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Valentin Fadeev

How comple^x can you get? Continued

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Edited by Valentin Fadeev, Thursday, 27 Mar 2014, 10:11

As a follow up thought I realized just how much easier it would have been to calculate the residue by definition, i.e. expanding the integrand in the Laurent series to get the coefficient a sub negative one . Let x equals negative one plus xi where xi is small:

equation sequence root of order four over four divided by left parenthesis one plus x times right parenthesis cubed equals root of order four over four divided by xi cubed equals two times e super pi times i divided by four divided by root of order four over four times left parenthesis one minus xi times right parenthesis super one divided by four left parenthesis one minus xi divided by two times right parenthesis super three divided by four divided by xi cubed

Of course, care is needed when choosing the value of the root. It depends on the value of the argument set on the upper bound of the cut. Since I chose it to be 0, the correct value of the root is e super pi times i divided by four .

Now

left parenthesis equation left hand side one minus xi times right parenthesis super one divided by four equals right hand side one minus one divided by four times xi minus three divided by 32 times xi squared plus cap o of xi cubed

left parenthesis equation left hand side one minus xi divided by two times right parenthesis super three divided by four equals right hand side one minus three divided by eight times xi minus three divided by 128 times xi squared plus cap o of xi cubed

left parenthesis one minus xi times right parenthesis super one divided by four left parenthesis equation left hand side one minus xi divided by two times right parenthesis super three divided by four equals right hand side one minus five divided by eight times xi minus three divided by 128 times xi squared plus cap o of xi cubed

Therefore, near x equals negative one the integrand has the following expansion:

equation left hand side root of order four over four divided by left parenthesis one plus x times right parenthesis cubed equals right hand side two times e super pi times i divided by four divided by root of order four over four times open one divided by left parenthesis x plus one times right parenthesis cubed minus five divided by eight left parenthesis x plus one times right parenthesis squared minus three divided by 128 times open x plus one close plus g of x close

where g of x is the regular part of the expansion which is of no interest in this problem.

Hence by definition:

equation sequence cap r sub negative one times f of x equals two times e super pi times i divided by four divided by root of order four over four times open negative three divided by 128 close equals negative three times e super pi times i divided by four divided by 64 times root of order four over four

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