X 2 sqrt a 2 x

• [PDF File]Trigonometric Substitutions Math 121 Calculus II

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  a 2 x, p a 2+ x 2, or p x a, where ais some constant. In each kind you substitute for xa certain trig function of a new variable . The substitution will simplify the integrand since it will eliminate the square root. Here’s a table summarizing the substitution to make in each of the three kinds. If use see use the sub so that and p a 2 2x x ...

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• [PDF File]INTERPOLATION

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  sqrt(2) ¶, ³ π 2,0 ´ Now find a quadratic polynomial ... P2(x)=mx for all x ThusthedegreeofP2(x) can be less than 2. HIGHER DEGREE INTERPOLATION We consider now the case of interpolation by poly-nomials of a general degree n.Wewanttofind a polynomialPn(x)forwhich deg(Pn) ≤n

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• [PDF File]5.2 Limits and Continuity

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  (x,y)!(0,0) x4 4y2 x2 +2y2 First, observe that (0,0) is not in the domain of x44y2 x2+2y2. So we answer NO for Step I and move onto Step II. Notice that we can simplify this expression. We get x4 4y2 x 2+2y = (x2 2y2)(x2 +2y2) x +2y2 = x2 22y . Now we return to Step I and observe that (0,0) is in the domain of x2 2y2. Therefore, we can plug in ...

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• [PDF File]Section 1.5. Taylor Series Expansions

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  and set x =a to obtain f00(a)=c 2¢2¢1=) c2= f00(a) 2!; take derivative again on (5) f(3)(x)= X1 n=3 cnn(n¡1)(n¡2)(x¡a)n¡3=c 33¢2¢1+c44¢3¢2(x¡a)+c55¢4¢3(x¡a) 2+::: and insert x =a to obtain f(3)(a)=c 33¢2¢1=) c3= f(3)(a) 3!: In general, we have cn = f(n)(a) n!; n =0;1;2;::: here we adopt the convention that 0!=1: All above process can be carried

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• [PDF File]18.06 Problem Set 9 - Solutions - MIT

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  plot(x, S(:,1:3), ’ro’, x, [sin(pi*x);sin(2*pi*x);sin(3*pi*x)]*sqrt(2*dx), ’k-’) to plot the numerical eigenfunctions (red dots) and the analytical eigenfunctions (black lines). (The sqrt(2*dx) is there to make the normalizations the same. You might need to flip some of the signs to make the lines match up.) You can compare the eigenvalues

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• [PDF File]Homework Assignment 4

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  (x y)2 p x2 + y2 sin 1 x y = lim (x;y)!(0;0) (x y) (x y) p x2 + y2 sin 1 x y = 0; because the rst factor, (x y), goes to 0, and everything else is bounded. This proves the di erentiability at the origin. On the other hand, the partial derivatives are not continuous. Ideed, for x6= ywe have f x(x;y) = 2(x y)sin 1 x y cos 1 x y and f y(x;y) = 2(x ...

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• [PDF File]Table of Basic Integrals Basic Forms

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  ln(x 2+ a) dx= xln(x2 + a2) + 2atan 1 x a 2x (50) Z ln(x2 a2) dx= xln(x2 a2) + aln x+ a x a 2x (51)Z ln ax2 + bx+ c p dx= 1 a 4ac b2 tan 1 2ax+ b p 4ac b2 2x+ b 2a + x ln ax2 + bx+ c (52) Z xln(ax+ b) dx= bx 2a 1 4 x2 + 1 2 x2 b2 a2 ln(ax+ b) (53) Z xln a 2 2bx 2 dx= 1 2 x + 1 2 x a2 b2 ln a2 b2x2 (54) Z (lnx)2 dx= 2x 2xlnx+ x(lnx)2 (55) Z (lnx ...

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• [PDF File]2. Waves and the Wave Equation - Brown University

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  argument from x to x-x 0, where x 0 is a positive number. If we let x 0 = v t, where v is positive and t is time, then the displacement increases with increasing time. So f(x-vt) represents a rightward, or forward, propagating wave. Similarly, f(x+vt) represents a leftward, or backward, propagating wave. v is the velocity of the wave.-4 -2 0 2 ...

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• [PDF File]z=8−x2−y2 S z=x2+y2 R x2 +y2

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  3 O plot3d({sqrt(9-x^2), sqrt(9-y^2)},x=-3..3,y=-3..3); Figure 2. Part of the region S bounded by x2+z2 = a2 and x2 +y2 = a2 for x ≥ 0 Note that the projection of region S1 on the y − z plane, call it R is a a square 0 ≤ y ≤ a, 0 ≤ z ≤ a. We break

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• [PDF File]Table of Integrals

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  Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=

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• [PDF File]Eigen Function Expansion and Applications.

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  > Int(f(x)*1/sqrt(b),x=0..b)=0; d = ⌠ ⌡ 0 1 f(x) x 0 d/ Solve the equation by undetermined constant method: The right hand side f(x) = x − 1 2 sin(π x). We check the solvability condition > f:=x->(x-1/2)*sin(Pi*x); eval(int(f(x),x=0..1)); f:= x → x − 1 2 sin(π x) 0 We looking for a particular solution of the form: u p (x) = (A x + B ...

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• [PDF File]1 The EM algorithm

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  sigma2_sq= (gamma[:,1] *X**2).sum() / (n*pi2) -mu2**2 sigma1=np.sqrt(sigma1_sq) sigma2=np.sqrt(sigma2_sq) values.append(likelihood) We can track the value of the likelihood and, since we have an EM algorithm, the likelihood should be monotone with iterations. plt.plot(values)

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• [PDF File]Math 104: Improper Integrals (With Solutions)

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  (x−1)2/3 dx, if it converges. Solution: We might think just to do Z 3 0 1 (x−1)2/3 dx= h 3(x− 1)1/3 i 3 0, but this is not okay: The function f(x) = 1 (x−1)2/3 is undefined when x= 1, so we need to split the problem into two integrals. Z 3 0 1 (x− 1)2/ 3 dx= Z 1 0 1 (x− 1)2/ dx+ Z 3 1 1 (x− 1)2/3 dx. The two integrals on the ...

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• [PDF File]Random Variate Generation

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  N(0,1) has pdf f(x) = 2/sqrt(2p) * exp(-x2/2) Bad news: Its CDF doesn’t have a convenient closed form Can’t do Inverse Transform (unless we approximate) Can do generalized accept/rejection Let g(x) be the an exponential distribution with l=1 Then c= sqrt(2e/p) Not as efficient Preferred method: Polar Coordinate Method

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• [PDF File]TRIGONOMETRY LAWS AND IDENTITIES

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  TRIGONOMETRY LAWS AND IDENTITIES DEFINITIONS sin(x)= Opposite Hypotenuse cos(x)= Adjacent Hypotenuse tan(x)= Opposite Adjacent csc(x)= Hypotenuse Opposite sec(x)= Hypotenuse Adjacent

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• [PDF File]Maxima by Example: Ch.7: Symbolic Integration

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  2 2 sqrt(b - x ) Example 3 The definite integral can be related to the ”area under a curve” and is the more accessible concept, while the integral is simply a function whose first derivative is the original integrand.

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• [PDF File]PLOTTING AND GRAPHICS OPTIONS IN MATHEMATICA

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  Plot Sqrt 1-x^2 ,-Sqrt 1-x^2 , x,-1, 1 , AspectRatio ÆAutomatic -1.0 -0.5 0.5 1.0-1.0-0.5 0.5 1.0 voila. Or, we can use the AspectRatio command to make an even more oblate shape (but the figure is still a circle):

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• [PDF File]Using Mathematica to solve ODEs

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  yplus@x_, yplus0_D := x + Sqrt@2 Hx^2-xL + yplus0^2D; yminus@x_, yminus0_D := x-Sqrt@2 Hx^2-xL + yminus0^2D Plot@8yplus@x, 10D, yminus@x, -10D

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• [PDF File]Chapter 5: The Normal Distribution and the Central Limit ...

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  169 Theorem (The Central Limit Theorem): Let X 1;:::;X n be independent r.v.s with mean and variance 2, from ANY distribution. For example, X i Binomial (n;p ) for each i, so = np and 2 = np (1 p): Then the sum S n = X 1 + :::+ X n = P n i=1 X i has a distribution that tends to Normal as n ! 1. The mean of the Normal distribution is E (S n) = P n i=1 E (X i) = n : The variance of the Normal ...

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• [PDF File]Math 431 - Real Analysis I Solutions to Homework due ...

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  2 + x k p 2 + 2 = 2: Thus, A(k+ 1) is true. So, by induction, x k is bounded above by 2. (d) By (b) and (c), x n is a bounded, monotone sequence; thus, by a theorem in class, it converges. Question 2. One very important class of sequences are series, in which we add up the terms of a given

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