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b) <math>\lim_{n\rightarrow\infty}n\int_{-\infty}^{\infty}e^{-(\frac{nx+1}{n})^2}-e^{-x^2}dx</math> | b) <math>\lim_{n\rightarrow\infty}n\int_{-\infty}^{\infty}e^{-(\frac{nx+1}{n})^2}-e^{-x^2}dx</math> | ||
| − | <math> = \lim_{n\rightarrow\infty}n\left[ \int_{-\infty}^{\infty}e^{-(\frac{nx+1}{n})^2}dx - \int_{-\infty}^{\infty}e^{-x^2}dx \right]</math> To do this we need both integrals to exist, which they do are just a scale/shift of the normal distribution. | + | <math> = \lim_{n\rightarrow\infty}n\left[ \int_{-\infty}^{\infty}e^{-(\frac{nx+1}{n})^2}dx - \int_{-\infty}^{\infty}e^{-x^2}dx \right]</math> To do this we need both integrals to exist, which they do since they are just a scale/shift of the normal distribution. |
Let <math>y = (\frac{nx+1}{n})</math> using the Riemann u-substitution. | Let <math>y = (\frac{nx+1}{n})</math> using the Riemann u-substitution. | ||
Revision as of 05:11, 10 July 2009
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 $
