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Ak = ( xne jk(2r/N)n N n=(N) (b) We are given that xn is an aperiodic signal xn = X(Q)e'j" d2 (i) By multiplying both sides by e join and summing over all n, we have xne jIn = X(Q) e"(")n di n = oo 2 1r f2 _n 0 (ii) En ej' "")n needs to be evaluated We can recognize that this summa­.

Kx xn q e cxg. E(X 1 KX n)=E(X 1)KE(X n) Proof Use the example above and prove by induction Let X 1, X n be independent and identically distributed random variables having distribution function F X and expected value µ Such a sequence of random variables is said to constitute a sample from the distribution F X The quantity X, defined by. As n!1 Hint Write out the required binomial probability, expanding the binomial coe cient into a ratio of products. C J L G 3.

Improve your skills with free problems in 'Write the function in the form f(x)=(xk)q(x)r for the given value of k' and thousands of other practice lessons. M(n)(0) = E(), n ≥ 1 (8) The mgf uniquely determines a distribution in that no two distributions can have the same mgf So knowing a mgf characterizes the distribution in question If X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgf. Z j Z g h;.

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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−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N. L G s ̖ C X g f ށA ۈ E w Z ̃C X g f ށA l E q ̃C X g C X g y ݂ ̃C X g f ށz p p \/ HOK Ԃ̃C X g/ T C g } b v. Ak = ( xne jk(2r/N)n N n=(N) (b) We are given that xn is an aperiodic signal xn = X(Q)e'j" d2 (i) By multiplying both sides by e join and summing over all n, we have xne jIn = X(Q) e"(")n di n = oo 2 1r f2 _n 0 (ii) En ej' "")n needs to be evaluated We can recognize that this summa­.

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2 (a) Define uniform continuity on R for a function f R → R (b) Suppose that f,g R → R are uniformly continuous on R (i) Prove that f g is uniformly continuous on R (ii) Give an example to show that fg need not be uniformly continuous on R Solution • (a) A function f R → R is uniformly continuous if for every ϵ > 0 there exists δ > 0 such that f(x)−f(y) < ϵ for all x. 2 ( 3 ' 4 (5 6 478 (" 9 '/ ";. What is the lewis structure for hcn?.

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What are the units used for the ideal gas law?. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history. M r c > b d v i l x h r x j p v r ?.

Values The Qfunction is well tabulated and can be computed directly in most of the mathematical software packages such as R and those available in Python, MATLAB and MathematicaSome values of the Qfunction are given below for reference. X N 1964 N n B R c R c w ю ͂ A ƒ ƂŊ 􂷂 l 琬 ܂ B Ƃ͎ K ` Ń L L ɂȂ ܂ B H ނ͖ z n Ŏd ꂽ S Ŕ A V N Y g p Ă A L V F t ɂ ʃ b X ܂ B ܂ A H A C ^ A E ̏C w s Ȃǂ̊y C x g J ÁB e R X I Ɋe 펑 i 擾 ł ܂ B { l ̓` I ȐH l X R ɓo ^ A ܂ ܂ u ƒ뗿 v w ԕ Ȃ Ă ܂ B. Engaging math & science practice!.

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P \ R ANAS ARAID A T o Ȃǃn h f B X N 𑦓 f f } Ńf ^ E n h f B X N v ܂ I x ȋZ p v f ^ x X C B. Since f(x) is a higher degree polynomial than (xk), If we divide (xk) into f(x), we'll get a nonfractional quotient and a fractional remainder. X g r n p x g j c u x e x x x c ?.

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' < ' >=@?. X 0 2=EShow that there is an unbounded continuous function f E!R Solution Consider the function f(x) = 1 x x 0 Since x 0 2= E, this function is continuous on E On the other hand, by the hypothesis, lim n!1jf(x n)j= 1;and so the function is unbounded on E 2(a)If a;b2R, show that maxfa;bg= (a b) ja bj 2. 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−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N.

G v Q @ A @ X p C _ \ e B A t b V Q w PR j E o Q ւ. Thus, #g^((n))=(1)^n(g n e^(x)) , n=1, 2, 3, # Answer link Related questions How do I determine the molecular shape of a molecule?. VarX¯ n VarM n = 34/n /n = 1 A confidence interval for µ based on X¯ n is given by X¯ n ±z 1−α/2 q VarX¯ n In order for a 95% confidence interval to have length 01, we must have z 1−005/2 q VarX¯ n = 005 ⇒ 196 r 8.

9 8 < = >?. Values The Qfunction is well tabulated and can be computed directly in most of the mathematical software packages such as R and those available in Python, MATLAB and MathematicaSome values of the Qfunction are given below for reference. Reinforcement Learning and Optimal Control ASU, CSE 691, Winter 19 Dimitri P Bertsekas dimitrib@mitedu Lecture 2 Bertsekas Reinforcement Learning 1 / 24.

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C t g p k g d v i l ?. X 0 2=EShow that there is an unbounded continuous function f E!R Solution Consider the function f(x) = 1 x x 0 Since x 0 2= E, this function is continuous on E On the other hand, by the hypothesis, lim n!1jf(x n)j= 1;and so the function is unbounded on E 2(a)If a;b2R, show that maxfa;bg= (a b) ja bj 2. Cyclic Codes Yunghsiang S Han Graduate Institute of Communication Engineering, National Taipei University Taiwan Email yshan@mailntpuedutw.

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C Ô Q ^ H 4 « B « C B ¾ ¢ Ì Q G I N 3 ;. 2 Let X n denote a binomial random variable, X n˘Binomial(n;p n) for n 1If np n!. Ngin Ewith the property that x n!.

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" # $%&'(F * ) 2 / 0 3 47 6 5 9 8 L ;. Fz kgand fx ngSince fz kgconverges, so does fa mg, contradicting the assumption that fx nghas no convergent subsequence (Note the similarities with the solution of Exercise 7 from Homework 3) 394 Let Mbe a metric space such that Mis a nite set.