Tutorial-4

# Tutorial-4 - Analysis II Tutorial Solution sketches...

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Analysis II Tutorial Solution sketches Alexandre Tomberg * 1 February 3, 2011 1.1 We will prove a famous theorem about continuous functions: Theorem 1.1 (Weierstrass’ Approximation Theorem) . Polynomials are dense in ( C ([ a, b ]) , k·k ) , where k f k = max x [ a,b ] | f ( x ) | . Actually, we will concentrate on the following explicit version of this theorem: Theorem 1.2. Let f C ([0 , 1]) and > 0 be given. Then there exists a polynomial P s.t. k f - P k < . Moreover, P can be chosen in the following concrete way. Set P n ( x ) = n X k =0 f ( k/n ) n k x k (1 - x ) n - k . Then n 0 > 0 such that for all n n 0 , k P n - f k < . Proof. We start with the binomial formula and differentiate it with respect to x . ( x + y ) n = n X k =0 n k x k y n - k = x ∂x ( x + y ) n = nx ( x + y ) n = n X k =0 k n k x k y n - k . * Please report any errors and typos to [email protected] All tutorial solutions in one pdf can be found at http://www.cs.mcgill.ca/ ~ atombe2/AnalysisTutorial.pdf . 1

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Taking y = 1 - x , we obtain nx = n X k =0 k n k x k (1 - x ) n - k . (1) Similarly, x 2 2 ∂x 2 ( x + y ) n = n ( n - 1) x 2 ( x + y ) n = n X k =0 k ( k - 1) n k x k y n - k , and n ( n - 1) x 2 = n X k =0 k ( k - 1) n k x k (1 - x ) n - k . (2) Adding ( 1 ) to ( 2 ), we get n ( n - 1) x 2 + nx = n X k =0 k 2 n k x k (1 - x ) n - k . Now, let r k ( x ) = ( n k ) x k (1 - x ) n - k . We already know that n X k =0 r k ( x ) = 1 n X k =0 kr k ( x ) = nx n X k =0 k 2 r k ( x ) = n ( n - 1) x 2 + nx. Consider n X k =0 ( k - nx ) 2 r k ( x ) = n X k =0 k 2 r k ( x ) - 2 nx n X k =0 kr k ( x ) + n 2 x 2 n X k =0 r k ( x ) = n ( n - 1) x 2 + nx - 2 n 2 x 2 + n 2 x 2 = nx - nx 2 = nx (1 - x ) . Recall that we defined P n ( x ) = n X k =0 f ( k/n ) n k x k (1 - x ) n - k = n X k =0 f ( k/n ) r k ( x ) . (3) And obviously, f ( x ) = f ( x ) n X k =0 r k ( x ) = n X k =0 f ( x ) r k ( x ) (4) 2
Now, since f C ([0 , 1]), f is continuous on [0 , 1], and therefore uniformly continuous there, i.e.

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