1804 Practice problems exam 2, Spring 18 Solutions Problem 1 Harmonic functions (a) Show u(x;y) = x3 3xy2 3x2 3y2 is harmonic and nd a harmonic conjugate It's easy to compute u x= 3x2 3y2 6x;Using the above identity taking a = x − y, b = y − z and c = z − x, we have a b c = x − y y − z z − x = 0 then the equation (x − y) 3 (y − z) 3 (z − x) 3 can be factorised as followsFind the Maclaurin series expansion for f = sin(x)/x The default truncation order is 6 T = x^5/1 x^3/6 x y^4/24 y^2/2 z^5/1 z^4/24 z^3/6 z^2/2 z 2 You can use the sympref function to modify the output order of a symbolic polynomial Redisplay the polynomial in ascending order
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What is the formula of (x+y+z)^3- (xy)^3 (yz)3 (zx)^3 = 3(xy)(yz)(zx) That is it no constraints etc It mentions "This can be done by expanding out the brackets, but there is a more elegant solution" Homework Equations The Attempt at a Solution First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agree⋅(x)3−k ⋅(y)k ∑ k = 0 3
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Using Properties of Determinants, Prove that (2y,Yzx,2y),(2z,2z,Zxy),(Xyz,2x,2x)=(XYZ)^3 CBSE CBSE (Arts) Class 12 Question Papers 17 Textbook Solutions Important Solutions 24 Question Bank Solutions Concept Notes & Videos 532 Time Tables 1811 Factoring x 3 y 3 z 3 Theory A sum of two perfect cubes, a 3 b 3 can be factored into (ab) • (a 2abb 2) Proof (ab) • (a 2abb 2) = a 3a 2 b ab 2 ba 2b 2 a b 3 = a 3 (a 2 bba 2)(ab 2b 2 a) b 3 = a 3 0 0 b 3 = a 3 b 3 Check x 3 is the cube of x 13 Answers3 For the nontrivial interpretation, you're looking for nonnegative solutions of a b c = n (each of these corresponds to a term x a y b z c ) Code each of these solutions as 1 a 0 1 b 0 1 c, for example ( 2, 3, 5) would be coded as Now it should be easy to see why the answer is ( n 2 n)
We obtain f (a)=c0 Next, we takeWe will see that for the expansion of a trinomial $(x y z)^n$, an analogous theorem holds For example, suppose that we want to expand the trinomial $(x y z)^3$ We will have there be $\binom{3 3 1}{3} = \binom{5}{3} = 10$ nonnegative integer solutions to this equationFind the sumofproducts expansion of the Boolean function F(w;x;y;z) that has the value 1 if and only if an odd number of w;x;y, and z have the value 1 Need to produce all the minterms that have an odd number of 1s The DNF is simply, wxyz wxyz wxyz wxyz wx yz wxy z wxy z wx y z
U yy= 6x 6 It's clear that r2u= u xx u yy= 0, so uis harmonic If vis a conjugate harmonic function to u, then uivis analytic and the CauchyRiemann Ex 42, 9 By using properties of determinants, show that 8 (x&x2&yz@y&y2&zx@z&z2&xy) = (x – y) (y – z) (z – x) (xy yz zx) Solving LHS 8 (𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Applying R1→ R1 – R2 = 8 (𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) ExTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Verify that `x^3y^3z^33x y z=1/2(xyz)(xy)^2(yz)^2(zx)^2`
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I is x i or x i 3 A sumofproducts expansion or disjunctive normal form of a Boolean function is the function written as a sum of minterms Discussion Consider a particular element, say (0,0,1), in the Cartesian product There is a unique Boolean product that uses each of the variables x, y, z or its complement (but not both) and hasCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyEvaluate the following java expression,if x=3,y=5 and z=10 zyyzx Posted on by Views Score Share EngineeringCS EngineeringIS mca JIT Davangere SEMVI Java
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X 0 0 = 3 {substituted 0 for y and z} x = 3 {combined like terms} coordinates are (3, 0, 0) Algebra 3 Section 35 Systems with Three Variables The graph of an equation in three variables, such as, Ax By Cz = D where A,B, and C are not all zero, is a plane yintercept x y z = 3 {the equation} 0 y 0 = 3 {substituted 0 for x and zExpand (xy)^3 (x y)3 ( x y) 3 Use the Binomial Theorem x3 3x2y3xy2 y3 x 3 3 x 2 y 3 x y 2 y 3Start your free trial In partnership with You are being redirected to Course Hero I want to submit the same problem to Course Hero Cancel
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Explanation (x −y)3 = (x − y)(x −y)(x −y) Expand the first two brackets (x −y)(x − y) = x2 −xy −xy y2 ⇒ x2 y2 − 2xy Multiply the result by the last two brackets (x2 y2 −2xy)(x − y) = x3 − x2y xy2 − y3 −2x2y 2xy2 ⇒ x3 −y3 − 3x2y 3xy2 Always expand each term in the bracket by all the otherLaurent Expansion Let f (z) = z(z − 1) −1 If we choose to make the Laurent expansion about z 0 = 0, then r > 0 and R < 1 These limitations arise because f (z) diverges both at z = 0 and z = 1 A partial fraction expansion, followed by the binomial expansion of (1 − zThe following are algebraix expansion formulae of selected polynomials Square of summation (x y) 2 = x 2 2xy y 2 Square of difference (x y) 2 = x 2 2xy y 2 Difference of squares x 2 y 2 = (x y) (x y) Cube of summation (x y) 3 = x 3 3x 2 y 3xy 2 y 3 Summation of two cubes x 3 y 3 = (x y) (x 2 xy y 2) Cube
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(x y) 3 = x 3 3x 2 y 3xy 2 y 3 Example (1 a 2 ) 3 = 1 3 31 2 a 2 31(a 2 ) 2 (a 2 ) 3 = 1 3a 2 3a 4 a 6 (x y z) 2 = x 2 y 2 z 2 2xy 2xz 2yzFree expand & simplify calculator Expand and simplify equations stepbystepThe binomial theorem (or binomial expansion) is a result of expanding the powers of binomials or sums of two terms The coefficients of the terms in the expansion are the binomial coefficients ( n k) \binom {n} {k} (kn ) The theorem and its generalizations can be used to prove results and solve problems in combinatorics, algebra, calculus
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It suffices to show that for any (x, y) ∈ p X − 1 V there exists an open ball A = B d X × Y ((x, y), r) centered at (x, y), with r > 0, such that A ⊆ p X − 1 V For any x ′ ∈ X Abelian subvarieties of a principally polarized abelian variety are principally polarizedIf X Y Z = 4 and X2 Y2 Z2 = 30, Then Find the Value of Xy Yz ZxThe consensus of XYZ and Y'Z'W' is (XZ)(ZW') Shannon Expansion Theorem The Shannon Expansion Theorem is used to expand a Boolean logic function (F) in terms of (or with respect to) a Boolean variable (X), as in the following forms
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The Taylor expansion is the standard technique used to obtain a linear or a quadratic approximation of a function of one variable Recall that the Taylor expansion of a continuous function f(x) is The value of ∂u(x i, y i)/Solve x y z = x 3 y 3 z 3 = 8 in Z First I tried to transform this equation, substituting x = 8 − y − z So I end up with Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got z ∈ Z, which implies z − 8 is an integer implying that the second term is also an integer4 Binomial Expansions 41 Pascal's riTangle The expansion of (ax)2 is (ax)2 = a2 2axx2 Hence, (ax)3 = (ax)(ax)2 = (ax)(a2 2axx2) = a3 (12)a 2x(21)ax x 3= a3 3a2x3ax2 x urther,F (ax)4 = (ax)(ax)4 = (ax)(a3 3a2x3ax2 x3) = a4 (13)a3x(33)a2x2 (31)ax3 x4 = a4 4a3x6a2x2 4ax3 x4 In general we see that the coe cients of (a x)n
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(xyz)^3 (x y z) (x y z) (x y z) We multiply using the FOIL Method x * x = x^2 x * y = xy x * z = xz y * x = xyU xx= 6x 6 u y= 6xy 6y;5!) Thus, (xyz) 10 = ∑(10!) / (P1
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first take it to be real, δz = δx Then f0(z) = lim δx→0 f(z δx)−f(z) δx = lim δx→0 u(xδx,y)iv(xδx,y)−u(x,y)−iv(x,y) δx = lim δxBinomial Expansions Binomial Expansions Notice that (x y) 0 = 1 (x y) 2 = x 2 2xy y 2 (x y) 3 = x 3 3x 3 y 3xy 2 y 3 (x y) 4 = x 4 4x 3 y 6x 2 y 2 4xy 3 y 4 Notice that the powers are descending in x and ascending in yAlthough FOILing is one way to solve these problems, there is a much easier wayIn mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression
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In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomialAccording to the theorem, it is possible to expand the polynomial (x y) n into a sum involving terms of the form ax b y c, where the exponents b and c are nonnegative integers with b c = n, and the coefficient a of each term is a specific positiveOutput is F • Logic Function F = 1 if and only if there is a 0 to the left of a 1 in the input • Truth Table Truth Table with Three Inputs X Y Z F Min term− y z 3 2 y − 2 z = 2 2 (− y z 3) − y 2 z = 1 5 Solve these equations for y and z respectively Solve these equations for y and z respectively
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Consider the expansion of (x y z) 10 In the expansion, each term has different powers of x, y, and z and the sum of these powers is always 10 One of the terms is λx 2 y 3 z 5 Now, the coefficient of this term is equal to the number of ways 2x′s, 3y′s, and 5z′s are arranged, ie, 10!QUESTION 3 Find the sumofproducts expansion of a Boolean function f (x;Z x2ndx = X1 n=0 (¡1)n 2n1 x2n1C By setting x =0above, we &nd C =0 So arctanx = X1 n=0 (¡1)n 2n1 x2n1 Taylor Series for General Functions Consider power series expansion f (x)= X1 n=0 cn (x¡a) n =c 0c1(x¡a)c2(x¡a) 2c 3(x¡a) 3 (3) for general function f (x)about x =a Setting x =a;
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This is the Solution of Question From RD SHARMA book of CLASS 9 CHAPTER POLYNOMIALS This Question is also available in R S AGGARWAL book of CLASS 9 You can FOur online expert tutors can answer this problem Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!This preview shows page 3 6 out of 11 pages 33 Find the sumofproducts expansion of the Boolean function f (x y z) that is 1 if and only if exactly two of the three variables have value 1 Ans xyz xyz xyz 34 Find the sumofproducts expansion of the Boolean function f (x y z) that is 1 if and only if either x z 1 and y 0, or x 0 and y z 1
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Answer by lenny460 (1073) ( Show Source ) You can put this solution on YOUR website! If x, y, z are different and Δ = (x, x2, 1 x3), (y, y2, 1 y3), (z, z2, 1 z3) = 0 then show that 1 xyz = 0 We have Now, we know that If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinantsSeparate f and z into real and imaginary parts f(z) = u(x,y)iv(x,y) where z = x iy and u, v are real functions Suppose that f is differentiable at z We can take δz in any direction;
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Click here👆to get an answer to your question ️ Using the identity and proof x^3 y^3 z^3 3xyz = (x y z)(x^2 y^2 z^2 xy yz zx)Y, z) that is 1 if and only if x = y = 1 and 2 = 0, or x = 0 and y= 2 = 1, or x=y=0 and z = 1 The use of' denotes the negation or NOT such that x' stands for NOT X O A xy z' x'y Zx'y'Z OBxy' z' x'y z x'y' z OC xyz x'y z x'y' z ODxyz x'y Z X'y' zExpand Using the Binomial Theorem (xyz)^3 (x y z)3 ( x y z) 3 Use the binomial expansion theorem to find each term The binomial theorem states (ab)n = n ∑ k=0nCk⋅(an−kbk) ( a b) n = ∑ k = 0 n n C k ⋅ ( a n k b k) 3 ∑ k=0 3!
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(xyz)^3 put xy = a (az)^3= a^3 z^3 3az ( az) = (xy)^3 z^3 3 a^2 z 3a z^2 = x^3y^3 z^3 3 x^2 y 3 x y^2 3(xy)^2 z 3(xy) z^2 =x^3 y^3 z^3 3 xTo find the tenth term, I plug x, 3, and 12 into the Binomial Theorem, using the number 10 – 1 = 9 as my counter 12C9 ( x) 12–9 (3) 9 = (2) x3 (196) = x3 Find the middle term in the expansion of (4x – y)8 Since this binomial is to the power 8, there will be nine terms in the expansion, which makes the fifth term the middle oneTherefore, F = m3 m4 m5 m6 m7, which is the same as above when we used term expansion x y z Minterms Notation 0 0 0 x' y' z' m0 0 0 1 x' y' z m1 0 1 0 x' y z' m2 0 1 1 x' y z m3 1 0 0 x y' z' m4 1 0 1 x y' z m5 1 1 0 x y z' m6 1 1 1 x y z m7 Table 39 F = x' y z x y' z x y z' x y z
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