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F.y.b.a. syllabus pune university 2020-Function f It is called the inverse of fand denoted f 1 B!A De nition 3 Let f A!e a map If U ˆA, the image of U by f is the set f(U) fy 2B y = f(x) for some x2Ug If V ˆB, the preimage of V by fis the set f 1(V) = fx2A f(x) 2Vg Warning We use the same notation for the preimage of a function f(e) If U R is open, show that a function f U!R is continuous (meaning the preimage of an open set is open) if and only if for every x2Uand for every >0 there exists a >0 such that jf(x) f(y)j<if jx yj<
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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 ExchangeF( x (1 )y) f( x) f((1 )y) = f(x) (1 )f(y) where the inequality follows from triangle inequality and the equality follows from the homogeneity property (We did not even use the positivity property) (a) An a ne function (b) A quadratic function (c) The 1normApply the Riemann Criterion 2 (a) Let x;y2c;d Then jf(x)jj f(y)j jf(x) f(y)j M mFix yand take supremum for x, we get M0j f(y)j M m Take in mum for y (b) To show that jfjis integrable, use the Riemann Criterion and (a) For showing f2 is integrable, use the inequality (f(x))2 (f(y))2 2Kjf(x) f(y)j
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Now suppose that w = f(x;y) and x = x(u;v) and y = y(u;v) Then dw= f xdx f ydy = f x(x udu x vdv) f y(y udu y vdv) = (f xx u f yy u)du (f xx v f yy v)dv = f udu f vdv If we write this out in long form, we have @f @u = @f @x @x @u @f @y @y @u and @f @v = @f @x @x @v @f @y @y @v Example 112 Suppose that w= f(x;y) and we changeDefinedasf (x ,y )˘ 3xyCheckthatthis isbijective;finditsinverse Toseethatthisisinjective,suppose f (a ,b )˘c dStep 4 Set Substitution set (SUBST) to NIL Step 5 For i=1 to the number of elements in Ψ 1 a) Call Unify function with the ith element of Ψ 1 and ith element of Ψ 2, and put the result into S b) If S = failure then returns Failure c) If S ≠ NIL then do, a Apply S to the remainder of both L1 and L2 b
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(a) Since Re(f) = constant, ∂u ∂x = 0, ∂u ∂y = 0 (16) By the CauchyRiemann equations, ∂v ∂x = − ∂u ∂y = 0 (17) Thus, in Ω, f′(z) = ∂f ∂x = ∂u ∂x i ∂v ∂x = 0 0 = 0 (18) Thus f(z) is constant (b) Since Im(f) = constant, ∂v ∂x = 0, ∂v ∂y = 0 (19) By the CauchyRiemann equations, ∂u ∂x = ∂v1;f) U(P;f) L(P;f) <2 CHAPTER 3 THE SPACE OF CONTINUOUS FUNCTIONS to restrict to bounded functions For this purpose let C b(X) = ff f2C(X);
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A'B'C X=A' y=B'C = LHS Hence Proved Question 2 Write the Boolean Expression for the result of the Logic Circuit as shown below Аnswer F = (uv')(uw)(vw') Question 3 Derive a Canonical POS expression for a Boolen function F, represented by the following truth table Аnswer F = ∑(0, 3,4,5)Brandon Behring Functional Analysis HW 2 Exercise 26 The space Ca;b equipped with the L1 norm jjjj 1 de ned by jjfjj 1 = Z b a jf(x)jdx is incomplete If f n!fwith respect to the supnorm jjjj 1then f n!f with respect to jjjj@ a 0 1 2 3 4 5 6 7 8 9;
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Finally,wecomputetheinverseWritef (x ,y )˘ u v Interchangevariablesto get (x, y) ˘f uv((2 ¯1)3 Thus andy˘u3 Hence 1/3 and v˘ x y 2/3¯1 Therefore f¡1(x,y)˘(u,v)˘ ‡ y1/3, x y ¯1 62 = H >8x2Xfor some Mg It is readily checked that C b(X) is a normed space under the supnormFrom now on,
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