نتایج جستجو برای: examination

تعداد نتایج: 246438  

2011

so ‖f‖p ≤ ‖f‖q · μ(X) 1 p − 1 q . Hence if f is in Lq, the left-hand side is finite hence so is the right-hand side, so f is in Lp. Also, the inequality shows that if ‖f‖p is small then ‖f‖q is also small, hence the inclusion Lq ↪→ Lp is continuous 2. Let X ⊂ Pn be an irreducible projective variety of dimension k, G(`, n) the Grassmannian of `-planes in Pn for some ` < n− k, and C(X) ⊂ G(`, n) ...

2011
Dennis N. Mehay

What empirical evidence is there that adding syntactic constraints to MT decoding in particular, PMT decoding will lead to improvements in translation quality? Your proposal claims that your method for adding syntactic constraints will result not only in a more complete search of the space of string permutations involved in PMT but also in an improved ability to discriminate between good and ba...

Journal: :The Ulster Medical Journal 1978
W. S. Saunders

THIS is one of the best cancer textbooks available today. The 800 pages cover much of the pathology, diagnosis, treatment and prognosis of malignant disease. The book is a multi-author work and many of the contributions are excellent. Inevitably, however, the presentation is uneven. Moreover, events have overtaken some of the chapters. For instance, the discussion on breast cancer contains scan...

2015
Michael Kearns Paul Erdos

This is a closed-book exam. You should have no material on your desk other than the exam itself and a pencil or pen. If you run out of room on a page, you may use the back, but be sure to indicate you have done so. You may also make annotations directly on any diagrams given.

2015

Solution: By the adjunction formula, the canonical divisor class is KC = OC(d− 3), that is, plane curves of degree d− 3 cut out canonical divisors on C. It follows that if d ≥ 4 then any two points p, q ∈ C impose independent conditions on the canonical series |KC |; that is, h(KC(−p − q)) = g − 2, so by Riemann-Roch h(OC(p+ q)) = 1, i.e., C is not hyperelliptic. Similarly, if d ≥ 5 then any th...

2016

Solution: Since V is locally trivializable and M is compact, one can find a finite open cover Ui, i = 1, . . . , n, of M and trivializations Ti : V |Ui → Rk. Thus, each Ti is a smooth map which restricts to a linear isomorphism on each fiber of V |Ui . Next, choose a smooth partition of unity {fi}i=1,...,n subordinate to the cover {Ui}i=1,...,n. If p : V → M is the projection to the base, then ...

2013

Solution. We can consider the integration from −∞ to ∞ instead. For R > 1, consider the contour that consists of the segment from −R to R and the arc {Reiθ | θ ∈ [0, π]}. Since z2+1 z4+1 decays as |z|−2 on the complex plane as |z| → ∞, this contour integral converges to twice of the original integral when R→∞. The contour encloses two simple poles eπi/4 and e3πi/4 of the function. At eπi/4 the ...

2010
Thomas T. Noguchi

Regardless of the suspected cause of death, a medical examiner or coroner's postmortem examination should always be thorough and comprehensive, for deaths which come under the jurisdiction of this office are often extremely complex, dealing as they do mostly with instances of sudden and unexpected demise. The postmortem examination should be approached with a certain degree of suspicion of ot.h...

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