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hw9sol

# hw9sol - Honors Introduction to Analysis I Homework IX...

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Honors Introduction to Analysis I Homework IX Solution April 21, 2009 Problem 1 Suppose f is C 1 on an interval and that f 0 satisfies the H¨older condition of order α , | f 0 ( x ) - f 0 ( y ) | ≤ M | x - y | α for all x, y in the interval, where α (0 , 1] is fixed. Show that | Δ 2 h f ( x ) | ≤ c | h | 1+ α . How does the constant c relate to M ? Solution. We apply the mean value theorem to g ( t ) = f ( t + h ) - f ( t ) on the interval [ x, x + h ]. Then Δ 2 h f ( x ) = f ( x + 2 h ) - 2 f ( x + h ) + f ( x ) = f ( x + 2 h ) - f ( x + h ) - ( f ( x + h ) - f ( x )) = g ( x + h ) - g ( x ) and g 0 ( t ) = f 0 ( t + h ) - f 0 ( t ). By the MVT we get that g ( x + h ) - g ( x ) = hg 0 ( x 0 ) for some x 0 ( x, x + h ), which is equivalent to saying Δ 2 h f ( x ) = h ( f 0 ( x 0 + h ) - f 0 ( x 0 )) so | Δ 2 h f ( x ) | = | h || f 0 ( x 0 + h ) - f 0 ( x 0 ) | ≤ | h | M | x 0 + h - x 0 | α = M | h | 1+ α c | h | 1+ α for c M . Hence | Δ 2 h f ( x ) | ≤ c | h | 1+ α for c M (the minimal value for c , for which this is true, is M ) Problem 2 Suppose that f has a zero of order j at x 0 and that g has a zero of order k at x 0 . What can you say about the order of zero of the function f + g , f · g and f/g at x 0 ? Solution. By definition f ( x ) = a j ( x - x 0 ) j + o ( | x - x 0 | j ) and g ( x ) = a k ( x - x 0 ) j + o ( | x - x 0 | k ), or f ( n ) ( x 0 ) = 0 for n = 0 , ..., j - 1 and f ( j ) ( x 0 ) 6 = 0 and g ( n ) ( x 0 ) = 0 for n = 0 , ..., k - 1 and g ( k ) ( x 0 ) 6 = 0. Let m = min { j, k } , then ( f + g ) ( n ) ( x 0 ) = 0 for n = 0 , ..., m - 1 and ( f + g ) ( m ) ( x 0 ) 6 = 0. Or ( f + g )( x ) = a m ( x - x 0 ) m + o ( | x - x 0 | m ), so f + g has a zero of order min { j, k } at x 0 . For any n , ( fg ) ( n ) ( x ) = n i =0 a i f ( i ) ( x ) g ( n - i ) ( x ) for a i = n i . From this we notice that ( fg ) ( n ) ( x 0 ) = 0 for n = 0 , ..., k + j - 1 and ( fg ) ( k + j ) ( x 0 ) = k + j j f ( j ) ( x 0 ) g ( k ) ( x 0 ) 6 = 0, so fg has a zero of order j + k . Equivalently one can see that ( fg )( x ) = a k a j ( x - x 0 ) j + k + o ( | x - x 0 | k + j ). Let’s first analyze the case when j < k . We write: f ( x ) g ( x ) = f ( j ) ( x 0 )( x - x 0 ) j + ... + a j f ( k ) ( x 0 )( x - x 0 ) k + o ( | x - x 0 | k ) g ( k ) ( x 0 )( x - x 0 ) k + o ( | x - x 0 | k ) or f ( x ) g ( x ) = f ( j ) ( x 0 ) + ... + a j f ( k ) ( x 0 )( x - x 0 ) k - j + o ( | x - x 0 | k - j ) g ( k ) ( x 0 )( x - x 0 ) k - j + o ( | x - x 0 | k - j ) where a j comes from the Taylor expansion, after dividing by 1 /k !. 1

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The first term will be f ( j ) ( x 0 ) g ( k ) ( x 0 )( x - x 0 ) k - j + o ( | x - x 0 | k - j ) , which gives us a singularity at x 0 , so the function is not defined there. Hence in order to be possible to have zeroes at x 0 we need that j k . In that case we have that f ( x ) g ( x ) = f ( j ) ( x 0 )( x - x 0 ) j + o ( | x - x 0 | j ) g ( k ) ( x 0 )( x - x 0 ) k + o ( | x - x 0 | k ) or f ( x ) g ( x ) = f ( j ) ( x 0 )( x - x 0 ) j - k + o ( | x - x 0 | j - k ) g ( k ) ( x 0 ) + o (1) o
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hw9sol - Honors Introduction to Analysis I Homework IX...

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