Revision as of 04:11, 10 July 2009 by Bbartle (Talk | contribs)

a) $ \lim_{n\rightarrow\infty}n\int_{-1}^{1}e^{-(\frac{nx+1}{n})^2}-e^{-x^2}dx $

$ = \lim_{n\rightarrow\infty}\int_{-1}^{1}\frac{e^{-(x+\frac{1}{n})^2}-e^{-x^2}}{\frac{1}{n}}dx $

The Mean Value Theorem implies $ \exist h \in (0,\frac{1}{n}) $ s.t. $ \frac{e^{-(x+\frac{1}{n})^2}-e^{-x^2}}{\frac{1}{n}} = \frac{d}{dx}(e^{-x^2})|_{h} $

$ = \lim_{n\rightarrow\infty}\int_{-1}^{1}-2he^{-h^2}dx $

$ = \lim_{n\rightarrow\infty}-4he^{-h^2} $

as $ n\rightarrow\infty $, $ h\rightarrow 0 $ so

$ = 0 $

b) $ \lim_{n\rightarrow\infty}n\int_{-\infty}^{\infty}e^{-(\frac{nx+1}{n})^2}-e^{-x^2}dx $

$ = \lim_{n\rightarrow\infty}n\left[ \int_{-\infty}^{\infty}e^{-(\frac{nx+1}{n})^2}dx - \int_{-\infty}^{\infty}e^{-x^2}dx \right] $ To do this we need both integrals to exist, which they do since they are just a scale/shift of the normal distribution.

Let $ y = (\frac{nx+1}{n}) $ using the Riemann u-substitution.

$ = \lim_{n\rightarrow\infty}n\left[ \int_{-\infty}^{\infty}e^{-y^2}dy - \int_{-\infty}^{\infty}e^{-x^2}dx \right] $

$ = \lim_{n\rightarrow\infty} 0 $

$ = 0 $

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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