F x x n 1 x
[PDF File]Real Analysis Math 125A, Fall 2012 Final Solutions 1. R
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1 n! xn. • (b) The expression for the Lagrange remainder is Rn(x) = 1 (n+1)! f(n+1)(ξ)xn+1 = 1 (n+1)! e˘ xn+1 for some ξ strictly between 0 and x. • (c) For n = 1, we get ex = 1+x+ 1 2 e˘x2. Since e˘ > 0, it follows that ex ≥ 1 + x, with equality if and only if x = 0. • (d) Take x = π e −1 in the inequality from (c). This gives ...
[PDF File]Chapter 1 Field Extensions - University of Washington
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Theorem 1.3 Let kbe a field and f∈k[x]. Then f has a splitting field, say K/k, and [K: k] ≤(degf)!. Proof. Induction on n= degf. If degf= 1, then f= αx+ βwith α,β∈k, so falready splits in k, so we can take K= k. Suppose that n>1. If f is already split we may take K= k, so we may assume that f has an ireducible factor, say g, of ...
[PDF File]Assignment-6 - University of California, Berkeley
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Assignment-6 (Due 07/30) 1.Let sequences f n and g n converge uniformly on some set EˆR to fand grespectively (a)Construct an example such that f ng n does not converge uniformly on E. Solution: Take f n = g n = x+ 1=nand E= R. Clearly f n;g n!xuniformly on R.Now f ng n = (x+ 1=n)2, and we claim that this does not converge uniformly to x2.To see this, we let h n(x) be the sequence of the ...
[PDF File]Chapter 5
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f(x) = 1 x. We have |fn(x)| < n for all x ∈ (0,1), so each fn is bounded on (0,1), but their pointwise limit f is not. Thus, pointwise convergence does not, in general, preserve boundedness. Example 5.3. Suppose that fn: [0,1] → R is defined by fn(x) = xn. If 0 ≤ x < 1, then xn → 0 as n → ∞, while if x = 1, then xn → 1 as n ...
[PDF File]1 Review of Probability - Columbia University
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A simple computation (utilizing X = X 1 +···+X n and independence) yields E(X) = np Var(X) = np(1−p) M(s) = (pes +1−p)n. Keeping in the spirit of (1) we denote a binomial n, p r.v. by X ∼ bin(n,p). 3. geometric distribution with success probability p: The number of independent Bernoulli p trials required until the first success yields ...
[PDF File]MATH 401 - NOTES Sequences of functions Pointwise and ...
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Applying the sandwich theorem for sequences, we obtain that lim n→∞ fn(x) = 0 for all x in R. Therefore, {fn} converges pointwise to the function f = 0 on R. Example 6. Let {fn} be the sequence of functions defined by fn(x) = cosn(x) for −π/2 ≤ x ≤ π/2. Discuss the pointwise convergence of the sequence.
[PDF File]3.2 The Growth of Functions - University of Hawaiʻi
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ICS 141: Discrete Mathematics I (Fall 2014) 3.2 pg 216 # 7 Find the least integer n such that f(x) is O(xn) for each of these functions. a) f(x) = 2x3 + x2 logx 2x 3+ x2 logx 2x + x3 for x > 0 2x3 + x2 logx 3x3 O(x3) with our witnesses C = 3 and k = 0. Therefore, n = 3. b) f(x) = 3x3 + (logx)4 3x3 + (logx)4 3x3 + x3 for x > 1 3x3 + (logx)4 4x3 O(x3) with our witnesses C = 4 and k = 1
[PDF File]MATH 461: Fourier Series and Boundary Value Problems ...
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f(x) ˘a0 + X1 n=1 h an cos nˇx L + bn sin nˇx L i; i.e., we can associate with f this Fourier series, butnot f is equal tothis Fourier series. The Fourier coefficientsof f, on the other hand,are never in doubt. They are given by a0 = 1 2L Z L L f(x)dx an = 1 L Z L L f(x)cos nˇx L dx; n = 1;2;::: bn = 1 L Z L L
[PDF File]Chapter 2. Order Statistics - 國立臺灣大學
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3 Now we prove that if U is uniformly distributed over the interval (0,1), then X = F−1 X (U) has cumulative distribution function F X(x).The proof is straightforward: P(X ≤ x) = P[F−1 X (U) ≤ x] = P[U ≤ F X(x)] = F X(x). Note that discontinuities of F become converted into flat stretches of F−1 and flat stretches of F into discontinuities of F−1.
[PDF File]Solutions to Assignment-3 - UCB Mathematics
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n(x) = 1 for all n>N and for n N, f n(x) = njxj 1. On the other hand, f n(0) = 0 for all n, and hence h(x) = (1; x6= 0 0; x= 0; and is discontinuous. 3.For each of the following, decide if the function is uniformly continuous or not. In either case, give a proof using just the de nition in terms of "and .
[PDF File]ECE 302: Lecture 5.1 Joint PDF and CDF
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c Stanley Chan 2020. All Rights Reserved. What are joint distributions? Joint distributions are high-dimensional PDF (or PMF or CDF). f X(x) |{z } one variable
[PDF File]FUNCTIONAL EQUATIONS - University of California, Irvine
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f(n+ 1) = f(n) + f(n 1): Example1.3(Radioactive decay)Let f(x) represent a measurement of the number of a specific type of radioactive nuclei in a sample of material at a given time x.
[PDF File]SOLUTIONS 1) (10 points)
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3) (9 points) Let f(x) be a function defined for x ≥ 1, such that 0 ≤ f(x) ≤ 1, for all x ≥ 1. What can be said about the series S1 = X∞ n=1 f(n) n, S2 = X∞ n=1 f(n) n2? A) S1 and S2 converge B) S1 diverges and S2 converges C) S1 converges and S2 diverges D) S1 and S2 diverge E) S2 converges, but S1 might converge or diverge.
[PDF File]Order Statistics 1 Introduction and Notation
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= n(n 1)[F(y) F(x)]n 2f(x)f(y): This hold for x < y and for x and y both in the support of the original distribution. For the sample of size 15 from the uniform distribution on (0;1), the joint pdf for the min and max
[PDF File]HW-Sol-5-V1 - Massachusetts Institute of Technology
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Defineafunctionk(x,y) h(x)/h(y) = 1, whichisboundedandnon-zero for any x ∈Xand y ∈X. Note that x and y such that n i=1 x i = n i=1 y i are equivalent because function k(x,y) satisfies the requirement of likelihood ratio partition. Therefore, T(x) n i=1 x i is a sufficient statistic. Problem 5: Let X1,X2,...,X m and Y1,Y2,...,Y n be two independent sam- ples from N(µ,σ2)andN(µ,τ2 ...
[PDF File]1 Definition and Properties of the Exp Function - UH
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• Atx = 0, f 0(x) = 0. Thus f(0) = e = 1 is the (only) local and absolute maximum. • Fromf0(x) = −xe−x 2 2, wehave f00(x) = −e− x+x2e−x = (x2 −1)e−. • At x = ±1, f00(x) = 0. Then, the graph is concave up on (−∞,−1) and (1,∞); the graph is concave down on (−1,1). • Thepoints (±1,f(±1)) = (±1,e−12) are points ...
[PDF File]Example 2. f x) = x n where n = 1 2 3 - MIT OpenCourseWare
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Δy n n ) n = ( x+Δ )n (−x = x+ n(Δ )(n−1 O(Δ 2 −x = nx n −1+O(Δx) Δx Δx Δx As it turns out, we can simplify the quotient by canceling a Δx in all of the terms in the numerator. When we divide a term that contains Δx2 by Δx, the Δx2 becomes Δx and so our O(Δx2) becomes O(Δx). When we take the limit as x approaches 0 we get:
[PDF File]1.1 CDF: Cumulative Distribution Function
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1-6 Lecture 1: CDF and EDF First, at each x, Fb n(x) is an unbiased estimator of F(x): bias Fb n(x) = E Fb n(x) F(x) = 0: Second, the variance converges to 0 when n!1. By Lemma 0.3, this implies that for a given x,
[PDF File]Lecture 15: Order Statistics - Duke University
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n!1 F(x)n = (1 if F(x) = 1 0 if F(x)
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