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Fundamental Theorem of Algebra Cis algebraically closed, ie, any polynomial anxn an 1x n 1a 0 with coe cients in C has a root in C This will be proved later, but at any rate the fact that C is algebraically closed is one of the most attractive features of working over C Example Find a square root of aib, ie, z= xiysuch that z2 = aX C V B 28,234 likes 3 talking about this UK Streewear brandC Stanley Chan All Rights Reserved What are joint distributions?

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6 Problems For Assignment 1 On Discrete Mathematics Structure Cmsc 250 Docsity

Setting the C (Carry), V (overflo w), N (negative) and Z (zero) bits How the C, V, N and Z bits of the CCR are changed Condition Code Register Bits N, Z, V, C N bit is set if result of operation in negative (MSB = 1) Z bit is set if result of operation is zero (All bits = 0) V bit is set if operation produced an overflowAnswer (1 of 5) (xa)(xb)(xx)(xy)(xz) (xx)=0 Since, 0 is there Therefore, whatever is multiplied with 0 is 0 Thus, it is 0N(x) = n Z x1/n x f(t)dt, x ∈ a,b Suppose f(x) ≤ K for all x ∈ a,b Then g n(x) ≤ K for all x ∈ a,b and n ∈ IN Since lim n→∞ g n(x) = F0(x) for almost every x ∈ a,b, by the Lebesgue dominated convergence theorem, we see that for each c ∈ a,b, Z c a F0(x)dx = lim n→∞ Z c a g n(x)dx But F is continuous

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4 = @ 3 7 5 > ?< 9 @ 3 4N 2 ˙2 ˘˜ 2 n 1 (c) b n and ˙b2 n are independent 7 A preview of the next few lectures Let us consider a simple experiment I toss a fair coin ntimes, and if the outcome is heads I record X i= 1, and if the outcome is tails I record X i= 1 These are called Rademacher random variables Now, let us consider the average b n= 1 n i=1 X• For p > n, lim x→∞ xp−n = ∞, then lim x→∞ ex xn = ∞ Quiz Quiz 1 domain of ln 1x2 (a) x > 1, (b) x > −1, (c) any x 2 domain of ln x p 4x2 (a) x 6= 0, (b) x > 0, (c) any x 2 Differentiation and Graphing 21 Chain Rule Differentiation Chain Rule Lemma 7 d dx eu = eu du dx Proof By the chain rule, d

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(b) We can substitute J x or J y for J z in (a) However a state cannot be simultaneously an eigenstate of J z and J x Derive the commutation relation for the angular momentum operators J x and J z, (ie J x,J z = iħJ y) from the definition of the linear momentum operatorZ!0¨Á (0 «0!È«³ ³ (0 çÈ( ç !Prove or disprove If ab = 0 in Z n, then a = 0 or b = 0 Disproof by CounterExample Consider multiplication in Z 4 as given in the previous problem One has 2 2 = 0, but 2 6= 0 in Z 4 223 Prove that if p is prime then the only solutions of x2 x = 0 in Z p are 0 and p 1 Proof

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Set Builder C = {x x ∈ Z ∧x is odd} Equality A = B ≡ ∀x(x ∈ A ↔ x ∈ B) This implies that {1,2,3} = {1,2,2,3,3,3} Subset A ⊆ B ≡ ∀x(x ∈ A → x ∈ B) Special Sets ∅ = { } The empty set ∀x(x 6∈ ∅) Z = {x x is an integer} R = {x x is a real number} N = {x x is an integer ∧ x ≥ 0} Z = {x x is anC was a Captain, all covered with lace;Solution The scaling u = x / a, v = y / b, w = z / c transforms the ellipsoid onto the unit sphere u 2 v 2 w 2 = 1 and the given plane onto the plane with equation = Let m u u m v v m w w = δ be the Hesse normal form of the new plane and = its unit normal vector Hence = is the center of the intersection circle and = its radius (see diagram) Where m w = ±1 (ie the plane

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B = 0 B B B B B B @ b 21 b 31 b n1 1 C C C C C C A Minimizing the sum of the squares of the residuals S = (b Ax)T(b Ax) requires solving thenormalequation ATAx = ATbTitle show_temppl Author WombleS Created Date 5/6/21 AMC x4 3x2 3 is irreducible according to Eisenstein's criterion with p = 3 d Consider x5 5x2 1 mod 2, which is x5 x2 1 It is easy to see that this polynomial has no roots in Z 2, and so to prove irreducibility in Z 2 it again suffices to show it has no quadratic factors The only quadratic polynomial in Z 2x that does not have a root in Z 2 is x 2x1 which does not divide x5 x 1

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(ax2 bx c)n where A, B, a,b,c are constants and m,n are positive integers Expressions such as the above can all be integrated using either logarithms or trigonometric substitutions Example With a little experimenting, you should be convinced that 3x2 2x 3 x3Page 5 x y z P x y z C P 316) Simplify the following expressions, and implement them with twolevel NAND gate circuits a) AB′ ABD ABD′ A′C′D′ A′BC′Proposition 17 Let C;C 1 and C 2 be convex sets in Rn and let 2R Then (a) The set C = fz2Rnjz= x;x2Cgis convex (b) The set C 1 C 2= fx2Rnjz= x 1 x 2;x 1 2C 1;x 2 2C 2gis convex Proof For part (a), let z 1 and z 2 both be elements or C Then there exists points x 1;x 2 2C such that z 1 = x 1 and z 2 = x 2Choose any ;1 and form the

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Are rightassociative For example, x = y = z is evaluated as x = (y = z)(c) In Rn the set H= fx2 Rn a 1x1 anxn = cg is a convex set For any particular choice of constants ai it is a hyperplane in Rn Its de ning equation is a generalization of the usual equation of a plane in R3, namely the equation axbyczd= 0 To see that His a convex set, let x(1);x(2) 2 H and de ne z2 R3 by z= (1 )x(1) x(2) ThenBq ef^c^bVb D^gigV, hd ZaVb efXq_ nVY cVXghfmi ^gedacc^ä eaVcV, efZdefZaccdYd Zaå cVg =dYdb o Zd gdhXdfc^å b^fV Hd ga^ bq dhXfYcb AYd, hd edZdWcd bcdY^b, Wq^b Zd cVg, efd\^Xb gXdä \^cr ^ ibfb, hV` ^ c gZaVX shdh efXq_ nVY – geVgc^ – ` Wd\ghXccd_ la^ cVnYd gioghXdXVc^å

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Finally, let's look at the case y= 0 Then either x= 1 and z= 0 or x= 1 and z= 2, from the equality constraints In the rst case, all the other rst order conditions hold as well, so (1;0;0) is another candidate solution In the second case, we get 2 = 0 in the second FOC and therefore this point cannot be a candidate solutionZ 0n I z( v ' v o u v v Ç & v v Ì 0XMHU RPEUH 7RWDO 0XMHU RPEUH 7RWDO d K d > í U ï ñ ï X ô ô ñ ó X ï î U î í í X í ñ U î ó ô ï U î î ô ô U ñ ì ò D Z µ } Ç µ U }A Practical Plan of Character Building, Volume I (of 17) Various He was a bookseller, but better known as a translator of the German contributor to the Gentleman's Magazine, &c

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By definition of complement, x ∈ implies that x 6∈B Hence, x ∈ A and x 6∈B By definition of set difference, x ∈ A− B Thus, A−B = A∩ Here are some basic subset proofs about set operations Theorem For any sets A and B, A∩B ⊆ A Proof Let x ∈ A∩B By definition of intersection, x ∈ A and x ∈ BAnswer (1 of 8) If you want to get proof of this question in an easy way then you may try the method given below take A= Ax iAy jAz k B= Bx iBy jBz k C= Cx iCy jCz k Now start with either L HS or R HS Lets start with LHS LHS = (A x(B x C)) First we will solve for (B x C) (B xD E D Ed K& ^K>/ /d d/KE l DK /&/ d/KE K& KEdZ d í X ^ } o } v E µ u W P } ( W P < r î ì î í r r ì ì î ñ í ï î X u v u v lD } ( } v E µ u

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> >~ ~ >w!x>y z > bwR >wRwR $}L fzy{ws~{ >wRzy{wR~ b 6 X S£lzy{®·6»E¸CÉY is the set of even numbers X \Y is the set of all even numbers in the interval ( 2;4), ie X \Y = f0;2g X Y = fx 2R j 2 < x < 4 or x = 2k for some k 2Zg (f) Let A be the xzplane in R3 and B be the yzplane in R3 Use set comprehension notation to describe the sets A\B and AB Answer A = 8 < 2 4 x y z 3 5 2R3X EX, the C's cancelling It is a desirable property that the spread should not be a ected by a change in location However, it is also desirable that multiplication by a constant should change the spread var(CX) = C2var(X) and sd(CX) = jCjsd(X), because (CX E(CX))2 = C2(X EX)2 In summary for constants aand b, var(a bX) = b2var(X) and sd

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X n x 1 y n y 1 z n z 1 1 C C C C C C A; For example, a b c is evaluated as (a b) c Rightassociative operators are evaluated in order from right to left The assignment operators, the nullcoalescing operators, and the conditional operator ?Subject to a b = 1 Or Minimize a2V ar(X) b2V ar(Y ) 2abCov(X;Y ) subject to a b = 1 Therefore our goal is to nd a and b, the percentage of the available funds that will be invested in each stock Substituting b = 1 a into the equation of the variance we get a2V ar(X) (1 a)2V ar(Y ) 2a(1 a)Cov(X;Y )

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Signals, Systems & Information Problem Set 7 Solutions PS 73 (b) nun!Z X1 n=1 nunz n= X(z) X(z) = z 1 2z 2 3z 3 4z 4 z 1X(z) = z 2 2z 3 3z 4 (1 z 1)X(z) = 1 1 z 1 z 2 z 3 z 4 = 1 1 1 z 1 z 1 1 z 1 X(z) = z 1 (1 z 1)2 ROC jz 1j11 05 0 05 1105 0 05 1 Real part Imaginary part 2 One zero at z=¥ Figure 2 zplane plot for (b) xn = nun PS 73X = 0 B B B @ x x 1 y y 1 z z 1 1 C C C A;

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