
Complex Analysis Complex Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 27th, 2017, 02:36 PM  #1 
Newbie Joined: Feb 2017 From: Maryland Posts: 3 Thanks: 0  Can someone help me get started on this proof?
Let f be analytic and bounded by M in z≤ r. Prove that f (n)(z) ≤ n!M/〖(rz)〗^n for (z < r).

February 27th, 2017, 02:50 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,965 Thanks: 2282 Math Focus: Mainly analysis and algebra 
I can't decipher your question.

February 27th, 2017, 03:00 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,560 Thanks: 376  That expression does not look right. What is $n$? Do you by any chance mean $f^n(z)$? Or $f^{(n)}(z)$ meaning the $n$th derivative? Also in my opinion and speaking only for myself, questions at this level should certainly be $\LaTeX$'d. If you can hack complex variables you can certainly hack $\LaTeX$ Last edited by Maschke; February 27th, 2017 at 03:03 PM. 
February 27th, 2017, 06:05 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,560 Thanks: 376 
ps  I hope I didn't frighten off the OP. It's ok if you don't use math markup. But what you wrote is not correct as given.


Tags 
proof, started 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
GETTING STARTED WITH CALCULUS.  mathmaniac  Academic Guidance  11  February 13th, 2013 05:46 PM 
Getting Started with Math  Bryson  Algebra  4  June 14th, 2009 09:40 AM 
Get me started  fred555  Calculus  1  February 27th, 2009 07:18 AM 
GETTING STARTED WITH CALCULUS.  mathmaniac  Calculus  0  December 31st, 1969 04:00 PM 