Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing...
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Transcript of Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing...
![Page 1: Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing Convergence at Endpoints.](https://reader035.fdocuments.in/reader035/viewer/2022081503/56649f535503460f94c78423/html5/thumbnails/1.jpg)
Ch 9.5Calculus Graphical, Numerical, Algebraic byFinney, Demana, Waits, Kennedy
Testing Convergence at Endpoints
![Page 2: Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing Convergence at Endpoints.](https://reader035.fdocuments.in/reader035/viewer/2022081503/56649f535503460f94c78423/html5/thumbnails/2.jpg)
![Page 3: Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing Convergence at Endpoints.](https://reader035.fdocuments.in/reader035/viewer/2022081503/56649f535503460f94c78423/html5/thumbnails/3.jpg)
![Page 4: Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing Convergence at Endpoints.](https://reader035.fdocuments.in/reader035/viewer/2022081503/56649f535503460f94c78423/html5/thumbnails/4.jpg)
![Page 5: Ch 9.5 Calculus Graphical, Numerical, Algebraic by Finney, Demana, Waits, Kennedy Testing Convergence at Endpoints.](https://reader035.fdocuments.in/reader035/viewer/2022081503/56649f535503460f94c78423/html5/thumbnails/5.jpg)
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Convergence of Two Series
2n=1 n=1
1 1Consider the two series, and
n n
1. What does the ratio test show about convergence of both series?
2. Use improper integrals to show the area of both curves over the interval 1 ≤ x ≤ ∞.
3. How does this relate to the ratio test?
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Convergence of Two Series
2n=1 n=1
1 1Consider the two series, and
n n
2 2
1 2 2n n n n
2
11n + 1n nn + 1
L = lim = lim = 1 L = lim = lim = 11 1n + 1 n + 1n n
3. How does this relate to the ratio test? Ratio Test is inconclusive when L = 1; but Integral Test works.
1. What does the ratio test show about convergence of both series?
2. Use improper integrals to show the area of both curves over the interval 1 ≤ x ≤ ∞.
k1k k
1
-1 k12 k k
1
1 dx = lim ln x| = lim ln k =
x
1 1 dx = lim -x | = lim - + 1 = 1
x k
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Using the Ratio Test gives a limit L =1 which is inconclusive.
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The p-Series Test
pn=1
pn=1
pn=1
11. Use the Integral Test to prove that converges if p > 1.
n
12. Use the Integral Test to prove that diverges if p < 1.
n
13. Use the Integral Test to prove that diverges if p
n
= 1.
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The p-Series Test
pn=1
k kk -p+1
p p p 1k k k1 1 11
p-1k
11. Use the Integral Test to prove that converges if .
n
1 1 x 1 1 dx = lim dx = lim = lim
x x -p + 1 1 p x
1 1 lim - 1
1- p
p
k
> 1
pn=1
kk -p+1
p pk k1 1 1
1 1 = - =
1-p p 1
12. Use the Integral Test to prove tha
The series converges by test.
0 < p < 1t diverges if .n
1 1 x dx = lim dx = lim
x x -p + 1
1- p
k
pn=1
k
p k1 1
1= lim k - 1 =
1- p
since (1 - p > 0).
13. Use the Integral Test to prove that diverges if .
n
The series diverges by test.
p = 1
1 1 dx = lim dx =
x x
k
1k k lim ln x = lim ln k =
The series diverges by test.
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Slow Divergence of Harmonic Series
n 1
n
1
Approximately
1 1 1 1 1
n 2 3 4 n
1
x
how many terms of the harmonic series
are required to form a partial sum > 20?
= 1 + + + + ... + > 20
< 1 + dx = 1 + ln n
19
1 + ln n > 20
ln n > 19
n > e 178,482,301 terms!!
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Example
21 x
Determine whether the improper integral converges or diverges:
1 + cosx dx
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Example
2 2 21 1 1
21
21
x x x
and
x
x
Determine whether the improper integral converges or diverges:
1 + cosx 1 + cosx 2 dx dx dx
2 dx converges since this is a p-series with p >1
Therefore,
1 + cosx d
x converges by comparison test and p-series test.
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Limit Comparison Test
nn=1
Deter min e whether the series converges or diverges:
1 1 1 1 1 + + + + ... +
1 3 7 15 2 - 1
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Limit Comparison Test
n n
nnn
n
n
nn=1
n
Deter min e whether the series converges or diverges:
1 1 1 1 1 + + + + ... +
1 3 7 15 2 - 1
For n large, 1/ (2 - 1) behaves like so we compare the se1/2 1/2
a 1lim
ries to
= l im b 2 -
n n
n n
nn 1
n n
n
=
Since 1/2 converges, the LCT guarantees that
1
2 2 1 = lim = lim = 1
1 1 2
also converges.2
- 1 1 - (
- 1
1/
2 )
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Limit Comparison Test
n=1
Deter min e whether the series converges or diverges:
1 1 1sin 1 + sin + sin + ... + sin
2 3 n
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Limit Comparison Test
n
n n
n=1
x 0
Deter min e whether the series converges or diverges:
1 1 1 sin 1 + sin + sin + ... + sin
2 3 n
sin xUsing lim = 1, it is useful to compare the series to (1/n)
x
toa
li get: mb
n
n=1
n 0
1Since (1/n) diverges, by
sin 1/n si
,
n n = lim = lim =
sin also diverges.n
11/n n
LCT
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Alternating Harmonic Series
Prove that the alternating harmonic series is convergent, but not absolutely convergent. Find a bound for the truncation error after 99 terms.
n 1
n=1
-11 1 11 - + - + ... +
2 3 4 n
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Alternating Harmonic Series
n 1
n=1
n
-11 1 1 1 - + - + ... +
2 3 4 n
Since the terms are alternating in sign and decrease in absolute value
1 1 1 1 1 > > > ..., and since lim 0
Alt
2 3 4 n
By the ernating Series
n
n=1
1
n=1
.
1However, the series is the harmonic series which diverges.
n
So, the alternating harmonic series converges but is not absolutely convergent.
-1Test, converges
n
Truncation error after 99
100
1 terms is < u =
100
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Rearranging Alternating Harmonic Series
The series of positive terms
1 1 1 1 + + + ... + diverges to
3 5 2n + 1While the series of negative terms
1 1 1 1 - - - - ... - diverges to -
2 4 6 2nSo, start by adding positive terms unti
l sum > 1, then add negative
terms until sum is less than -2, then add terms until sum > 3 etc so the sum
swings further in both directions and thus diverges.
To get the sum of , add positive terms unt il the partial sum is greater
than , then add negative terms until the sum is less than . Continue
indefinitely always getting closer to . Since the partial sums go to 0,
the sum gets converges to
.
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Word of Caution
Although we can use the tests we have developed to find where a given power series converges, it does not tell us what function that power series is converging to. That is why it is so important to estimate the error.
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Maclaurin Series of a Strange Function
21-
x
(n)
0, x = 0Let f(x) =
e , x 0
f (x) has derivatives of all orders at x = 0 and f (0) = 0 for all n
1. Construct the Maclaurin series for f.
2. For what values of x does this series converge?
3. Find all values of x for which the series actually converges to f(x).
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Maclaurin Series of a Strange Function
21-
x
(n)
0, x = 0Let f(x) =
e , x 0
f (x) has derivatives of all orders at x = 0 and f (0) = 0 for all n
1. Construct the Maclaurin series for f.
2. For what values of x does this series converge?
The series converges to 0 for all values of x.
3. Find all values of x for which the series actually converges to f(x). The only place that this series actually converges to its f-value is at x = 0
n
n 0
x0 = 0
n!
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Series DivergesIs lim an = 0?nth-Term Test
no
Geometric
Series TestIs Σ an = a + ar + ar2 + …? Converges to a/(1 - r) if |r| < 1.
Diverges if |r| ≥ 1.
p-series Test Does the series have the form
p
n 1
1
n ?
Series converges if p > 1Series diverges if p ≤ 1
yes
yes
Absolute
convergenceDoes Σ |an| converge? Apply 1 of the Comparison tests, IntegralTest, Ratio Test or nth-Root Test
Original series convergesyes
Alternating
Series TestIs Σ an = u1 – u2 + u3 - …?
Is there an integer N such thatun ≥ un+1 ≥… ?
Series converges if un →0.Otherwise series diverges.
no
no
no
no
Try partial sums
yes
yes