Can somebody explain to me parameterization in layman's terms?
915931
There can be several parameterization\n\n1. The specification of a curve, surface, etc., by means of one or more variables which are allowed to take on values in a given specified range. \n\n2. A parameterization of a surface  X(u,v) in uand v is regular if the tangent vectors \ndX/du and dX/dv are always linearly independent. \n[THE DERIVATIVE ARE PARTIAL HERE.]\n\n3. A surface in 3-space can be parameterized by two variables (or coordinates) u and u such that \n\n   (1)  x= x(u,v)\n   (2)  y= y(u,v)\n   (3)  z= z(u,v)\nIf a surface is parameterized as above, then the tangent vectors \n\n (4) Tu =   (dX/du)x^ + (dY/du)x^ + (dZ/du)x^ \n\n (5)  Tv =  (dX/dv)x^ + (dY/dv)x^ + (dZ/dv)x^ \nare useful in computing the surface area and surface integral. \n\n[Tv means T with suffix v, (dX/dv) means partial derivative of X with v and x^ means x with a cap].\n\nSorry, but I can not write euqations so well without equation editor.
1049307
reply:\n\nConstants are not variables you can not write to them they are constants for one. For example having a constant called pie you could not reassign it. Now you can use constants in what ever you want provided they dont break the code security.\n\n- Mike Roth
908586
GB >= 0.6*(CS + GB)\n\n"cushion" variable U introduced : \n\nU >= 0 \nGB + U = 0.6 * (CS + GB)
238393
I will Answer On A Sample :\n\n    Delegate Function fns_Operation(ByVal par1 As Double, ByVal par2 As Double) As Double\n\n    Function fns_01(ByVal par1 As Double, ByVal par2 As Double) As Double\n        Return par1 + par2\n    End Function\n\n    Function fns_02(ByVal par1 As Double, ByVal par2 As Double) As Double\n        Return par1 * par2\n    End Function\n\n    Function fns_Operate(ByVal par1 As Double, ByVal par2 As Double, ByVal fns As fns_Operation) As Double\n        Return fns(par1, par2)\n    End Function\n\n    Sub Test()\n        MsgBox(fns_Operate(123, 343, AddressOf fns_01))\n        MsgBox(fns_Operate(3453, 24234, AddressOf fns_02))\n    End Sub
200031
p is actual parameter of distribution, and is unknown to you.\np-hat is it's estimate, based on observed data.
1147478
Memory Handling.\nWhatever memory space you occupy by instantiating the objects should be released after your use.\nType Casting should also be taken care.
532980
I'm sorry but did you see this in My_Computer properties?\n\n\n\n--- ► λ ◄ \n\nN.B. I'm still a learner.
128537
It's arbitrary.\n\nWe are used to "x" and "y" (and perhaps "z") being used as the variables, but we also need to have a representation of constants. Using a specific number would make it seem as though only that number works in the equation.
573927
single replacement, double replacement, decomposition, synthesis
531852
you have to use instinct, there is no clear rules that states what the expressions must look like (except if the problem insist you on giving the answer in a specific form). \n\nPerhaps one of the guidelines is make variables that have same indices grouped into one. So x + 3x becomes 4x, group them out don't leave them straying.
208594
it means that there is a method or function that includes a parameter (or argument) called Start.  the number you pass into that method or function must be greater than zero.\n\nmyMethod(int Start) {\n...\n}\n\n//now to call method do this\nmyMethod(1);\n\n//but don't do this\nmyMethod(0);
243664
I'm Matlab illiterate; however, it's obvious that lines (or line segments) starting with % are comments. Now, I'll try with the working lines:\nThe definition of the function, with 4 arguments (y,X,intercept and name) and 4 outputs. (b,resid,yh and dH)\n\nfunction [b, resid, yh, dH] = lsregr(y, X, intercept, name)\n\nThese are housekeeping chores: nargin is the number of arguments. If you have 2 arguments, you default intercept to 1 and namte to empty; if you have 3 arguments, you just default name to empty:\n\nif nargin == 2, intercept = 1; name = []; end;\nif nargin == 3, name = []; end;\nY size must be one:\n\nif size(y, 1) == 1\nerror('The input y must be a column vector');\nend;\nn = size(y, 1);\n\nX rows must be equal to Y rows and X must be square\n\nif size(X, 1) ~= n & size(X, 1) ~= size(X, 2)\n\ntranspose X (X(i,j)=X(j,i))\nX = X';\nend;\nexplaint by itself\n\nconst = intercept;\n\nif intercept is 1 (thus const is 1) then add a column of ones to X\nif const == 1, X = [ones(n, 1), X]; end;\np is the (new) number of columns\np = size(X, 2);\ncompute the mean of Y\nybar = mean(y);\n\ncompute X transposed multiplied by X; compute the inverse of X'X\nXtX = X'*X; invXtX = inv(XtX);\nDefine "Hat MAtrix" as X multiplied by X'^X^-1 and then by X'\n\nH = X * invXtX * X'; %HAT matrix\n\ngenerate the diagonal of H\ndH = diag(H); %diagonal of the HAT
978671
[i]  = I\n[d]s = does (without the s it would be do)\n[t]  = to or the\n[o] = of\n\nAnyway I think all the [ ] are suppose to be abbreviations and these are the most likely.◘
115853
#include <iostream>\n#include <vector>\nusing namespace std;\n#define rep(i, n) for((i) = 0; (i) < (n); (i)++)\n#define rep2(i, k, n) for((i) = k; (i) < (n); (i)++)\nvoid f21();\nint n;\ndouble mat[100][100];\nint main()\n{\nwhile(true)\n{\ncin >> n;\nif(!n) break;\nmemset(mat, 0, sizeof(mat));\nf21();\n}\nreturn 0;\n}\n\nvoid f21()\n{\ncout << "****************************************" << endl;\ncout << "*          INVERSE OF MATRIX           *" << endl;\ncout << "****************************************" << endl;\n\ncout << "ORDER OF THE MATRIX: " << n << endl;\ncout << "ENTER ELEMENT ROW-WISE" << endl;\nint m = 2*n;\nint i, j, k;\nrep(i, n) {\nrep(j, n) {\ncin >> mat[i][j];\ncout <<"&#92;t";\nprintf("%5.3lf", mat[i][j]);\n}\nrep2(j, n, m) {\nif(i == (j-n)){\nmat[i][j] = 1;\ncontinue;\n}\nmat[i][j] = 0;\n}\ncout << endl;\n}\n\ncout << endl << "INVERSE MATRIX: " << endl;\ndouble pivot, comm;\nrep(i, n) {\npivot = mat[i][i];\nif(mat[i][i] == 0){\ncout << "DIVISION BY ZERO." <<endl;\nreturn;\n}\nrep(j, m) mat[i][j] = mat[i][j]/pivot;\nrep(j, n) {\nif(i == j) continue;\ncomm = mat[j][i];\nrep2(k, i, m) {\nmat[j][k] = mat[j][k] - (comm*mat[i][k]);\n}\n}\n}\n\nrep(i, n) {\nrep2(j, n, m) {\ncout <<"&#92;t";\nprintf("%6.4lf", mat[i][j]);\n}\ncout << endl;\n}\ncout << endl;\n}\n\n/*This is writen in C++ this will output the inverse of a Matrix\ninput:\n4\n1 2 3 4\n2 1 4 2\n2 5 7 12\n1 -1 3 11\n\n5\n1.173 -0.634 0 0 0\n-0.539 1.173 -0.634 0 0\n0 -0.539 1.173 -0.634 0\n0 0 -0.539 1.173 -0.634\n0 0 0 -0.539 1.173\n\n0\n*/
1144513
1) pass by value\n2) pass by reference\n3) pass by value return\n4) pass by name\n\n\nAs the other answers mention, the most common mechanisms\nare pass by value and pass by reference.\n\nIn pass by value, a copy of the argument is passed to\nthe receiving function.  This is the mechanism used by\nC for scalars.\n\nIn pass by reference, the address of the argument is\npassed to the receiving function.  This is the mechanism\nused by C for arrays.\n\nHowever, that's not the end of the story.  At the very\nleast, you might want to consider other mechanisms that\nhave been used for other languages.\n\nFortran passes values by reference, but historically this\nlead to some "interesting" results where a subroutine could\nactually alter the value of a *constant* which had been\npassed as a parameter.  Because of this, some compilers\nimplemented "pass by value return".  This was a mechanism\nwhere  the caller would copy the argument to a temp\nlocation and pass *that* to the callee by reference.  Then\nupon return, the compiler could copy the result back to the\noriginal argument if-and-only-if it made sense (ie.  the\nargument was neither an expression nor a constant).\nThis mechanism has also been used for Ada.\n\nAlgol implemented an extraordinary mechanism called\n"call by name".  The simplest way to think of it is to\npretend that the function was implemented as a macro --\nwith the parameters expanded inline.  The actual\nimplementation, however was that instead of passing the\nparameter, the compiler would pass the address of a\nprocedure, called a "thunk", which would evaluate the\nparameter whenever it was actually needed in the called\nfunction.  This mechanism was complicated to implement and\nsurprising to use -- for example, it was not possible to\nimplement a correctly working version of a swap routine\nusing "call by name".\n\nIf the called sequence for parameters x and y was:\n\n    temp = x\n    x = y\n    y = temp\n\nThe function would not work correctly for arguments\n\n    i\n    a[i]\n\nbecause the arguments would be evaluated within the\ncalled procedure, (via a thunk), as if the code were:\n\n    temp = i\n    i = a[i]\n    a[i] = temp\n\nbut as you can see, the wrong element of a[] is updated\nbecause the index, "i", is modified by the preceeding\nstatement.  [Maybe that's why current languages haven't\nused this mechanism.]
41764
you can find a very good tutorial at the link.\n\nNo need to write whats already writtenm =)
120442
Because of the way you would call the function \nsomefunct(param1, param2, param3); // canonical\nsomefunct(param1, param2); // compiler knows you mean to use the default param for 3 because it´s the third param you don´t provide.