Arrays sorted in alphabetical order C Elements of the array are placed at equally spaced addresses in the memory D All of the mentioned
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2021-06-13 08:50:35
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