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### Exercices: hints, solutions, comments Second course

1.Prove the two lemmas on entire functions p. 16.

×

2iπz/ω

iπz

iπz

iπz0

0

×

2iπz

e2iπz

f

×

g

×

2iπz

0

1

0

×

2iπz/ω

N

2iπz/ω

2iπz/ω

×

2iπz/ω

1

1

2

3

r

αr

n∈Z

n

n

n

|t|=r

n+1

(2)

n

n

r

n

N

### A(t) + B(t) where A and B are polynomials of degree ≤ N.

2.Checkc00n=cn−1forn≥1p. 22.

tz

−tz

t

−t

n≥0

n

2n

2

2

2

n≥0

00n

2n

000

001

2

002

4

2

n≥0

n

2n

0

2

1

4

2

6

000

00n

n−1

0

0

n

n

### (z) for n ≥ 0.

3.LetSbe a positive integer and letz∈C. Using Cauchy’s residue Theorem, compute the integral (see p. 26)

1 2πi

Z

|t|=(2S+1)π/2

t−2n−1sh(tz) sh(t)dt.

tz

−tz

t

−t

2t

−2n−1 sh(tz)

sh(t)

−2n

sh(tz)sh(t)

n

t

s

−t

s

t

−t

s

tz

iπsz

−tz

−iπsz

(3)

tz

−tz

t

−t

s

−2n−1 sh(tz)

sh(t)

n+s

−2n−1

n

n

2n+1

S

s=1

s

2n+1

|t|=(2S+1)π/2

−2n−1

### for S = 1, 2, . . . and z ∈ C .

4.Prove the proposition p. 31 :

Letf be an entire function. The two following conditions are equivalent.

(i)f(2k)(0) =f(2k)(1) = 0for all k≥0.

(ii)f is the sum of a series

X

n≥1

ansin(nπz)

which converges normally on any compact.

Prove also the following result :

Letf be an entire function. The two following conditions are equivalent.

(i)f(2k+1)(0) =f(2k)(1) = 0for all k≥0.

(ii)f is the sum of a series

X

n≥1

ancos

(2n+ 1)π

2 z

which converges normally on any compact.

(2k)

(2k)

(2k)

(2k)

×

iπz

n∈Z

n

n

×

−n

n

0

n≥1

n

n

−n

n

n

(2n+1)π

2

(2k+1)

(2k)

(4)

×

iπz/2

n≥1

n

n

−n

### which implies condition (ii).

5.Complete the three proofs of the Lemma p. 33.

(2n+1)

(2n)

(2n+1)

(2n)

0

1

0

0

1

00

00

(2n+1)

(2n)

0

2

2

4

4

6

6

8

6

2n

2n

00

(iv)

### (1) = · · · = 0 :

a0 +a2 +a4 +a6 +· · · +a2n +· · ·= 0

2a2 +12a4 +30a5 +· · · +2n(2n−1)a2n +· · ·= 0

24a4 +360a6 +· · · +(2n−4)!(2n)! a2n +· · ·= 0

. .. ...

### The matrix of this system is triangular with maximal rank.

6.Let(Mn(z))n≥0 and(fMn(z))n≥0 be two sequences of polynomials such that any polynomialf ∈C[z]has a finite expansion

f(z) =

X

n=0

f(2n)(1)Mn(z)+f(2n+1)(0)Mfn(z) , with only finitely many nonzero terms in the series (see p. 34). Check

Mfn(z) =−Mn+10 (1−z)

(5)

forn≥0.

Hint:Considerf0(1−z).

0

n=0

(2n)

n

(2n+1)

n

0

n=0

(2n)

n0

(2n+1)

n0

0

n=0

(2n)

n0

(2n+1)

n0

(2n+2)

n+10

n=0

(2n)

n

(2n+1)

n

(2n)

(2n+1)

(2n+1)

(2n+2)

n=0

(2n+1)

n

(2n+2)

n

(2n+2)

n

n

n+10

00

### = 0).

7.LetSbe a positive integer and letz∈C. Using Cauchy’s residue Theorem, compute the integral (see p. 39) 1

2πi Z

|t|=Sπ

t−2n−1ch(tz) ch(t)dt.

tz

−tz

t

−t

2t

12

12

12

### The residue at t = 0 of t

−2n−1 ch(ch(t)tz)

−2n

ch(tz)ch(t)

n

12

t

s

iπ/2

s

−t

s

−iπ/2

s

s

(6)

t

−t

s

tz

−tz

12

### iπ of t

−2n−1 ch(ch(t)tz)

n+s

−2n−1

−2n−1

12

n

n

2n+2

2n+1

S−1

s=0

s

2n+1

|t|=Sπ

−2n−1

### for S = 1, 2, . . . and z ∈ C .

8.Give examples of complete, redundant and indeterminate systems in Whittaker classification p. 43.

0

1

2

0

1

2

(7)

0

1

2

0

1

2

## Références

### [Whittaker, 1933] Whittaker, J. M. (1933). On Lidstone’s series and two-point expansions of analytic functions. Proc. Lond. Math. Soc. (2), 36 :451–469.

https://doi.org/10.1112/plms/s2-36.1.451

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