Exercise 15.4.15

Show carefully that (15.58) follows from (15.57).

Answers

Proof. To know the first coefficients of the power series expansion of φ ( βz ) , it is sufficient to do an asymptotic expansion in the neighborhood of 0 up to degree 9.

By (15.54),

φ ( βz ) = b 0 ( β ) ( z + c 1 z 5 + c 2 z 9 + ) + b 1 ( β ) ( z + c 1 z 5 + c 2 z 9 + ) 5 + b 2 ( β ) ( z + c 1 z 5 + c 2 z 9 + ) 9 + = b 0 ( β ) ( z + c 1 z 5 + c 2 z 9 + o ( z 9 ) ) + b 1 ( β ) ( z + c 1 z 5 + o ( z 5 ) ) 5 + b 2 ( β ) ( z + o ( 1 ) ) 9 + o ( z 9 )

The second line gives

b 1 ( β ) ( z + c 1 z 5 + c 2 z 9 + o ( z 9 ) ) 5 = b 1 ( β ) z 5 ( 1 + c 1 z 4 + o ( z 4 ) ) 5

Since ( 1 + w ) 5 = 1 + 5 w + o ( w ) , where w = c 1 z 4 + o ( z 4 ) 0 when z 0 , and w c 1 z 4 , thus o ( w ) = o ( z 4 ) , we obtain

( 1 + c 1 z 4 + o ( z 4 ) ) 5 = 1 + 5 c 1 z 4 + o ( z 4 ) ,

so that

b 1 ( β ) ( z + c 1 z 5 + c 2 z 9 + o ( z 9 ) ) 5 = b 1 ( β ) ( z 5 + 5 c 1 z 9 ) + o ( z 9 ) .

Moreover

b 2 ( β ) ( z + o ( 1 ) ) 9 = b 2 ( β ) z 9 + o ( z 9 ) .

Therefore

φ ( βz ) = b 0 ( β ) ( z + c 1 z 5 + c 2 z 9 ) + b 1 ( β ) ( z 5 + 5 c 1 z 9 ) + b 2 ( β ) z 9 + o ( z 9 ) = b 0 ( β ) z + ( b 0 ( β ) c 1 + b 1 ( β ) ) z 5 + ( b 0 ( β ) c 2 + 5 b 1 ( β ) c 1 + b 2 ( β ) ) z 9 + o ( z 9 ) ,

thus the power series expansion of φ ( βz ) begins with

φ ( βz ) = b 0 ( β ) z + ( b 0 ( β ) c 1 + b 1 ( β ) ) z 5 + ( b 0 ( β ) c 2 + 5 b 1 ( β ) c 1 + b 2 ( β ) ) z 9 + .

This gives (15.58). □

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2022-07-19 00:00
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