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Unformatted text preview: C v2 w2 =kvkkwk D v w=kvkkwk. This is cos because ˇ ˛ D .
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.u2 C U1 / and ju2 jjU2 j 1 .u2 C U2 /. The whole line
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1
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becomes :96 .:6/.:8/ C .:8/.:6/ 2 .:6 C :8 / C 2 .:8 C :6 / D 1. True: :96 < 1.
p
The cosine of is x= x 2 C y 2 , near side over hypotenuse. Then j cos j2 is not greater
than 1: x 2 =.x 2 C y 2 / 1.
The vectors w D .x; y/ with .1; 2/ w D x C 2y D 5 lie on a line in the xy plane.
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The shortest w on that line is .1; 2/. (The Schwarz inequality kwk v w=kvk D 5
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is an equality when cos D 0 and w D .1; 2/ and kwk D 5.)
The length kv wk is between 2 and 8 (triangle inequality when kvk D 5 and kwk D
3). The dot product v w is between 15 and 15 by the Schwarz inequality.
Three vectors in the plane could make angles greater than 90ı with each other: for
example .1; 0/; . 1; 4/; . 1; 4/. Four vectors could not do this (360ı total angle).
How many can do this in R3 or Rn ? Ben Harris and Greg Marks showed me that the
answer is n C 1: The vectors from the center of a regular simplex in Rn to its n C 1
vertices all have negative dot products. If n C 2 vectors in Rn had negative dot products,
project them onto the plane orthogonal to the last one. Now you have n C 1 vectors in
Rn 1 with negative dot products. Keep going to 4 vectors in R2 : no way!
For a speciﬁc example, pick v D p 2; 3/ and then w D . 3; 1; 2/. In this example
.1; p
cos D v w=kvkkwk D 7= 14 14 D 1=2 and D 120ı . This always
happens when x C y C z D 0:
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v w D xz C xy C yz D .x C y C z/2
.x C y 2 C z 2 /
2
2
1
1
This is the same as v w D 0
kvkkwk: Then cos D :
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2
p
Wikipedia gives this proof of geometric mean G D 3 xyz arithmetic mean
A D .x C y C z/=3. First there is equality in case x D y D z . Otherwise A is
somewhere between the three positive numbers, say for example z < A < y .
Use the known inequality g a for the two positive numbers x and y C z A. Their
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mean a D 2 .x C y C z A/ is 2 .3A A/ D same as A! So a g says that
A3 g 2 A D x.y C z A/A. But .y C z A/A D .y A/.A z / C yz > yz .
Substitute to ﬁnd A3 > xyz D G 3 as we wanted to prove. Not easy! 24 Example 6 gives ju1 jjU1 j
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28 29 30 There are many proofs of G D .x1 x2 xn /1=n A D .x1 C x2 C C xn /=n. In
calculus you are maximizing G on the plane x1 C x2 C C xn D n. The maximum
occurs when all x ’s are equal.
31 The columns of the 4 by 4 “Hadamard matrix” (times 1 ) are perpendicular unit
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vectors:
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3
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1
1
1
1 61
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HD 4
:
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1
15
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32 The commands V D randn .3; 30/I D D sqrt .diag .V 0 V //I U D V \D I will give 30 random unit vectors in the columns of U . Then u 0 U is a row matrix of 30 dot
products whose average absolute value may be close to 2= . Solutions to Exercises 6 Problem Set 1.3, page 29
1 2s1 C 3s2 C 4s3 D .2; 5; 9/. The same vector b comes from S times x D .2; 3; 4/: " 100
110
111 #" # "
# "#
.row 1/ x
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2
3 D .row 2/ x D 5 :
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.row 2/ x
9 2 The solutions are y1 D 1, y2 D 0, y3 D 0 (right side D column 1) and y1 D 1, y2 D 3, y3 D 5. That second example illustrat...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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