Results 281 to 290 of about 24,345 (312)

Existence Theorems: Problems of Slow Growth

1983
We discuss here existence theorems for the usual integrals \( I\left[ x \right] = {\text{ }}\int_{{{{t}_{1}}}}^{{{{t}_{2}}}} {{{F}_{0}}(t,x(t),x'(t))dt} {\text{ }} \) as in Section 11.1, but where F0(t, x, x′) does not satisfy any of the growth conditions we have considered in Chapters 11, 12, 13. Well known problems are of this kind (cf. Section 3.12).
openaire   +1 more source

The Fundamental Duality Theorem of Balanced Growth [PDF]

, 2007
In this paper we demonstrate that a simple duality relation underlies balanced growth models with non-joint production. Included in this class of models is the standard neoclassical growth model and endogenous growth models that admit balanced growth paths.
Ian King, Don Ferguson
openaire  

Cancer treatment and survivorship statistics, 2022

Ca-A Cancer Journal for Clinicians, 2022
Kimberly D Miller, Leticia M Nogueira, Angela B Mariotto   +2 more
exaly  

Systemic therapy for hormone receptor‐positive/human epidermal growth factor receptor 2‐negative early stage and metastatic breast cancer

Ca-A Cancer Journal for Clinicians, 2023
Laura A Huppert, Hope S Rugo
exaly  

Current treatment and recent progress in gastric cancer

Ca-A Cancer Journal for Clinicians, 2021
Smita S Joshi, Brian D Badgwell
exaly  

Existence Theorems: Weak Convergence and Growth Conditions

1983
Let A be a subset of the tx-space Rn+1 and let A(t) denote its sections, that is, A(t) = [x ∈ R n | (t, x) ∈ A]. For every (t, x) ∈ A let Q(t, x) be a given subset of the z-space R n , x = (x1,…x n ), z = (z1,…,z n ). Let Mo be the set Mo = [(t,x, z) | (t, x) ∈ A, z ∈ Q(t, x)] ⊂ R1+2n and let F o (t, x, z) be a given real valued function defined on Mo.
openaire   +1 more source

The growth theorem for holomorphic convex mappings

1998
In the case of one complex variable, the following growth theorem is well known. If f (z) = z + … is holomorphic and univalent in the unit disk \( \Delta = \left\{ {z \in \mathbb{C}:\left| z \right| < 1} \right\} \), then $$\frac{r}{{{{(1 + r)}^2}}} \leqslant \left| {f(z)} \right| \leqslant \frac{r}{{{{(1 - r)}^2}}},\left| z \right| = r < 1.$$
openaire   +1 more source

Clinical insights into small cell lung cancer: Tumor heterogeneity, diagnosis, therapy, and future directions

Ca-A Cancer Journal for Clinicians, 2023
Balazs Dome
exaly  

Treatment of muscle‐invasive and advanced bladder cancer in 2020

Ca-A Cancer Journal for Clinicians, 2020
Vaibhav G Patel, William K Oh, Matthew D Galsky   +2 more
exaly  

Growth bound theorem for semigroups

2020
openaire   +1 more source