as a geometric sexes. -32-2 2² + | -32 2 22+1 We know that I can be expressed ни n 2 (-1)^u^ n=0 => = ни =>-32 21 0-32. 관세 = -3z. 2 (-1)^ z ги n=0 -₤(-1)^(-22") = n=0 ги -32-2 = (-1)" (-32+4+1) + 2(-1) " (-22 *^) 2+1 = = n=0 ₤1-17ª (-322+1 1=0 n=0 222) Σ (-1)^(-3224+1-222) a=0 0:22 2 3 크 = 1=0 00 322 n=0 -2-1 -1-n Now combining all terms, the lawrent series له -1-n of 27724 2432 for 0 2+D=0 => C=~3 A+C-0 D- -37-2 f(x) = 3/2 + 1/2 ²² + टे 2² + 1
as a geometric sexes. -32-2 2² + | -32 2 22+1 We know that I can be expressed ни n 2 (-1)^u^ n=0 => = ни =>-32 21 0-32. 관세 = -3z. 2 (-1)^ z ги n=0 -₤(-1)^(-22") = n=0 ги -32-2 = (-1)" (-32+4+1) + 2(-1) " (-22 *^) 2+1 = = n=0 ₤1-17ª (-322+1 1=0 n=0 222) Σ (-1)^(-3224+1-222) a=0 0:22 2 3 크 = 1=0 00 322 n=0 -2-1 -1-n Now combining all terms, the lawrent series له -1-n of 27724 2432 for 0 2+D=0 => C=~3 A+C-0 D- -37-2 f(x) = 3/2 + 1/2 ²² + टे 2² + 1
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 23E
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Transcribed Image Text:as a geometric sexes.
-32-2
2² +
|
-32
2
22+1
We know that I can be expressed
ни
n
2 (-1)^u^
n=0
=>
=
ни
=>-32
21
0-32. 관세
=
-3z. 2 (-1)^ z
ги
n=0
-₤(-1)^(-22")
=
n=0
ги
-32-2 = (-1)" (-32+4+1) + 2(-1) " (-22 *^)
2+1
=
=
n=0
₤1-17ª (-322+1
1=0
n=0
222)
Σ (-1)^(-3224+1-222)
a=0
0:22 2
3
크
=
1=0
00
322
n=0
-2-1
-1-n
Now combining all terms, the lawrent series
له
-1-n
of 27724
2432
for 0</2/<113
1 2 2 - 2-4 + 3 2 2 114 + 2 (-)" (32-28*^)
=22Z
1=0
1-0
Transcribed Image Text:d) By Thm 13.2 (Laurent Series): If f (I) is analytic on the annuly A centred at
20: R₁ </2-20 | <R2, then fis) has a mique havent Series expansion convergy
absolutely
fiz) at every ZEA.
8
ماد
flx) = Σ an (z-zo) ^ + Σ bn (2-Zo)", an, bn € C V n
n=0
n =
o<
土
Now we obseme that o</2/< | which implies that D</±² | < 1, nd therefore
|0 < | 1+24 | <2 in this
region.
lex fix)=
2+32
0417151
4
,
So, we stent by factorising the denominator
fist-
2432
2+32
=
(2²+24) (2-(1+2²))
a Jum of 2 functions using a partial function expansion:
0840
Then
it will
f
B
+
2'
22+1
A2 (2+1)
+
22 (2²+1)
Bz² (2²+1)
22
z² (CZ+D) (841)
+
(2)- A
टे
2'(33-12) (+1)=
2" (24+1)
2+32= Az (2² + 1) + B (2²+1) + 7' (CZ+D)
B=2
3
2+32= Az³ +AZ+ 22² + 2 + C =³ +DZ²
32 12 - 2³ (ATC) + 2² (2 x D) + AZ +2
CA = 3
A = 3
=>
2+D=0
=>
C=~3
A+C-0
D-
-37-2
f(x) = 3/2 + 1/2 ²²
+
टे
2² + 1
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