(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
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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
Transcribed Image Text: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?
Transcribed Image Text:(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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