Loading [MathJax]/jax/output/HTML-CSS/jax.js
Toronto Math Forum
Welcome,
Guest
. Please
login
or
register
.
1 Hour
1 Day
1 Week
1 Month
Forever
Login with username, password and session length
News:
Home
Help
Search
Calendar
Login
Register
Toronto Math Forum
»
MAT334-2018F
»
MAT334--Tests
»
Term Test 2
»
TT2B Problem 2
« previous
next »
Print
Pages: [
1
]
Author
Topic: TT2B Problem 2 (Read 9463 times)
Victor Ivrii
Administrator
Elder Member
Posts: 2607
Karma: 0
TT2B Problem 2
«
on:
November 24, 2018, 05:20:09 AM »
(a)
Find the decomposition into power series at
z
=
0
of
f
(
z
)
=
(
1
−
z
)
−
1
.
What is the radius of convergence?
(b)
Plugging in
−
z
2
instead of
z
and integrating, obtain a decomposition at
z
=
0
of
arctan
(
z
)
.
«
Last Edit: November 29, 2018, 07:27:48 AM by Victor Ivrii
»
Logged
Yifei Wang
Jr. Member
Posts: 10
Karma: 3
Re: TT2 Problem 2
«
Reply #1 on:
November 24, 2018, 05:38:47 AM »
I think the power is -1/2 inside of the -1
Correct V.I. It was actually Test2B
«
Last Edit: November 29, 2018, 07:28:07 AM by Victor Ivrii
»
Logged
Wanying Zhang
Full Member
Posts: 21
Karma: 6
Re: TT2 Problem 2
«
Reply #2 on:
November 24, 2018, 12:15:18 PM »
Here is the solution to problem 2.
You need to know decomposition of
(
1
−
z
)
−
1
. The rest is simply wrong. V.I.
«
Last Edit: November 29, 2018, 07:19:35 AM by Victor Ivrii
»
Logged
Huanglei Ln
Jr. Member
Posts: 8
Karma: 7
Re: TT2 Problem 2
«
Reply #3 on:
November 25, 2018, 01:34:47 AM »
a
)
f
(
z
)
=
1
1
−
z
=
∞
∑
n
=
0
z
n
1
R
=
lim
n
→
∞
|
1
1
|
=
1
⇒
R
=
1
b
)
f
(
−
z
2
)
=
1
1
+
z
2
=
∞
∑
n
=
o
−
z
2
n
=
Σ
∞
(
−
1
)
n
z
2
n
⇒
∫
f
(
−
z
2
)
d
z
=
∞
∑
n
=
o
(
−
1
)
n
∫
2
n
d
z
⇒
∫
1
1
+
z
2
d
z
=
∞
∑
(
n
=
0
)
(
−
1
)
n
∫
z
2
n
d
z
⇒
arctan
(
z
)
+
c
=
∞
∑
n
=
0
(
−
1
)
n
z
2
n
+
1
2
n
+
1
⇒
arctan
(
z
)
=
∞
∑
n
=
0
(
−
1
)
n
2
n
+
1
z
2
n
+
1
+
c
«
Last Edit: November 29, 2018, 07:24:02 AM by Victor Ivrii
»
Logged
Huanglei Ln
Jr. Member
Posts: 8
Karma: 7
Re: TT2 Problem 2
«
Reply #4 on:
November 25, 2018, 01:39:58 AM »
a
)
f
(
z
)
=
1
1
−
z
=
∞
∑
n
=
o
z
n
1
R
=
l
i
m
n
→
∞
|
1
1
|
=
1
⇒
R
=
1
\end{displaymath}
\begin{displaymath}
b
)
f
(
−
z
2
)
=
1
1
+
z
2
=
∞
∑
n
=
o
−
z
2
n
=
Σ
∞
(
−
1
)
n
z
2
n
⇒
∫
f
(
−
z
2
)
d
z
=
∞
∑
n
=
o
(
−
1
)
n
∫
2
n
d
z
⇒
∫
1
1
+
z
2
d
z
=
∞
∑
(
n
=
0
)
(
−
1
)
n
∫
z
2
n
d
z
⇒
a
r
t
a
n
(
z
)
+
c
=
∞
∑
n
=
0
(
−
1
)
n
z
2
n
+
1
2
n
+
1
⇒
a
r
t
a
n
(
z
)
=
∞
∑
n
=
0
(
−
1
)
n
2
n
+
1
z
2
n
+
1
+
c
Logged
Victor Ivrii
Administrator
Elder Member
Posts: 2607
Karma: 0
Re: TT2 Problem 2
«
Reply #5 on:
November 29, 2018, 07:23:37 AM »
Huanglei
I fixed your LaTeX. Don't learn it from crappy sources!
Also as
z
=
0
you'll see that
c
=
0
. In actual test missing this will lead to the mark reduction
«
Last Edit: November 29, 2018, 07:25:10 AM by Victor Ivrii
»
Logged
Print
Pages: [
1
]
« previous
next »
Toronto Math Forum
»
MAT334-2018F
»
MAT334--Tests
»
Term Test 2
»
TT2B Problem 2
Jump to:
===> Term Test 2