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If f is not oneone, then the proof ends there itself But whatever be the case of f, f o g cannot be oneone This can be proved by the following argument Let g A →B and f B →C By assumption, since g is not oneone, there exists 2 distinct elements x1 and x2 such that g(x1) = g(x2) = y where y belongs to B Let f(y) = z for some zDelivery & Pickup Options 143 reviews of F & B Atlanta "I am torn about this rating It's either three or four, so see what I say and decide for yourselves My birthday dinner with the fam this year at a newly opened spot On the negative/three side This place is like Terminus in that it's a pain to get into You either have to uturn at P'tree and P'tree Dunwoody if you are comingC 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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B F(A,B,C,D) = D (A' C') 6 a Since the universal gates {AND, OR, NOT can be constructed from the NAND gate, it is universalUsing 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(−3x3 Where Lα = lim x→0 αx −1 x And, we now have two different ways of calculating this limit Method1 Using the limit definition of the derivative we have f '(a) = lim h→0 f (x) −f (a) x −a We note that Lα = lim x→0 αx −1 x = lim x→0 αx − α0 x − 0

Run with cmptrap(1,2,4,'f') where 'f' is the name of the function definition file function y = f(t) y = t*log(t);P yare quadratic surds and if a p x= p y,thena= 0 and x= y 22 If p x;L 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 semi

2 F (b) if B 2 F then its complement is also in F (c) if B1;B2;You have only shown the former inclusion Moreover, you have not used the hypothesis that f is injective anywhere;B− a n (f(xn)− f(x0)) = b− a n (f(b)− f(a)) Nowletε > 0bearbitrary Choosen sufficiently largethat (b−a)(f(b)−f(a)) n < ε Then U(f;Pn)− L(f;Pn) < ε By the Cauchy criterion for integrability, f is integrable 6 Example 42 We will evaluate the integral R1 0 x Note first that, since the

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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 equationTPain's "FBGM" feat Young MA available nowApple Music http//smarturlit/iFBGMSpotify http//smarturlit/sFBGM Amazon http//smarturlit/aFBGThe coefficients an and bn are called the Fourier coefficients of f and the series a0 2 X∞ n=1 an cosnxbn sinnx the Fourier series of f We denote this fact by f(x) ∼ a0 2 X∞ n=1 an cosnxbn sinnx The symbol ∼ should be read as f "has Fourier serier"

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That is, you must show that f(A∩B)⊆f(A)∩f(B) and f(A)∩f(B)⊆f(A∩B); $\begingroup$ @mlainz You might find it amusing to learn that the graph of any such function that is not continuous is dense in ${\mathbb R}^m \times {\mathbb R}^{n}$ In particular, the graph of any such function from the reals to the reals that is not continuous has a graph that is dense in the plane This means that the graph comes arbitrarily close to every point Stack 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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 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﷮𝑏# * ' * # ) $ # * /' # * 0 1 !If an= bnwhere n6=0,then a=b 21 If p x;

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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 iMATH 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(x 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 satisfied

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0 ' ( 2 3 4 5 4 6 7 8 9 & ' !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 calculusR R E v j O E A h E R E F u E z y W E O t B b N E f U C E B S A j X ^ A ` V A h A Ŕ A o i 쁗 O A ^ { @Ko'Ko' Planning & Co Makes a plan, design,

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Restriction of a convex function to a line f Rn → R is convex if and only if the function g R → R, g(t) = f(xtv), domg = {t xtv ∈ domf} is convex (in t) for any x ∈ domf, v ∈ Rn can check convexity of f by checking convexity of functions of one variableBank of America Private Bank is a division of Bank of America, NA, Member FDIC and a wholly owned subsidiary of Bank of America Corporation Trust and fiduciary services are provided by Bank of America, NA and US Trust Company of Delaware Both are indirect subsidiaries of Bank of America Corporation0 if f(x) > n It is easily seen that F = F n G n, where F n(x) = Z x a f n(t)dt and G n(x) = Z x a f(t) − f n(t)dt, a ≤ x ≤ b Since f(t)−f n(t) ≥ 0 for all t ∈ a,b, G n is an increasing function on a,b Moreover, by what has been proved, F0 n (x) = f n(x) for almost every x ∈ a,b Thus, we have F0(x) = F0 n (x) G0 n

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The 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 expansion⇐⇒ f(x) ∈ B1, but f(x) ∈/ B2 ⇐⇒ x ∈ f−1(B 1), but x /∈ f−1(B 2) ⇐⇒ x ∈ f−1(B 1)−f−1(B 2) 2 Let f A → B be a function and let A1 ⊆ A be arbitrary Show that f−1(f(A 1)) ⊇ A1 and that equality holds whenever the function f is injective To prove the inclusion for any function f,Answer (1 of 5) Basically,the set A is the collection of numbers,which are divisible by 3Similarly,the set B is the collection of numbers,which are divisible by 4 So,A∩B is the collection of those positive numbers,which are divisible by both 3 and 4,ie which are divisible by LCM of 3 and 4 =1

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Is 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 1x2Graph 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 discontinuity4 = @ 3 7 5 > ?< 9 @ 3 4

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N;a;b = sup Xk i=1 f n(x i) f n(x i 1) M for all n Furthermore, because f k!fpointwise, Vf;a;b = sup Xk i=1 f(x i) f(x i 1) = sup lim n!1 Xk i=1 f n(x i) f n(x i 1) M As for the example of a convergent series of functions of bounded variation whose limit is not of bounded variation, taking a hint from problem 1, consider a function f n(x(10) Z x a2 x2 dx= 1 2 lnja2 x2j (11) Z x2 a 2 x dx= x atan 1 x a (12) Z x3 a 2 x dx= 1 2 x2 1 2 a2 lnja2 x2j (13) Z 1 ax2 bx c dx= 2 p 4ac b2 tan 1 2ax b p 4ac b2 (14) Z 1 (x a)(x b) dx= 1 b a ln a x b x;X= b Clearly Fis an extension of f Claim Fis continuous on a;b Proof Clearly F is continuous on (a;b)To prove continuity at a, let fx ngbe a sequence in (a;b) converging to a We need to show that F(x n) = f(x n) !F(a) = A Let ">0 There exists N 1 such that for all n

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F (x)= X1 n=0 dn (x¡b) n if a 6= b;Presumably that must be used at some point, or else you could prove the desired equality for all functions f% pay attention to the dot The result is 2a Matlab code for the Composite Simpson's rule function integral = cmpsimp(a,b,n,f) h

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 We know that f'(x) = lim﷮h→0﷯ f x h﷯ − f(x)﷮h﷯ Here, f(x) = (ax b)n So, f(x h) = (a(x h) b)n Putting values f'(x) = lim﷮h→0﷯ 𝑎 𝑥ℎ﷯𝑏﷯𝑛 − 𝑎𝑥 𝑏﷯𝑛﷮h﷯ f'(x) = lim﷮h→0﷯ 𝑎 𝑥ℎ﷯𝑏﷯𝑛 − 𝑎𝑥 𝑏﷯𝑛﷮h﷯ = lim﷮h→0﷯ 𝑎𝑥 𝑎ℎB= lim n!1 f(b n); 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)(x

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Yingwei Wang Real Analysis Then f ∈ l1(N) and (l1(N),k ·k1) is a Banach space 3 Application of Baire Theorem Question Let f R → R be a smooth function (ie C∞)Suppose that for each t ∈ R there isChapter 8 Integrable Functions 81 Definition of the Integral If f is a monotonic function from an interval a,b to R≥0, then we have shown that for every sequence {Pn} of partitions on a,b such that {µ(Pn)} → 0, and every sequence {Sn} such that for all n ∈ Z Sn is a sample for Pn, we have {X (f,Pn,Sn)} → Abaf 81 Definition (Integral) Let f be a bounded function from an interval9 < 7 8 = > 3 ?;

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 So, from the basic properties of limits we then have, lim n→∞ n ∑ i=1f (x∗ i)Δx ≥ lim n→∞ 0 = 0 lim n → ∞ ⁡ ∑ i = 1 n f ( x i ∗) Δ x ≥ lim n → ∞ ⁡ 0 = 0 But the left side is exactly the definition of the integral and so we have, ∫ b a f (x) dx = lim n→∞ n ∑ i=1f (x∗ i)Δx ≥ 0 ∫ a b f ( x) d xA6=b (15) Z x (x a)2 dx= a a xConsider a set X A ˙{algebra F of subsets of X is a collection F of subsets of X satisfying the following conditions (a) ;

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Math 432 Real Analysis II Solutions to Homework due March 11 Question 1 Let f(x) = k be a constant function for k 2R 1 Show that f is integrable over any a;b by using Cauchy's " P condition for integrabilityI) 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 jfSection 61 Sets and Set Operations Sets A set is a welldefined collection of objects The objects in this collection are called elements of

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Definition 13 A bounded function f a,b → Ris Riemann integrable on a,b if its upper integral U(f) and lower integral L(f) are equal In that case, the Riemann integral of f on a,b, denoted by Zb a f(x)dx, Zb a f, Z a,b f or similar notations, is the common value of U(f) and L(f) An unbounded function is not Riemann integrableSets 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 funcAnd de ne F(x) = 8 >< > A;

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