{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

taylor - ACM95a/100a Lecture Notes Niles A Pierce Caltech...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ACM95a/100a Lecture Notes Niles A. Pierce Caltech 2008 Taylor Series Theorem 1 If f ( z ) is analytic for | z - z 0 | < R then f ( z ) = X k =0 f ( k ) ( z 0 ) k ! ( z - z 0 ) k (1) uniformly in | z - z 0 | ≤ ρ < R ( i.e. f ( z ) is represented uniformly by its Taylor series in the closed disc | z - z 0 | ≤ ρ ) . Proof: For convenience substitute z z + z 0 and let h ( z ) f ( z + z 0 ) = X k =0 f ( k ) ( z 0 ) k ! z k (2) so h ( z ) is analytic for | z | < R and we must show uniform convergence to h ( z ) for | z | ≤ ρ < R . Consider the positive circle C 0 : | z | = ρ 0 with ρ < ρ 0 < R . R ρ 0 ρ 0 C 0 By the Cauchy integral formula h ( z ) = 1 2 πi Z C 0 h ( ξ ) ξ - z dξ, for | z | < ρ 0 . (3) If n is a positive integer and w 6 = 1 then note that 1 1 - w = 1 + w + w 2 + · · · + w n - 1 + w n 1 - w (4) as can be readily verified by multiplying by 1 - w . Setting w = z/ξ (and dividing through by ξ ) we have 1 ξ - z = 1 ξ + z ξ 2 + · · · + z n - 1 ξ n + z n ξ n 1 ξ - z . (5) Hence, h ( z ) = 1 2 πi Z C 0 h ( ξ ) ξ + · · · + z n - 1 2 πi Z C 0 h ( ξ ) ξ n + z n h n ( z ) (6) where h n ( z ) = 1 2 πi Z C 0 h ( ξ ) ξ n ξ - z . (7) Then by the generalized Cauchy integral formula 1 2 πi Z C 0 h ( ξ ) ξ k
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern