Name: Lily Tsai

Section instructor: Tyler Section time: Friday 1-2

MATHEMATICS 23b/E-23b, SPRING 2014Practice Quiz # 1February 14, 2014

Problem 1st Try 2nd Try Points Score1 e 22 c 23 xxx xxx 34 xxx xxx 35 xxx xxx 36 xxx xxx 47 xxx xxx 4

Total 20

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To speed up grading, please transcribe your answers to questions 1 and 2into the “1st Try” column on the front page.

1. Which one of the following matrices has a determinant that is greaterthan zero?

(a)

0 1 0 01 0 0 00 0 1 00 0 0 1

(b)

1 1 0 00 0 1 11 0 1 00 1 0 1

(c)

1 2 1 02 0 2 13 0 3 00 1 0 1

(d)

1 0 0 20 1 0 00 0 1 02 0 0 1

(e)

0 0 0 10 0 1 10 1 1 11 1 1 1

2. If f(x) = 1 for rational x, f(x) = 0 for irrational x, and you attempt to

evaluate I =∫ 1

−1 f(x)dx, then

(a) Simpson’s rule, with any number n of subdivisions, gives 0 as theapproximate value of I.

(b) A Gaussian rule, using two points, gives 1 as the approximate valueof I.

(c) The average of the upper and lower Riemann sums equals 1, butthe integral does not exist.

(d) The integral exists and is equal to 2.

(e) The integral exists and is equal to 0.

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3. (3 points) A thin metal sheet is in the form of a quarter of a disc ofradius 1 with uniform density. Its area is of course π

4. By evaluating

an appropriate double integral, determine the location of its center ofmass. By symmetry, both coordinates of the center of mass are equal;so finding one of them will suffice.∫ 1

0dx

∫ √1−x20

ydy = .5y2 evaluated from 0 to√

1− x2

=∫ 1

0.5(1− x2)dx

= .5− (1/6)

= 1/3

1/3 ∗ 4/π = 4/(3π) = coordinate of y = coordinate of x

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4. (3 points - section problem 14.3) Sometimes you can evaluate a difficult-looking limit of a sum by noticing that the sum is an upper or lowerRiemann sum for an integral that is easy to evaluate.

Evaluate limN→∞

1

N2

N∑k=1

2N∑l=1

ek+lN .

The function for which this is a Riemann sum is of the form f1(x1)f2(x2).

Analyze the problem as if N = 2N , as with dyadic cubes. Then we get:

limN→∞

1

(2N)2

2N∑k=1

2(2N )∑l=1

ek+l

(2N ) .

= limN→∞

1

(2N)2

2N∑k=1

2(2N )∑l=1

ek

(2N ) ∗ el

(2N ) .

= limN→∞

1

(2N)2

2N∑k=1

ek

(2N )

2(2N )∑l=1

el

(2N ) .

This is essentially an upper Reimann sum of the function ekel (we canrecognize the familiar multiplication by the cubes’ volume).

To figure out the bounds of the integral, we substitute in ∞ for N and

evaluate ek

2N for when k = 1 and k = 2N and see that the bounds of krange from 0 to 1. Similarly, the bounds of l range from 0 to 2. (Wecan also use the fact that this is similar to dyadic cubes to figure out thebounds of the functions). ∫ 1

0

ekdk

∫ 2

0

eldl

We then evaluate this integral and get (e− 1)(e2 − 1)

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5. (3 points) Let

F (x) =

∫ 1

x

2ey2

dy.

(There is no formula for the value of this integral.)

By applying Fubini’s theorem to an appropriate double integral, provethat

∫ 1

0

F (x)dx = e− 1.

Please draw a diagram to show the region over which you are evaluatingthe double integral.∫ 1

0dx

∫ 1

x2ey

2dy =

∫ 1

0dy

∫ y0

2ey2dx

=∫ 1

02yey

2dy

= ey2

from 0 to 1

= e− 1

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6. (3 points) Suppose that you are evaluating the integral of the functionwhose value is 1 inside the square 1

5≤ x, y ≤ 3

5, 0 elsewhere, using dyadic

cubes. With the aid of a diagram, explain how to calculate to illustratethe upper and lower sums for N = 2, where you have 4 subdivisionsalong each axis for a total of 16 dyadic cubes. Then, by consideringupper and lower Riemann sums, show that the integral exists. A carefulexplanation of what happens for large N is sufficient; you need not do arigorous proof.

For cubes labeled ”X”, these add 0 to both UN and LN because themin and max values are both 0.

For cubes labeled ”U”, these add 1*their volume to UN and 0 to LNbecause the min = 0 and the max value = 1.

For cubes labeled ”B”, these add 1*their volume to UN and 1*theirvolume to LN because the min = max = 1.

Thus, U2 = 9 ∗ (1/22)2 = 9/8L2 = 1 ∗ (1/22)2 = 1/8

As N approaches infinity, the dyadic cubes that have oscillation notequal to 0 eventually include only the boundary of the square (the fourlines). This boundary is a set of volume zero (in 2 dimension) becauseit is one dimensional. Thus, since f is continuous everywhere except theboundary, it is integrable. Because adding the volume of the boundary(1, the oscillation, multipled by the volume of the ”U” cubes that includethe boundary of the square) to UN approaches adding 0, UN approachesLN and f is integrable.

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7. (4 points, - proof 15.1) Using only the linearity, antisymmetry and nor-malization properties of the determinant of an n× n matrix, prove thatdet(AB) = detA detB.

C = AB, A is linear.

~c1 = A~b1 = A(n∑i=1

bi,1~ei) =n∑i=1

bi,1A(~ei)) =n∑i=1

bi,1~ai.

This is true for all n columns of C. Thus:

detC = det(n∑

i1=1

bi1,1~ai1)(n∑

i2=1

bi2,2~ai2) · · · (n∑

in=1

bin,n~ain)

By linearity of the determinant function:

detC =∑

i1,i2,···in

bi1,1bi2,2 · · · bin,n det(~ai1 , ~ai2 , · · ·~ain)

nn terms in this sum, only n! of them are nonzero because they donot have two subscripts that are equal. Having two subscripts that areequal means the determinant is 0 because of antisymmetry (swapping twocolumns must change the sign of the determinant unless the determinantis 0).

These n! terms all have some factor of detA (permutations of A, differonly by sign)

Therefore detC = f(B) det( ~a1, ~a2, · · · ~an)

f(B) = function of products of all the entries of B.

This formula is valid for any A. Set A = identity matrix, then C = B,and det(A) = 1.

So detB = f(B) det(I) = f(B) and C = AB

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