Selected Answers to Problem Set 6

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1. Using the leastsqrs method in Maple, we create a matrix A and vector v such that

Then by executing leastsqrs(A,v); we get:

which equates to a = 2.0066, b = -1.08, and c = 1.033.

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2. First, multiply both sides by (a + bx²) then factor the y through. This yields:

In part (b), perform a similar operation; multiply both sides by (1 + e^{(a - bx)}, divide both sides by y, subtrace one, then find the natural log. This yields:

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3. Here are the matrices and their inverses. Notice that if you want to compose two functions (that is, rotate by +45 degrees, then mirror) you multiply them in reverse order! This is the standard definition of composition of functions. That is, (f o g)(x) = g(f(x)), because you first do f(x), then you do g(x). Included in this table is F and Finv, where F mirrors over the horizontal axis (and Finv does the same thing, so F * Finv = I)

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4. a)
b)

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5.

a)	b)
c)	d)

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6. We know that the dot product will yield zero iff the vectors are perpendicular (aka orthogonal). So:

(1+a)(2) + (-2b)(1) + (4)(-1) = 0
2a - 2b = 2
a - b = 1
a = b + 1

If the vector U is perpendicular to W as well, then we solve the equation again, knowing that a = b + 1:

(1+a)(1) + (-2b)(1) + (4)(0) = 0
a - 2b + 1 = 0
a = 2b - 1

Solving these linearly give us:

a = b + 1
a = 2b - 1
a = 3, b = 2

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7.  Since V is orthogonal to X, then  = 0:

(v1)(3) + (v2)(4) + (v3)(0) = 0
(v2)(4) = -(v1)(3)
v2 = -(3/4)(v1)

Second, we can put this equation into a linear form:

 1 = c(3) + v1
-2 = c(4) + v2
 1 = c(0) + v3

Now we can substitute for v2 to find c:

 1 = 3c + v1
-2 = 4c - (3/4)(v1)

 3 =  9c + 3v1
-8 = 16c - 3v1
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-5 = 25c  =>  c = -(1/5)

Knowing c we can find v1 and v2

 1 = 3(-1/5) + v1
 8/5 = v1

 v2 = -(4/3)(8/5)
    = -(6/5)

c = -1/5, v1 = -6/5, v2 = 8/5

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8.

||X||2 = x12 + x22
||Y||2 = y12 + y22
||X+Y||2 = (x1 + y1)2 + (x2 + y2)2
         = x12 + x22 + y12 + y22 + 2x1y1 + 2x2y2
         = ||X||2 + ||Y||2 + 2x1y1 + 2x2y2

If ||X+Y||2 = ||X||2 + ||Y||2, then 2x1y1 + 2x2y2 = 0
x1y1 + x2y2 = 0 iff the angle between them is 90 degrees (cos x = 0).
That is, x1y1 + x2y2 = < X,Y >

Without components, this looks like:
||X+Y|| = < X,X > + < Y,Y > + 2< X,Y >

2< X,Y > = 0 if the angle is Pi/2.

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