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Spring 2019 - Problem 8

Hadamard factorization

Show that

sin⁡z−zcos⁡z=z33∏n=1∞(1−z2λn2),z∈C,\sin{z} - z\cos{z} = \frac{z^3}{3} \prod_{n=1}^\infty \p{1 - \frac{z^2}{\lambda_n^2}}, \quad z \in \C,

where {λn}n\set{\lambda_n}_n is a sequence in C\C, λn≠0\lambda_n \neq 0 for all nn, such that

∑n=1∞1∣λn∣2<∞.\sum_{n=1}^\infty \frac{1}{\abs{\lambda_n}^2} < \infty.
Solution.

Let f(z)=sin⁡z−zcos⁡zf\p{z} = \sin{z} - z\cos{z}. Notice that f(−z)=−sin⁡z+zcos⁡z=−f(z)f\p{-z} = -\sin{z} + z\cos{z} = -f\p{z}. Thus, if λ\lambda is a root of ff, then so is −λ-\lambda. Let {−λn,λn}n\set{-\lambda_n, \lambda_n}_n be the non-zero zeroes of ff counted with multiplicity. At the origin, we have

f(0)=0f′(z)=zsin⁡z  ⟹  f′(0)=0f′′(z)=sin⁡z+zcos⁡z  ⟹  f′′(0)=0f′′′(z)=2cos⁡z−zsin⁡z  ⟹  f′′′(0)=2,\begin{aligned} f\p{0} &= 0 \\ f'\p{z} = z\sin{z} \implies f'\p{0} &= 0 \\ f''\p{z} = \sin{z} + z\cos{z} \implies f''\p{0} &= 0 \\ f'''\p{z} = 2\cos{z} - z\sin{z} \implies f'''\p{0} &= 2, \end{aligned}

so ff has a zero of order 33 at the origin. ff is also entire of first order since z,sin⁡z,cos⁡zz, \sin{z}, \cos{z} all are, so by Hadamard factorization,

f(z)=eaz+bz3∏n=1∞(1−z2λn2)f\p{z} = e^{az+b} z^3 \prod_{n=1}^\infty \p{1 - \frac{z^2}{\lambda_n^2}}

with

∑n=1∞1∣λn∣2<∞.\sum_{n=1}^\infty \frac{1}{\abs{\lambda_n}^2} < \infty.

To complete the proof, we need to calculate a,b∈Ca, b \in \C. Notice that if a3a_3 is the coefficient of z3z^3 in the Taylor series of ff, then

eb=lim⁡z→0f(z)z3=a3=f′′′(0)3!=13.e^b = \lim_{z\to0} \frac{f\p{z}}{z^3} = a_3 = \frac{f'''\p{0}}{3!} = \frac{1}{3}.

For aa, notice that

−1=f(z)f(−z)=−e−az  ⟹  a=0,-1 = \frac{f\p{z}}{f\p{-z}} = -e^{-az} \implies a = 0,

and so

sin⁡z−zcos⁡z=z33∏n=1∞(1−z2λn2).\sin{z} - z\cos{z} = \frac{z^3}{3} \prod_{n=1}^\infty \p{1 - \frac{z^2}{\lambda_n^2}}.