T is real nxn matrix s.t T^mx=0,T^(m-1) not 0 for some x.Prove x,Tx,T^2x,...,T^(m-1)x are linearly independent
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Homework.\n\nSuppose a1*x+a2*Tx+..+am*T^(m-1)x=0.\n\nApply T^(m-1) to get a1=0.\nApply T^(m-2) to get a2=0, etc.
1008487
x = t + 1/t\ny = t - 1/t\nAdd the two equations together.\nx + y = 2t\nThus, t = (x+y)/2\nNow replace t with (x+y)/2 in either one of the original equations to get the rectangular equation.\nx = t + 1/t\nx = (x+y)/2 + 2/(x+y)\nMultiply everything by 2(x+y)\n2x^2 + 2xy = x^2 + 2xy + y^2 + 4\nMove everything except the 4 to the left side\nx^2 - y^2 = 4\nIf you use the other equation, y = t - 1/t, you should get the same result.
949909
So Im guessing your have a transformation matrix.  \n\nYou have an initial state specified by matrix [A] and the linear transform specified by matrix [T].\n\nTo find the state after iteration 1 you multiply\n[A]x[T]=[y1]\n\nIf [y1] was your initial matrix you would multiply it by [T] to get the state after iteration.\n[y1]x[T]=[y2]\nby substitution\n([A] x [T]) x [T] = [y2]\nor \n[A] x [T]^2 = [y2]\n\nso what is T at the nth iteration?\n[A] x [T]^n = [yn]\n\nso what is the limit as n goes to inifinity?\n[A] x lim([T]^n,n->inf) = [y_inf]\n\nNumerical results break down, consider a rotation.  The limit as it goes to infinity is undefined.\nT(1,1)=sin(x)\nT(1,2)=cos(x)\nT(2,1)=-1*cos(x)\nT(2,2)=sin(x)\n\nWhen you square it you get\nT2(1,1)=sin(x)^2-cos(x)^2 = -cos(2*x)\nT2(1,2)=2*cos(x)*sin(x) = sin(2*x)\nT2(2,1)=-2*cos(x)*sin(x) = -sin(2*x)\nT2(2,2)=sin(x)^2-cos(x)^2 = -cos(2*x)\n\nWhen you cube it you get\nT3(1,1)=-sin(3*x)\nT3(1,2)=-cos(3*x)\nT3(2,1)=cos(3*x)\nT3(2,2)=-sin(3*x)\n\nWhen you raise it to the power of 100 (come on MatLab, dont die on me now, darnit) ... ahem.  When you raise it to the power of 10 (?)  aha\nT10(1,1)=-cos(10*x)\nT10(1,2)=sin(10*x)\nT10(2,1)=-sin(10*x)\nT10(2,2)=-cos(10*x)\n\nSee how that works?  Not bad huh?\n\nFor the pure rotation it never ends.\nWhat about a shrinking, a scaling by a matrix whose norm is less than 1? (note if you use 0.9, the norm is 1.8.)\n\nT(i,j)=0.45\nyields\nT2(i,j)=0.4050\nT3(i,j)=0.3645\nT10(i,j)=0.1743\nT1000(i,j)=0.1328*10^-4\n\nIt disappears to zero.\n\nintuitively you can also figure out that for a scaling by a matrix whose norm is larger than one the transform eventually goes to infinity as the limit is taken to infinity.\n\nyou could make a non-trigonometric rotation, like the sin(1/x)  Thats a fun expression.\n\nT(i,i)=sin(1/x) \nT(1,2)=cos(1/x)\nT(2,1)=-1*cos(1/x)\n\nAs x goes to infinity what happens?\nit becomes a reflection\nT_inf(i,i)=0\nT_inf(1,2)=1\nT_inf(2,1)=-1\n\nIm digressing.  Its a fun field with interesting applications.  Weather forecasting, deformation of solids, heat transfer, optimization, all use linear transformations to do their work.\n\nHave fun.
864701
Prove by induction.\n\nWhen k=1, then (n+a)^1 = n+a = Θn, true\n\nAssume true for k.\n\nFor k+1,\n(n+a)^(k+1) = (n+a) * (n+a)^k = Θn * Θ(n^k) = Θ(n^(k+1))\n\nTherefore it holds for all k>0
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Matrix Transpose means interchanging of rows and columns.\n\nIf A = \n[ 1 3 ]\n[ -2 -1]\n\nthen A^T =\n[ 1 -2 ]\n[ 3 -1 ]\n\nTransposing this last matrix, (A^T)^T =\n[ 1 3 ]\n[ -2 -1 ]\n= A.\n\nThis will be true no matter what the numbers are, or the number of rows or columns, so in general: (A^T)^T = A.\n\nTo show that (A+B)^T = A^T + B^T:\n\nA + B =\n[ 1+3 3+(-4) ]\n[ (-2)+2 (-1)+(-2) ] =\n\n[ 4 -1 ]\n[ 0 -3 ]\n\nso, (A+B)^T =\n[ 4 0 ]\n[ -1 -3 ]\n\nA^T =\n[ 1 -2 ]\n[ 3 -1 ]\n\nB^T =\n[ 3 2 ]\n[ -4 -2 ]\n\nAdding, A^T + B^T =\n\n[ 1+3 -2+2 ]\n[ 3+(-4) -1+(-2) ] =\n\n[ 4 0 ]\n[ -1 -3 ]\n\nSo, left side = right side. Again, the result is true for any matrices A and B, provided that they have the same number of rows and the same number of columns.
750249
firt let me tell you one of its property.\n\nLn M - Ln N = Ln (M/N)\n\nLn (3t-4) - Ln (t+1) = Ln 2 (apply the law)\nLn [(3t-4)/(t+1)] = Ln 2 (now, raise both side of the eqn by e)\n(3t-4)/(t+1) = 2 (both Ln was cancel)\n3t-4 = 2 * (t+1)\n3t - 4 = 2t + 2\n3t-2t = 2 + 4\nt = 6\n\nThat's it.
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x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
744242
let y=4x(x+x^1/2)^1/2\nlny=ln4+lnx+1/2ln(x+x^1/2)\ny'/y=1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)\ny'=y[1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)]
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Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
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false,\nuse the identity and -identity,\nboth are invertible, but the sum is the zero matrix, that is NOT invertible.....
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
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r = a_{n+1}/a_n -> 1 if <a_n> is a sequence of Real values that converges to a non-zero L.\n\na_{n+1}/a_n = (L + (a_{n+1} - L))/(L + (a_n - L)).  But (a_{n+1} - L) -> 0 and (a_n - L) -> 0 since a_n -> L so that (L + (a_{n+1} - L))/(L + (a_n - L)) -> (L+0)/(L+0) = L/L = 1
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Hello again dear Math student \n\nas you know ;\n▪sgn(x) = \n+1 , x >0\n 0  , x = 0\n- 1 , x <0\n\nso \n▪ D sgn(x)  = IR   \n▪ R sgn(x) = { -1 , 0 , +1 }\n\nGood Luck
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domain\nall real x\nx intercept\nputting y=0\n0=x+rt{x]\n-x=rtx\nx^2=x\nx(x-1)=0\nx=0 or 1\nx intercept=(0,0) and (1,0)\ny intercept\nsetting x=0\ny=0\ny intercept=(0,0)\nno symmetry exhibited
1183592
If your matrix is not square, then you can compute a generalized inverse matrix.  There are some links below.  There are many generalized inverses for matrices (other than those that are square and invertible; those only have one generalized inverse, which is its inverse).  But there is only one Moore-Penrose inverse for any given matrix.\n\nedit: reetz, I have already reported you for copying.  I suggest you back off.
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Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
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For any linear network which may be linear or non linear, active or passive, time varying or time invarient, the algebraic sum of powers delievered by all branches at any instant is zero. All currents and voltages in that nework must satisfy Kirchoff's laws