Sentence examples for be a polynomial and from inspiring English sources

Theorem I. Let be a polynomial, and let be an intege; let be a transcendental entire function, and let.

Let (q z)) be a rational function, (p z)) be a polynomial and (m,ninmathbb{N}), (tin mathbb{N}cup{0}), (ainmathbb{C}backslash{0}).

That is, Theorem 1.1 generalizes Theorems B and D. Let (Q z)) be a non-zero polynomial, (alpha z)) be a polynomial and (cinmathbb{C}backslash{0}).

Let f be a nonconstant meromorphic function of finite order, p (≢0) be a polynomial, and (nge 2) be an integer.

Let (ngeq4) be an integer, q be a polynomial, and (p_{1}), (p_{2}), (alpha_{1}), (alpha_{2}) be nonzero constants such that (alpha_{1}neqalpha_{2}).

Theorem 1.4 Let k be a positive integer, b (≠0) be a complex number, h ( z ) be a polynomial, and let H ( f, f ′, …, f ( k ) ) be a differential polynomial with Γ γ | H < k + 1.

where P ( z ) is a polynomial and b ( z ) = Δ η n a ( z ) − a ( z ).

If k ≤ 2, then it follows from (3.6) and Lemma 2.3 that α' is a polynomial, and so α is a nonconstant polynomial.

This is certainly the case for polynomial growth since polynomial growth (either in the entire or in the harmonic case) implies that the function is a polynomial and, obviously, the translation of a polynomial is another polynomial of the same degree.

Then we have P ( f ) f ( z + c ) − α ( z ) = R ( z ) e Q ( z ) / β ( z ), where Q ( z ) is a polynomial, and R ( z ) ≢ 0 is a rational function.

Case 2. If g is a polynomial and the zeros of (g xi)) are at least k multiple, and (ngeqfrac{1+sqrt{1+4k(k+1)^{2}}{2k}}), then (g^{n}(xi)(g^{ k)}(xi -a=0) must have zeros, which is a contradiction.