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Therefore g is rational but not a polynomial.
We know f is rational but not a polynomial, then (f^{n}f^{ k)}) is rational but not a polynomial.
If (f^{n}f^{ k)}-aneq0) and we know f is rational but not a polynomial, then (f^{n}f^{ k)}) also is rational but not a polynomial.
Case 2. If is a nonconstant rational function but not a polynomial.
If f is a rational but not a polynomial meromorphic function and f has only zeros of multiplicity at least k, then (f^{n}f^{ k)}-a) has at least two distinct zeros.
If g is a rational function but not a polynomial, by Lemma 2.2, then (g^{n}g^{ k)}-a=0) at least has two distinct zeros, which is a contradiction.
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where ϕ j ( x, λ ) = − 1 β j 〈 G k ( x, …, x ), e j ∗ 〉 H. Remark 4.2 If g ( s ) in (1.1) is not a polynomial but a C ω with Taylor's expansion in s = 0 as g ( s ) = ∑ k = 2 ∞ a k s k ; if 2 a 2 2 + 45 μ a 3 + 9 α a 3 < 0 is satisfied, then the conclusions of Theorem 4.1 also hold true.
Choose u=ln(x) and dv=1 dx (note ln(x) is not a polynomial, but x^0=1 is; dv is the polynomial in this case).
These polygons are helpful in determining whether or not a polynomial can be reduced to a product of smaller degree polynomials.
The associated linear equation is characterized by a rational symbol which is not a polynomial, except when the energy parameter is zero.
Also, L is not a polynomial, since P ≠ 0. Let L = e C with C an entire function.
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CEO of Professional Science Editing for Scientists @ prosciediting.com