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Unformatted text preview: In Lab 2, our mission is to explore the values of the integral Gamma(P) = x x px1 xE xx xxx . We did this by inputting it into the program Joy of Mathematica and evaluating for the integers 110. Our results saw that as p increased, so did Gamma(P). Looking even closely at the data, we noticed that Gamma(P) was equal to Gamma(P1) * (P1). Simplifying the equation gave us Gamma(P+1)=Gamma(P)*P. Next it was time to try this out for nonintegers. Using numbers such as ½, 3/2, etc., we calculated the equation and saw similar results. Once again Gamma(P+1)=Gamma(P)*P. Even when using exotic values such as 2.3457, the formula held true as long as the intervals were being increased by one each time. Now, the function, first evaluated by the Swiss mathematician Leonhard Euler in the 16 th century, must serve some purpose. If one was to look closely at the data, this is easy to see. Below are the charts of Gamma(P) and P!...
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 Spring '08
 Pietraho
 Calculus, Integers, Leonhard Euler, GAMMA, p+1, mathematician Leonhard Euler

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