(c) 1 3 n un!Z X1 n=1 1 3 n unz n = X1 n=0 1 3 n z n = X1 n=0 1 3 z 1 n = 1 1 1 3z 1 ROC j1 3z 1j1 31 05 0 05 1105 0 05 1 Real part Imaginary part Figure 3 zplane for (c) xn = 1 3 n un= 1 2ˇ Rˇ ˇ X(!)e!n d!Actually this question I am getting since long back With a little bit zooming we see prior to (xz) there is (xy ) and just before (xy) there is (x x) which is equals to 0 and hence product of entire sequence will be zero (0) So the value of
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So xis in none of the A n for n N But then this means that xcan be in A n only for nX n c n jni c n = hnj i (416) By inserting a continous CONS of position eigenstates (Eq (332)) into the transition amplidute the expansion coe cients c n can be rewritten as c n = hnj i= Z dxhnjxihxj i= Z dx n (x) (x) (417) We can now extend the expansion from the time independent case to the time dependent oneWhere c 1(λ) and c 2(λ) are arbitrary functions of λ Q Show that (9) is a solution of the equation (1) for any c 1(λ) and c 2(λ) If we let λ = ω2 then (9) becomes u(x,t) = Z ∞ 0 A(ω)cosωxe−kω2t B(ω)sinωxe−kω2tdω (10) where A(ω) = 2ωc 1(ω2),B(ω) = 2ωc 2(ω2) are arbitrary functions To satisfy the initial condition (2) we must have
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(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) ThenUniform TimeDomain Sampling xn = xa(nTs) X(!) = 1 Ts P1 k=1 Xa !=(2ˇ) k Ts (sum of shifted scaled replicates of Xa()) Recovering xa(t) from xn for bandlimited xa(t), where Xa(F) = 0 for jFj Fs=2 Xa(F) = Ts rect F Fs X(2ˇFTs) (rectangular window to pick out centerN)= Lcn 2π = 1 2π Z −L 2 L 2 dxf(x)e−ikn x (7) The factor of 2π in this equation is just a convention Now we can take L → ∞ so that kn can get arbitrarily close to zero This gives f˜(k)= 1 2π Z −∞ ∞ dxf(x)e−ikx (8) where now k can be any real number This is the Fourier transform It is a continuum generalization of the
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And more generally 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 mgfThe partition theorem says that if Bn is a partition of the sample space then EX = X n EXjBnP(Bn) Now suppose that X and Y are discrete RV's If y is in the range of Y then Y = y is a event with nonzero probability, so we can use it as the B in the aboveTitle Microsoft Word Comment 23 Author samah Created Date 1230 PM
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> >~ ~ >w!x>y z > bwR >wRwR $}L fzy{ws~{ >wRzy{wR~ b 6 X S£lzy{®·6»E¸CÉThe mean or expected value of g(X) is E(g(X)) = Z g(x)dF(x) = Z g(x)dP(x) = (R 1 1 g(x)p(x)dx if Xis continuous P j g(x j)p(x j) if Xis discrete Recall that 1 Linearity of Expectations E P k j=1 c jg j(X) = P k j=1 c jE(g j(X)) 2 If X 1;;X n are independent then E Yn i=1 X i!4 Convolution Solutions to Recommended Problems S41 The given input in Figure S411 can be expressed as linear combinations of xin, x 2n, X3n x, n
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Numerical Analysis Trapezoidal and Simpson's Rule Natasha S Sharma, PhD General Trapezoidal Rule T n(f) 1 We saw the trapezoidal rule T 1(f) for 2 points a and b 2 The rule T 2(f) for 3 points involves three equidistant points a, ab 2 and b 3 We observed the improvement in the accuracy of T 2(f) over T 1(f) so inspired by this, we would like to apply this rule to n 1 equally spaced24 c JFessler,May27,04,1310(studentversion) 212 Classication of discretetime signals The energy of a discretetime signal is dened as Ex 4= X1 n=1 jxnj2 The average power of a signal is dened as Px 4= lim N!1 1 2N 1 XN n= N jxnj2 If E is nite (E < 1) then xn is called an energy signal and P = 0 If E is innite, then P can be either nite or inniteIf you like this Site about Solving Math Problems, please let Google know by clicking the 1 button If you like this Page, please click that 1 button, too Note If a 1 button is dark blue, you have already 1'd it Thank you for your support!
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Z x2ex dx= x2 2x 2 ex (63) Z x2eax dx= x2 a ax 2x a2 2 a3 e (64) Z x3ex dx= x3 3x2 6x 6 ex (65) Z xneax dx= xneax a n a Z xn 1eaxdx (66) Z xneax dx= ( n1) an1 1 n;–1 C i≡ 37 x 1010 decays/s • The SI unit of activity is the becquerel (Bq) –1 Bq ≡ 1 decay/s • Therefore, 1 Ci = 37 x 1010 Bq • The most commonly used units of activity are the millicurie and the microcurie • There have been around 2,000 nuclear test explosions • Atomic Tests released approximately 9 MCi of Sr90Example 5 X and Y are jointly continuous with joint pdf f(x,y) = (e−(xy) if 0 ≤ x, 0 ≤ y 0, otherwise Let Z = X/Y Find the pdf of Z The first thing we do is draw a picture of the support set (which in this case is the first
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Z C Fdr = Z C (M;N) (dx;dy) = Z C Mdx Ndy Comment The notation Fdr is common in physics and MdxNdyin thermodynamics (Though everyone uses both notations) We'll see what these notations mean in practice with some examples Example GT3 Let F(x;y) = x2y;x 2y and let Cbe the curve r(t) = t;t2, with t running from 0 to 1 Compute the lineI = P n i=1 ˆ~ i!~ i 2X , then the components of ˆ transform under a change of basis according toCutaneous lesions on hands of casepatient 3 (A, B) and casepatient 5 are shown Negative staining electron microscopy of samples from casepatient 3 (D) and casepatient 5 (E, F) show ovoid particles ( ≈250 nm long, 150 nm
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Then f(n)(0) = 0 for all n∈N, meaning that the Taylor series centred at x = 0 is identically zero, but f is not identically zero near 0, so f ( x ) has no power series there (or, another way ofJ ALKNM>HM>OPM>@Ho & TGd M>HO_ACDLI5AC LO_M>@Ho C@H A C∩ ^@\>aG^ cC^@d&CW C1 XCd@e C2 V>WYe@L_V>d&CehXLW C∩={x x∈C1 ∧x∈C2} Exercise `_^@ &zChapter 4 Vector Norms and Matrix Norms 41 Normed Vector Spaces In order to define how close two vectors or two matrices are, and in order to define the convergence of sequences
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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 ∈ NDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US~ zxr\~ u«rPn@¡q~ n q Í Z X Z X (b) Y (a) Y W W Êwyk¦\mx~ ç 43 w¢w²¡~Bmxw ~ z´n ¨ £
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Sinhz = 1 2 (ez −e−z) Remark In all of what follows we will make use of the fundamental identity (∗) ezw = ezew, z,w ∈ C Here is a proof of (*) in which we leave out the nontrivial justification of some of the steps We have ezw = X∞ n=0 (z w)n n!=}}@ =} =} ~ A ?X(!) = P1 n=1 xne !n = X(z)j z=ej!
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