MATH 141 CALC ANLY GEOM II In-Class Worksheet Week 12 . Solutions. Winter 2023.

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In-Class Worksheet Week 12

Math 141

Page 1 of 7

Question 1 Determine the interval of convergence of the following power series. (a) (b) (c) (d) (e)

∞ X (x + 2)n n=1 ∞ X

n 3n (−1)n

n=1 ∞ X

(−1)n

n=1 ∞ X

(x + 2)n n 3n (x + 2)2n n 3n

(x + 2)2n n 3n n=1 ∞ X (x + 2)3n n=1

n 3n

Recall that to calculate the radius of convergence, we evaluate the Ratio Test limit and set the result < 1, and then rewrite into the form |x − c| < R. Then, • absolute convergence is guaranteed for c − R < x < c + R, • divergence is guaranteed for x < c − R or x > c + R, • the endpoints x = c ± R are inconclusive.

Therefore, the interval of convergence will be c − R < x < c + R, along with either or both endpoints after they’re considered separately using a test other than the Ratio Test. If you get a series for an endpoint on which the Ratio Test would work, you made a mistake since they’re precisely where the Ratio Test is inconclusive. (a) For the Ratio Test limit we have

|x + 2|n+1 n 3n (x + 2)n+1 (x + 2)n lim = lim · n+1 n n+1 n→∞ (n + 1)3 n→∞ (n + 1)3 n3 |x + 2|n

n |x + 2|n+1 n 3 = lim n→∞ |x + 2|n n+1 3n+1

1 n = lim |x + 2| n→∞ n+1 3 | {z } →1

|x + 2| = .

3 We set this < 1 and manipulate: |x + 2| <1 3 |x + 2| < 3, −3 < x + 2 < 3, −5 < x < 1.

Therefore the interval of convergence is −5 < x < 1 along with either or both endpoints, but those need to be considered separately since they’re precisely when the Ratio Test limit is 1.

For x = −5, we have ∞ X (−5 + 2)n n=1

n 3n

=

∞ X (−3)n n=1

n3n

=

∞ X (−1)n 3n n=1

n3n

=

∞ X (−1)n n=1

n

,

1 1 which converges by the Alternating Series Test (since is decreasing and lim = 0). n→∞ n n For x = 1, we have ∞ ∞ ∞ X X X (1 + 2)n 3n 1 = = , n n n3 n3 n n=1 n=1 n=1

In-Class Worksheet Week 12

Math 141

Page 2 of 7

which we know diverges.

Therefore, the interval of convergence is −5 ≤ x < 1 . (b) Here the first steps of finding the radius and open interval of convergence are unchanged compared to (a) — since we take absolute values when doing the ratio test and |(−1)n | = 1 the calculations are the same.

However, the endpoints will not be the same. For x = −5 we have ∞ X

(−1)n

n=1

∞ ∞ ∞ n X X X (−1)2n 3n 1 (−5 + 2)n n (−3) = = = (−1) , n n n 3n n3 n3 n n=1 n=1 n=1

which diverges (for any integer n, 2n is even thus (−1)2n = 1).

For x = 1 we have ∞ ∞ X X (−1)n 3n (1 + 2)n = (−1)n n = , (−1)n n n3 n3 n n=1 n=1 1 is decreasing with limit 0). n Therefore, the interval of convergence is −5 < x ≤ 1 . which converges by the Alternating Series Test (

(c) For the Ratio Test limit we have

2(n+1) 2n |x + 2|2n+2 n 3n n+1 (x + 2) n (x + 2) lim (−1) (−1) = lim · n→∞ n→∞ (n + 1)3n+1 |x + 2|2n (n + 1)3n+1 n 3n n

3 |x + 2|2n+2 n = lim n→∞ |x + 2|2n n+1 3n+1

1 n 2 = lim |x + 2| n→∞ n+1 3 | {z } →1

2

=

|x + 2| .

3

We set this < 1 and manipulate: |x + 2|2 <1 3 2 |x + 2| < 3, √ |x + 2| < 3, √ √ − 3 < x + 2 < 3, √ √ −2 − 3 < x < −2 + 3. √ √ Therefore the interval of convergence is −2 − 3 < x < −2 + 3 along with either or both endpoints, but those need to be considered separately since they’re precisely when the Ratio Test limit is 1. √ For x = −2 − 3, we have √ √ 2n √ 2 n ∞ ∞ ∞ ∞ ∞ 2n n X X X X X 3 + 2)2n (−1)n n (−2 − n (− 3) n (−1) ( 3 ) n 3 (−1) = (−1) = (−1) = (−1) = , n 3n n3n n3n n 3n n n=1 n=1 n=1 n=1 n=1 1 1 which converges by the Alternating Series Test (since is decreasing and lim = 0). n→∞ n n √ For x = −2 + 3, we have √ √ 2n ∞ ∞ ∞ X X X (−1)n 3 + 2)n n (−2 + n ( 3) (−1) = (−10 = , n 3n n 3n n n=1 n=1 n=1 which converges.

Therefore, the interval of convergence is −2 −

√

3 ≤ x ≤ −2 +

√

3.

In-Class Worksheet Week 12

Math 141

Page 3 of 7

(d) The radius and open interval of convergence calculations are the same as in (c), the only thing that changes is that neither endpoint series will be alternating — they will both simplify to the √ √ harmonic series. Therefore the interval of convergence is −2 − 3 < x < −2 + 3 . (e) The calculation is very similar to that in (a), except the ratio test limit will be we have

|x + 2|3 . Then, 3

|x + 2|3 < 1, 3 3 |x + 2| < 3, |x + 2| < 31/3 , −31/3 < x + 2 < 31/3 , −2 − 31/3 < x < −2 + 31/3 .

For x = −2 − 31/3 , we have ∞ X (−2 − 31/3 + 2)3n n=1

n3n

=

∞ X (−31/3 )3n n=1

n3n

=

∞ X (−1)3n 3n n=1

n3n

=

n ∞ (−1)3 X n

n=1

=

∞ X (−1)n n=1

n

,

which converges.

For x = −2 + 31/3 , we have ∞ X (−2 + 31/3 + 2)3n

n3n

n=1

=

∞ X 33n/3 n=1

n3n

=

∞ X 1 n=1

n

,

which diverges.

Therefore the interval of convergence is −2 − 31/3 ≤ x < −2 + 31/3 .

Question 2 (a) Suppose that the power series

∞ X

cn (x − 1)n converges for x = −1 and diverges for x = 5. Does

n=0

the series converge for x = 2? Does the series converge for x = −5? Does the series converge for x = −2? ∞ X (b) Suppose that the power series dn (x + 2)n converges conditionally for x = 8. Does the series n=0

converge for x = 7? Does the series converge for x = 9?

Hint: Does the ratio test ever guarantee conditional convergence? What must be true about the ratio test limit if the series converges conditionally?

For these, recall that the power series must converge for |x − c| < R, must diverge for |x − c| > R, and that anything is possible at the endpoints which are characterized by |x − c| = R. (a) Here we see the series is centered at 1. Since we’re told the series converges for x = −1, we know that if −1 is on the interior of the interval of convergence that | − 1 − 1| < R, meaning R > 2.

Since convergence at an endpoint is also possible, it could that R = 2 as well. Together, if a power series centered at 1 converges for x = −1 then R ≥ 2.

Similarly, since we’re told the series diverges for x = 5, then either |5 − 1| > R or |5 − 1| = R. Therefore, R ≥ 4.

Visualizing these distances and intervals on the number line is helpful.

In-Class Worksheet Week 12

Math 141

3 converges for x

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tho Interval centered

here

rn Any values of x in the red interval must lead to convergent series, and any values outside the ai blue must lead to divergent series. That is, the red interval is the smallest possible interval of convergence, and the blue interval is the largest possible interval of convergence. Any points in the blue interval but not the red interval could go either way.

Since |2 − 1| = 1 and R ≥ 2, x = 2 must be inside the interval of convergence. Therefore the power series must converge for x = 2 .

Since | − 5 − 1| = 6 and R ≤ 4, x = −5 must be outside the interval of convergence. Therefore the power series must diverge for x = −5 .

Since | − 2 − 1| = 3, we cannot tell what happens for x = −2 . This is because we only know 2 ≤ R ≤ 4 — if R = 2.5 for example then the series would diverge for x = −2, whereas if R = 3.5 then the series would converge for x = −2.

r

tho interval

(b) Remember that the ratio test never shows conditional convergence — it either shows absolute convergence (limit < 1), divergence (limit > 1 or ∞), or nothing (limit = 1). Therefore, if a here power series converges conditionally for some value of x, it must have a ratio test limit of 1 for 3 that value of x, and this was precisely where |x − c| = R. Therefore, if a power series centered at −2 converges conditionally = 8, then we know R = 10 since 8 needs to be an endpoint of converges for x for xInterval the interval of convergence. centered

here

rn

ai

r

interval here Any values of x in the green interval must lead to convergent series, and any values outside must lead to divergent series.

Therefore, since |7 − (−2)| = 9 < 10, x = 7 must be inside the interval of convergence. Therefore the power series must converge for x = 7 .

Similarly, since |9−(−2)| = 11 > 10, x = 9 must be outside the interval of convergence. Therefore the power series must diverge for x = 9 .

In-Class Worksheet Week 12

Math 141 Question 3 Suppose that f (x) =

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∞ ∞ X X xn x2n and g(x) = . n (n!) n! n=3 n=3

f (x) . x→0 g(x)

(a) Calculate lim

x4 f (x) . x→0 g(x)

(b) Calculate lim

x3 f (x) . x→0 g(x)

(c) Calculate lim

It will be helpful to write out the first few terms. x3 x4 x5 + + + (terms with at least x6 ), 3(3!) 4(4!) 5(5!) x6 x8 x10 g(x) = + + + (terms with at least x12 ).

3!

4!

5!

f (x) =

(a) One way to approach these is by using L’Hospital’s Rule and the fact that power series can be differentiated term-by-term (with the power rule) without changing their radius of convergence.

We have x4 x5 x3 6 →0 3(3!) + 4(4!) + 5(5!) + (terms with at least x ) is lim x6 8 10 x x 12 x→0 →0 3! + 4! + 5! + (terms with at least x ) 4x3 5x4 3x2 5 →0 H 3(3!) + 4(4!) + 5(5!) + (terms with at least x ) is = lim 6x5 8x7 10x9 11 ) x→0 →0 + + + (terms with at least x 3!

4!

5!

2 3 12x 20x 6x 4 →0 H 3(3!) + 4(4!) + 5(5!) + (terms with at least x ) = lim 30x4 is 6 8 56x 90x 10 x→0 →0 3! + 4! + 5! + (terms with at least x ) 2 6 24x 60x 3 H 3(3!) + 4(4!) + 5(5!) + (terms with at least x ) = lim 120x3 5 7 336x 720x 9 x→0 3! + 4! + 5! + (terms with at least x ) →6/18

z

}| { 6 24x 60x2 3 + + + (terms with at least x ) 3(3!) 4(4!) 5(5!)

= lim x→0 120 336x2 720x4 3 6 x + + + (terms with at least x ) |{z} 3!

4!

5! →0 | {z }

is

→ 1/3 →0

→20

does not exist.

As x → 0+ , x3 → 0+ and so the entire limit is +∞. As x → 0− , x3 → 0− , and so the entire limit is −∞. Therefore we just say the limit does not exist. (b) Another way to approach these is to cancel common factors in the numerator and denominator.

In-Class Worksheet Week 12

Math 141

Page 6 of 7

We have x4

lim

x3 x4 x5 6 3(3!) + 4(4!) + 5(5!) + (terms with at least x ) x6 x8 x10 12 3! + 4! + 5! + (terms with at least x )

x→0

x7 = lim

x x2 1 3 3(3!) + 4(4!) + 5(5!) + (terms with at least x ) 1 x2 x4 6 3! + 4! + 5! + (terms with at least x ) →1/18

x6

x→0

→0

z

z}|{ x = lim

x→0

}| { 1 x x2 + + + (terms with at least x3 ) 3(3!) 4(4!) 5(5!)

1 x2 x4 + + + (terms with at least x6 ) 3!

4!

5! {z } | →1/6

= 0. (c) We have x3 lim

x→0

x3 x4 x5 6 3(3!) + 4(4!) + 5(5!) + (terms with at least x )

x8 x10 x6 12 3! + 4! + 5! + (terms with at least x )

x6 = lim

1 x x2 3 3(3!) + 4(4!) + 5(5!) + (terms with at least x )

x6

x→0

x2 x4 1 6 3! + 4! + 5! + (terms with at least x ) →1/18

z

}| { 1 x x2 + + + (terms with at least x3 ) 3(3!) 4(4!) 5(5!) = lim x→0 1 x2 x4 + + + (terms with at least x6 ) 3!

4!

5! | {z } →1/6

1/18 = 1/6 6 = 18 1 = .

3

Question 4 Find a power series representation for f (x) = Since

3x5 centered at 0. For which values of x is it valid?

1 + 2x3

∞ X 1 = yn = 1 + y + y2 + · · · 1 − y n=0

for |y| < 1,

we can substitute y = −2x3 to obtain ∞ X 1 1 = = (−2x3 )n = 1 + (−2x3 ) + (−2x3 )2 + · · · 1 + 2x3 1 − (−2x3 ) n=0

= =

∞ X

(−1)n 2n x3n = 1 − 2x3 + 4x6 + · · ·

n=0 ∞ X

(−1)n 2n x3n = 1 − 2x3 + 4x6 + · · ·

n=0

for |x3 | <

1 , 2

1 for |x| < √ .

3 2

for | − 2x3 | < 1,

In-Class Worksheet Week 12

Math 141

Page 7 of 7

We can then multiply by 3x5 to obtain ∞ X

3x5 5 = 3x (−1)n 2n x3n = 3x5 1 − 2x3 + 4x6 + · · · 3 1 + 2x n=0

1 for |x| < √ , 3 2

which can be rewritten as ∞ X 3x5 = 3(−1)n 2n x3n+5 = 3x5 − 6x8 + 12x11 + · · · 1 + 2x3 n=0

1 . for |x| < √ 3 2

1 , find a power series representation for 1−x arctan x centered at 0. What do we know about the radius of the interval on which this expansion is valid? You likely saw this done in class, but work through the steps to make sure you understand.

We know that 1 that equals 0 for x = 0. arctan x is the antiderivative of 1 + x2 We know that ∞ X 1 = y n = 1 + y + y 2 + · · · for |y| < 1, 1 − y n=0

Question 5 Starting from the power series centered at 0 for g(x) =

and so ∞ ∞ X X 1 1 2 n = = (−x ) = (−1)n x2n = 1 − x2 + x4 − · · · for | − x2 | < 1 ⇐⇒ |x| < 1.

1 + x2 1 − (−x2 ) n=0 n=0

If we use the first few terms, we have that Z

x3 x5 arctan x = 1 − x2 + x4 − · · · dx = C + x − + at least for all |x| < 1, 3 5 for some C. (Term-by-term antidifferentiation of a power series does not change the radius of convergence, but endpoints may be added to the interval of convergence. Whether or not this happens is not something we’ll be that concerned with in this course.) Using arctan(0) = 0 quickly yields C = 0.

To get the general nth term, we can do !

Z X ∞ n 2n arctan x = (−1) x dx n=0

= =

∞ Z X

(−1)n x2n dx

n=0 ∞ X

(−1)

n

Z x

2n

dx

(term-by-term antidifferentiation) (n is a constant as far as the x-antifferentiation is concerned)

n=0

=C+

∞ X n=0

(−1)n

x2n+1 2n + 1

(power rule),

and we can similarly obtain C = 0. Therefore, ∞ X x3 x5 x2n+1 arctan x = x − + − ··· = (−1)n on an interval centered at x = 0 with radius 1.

3 5 2n + 1 n=0

(The radius of an interval is half its length — the word radius makes more sense if x were a complex number where instead of an interval we would have a disk.)