Find the Taylor series expansion of this expression By default, taylor uses an absolute order, which is the truncation order of the computed series T = taylor (1/ (exp (x)) exp (x) 2*x, x, 'Order', 5) T = x^3/3 Find the Taylor series expansion with a relative truncation order by using OrderModeGiven that x 3 4y 3 9z 3 = 18xyz and x 2y 3z = 0 x 2y = 3z, 2y 3z = x and 3z x = 2y Now `( x 2y )^2/(xy) (2y 3z)^2/(yz) (3x x)^2/(zxNumber of terms is equal to nr−1C n For (xy z)n r =3 Hence number of terms are n3−1C n = n2C n
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(x+y)^3 expansion-It can never be $3^n$ As you can see for $(ab)^n$ contains just $n1$ terms Note that we have to keep the sum of powers in each of the combinations of $x,y,z$ to $n$, so it will be reduced Now replace $a$ and $b$ by $x$ and $(yz)$ respectively So total number of terms should be $123\cdots(n1)=\dfrac{(n1)(n2)}{2}$You can use f(x,y) for the function if you prefer, but that's one more letter to keep track of
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If the coefficients of x 3 and x 4 in the expansion of ( 3 k x) 9 are equal, then the value of k is KEAM 16 5 The coefficient of x 4 in the expansion of ( 1 − 2 x) 5 is equal to KEAM 17 6 If the 7 t h and 8 t h term of the binomial expansion ( 2 a − 3 b) n are equal, then 2 a 3 b 2 a − 3Consider 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!Obtain The Taylor Expansion Of Z(x, Y) Near (0,0) Upto The Second Order Using The Gradient And Hessian Matrix Z(x, Y) Is Given In Implicit Form As F(x, Y, Z) = Z Cos X X2 – Eyz = 0 This problem has been solved!
Share It On Facebook Twitter Email 1 Answer 1 vote answered by Juhy (631k points) selected by Vikash Kumar Best answerX 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 zAnswer by lenny460 (1073) ( Show Source ) You can put this solution on YOUR website!
Binomial 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 wayBinomial Expansion Calculator is a handy tool that calculates the Binomial Expansion of (xyz)^10 & displays the result ie, x^10 10x^9y 10x^9z 45x^8y^2 90x^8yz 45x^8z^2 1x^7y^3 360x^7y^2z 360x^7yz^2 1x^7z^3 210x^6y^4 840x^6y^3z 1260x^6y^2z^2 840x^6yz^3 210x^6z^4 252x^5y^5 1260x^5y^4z 25x^5y^3z^2 25x^5y^2z^3 1260x^5yz^4The number of distinct terms in the expansion of (x y – z)16 is (a) 136 (b) 153 (c) 16 (d) 17 binomial theorem;
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SOLUTION Since the number of terms in the expansion of (xyz) n is (n1) (n2)/2 The number of terms in the expansion of (x y z) 10 is (101) (102)/2=11X12/2 =11X65!) Thus, (xyz) 10 = ∑(10!) / (P1About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us Creators
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history (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 agreeThe 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)/
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Therefore, 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 zThis 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 F The number of terms in the expansion of (xyz) n Related questions 0 votes 1 answer If the integers r > 1, n > 2 and coefficients of (3r)th and (r 2)nd terms in the binomial expansion of (1 x)2n are equal, then asked in Class XI Maths by rahul152
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Adding 3abc both side a^3 b^3 c^3 = (a b c) (a^2 b^2 c^2 ab ac bc) 3abc Now, put a=x, b=y, and c=z x^3 y^3 z^3 = (x y z) (x^2 y^2 z^2 xy xz yz) 3xyz Put xyz =0 (Given) x^3 y^3 z^3 = (0) (x^2 y^2 z^2 xy xz yz) 3xyz x^3 y^3 z^3) rr Adopt the standard subscriptcomponent conventions x' = x 1 ', y' = x 2 ' and z' = x 3 'Question from Binomial Theorem,cbse,math,class11,ch8,binomialtheorem,exemplar,q26,secb,medium
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⋅(x)3−k ⋅(y)k ∑ k = 0 3 Z=z (x,y) R S dx dy dS A B dx dy n dS B dy In the first we use z both for the dependent variable and the function which gives its dependence on x and y;SolutionShow Solution ( x y z ) 2 = x 2 y 2 z 2 2 (x) (y) 2 (y) (z) 2 (z) (x) = x 2 y 2 z 2 2xy 2yz 2zx Concept Expansion of Formula
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Similar Problems from Web Search (xyz) (xzxyyz)xyz Final result x2y x2z xy2 2xyz xz2 y2z yz2 Step by step solution Step 1 Equation at the end of step 1 (x y z) • (xy xz yz) xyz Step 2 Final 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) ExThe 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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Expand (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 3 True or False True False 3 Determine if x x = x when x is a Boolean variable Question 1 Find the sumofproducts expansion of the following Boolean function F(x, y, z) = x 2 (xyz) xyz) (xyz) = 1 when x, y and z are Boolean variables True or False True False 3 Determine if x x = x when x is a Boolean variableHere is the question What is the coefficient of w˛xłyzł in the expansion of (wxyz) 9 There are 9 4term factors in (wxyz) 9 (wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz) To multiply it all the way out we would choose 1 term from each factor of 4 terms To get w˛xłyzł,
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In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial According to the theorem, it is possible to expand the polynomial n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b c = n, and the coefficient a of each term is a specific positive integer depending on n and b For example, 4 = x 4 4 x 3 y 6 x 2 y 2 4 x y 3 y 4 {\displaystyle ^{4}=x^{4}4x^{3}y6x^{2}y^{2}4xy^{3}y(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 = xyExpand (xy)^3 full pad » x^2 x^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot \msquare {\square} \le \ge
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In mathematics, the Taylor series of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point For most common functions, the function and the sum of its Taylor series are equal near this point Taylor's series are named after Brook Taylor, who introduced them in 1715 If zero is the point where the derivatives areWe know the corollary if abc = 0 then a3 b3 c3 = 3abcUsing the above identity taking a = x−y, b = y−z and c= z−x, we have abc= x−yy−zz −x= 0 then the equation (x− y)3 (y−z)3 (z−x)3 can be factorised as follows(x−y)3 (y−z)3 (z−x)3 = 3(x−y)(y−z)(z−x)Hence, (x−y)3 (y−z)3 (z −x)3 = 3(x−y)(y −z)(z −x)243x 5 810x 4 y 1080x 3 y 2 7x 2 y 3 240xy 4 32y 5 Finding the k th term Find the 9th term in the expansion of (x2y) 13 Since we start counting with 0, the 9th term is actually going to be when k=8 That is, the power on the x will 138=5 and the power on the 2y will be 8
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Expand 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!Explanation In the above expression, z means that the value will first increment by 1, then used Now, evaluate the statement by putting the values of x, y, and z On calculating the final answer is 24, as shown below z y y z x 11 5 5 10 3 = 24 Hence, the correct answer is option (a) answered by Editorial The sum of the coefficient of all the terms in the expansion of (2x – y z) in which y do not appear at all while x appears in even powers and z appears in odd powers is – (a) 0 (b) (2 1)/2 (c) 2 19 (d) (3 1)/2
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So $n$ is equal to 9 times the coefficient of $x^3y^2z^3$ in the expansion of $(2x3y4z)^8$ From Brian M Scott's answer, we see that $n$ is equal to $$9 \times \frac{8!}{3!2!3!}$$ which I will let you calculate yourself Also note that this is equal to $$\frac{9!}{3!2!3!1!}$$ which is the coefficient of $x^3y^2z^3w$, as expected Example 3 Example 4 Example 5 Example 6 Deleted for CBSE Board 21 Exams only Example 7 Deleted for CBSE Board 21 Exams only Example 8X, y, z are fixed during the integrations, a Taylor's series expansion in the source point coordinates x ', y ', z ' about (0,0,0) provides an approximation of the source coordinate dependence of D (;
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CognizantChoose the correct optionOPTIONSIf LCM and HCF of two numbers are 294 and 49 respectively, product of two numbers canbe expressed as2x3 x 7 42 x 32 x 224 x 32x722 x 34 x 7 Without solving the following quadratic equation, find the value of 'p' for which the given equation has real and equal roots 4 x2 (p–3)xp=0 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 3xy2The coefficient of x n in the expansion of (1 x) (1 – x) n is The coefficients a,b and c of the quadratic equation, ax 2 bx c = 0 are obtained by throwing a dice three times The probability that this equation has equal roots is The combined equation of the bisectors of the angle between the lines represented by (x 2 y 2)√3 = 4xy is
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To 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 one1 If $2^x2^y = 2^{xy}$, then $\frac {dy}{dx}$ is 2 The probability of solving a problem by three persons $A, B$ and $C$ independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively Then the probability of the problem is solved by any two of them is 3Find 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
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