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9 May 2024: R Hložek, AI Malz, KA Ponder, M Dai, G Narayan, EEO Ishida, TA AllamJr, A Bahmanyar, X Bi, R Biswas, K Boone, S Chen, N Du, A Erdem, L Galbany, A

21 Oct 2020: KA.Kuulasmaa J.L.Gilbert C.T.Sparrow A.S.H.Basinski R.A.Sutherland C.G.Small J.E.Forster. A.WoodR.E.Lee D.J.Steinitz A.C.Kowalski

17 Aug 2010: Since K H+ {0},we have θHa θ. Ka and p. Ha p. ... Ka , and therefore pg p. Ha by Theorem 5. We pose two questions.

27 Aug 2002: 3 YYY4 ¿À( 331£3¡£3¥¤¡Àée£e@¢ÂY¿AºÛz1º@£I  d«Ë & &(()pÀ( ðN¢.1ñ9º.É«KÅ@ÇseÀ Ù¢Âz¿wº.Â,»65À ÙÙIº.¢»é» Å3EÅZ ) }Çe«7Ë ªÕ ÅZË ÿ ªÑ ... 3 YY@KMLONSP Q 4 _ SR;T U £I¡s«Ë

10 Nov 2012: Nonparametric regres-sion estimates were analyzed with a sliding windowlength of 1500 ka and a step size of 125 ka. ... Theshorter %NCW record was analyzed with a windowlength of 625 ka and a step size of 62.5 ka.

17 Mar 2017: Let hA(x) = Px(TA < ) and kA(x) = Ex[TA]. Theorem 2.2. ... kA(x) = 0 x A. QkA(x) =y. qxykA(y) = 1 x / A.

22 May 2013: Theorem 4.2. The vector of mean hitting times kA = (kAi : i I) is the minimalnon-negative solution to the system of linear equations. {. ... 4.1). Proof. First we show that kA satisfies (4.1). If X0 = i A, then HA = 0, so kAi = 0.If X0 6 A, then HA 1, so

24 Sep 2020: Math. Stat. Learn. 2 (2019), 165–216DOI 10.4171/MSL/14. Mathematical Statistics and Learning European Mathematical Society. On statistical Calderón problems. Kweku Abraham and Richard Nickl. Abstract. For D a bounded domain in Rd;d 2; with smooth

2 Oct 2001: O1[2' ),5' )6?'_$&1Ë.%4A+ ,5O'g?C56#P+?ì1'M?'. Ó F>Å+!"eê ¤ ê Ò Ka)6?@-2a¢ê £9¤ =Ãh)6,!" ,/ F F7?"56#N¢æ £ ... 1.'.)6,5'.)? 'e$;561['.0?-#!CBEÆ «! "= 1[0.@?' $:!-561(Ka/2?"S"a05+!"A+-)!"=61., @?'.$;!"5 13'!ì,5 1[A+ '.)6_%M)6?"$:5ì8;0?@

24 Apr 2005: z D {. KÃ ÈÉ nÇLÌ Ç ÈL « /¡¢ N[@ D  ND/@Dj£w DzH« z@ ¥ @zR@ | g.