n = 1 18 (−1)”+ ¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)” · - 1)(x-3)" 4n
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
Find the interval of convergence of the power series. Also include a check for convergence at the endpoints of the interval.
![.
33.
(− 1)”+¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)"
•
-
-
4n
11=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78a03948-fc7d-4d79-bfea-4ad5527b5a3c%2F3dab0959-6a8e-4812-ae21-62258439de81%2Fz5w7byn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:.
33.
(− 1)”+¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)"
•
-
-
4n
11=
Expert Solution
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Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
In this answer to the previous follow-up question, why do we multiply the (4n - 1) at the top? This was one of my previous follow-up questions, and the answer to that was "...no need to write (4n -1) in the top." And it is still confusing me.
![Now,
을(-1)+1 3.7.11.
-
(4n-1)(x-3)^
n=1
կո
an =
1-117+1 3.7. 11.
կո
= land =
3.7.11
կո
(4n-1) (x-3)"
(4n-1)/(x-3)^
&lantıl
3. 7. 11.
-
-
4n+1
Σ an (say)
n=1
(4n-1) (4n+3) (x-3)/2+1
Series, Eland converges if
M=1
By Ratio
test
lim lantil
lanl
< 1
L
1.
at lim lantil
1,
test fails.
n-p
1 anl
Now, lim lantil
nto lanl
=
lim
ntoo
37.11.
3.7.11.
-
=) lim lantil
Flim
(4n+3)1x-317+1
ntoo
lanl
7-7700
4n+1
= lim lantıl
n700
Ian'
= lim (4n+3) 1x-31
n200
4
(4n-1) (4n+3) 1x-317+1
4n+1
(49-1) 1x-3/m
կո
x կո
1x-31"](https://content.bartleby.com/qna-images/question/78a03948-fc7d-4d79-bfea-4ad5527b5a3c/2e4aaf7c-1f84-4d5b-89d4-9fb538ee74d1/lyeu9g4_thumbnail.jpeg)
Transcribed Image Text:Now,
을(-1)+1 3.7.11.
-
(4n-1)(x-3)^
n=1
կո
an =
1-117+1 3.7. 11.
կո
= land =
3.7.11
կո
(4n-1) (x-3)"
(4n-1)/(x-3)^
&lantıl
3. 7. 11.
-
-
4n+1
Σ an (say)
n=1
(4n-1) (4n+3) (x-3)/2+1
Series, Eland converges if
M=1
By Ratio
test
lim lantil
lanl
< 1
L
1.
at lim lantil
1,
test fails.
n-p
1 anl
Now, lim lantil
nto lanl
=
lim
ntoo
37.11.
3.7.11.
-
=) lim lantil
Flim
(4n+3)1x-317+1
ntoo
lanl
7-7700
4n+1
= lim lantıl
n700
Ian'
= lim (4n+3) 1x-31
n200
4
(4n-1) (4n+3) 1x-317+1
4n+1
(49-1) 1x-3/m
կո
x կո
1x-31"
Solution
Follow-up Question
So I worked out the question. And I got the below setup. And I got the interval of convergence as (-1, 7). But when I checked the back of the book, it said the answer was x = 3. Could anyone help with what I did wrong?
![3315
T
307-11---Un-1/x-3)
we enforce the condition of converg
ence of the ratio test
fim 3-7-7-3) (x3)
☑
い 。37 11-441-163
14/4731
41
4 un- 4T
⇒ x 344
Interval of convergenc
Endpoints
TWE
7844
h=|
(-1) 3070/1000 (n-1)/4"
14
diverges
(3.7-11-0
4
diverges](https://content.bartleby.com/qna-images/question/78a03948-fc7d-4d79-bfea-4ad5527b5a3c/b565988f-4426-4f8d-be22-4d724d381868/z3zrene_thumbnail.jpeg)
Transcribed Image Text:3315
T
307-11---Un-1/x-3)
we enforce the condition of converg
ence of the ratio test
fim 3-7-7-3) (x3)
☑
い 。37 11-441-163
14/4731
41
4 un- 4T
⇒ x 344
Interval of convergenc
Endpoints
TWE
7844
h=|
(-1) 3070/1000 (n-1)/4"
14
diverges
(3.7-11-0
4
diverges
Solution
Follow-up Question
Why is there an extra (4n - 1) in the top in u(n+1)?
![Оли =
(+372
3.7.11 (4nes (4n+3)
(x-3)
4741
7+1](https://content.bartleby.com/qna-images/question/78a03948-fc7d-4d79-bfea-4ad5527b5a3c/f663155c-e7fe-4317-96e5-f85d2b5924cb/9kql6pk_thumbnail.png)
Transcribed Image Text:Оли =
(+372
3.7.11 (4nes (4n+3)
(x-3)
4741
7+1
Solution
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