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Note that F has a unique root in ((0, mu)) if (textbf{u} ^{k} in 0, mu)).
Thus by the fixed point theorem, each of (60) and (61) has a unique root λ 1 and λ 2 in U ( n ) respectively.
By using the Sturm sequence for the polynomial G 2 ( s ) as stated at the end of the Introduction, both functions (in Maple and in Maxima) return two roots in ( 0, 1 ]. Since G 2 ( 1 ) = 1, we find that G 2 ( s ) has a unique root in ( 0, 1 ).
In consequence the equation h' = 0 has a unique root α * ∈ ( ȳ 2 ∕ ( 2 ȳ 2 ), ( 1 - ȳ 2 ) ∕ ( 2 ȳ ( 1 - ȳ ) ) ), h' is negative at the left-hand side of α* and positive at the right-hand side of α*.
It is enough to show that Ψ ( R ) = 1 + R ln ( R ) 2 ( R − 1 ) < 2, and this holds true if and only if F ( R ) = R ln ( R ) − 2 R + 2 < 0. By simple computations one can show that the function F ( R ) has a unique root in ( 1, ∞ ).
Finally, assume that γ i, i = 1, …, p, satisfy (24), (25), and choose u ∈ B ¯. Then the function σ ( t ) = γ i ( u ( t ), u ′ ( t ), …, u ( n − 2 ) ( t ) ) − t, t ∈ [ a, b ], (29). is continuous and decreasing on [ a, b ] and it has a unique root in the interval ( a, b ), i.e. there exists a unique solution of the equation t = γ i ( u ( t ), …, u ( n − 2 ) ( t ) ). (30).
Similar(52)
Further, each
As in tree structures the pathways from each input variable to the output are unique, all pairs of input variables will have a unique root.
has a unique solution.
By a similar analysis, we derive that Eq. (9) has a unique zero root and all other roots with negative real parts for any τ > 0 if R 0 = 1.
According to Lemma 4.2, (20) has a unique positive root, denoted by (z_{0}), and thus (19) has a unique positive root (omega_{0}= sqrt{z_{0}}).
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com