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‰Âˆ¤‚¢ ƒŠƒ…ƒbƒN ƒŒƒfƒB[ƒX-(a) The function f = f(x;3x) x 2Zgis injective but not surjective Proof Let a;b 2Z Then if f(a) = f(b), we have 3a = 3b Dividing both sides by 3 gives a = b Therefore, f is injective Consider 2 2Z Then 3x = 2 only if x = 2 3 2= Z Therefore, there is a point y 2Z for which there is no x such that f(x) = y and so f is not surjective 1Fn(x)dx =0 Theorem 93G If fn ∈Ra,b and if fn → f uniformly on a,b, then Zb a f → Zb a f Remark Let fn(x)=x n n on 0,1 and let f(x) = 0 Then fn → f uniformly but f0 (1) = 1 while f0(1) = 0 Thus lim n→∞ f0 n(x)=f 0(x) does not hold at x =1 Theorem 93I If f0 n(x) exists for each n and each x ∈ a,b, if f0 n is

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4 = @ 3 7 5 > ?< 9 @ 3 4Quadrature for these points Then for any interval a;b, Z b a P(x)dx = b a 2 i=0 a i f b a 2 x i a b 2 Proof We will give the proof just for the interval 1;1 Adapting the proof to a general interval just requires a careful change of variables Let P n1(x) be the Legendre polynomial with its roots fx 0;x 1;;x ng Assume that PExplain your answer e) Let (,b = 25,0) To which zero of does the bisectžon method converge to?

0 ' ( 2 3 4 5 4 6 7 8 9 & ' !F (x)= X1 n=0 dn (x¡b) n if a 6= b;Theorem 24 Let {f n} be a sequence of measurable functions defined on EFor x ∈ E, set g(x) = sup n f n(x) and h(x) = limsup n f n(x) Then g and h are measurable Proof We need only to show that g is measurable since the measurability of h comes

(a) The set of all diagonal n x n matrices (b) The set of all n x n matrices A such that det(A) = 0 (c) The set of all n x n matrices A such that tr(A) = 0 (d) The set of all symmetric n x n matrices (e) The set of all n x n matrices A such that AT = A (f) The set of all n x n matrices A for which Ax = 0 has only the trivial solutionThe corrected trapezoidal rule is illustrated in the following table n I T n Ratio I CT n Ratio 2 5319 3552E 1 4 1266 4 2474E 2 144 8 3118E 1 406 15E 3 156F(x) = ax b, g(x) = cx d

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 Given f(x) = (x – a) m (x – b) n As we know that every polynomial function is continuous and differentiable everywhere So, f(x) is continuous and differentiable on the given indicated interval Also, f(a) = 0 = f(b) Thus, all the conditions of Rolle's theorem are satisfiedTranscribed image text Let f (x) = (x 2) (x 2)2 1) cos?n a) Let a, b = 0,1,5) Use the Bisection method to find p2 b) Let a, b = 15,3) To which zero of does the Bisection method converge to?Sets A and B are called equinumerous, written A ∼ B, if there is a bijection f X → Y A set A is called countably infinite if A ∼ N We say that A is countable if A ∼ N or A is finite Example 31 The sets (0,∞) and R are equinumerous Indeed, the func

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Using the Properties of the Definite Integral Use the properties of the definite integral to express the definite integral of f(x) = −3x3 2x 2 over the interval −2, 1 as the sum of three definite integrals Solution Using integral notation, we have ∫1 −2(−3x3The concept of (AB)^n and (AB)^n formula expander is used to describe the expression for the given nth value of formula The binomial theorem is applied here to expand the formula Any algebraic expression consisting of only two terms is known as a binomial expression It's expansion in powers of n is known as the binomial expansionDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US

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Departure of evil men by their return Such recurrences should not constitute occasions for sadness but realities for awareness, so Misc 43Choose the correct answerIf 𝑓﷐𝑎𝑏−𝑥﷯=𝑓﷐𝑥﷯, then ﷐𝑎﷮𝑏﷮𝑥 𝑓﷐𝑥﷯﷯𝑑𝑥 is equal to (A) ﷐𝑎𝑏﷮2﷯﷐𝑎﷮𝑏﷮ 𝑓﷐𝑏−𝑥﷯﷯𝑑𝑥 (B) ﷐𝑎𝑏﷮2﷯﷐𝑎﷮𝑏﷮ 𝑓﷐𝑏𝑥﷯﷯𝑑𝑥 ﷐𝑏 −𝑎﷮2﷯﷐𝑎﷮𝑏﷮ 𝑓﷐𝑥﷯﷯𝑑𝑥 (D) ﷐𝑎𝑏﷮2﷯﷐𝑎﷮𝑏Edelt (31) d(x,y) < δ =⇒ d0(f(x),f(y)) < ε Let us use the notation B(x,δ) = {y d(x,y) < δ} For a subset A ⊂ X, we also use the notation f(A) = {f(x) x ∈ A} Similarly, for B ⊂ Y f−1(B) = {x ∈ X f(x) ∈ B} Then (edelt 31) means f(B(x,δ)) ⊂ B(f(x),ε) Or in a very nonformal way f

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% pay attention to the dot The result is 2a Matlab code for the Composite Simpson's rule function integral = cmpsimp(a,b,n,f) hL f(x), which proves that fis lower semicontinuous Conversely, suppose that f is lower semicontinuous Let (z n) be a sequence in epif that converges to z2X R Writing z n = (x n;t n) and z = (x;t), we have x n!x, t n!t, and t n f(x n) Taking the limit of this inequality at n!1, we get that t liminf f(x n), and then the lower semiIf an= bnwhere n6=0,then a=b 21 If p x;

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Consider a set X A ˙{algebra F of subsets of X is a collection F of subsets of X satisfying the following conditions (a) ;Solution 1) Let F n= B n 1 j=n1 j B n 2A 1 A n, n2N Using that n j=1 A j is a ˙algebra, it can be easily veri ed that F n is a ˙algebra as well Moreover, since B n n1 2 n1 j=1 A j for any set B n 2 j=1 A j we trivially obtain F n F n1 2) Now de ne F= S n2N F n (which is not a ˙algebra in general) Then Fis an algebra iNumerical 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 spaced

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% " % ( % ) ) / & , 0 % 0 % 1 2 3 4 5 4 6 7 8 9 7 ;Algorithms AppendixIISolvingRecurrencesFa'13 Change is certain Peace is followed by disturbances;Exercise 08 Chapter 2, # 10 If a 1 < 2 < < n bis a nite sequence and jsj

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9 < 7 8 = > 3 ?; Question Show that the derivative of f(x) = (xa)m (xb)n vanishes at some point between a and b if m and n are positive integers My attempt f(x) = (xa)m (xb)n f '(x) = m(xa)m1 (xb)n n(xa)m (xb)n1 f '(x) = (xa)n1 (xb)n1 (m)(xb) (n)(xClearly, f(n) and S will di er on whether they contain n, thus, 8n 2 N f(n) 6= S, and so fis not a surjection In general, unattributed Theorems are due to Cantor Theorem 112 If Ais any set, then A6˘P(A) Proof Suppose that f A!P(A) is any function Consider the set S= fa2

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Since f(x) is continuous, by the Intermediate Value Theorem it takes every possible value between m and M In particular, there is atleast one place c at which the function f(x) hasa value equal to f(c) = ∫b a f(x)dx b a Multiplying bothsides by b a proves the result 4The first fundamental theorem of integral calculusWhere G0 is an operator which does not depend on x Checking the special case x= 0, we find G= 1 and F=0 Therefore G0 =1 Eliminating Fand Gin favor of nd B, we find exA exB = exAxB 1 2 x2 A,B Setting x=1, this becomesConsider f(x)=x^2*exp(x), k(x)=x^2, g(x)=exp(x) I(f)=Int_{0}^{infinity}(f(x))=Gamma(3)=2 k < function(x){x^2} #Function to be integrated wrt g(x)=exp(x)

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P yare quadratic surds and if a p x= b p ythen a= band x= y 23 If a;m;nare positivereal numbersanda6=1,thenlog a mn=log a mlog a n 24 If a;m;nare positive real numbers, a6=1,thenlog a m n =log a m−log a n 25 If aand mare22 FUNCTIONS 30 bijection, if it is onetoone and onto (fig 213)For instance, f(x) = x3 from Z to Z is a bijection A B Figure 213 Bijection 224 Identity Function Given a set A, the function 1A A → A defined by 1A(x) = x for every x in A is called the identity function for A 225 Function Composition Given two functions f A → B and g B → C, the composite functionI) f(x i 1)j either does not change, or increases This is useful when we are trying to prove things about the supremum of such sums, as we need to when looking at functions of bounded variation Now we prove parts of Theorem 24 Proof To prove (2) we begin by assuming that fis of bounded variation on a;b Thus V(f;a;b) = supf P n i=1 jf

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2 F (b) if B 2 F then its complement is also in F (c) if B1;B2;MATH 140B HW 5 SOLUTIONS Problem1(WR Ch 7 #8) If I(x) ˘ 0 (x •0),1 (x ¨0),if {xn} is a sequence of distinct points of (a,b), and ifP jcnj converges, prove that the series f (x) ˘ X1 n˘1 cnI(x¡xn) (a •x •b) converges uniformly, and that f is continuous for every x 6˘xn Solution Let fk(x) ˘ Xk n˘1 cnI(x¡xn) By the Weierstrass Mtest (Theorem 710) with Mn ˘jcnj, {fk(x4 6 Sequences and Series of Functions 62 Uniform Convergence and Di erentiation Theorem 63 Let f n(x) !f(x) pointwise on a;b and assume each f n is di erentiable on an open interval containing a;bIf f0 n converges uniformly on a;b to a function g, then fis di erentiable and f0= gon a;b

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 where l and k are nonnegative integers, a and b are nonnegative real numbers, the initial values \(x_{s}, x_{s1},\ldots, x_{0}\) are positive real numbers, \(s=\max\{l,k\}\), and \(f (u,v ) ( 0,\infty ) ^{2}\rightarrow ( 0,\infty ) \) is a continuous and homogeneous real function of degree zero We consider the stability, boundedness, and periodicity of the solutions of this equationGraph f(x)=b^x Find where the expression is undefined The domain of the expression is all real numbers except where the expression is undefined In this case, there is no real number that makes the expression undefined The vertical asymptotes occur at areas of infinite discontinuityStack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange

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F n(x)dx Z b a f(x)dxj kf n fk 1(b a) That's the end of the proof Corollary 15 Let P= ff2Ba;b 9fp ngsuch that kp n fk 1!0gThen P is contained in Ca;b A remarkable result of Weierstrass asserts indeed that P= Ca;bIn particular, any continuous function on a;b can be approximated uniformly by a sequence ofX f nd R X fd for all n, we get that R X fd Next we prove the reverse inequality For this, let 0 s f be a simple measurable function It su ces to prove that R X sd Let 0Review Intro to Power Series A power series is a series of the form X1 n=0 a n(x x 0)n= a 0 a 1(x x 0) a 2(x x 0)2 It can be thought of as an \in nite polynomial" The number x 0 is called the center A power series may converge for some values of x, but diverge for other

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# * ' * # ) $ # * /' # * 0 1 !The trapezoid rule for N subintervals of a, b of equal length is T N ( f) = Δ x 2 ∑ i = 1 N ( f ( x i) f ( x i − 1)) where Δ x = ( b − a) / N is the length of the subintervals and x i = a i Δ x Notice that the trapezoid is the average of the left and right Riemann sums7 Question Details SCalc7 ­ (a) Let An be the area of a polygon with n equal sides inscribed in a circle with radius r By dividing the polygon into n congruent triangles with central angle show that the following is true

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Each element y2Bwith the unique element a2Asuch that f(x) = yThe subset gis a map because fis bijective, and by its very de nition we see that g f= id nd f g= id B (() Suppose now that the map gexists To see fis injective, assume that f(x) = f(z) and apply the map gto obtain x= g(f(x)) = g(fN b a f(x) n (x)dx b a 2 n (x)dx, n 0, 1, 2, b a f(x) n(x) dx c n b a 2(x) dx where f n(x) 2 (11) The series (7) with coefficients given by either (8) or (10) is said to be an orthogonal series expansion of f or a generalized Fourier series COMPLETE SETS TheIs completely di¤erent fromthe Taylorseries expansionabout x =a Generally speaking, the interval of convergence for the representing Taylor series may be di¤erent from the domain of the function Example 51 Find Taylor series expansion at given point x =a (a) f (x)= 1 1x2

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3For any two numbers aand bwith a Section 75 Proof of Various Integral Properties In this section we've got the proof of several of the properties we saw in the Integrals Chapter as well as a couple from the Applications of Integrals Chapter Proof of ∫ kf(x)dx = k∫ f(x)dx ∫ k f ( x) d x = k ∫ f ( x) d x

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