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| 18 Jun 2012 09:37 PM |
When a true group member has spent a long time in a group they will find the truth behind being in a group. I will give out the truth of this in a short paragraph. The truth behind being a true group member is when you realize that having a rank doesn't mean anything. We actually care about the ranks because we think that those ranks will earn us respect. But when you become a true group member you realize that you actually earned all of that respect yourself and not because of a rank. This is when you won't care about being a high rank. That is the time when you will be having more friends in the group. You will also start thinking about what would happen If out of no where the owner of that group would announce you as the new owner. You will be thinking about If you can handle such a successful group with your own hands. Thats the time when you will have your greatest fear being revealed. As long as your a true member with many friends around you that will help you get more confident in yourself then you will be fine. Thats a true group member in my own way.
[Sorry that I made it way longer] [Also sorry for my horrible grammar] [Don't feel like reading it then fine by me] [Hope this gave you some advices] |
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| 18 Jun 2012 09:40 PM |
| It would be nice to have atleast 1 comment about my thread right? Totally ignored :( |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 18 Jun 2012 09:41 PM |
the anti-derivative of x^2 + 3x + 9 from 0 to 10 times 3/20 then - 59
anti derivative is a big S but longer, 0 goes on the bottom of the S, 10 on the top, put 3/20 on the left side of the x, then right x^2 + 3x + 9 with parenthesis around the entire thing, put "dx" after the equation surrounded by parenthesis, then seperate it a from -59, you can put more parenthesis around the anti-derivative and equation, and that equals 0
Therefore
[(3!)!]! ==> [(3 * 2)!]! ==> [6 * 5 * 4 * 3 * 2]! ==> 720!
720 * 719 * 718 * ... * 3 * 2 * 1
720 / 5 = 144 factors of 5 720 / 25 = 28 factors of 25 720 / 125 = 5 factors of 125 720 / 625 = 1 factor of 625
Therefore, [(3!)!]! has 144 + 28 + 5 + 1 = 178 trailing zeroes
Then
a = c = 0, b = 5, and d = 4. Thus, p = 0. The cubic equation (6) is
The three real roots are -2, 2, 5/2. However, we must select q = 5/2 in order to satisfy equation (5). Then e2 = 5 - 5 = 0 and f2 = 25/4 - 4 = 9/4, giving e = 0 and f = 3/2. The quadratic equations (7) become x2 + 4 = 0 and x2 + 1 = 0, yielding the four roots to the original polynomial, x = -i, i, -2i, 2i.
Then -2, 2, -5/2, say q = 2. Then e2 = 4 + 5 = 9 and f2 = 4 - 4 = 0, giving e = 3 and f = 0. The quadratic equations (7) become
therefore
procedure Quadratic_Equation is type Roots is array (1..2) of Float; function Solve (A, B, C : Float) return Roots is SD : constant Float := sqrt (B**2 - 4.0 * A * C); X : Float; begin if B < 0.0 then X := (- B + SD) / 2.0 * A; return (X, C / (A * X)); else X := (- B - SD) / 2.0 * A; return (C / (A * X), X); end if; end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0); begin Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2))); end Quadratic_Equation;
So..
The notation is due to Legendre. If the real part of the complex number z is positive (Re(z) > 0), then the integral
converges absolutely. Using integration by parts, we see that the gamma function satisfies the functional equation:
Combining this with , we get:
for all positive integers n.
The absolute value of the gamma function on the complex plane. The identity Γ(z) = Γ(z+1) / z can be used (or, yielding the same result, analytic continuation can be used) to extend the integral formulation for Γ(z) to a meromorphic function defined for all complex numbers z, except z = −n for integers n ≥ 0, where the function has simple poles with residue (−1)n/n!. It is this extended version that is commonly referred to as the gamma function. [edit]Alternative definitions The following infinite product definitions for the gamma function, due to Euler and Weierstrass respectively, are valid for all complex numbers z, except the non-positive integers:
where is the Euler–Mascheroni constant. It is straightforward to show that the Euler definition satisfies the functional equation (1) above. A somewhat curious parametrization of the gamma function is given in terms of generalized Laguerre polynomials, which converges for Re(z) < 1/2.
[edit]The gamma function in the complex plane The behavior of for an increasing positive variable is simple: it grows quickly — faster than an exponential function. Asymptotically as , the magnitude of the gamma function is given by Stirling's formula
where the symbol ~ means that the quotient of both sides converges to 1. The behavior for nonpositive z is more intricate. Euler's integral does not converge for z ≤ 0, but the function it defines in the positive complex half-plane has a unique analytic continuation to the negative half-plane. One way to find that analytic continuation is to use Euler's integral for positive arguments and extend the domain to negative numbers by repeated application of the recurrence formula,
choosing n such that z + n is positive. The product in the denominator is zero when z equals any of the integers 0, −1, −2,... . Thus, the gamma function must be undefined at those points due to division by zero; it is a meromorphic function with poles at the nonpositive integers. The following image shows the graph of the gamma function along the real line:
The gamma function is nonzero everywhere along the real line, although it comes arbitrarily close as . There is in fact no complex number z for which , and hence the reciprocal gamma function is an entire function, with zeros at z = 0, −1, −2,.... We see that the gamma function has a local minimum at where it attains the value . The gamma function must alternate sign between the poles because the product in the forward recurrence contains an odd number of negative factors if the number of poles between and is odd, and an even number if the number of poles is even. Plotting the gamma function in the complex plane yields:
Absolute value
Real part
Imaginary part
Which means that saidfaic(2) multiplied by the variable pull of a black hole, + :( to the fifth power of a gregarian equations along with the ultimate mass of a fat guy = SADFAICx5
Therefore C&G members don't exist in the same dimension as a 15th dimension SADFAIC (multiplied by 5 times the quadratic equation of a horse, of course.) |
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| 18 Jun 2012 09:44 PM |
| Uhhh.. Sorry I'm horrible at math thats why I have summer school hehe... |
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| 18 Jun 2012 09:45 PM |
| He is kind of right. We need to stop worrying about the ranks and start representing our groups more as a team player. Not just for the ranks. |
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| 18 Jun 2012 09:47 PM |
| Yeah I know this group I use to be in made it just because the person was a higher rank he has control over everything. The person wasnt even the 3rd rank . |
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| 18 Jun 2012 09:50 PM |
| Well, Now I finally know a group its not about ranks. A group for me is about having fun with your friends and earning your respect within the group. |
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