{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW1 - MATH 108B HW 1 SOLUTIONS RAHUL SHAH Problem 1[6.1...

This preview shows pages 1–3. Sign up to view the full content.

MATH 108B HW 1 SOLUTIONS RAHUL SHAH Problem 1. [ § 6.1] Solution. By the law of cosines, k x k 2 + k y k 2 - k x - y k 2 2 = k x kk y k cos( θ ) . Thus k x kk y k cos( θ ) = x 2 1 + x 2 2 + y 2 1 + y 2 2 - ( x 2 1 - 2 x 1 y 1 y 2 1 ) - ( x 2 2 - 2 x 2 y 2 y 2 2 ) 2 = 2 x 1 y 1 + 2 x 2 y 2 2 = x 1 y 1 + x 2 y 2 = h x,y i . ± Problem 2. [ § 6.2] Solution. ’ Since h u,v i = 0, h u,av i = 0 for any a F . Thus k u + av k 2 = k u k 2 + k av k 2 by the Pythagorean theorem. Thus k u + av k 2 - k u k 2 = k av k 2 . Since k av k 2 0, k u k 2 ≤ k u + av k 2 and thus k u k ≤ k u + av k . ’ If either u,v = 0, this is obvious. So assume otherwise. Write u = av + ( u - av ) as the sum of a scalar multiple of v and a vector orthogonal to v . Thus k u k 2 = k av k 2 + k u - av k 2 (Pythagorean theorem). However, k u k 2 ≤ k u - av k 2 and thus k av k 2 = 0 and thus a = 0 (since v 6 = 0). However a = h u,v i k v k 2 and thus h u,v i = 0. ± Problem 3. [ § 6.4] Solution. By a simple application of the parallelogram law, we ﬁnd that k v k = 17. ± Problem 4. [ § 6.5] 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 RAHUL SHAH Solution. If there was an inner product that gave rise to the given norm, then the norm would satisfy the parallelogram equality. However, k (1 , - 1) + (1 , 1) k 2 + k (1 , - 1) - (11 , 1) k 2 = k (2 , 0) k 2 + k (0 , 2) k 2 = 4 + 4 = 8 6 = 16 = 2(4 + 4) = 2 k (1 , - 1) k 2 + 2 k (1 , 1) k 2 Thus the given norm cannot have come from an inner product. ± Problem 5. [ § 6.6] Solution. Notice that h u + v,u + v i = k u + v k 2 . However we also have that h u + v,u + v i = h u,u i + h u,v i + h u,v i + h v,v i = k u k 2 + k v k 2 + 2 h u,v i ; since V is a real inner-product space k u + v k 2 = k u k 2 + k v k 2 + 2 h u,v i h u,v i = | u + v k 2 - ( k u k 2 + k v k 2 ) 2 = k u + v k 2 2 - k u + v k 2 + k u - v k 2 4 ; by the parallelogram law = k u + v k 2 - k u - v k 2 4 ± Problem 6. [ § 6.9] Solution. It is easy to see that 1 2 π R π - π sin( nx ) dx = 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

HW1 - MATH 108B HW 1 SOLUTIONS RAHUL SHAH Problem 1[6.1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online