I find the Lautent Series of expansion ofthe following.if(₂)=1202²413@0212121612121li☺ 1-2 Valid in the annulus 02-il <√2ízValid in 0</2-4/<4.that is valid for(2+1)202-4)³

Answered: I find the Lautent Series of expansion…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

I find the Lautent Series of expansion of
the following.
if(₂)=
1
202²413
@0212121
612121
li
☺ 1-2 Valid in the annulus 0</2-il <√2
íz
Valid in 0</2-4/<4.
that is valid for
(2+1)
202-4)³

Expert Solution

Step 1: Laurent series expansion

1) (i)

(a) $f left parenthesis z right parenthesis equals fraction numerator 1 over denominator z open parentheses z squared plus 1 close parentheses end fraction comma 0 less than open vertical bar z close vertical bar less than 1$

$equals 1 over z open parentheses 1 plus z squared close parentheses to the power of negative 1 end exponent equals 1 over z open parentheses 1 minus z squared plus z to the power of 4 minus z to the power of 6 plus......... close parentheses$

$equals 1 over z minus z plus z cubed minus z to the power of 5 plus..........$

(b) $f left parenthesis z right parenthesis equals fraction numerator 1 over denominator z open parentheses z squared plus 1 close parentheses end fraction comma open vertical bar z close vertical bar greater than 1$

$equals 1 over z cubed open parentheses 1 plus open parentheses 1 over z close parentheses squared close parentheses to the power of negative 1 end exponent equals 1 over z cubed open parentheses 1 minus 1 over z squared plus 1 over z to the power of 4 minus 1 over z to the power of 6 plus............. close parentheses$

$equals 1 over z cubed minus 1 over z to the power of 5 plus 1 over z to the power of 7 minus 1 over z to the power of 9 plus.........$