(New page: MA_598R_pweigel_Summer_2009_Lecture_4 <math>\text{Prove that the sum}</math> <math>\sum_{n=0}^{\infty}{\int_{0}^{\pi/2}{\left(1-\sqrt{\sin x}\right)^n\cos x }dx }</math> <math>\text...) |
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[[MA_598R_pweigel_Summer_2009_Lecture_4]] | [[MA_598R_pweigel_Summer_2009_Lecture_4]] | ||
| − | <math>\text{Prove that the sum}</math> | + | <math>\text{4.6) Prove that the sum}</math> |
<math>\sum_{n=0}^{\infty}{\int_{0}^{\pi/2}{\left(1-\sqrt{\sin x}\right)^n\cos x }dx }</math> | <math>\sum_{n=0}^{\infty}{\int_{0}^{\pi/2}{\left(1-\sqrt{\sin x}\right)^n\cos x }dx }</math> | ||
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<math>\int_{0}^{\pi/2}{\frac{\cos x}{\sqrt{\sin x}} } dx = \int_{0}^{1}{\frac{1}{\sqrt{x}} } dx = 2</math> | <math>\int_{0}^{\pi/2}{\frac{\cos x}{\sqrt{\sin x}} } dx = \int_{0}^{1}{\frac{1}{\sqrt{x}} } dx = 2</math> | ||
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| + | -Ben Bartle | ||
Latest revision as of 18:10, 5 July 2009
MA_598R_pweigel_Summer_2009_Lecture_4
$ \text{4.6) Prove that the sum} $
$ \sum_{n=0}^{\infty}{\int_{0}^{\pi/2}{\left(1-\sqrt{\sin x}\right)^n\cos x }dx } $
$ \text{converges to a finite limit, and find its value.} $
$ \text{Solution: } $
$ \sum_{n=0}^{\infty}{\int_{0}^{\pi/2}{\left(1-\sqrt{\sin x}\right)^n\cos x }dx } $
$ = \lim_{ N \rightarrow\infty}{\sum_{n=0}^{N}{\int_{0}^{\pi/2}{\left(1-\sqrt{\sin x}\right)^n\cos x }dx }} \text{ by definition} $
$ = \lim_{ N \rightarrow\infty}{\int_{0}^{\pi/2}{\cos x\sum_{n=0}^{N}{\left(1-\sqrt{\sin x}\right)^n } }dx} \text{ because the sum is finite} $
$ = \int_{0}^{\pi/2}{\cos x\sum_{n=0}^{\infty}{\left(1-\sqrt{\sin x}\right)^n } dx} \text{ by the Monotone Convergence Theorem} $
$ \left(1-\sqrt{\sin x}\right)^n <1 \text{ on } \left(0,\pi/2\right] \text{ so } \cos x \sum_{n=0}^{\infty}{\left(1-\sqrt{\sin x}\right)^n} = \frac{\cos x}{\sqrt{\sin x}} \text{ is finite a.e. on a bounded domain, so the integral exists} $
$ \int_{0}^{\pi/2}{\frac{\cos x}{\sqrt{\sin x}} } dx = \int_{0}^{1}{\frac{1}{\sqrt{x}} } dx = 2 $
-Ben Bartle
