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Unformatted text preview: Math 8601. November 17, 2010. Midterm Exam 2. Problems and Solutions. Problem 1. Find lim n 1 n + 1 n + 1 + + 1 2 n 1 . Solution. After rewriting S n := 1 n + 1 n + 1 + + 1 2 n 1 = 1 n n 1 X k =0 1 1 + k n , one can see that this is a Riemann sum for the integral 1 R (1 + x ) 1 dx . Therefore, lim n S n = Z 1 dx 1 + x = ln(1 + x ) 1 = ln2 . Problem 2. Let f ( x,y ) be a function defined on the unit square { ( x,y ) R 2 : 0 x 1 , y 1 } , which is continuous in each variable separately. Show that f is a Borel measurable function of ( x,y ). Proof. For integers n = 1 , 2 ,..., define the functions F k,n ( x,y ) := f ( x, 0) { } ( y ) + 2 n X k =1 f x, k 2 n I k,n , where I k,n := k 1 2 n , k 2 n , For fixed y , the function f ( x,y ) is continuous in x , therefore it is Borel measurable in x (Corollary 2.2), hence F k,n ( x ) are Borel measurable in ( x,y ). We obviously have F n ( x, 0) f ( x, 0) for all x [0 , 1], and...
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 Fall '08
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 Math

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