What is the commutator of (x)op and (H)op, i.e. [x(op), H(op)]?
126249
commutator: [x, H] = (xH - Hx)\n\n\napply to Hilbert vector ψ\n\n(xH - Hx)ψ\n\n\n= x(p²/2m + V)ψ - (p²/2m + V)xψ\n\n= x*p²ψ/2m + x*V*ψ - p²(x*ψ)/2m - V*x*ψ\n\n= x*p²ψ/2m - p²(x*ψ)/2m \n\n= (x*p²ψ - p²(x*ψ))/2m\n\n\n1-dim case: p = -ih∂\n\n\n= -h²(x*∂²ψ - ∂²(x*ψ))/2m\n\n= -h²(x*∂²ψ - ∂(∂x*ψ + x*∂ψ))/2m\n\n= -h²(x*∂²ψ - (∂∂x*ψ + ∂x*∂ψ + ∂x*∂ψ + x*∂∂ψ))/2m\n\n= -h²(x*∂²ψ - 2*∂ψ - x*∂²ψ)/2m\n\n= h²(∂ψ)/m\n\n= ih(-ih∂ψ)/m\n\n= ih*pψ/m\n\n\n=> [x, H] = -ihp/m
359931
Its substitution.\n\nYou are given p=i*v\n\nand you need to express it in terms of one variable using ohms law v=i*r\n\nYou want p, so you need to trade out either the i or the v from p=i*v\n\nif you trade out the v using ohms law as (i*r) then your power equation becomes p=i*(i*r) = i^2 *r\n\nif you trade out the i using ohms law, first you solve for i getting i=v/r then substitute into the p-equation.  \nThe substitution yields p=(v/r)*v  = v^2 /r\n\nThis is straightforward algebra.
488168
You reasoning is faulty.  \n\nSR (-1) = SR (i * i) = i\n\n-i = ( SR (-1) ) * -1\n\ni^2 = -1 by definition\n\n(-i)^2 = (-i * -i) = ( (-1 * i) * (-1 * i) ) = (-1 * -1) * (i * i) = (+1) * (-1) = -1
54048
here's an output step by step:\n\nthis is the initial state of deque D\nD :: first -> <- last\n\naddFirst(3)\nD :: first -> 3 <- last\n\naddLast(8)\nD :: first -> 3 8 <- last\n\naddLast(9)\nD :: first -> 3 8 9 <- last\n\naddFirst(5)\nD :: first -> 5 3 8 9 <- last\n\nremoveFirst()\nD :: first -> 3 8 9 <- last\n\nremoveLast()\nD :: first -> 3 8 <- last\n\nfirst()\nfirst = 3\n\naddLast()\nD :: first -> 3 8 <- last\n\nremoveFirst()\nD :: first -> 8 <- last\n\nlast()\nlast = 8\n\nremoveLast()\nD :: first -> <- last\n\nhope this helps...
476030
we have V*T = const\nV1*T1 = V2*T2\n=> V2 = V1*T1/ T2 = 3.8 * (-45 + 273) / (45 + 273) = 2.72L
418375
If h(x) is I-able, then h(x) is defined and exists for any x in between [a,b]. So Int{x=a until b} of h(x)*dx=H(b)-H(a). And back: If h(x) is defined in [a,b], then h(x) is I-able, being |H(b)-H(a)| a definite number.\nNow if both f(x) & g(x) are defined, then h(x)=f(x)*g(x) is also defined, thus h(x) is I-able.\nIf g(x)=f(x), then f^2(x) is also I-able.\nmay it help you Kevin!
565431
2x<1 , x<0.5\n -1<x\n\n     -1< x<0.5
83429
for (i=1,t=1,cosx=1;i<=n;++i)\n{\n  t=(-1)*(t*x*x)/((2*i)*(2*i-1);\n  cosx += t;\n}
365900
2points answer.\n\nTry again.  Your question is unreadable.\n\nDon't use x as both a variable and an operator.  On that first one, which side of the multiplication is the variable x on?\n\nI don't understand what the third one is trying to say at all.\n\nUse * for multiplication, use / for division.\n\nAnd most importantly, what answer do you want from this?  You aren't showing any equalities or inequalities.  The second one shows five variables, so there aren't enough expressions to solve them anyway.
620069
([O2]^(O2 coefficient)*[H2]^(H2 coefficient))/\n[H2O2]^(H2O2 coefficient)
348515
A projection operator is defined by the property (idempotence):\n\nPP = P\n\nTo prove that for:\n\nP = |j><j|\n\nyou need to assume that |j> is normalized so that <j|j> = 1. In that case:\n\nPP = |j><j|j><j|\n= |j><j|\n= P\n\nAlso, if |v> is any vector:\n\nP |v> = |j><j|v> = u |j>\n\nwhere u=<j|v> (some scalar). This is an eigenvector of P with eigenvalue 1 if and only if:\n\nu |j> = |v>\n\nin other words if and only if |v> is any scalar multiple of |j>. (In general, eigenvectors are unique only to within multiplication by a scalar). The operator |j><j| projects any vector into the 1 dimensional subspace spanned by |j>. That is, it preserves any scalar multiple of |j> and kills any vector orthogonal to |j>
440701
df/dx=(df/dg)*(dg/dx)=x’=1, dg/dx=(dg/df)*(df/dx)=x’=1; both are const\nIt means g’(b)=1
45758
#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*/
415329
do you mean 6x([1/2]x^2)?\nif so 3x^3\nor 6x(1/(2x^2))?\nthen its 3x^-1
161015
-> is the in-place dereference operator. It is most commonly used in OOP.\n\n$a->{'b'} is exactly the same as $$a{'b'}\n\n-> can also be used for 'method' calls with blessed objects.\n$a->function(); is exactly the same as module::function($a);\n\nAn 'object' is Perl is actually just a reference to a variable, usually a hash (associative array). As such, whenever you want to use an object, you have to dereference it (othewise perl doesn't know what to do).\n\nSee the following in perldoc:\nhttp://perldoc.perl.org/perlop.html - Perl operators\nhttp://perldoc.perl.org/perlref.html - Perl references\nhttp://perldoc.perl.org/perlobj.html - Perl objects\nhttp://perldoc.perl.org/perlmod.html - Perl modules
394202
The exponent is not divided by (1 + 1) in 2*x^(1+1)/(1+1), the whole expression is. ie 2x²/2 = x². Its not to be confused with 2x^(2/2) ie 2x^1 = 2x\n\nWhen there are limits it makes no difference\n\neg ∫(from x = a to x = b) 2x dx \n= [x² + c](from x = a to x = b)\n= (b² + c) - (a² + c)\n=  b² - a² (irrespective of the value of c)
163332
it will be choice (c)\n(x+1)(x-1)\nx^2+x-x-1\nx^2-1