Basic Integral BoundASSUMPTION FOR PART bLet f(2) be continuous along the curve y and have |f(z)| < M along y.Let f(z) be an entire function that satisfies(Inequality 1)(-) d: < M Length().ThenIftz)| SA+ B|z| for some positive constants.Then f(2) is a linear function f(2) = c, + C,z for some complex constantsQUESTIONProve this as follows. As f(2) is entire, it hasa series expansion about z = 0-Σf(2) = co + c1z + c22² + c22³ + •n=0To prove the result we just need to proveCn = 0 for all n > 2for then the series expansion will reduce to just f(z) = co+cız. Becausef(2) is entire its radius of convergence is infinite and therefore we havethat the coefficients in the series are given by1f(2)Cndz2ni|z|=rand this holds for any r > 0. To make this look a little more like someof the notation above letTr = the circle of radius r defined by |z| =rand then our formula for c, becomesCndz2πί(a) What is Length(,)?(b) Use the assumption of inequality 1 to show|f(2)| < Ar + Bon Yr (c) Use part (b) to showf(z)Ar + Bon Yrzn+1pn+1(d) Use part (b) and the Basic Integral Bound to showf(2)dz <zn+1Ar + B|cn|2ni(e) Now take a limit as r → o to show cn0 for n > 2. Why doesn'tthis work for n = 1?

aγr=Circle of radius r, defined by z=r. Writing it as z-0=r It is a circle having center (0 ,0) and…

(c) Use part (b) to show |f(2)| zn+1 Ar + B on Yr pn+1 (d) Use part (b) and the Basic Integral Bound to show 1 |cn| = 2ni f(2) dz zn+1 Ar + B sn (e) Now take a limit as r → o to show c, = 0 for n> 2. Why doesn't this work for n = 1? %3D

(c) Use part (b) to show |f(2)| zn+1 Ar + B on Yr pn+1 (d) Use part (b) and the Basic Integral Bound to show 1 |cn| = 2ni f(2) dz zn+1 Ar + B sn (e) Now take a limit as r → o to show c, = 0 for n> 2. Why doesn't this work for n = 1? %3D

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

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Basic Integral Bound
ASSUMPTION FOR PART b
Let f(2) be continuous along the curve y and have |f(z)| < M along y.
Let f(z) be an entire function that satisfies
(Inequality 1)
(-) d: < M Length().
Then
Iftz)| SA+ B|z| for some positive constants.
Then f(2) is a linear function f(2) = c, + C,z for some complex constants
QUESTION
Prove this as follows. As f(2) is entire, it has
a series expansion about z = 0
-Σ
f(2) = co + c1z + c22² + c22³ + •
n=0
To prove the result we just need to prove
Cn = 0 for all n > 2
for then the series expansion will reduce to just f(z) = co+cız. Because
f(2) is entire its radius of convergence is infinite and therefore we have
that the coefficients in the series are given by
1
f(2)
Cn
dz
2ni
|z|=r
and this holds for any r > 0. To make this look a little more like some
of the notation above let
Tr = the circle of radius r defined by |z| =r
and then our formula for c, becomes
Cn
dz
2πί
(a) What is Length(,)?
(b) Use the assumption of inequality 1 to show
|f(2)| < Ar + B
on Yr

(c) Use part (b) to show
f(z)
Ar + B
on Yr
zn+1
pn+1
(d) Use part (b) and the Basic Integral Bound to show
f(2)
dz <
zn+1
Ar + B
|cn|
2ni
(e) Now take a limit as r → o to show cn
0 for n > 2. Why doesn't
this work for n = 1?

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