X sqrt x 2 1
[PDF File]Lecture1.TransformationofRandomVariables
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7 2.3ATypicalApplication Let Xand Ybe independent,positive random variables with densitiesf X and f Y,and let Z= XY.We find the density of Zby introducing a new random variable W,as follows: Z= XY, W= Y (W= Xwould be equally good).The transformation is one-to-one because we can solve for X,Yin terms of Z,Wby X= Z/W,Y= W.In a problem of this type,we must always
[PDF File]NUMERICAL INTEGRATION: ANOTHER APPROACH
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The case n=2.Wewantaformula w1f(x1)+w2f(x2) ≈ Z 1 −1 f(x)dx The weights w1,w2 and the nodes x1,x2 aretobeso chosen that the formula is exact for polynomials of as large a degree as possible. We substitute and force
[PDF File]Trig Substitution
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Example 2. Compute Z 1 (x2 9)2 dx Soluion: This is almost identical to the rst example. Again, no u-substitution will work, and even though we have no square roots, we can use trig sub with x= 3sec and dx= 3sec tan d . Hence, the integral becomes: Z 1 (x 2 9) dx = Z 1 (9tan2 )2 (3sec tan d ) = Z sec 27tan3 ;d :
[PDF File]Chapter 1 Iteration - MathWorks
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x = sqrt(1+x) end This produces the same 32 lines of output as the for loop. However, this code is open to criticism for two reasons. The first possible criticism involves the termi-nation condition. The expression x ~= sqrt(1+x) is the Matlab way of writing x ̸= √ 1+x. With exact arithmetic, x would never be exactly equal to sqrt(1+x),
[PDF File]Integration by substitution
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1+x2 dx = 2 3 (1+x2)3/2 +c We have completed the integration by substitution. Let us analyse this example a little further by comparing the integrand with the general case f(g(x))g′(x). Suppose we write g(x) = 1+x2 and f(u) = √ u Then we note that the composition1of the functions f and g is f(g(x)) = √ 1+x2. 1when finding the composition ...
[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):
[PDF File]Gradients and Directional Derivatives
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(x2 +1), (f) f(x,y,z) = sin(x)ey ln(z). Section 3: Directional Derivatives 7 3. Directional Derivatives To interpret the gradient of a scalar field ∇f(x,y,z) = ∂f ∂x i+ ∂f ∂y j + ∂f ∂z k, note that its component in the i direction is the partial derivative of f with respect to x. This is the rate of change of f in the x direction
[PDF File]Calculus I Homework: The Tangent and Velocity Problems Page 1
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Calculus I Homework: The Tangent and Velocity Problems Page 1 Questions Example The point P(1,1/2) lies on the curve y = x/(1+x). a) If Q is the point (x,x/(1 + x)), use Mathematica to find the slope of the secant line PQ correct to six decimal places
[PDF File]TheChainRule g (h(x))h (x) Example1
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1 x 3 +2 = 1 x3 +2 Try a Javaapplet. The derivative of the composition of two non-constant functions is equal to the product of their derivatives, evaluated appropriately. TheChainRule We have the Chain Rule: g(h(x)) = g (h(x))h (x) Example1: Using g(x)= 1 x = x−1 and h(x) = x3 +2, we have g (x) = (−1)x−2 and h (x) = 3x2, g (h(x)) = (−1 ...
[PDF File]Chapter 5 4ed
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20 Chapter 5: Solved Problems Problem 19 Script File: F=[0 13345 26689 40479 42703 43592 44482 44927 45372 46276 47908 49035 50265 53213 56161]; L=[25 25.037 25.073 25.113 25.122 25.125 25.132 25.144
[PDF File]Disciplined Convex Programming and CVX
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sqrt(x) √ x (x ≥ 0) ccv, nondecr ... x2, |x| ≤ 1 2|x|−1, |x| > 1 cvx Convex Optimization, Boyd & Vandenberghe 11. Composition rules • can combine atoms using valid composition rules, e.g.: – a convex function of an affine function is convex – the negative of a convex function is concave
[PDF File]Power Series
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1 n! xn = 1+x+ 1 2! x+ 1 3! x3 +... has radius of convergence R= lim n→∞ 1/n! 1/(n+1)! = lim n→∞ (n+1)! n! = lim n→∞ (n+1) = ∞, so it converges for all x ∈ R. Its sum provides a definition of the exponential function exp : R → R. (See Section 6.5.) Example 6.9. The power series ∑∞ n=0 (−1)n (2n)! x2n = 1− 1 2! x2 + 1 ...
[PDF File]Table of Integrals - UMD
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1 2 secxtanx+ 1 2 ln|secxtanx| (76)!secxtanxdx=secx (77)!sec2xtanxdx= 1 2 sec2x (78)!secnxtanxdx= 1 n secnx, n!0 2 (79)!cscxdx=ln|cscx"cotx| (80)!csc2xdx="cotx (81)!csc3xdx=" 1 2 cotxcscx+ 1 2 ln|cscx"cotx| (82)!cscnxcotxdx=" 1 n cscnx, n!0 (83)!secxcscxdx=lntanx TRIGONOMETRIC FUNCTIONS WITH xn (84)!xcosxdx=cosx+xsinx (85)!xcos(ax)dx= 1 a2 ...
[PDF File]Exponential Distribution
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Interarrival and Waiting Time • Define T n as the elapsed time between (n − 1)st and the nth event. {T n,n = 1,2,...} is a sequence of interarrival times. • Proposition 5.1: T n, n = 1,2,... are independent identically distributed exponential random variables
[PDF File]Volumes by Cylindrical Shells: the Shell Method
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Ex. Find the volume of the solid generated by revolving the region bounded by y = x2, y = 0, x = −1, and x = 1, about the line x = 2. The axis of rotation, x = 2, is a line parallel to the y-axis, therefore, the
[PDF File]Techniques of Integration - Whitman College
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(1−x 2)) p 1− x dx. This looks messy, but we do now have something that looks like the result of the chain rule: the function 1 − x2 has been substituted into −(1/2)(1 − x) √ x, and the derivative 8.1 Substitution 165 of 1−x2, −2x, multiplied on the outside. If we can find a function F(x) whose derivative is −(1/2)(1− x) √
[PDF File]Programming Iterative Loops - Stanford University
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x = sqrt(x) n = n + 1 end println(n) end The number n is a counter –it just counts the loops. While Loops with Counters, 2 Let’s say you wanted to run the reproot program through 20 loops, then print the answer. This can also be done using counters: function reproot(x) n = 0 while n < 20 x = sqrt(x) n = n + 1 end
[PDF File]Math 504, Fall 2013 HW 3
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The minimal polynomial of x1/2 over F is f(y) = y2 x. Then f0(y) = 2y = 0 so x1/2 is not separable. Because it is a polynomial of degree two this implies that f has only one root. Furthermore, because (x1/2)2 = x 2F(x), we can reduce any polynomial of larger degree to a polynomial of degree 1 or 2, so no non-linear polynomial will split into
[PDF File]Left endpoint approximation
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2 = 2=4; x 3 = 3=4; x 4 = 1: 2.Above each subinterval draw a rectangle with height equal to the height of the function at the left end point of the subinterval. The values of the function at the endpoints of the subintervals are x i x 0 = 0 x 1 = 1=4 x 2 = 1=2 x 3 = 3=4 x 4 = 1 f(x i) = 1 x2 i 1 15=16 3=4 7=16 0 3.We use the sum of the areas of ...
[PDF File]Working a difference quotient involving a square root
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x+h x h = p x+h x h p p x+h+ x p x+: The key idea is that the numerators multiply in a nice way. Note that the two numerators together have the form (A B)(A+B) which is equal to A2 B2 (you might recall the phrase difference of squares). The squaring eliminates the square roots from the numerator. As a result, our expression above becomes p x+h ...
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