HW 03 - Mathematics 218: Analysis in Several Variables...

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Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Third Problem Set Rik Sengupta October 14, 2009 1 Group 1 1. Find 1 2 sin ( x 1 x 2 ) and 2 1 sin ( x 1 x 2 ). Solution. We have 1 2 sin ( x 1 x 2 ) = 1 { x 1 cos ( x 1 x 2 ) } = x 1 1 cos ( x 1 x 2 ) + cos ( x 1 x 2 ) 1 x 1 = - x 1 x 2 sin ( x 1 x 2 ) + cos ( x 1 x 2 ) . Similarly, we also get 2 1 sin ( x 1 x 2 ) = 1 { x 2 cos ( x 1 x 2 ) } = x 2 1 cos ( x 1 x 2 ) + cos ( x 1 x 2 ) 1 x 2 = - x 2 2 sin ( x 1 x 2 ) . 2. Find 2 1 e x 1 + x 2 + x 3 and 1 2 3 e x 1 + x 2 + x 3 . Solution. We have 2 1 e x 1 + x 2 + x 3 = 1 e x 1 + x 2 + x 3 = e x 1 + x 2 + x 3 . Similarly, 1
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1 2 3 e x 1 + x 2 + x 3 = 1 2 e x 1 + x 2 + x 3 = 1 e x 1 + x 2 + x 3 = e x 1 + x 2 + x 3 . 3. Find the Taylor expansion at the origin up to three terms plus the re- mainder for the function f ( x 1 ,x 2 ,x 3 ) = e x 1 x 2 sin ( x 2 x 3 ). You may use your previous knowledge of the Taylor expansions of the functions e x and sin x of one variable if you prefer. Solution. We have e x 1 x 2 = 1 + x 1 x 2 + 1 2 ( x 1 x 2 ) 2 + 1 6 ( x 1 x 2 ) 3 + ... and sin ( x 2 x 3 ) = x 2 x 3 - 1 6 ( x 2 x 3 ) 2 + 1 5! ( x 2 x 3 ) 5 - .... Multiplying these series together and retaining only the non-negligible terms, we get e x 1 x 2 sin ( x 2 x 3 ) = x 2 x 3 + x 1 x 2 2 x 3 + ( t 2 1 t 3 2 t 3 2 - t 3 2 t 3 3 6 ) for some t = ( t 1 ,t 2 ,t 3 ) between 0 and x = ( x 1 ,x 2 ,x 3 ) on the line con- necting them. The remainder is ± , where ± consists of the third order terms in the Taylor expansion. 4. Find the directional derivative u ( x 2 1 x 2 +sin ( πx 1 x 3 )) at the point (0 , 1 , 2) in the direction of the unit vector u = 1 14 (1 , 2 , 3). Solution. We have, putting f ( x 1 ,x 2 ,x 3 ) = ( x 2 1 x 2 + sin ( πx 1 x 3 )), 2
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u ( x 2 1 x 2 + sin ( πx 1 x 3 )) = 3 X i =1 i f (0 , 1 , 2) u i = 1 f (0 , 1 , 2) u 1 + 2 f (0 , 1 , 2) u 2 + 3 f (0 , 1 , 2) u 3 = (2 x 1 x 2 + πx 3 cos ( πx 1 x 3 )) 1 14 + x 2 1 2 14 + πx 1 cos ( πx 1 x 3 ) 3 14 = 2 π 1 14 + 0 + 0 = 2 π 14 . 2 Group 2 5. Find 2 ∂r 2 f ( r cos θ,r sin θ ) for a twice differentiable function f ( x 1 ,x 2 ) of two variables. Solution. We have, ∂r f ( x 1 ,x 2 ) = 1 f ∂x 1 ∂r + 2 f ∂x 2 ∂r . So, differentiating with respect to r again, we get 2 ∂r 2 f ( x 1 ,x 2 ) = ∂r ( 1 f ∂x 1 ∂r + 2 f ∂x 2 ∂r ) = 1 f 2 x 1 ∂r 2 + ∂x 1 ∂r ( 2 1 f ∂x 1 ∂r + 2 1 f ∂x 2 ∂r ) + 2 f 2 x 2 ∂r 2 + ∂x 2 ∂r ( 2 2 f ∂x 2 ∂r + 1 2 f ∂x 1 ∂r ) Here, x 1 = r cos θ and x 2 = r sin θ , so 2 x i ∂r 2 = 0 for i = 1 , 2. Hence we get 2 ∂r 2 f ( r cos θ,r sin θ ) = cos θ (cos θ∂ 2 1 f + 2 1 f sin θ ) + sin θ (sin θ∂ 2 2 f + 1 2 f cos θ ) = cos 2 θ∂ 2 1 f + cos θ sin θ ( 1
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This note was uploaded on 05/03/2010 for the course MAT 218 at Princeton.

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HW 03 - Mathematics 218: Analysis in Several Variables...

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