# HW 02 - Mathematics 218 Analysis in Several Variables...

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Unformatted text preview: Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Second Problem Set Rik Sengupta October 7, 2009 1 Group 1 1. f : R 2 → R 1 , where f ( x ) = x 2 1 + x 3 2 + x 1 x 2 . Solution. The derivative is the 1x2 matrix ( 2 x 1 + x 2 3 x 2 2 + x 1 ) The partial derivatives are simply the elements of this matrix, where a ij = ∂ j f i . Here, since there are no different components f i , our task is easier, and the two elements represent in order ∂ 1 f and ∂ 2 f . 2. f : R 3 → R 3 , where f ( x ) = x 1 + x 2 x 3 2 + x 2 3 x 1 x 3 . Solution. The derivative is the 3x3 matrix 1 1 3 x 2 2 2 x 3 x 3 x 1 The partial derivatives are simply the elements of this matrix, where a ij = ∂ j f i . 3. f : R 2 → R 1 , where f ( x ) = sin( x 2 1 + x 2 ). Solution. The derivative is the 1x2 matrix ( 2 x 1 cos( x 2 1 + x 2 ) cos( x 2 1 + x 2 ) ) 1 The partial derivatives are simply the elements of this matrix, where a ij = ∂ j f i . Here, since there are no different components f i , our task is easier, and the two elements represent in order ∂ 1 f and ∂ 2 f . 4. f : R 1 → R 2 , where f ( x ) = cos(sin x ) e sin x . Solution. The derivative is the 2x1 matrix- sin(sin x )cos x e sin x cos x The partial derivatives are simply the elements of this matrix, where a ij = ∂ j f i . 2 Group 2 5. Show that the function f ( x ) = p | x 1 x 2 | (the non-negative square root) is not differentiable at the origin. Show that a function f ( x ) such that | f ( x ) | ≤ k x k 2 2 in an open neighborhood of the origin is differentiable at the origin, and find its derivative at the origin. If a function f ( x ) satisfies | f ( x ) | ≤ k x k 2 in an open neighborhood of the origin is it necessarily differentiable at the origin? (Either show that it is differentiable or give a counterexample.) Solution. We can approach the origin from two sides – from the right side along the line x = y and from the left along the line x + y = 0. Taking the partials ∂ 1 f ( x ) on both sides, we get 1 2 q | x 2 | | x 1 | ∂ | x 1 | ∂x 1 , which equals + 1 2 on the right side and- 1 2 on the left side. Since these two partials are not the same at the origin, it follows that the function f ( x ) is not differentiable at the origin. For the second part, take some h within the open neighborhood of the origin. Then we have | f ( h )- f (0) | ≤ | f ( h ) | + | f (0) | ≤ k h k 2 2 + k k 2 2 ≤ k h k 2 2 , so that | f ( h )- f (0) | k h k 2 ≤ k h k 2 . Now if we take the limit as k h k 2 goes to zero, we get | f (0) | ≤ 0 = ⇒ f (0) = 0. Thus, the function f ( x ) is differentiable at the origin, and its derivative is 0....
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HW 02 - Mathematics 218 Analysis in Several Variables...

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