A Transcendental Julia Set of Dimension 1 Jack Burkart 2 October - PowerPoint PPT Presentation

A Transcendental Julia Set of Dimension 1 Jack Burkart 2 October 2017 Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

Some History (Baker, 1975): Julia set of a transcendental entire function 1 contains a continuum. So we always have dim ( J ( f )) ≥ 1. (McMullen, 1987): Studied two families of transcendental 2 entire functions: { f ( z ) = λ e z : λ � = 0 } , dim J ( f ) = 2 { g ( z ) = sin ( az + b ) : a � = 0 } , J ( f ) has positive area . (Stallard, 1997-2000): Constructed examples in B with 3 Hausdorff dimension d for all d ∈ ( 1 , 2 ] . Jack Burkart

The Main Theorem Theorem (Bishop, 2011) There exists a transcendental entire function f so that J ( f ) has Hausdorff dimension 1 . Jack Burkart

The Main Theorem Theorem (Bishop, 2011) There exists a transcendental entire function f so that J ( f ) has Hausdorff dimension 1 . Jack Burkart

What is f ? The function f is a family of infinite products ∞ � f ( z ) = F 0 ( z ) · F k ( z ) . k = 1 Each f is determined by fixed parameters { N ∈ N , λ > 1 , R > 1 , S ⊂ N } . F 0 ( z ) = N th iterate of p λ ( z ) = λ ( 2 z 2 − 1 ) , � z � n k � � 1 − 1 F k ( z ) = . 2 R k Here, { R k } and { n k } are defined in terms of { N , λ, R , S } and increase rapidly to ∞ . Jack Burkart

What is f ? The function f is a family of infinite products ∞ � f ( z ) = F 0 ( z ) · F k ( z ) . k = 1 Each f is determined by fixed parameters { N ∈ N , λ > 1 , R > 1 , S ⊂ N } . F 0 ( z ) = N th iterate of p λ ( z ) = λ ( 2 z 2 − 1 ) , � z � n k � � 1 − 1 F k ( z ) = . 2 R k Here, { R k } and { n k } are defined in terms of { N , λ, R , S } and increase rapidly to ∞ . Jack Burkart

What is f ? The function f is a family of infinite products ∞ � f ( z ) = F 0 ( z ) · F k ( z ) . k = 1 Each f is determined by fixed parameters { N ∈ N , λ > 1 , R > 1 , S ⊂ N } . F 0 ( z ) = N th iterate of p λ ( z ) = λ ( 2 z 2 − 1 ) , � z � n k � � 1 − 1 F k ( z ) = . 2 R k Here, { R k } and { n k } are defined in terms of { N , λ, R , S } and increase rapidly to ∞ . Jack Burkart

What is f ? The function f is a family of infinite products ∞ � f ( z ) = F 0 ( z ) · F k ( z ) . k = 1 Each f is determined by fixed parameters { N ∈ N , λ > 1 , R > 1 , S ⊂ N } . F 0 ( z ) = N th iterate of p λ ( z ) = λ ( 2 z 2 − 1 ) , � z � n k � � 1 − 1 F k ( z ) = . 2 R k Here, { R k } and { n k } are defined in terms of { N , λ, R , S } and increase rapidly to ∞ . Jack Burkart

What is f ? The function f is a family of infinite products ∞ � f ( z ) = F 0 ( z ) · F k ( z ) . k = 1 Each f is determined by fixed parameters { N ∈ N , λ > 1 , R > 1 , S ⊂ N } . F 0 ( z ) = N th iterate of p λ ( z ) = λ ( 2 z 2 − 1 ) , � z � n k � � 1 − 1 F k ( z ) = . 2 R k Here, { R k } and { n k } are defined in terms of { N , λ, R , S } and increase rapidly to ∞ . Jack Burkart

What is f ? The function f is a family of infinite products ∞ � f ( z ) = F 0 ( z ) · F k ( z ) . k = 1 Each f is determined by fixed parameters { N ∈ N , λ > 1 , R > 1 , S ⊂ N } . F 0 ( z ) = N th iterate of p λ ( z ) = λ ( 2 z 2 − 1 ) , � z � n k � � 1 − 1 F k ( z ) = . 2 R k Here, { R k } and { n k } are defined in terms of { N , λ, R , S } and increase rapidly to ∞ . Jack Burkart

What is f ? To illustrate, choose parameters N = 5, R = λ = 10. Then F 0 ( z ) = ( 2 λ ) 2 N − 1 z 2 N + lower order terms We define { n k } in terms of N by n k = 2 N + k − 1 We define { R k } so that we have growth at least R k + 1 ≥ 2 R 2 k . Then, for example � � 512 � � z F 4 ( z ) = 1 − . 1 , 600 , 000 , 000 Jack Burkart

What is f ? To illustrate, choose parameters N = 5, R = λ = 10. Then F 0 ( z ) = ( 2 λ ) 2 N − 1 z 2 N + lower order terms We define { n k } in terms of N by n k = 2 N + k − 1 We define { R k } so that we have growth at least R k + 1 ≥ 2 R 2 k . Then, for example � � 512 � � z F 4 ( z ) = 1 − . 1 , 600 , 000 , 000 Jack Burkart

What is f ? To illustrate, choose parameters N = 5, R = λ = 10. Then F 0 ( z ) = ( 2 λ ) 2 N − 1 z 2 N + lower order terms We define { n k } in terms of N by n k = 2 N + k − 1 We define { R k } so that we have growth at least R k + 1 ≥ 2 R 2 k . Then, for example � � 512 � � z F 4 ( z ) = 1 − . 1 , 600 , 000 , 000 Jack Burkart

What is f ? To illustrate, choose parameters N = 5, R = λ = 10. Then F 0 ( z ) = ( 2 λ ) 2 N − 1 z 2 N + lower order terms We define { n k } in terms of N by n k = 2 N + k − 1 We define { R k } so that we have growth at least R k + 1 ≥ 2 R 2 k . Then, for example � � 512 � � z F 4 ( z ) = 1 − . 1 , 600 , 000 , 000 Jack Burkart

What is f ? To illustrate, choose parameters N = 5, R = λ = 10. Then F 0 ( z ) = ( 2 λ ) 2 N − 1 z 2 N + lower order terms We define { n k } in terms of N by n k = 2 N + k − 1 We define { R k } so that we have growth at least R k + 1 ≥ 2 R 2 k . Then, for example � � 512 � � z F 4 ( z ) = 1 − . 1 , 600 , 000 , 000 Jack Burkart

Why F 0 ( z ) ? The Julia set of p λ ( z ) , and therefore of F 0 ( z ) , is a Cantor set in [ − 1 , 1 ] . It’s dimension tends to 0 as λ → ∞ . { R k } and { n k } are chosen to increase sufficiently quickly, so that on D = B ( 0 , 1 / 2 R ) , ∞ � F k ( z ) ≈ 1 . k = 1 Therefore on D f ( z ) ≈ F 0 ( z ) = ( p λ ( z )) ◦ N . Jack Burkart

Why F 0 ( z ) ? The Julia set of p λ ( z ) , and therefore of F 0 ( z ) , is a Cantor set in [ − 1 , 1 ] . It’s dimension tends to 0 as λ → ∞ . { R k } and { n k } are chosen to increase sufficiently quickly, so that on D = B ( 0 , 1 / 2 R ) , ∞ � F k ( z ) ≈ 1 . k = 1 Therefore on D f ( z ) ≈ F 0 ( z ) = ( p λ ( z )) ◦ N . Jack Burkart