Abstract
Let $\mathcal{S}_H^0$ denote the class of all functions $f(z)=h(z)+\overline{g(z)}=z+\sum^\infty_{n=2} a_nz^n +\overline{\sum^\infty_{n=2} b_nz^n}$ that are sense-preserving, harmonic and univalent in the open unit disk $|z|<1$. The coefficient conjecture for $\mathcal{S}_H^0$ is still \emph{open} even for $|a_2|$. The aim of this paper is to show that if $f=h+\overline{g} \in \mathcal{S}^0_H$ then $ |a_n| < 5.24 \times 10^{-6} n^{17}$ and $|b_n| < 2.32 \times 10^{-7}n^{17}$ for all $n \geq 3$.
Results & Lemmas (8)
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Theorem 1.
Theorem 1. Let f = h + g ∈U 0 H(16.5) with series representation as in (1). Then we have (4) |an| < 5.24 × 10−6n17 and |bn| < 2.32 ×…
Theorem 1. Let f = h + g ∈U 0 H(16.5) with series representation as in (1). Then we have (4) |an| < 5.24 × 10−6n17 and |bn| < 2.32 × 10−7n17 for all n ≥3.
Theorem 1
Theorem 1 relies on the bound |a2| ≤16.5 for f ∈U 0 H(16.5). If we use the conjectured bound |a2| ≤5/2, then Theorem 1 takes an improved…
Theorem 1 relies on the bound |a2| ≤16.5 for f ∈U 0 H(16.5). If we use the conjectured bound |a2| ≤5/2, then Theorem 1 takes an improved version which is stated in Section 4. 1.2. Injectivity of sections of univalent harmonic functions. For an analytic func- tion h(z) = P∞ k=1 akzk in the unit disk D, the n-th section/partial sum sn(h) of h is defined by (5) sn(h)(z) = n X k=1 akzk. In [25], Szeg¨o proved that every section sn(h) of h ∈S is univalent in |z| < 1/4 for all
Theorem 2.
Theorem 2. Suppose that f = h + g ∈S0 H with the series representation as in (1). For r ∈(0, 1), define U(r) = 1 log r ( −28.5 +…
Theorem 2. Suppose that f = h + g ∈S0 H with the series representation as in (1). For r ∈(0, 1), define U(r) = 1 log r ( −28.5 + log(r(log(1/r))19) + log "1 −r 1 + r 17 − 1 −r 1 + r 51#)
Corollary 1.
Corollary 1. For f ∈S0 H, we have (1) sn,n(f) is univalent in the disk |z| < 1/4 whenever n ≥81. (2) sn,n(f) is univalent in the disk |z| <…
Corollary 1. For f ∈S0 H, we have (1) sn,n(f) is univalent in the disk |z| < 1/4 whenever n ≥81. (2) sn,n(f) is univalent in the disk |z| < 1/e ≈0.36788 whenever n ≥131. (3) sn,n(f) is univalent in the disk |z| < 1/2 whenever n ≥220. From the proof of Theorem 2, it is also clear that the result could be improved, if we knew the exact upper bounds on |an| and |bn| for f ∈S0 H. Therefore it is natural to state an improved form of this result with the assumption on the order of the family considere
Lemma 1
Lemma 1] it was shown that r 7→Cα(r) is strictly decreasing on (0, 1). This fact implies that ψ(n, m, r) is decreasing in (0, 1) and thus…
Lemma 1] it was shown that r 7→Cα(r) is strictly decreasing on (0, 1). This fact implies that ψ(n, m, r) is decreasing in (0, 1) and thus ψ(n, m, r) > 0 for all r ∈(0, rn,m), where rn,m is the unique positive root of the equation ψ(n, m, r) = 0 which is less than 1. It is easy to see that sn,m(Fr)(z) is sense-preserving in D provided r ∈(0, rn,m) (see e.g [19]) and hence, sn,m(f) is univalent in |z| < rn,m. Now, let us consider the special case m = n. In this case ψ(n, m, r) reduces to ψ(n, n, r
Theorem 3.
Theorem 3. Suppose that f = h + g ∈U 0 H(5/2) with series representation as in (1). Then for all n ≥3, |an| ≤8(2 + n)3( √ 4n2 + 4n + 17 +…
Theorem 3. Suppose that f = h + g ∈U 0 H(5/2) with series representation as in (1). Then for all n ≥3, |an| ≤8(2 + n)3( √ 4n2 + 4n + 17 + 2n −1) n( √ 4n2 + 4n + 17 −2n −9)4 √ 4n2 + 4n + 17 −5 2(2 + n) !1−n
Theorem 4.
Theorem 4. Suppose that f = h + g ∈U 0 H(5/2) ∩S0 H. Then the partial sums sn,m(f) is univalent in the disk |z| < rn,m. Here rn,m is the…
Theorem 4. Suppose that f = h + g ∈U 0 H(5/2) ∩S0 H. Then the partial sums sn,m(f) is univalent in the disk |z| < rn,m. Here rn,m is the unique positive root of the equation ϕ(n, m, r) = 0, where (14) ϕ(n, m, r) = 1 12r 1 −r 1 + r 3 " 1 − 1 −r 1 + r
Corollary 2.
Corollary 2. Suppose that f = h + g ∈U 0 H(5/2) ∩S0 H. Then sn,n(f) is univalent in the disk |z| < r, where (i) r = 1/4 whenever n ≥10;…
Corollary 2. Suppose that f = h + g ∈U 0 H(5/2) ∩S0 H. Then sn,n(f) is univalent in the disk |z| < r, where (i) r = 1/4 whenever n ≥10; (ii) r = 1/2 whenever n ≥29; and (iii) r = 3/4 whenever n ≥98. Better lower bounds for the radius of univalence rn,n of sn,n(f) (under the assumptions of Theorem 4) for certain values of n are listed in Table 1. They are obtained by solving the equation ϕ(n, n, r) = 0. Value of n Lower bound for rn,n Value of n Lower bound for rn,n 2 0.0635798 10
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