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George Mason UniversityConor Philip NelsonSource: Home > Classes > Math 400 > Final Project > IntroductionMath 400: History of MathematicsConsider these instances of the factorial: $$3!?$$ $$2!?$$ $$1!?$$ What about \(0!\)? By defintion, this is equal to 1. However, how is this derrived? "Consider a particle of mass \(m\) which is moving under the influence of an attractive froce whose intensity varies inversely as the distance of the particle ofrom the center of attraction. The equation of such a motion is obtainable immediately form the defintion of force (Newton's second law). Denoting the distance from the center of attraction: $$m\frac{d^2y}{dt^2}=-\frac{k}{y}$$ or $$\frac{d^2y}{dt^2}=-\frac{a}{y}$$ where \(a=\frac{k}{m}\). This is a non-linear equation of the type $$\frac{d^2y}{dt^2}=F(y)$$ which always can be solved by multiplying both sides of the equation by \(2\frac{dy}{dt}\) and integrating. THus, $$2\frac{dy}{dt}\frac{d^2y}{dt^2}=-2\frac{dy}{dt}\frac{a}{y}$$ and integrating with respect to \(t\) gives $$\left(\frac{dy}{dt}\right)^2=-2a\log(y) +c$$ if the velocity of the particle is zero when \(y=y_0\), then \(c=2a\log(y_0)\) and $$\frac{dy}{dt}=-\sqrt{2a\log(\frac{y_0}{y})}$$ The negative sign was chosen for the square root because \(y\) is a decreasing functino of \(t\). Solving for \(dt\) and integrating, $$t=-\frac{1}{\sqrt{2a}}\int^{y}_{y_0} \frac{dy}{\sqrt{\log(\frac{y_0}{y})}}$$ This integral can be put in a simpler form by making the obvious transformation \(\log(\frac{y_0}{y})=x\) or \(y=y_0e^{-x}\). The integral becomes $$t=\frac{y_0}{\sqrt{2a}}\int^{y=y_0e^{-x}}_0x^{-\frac{1}{2}}e^{-x}dx$$ This integral cannot be evaluated in terms of elementary functions. In fact, an integral of this type led Euler to the discovery of the so-called Gamma functions. For purposes of computation the integrand can be expanded in an infinite seires, and term-by-term integration will give the series solution for \(t\)." $$\Gamma(n) = \int^{\infty}_0e^{-x}x^{n-1}dx$$
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