what is big-theta notation?
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(m)\n\nTwo important ways to characterize the effectiveness of an algorithm are its space complexity and time complexity. Time complexity of an algorithm concerns determining an expression of the number of steps needed as a function of the problem size. Since the step count measure is somewhat coarse, one does not aim at obtaining an exact step count. Instead, one attempts only to get asymptotic bounds on the step count. Asymptotic analysis makes use of the O (Big Oh) notation. Two other notational constructs used by computer scientists in the analysis of algorithms are Θ (Big Theta) notation and Ω (Big Omega) notation.\nThe performance evaluation of an algorithm is obtained by totaling the number of occurrences of each operation when running the algorithm. The performance of an algorithm is evaluated as a function of the input size n and is to be considered modulo a multiplicative constant.
1052728
the c actually represents combination\nnow 5c2(actually, the 5 is above c on the left side and the 2 is below c on the right) means selection of 2 articles from a grp of 5\nin an exp say, (a+b)^n\nwe have the rth term as nc(r-1)a^r*b^(n-r)\nso in (x+3)^5 the third term is(n=5, r=3, a=x, b=3)\n5c(3-1)x^3*3^2=10*9*x^3=90x^3
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1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
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-\na=(2, 1, 1)\nb=(1, root3-1, -root3 -1)\n\ncos(theta)= a.b/|a||b|\n= (2 +sqrt(3)-1 -sqrt(3)-1 )/|a||b| = 0\nso theta = pi/2 (90 degrees)
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
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I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
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Polar\nits use to converse from dimensional to polar.\nexp\n(3,4) --> (5;53)\nits mean 5 for radius and 53degree
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printf("%-15u&#92;n",40000);\n\nprovided I understood it correctly.
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!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
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simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
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There are several reasons. One of the most basic ones is that this representation provides a "polar" representation of complex numbers. Any complex number z=a+ib can be written as\n\nz= r exp(i theta)\nr= sqrt(a^2+b^2)\ntheta = atan(b/a)\n\nThis helps because, for example, a product of z1 and z2 would give\n\nz1 z2 = R exp(i phi)= r1*r2 exp (theta1 + theta2)\nR=r1*r2\nPhi = theta1 + theta2\n\nSo multiplying by z2 "stretches" z1 by amount r2 and "rotates" z1 by theta2. Such interpretations are convenient in many instances. Also, exponential functions are easier to deal with than regular a+ib type expressions in other cases.
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When converting from rectangular coordinates to polar coordinated, remember the following.\n\n@ = Theta\nx = r*cos(@)\ny =r*sin(@)\n\nr^2 = x^2 +y^2\n\ntan(@) = y/x\n\nso your equation\nx^2 +y^2 =36 becomes\nr^2 = 36   OR\nr =6 (which is a circle with radius 6)
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See the explanation in the Wikipedia here:\n\nhttp://en.wikipedia.org/wiki/Methods_of_computing_square_roots\n\nScroll down the the section marked "An exact long-division-like algorithm".
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You have not provided adequate information about the quation to find an answer.\n\n\nBTW, it's "Ramanujan"
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This is generally used for answering or solving problems.\nFor example if you do something like:\n4sin^2(theta)-9cos^2(theta)=(2sin(theta)+3cos(theta))(2sin(theta)-3cos(theta)) \nYou may write "Using x^2-y^2=(x+y)(x-y)"\nHope this helps...
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take the log\nso log b= log (1+(r/n))^n*t\nlog b = nt log (1+(r/n))\n{(log b)} / {log (1+(r/n))} = nt\n[{(log b)} / {log (1+(r/n))}] / n = t\ni hope those parenthesis dont mess u up lol\ni just learned this in school
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Delta = Theta x Pi\nGamma = Omega / Alpha ^ Delta\nTheta = Zeta Reticuli * Omega * c^2