MATH 141 In-Class Worksheet Week 14 Solutions. Winter 2023

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In-Class Worksheet Week 14

Math 141

Page 1 of 11

“Known” Maclaurin series expansions and the interval on which they’re valid: • ex =

∞ X xn n=0

• sin x =

n!

∞ X (−1)n x2n+1 n=0

• cos x =

−∞ < x < ∞

,

−∞<x<∞

(2n + 1)!

∞ X (−1)n x2n

• tan−1 x =

−∞<x<∞

(2n)!

n=0

∞ X (−1)n x2n+1

• ln(1 + x) =

−1≤x≤1

2n + 1

n=0

∞ X (−1)n−1 xn

−1<x≤1

n

n=1

p(p − 1) 2 p(p − 1)(p − 2) 3 x + x + ··· 2!

3! (this is true for more values of x for certain values of p)

• (1 + x)p = 1 + px +

−1<x<1

Question 1 Find the Maclaurin series of each of the following functions. Write the first three nonzero terms as well as the full series using summation notation. (a) h(x) = x3 cos(x3 ) Z (b) i(x) = x3 cos(x3 ) dx.

(a) We have cos y = 1 −

∞ X (−1)n y 2n y2 y4 + − ··· = , 2 4!

(2n)! n=0

for all y,

and so cos(x3 ) = 1 − =1−

∞ X (−1)n (x3 )2n (x3 )2 (x3 )4 + − ··· = 2 4!

(2n)! n=0 ∞ X x12 (−1)n x6n x6 + − ··· = 2 24 (2n)! n=0

x3 cos(x3 ) = x3 −

for all x,

for all x,

∞ X x9 x15 (−1)n x6n+3 + − ··· = for all x .

2 24 (2n)! n=0

(b) Using our answer from the previous part, we have Z

Z Z x9 x15 3 3 3 x cos(x ) dx = x − + − · · · dx = 2 24

∞ X (−1)n x6n+3 n=0

(2n)!

! dx

∞ Z X x4 x10 x16 (−1)n x6n+3 =C+ − + − ··· = dx 4 2 · 10 16 · 24 (2n)! n=0 = C+

∞ X

(−1)n x6n+4

. n=0 (6n + 4) (2n)!

In-Class Worksheet Week 14

Math 141

Page 2 of 11

Question 2 Z Write the first three nonzero terms (not counting the integration constant) of the Maclaurin series of 1 √ dx.

4 + x2 We have

−1/2

−1/2 x2 x2 1 1 √ = (4 + x2 )−1/2 = 4−1/2 1 + = 1+ , 4 2 4 4 + x2 1 and so we can use the binomial series with p = − : 2

− 12 − 32 2 1 −1/2 (1 + y) =1+ − y+ y − ··· 2 2 1 3 = 1 − y + y2 − · · · for |y| < 1, 2 8

2 ! −1/2 1 x2 1 1 x2 3 x2 1+ = 1− + − ··· 2 4 2 2 4 8 4

for |y| < 1,

for

x2 < 1, 4

3x4 1 x2 − + − ··· for |x2 | < 4, 2 16 256

Z Z 1 3x4 1 x2 − + − · · · dx for |x| < 2, dx = 4 + x2 2 16 256 =

= C+

x x3 3x5 − + − ··· 2 48 1280

for |x| < 2 .

Question 3 Find the sum of each of the following series if they converge. (a)

∞ X (−1)n π 2n+1 2n+1 4 (2n + 1)! n=0 We recognize this as ∞ X (−1)n (π/4)2n+1 n=0

(2n + 1)!

= sin

π 4

√ =

2 .

2

(The Maclaurin series of sin x equals sin x for all x.) ∞ X (−1)n √ (b) ( 3)2n+1 (2n + 1) n=0 We recognize this as √

∞ X 1 (−1)n (1/ 3)2n+1 π = arctan √ = .

2n + 1 6 3 n=0

(c)

(The Maclaurin series of arctan x equals arctan x for −1 ≤ x ≤ 1.) √ ∞ X (−1)n ( 3)2n+1 2n + 1

√ This series diverges . We are cautious since 3 > 1, so our “known” series table does not assert anything about this series, and indeed the Ratio Test Yields , √ √ (−1)n ( 3)2n+1 (−1)n+1 ( 3)2n+3 = · · · = 3 > 1. lim n→∞ 2n + 3 2n + 1 n=0

Question 4 In statistics, the normal distribution with mean µ and standard deviation σ has the density function e−(x−µ) /2σ √ f (x) = . σ 2π 2

2

In-Class Worksheet Week 14

Math 141

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The normal distribution corresponds to the “bell curve” that you may have seen. For an example of how they are used, if the SAT score, x, of PSU applicants follows a normal distribution with mean µ and standard deviation σ, then Z b

e−(x−µ) /2σ √ dx = the fraction of PSU applicants with SAT score between a and b. σ 2π a 2

2

Unfortunately, we are unable to write antiderivatives of these types of density functions using our usual functions and combinations thereof, so historically this type of integral would be written into a “standard form”, after which you would consult a table of appropriate numerical values.

The standard normal distribution has mean µ = 0 and standard deviation σ = 1, and has density function 2 e−x /2 √ . φ(x) = 2π (a) Find the Maclaurin series for φ(x). Write out the first three nonzero terms as well as the full series using summation notation.

We know ∞ X y2 yn ey = 1 + y + + ··· = for all y, 2 n! n=0 and so

2 2 2 n X

2 ∞ − x2 − x2 1 x + + ··· = , φ(x) = √ 1+ − 2 2 n!

2π n=0

which can be simplified to ∞ X 1 (−1)n x2n x2 x4 √ φ(x) = √ − √ + √ + · · · = .

2π 2 2π 8 2π 2π(n!) n=0

(b) Use the first three nonzero terms from (a) to approximate Z 1 φ(x) dx. −1

Since φ(−x) = φ(x) for all x, φ is an even function. Therefore, Z 1

Z 1 φ(x) dx = 2

−1

φ(x) dx.

0

(If you don’t notice this it’s not a big deal — it will take a little more time to plug in the bounds later but will give you the same answer.) Then,

Z 1 Z 1 1 x2 x4 √ − √ + √ φ(x) dx ≈ 2 dx 2π 2 2π 8 2π −1 0

1 x x3 x5 =2 √ − √ + √ 2π 6 2π 40 2π 0 2 1 1 =√ − √ + √ 2π 3 2π 20 2π

1 1 1 =√ 2− + 3 20 2π =

103 √ .

60 2π

In-Class Worksheet Week 14

Math 141

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(c) You may have heard “In a normal distribution, about 68% of the values are within 1 standard deviation of the mean.” Confirm that your answer to (b) is consistent with this well known statistical fact. (For example, the worldwide SAT score distribution has mean 1060 and standard deviation 217, meaning that about 68% of test takers have scores between 1060 − 217 = 843 and 1060 + 217 = 1277.)

We have 103 √ ≈ 68.49%, 60 2π which is consistent with the “68-95-99.7 Rule” — in normally distributed data, about 68.27% of the samples are within one standard deviation of the mean; about 95.45% are within two standard deviations of the mean, and about 99.73% are within three standard deviations of the mean.

Problems to work on throughout the last week of class: Question 5 For this problem consider the complex numbers z = 3 + 2i and w = 5 − i. For the following, give your answer in standard form a + bi, where a and b are real. (a) Calculate z + w.

We have z + w = (3 + 2i) + (5 − i) = 8 + i . (b) Calculate z − w.

We have z − w = (3 + 2i) − (5 − i) = 3 + 2i − 5 + i = −2 + 3i . (c) Calculate zw.

We have zw = (3 + 2i)(5 − i) = 15 + 10i − 3i − 2i2 = 15 + 7i + 2 = 17 + 7i . z (d) Calculate . w For this we need the “conjugate trick”:

3 + 2i 5 + i 15 + 13i − 2 1 1 z 35 + 10i + 3i + 2i2 = = 2 = + i. = w 5−i 5+i 52 − (i2 ) 5 − (−1) 2 2 This trick always works to rewrite a fraction with a complex denominator into a fraction with a real denominator since any complex number times its complex conjugate is real: (a + bi)(a − bi) = a2 − abi + abi − b2 i2 = a2 − b2 i2 = a2 + b2 .

π

π

Question 6 Consider the complex numbers z = 2e 4 i and w = 4e 2 i . (a) Use Euler’s identity separately for z and w, and then calculate zw. Give your answer in standard a + bi form.

We have ! √ √ ! π π √ √ π 2 2 i z = 2e 4 = 2 cos + i sin =2 + i = 2 + 2 i, 4 4 2 2 ! π π π w = 4e 2 i = 4 cos + i sin = 4 (0 + i)) = 4i, 2 2 √ √ √ √ √ √ zw = 2 + 2 i (4i) = 4 2 i + 4 2 i2 = −4 2 + 4 2 i .

In-Class Worksheet Week 14

Math 141

Page 5 of 11

(b) Do the same calculation but first calculate zw while still in exponential form (using familiar exponential arithmetic), then use Euler’s identity on the result. Give your answer in standard a + bi form, making sure your answer agrees with (a).

We have π π zw = 2e 4 i 4e 2 i π

π

= 2(4)e 4 i+ 2 i 3π

= 8e 4 i

= 8 cos

3π 4

+ i sin

√ !

2 2 =8 − + i 2 2 √ √ = −4 2 + 4 2 i .

3π 4

!

√

Question 7 Here we use Euler’s identity to derive the double angle identities for cosine and sine. On one hand, e2θi = cos(2θ) + i sin(2θ).

On the other hand, 2 e2θi = eiθ = (cos θ+i sin θ)2 = cos2 θ+sin θ cos θ i+sin θ cos θ i+sin2 θ i2 = (cos2 θ−sin2 θ)+(2 sin θ cos θ) i.

When written in standard form (a, b, c, d all real), a + bi = c + di guarantees both a = c and b = d, we have the familiar double angle identities cos(2θ) = cos2 θ − sin2 θ,

sin(2θ) = 2 sin θ cos θ.

Use similar reasoning to discover identities for cos(3θ) and sin(3θ).

We have e3θi = cos(3θ) + i sin(3θ).

On the other hand, e3θi = eiθ

3 3

= (cos θ + i sin θ)

= cos3 θ + 3(cos θ)2 (i sin θ) + 3(cos θ)(i sin θ)2 + (i sin θ)3 = cos3 θ + 3 cos2 θ sin θ i + 3 cos θ sin2 θ i2 + sin3 θ i3 = cos3 θ + 3 cos2 θ sin θ i − 3 cos θ sin2 θ − sin3 θ i, where we used i2 = −1 and i3 = i2 i = −i. We can organize this into standard form

e3θi = cos3 θ − 3 cos θ sin2 θ + (3 cos2 θ sin θ − sin3 θ)i.

The only way a + bi could equal c + di with a, b, c, d real is if both a = c and b = d, and so our two expressions for e3θi yield the triple angle identities cos(3θ) = cos3 θ − 3 cos θ sin2 θ,

sin(3θ) = 3 cos2 θ sin θ − sin3 θ .

Question 8 Find polar coordinates (with 0 ≤ θ < 2π) for each of the following points in the xy-plane.

In-Class Worksheet Week 14

Math 141 (a) (1, 1) We have

r= tan θ =

p

x2 + y 2 =

Page 6 of 11

p √ 12 + 12 = 2,

y = 1, x

and since we know (1, 1) is in the first quadrant we can deduce r =

(b) (1, −1) We have

√

2, θ =

π .

4

p p √ x2 + y 2 = 12 + (−1)2 = 2, y tan θ = = −1, x r=

ti

ait

and since we know (1, −1) is in the fourth quadrant we can deduce r =

√

2, θ =

7π .

4

ti ait

(c) (−1, 1) We have

ti

p p √ x2 + y 2 = (−1)2 + 12 = 2, y tan θ = = −1, x r=

and since we know (−1, 1) is in the second quadrant we can deduce r =

ti

unify

√

2, θ =

3π .

4

In-Class Worksheet Week 14

Math 141

Page 7 of 11

(d) (−1, −1) We have r=

p

x2 + y 2 =

p

(−1)2 + (−1)2 =

√

2,

y tan θ = = 1, x and since we know (−1, −1) is in the third quadrant we can deduce r =

√

2, θ =

5π .

4

ait

√ (e) (1, 3) We have ti

q p √ √ x2 + y 2 = 12 + ( 3)2 = 4 = 2, ait √ y tan θ = = 3, x r=

√ π and since we know (1, 3) is in the first quadrant we can deduce r = 2, θ = . (You should 3 be very familiar with the cosine and sine special angle values, and a nice way to remember the π π π √ 1 ordering of tan = √ , tan = 1, tan = 3 is to keep in mind that f (x) = tan x is 6 4 3 3 an increasing function.)

ti unify

√ (f) (−5 3, 5) We have q p √ 2 √ x2 + y 2 = (−5 3 + 52 = 100 = 10, y 1 tan θ = = − √ x 3 r=

unify

√ 5π . and since we know (−5 3, 5) is in the second quadrant we can deduce r = 10, θ = 6

In-Class Worksheet Week 14

Math 141

Page 8 of 11

unify

Question 9 Express the following complex numbers in the form reiθ with 0 ≤ θ < 2π. (a) i The complex number i = 0 + 1i corresponds to the point (0, 1) in the xy-plane, which we immeπ diately know has polar coordinates r = 1 and θ = . Therefore, 2 π

π

i = 1e 2 i = e 2 i .

You can check your answer by using Euler’s identity.

t

(b) −i The complex number −i = 0 + (−1)i corresponds to the point (0, −1) in the xy-plane, which we 42.21 3π . Therefore, immediately know has polar coordinates r = 1 and θ = 2

Éo

t

i i =0 1e 2 2 = e2i.

3π

You can check your answer by using Euler’s identity.

3π

É

t

42.21

Éo

(c) 2 + 2i This complex number corresponds to the point (2, 2) in the xy-plane, which corresponds to r =

0 2

t

Math 141

In-Class Worksheet Week 14 Page 9 of 11 p p √ t y π x2 + y 2 = 22 + 22 = 2 2, and tan θ = = 1. The point is in the first quadrant so θ = . x 4 Therefore, √ πi 2 + 2i = 2 2e 4 .

You can check your answer by using Euler’s identity.

42.21

Éo

0 2

t

É

√ (d) 4 − 2 3 i √ This complex number corresponds to the point (4, −2 3) in the xy-plane, which corresponds to q p √ √ √ √ r = x2 + y 2 = 42 + (−2 3)2 = 16 + 12 = 28 = 2 7, and

√ √ y 2 3 3 tan θ = = − =− . x 4 2 Be careful! This is a special value for sine and cosine, but not a special value for tangent.

Therefore, the best we can do is say √ ! √ !

3 3 θ = arctan − = − arctan (tangent and therefore arctangent are odd functions), 2 2 π but arctangent outputs values between − and 0 for negative inputs. Therefore, to obtain an 2 angle in 0 ≤ θ < 2π that is coterminal we simply add 2π and use t √ !

3 θ = 2π − arctan .

2 √ √ (2π−arctan √3 )i 2 4 − 2 3 i = 2 7e .

42.21

Éo

0 2

t

É

Question 10 Provide inequalities for r and θ that precisely describe the following regions in the xy-plane. (a) the region bounded by the circle x2 + y 2 = 9 This region consists of points at most 3 away from the origin, and since we think of r as a distance we have 0 ≤ r ≤ 3 . There are no restrictions on θ but to make sure we count points only once,

É

t

In-Class Worksheet Week 14

Math 141

Page 10 of 11

we restrict 0 ≤ θ < 2π .

42.21

Éo

t

0 2

t

É

(b) the region bounded by the circle 42.21 x2 + y 2 = 9 in the 2nd quadrant This region consists of points at most 3 away from the origin, and since we think of r as a distance π we have 0 ≤ r ≤ 3 . To obtain second quadrant points only, we restrict ≤θ≤π .

2 2

Éo

0

t

É

x (c) the region in the first quadrant bounded by the x-axis, the line y = √ , and the circle x2 + y 2 = 2 3 √ √ Here we have 0 ≤ r ≤ 2 since the radius of the circle is 2. Thinking of the picture and using x the fact that the polar angle of the line y = √ satisfies 3 tan θ = we know the line makes an angle of

1 y =√ , x 3

π π with the positive x-axis. Therefore, we choose 0 ≤ θ ≤ .

6 6

(d) the triangular region with vertices (0, 0), (0, 1), and (1, 1) Hint: this can be done in polar coordinates, but rectangular coordinates are better suited. This example is so you learn that polar coordinates aren’t always the best choice.

The issue here is that the ”outer bounding curve” (with respect to the origin) is NOT a circle centered at the origin — in order to precisely describe the region we must have our r bounds π π depend on θ. Thinking of the picture, the θ values satisfy ≤ θ ≤ . The lower bound for 4 2

Math 141

In-Class Worksheet Week 14

Page 11 of 11

r is 0 regardless of θ, but the upper bound for r must be obtained from the “outer bounding curve” y = 1. Recalling that y = r sin θ, we can obtain the polar equation of the horizontal line √ π π is r = csc θ. Therefore, 0 ≤ r ≤ csc θ . (The inequalities 0 ≤ r ≤ 2 2, ≤ θ ≤ are true for all 4 2 points in the region — however we cannot draw the region from these bounds! We would obtain an eighth-disk. To precisely describe certain regions, more complicated inequalities are needed!)

It makes more sense to think of the region as lying above the “lower bounding cruve” y = x and below the “upper bounding curve” y = 1 for 0 ≤ x ≤ 1, and so the rectangular bounds would be 0 ≤ x ≤ 1, x ≤ y ≤ 1 .