1. | y=5x | 2. | fx=5x-1 | | Differentiate the function | | | 1. | Vx=2x3+3x4-2x | 2. | Fy=1y2-3y4y+5y3 | 3. | y=x31-x2 | 4. | y=t23t2-2t+1 | 5. | Suppose that f5=1‚ f’5=6‚ g5=-3 and g’5=2Find the following values a) fg’5 b) fg’5 c) gf’5 | 6. | Fx=x4+3x2-25 | 7. | y=cos(a3+x3) | 8. | fx=2x-34x2+x+15 | 9. | y=sin2x+1 | 10. | y=tan4x | 11. | y=sinxcosx | 12. | y=sintan2x | 13. | y=x+x | 14. | y=sinsinsinx | 15. | y=a3+cos3x ‚ where a is constant | 16. | y=xsin1x
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Using the recursive least squares method‚ obtain parameter estimates for different model structures. The program is below: clear all close all N=1500; % set lenghth of data y=zeros(1‚N); u=zeros(1‚N); yhat=zeros(1‚N); e=zeros(1‚N); noise=rand(1‚N); % generate noise %ny=2; true value %nu=3; true value %ny=2; good// model with different order :test 1 %nu=4; %ny=3; bad %nu=5; %ny=4; bad %nu=5; %ny=3; good %nu=3; %ny=2; with more numerator terms:test 2 %nu=4; ny=2; %test 3 nu=3; m=ny+nu; P=eye(m‚m)*1000;
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where X is N [µ‚ σ 2 ]. Proposition 4: If X is N [µ‚ σ 2 ] then E(X) = µ and Var(X) = σ 2 . 1 2 Proposition 5: If Y is LN [µ‚ σ 2 ] then E(Y ) = eµ+ 2 σ and 2 2 Var(Y ) = e2µ+σ (eσ − 1). Proposition 6: If X is N [µ‚ σ 2 ] then aX + b is N [aµ + b‚ a2 σ 2 ]. 2 2 Proposition 7: If X is N [µ1 ‚ σ1 ]‚ Y is N [µ2 ‚ σ2 ]‚ and X and Y are indepen2 2 dent‚ then X + Y is N [µ1 + µ2 ‚ σ1 + σ2 ]. n Corollary 1: 2 If Xi are independent N [µ‚ σ ] for i = 1 . . . n then Xi is i=1
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P r ob lem s in M alays ia Mala y si a is ca te g o riz e d a s thi rd w or ld countr y a nd ha s re c e ived r a pid growt h in s oc ioec onomi c a nd a dva n c e te c hnolog i e s. The g lob a li z a ti on make s the w or ld be c ome sma ll e r a nd a ll the infor mation could be obt a ined e a sil y b y c li c kin g o n the c omput e rs. W or ld wi thout a n y ba rr ier s a ll ows c ult ur e
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Calculus in 3D Geometry‚ Vectors‚ and Multivariate Calculus Zbigniew H. Nitecki Tufts University August 19‚ 2012 ii This work is subject to copyright. It may be copied for non-commercial purposes. Preface The present volume is a sequel to my earlier book‚ Calculus Deconstructed: A Second Course in First-Year Calculus‚ published by the Mathematical Association in 2009. I have used versions of this pair of books for severel years in the Honors Calculus course at Tufts‚ a two-semester
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level of output Components of Aggregate Demand Two-sector Model AD = C + I Consumption Function (C) • A functional statement of the relationship between disposable income (Y) and consumption expenditure (C) C = f(Yd) • Consumption is a positive linear function of income C = a + bYd Note: In a two-sector model Y = Yd. (Why?) a is a positive constant‚ (a>0) showing the level of consumption at zero level of income‚ also known as autonomous consumption b represents the slope of the consumption
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Integration Differential Equation Solution dy f x dx y f x dx C dy f y dx 1 dy f y dx 1 f y dy 1 f y dy d2 y f x dx 2 1 1 dx dy dx F x C y f x dx C F x C dx G x Cx D xC 2. Substitution Use the substitution v x y to find the general solution of the differential equation dy 2 x y . dx Step 1: Apply product rule/quotient rule/chain rule
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edu/~busch/courses/theorycomp/fall2008/ 1 The Pumping Lemma: • Given a infinite regular language L • there exists an integerm | w | m with length • for any string w L • we can write w x • with |x y| m • such that: Fall 2006 (critical length yz and | i xy z L Costas Busch - RPI y | 1 i 0‚ 1‚ 2‚ ... 2 Non-regular languages R L {vv : v *} Regular languages Fall 2006 Costas Busch - RPI 3 Theorem:The language R L {vv : v *} {a‚ b} is not regular Proof: Fall
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Consider the figure shown below y x A dA y x Product of Inertia of A wrt x and y axis: Product of Inertia of Element dA: 2 5/3/2011 Product of Inertia for an Area Consider the figure shown below y x A dA y x Unit: length4 – m4‚ mm4‚ ft4‚ in4 g NOTE: 1. Ixy can be positive‚ negative or zero. 2. The product of inertia of an area wrt any two orthogonal axes is zero when either of the axes is an axis of symmetry. Product of Inertia of A wrt x and y axis: Product of Inertia for an Area
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problem for an ordinary differential equation involves finding a function y(t) that satisfies dy(t) = f (t‚ y(t)) dt together with the initial condition y(t0 ) = y0 . A numerical solution to this problem generates a sequence of values for the independent variable‚ t0 ‚ t1 ‚ . . . ‚ and a corresponding sequence of values for the dependent variable‚ y0 ‚ y1 ‚ . . . ‚ so that each yn approximates the solution at tn : yn ≈ y(tn )‚ n = 0‚ 1‚ . . . . Modern numerical methods automatically determine
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