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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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Valentin Fadeev

How comple^x can you get?

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Edited by Valentin Fadeev, Thursday, 21 Apr 2011, 23:58

I found this example in a textbook dated 1937 which I use as supplementary material for M828. It gave me some hard time but finally sorted out many fine tricks of contour integration. Some inspiration was provided by the discussion of the Pochhammer's extension of the Eulerian integral of the first kind by Whittaker and Watson.

Evaluate the following integral:

integral over zero under one root of order four over four divided by open one plus x close cubed d x

Consider the integral in the complex plane:

cap j equals contour integral over cap c root of order four over four divided by open one plus z close cubed d z

where contour C is constructed as follows. Make a cut along the segment open zero comma one close of the real axis. Let arg of z equals zero on the upper edge of the cut. Integrate along the upper edge of the cut in the positive direction. Follow around the point z equals one along a small semi-circle equation left hand side cap c sub gamma super one equals right hand side open z colon absolute value of one minus z equals gamma comma normal black letter cap i times z greater than zero close in the clockwise direction. The argument of the second factor in the numerator will decrease by three times pi divided by four . Then proceed along the real axis till z equals cap r and further round the circle equation left hand side cap c sub cap r equals right hand side open z colon absolute value of z equals cap r close in the counter-clockwise direction.

This circle will enclose both branch point of the integrand z equals zero and z equals one , however since the exponents add up to unity the function will return to its initial value:

equation sequence z super one divided by four times e super two times pi times i times one divided by four times open one minus z close super three divided by four times e super two times pi times i times three divided by four equals z super one divided by four times open one minus z close super three divided by four times e super two times pi times i equals z super one divided by four times open one minus z close super three divided by four

This is the reason why we only need to make the cut along the segment open zero comma one close and not along the entire real positive semi-axis. Then integrate from z equals cap r in the negative direction, then along the second small semi-circle equation left hand side cap c sub gamma squared equals right hand side open z colon absolute value of one minus z equals gamma comma normal black letter cap i times z less than zero close around the point z equals one where the argument of the second factor will again decrease by three times pi divided by four . Finally integrate along the lower edge of the cut and along a small circle equation left hand side cap c sub delta equals right hand side open z colon absolute value of z equals delta close around the origin where the argument of the first factor will decrease by equation left hand side two times pi times one divided by four equals right hand side pi divided by two .

contour

As the result of this construction the integral is split into the following parts:

multiline equation row 1 cap j equals sum with, 4 , summands integral over delta under one minus gamma f of z d z plus integral over cap c sub gamma super one f of z d z plus e super negative three times pi times i divided by four times integral over one plus gamma under cap r f of z d z plus integral over cap c sub cap r f of z d z plus row 2 sum with, 4 , summands prefix plus of e super five times pi times i divided by four times integral over cap r under one plus gamma f of z d z plus integral over cap c sub gamma squared f of z d z plus e super pi times i divided by two times integral over one minus gamma under delta f of z d z plus integral over cap c sub delta f of z d z

Two integrals around the small circles add up to an integral over a circle:

equation sequence integral over cap c sub gamma super one f of z d z plus integral over cap c sub gamma squared f of z d z equals integral over cap c sub gamma f of z d z equals integral over zero under two times pi root of order four over four divided by open two minus gamma times e super i times theta close cubed times i times gamma times e super i times theta d theta

lim over gamma right arrow zero of integral over cap c sub gamma f of z d z equals zero

Similarly, the integral around a small circle around the origin vanishes:

equation left hand side integral over cap c sub delta f of z d z equals right hand side integral over zero under two times pi root of order four over four divided by open one plus delta times e super i times theta close cubed times i times delta times e super i times theta d theta

lim over delta right arrow zero of integral over cap c sub delta f of z d z equals zero

Integrals along the segment open one plus gamma comma cap r close cancel out:

multiline equation row 1 equation left hand side e super negative three times pi times i divided by four times integral over one plus gamma under cap r f of z d z plus e super five times pi times i divided by four times integral over cap r under one plus gamma f of z d z equals right hand side e super negative three times pi times i divided by four times integral over cap r under one plus gamma f of z d z plus e super two times pi minus three times pi times i divided by four times integral over cap r under one plus gamma f of z d z equals row 2 equation sequence equals e super negative three times pi times i divided by four times integral over cap r under one plus gamma f of z d z minus e super negative three times pi times i divided by four times integral over cap r under one plus gamma f of z d z equals zero

The integral over the large circle also tends to 0 as R increases. This can be shown using the Jordan's lemma, or by direct calculation:

equation sequence integral over cap c sub cap r f of z d z equals integral over zero under two times pi root of order four over four divided by open one plus cap r times e super theta close cubed times i times cap r times e super i times theta d theta equals integral over zero under two times pi root of order four over four divided by open one divided by cap r plus e super i times theta close cubed times i times e super i times theta divided by cap r d theta

lim over cap r right arrow normal infinity of integral over cap c sub cap r f of z d z equals zero

Finally the only two terms that survive allow us to express the contour integral in terms of the integral along the segment of the real axis:

multiline equation row 1 equation sequence cap j equals integral over zero under one f of z d z minus e super pi times i divided by two times integral over zero under one f of z d z equals open one minus e super pi times i divided by two close times integral over zero under one f of z d z equals e super pi times i divided by four times open e super negative pi times i divided by four minus e super pi times i divided by four close times integral over zero under one f of z d z equals row 2 equation left hand side equals right hand side negative two times i times e super pi times i divided by four times sine of pi divided by four times integral over zero under one f of z d z

The contour cap c encloses the only singularity of the integrand which is the pole of the third order at z equals negative one . Hence, by the residue theorem:

cap j equals two times pi times i times cap r sub negative one times f of z

equation left hand side integral over zero under one f of z d z equals right hand side negative Square root of two times pi times e super negative pi times i divided by four times cap r sub negative one times f of z

The residue can be calculated using the standard formula:

equation sequence cap r sub negative one times f of z equals one divided by two times lim over z right arrow negative one of open open one plus z close cubed times root of order four over four divided by open one plus z close cubed close super double prime equals one divided by two times lim over z right arrow negative one of open root of order four over four close super double prime

Calculation of the derivative can be facilitated by taking logarithm first:

g equals root of order four over four

natural log of g equals one divided by four times natural log of z plus three divided by four times natural log of one minus z

equation sequence g super prime divided by g equals one divided by four times z minus three divided by four times open one minus z close equals one minus four times z divided by four times z times open one minus z close

equation left hand side g super prime equals right hand side one minus four times z divided by four times z super three divided by four times open one minus z close super one divided by four

natural log of g super prime equals natural log of one divided by four plus natural log of one minus four times z minus three divided by four times natural log of z minus one divided by four times natural log of one minus z

equation sequence g super double prime divided by g super prime equals negative four divided by one minus four times z minus three divided by four times z plus one divided by four times open one minus z close equals negative three divided by four times open one minus four times z close times z times open one minus z close

equation left hand side g super double prime equals right hand side negative three divided by 16 times z super seven divided by four times open one minus z close super five divided by four

equation sequence cap r sub negative one times f of z equals one divided by two times lim over z right arrow negative one of negative three divided by 16 times z squared times open one minus z close times z super one divided by four divided by open one minus z close super one divided by four equals negative three times e super pi times i divided by four divided by 64 times root of order four over four

equation sequence integral over zero under one root of order four over four divided by open one plus x close cubed d x equals negative Square root of two times pi times e super negative pi times i divided by four of negative three times e super pi times i divided by four divided by 64 times root of order four over four equals three times pi times root of order four over four divided by 64

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Valentin Fadeev

Vector horror

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Edited by Valentin Fadeev, Sunday, 23 Jan 2011, 21:49

Had to do some revision of vector calculus/analysis before embarking on M828.

One point which I was not really missing, but did not quite get to grips with was the double vector product. I remembered the formula:

equation left hand side a right arrow multiplication open b right arrow multiplication c right arrow close equals right hand side b right arrow times open a right arrow dot operator c right arrow close minus c right arrow times open a right arrow dot operator b right arrow close ,

but nevertheless had difficulties applying it in excercises.

The reason whas that that the proof I saw used the expression of vector product in coordinates and comparison of both sides of the equation. However, I was aware of another, purely "vector" argument with no reference to any coordinate system.

Eventually I was able to reproduce only part of it, consulting one old textbook for some special trick. So here's how it goes.

For b right arrow multiplication c right arrow is perpendicular to the plane of b right arrow and c right arrow , a right arrow multiplication open b right arrow multiplication c right arrow close must lie in this plane, therefore:

equation left hand side a right arrow multiplication open b right arrow multiplication c right arrow close equals right hand side lamda times b right arrow plus mu times c right arrow

Dot-multiply both parts by a right arrow :

equation left hand side open a right arrow multiplication open b right arrow multiplication c right arrow close dot operator a right arrow close equals right hand side lamda times open b right arrow dot operator a right arrow close plus mu times open c right arrow dot operator a right arrow close

Since a right arrow multiplication open b right arrow multiplication c right arrow close up tack a right arrow , left-hand side is 0, so:

lamda times open b right arrow dot operator a right arrow close plus mu times open c right arrow dot operator a right arrow close equals zero reverse solidus qquod open asterisk operator close

Now define vector c prime right arrow lying in the plane of b right arrow and c right arrow , perpendicular to c right arrow and directed so that c prime right arrow , c right arrow and b right arrow multiplication c right arrow form the left-hand oriented system. This guarantees that the angle between b right arrow and c prime right arrow , open times times b right arrow c prime right arrow hat close less than pi divided by two .

Dot-multiply both parts by c prime right arrow :

equation left hand side open a right arrow multiplication open b right arrow multiplication c right arrow close dot operator c prime right arrow close equals right hand side lamda times open b right arrow dot operator c prime right arrow close

However, equation left hand side open a right arrow multiplication open b right arrow multiplication c right arrow close dot operator c prime right arrow close equals right hand side open c prime right arrow multiplication open b right arrow multiplication c right arrow close dot operator a right arrow close

equation sequence absolute value of b right arrow multiplication c right arrow equals absolute value of b right arrow times absolute value of c right arrow times sine of times times b right arrow c right arrow hat equals absolute value of b right arrow times absolute value of c right arrow times cosine of times times b right arrow c prime right arrow hat

MathJax failure: TeX parse error: Missing or unrecognized delimiter for \right therefore

sine of multiplication multiplication times times c prime right arrow b right arrow c right arrow hat equals one

and

equation sequence absolute value of c prime right arrow multiplication open b right arrow multiplication c right arrow close equals absolute value of c prime right arrow times absolute value of b right arrow times absolute value of c right arrow times cosine of times times b right arrow c prime right arrow hat equals absolute value of c right arrow times open b right arrow dot operator c prime right arrow close

Hence equation left hand side c prime right arrow multiplication open b right arrow multiplication c right arrow close equals right hand side c right arrow times open b right arrow dot operator c prime right arrow close

equation left hand side open a right arrow multiplication open b right arrow multiplication c right arrow close dot operator c prime right arrow close equals right hand side open b right arrow dot operator c prime right arrow close times open a right arrow dot operator c right arrow close

equation left hand side open b right arrow dot operator c prime right arrow close times open a right arrow dot operator c right arrow close equals right hand side lamda times open b right arrow dot operator c prime right arrow close

lamda equals open a right arrow dot operator c right arrow close

mu can be calculated in a similar manner, however, it is easier achieved using equation open asterisk operator close .

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