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SOLUTIONS TO SECTION 1 5 Solution We have u x=u ss x u tt x= u s 2u t u xx=u sss x u stt x 2u sts x 2u ttt x= u ss 4u st 4u tt u xy=u sss y u stt y 2u sts y 2u ttt y= u ss 2u st u y=u ss y u tt y= u s u yy=u sss y u stt y= u ss Substituting10/3/18 1 (a) The case r= 1 of Chebychev's Inequality is known as Markov's Inequality and is usually written P(jXj ) E(jXj)= for an arbitraryClearly f(x;y) is nonnegative as it is zero outside of the range indicated above, and positive inside In order to check the second condition we need to integrate the pdf over the shaded region that is, the region where y
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J a funk funeral home-J=1 u iv i= j=1 p ja j b j p j = j=1 a jb j kuk2 = j=1 ja2 j;Thus (fx jf(x)j= 1g) lim n!1 (E n) " 0 and then kfk r = 1, a contradiction Conversely, suppose there is a sequence of measurable sets fE ngwith 0 < (E n) > >< >> > X1 n=1 a 1=s n ˜ En;if s
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MATH 106 HOMEWORK 3 SOLUTIONS 1 Using the CauchyRiemann equations, show that if f and f are both holomorphic then f is a constant Solution Let f = uiv,so f = u iv Since they are holomorphic, we can use the CauchyRiemannTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Suppose that` h (x)=f(x)g(x) `and `F(x) =f(g(x))`, where `f (2)=3 ;F x(x;y) = x y2 f y= x y2 SOLUTION False If the function has continuous second partial derivatives, then Clairaut's Theorem would apply (and f xy= f yx) However, in this case f xy= 2y f yx= 2y (b) The function fbelow is continuous at the origin f(x;y) =
Kvk2 = j=1 b2 j j By Schwartz's inequality, we have jhu;vij2 kuk2kvk2 Plug in the Schwartz's inequality the expression for hu;viand the norms for u;v, we get the required inequality Problem 3 Chapter 6 ex 5 Prove or disprove existence of an inner product on R2Statistics 581, Problem Set 1 Solutions Wellner;14 Let Y be a random variable having the density function f given by f(y) = y/2 for 0 < y < 2 and f(y) = 0 otherwise (a) Determine the distribution function of Y (b) Let U be uniformly distributed on (0,1) Determine an increasing function g on (0,1) such that g(U) has the same distribution as Y
(b) Write down the joint pdf of g(x,y) of X and Y f(x,y) = 2e−2x2e−2y = 4e−2(xy), for x > 0, y > 0 (c) Define U = X 3Y, and V = Y, then find the joint pdf of U and V u = x3y,v = y;In order for a function f(x,y) to be a joint density it must satisfy f(x,y) ≥ 0 Z ∞ −∞ Z ∞ −∞ f(x,y)dxdy = 1 Just as with one random variable, the joint density function contains all the information about the underlying probability measure if we only look at the random variables X and Y In particular, we can compute the probabilityExercise 2214 let V = P(R), and for j ≥ 1 define T j(f(x)) = f(j)(x), where f(j)(x) is the jth derivative of f(x) Prove that the set {T 1,T 2,,T n} is a linearly independent subset of L(V) for any positive integer n Solution Let n be arbitrary Let a 1,a 2,,a n be scalars such that a 1T 1
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X = u−3v, y = v, J = 1 u−3v > 0,v > 0 → u > 3v > 0 h(u,v) = 4e −2(u 3 v) = 4e −2(u 2v) u > 3v > 0 (d) Find the pdf of U h(u) = 4The marginal probability density function of Xis f X(x) = Z 1 1 f(x;y)dy = Z 1 jxj 1 8 (y2 yx2)e dy Z 1 jxj 1 4 ye ydy using integration by parts 1 4 jxje jx Z 1 jxj 1 4 e ydy using integration by parts 1 4 jxje jx 1 4 e jx 1 4 e jx jxj 1 Let f Y be the marginal probability density function of Y For y < 0 we have f Y(y) = 0, and for y 0 we have f Y(y) = Z 1And the lower Riemann sum of fwith respect to the partition Pby L(f;P) = k=1 m kjI kj= k=1 m k(x k x k 1) 210 11 The Riemann Integral Geometrically, U(f;P) is the sum of the signed areas of rectangles based on the intervals I
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(iii) The maximum rate of change occurs in the direction of the gradient at the given point ie, {eq}\nabla f(x_1,y_1) {/eq} Answer and Explanation 1 Become a Studycom member to unlock thisU(f;P) = k=1 M kjI kj= k=1 M k(x k x k 1);
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Question Consider the following f(x, y) = x/y, P(2, 1), u = 3 5 i 4 5 j (a) Find the gradient of f ∇f(x, y) = (b) Evaluate the gradient at the point P ∇f(2, 1) = (c) Find the rate of change of f at P This problem has been solved!G Z k l h y s b c l _ d k l y \ e y _ l k y h t _ d l h f Z \ l h j k d h h i j Z \ Z K \ h h ^ g h _ b _ a \ h a f _ a ^ g h _ b k i h e v a h \ Z g b _ e x Consider the following f (x, y) = x/y, P (8, 1), u = 3/5 i 4/5 j (a) Find the gradient of f ∇f (x, y) = (b) Evaluate the gradient at the point P ∇f (8, 1) = (c) Find the rate of change of f at P in the direction of the vector u Duf (8, 1) =
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Now we need to differentiate the cdf, fZ(z) = dFZ(z)/dz = Z ∞ −∞ fX(x)fY (x −z)dx Method C OK, we know that fXY (z) = R∞ −∞ fX(x)fY (z − y)dy Let us use it and get the wished pdf Introduce U = −Y and note that fU(u) = fY (−u) (you should know how to prove the last relation)J j −1 f(x) dx, let us ex pand f in a (short, truncated) Taylor series near x = x j−1 f(x) = f(x j−1) f " (y(x))h, where y(x) ∈ x j−1,x We've written y(x) to highlight that it depends on x This works as long as f has one derivative Integrating on both sides, andFind the type of the business you need, using the BBB Business Category listing for Ashburn, VA
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Staining electron microscopy of samples from casepatient 3 (D) and casepatient 5 (E, F) show ovoid particles ( ≈250 nm long, 150 nm wide) with a crisscross appearance;(f) Show that the map F R2R2 given by F(x,y)=(xy,x1) is not linear Solution (a) For T to be linear it must satisfy the equality T(uav)=T(u)aT(v) Itestthiswithdirect computation F(x 1 x 2,y 1 y 2)=(x 1 x 2 y 1 y 2,x 1 x 2) =(x 1 y 1,x 1)(x 2 y 2,x 2) (b) I must show that for each u 2 R2 there exists v 2 R2 such that T(v)=u IU o u n F a i r f a x L o u d o u n P r i n c e W i l l i a m F a i r f a x P r i n c e W i l l i m 44LD0459 4FX 238 4FX 237 M a t c h L i n e P a g e 7 1 / 2 IIa Dominion Energy Virginia Lo u dnOx 230 kV T r asm ieP t lR b 0 2,500 5,000 F et!(Cemetery j Park 5 School Æ Place of Worship #* Substation o ud nB l R S egm t Other Partial
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Therefore, U() and U(24) are not isomorphic Chapter 7 8) Let H = ha5i be the subgroup of the group G = hai given in the question Note that two coset a iH and ajH of H are the same if ai−j ∈ H So, a and aj are in the same coset if ai−j = e or a i−j= a5 or a = a10 In other words, aiσ= E(εT ε)o a o = E(ε εb – ε) = Eu(,x – yβ,x – ε)o ∴ F = σdA F = Eu(,x – yβ,x – ε)o dA F = u,x EA d β,x yE A d εoEAd M = yσdA M = yE u(,x – yβ,x – ε)o dA M = u,x yE A β,x y d 2 EA d yε d oEA X$ yE A d = 0 % F = u,x EA d εoEAd M = β,x y d yεoEA 2 EA dSolution For x 6˘0, jxj is a differentiable function with derivative sgn(x) ˘1 if x ¨0 ¡1 if x ˙0 Thus by the chain rule in the first line and by the product rule in the second line, f 0(x) ˘3jxj2 sgn(x) ˘3xjxj f 00(x) ˘3jxj¯3x sgn(x) ˘3jxj¯3jxj˘6jxj Checking the cases for x ˘0 by hand, we have f 0(0) ˘ lim h!0 f (x¯h)¡ f (x) h ˘ lim
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1 Chap 5 Joint Probability Distributions • Probability modeling of several RV‟s • We often study relationships among variables – Demand on a system = sum of demands from subscribers (D = S 1 S 2 S n) – Surface air temperature & atmospheric CO 2 – Stress & strain are related to material properties;MATH 42 (1516) partial diferential equations CUHK 8Note that u(x;y) = ex2y=4 is a special solution of te inhomogeneous equation, and by the result of 128 above, the general solution of the corresponding homogeneous equation is f(x y)e (xy)=2Thus the1 Math 113 Homework 3 Solutions By Guanyang Wang, with edits by Prof Church Exercises from the book Exercise 3B2 Suppose V is a vector space and S;T2L(V;V) are such that
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See the answer See the answer See theDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the USMath 35 Real Analysis Winter 18 Monday 02/26/18 It remains to show that the integral is wellde ned This means that the integral is independent of the chosen admissible partition for f oTshow this we rst prove the following lemma
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Summary First Quarter 07 Results Three Month Period Ended (in thousands) Change % Change TOTAL REVENUE $77,844 $59,161 $18,6 316% TOTAL EXPENSE 47,557 50,819 (3,262) 64% INCOME BEFORE TAXES 30,287 8,342 21,945 2631% PROVISION FOR INCOME TAXES 12,722 3,335 9,387 2815% NET INCOME $17,565 $5,007Also nd the marginal density f U(u) We have x= vand y= v=u Thus, for the Jacobian we have 1 = 1 @u @x @u @y @v @x @v @y = =y2 1 0 =y 2 = u2=v For limits on uand vnotice that uis equal to vdivided by a number bigger than 1 Consequently, 1 James A DeVita & Associates P (703) E info@jamesadevitacom Fairfax 14 Springhill Rd Arlington 2111 Wilson Blvd Montgomery 1700 Rockville Pike James A DeVita and Associates is a full service law firm providing legal representation in Virginia, Maryland, and Washington, DC in seven areas of the law Immigration, Bankruptcy, Divorce,
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By Theorem 42, if f(x) jg(x), then deg f(x) deg g(x) To make greatest common divisors unique, we need a new condition (analogous to assuming they were positive in Z) De nition p 91 Let f(x);g(x) 2Fx, not both 0 The greatest common divisor of f(x) and g(x) is the monic polynomial of highest degree that divides them both TheThe density of Mis f(x) = F0(x) = 2(1 x) for x2(0;1) Problem 3 (p345 #9) Suppose a straight stick is broken in three at two points chosen independently at random along its length What is the chance that the three sticks so formed can be made into the sides of a triangle?Proofs 1 (10 points) Let fbe de ned on R and suppose that jf(x) f(y)j (x y)2 8x;y2R Prove that fis a constant function Solution It su ces to prove that f0(x) = 0 for all x2RLet >0 be given
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The size and appearance of these particles are highly suggestive of parapoxvirus virionsLet g ( x, y) = ( x y, x − y) Then ( U, V) = g ( X, Y) You know that f U V ( u, v) = f X Y ( h ( u, v)) J h ( u, v) where h = g − 1, J h is the jacobian matrix of h So h ( u, v) = ( u v 2, u − v 2) J h = 1 2, f X Y ( x, y) = f X ( x) f Y ( y) (because X and Y are independent) Therefore f X YG(2) = 5 ;
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M= bgof a;b such that fis constant on each interval (u j 1;u j), say f(x) = c j for xin (u j 1;u j) (a)Show that a step function fis integrable and evaluate Z b a f (b)Evaluate the integral Z 4 0 P(x)dxfor the postagestamp function Pin Exercise 1716 Solution (a)If fis a step function, then fis bounded, as the range of fis the nite set The function f X!Y is injective if it satis es the following For every x;x02X, if f(x) = f(x0), then x= x0 In words, fis injective if whenever two inputs xand x0have the same output, it must be the case that xand x0are just two names for the
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