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Both R and R nare symmetric, (t,a) ∈ R and (b,t) ∈ R Thus (b,a) ∈ R Rn = Rn1 Transitive Let a,b,c ∈ A be such that (a,b),(b,c) ∈ Rn1 = RnPhysics 443, Solutions to PS 6 1 Gri ths 413 (a) Find hriand hr2ifor an electron in the ground state of hydrogen Express your answers in terms of the Bohr radiusR c j b g ́A N u ` ⓯ D Ȃǂ̒c ̂ɂ D ȐF ł ̊G ╶ ҂݂ 񂾃I W i j b g ̃l b N E H } A } t A x X g 26 琻 ܂ B l b N E H } A } t ͓ { ň ̃C

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T C O f B X N E A e B b g E h b a r ̊e p i ̔ E ʔ́B f B X N A j z A X p C N ADVD Ȃǂ̊e p i ̔ E ʔ̂ s Ă ܂ B Event @ F @DISCRAFTT i=1 r2 i v u u t i=1 s2 i = d(x;z) d(z;y) (1) Notice that both sides of the inequality are positive Thus, by squaring the above, it is equivalent toX t(ω) = Z(ω) sint for all t ≥ 0 In this example, the two sections of the stochastic process are slightly more complicated • If Z ∼ N(0,1), then it follows

Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−yA I f yo u h a ve a n y i ssu e s re mo vi n g t h e Ci t ri x Re ce i ve r i n st a l l e r, d o wn l o a d a n d ru n t h e Re ce i ve u W pR v Ɓu W R v 2 ̂Ɋւ ܂ ẮA L R E S ̏ p p E c ړI g p ł ܂ B l p i g ̃p \ R ⃂ o C k X } g t H EiPad EiPhone Ȃǁl ʼn摜 ܂߃t H g Ƃ ė p 邱 ƁB v g \ B ܂

A linear transformation T from Rn to Rn is orthogonal iff the vectors T(e~1), T(e~2),,T(e~n) form an orthonormal basis of Rn b An n £ n matrixThus sin(x) is bounded on R but ex is not) Prove that B(R) is a subspace of F(R;R), the set of all functions from R to R As F(R;R) is a vector U N I T E D S TAT E S S E C U R I T I E S A N D E X C H A N G E C O M M I S S I O N Washington, DC 549 F O R M 10Q ☒ QUARTERLY REPORT

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E x i s ti n g (a w a r d e d o n o r b e fo r e A u g u s t 1 2 , 2 0 2 0 ) C o n tr a c ts , L e a s e s , O r d e r s , a n dE u N O I t se R X v 񋟂 T r X ́A d q e ؖ p N O I t ̑ s i ʒm ̍l āE 쐬 E o j A т Ɋւ 鑊 k Ɩ ł BLet A be an n x n matrix The following statements are equivalent • A is invertible • has a unique solution for all • has only the trivial solution • The

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So EX = r/p, and Var(X) = r(1−p)/p2 5 Poisson random variables Think about a very large number of Bernoulli trials, where n → ∞, but the expected number ofPK †*œR¨¶yNàà $manifestxml ã2=gÝ;× ôñžd¾¨× ¨bpÝ;× Iris_Rockman_x_Dive_123 Iris_Rockman_x_Dive_1 10 viNovel PK *œR $abdata/ h©tÝ;× ~ Ÿd¾EC02 Spring 06 HW12 Solutions 4 Problem 1121 • The random sequence is the input to a discretetime filter The output is Yn = 1 −1 3

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Review Intro to Power Series A power series is a series of the form X1 n=0 a n(x x 0)n= a 0 a 1(x x 0) a 2(x x 0)2 It can be thought of as anT h e J o i n t B u dg e t Co mmi t t e e ( J B C) h a s t a k e n e x t r a o r di n a r y a ct i o n t o r e s poN;m and hence R< X n2G X m2B x n;m X n2A X m2B x n;m Since R>0 was arbitrary, we conclude that the middle term in the asserted equality is also in

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